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Home > Error Bound > Error Bound For Taylor Polynomials # Error Bound For Taylor Polynomials ## Contents But if you took a derivative here, this term right here will disappear, it will go to zero, I'll cross it out for now, this term right over here will be And we already said that these are going to be equal to each other up to the nth derivative when we evaluate them at "a". Use a Taylor expansion of sin(x) with a close to 0.1 (say, a=0), and find the 5th degree Taylor polynomial. and what I want to do is approximate f of x with a Taylor Polynomial centered around "x" is equal to "a" so this is the x axis, this is the Check This Out The error function at "a" , and for the rest of this video you can assume that I could write a subscript for the nth degree polynomial centered at "a". It's going to fit the curve better the more of these terms that we actually have. If x is sufficiently small, this gives a decent error bound. Paul Seeburger 4 650 visningar 11:13 Estimating error/remainder of a series - Längd: 12:03. ## Error Bound For Taylor Polynomials I'm just going to not write that every time just to save ourselves some writing. So the n+1th derivative of our error function, or our remainder function you could call it, is equal to the n+1th derivative of our function. If is the th Taylor polynomial for centered at , then the error is bounded by where is some value satisfying on the interval between and . Kommer härnäst 9.3 - Taylor Polynomials and Error - Längd: 6:15. And so it might look something like this. Taking a larger-degree Taylor Polynomial will make the approximation closer. Läser in ... Lagrange Error Bound Calculator Visningskö Kö __count__/__total__ Ta reda på varförStäng Find the error bound for a Taylor polynomial Bob Martinez PrenumereraPrenumerantSäg upp136136 Läser in ... Thus, we have a bound given as a function of . Suppose you needed to find . And this polynomial right over here, this nth degree polynimal centered at "a", it's definitely f of a is going to be the same, or p of a is going to https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation and maybe f of x looks something like that... Please try the request again. Lagrange Error Bound Problems patrickJMT 147 176 visningar 11:35 Example of Trapezoid Rule with Error Bound - Längd: 6:04. This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation. We differentiated times, then figured out how much the function and Taylor polynomial differ, then integrated that difference all the way back times. ## Use The Error Bound For Taylor Polynomials To Find A Reasonable So because we know that p prime of a is equal to f prime of a when we evaluate the error function, the derivative of the error function at "a" that http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/PowerSeries/error_bounds.html So let me write that. Error Bound For Taylor Polynomials Läser in ... Error Bound Taylor Series Calculator fall-2010-math-2300-005 lectures © 2011 Jason B. The derivation is located in the textbook just prior to Theorem 10.1. http://megavoid.net/error-bound/error-bound-taylor.html Språk: Svenska Innehållsplats: Sverige Begränsat läge: Av Historik Hjälp Läser in ... And that polynomial evaluated at "a" should also be equal to that function evaluated at "a". So, we have . Lagrange Error Bound Formula Example The third Maclaurin polynomial for sin(x) is given by Use Taylor's Theorem to approximate sin(0.1) by P3(0.1) and determine the accuracy of the approximation. Hence, we know that the 3rd Taylor polynomial for is at least within of the actual value of on the interval . Thus, we have What is the worst case scenario? this contact form Return to the Power Series starting page Representing functions as power series A list of common Maclaurin series Taylor Series Copyright © 1996 Department of Mathematics, Oregon State University If you So it's literally the n+1th derivative of our function minus the n+1th derivative of our nth degree polynomial. Lagrange Error Bound Khan Academy You may want to simply skip to the examples. of our function... ## Bob Martinez 517 visningar 6:02 Taylor's Remainder Theorem - Finding the Remainder, Ex 1 - Längd: 2:22. Finally, we'll see a powerful application of the error bound formula. And then plus go to the third derivative of f at a times x minus a to the third power, (I think you see where this is going) over three factorial, Visa mer Läser in ... Lagrange Error Bound Proof Get it on the web or iPad! All Rights Reserved. If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. Generated Mon, 10 Oct 2016 15:52:27 GMT by s_wx1131 (squid/3.5.20) http://megavoid.net/error-bound/error-bound-formula-for-taylor-polynomials.html It considers all the way up to the th derivative.

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# 2.4.1: Sankey Diagram Questions $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ Please refer to the Sankey diagrams for Washingto and Pennsylvania to answer the knowledge check questions. For consistency, the 2012 Sankey diagram for the U.S. is also provided for comparison. Q1: Compared to Washington State, is Pennsylvania’s energy mix more or less similar to the energy mix of the U.S. as a whole? Compared to the U.S. as a whole, Pennsylvania is very similar. Like the U.S. as a whole, Pennsylvania relies primarily on a mix of fossil fuels (natural gas and coal) and nuclear energy for electric power generation and oil for transportation. Washington on the other hand relies principally on hydropower for electricity generation. Q2: Identify the biggest source of energy used in electricity production in Pennsylvania, Washington and the US as a whole?

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× INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS Are you an Engineering professional? Join Eng-Tips Forums! • Talk With Other Members • Be Notified Of Responses • Keyword Search Favorite Forums • Automated Signatures • Best Of All, It's Free! *Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. #### Posting Guidelines Promoting, selling, recruiting, coursework and thesis posting is forbidden. # Tension Ring ## Tension Ring (OP) I've been looking at a turret design and trying to determine if I could run a strap around the tails of the rafters so that I could eliminate the internal bracing: I'm trying to figure out how to calculate the tension in the strap given a uniform (snow load) on the turret roof. The steeper the pitch the less the tension in the strap. This reminds me of some of my calculus problems from my college days, but it is probably much simpler than that. A larger overhang would counter the flattening effect of the load on the roof and reduce the tension in the strap. If you take the tributary load at the center of the roof as half the distance from the center to the outside wall the point load at the center would be given as P = (S + D) * pi * (d/4)^2. The horizontal force exerted by each rafter would then be: Fh = P/tan(Theta) * (1/n) where "n" equals the number of rafters. The tension in the strap from basic statics would be T = (Fh/2) / cos(omega) where omega = 90 * (1 - 1/n). Anyone care to check my math and see if it adds up? I'm assuming an overhang of zero or at least conservatively ignoring the countering affect of the overhang. A confused student is a good student. Nathaniel P. Wilkerson, PE www.medeek.com ### RE: Tension Ring this is a puzzle to me: haven't worked through your formulae but it looks like you're assuming rafter thrust is the same for both the 'non-continuous' rafters as well as the rafters that are continuous to the ridge. the framing system is something i might use in a pyramid shaped roof, which I've modeled in the past as intersecting 3 hinged arches, where the horizontal thrust of the 4 ridge beams is supported by a 'tension ring' with 4 'straight sides' each connection resolvable by 'basic statics'. the non-continuous rafters i would model as simply supported beams with no lateral thrust. in the present case, if i looked at it as 4 intersecting 3 hinged arches, 8 ridge beams, but with tension ring consisting of 8 'curved sides'...the tendency to 'straighten that curve' will induce a thrust into the non-continuous rafters. (and yes, a diaphragm can be installed which will 'cover a multitude of sins', but to me the puzzle is to see how the system truly works) as the model is extended further, at some point, the tension ring is perpendicular to the thrusting rafter and one is dividing by cos90-ish, as in your equation, and this is no longer basic, at least my kind of basic, statics. Seems like a hoop stress condition, which I've seen but never worked through the math on.... ### RE: Tension Ring (OP) Yes, for simplicity I'm assuming the same rafter thrust for all rafters. A confused student is a good student. Nathaniel P. Wilkerson, PE www.medeek.com ### RE: Tension Ring then i suspect hoop stress is your model and the simple beam supporting the upper end of your non-continuous rafters needs to be designed for some lateral bending. ### RE: Tension Ring Reminds me of Frank Lloyd Wright's studio near Chicago where he used a great big chain looped around to counteract thrust. not that his designed were necessarily what you call "Engineered". ### RE: Tension Ring (OP) Great minds think alike. A confused student is a good student. Nathaniel P. Wilkerson, PE www.medeek.com ### RE: Tension Ring Except for the overhang, the structure should act like a conical shell. If the total weight is W, the vertical reaction is also W. The vertical reaction per unit length of circumference is W/(2*pi*r). If the roof slopes at angle θ, the radial reaction is W/(2*pi*r*tanθ) per unit length of circumference and the tension in the strap is W/(2*pi*tanθ). The overhang does not contribute to the strap tension but adds to the vertical reaction. BA #### Red Flag This Post Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework. #### Red Flag Submitted Thank you for helping keep Eng-Tips Forums free from inappropriate posts. The Eng-Tips staff will check this out and take appropriate action. #### Resources White Paper: Industrial Control Basics: Contactors A contactor is an electrical device used for switching an electrical circuit on or off. Considered to be a special type of relay, contactors are used in applications with higher current carrying capacity, while relays are used for lower current applications. Download Now Research Report: State of IoT Adoption in Product Development 2019 This research report, based on a survey of 234 product development professionals, examines the current state of Internet of Things (IoT) adoption by product design teams, its perceived importance, and what features and capabilities teams consider important when making decision about adding IoT functionality to their products. Download Now Research Report: Augmented Reality for Maintenance, Repair and Overhaul (MRO) The term Industry 4.0 denotes a cluster of technologies that’s poised to fundamentally reshape manufacturing and bring about a new industrial revolution. These include 3D printing (AM), the Industrial Internet of Things (IIoT), artificial intelligence (AI) and mixed reality technologies, more commonly known as virtual reality (VR) and augmented reality (AR). Download Now Close Box # Join Eng-Tips® Today! Join your peers on the Internet's largest technical engineering professional community. It's easy to join and it's free. Here's Why Members Love Eng-Tips Forums: • Talk To Other Members • Notification Of Responses To Questions • Favorite Forums One Click Access • Keyword Search Of All Posts, And More... Register now while it's still free!

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# The absorbance of 2 x 10^-4 M solution was found to be 0.123 when placed in a cell of 1 cm length. Calculate the transmittance, percent transmittance and molar absorptivity of a cell of length 2 cm? Feb 12, 2016 $T = 0.568$ %T=56.8% $\epsilon = 615 {M}^{- 1} \cdot c {m}^{- 1}$ #### Explanation: $C = 2 \times {10}^{- 4} M$ $A = 0.123$ $l = 2 c m$ ?T, ?%T, ?epsilon The relationship between transmittance $T$ and absorbance is: $A = - \log T \implies T = {10}^{- A}$ Since the length of the cell was increased to $l = 2 c m$, the absorbance will double and therefore, ${A}_{2} = 2 \times 0.123 = 0.246$ $\implies T = {10}^{- 0.246} = 0.568$ Therefore, %T=100xxT=100xx0.568=56.8% In order to solve for the molar absorptivity $\epsilon$, we will use the Beer-Lambert law: $A = \epsilon \times l \times C \implies \epsilon = \frac{A}{l \times C}$ $\epsilon = \frac{0.123}{1 c m \times 2 \times {10}^{- 4} M} = 615 {M}^{- 1} \cdot c {m}^{- 1}$ Here is a video that explains Beer-Lambert law and it use experimentally: AP Chemistry Investigation #1: Beer-Lambert Law.

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# When it comes to motors, how hot is hot? ## Temperatures that are too high affect machine performance and lifespan. 09/10/2013 It's not uncommon for a maintenance technician to call a service center about a repaired motor that’s now “running hot.” When asked how hot, the reply frequently is: “Well, I can’t hold my hand on it!” Think about that answer for a minute. The typical human can tolerate touching something that’s about 60 to 65 C (140 to 150 F), depending on calluses, threshold of pain, or how many people are watching. Now keep those numbers in mind as you read the following discussion of typical motor operating temperatures. NEMA Standards MG 1-2011, 12.43 defines temperature rise for motors in a maximum ambient of 40 C. Two points are noteworthy here. First, in keeping with the NEMA standard, temperature is given here in Celsius; Fahrenheit equivalents are provided only where appropriate. Second, ambient temperature refers to the temperature of surrounding air. Some people confuse this with the expected temperature rise of the motor, which it is not. The Class F (155 C) temperature rating is popular these days, so it’s a good one to discuss. With Class F insulation, the maximum allowable temperature rise ranges from 105 to 115 C for the motor’s winding insulation, depending on how the motor is configured. For example, if the motor that was reported to be “running hot” has a 1.15 service factor, its maximum temperature rise would be 115 C plus the 40 C ambient. The winding embedded in the slot is almost always the hottest part, so the winding temperature for this motor could reach 155 C. Motor construction The surface that our maintenance technician tried to touch will be somewhat cooler than the winding temperature, depending on the motor construction. On a large cast iron, totally enclosed, fan cooled (IP 54) motor, for instance, the surface may be 20 to 25 C cooler than the winding hot spot, but only 10 to 15 C cooler on a rolled steel frame motor, where the surface is much closer to the winding. Temperature differentials are often much greater—as much as 60 C—for open drip-proof motors (IP 12) and Weather-Protected I (WP I) or Weather-Protected II (WP II) enclosures. Differences in flat rolled steel and ribbed cast iron frame surfaces also affect the amount of heat our technician would feel. Of course, motors are not designed to run at the maximum allowable temperature, because that would drastically decrease motor life. In fact, every 10 C rise in operating temperature reduces insulation life by half. Therefore the ultimate design is one that optimizes motor life and function, while keeping the cost of production, maintenance and efficient operation as low as possible. As an example, suppose that a motor with a 65 C rise (very conservative by most standards) is put in service on a hot summer day. If the ambient temperature is 35 C (95 F), the winding total temperature would be: 65 + 35 = 100 C. If motor is so constructed that its surface is about 20 C cooler than the winding, the surface temperature would be: 100 – 20 = 80 C (176 F)—i.e., much too hot to touch safely! Remember that this is a conservative design, so the surface temperatures of many motors will be much warmer. ### Thermally protected motors At first glance, the maximum motor temperature rise limits given in NEMA Stds. MG 1, 12.43 appear to contradict those given for the same insulation classes in NEMA Stds. MG 1, 12.56 and Table 12-8. Actually, the maximum temperatures shown in Figure 3 apply only to “thermally protected” motors. The words “Thermally Protected” on a motor nameplate indicate that a thermal protection device is an integral part of the machine and, when properly applied, will protect it from dangerous overheating. In other words, thermally protected motors are an exception to the rule. If the motor has this added, special layer of protection, the higher temperatures may be allowed. Based on our earlier example, you can see how these higher winding temperatures would impact the surface temperature of the motor. Thermal protection and the correspondingly higher temperature limits are generally reserved for smaller motors. Even then, however, motor manufacturers will not design a motor to run at the maximum temperature allowed unless application considerations make it necessary. ### Special cases Some applications require that a motor be housed in an enclosure for noise abatement or other reasons. In such cases, special care should be taken to control the ambient temperature inside the enclosure where the motor is located. Cooling is usually adequate if the auxiliary cooling supplies the same volume of air as the motor’s integral fan. If the driven equipment generates heat (e.g., a compressor) and is contained in the motor enclosure, it can contribute to the temperature rise of the motor. One example that comes to mind (details vague to protect the innocent) involved more than 100 compressors, each of which was housed in an enclosure with its drive motor and equipped with a radiator to cool the gas as it was compressed and liquified. Unfortunately, air for cooling was drawn through the radiator and then exhausted by a fan on the drive motor. With this arrangement, ambient temperatures in the enclosures reached 70 C, thermally stressing the motor winding insulation to its limit. Overheated lubricant also evacuated bearing housings, causing numerous bearing failures. Temperature control is an important factor in successful motor operation. Care must be taken in the design, application and maintenance of these machines to optimize their performance and life. Having said all that, it is unsafe to place your hand on a motor to see if it is too hot; get a thermometer instead. Jim Bryan a technical support specialist at the Electrical Apparatus Service Association (EASA). The Engineers' Choice Awards highlight some of the best new control, instrumentation and automation products as chosen by... The System Integrator Giants program lists the top 100 system integrators among companies listed in CFE Media's Global System Integrator Database. Each year, a panel of Control Engineering and Plant Engineering editors and industry expert judges select the System Integrator of the Year Award winners in three categories. This eGuide illustrates solutions, applications and benefits of machine vision systems. Learn how to increase device reliability in harsh environments and decrease unplanned system downtime. This eGuide contains a series of articles and videos that considers theoretical and practical; immediate needs and a look into the future. Additive manufacturing benefits; HMI and sensor tips; System integrator advice; Innovations from the industry Robotic safety, collaboration, standards; DCS migration tips; IT/OT convergence; 2017 Control Engineering Salary and Career Survey Integrated mobility; Artificial intelligence; Predictive motion control; Sensors and control system inputs; Asset Management; Cybersecurity Featured articles highlight technologies that enable the Industrial Internet of Things, IIoT-related products and strategies to get data more easily to the user. This article collection contains several articles on how automation and controls are helping human-machine interface (HMI) hardware and software advance. This digital report will explore several aspects of how IIoT will transform manufacturing in the coming years. Infrastructure for natural gas expansion; Artificial lift methods; Disruptive technology and fugitive gas emissions Mobility as the means to offshore innovation; Preventing another Deepwater Horizon; ROVs as subsea robots; SCADA and the radio spectrum Future of oil and gas projects; Reservoir models; The importance of SCADA to oil and gas Automation Engineer; Wood Group System Integrator; Cross Integrated Systems Group Jose S. Vasquez, Jr. Fire & Life Safety Engineer; Technip USA Inc. This course focuses on climate analysis, appropriateness of cooling system selection, and combining cooling systems. This course will help identify and reveal electrical hazards and identify the solutions to implementing and maintaining a safe work environment. This course explains how maintaining power and communication systems through emergency power-generation systems is critical.

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# Lorentz Transformations and Photon Delay For fun, I'm writing a simple special relativity simulator with a much smaller speed of light so that relativistic effects are clear even at low speeds. I already have time dilation and speed of light delay working. However, right now, the speed of light does NOT always appear to be the same for all observers. The main problem is computing where something appears to be from the perspective of the user. The speed of light is not infinite. So when you look at something, you see it where it used to be. Ignoring special relativity, but still including a finite speed of light, let's compute where you see it. Let p(T) be the location of the object at time T. Let u(T) be your location at time T. Let T be how "late" the image of the object is: right now, you see the object where it was at time (now-T). Let c be the speed of light in meters per second. Assume that all distance/location units are in meters. Light travels at the speed of light. Therefore, T seconds ago, the light that you are seeing now was T*c light-seconds away. distance( p(now - T), u(now) ) = c * T Assuming that the object in question never moves faster than the speed of light, this equation always yields a unique satisfying T. For my purposes, this is enough to compute T. Suppose you're in a spaceship moving near the speed of light and you emit a flash of light in all directions around you. Under special relativity, from your perspective, the light beams should always look like they are centered around you, even as you move, because that's what it would look like if you were stationary. However, in the simulation, they do what you would expect in Newtonian mechanics: they form a sphere around the location that you emitted them from. So it's not right yet. ---- I compute everything relative to a nonaccelerating global reference frame. When I want to draw something, I compute where it should appear to be in the reference frame of the user. Similarly, I alter the time-step of the program so that when the user is travelling close to the speed of light, stationary clocks appear to move faster. Let's have a short thought experiment to defend this model. Suppose there's a spaceship A, a spaceship B, and an observer O. A and B are both moving near the speed of light. Suppose we want to know where B appears to be from A's perspective. All we have to do is use O's information about A and B's locations over time: the time that A sees the image of B is the time that O sees A seeing the image of B. To convert that time into A's perspective, all we have to do is compute how much time A has observed passing since T_0. Keep in mind that all of that holds even if A and B are acclerating. If you have an event E that happens at time T and location x,y,z in the global reference frame, and the user's velocity (as observed by the global reference frame) is Vx, Vy, Vz, then I understand that you apply the Lorentz transformation to get T', x', y', z', the location of E in spacetime according to the user's reference frame. So here's the question: given the posititions of A and B for all T up to now (in the global reference frame), where does B appear to be from A's perspective in A's reference frame, taking both special relativity and speed of light delay into account? Does the Lorentz transformation hold for an accelerating observer? ghwellsjr Gold Member Can you show us what you've go so far? I'm not sure your approach is the best way to go. The Lorentz Transformation only works for Frames of Reference moving at a constant relative speed. Why don't you just do it for your global reference frame and then you won't have to worry about any problems with acceleration? I will post source code if you insist, but it's a bit long and I don't want to distract between coding problems / problems of efficiency and problems about the correctness of the algorithm. Furthermore, pseudocode / descriptions of algorithms are much easier to read than source code. On each frame, I compute how much time should pass. This amount of time is (1/60) / sqrt(1-v^2/c^2) where v is the velocity of the user. For each ship, if it is accelerating, I use F=ma to compute its new velocity, while keeping in mind that mass is not constant during acceleration. If the speed of any object is > .99c, I scale it down to .99c. For each ship, on each frame, I store its current location, rotation, velocity, global seconds passed since the simulation began, and local seconds passed since the simulation began. To determine where to draw an object, I compute the "light time" of the object relative to the user, then draw it where it was at that time. The light time is now - T, where T minimizes |distance(objectLocation(now-T), userLocation(now)) - c * T| I compute this error term for all recorded times to find T. Why don't you just do it for your global reference frame and then you won't have to worry about any problems with acceleration? Do what from the global reference frame? All of the data I store (location, velocity, etc) I store in the global reference frame. This makes things significantly easier than storing everything in the reference frame of the user. Currently, when the user accelerates, I only have to update data about the user. If I stored everything in the reference frame of the user, then I'd have to update every object when the user accelerates. Furthermore, even ignoring this problem, I'd eventually like to be able to support multiple users with different reference frames. Eventually, I'm going to have to solve the problem I described. Also, while the global reference frame does not accelerate, objects in the world DO accelerate relative to it, so acceleration is an inherent problem to the system. ghwellsjr Gold Member So here's the question: given the posititions of A and B for all T up to now (in the global reference frame), where does B appear to be from A's perspective in A's reference frame, taking both special relativity and speed of light delay into account? Does the Lorentz transformation hold for an accelerating observer? I thought your problem was being able to transform events from your global reference frame to an accelerating reference frame. Did I misunderstand? I don't want to see any source code but I thought you had parts of your simulation working. I thought your simulation was going to display something and since I have no idea what to expect, I thought it might be helpful to see the display of what you have working so far. I got the impression that you were planning on a display that would show us what an observer would see, is that correct? If so, that won't really show us what Special Relativity is all about since SR allows us to "see" things that observers cannot see and that's why I made my suggestion. Also, if you just use your one global reference frame, you won't have to concern yourself with the fact that the Lorentz Transform does not work on accelerating frames. I thought your problem was being able to transform events from your global reference frame to an accelerating reference frame. Did I misunderstand? That's part of it. The other part is deciding what event needs to be translated. Is it the emission of light from the observed object? From when? Or is it the arrival of the light emitted from the observed object, and again, from when? http://imgur.com/DdGVF" [Broken] The triangle on the left is moving up. The one in the middle is the user. The user is moving down. The one on the right is stationary. Each has a crude clock drawn over it, but they're a little hard to see. Before taking the screenshot, the user went far up, then back, so the one on the right has observed more time passing than the user. The dim triangle to the left is where the triangle on the left "actually" is, rather than where it appears to be to the user. I got the impression that you were planning on a display that would show us what an observer would see, is that correct? If so, that won't really show us what Special Relativity is all about since SR allows us to "see" things that observers cannot see and that's why I made my suggestion. Also, if you just use your one global reference frame, you won't have to concern yourself with the fact that the Lorentz Transform does not work on accelerating frames. Then maybe special relativity isn't sufficient for what I want. I'd like to show what the user would actually see. The "camera", so to speak, has to follow the user. I don't want the user to be able to see events before the light from the event would actually reach them. Last edited by a moderator: ghwellsjr Gold Member When you say you'd like to show what the user would actually see, I think of the kind of video game where you never see yourself, the display depicts what you would actually see with your eyes, as if you were wearing a videocam on your forehead. But that's not what you presented in your screenshot. That depiction is like some of the really old video games like asteroids where you saw yourself and your surroundings and if you fired a missle at something, you could watch its trajectory and you could easily determine how fast it approached the target and how close it was getting to the target. But that's not what a user would see watching a missile. He would only see the missle in the middle of his field of view getting smaller and smaller until it impacted the target. But I think I see what you are trying to do. Let's say you want to illustrate the Twin Paradox. You would show the stationary twin as the "user" who would have a clock running at normal speed. Then you would show two images of his twin traveling away from him, one showing where he actually was with a clock showing his dilated time and another one showing where he would appear to be to the user showing his relativistic doppler time. Then when the twin reached his destination, he would turn around and head back home which the first image would show immediately but the other one would be lagging behind. It wouldn't be until the first image got most of the way home that the second image finally reaches the turnaround point and then rushes back home with the clock now running way faster and then the two images finally converge at the home position with the same time on both clocks (for the twin) while the user's clock has a much more advanced time on it. Is this the sort of thing you have in mind? Yes! That's exactly what I meant. I'm keeping things in two (spatial) dimensions for simplicity. When I said "what the user would actually see" what I meant was that you see the same information that the user would actually see*, although it would be presented in a different way than two-dimensional eyes might present it to a two-dimensional brain. If and when I switch to 3D, I'll probably switch over to either a "forehead" camera or a following camera. *ignoring the fact that you can't see through solid objects, etc

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# 2.23 HIV in Swaziland, Swaziland has the highest HIV prevalence in the world: 25.9% of this... 2.23 HIV in Swaziland, Swaziland has the highest HIV prevalence in the world: 25.9% of this country's population is infected with HIV.66 The ELISA test is one of the first and most accurate tests for HIV. For those who carry HIV, the ELISA test is 99.7% accurate. For those who do not carry IllV. Lhe ksı is 9%.6% aocurate. If an İndividual frun Swaziland has ksiexl IX2ilLive, whai. is the probability that he carries HIV? Given, P(HIV) = 0.259 So, P(Not HIV) = 1 - 0.259 = 0.741 P(Test positive |HIV) = 0.997 P(Test negative |Not HIV) = 0.926 Hence, P(Test positive |Not HIV) = 1 - 0.926 = 0.074 Using Baye's theorem, P(HIV |Test positive) $= \frac{P(Test Positive |HIV)P(HIV)}{P(Test Positive |HIV)P(HIV) + P(Test Positive |Not HIV)P(Not HIV)}$ $= \frac{0.997*0.259}{0.997*0.259 + 0.074*0.741}$ = 0.8248 ##### Add Answer to: 2.23 HIV in Swaziland, Swaziland has the highest HIV prevalence in the world: 25.9% of this... Similar Homework Help Questions • ### The department of health of a certain state estimates a 10% rate of HIV The Department... The department of health of a certain state estimates a 10% rate of HIV The Department of Health of a ce a state estimates a 10% ate of HN for the at sk population and ล 0.3% rate for the general population. Tests for HM are 95% accurate in detect ng bath true negatives and 20,000 people frorn the general papulation resuts in the follawing table. Use the table below to complete parts (a) through (e) ue pasitives Random selection... • ### Since the 1980s, HIV (Human Immunodeficiency Virus) has been infecting humans around the world causing the... Since the 1980s, HIV (Human Immunodeficiency Virus) has been infecting humans around the world causing the condition known as AIDS (Acquired Immune Deficiency Syndrome). HIV, like all viruses, needs to enter cells and use their machinery to reproduce and spread. During HIV infection, the virus enters specific cells of the immune system (T-cells) by "docking" onto cell surface proteins, including one called CCR5 Genetic analysis of individuals who are naturally immune (resistant) to HIV have revealed that resistance to HIV... • ### 1. For the choirmaster. A psalm of David. 2. Hear.my.troubles, O God. Kesp.me.safe from terror, T... 1. For the choirmaster. A psalm of David. 2. Hear.my.troubles, O God. Kesp.me.safe from terror, The Department of Health of a certain state estimates a 10% rate of HIV for the general population Tests for HIV are 95% accurate in detecting both true negatives and true positives. Random see 5000 "at risk people and 20,000 people from the general population results in the following table. Use the table below to complete parts (a) through (e). at risk population and a... • ### Read the book "The Drunkard's Walk - How Randomness Rules Our Lives" by Mlodinow and pay... Read the book "The Drunkard's Walk - How Randomness Rules Our Lives" by Mlodinow and pay special attend to the following questions. Some of these questions may appear on quizzes and exams. Chapter 1 Peering through the Eyepiece of Randomness 1. Explain the phenomenon "regression toward the mean." 2. What factors determine whether a person will be successful in career, investment, etc.? 3. Was Paramount's firing of Lansing the correct decision? After she was fired, Paramount films market share rebounded.... • ### Questions 3 and 5 Frozen Coke and Burger King and the Richmond Rigging 12 Case 8.17 president of Coca-Cola's Foodservice and Hospitality Division, was looking on sells fountain-dispensed soda... Questions 3 and 5 Frozen Coke and Burger King and the Richmond Rigging 12 Case 8.17 president of Coca-Cola's Foodservice and Hospitality Division, was looking on sells fountain-dispensed soda to restaurants, convenience marts fountain division, a division responsible for one-third of all of Coke's revenues , and Tom Moore, president of sales in the The fountain division fourn theaters. Sales were stagnant, and he knew from feedback from the salespeople that Pepsi ias moving aggressively in the area. In 1999,... • ### The experimental rats were being raised in separate environments 1. The experimental rats were being raised in separate environments. One group lived in the traditional cage in a quiet, but well lighted room. They were supplied with plenty of food and water. The second group inhabited "Rat Disney Land." There were colorful toys, exercise wheels, and mirrors. The experimenters talked to the rats in the second group and demonstrated concern. Based on the results of previous research the researchers expected __________________. A. The rats in the first group to exhibit depression.B. The rats... • ### The experimental rats were being raised in separate environments 1. The experimental rats were being raised in separate environments. One group lived in the traditional cage in a quiet, but well lighted room. They were supplied with plenty of food and water. The second group inhabited "Rat Disney Land." There were colorful toys, exercise wheels, and mirrors. The experimenters talked to the rats in the second group and demonstrated concern. Based on the results of previous research the researchers expected __________________. A. The rats in the first group to... • ### All of the following questions are in relation to the following journal article which is available... All of the following questions are in relation to the following journal article which is available on Moodle: Parr CL, Magnus MC, Karlstad O, Holvik K, Lund-Blix NA, Jaugen M, et al. Vitamin A and D intake in pregnancy, infant supplementation and asthma development: the Norwegian Mother and Child Cohort. Am J Clin Nutr 2018:107:789-798 QUESTIONS: 1. State one hypothesis the author's proposed in the manuscript. 2. There is previous research that shows that adequate Vitamin A intake is required... • ### Vitamin A and D intake in pregnancy, infant supplementation, and asthma development: the Norwegian Mother and... Vitamin A and D intake in pregnancy, infant supplementation, and asthma development: the Norwegian Mother and Child Cohort .5.6 Oystein Karilstad Kristin Holvik Nicolai A Land-BlixMargareta Haugen 1.7 Christian M Page. Per NdPer M Ueland. Stephanie JLondon Siri E Haberg.and Wenche Nysad Division of Mental and Physical Health. 2Department of Expount and Risk Assesort and Oate fr Ftlay and Health, Norwegian Intitute of Public Health, Oslo, Norway. ●Department of Nunung and Health Prometne akMet-ab Metropdaan Usoruty. Odo, Norway, ,Modical R... • ### SYNOPSIS The product manager for coffee development at Kraft Canada must decide whether to introduce the... SYNOPSIS The product manager for coffee development at Kraft Canada must decide whether to introduce the company's new line of single-serve coffee pods or to await results from the product's launch in the United States. Key strategic decisions include choosing the target market to focus on and determining the value proposition to emphasize. Important questions are also raised in regard to how the new product should be branded, the flavors to offer, whether Kraft should use traditional distribution channels or... Free Homework App

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# Homework Help: Prove, if p and q are distinct prime numbers 1. Jan 27, 2010 ### ayusuf Prove, if p and q are distinct prime numbers, then sqrt(p/q) is irrational. I know how to prove that if p is a distinct prime number, then sqrt(p) is irrational. From there let sqrt(p) = q/r and then prove but for this I'm stuck. Do we let sqrt(p/q) = (a/b)/(c/d).Thanks. Last edited by a moderator: Jan 7, 2014 2. Jan 27, 2010 ### VeeEight Re: proof Assume not and that sqrt(p/q) is equal to a/b, for integers a, b and in lowest terms.. Then you can say a2q = b2p. This implies that p divides a2q, but q is prime, so p divides a2. Since p is prime and divides a2=(a)(a), we know it divides a. So write a as a multiple of p and write out your equation and rearrange to isolate b2 (or b). Can you find a contradiction here? 3. Jan 27, 2010 ### ayusuf Re: proof So from a2q = b2p which implies that p divides a2. I understand up to there but when you say write a as a multiple of p I get confused. Could you elaborate a little bit more on that? Thanks. 4. Jan 27, 2010 ### VeeEight Re: proof If p divides a2 then p divides a (why? There is the famous fact that if p is prime and p divides ab then p divides a or p divides b). So since p divides a, we can write out a=pc for some integer c. Now, back to the original equation a2q = b2p a=pc implies that p2c2q = b2p Now, isolate b2 in this equation. Can you say anything about p dividing b here? 5. Jan 27, 2010 ### ayusuf Re: proof Okay now I understand what you meant by multiple but where are you getting that famous fact from? Okay so now if we isolate b2 in this equation we will have b2 = pc2q and by taking the square root of both sides we'll have b = sqrt(pc2q) = csqrt(pq). Why would p divide b? 6. Jan 27, 2010 ### VeeEight Re: proof You have correctly isolated b2 but you do not need to reduce it. b2 = pc2q means that b2 is a multiple of p, which means that p divides b2. This implies that p divides b. Now p divides a AND p divides b, contradicting that a/b was in lowest terms. The famous fact (that prime p divides ab implies p divides a or p divides b) will be in the first few chapters of any number theory book. Here is a link: http://en.wikipedia.org/wiki/Euclid's_lemma. This fact is very important in number theory and even in group theory & ring theory so I would advise that you understand it. 7. Jan 27, 2010 ### ayusuf Re: proof The reason your saying b2 is a multiple of p is because c/d = k and if we isolate c we get c = dk. So in this c is b2, d is p and k is c2q right? also how do go from p dividing b2 to p dividing b? Thanks. 8. Jan 27, 2010 ### VeeEight Re: proof You yourself found the equation b2 = pc2q, which is the same thing as b2 = p(c2q), which means that b2 is equal to p times something. It doesn't get much clearer than that. b2 = bb and using the famous fact we see that p divides b. 9. Jan 27, 2010 ### ayusuf Re: proof Okay for the first part kind of what I was saying that c = dk. It's like saying 12 = 6*k where k = 2 right? I know it's pretty clear but I just want to make sure I fully digest this stuff. Okay now for p divides b2 because according to Euclid's lemma it must divide one of the factors right? Thanks. I really appreciate your help. 10. Jan 27, 2010 ### VeeEight Re: proof Yup. Yes. p is prime and divides bb so by the lemma it divides b.

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EN English Italian German Languages # Sound pressure measurement Sound pressure or acoustic pressure is the local pressure deviation from the ambient atmospheric pressure, caused by a sound wave. In air, sound pressure can be measured using a microphone, and in water with a hydrophone. ## Sound pressure and sound pressure level Sound is a wave motion in air or in other elastic media. It is caused by the objects vibrating at specific frequencies (loudspeakers, reeds, machinery). If an air particle is displaced from its original position, elastic forces of the air tend to restore it to its original position. Because of the inertia of the particle, it overshoots the resting position, bringing into play elastic forces in the opposite direction, and so on. Sound is readily conducted in gases, liquids, and solids such as air, water, steel, concrete and so on, which are all elastic media. Sound cannot propagate without a medium - it propagates through compressible media such as air, water and solids as longitudinal waves and also as transverse waves in solids. The sound waves are generated by a sound source (vibrating diaphragm or a stereo speaker), which creates vibrations in the surrounding medium. As the source continues to vibrate the medium, the vibrations are propagating away from the source at the speed of sound and are forming the sound wave. At a fixed distance from the sound source, the pressure, velocity and displacement of the medium vary in time. ## Wavelength and frequency In the picture below a sine wave is illustrated. The wavelength λ is the distance a wave travels in the time it takes to complete one cycle. A wavelength can be measured between successive peaks or between any two corresponding points on the cycle. This is also true for periodic waves other than the sine wave. The frequency f specifies the number of cycles per second, measured in hertz (Hz). ## Sound pressure Sound pressure or acoustic pressure is the local pressure deviation from the ambient (average, or equilibrium) atmospheric pressure, caused by a sound wave. In air, sound pressure can be measured using a microphone, and in water with a hydrophone. The SI unit for sound pressure p is the pascal (symbol: Pa). ## Sound pressure level Sound pressure level (SPL) or sound level is a logarithmic measure of the effective sound pressure of a sound relative to a reference value. It is measured in decibels (dB) above a standard reference level. The standard reference for the sound pressure in an air or other gases is 20 µPa, which is usually considered the threshold of human hearing (at 1 kHz). The following equation shows us how to calculate Sound Pressure level (Lp) in decibels [dB] from sound pressure (p) in Pascal [Pa]. where is the reference sound pressure and is the RMS sound pressure being measured. Most sound level measurements will be made relative to this level, meaning 1 pascal will equal an SPL of 94 dB. In other media, such as underwater, a reference level () of 1 µPa is used. The minimum level of what the (healthy) human ear can hear is SPL of 0 dB, but the upper limit is not as clearly defined. While 1 bar (194 dB Peak or 191 dB SPL) is the largest pressure variation an undistorted sound wave can have in Earth's atmosphere, larger sound waves can be present in other atmospheres or other media such as underwater, or through the Earth. Ears detect changes in sound pressure. Human hearing does not have a flat frequency response relative to frequency versus amplitude. Humans do not perceive low- and high-frequency sounds as well as they perceive sounds near 2000 Hz. Because the frequency response of human hearing changes with amplitude, weighting curves have been established for measuring sound pressure. We will learn in this course how to make a measurement and how to "distort" the results that they will match with what the human ear is "measuring". ## Frequency weighting curves A human ear doesn't have an equal "gain" at different frequencies. We will perceive the same level of sound pressure at 1 kHz louder than at 100 Hz. To compensate for this "error", we use frequency weighting curves, which give the same response as the human ear has. The most commonly known example is frequency weighting in sound level measurement where a specific set of weighting curves known as A, B, C and D weighting as defined in IEC 61672 are used. Unweighed measurements of sound pressure do not correspond to perceived loudness because the human ear is less sensitive at too low and high frequencies. The curves are applied to the measured sound level, by the use of a weighting filter in a sound level meter. • A-weighting: A-weighting is applied to measured sound levels in effort to account for the relative loudness perceived by the human ear. The human ear is less sensitive to low and high audio frequencies. • B-weighting: B-weighting is similar to A, except for the fact that low-frequency attenuation is a less extreme (-10 dB at 60 Hz). This is the best weighting to use for musical listening purposes. • C-weighting: C-weighting is similar to A and B as far as the high frequencies are concerned. In the low-frequency range, it hardly provides attenuation. This weighting is used for high-level noise. • D-weighting: D-weighting was specifically designed for use when measuring high-level aircraft noise in accordance with the IEC 537 measurement standard. The large peak in the D-weighting curve reflects the fact that humans hear random noise differently from pure tones, an effect that is particularly pronounced around 6 kHz. • Z-weighting (linear): Z-weighting is linear at all frequencies and it has the same effect on all measured values. ## CPB (Constant Percentage Bandwidth) analysis As opposed to the FFT analysis, which has specific number of lines per linear frequency (x axis), CPB (constant percentage bandwidth, called also an octave) has a specific number of lines if a logarithmic frequency x-axis is used. Therefore, lower frequencies have a higher number of lines and higher frequencies have a lower number of lines. CPB analysis is traditionally used in sound and vibration field. CPB filter is a filter whose bandwidth is a fixed percentage of the centre frequency. The width of the individual filters is defined relative to their position in the range of interest. The higher the center frequency of the filter, the wider the bandwidth. The bandwidth is defined in octaves or as a fixed percentage of the center frequency of the filter. Filters, which all have the same constant percentage bandwidth (CPB filters) e.g.1/1 octave, are normally displayed on a logarithmic frequency scale. Sometimes these filters are also called relative bandwidth filters. Analysis with CPB filters (and logarithmic scales) is almost always used in connection on with acoustic measurements, because it gives a fairly close approximation to how the human ear responds. The widest octave filter used has a bandwidth of 1 octave. However, many subdivisions into smaller bandwidths are often used. The filters are often labeled as “Constant Percentage Bandwidth” filters. A 1/1 octave filter has a bandwidth of close to 70% of its center frequency.The most popular filters are perhaps those with 1/3 octave bandwidths. One advantage is that this bandwidth at frequencies above 500 Hz corresponds well to the frequency selectivity of the human auditory system. Filter bandwidths down to 1/96 octave have been realised. A detailed signal with many frequency components shows up with a filter shape as the dotted curve when subjected to an octave analysis. The solid curve shows the increased resolution with more details when a 1/3 octave analysis is used. ## What is a microphone? A microphone is an acoustic-to-electric transducer or a sensor that converts sound in the air into an electrical signal. Microphones are basically pressure sensors, but they are dedicated to the measurement of very small variations of pressure around the atmospheric pressure. All microphones convert sound energy into electrical energy, but there are many different ways of doing the job, using electrostatics, electromagnetism, piezo-electric effects or even the change in resistance of carbon granules. The majority of microphones used in applications are either capacitor (electrostatic) or dynamic (electromagnetic) models. Both types employ a moving diaphragm to capture the sound, but they use a different electrical principle for converting the mechanical energy into an electrical signal. The efficiency of this conversion is very important because the amount of acoustic energy produced by voices and musical instruments are small. ## How does microphone work? Different types of microphone have different ways of converting sound energy, but they all share one thing in common: a diaphragm. This is a thin piece of material (such as paper, plastic or aluminium) which vibrates when it is struck by sound waves. In a typical hand-held microphone, the diaphragm is located in the head of the microphone. When the diaphragm vibrates, it causes other components in the microphone to vibrate. These vibrations are converted into an electrical current which becomes the audio signal. ## Types of microphones The way in which a microphone transforms an input signal from the acoustic to the electrical signal is known as its transducer principle. 1. The condenser microphone is the most widely used type. It is based on the principle that the capacitance of two electrically charged plates will change with their separation distance - it transforms an acoustic signal into an electric one by using a capacitor. In microphones, this capacitor is formed by the back plate and the light diaphragm, which moves in response to acoustic pressure variations. The diaphragm of the microphone is actually used as one element of this capacitor; when the acoustical signal causes the diaphragm to move, the distance between it and the other half of the capacitor is affected, creating the electrical output signal A condenser microphone is essentially a capacitor, with one plate of the capacitor moving in response to sound waves. The movement changes the capacitance of the capacitor, and these changes are amplified to create a measurable signal. Condenser microphones usually need an external power supply or a small battery to provide a voltage across the capacitor. A capacitor has two plates with a voltage between them. The diaphragm vibrates when struck by sound waves, changing the distance between the two plates and therefore changing the capacitance. Specifically, when the plates are closer together, capacitance increases and a charge current occurs. When the plates are further apart, capacitance decreases and a discharge current occurs. Condenser microphones usually need a small battery or external power to provide a voltage across the capacitor. 2. A dynamic microphone takes advantage of the electromagnetic effects. When a magnet moves past a wire (or coil of wire), the magnet induces an electrical current to flow in the wire. In a dynamic microphone, the diaphragm moves either a magnet or a coil when the sound waves hit the diaphragm, and the movement creates a small current. Using this electromagnet principle, the dynamic microphone uses a wire coil and magnet to create the audio signal. 3. Piezoelectric microphone - piezoelectric microphones feature sensitive crystals that respond to the physical vibration of the acoustic signal coming into the microphone and create the electrical output. While this technology was once widely used in tape recorders, it is most commonly used today in pickup devices for acoustic instruments. 4. Loudspeakers perform the opposite function of microphones by converting electrical energy into sound waves. This is demonstrated perfectly in the dynamic microphone which is basically a loudspeaker in reverse. ## Microphone's directionality A microphone's directionality or the polar pattern indicates how sensitive it is to sound arriving at different angles about its central axis. The pattern of microphones describes a three-dimensional orientation in space relative to sound sources in the ambient environment. Microphones can be divided into omnidirectional and unidirectional devices. • Omnidirectional microphone (non-directional microphone) can absorb acoustic signals from any input direction. • Unidirectional microphone can use several different polar patterns or shapes that indicate from which direction they will best receive an acoustic signal. Unidirectional audio pattern is great for focusing in on a specific sound and blocking out ambient noise. Unidirectional microphones are highly directional and must be pointed directly at the subject to capture the best sound quality. Cardioid microphones are unidirectional microphones that absorb an acoustical signal in a heart-shaped pattern. This sound pattern makes these microphones good vocal microphones, as they can absorb the dynamic range of a vocal performance while not capturing signals from other directions. Omnidirectional microphone is gathering a wide range of sounds from all directions. This kind of microphone is most accurate at representation of the total environment. It collects sound equally well from all directions. Bidirectional microphones receive a balanced signal from both the front and back of the microphonic element and are the best to use when we must capture the interplay of two sound sources. ## How to select the right microphone? Selecting a microphone involves a number of choices: • externally polarized or pre-polarized, • free-field, pressure or random incidence, • dynamic range, • frequency range. ## Externally polarized or pre-polarized Condenser type microphones require a polarization voltage which can either be supplied from an external power supply or the microphone itself can be polarized by injecting a permanent electrical charge into a thin PTFE layer on the microphone backplate. ## Externally polarized microphones These microphones are used with standard preamplifiers. The preamplifier must be connected to a power module or an analyser input which can supply the preamplifier with power as well as 200 V for polarization. Externally polarized microphones are the most accurate and stable and are preferred for very critical measurements. ## Pre-polarized microphones These microphones are used typically with constant current power preamplifiers. Pre-polarized microphones must be connected to an input stage for constant current power transducers or be powered by a constant current power supply. Constant current power preamplifiers use standard coaxial cables.The long-term stability and high-temperature stability of pre-polarized microphones are not as good as for externally polarized microphones. ## Free-field, pressure or random incidence Measurement microphones can be divided into three groups: free-field, pressure, and random incidence. The differences between microphones from group to group are at the higher frequencies, where the size of a microphone becomes comparable with the wavelengths of the sound being measured. ## Free-field microphones A free-field microphone is designed essentially to measure the sound pressure as it was before the microphone was introduced into the sound field. At higher frequencies, the presence of the microphone itself in the sound field will disturb the sound pressure locally. The frequency response of a free field microphone has been carefully adjusted to compensate for the disturbances to the local sound field. Free-field microphones are recommended for most sound pressure level measurements for example with a sound level meters and sound power measurements. ## Pressure microphones A pressure microphone is for measuring the actual sound pressure on the surface of the microphone’s diaphragm. A typical application is in the measurement of sound pressure in a closed coupler or, as shown below, the measurement of sound pressure at a boundary or wall; in which case the microphone forms part of the wall and measures the sound pressure on the wall itself. Pressure microphones are recommended for studies of sound pressures inside closed cavities. ## Random incidence microphones A random incidence microphone is for measuring in sound fields, where the sound comes from many directions e.g. when measuring in a reverberation chamber or in other highly reflecting surroundings. The combined influence of sound waves coming from all directions depends on how these sound waves are distributed over the various directions. For measurement microphones, a standard distribution has been defined based on statistical considerations; resulting in a standardized random incidence microphone. Random incidence is used typically for sound pressure level measurements according to ANSI standards. ## Dynamic range of a microphone The dynamic range of a microphone can be defined as the range between the lowest level and the highest level that the microphone can handle. This is not only a function of the microphone alone, but also of the preamplifier used with the microphone. The dynamic range of a microphone is, to a large extent, directly linked to its sensitivity. In general, a microphone with a high sensitivity will be able to measure very low levels, but not very high levels, and a microphone with low sensitivity will be able to measure very high levels, but not very low levels. The sensitivity of a microphone is determined by the size of the microphone and the tension of its diaphragm. A large microphone, with a loose diaphragm, will have a high sensitivity and a small microphone, with a stiff diaphragm, will have a low sensitivity. ## Upper limit of dynamic range The highest levels that can be measured are limited by the amount of movement allowed for the diaphragm before it comes into contact with the microphone’s back plate. As the level of the sound pressure on a microphone increases, the deflection of the diaphragm will accordingly be greater and greater until, at some point, the diaphragm strikes the back plate inside the body of the microphone. This is ultimately at the highest level the microphone can measure. ## Lower limit of dynamic range The thermal agitation of air molecules is sufficient for a microphone to generate a very small output signal, even in absolutely quiet conditions. This “thermal noise” lies normally at around 5 μV and will be superimposed on any acoustically excited signal detected by the microphone. Because of this, no acoustically excited signal below the level of the thermal noise can be measured. ## Frequency range of a microphone The frequency range of a microphone is defined as the interval between its upper limiting frequency and its lower limiting frequency. With today’s microphones, it is possible to cover a frequency range starting from around 1Hz and reaching up to 140 kHz. Low-frequency measurements require a microphone with a well controlled static pressure equalization with a very slow venting. Special versions are available for infrasound measurements. High-frequency measurements are very sensitive to diaphragm stiffness, damping, and mass, as well as diffraction. ## Upper limiting frequency The upper limiting frequency is linked to the size of the microphone, or more precisely, the size of the microphone compared with the wavelength of sound. Since wavelength is inversely proportional to frequency, it gets progressively shorter at higher frequencies. Hence, the smaller the diameter of the microphone, the higher are the frequencies it can measure. On the other hand, the sensitivity of a microphone is also related to its size which also affects its dynamic range. ## Lower limiting frequency The lower limiting frequency of a microphone is determined by its static pressure equalization system. Basically, a microphone measures the difference between its internal pressure and the ambient pressure. If the microphone was completely airtight, changes in barometric pressure and altitude would result in a static deflection of its diaphragm and, consequently, in a change of frequency response and sensitivity. To avoid this, the microphone is manufactured with a static pressure equalization channel for equalising the internal pressure with ambient pressure. On the other hand, equalization must be slow enough to avoid affecting the measurement of dynamic signals. ## How to connect a condenser microphone? We can connect the condenser microphone directly to the ACC or the ACC+ module. We used a prepolarized free-field microphone which already has an integrated preamplifier. A condenser microphone needs a power supply. When we select the IEPE mode, the excitation for the microphone will be provided. We can choose between 4 and 8 mA excitation. ## Microphone calibration In order to take a scientific measurement with a microphone, its precise sensitivity must be known (in volts per pascal - V/Pa). Since this may change over the lifetime of the device, it is necessary to regularly calibrate measurement microphones. Microphones can be calibrated in two ways. First, we have to know that the direct value of measurement from the microphone is the sound pressure in Pa. Therefore, we need to scale it to the physical quantity. ## Scaling with a calibration certificate If we don't use the calibrator but have the sensitivity of microphones, we can define it directly in the Channel setup. First, Pa is defined as the Physical unit of measurement. Next, we go to Scaling by function, check the Sensitivity and enter the value in mV/Pa, which can be found on the calibration certificate of the microphone. ## Calibrating the microphone with calibrator Another way to calibrate the microphone is with the calibrator. In this case, the known parameter is the sound level emitted by the calibrator. In our case, it is 94 dB (at 1000 Hz). First we have to enter the channel setup of the microphone. The sensitivity is set to 1 by default. In the right side of the microphone scaling section, we can see information from the microphone, which is plugged in the calibrator. The calibration frequency is set to 1000 Hz and the current value detected by the microphone is 127.4 dB. This is, of course, wrong because our calibrator has the output value of 94 dB. After we press Calibrate, the microphones sensitivity will be measured from the highest peak in the frequency spectrum, usually at 1000 Hz (using of course amplitude correction to get the right amplitude). Microphones sensitivity can be also read from TEDS. In that case, there is no need for calibration, because sensitivity is written on TEDS. After we press the Calibrate button, we can see that the sensitivity has changed. Also, under the current value we can see the number 94 dB. This means that our microphone is now calibrated. ## Sound level The Sound level Math section allows calculation of the typical parameters for sound level measurements from a single microphone. It allows for Dewesoft X to be used as the typical sound level meter. With appropriate hardware (Sirius ACC) it can easily fulfill all the requirements for a Class I sound level meter. It supports also different standards: IEC60651, IEC60804, IEC61672. Required hardware SIRIUS ACC, MULTI, STG, DEWE-43 with MSI-BR-ACC Required software Dewesoft X2, SE or higher + SoundLevelMeter option, DSA or EE Setup sample rate At least 10 kHz Sound level measurements are available selecting the Sound level meter checkbox under Math section. After selecting this option, a tab labeled Sound levels appears in the Dewesoft X Setup screen. Basic procedures of Sound level measurement are: • channel setup • microphone calibration • measurement ## Calibrating the microphone with the calibrator in Sound level This value is calculated directly in the Medium & Calibration field of the sound level module channel setup. We connect the calibrator to the microphone and turn it on. We can see the signal directly in the small overview. In our case, it should be a sine wave with a frequency of 1000 Hz. Since all the frequency weighted curves are referenced to 1000 Hz, this is a very usual frequency for calibrating microphones. We can also choose the medium in which we are measuring. It can be chosen from Air or Water, the difference is in reference sound pressure. After we see that the sound is correctly recognized as the sine wave at 1000 Hz, we can click the Calibrate button to perform a calibration. The sound module will calculate the Sensitivity of microphone from the highest FFT amplitude and reference value. The sensitivity will already be directly corrected in the source channel and, therefore, no additional analog scaling is necessary. We can directly check the calibrated sensitivity of the information found on the calibration certificate. Now we have to check if the calibration was successful. Set a sampling rate of at least 5 kS/s - we would recommend 20 - 50 kS/s - and enter the FFT analysis. Set the FFT options to Flat top filter and the Y scale type to dB Noise. Now press the RMS icon to display the RMS values within the FFT graph. Switch on again your microphone calibrator and the RMS values should display 94 dB. If there are mismatches you should do the calibration again. ## Measurement with microphone Because of differences in their construction, microphones have their own characteristic responses to sound. This difference in response produces non-uniform phase and frequency responses. The dynamic range of a microphone is the difference in SPL (sound pressure level) between the noise floor and the maximum sound pressure level. Sensitivity of a microphone indicates how well the microphone converts acoustic pressure to output a voltage (unit: mV/Pa). A high sensitivity microphone creates more voltage and vice versa. Microphones are not uniformly sensitive to sound pressure and can accept differing levels without distorting. For scientific applications, microphones with a more uniform response are desirable, but this is often not the case for music recording, as the non-uniform response of a microphone can produce a desirable coloration of the sound. This is why comparison of published data from different manufacturers is difficult because different measurement techniques are used. Frequency response diagram plots the microphone sensitivity in decibels over a range of frequencies (typically 20 Hz to 20 kHz). Frequency response may be stated as: "30 Hz–16 kHz ±3 dB". This is interpreted as meaning a nearly flat, linear, plot between the stated frequencies, with variations in amplitude no more than plus or minus 3 dB. Commonly made statements such as "20 Hz–20 kHz" are meaningless without a decibel measure of tolerance. Directional microphones frequency response varies mostly with distance from the sound source, and with the geometry of the sound source. Noise level is the sound pressure level that creates the same output voltage as the microphone does in the absence of sound. This represents the lowest point of the microphone's dynamic range (it is important if you wish to record quiet sounds). The measure is often stated in dB(A), which is the equivalent loudness of the noise on a decibel scale, frequency-weighted for how the ear hears (A-weighting). ## Example 1 In the example below we measured the sound that comes from an accordion. The condenser microphone was placed near the accordion and we measured the beating frequency of the signal. Signal from the microphone clearly shows that the accordion has more than one clean tone. We hear two closely spaced frequencies as a beating. ## Example 2 The next example with the condenser microphone was made with a cantilever beam. We measured the beams natural frequency. The microphone was plugged in ACC module and it was set to IEPE mode. It was calibrated as it was described on the previous pages. We excite the beam with a hammer and then it vibrates with its own frequency. The signal from the microphone was put into the FFT analyser. The first peak in FFT spectrum of the signal from the microphone was at 91.55 Hz. That was also very close to the beam's natural frequency measured with the frequency response function math (91 Hz).

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typodupeerror # Slashdot videos: Now with more Slashdot! • View • Discuss • Share We've improved Slashdot's video section; now you can view our video interviews, product close-ups and site visits with all the usual Slashdot options to comment, share, etc. No more walled garden! It's a work in progress -- we hope you'll check it out (Learn more about the recent updates). × ## Comment: let the punishment fit the excess kinetic energy (Score 1)760 It would help focus peoples' minds on the harmfulness of speed if, rather than setting arbitrary thresholds, the fines were proportional not only to income, but also to the excess kinetic energy of the vehicle: Ek = 1/2 . m . v^2 Kuisla was doing 65 in a 50 zone, so his "kinetic tariff" would be (65x65) - (50x50) = 4225 - 2500 = 1725 units. (It's the exponential nature of the velocity squared factor that yields the disincentive, but, by all means introduce the weight of the miscreant's vehicle too) This would be a rational way of reflecting the a) the risk of injury in the event of a collision; b) the undesirable environmental effects (noise, gaseous emissions), both of which rise exponentially with the kinetic energy of the vehicle. This kinetic tariff can then be applied to income data to calculate the actual fine for the individual. ## Comment: A practical DoS attack on the web (Score 1)396 by ColMstrd (#48624325) Attached to: Google Proposes To Warn People About Non-SSL Web Sites The choice about whether or not to encrypt traffic should be left to each website's administrator. Many sites--shock!--use the web to disseminate information they wish to be public, and the site's users have no problem with their access to it being public either. So get out of their faces! Using the browser to deprecate admin's particular choices is contrary to the spirit of the web, which should always do its damndest to serve something, and degrade gracefully when it's in difficulty, not pop annoying dialog boxes in the user's way. Self-certificates are already a fairly effective denial of service attack when Firefox is used to access many independent sites that try to implement https, but who fail to do so in a way that offers a smooth user experience to J. Random User (I'm thinking particularly of IndyMedia). Please note: in China, the censorship does not rely on blocking everything; just on blocking enough that all but the very motivated fail to access it. This troublesome minority can presumably be picked off at leisure later. Keep it simple, stupid! ## Comment: President was educated to do his job: it shows (Score 5, Interesting)139 by ColMstrd (#41538871) Attached to: French Science and Higher Education Programs Avoid Austerity yeh, the French know that sacking the public sector in times of crisis does not help the economy; quite the reverse in fact. M. Hollande is old school ENA (Ecole Nationale d'Administration) which turns out highly-educated senior French bureaucrats and politicians, who, whatever else they may be, are not daft. ## Comment: Re:Yahoo has TWO things that don't suck... (Score 1)311 by ColMstrd (#34585758) Attached to: Yahoo! To Close Delicious Diigo seems to work quite handily, but I still prefer Delicious and will mourn its passing. It's easy to export your Delicious bookmarks to an html file and import that into Diigo. ## Comment: Seed bank? Bah! Apocalyptic spectacle more like (Score 3, Informative)115 by ColMstrd (#34075022) Attached to: How the Global Seed Vault Aims To Fight Future Famine A single seed bank like this doesn't make any kind of biological sense. It is remarkably unlikely to be useful in the event of catastrophe: it's a long old road up there to Norway to replenish stocks of some ancient carrot variety from most parts of Europe. If you actually wanted to guard biodiversity, you would encourage social networks of gardeners to replant varieties each season and share the ensuing seeds. The French organisation Kokopelli does this, but seems to suffer from legal harassment rather than incur the subsidies it would receive in any sane world. An analogy for the slashdot crowd might be Napster (centralised) vs. BitTorrent (distributed). ## Comment: Re:Freecycle (Score 1)249 by ColMstrd (#32783954) Attached to: What To Do With Old 802.11b Equipment? that moldy alfafa will have made really nice mulch/compost for growing veg. Your old computers on the other hand are destined to pollute a water table somewhere. "Mr. Watson, come here, I want you." -- Alexander Graham Bell Working...

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A066742 Square numbers not divisible by 10 whose reverse is pentagonal. 0 0, 1, 529, 1089, 2116, 6241, 103041, 26222944225, 2648772525025, 6262611355441, 62443279235236, 1520130841269290490409 (list; graph; refs; listen; history; text; internal format) OFFSET 1,3 LINKS Table of n, a(n) for n=1..12. EXAMPLE 925 is pentagonal and 529 is square MATHEMATICA dtn[L_] := Fold[10#1+#2&, 0, L] A={0}; For[i=1, i>0, i++, t=i(3i-1)/2; r=dtn[Reverse[IntegerDigits[t]]]; If[IntegerQ[Sqrt[r]]&&Mod[t, 10]>0, AppendTo[A, r]; Print[A]]] CROSSREFS Sequence in context: A020289 A274932 A082409 * A067475 A052074 A112079 Adjacent sequences: A066739 A066740 A066741 * A066743 A066744 A066745 KEYWORD base,more,nonn AUTHOR Erich Friedman, Jan 16 2002 EXTENSIONS Description clarified, offset corrected, and a(12) added by Lars Blomberg, May 29 2011 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified July 15 02:08 EDT 2024. Contains 374323 sequences. (Running on oeis4.)

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It is important that you memorize the below three addition formulas as it is necessary to establish your foundations in trigonometry. From these addition formulas, you will be able to derive the double angle formula. Let A and B be any two angles. Then 1) s 2) 3) Without the use of tables or calculator, find the value of and in surd form. We have to figure out the common angles like and that add up to . We do this because we know the numerical values of cosine and sine of these common angles. Then we expand it out using addition formulas and solve for theanswer. = = This is similar as well. It's just that in this example, subtraction is required to obtain . Now, we are using the addition formula for tangent. = = = = = If we had used the calculator instead of addition formulas , it would have been very difficult to arrive at the exact value that was required. No credit would have been rewarded.

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## Friday, May 15, 2009 ### Is 1 = 0.999999... ? Some years ago a computer science guy who looked eager to pick a fight crossed my way. He asked me if I think 1 equals 0.99999... or doesn't. Notice the three dots supposed to say the nines continue infinitely. This infinite continuation is commonly abbreviated with a bar and looks like this Searching in what was left from my 6 semesters maths I said I think so, upon which the guy looked disappointed. I figured it was the right answer - right at least for my purposes. We then had a pleasant conversation about graphic cards I believe. Anyway, recently this question crossed my path again. It seems to have some eternal fascination, so let me just offer my take on it. The left side is the limit for n to infinity of  the sum over 9 x 10-m from m=1 to n. The limit of this sum for taking n to infinity is one, according to the most straightforward proof you can think of, namely if you pick any number only an epsilon smaller than 1 there will always be an n0 such that the sum is closer to 1 than that number for all larger n. Thus, both are equal. You find a furious discussion of the topic here. This post was created by random association to Mark's post You can't write that number; in fact, you can't write most numbers even though you can write that number. Wishing you all a good start into the weekend! PS: As was pointed out in the comments, Wikipedia has an excellent site on the matter. Anonymous said... 1/3 = 0.333333..... 3*(1/3) = 3* 0.333333333 1 = 0.99999999 Bee said... which merely states that the first statement is identical to the last, not that either is correct (and you forgot the dots...) Blaise Pascal said... I find the "1 = 0.9999..." bit annoying because it means every rational number of the form a/(b*2^m*5^n), where m,n are nonnegative integers not both 0, has two decimal representations. It's a complication which makes Cantor's Diagonalisation argument a bit tricky to get right. Tiger said... Hello Bee, There are lots of ways to see this result. I like the observation that all periodic numbers are equal to their period divided by as many nines as the number of digits in the period. For example 0.48484848...=48/99 In this case 0.999...=9/9=1 Or else, one must agree that there exist an infinity of numbers between two real numbers. Since there is no X so that 1 > X > 0.99999..., then the numbers are equal. We could go on and on with descriptions of this fact, in rather different ways. But from a simple depiction to a convincing proof, there is quite a leap. "The reasons one need in order to be convinced by a reasoning are psychological in nature, in mathematics like anywhere else." Henri L. Lebesgues (1875 - 1941), "Courses on integration and research of antiderivative functions", p. 328 (traduction from french, sorry if it isn't perfect). The reasons why it is so difficult to accept for non specialists, are the same as in Zeno's paradoxes. The concept of infinity is startling, even when only considered as potentially, merely trough the concept of limit. And because one can live an entire life without coming across such a dilemma, most people still think like the ancient Greeks... Best, Johann. Bee said... Hi Johann, Yes, the issue is of course here as in many other cases that infinity is startling. It's a concept that seems to be difficult to grasp. There must be an infinite amount of 'proofs' where people treat infinity like a number and come to contradictions. I am not entirely sure how come though. At least we learned how to take limits already in high school. I think a lot of damage can be done by careless statements (as unfortunately many physicists like to use them). Either way, I've found in several cases that one is likely to find online a correct rather than a wrong answer to questions like this. (That is not to say you wont find discussions about it with many misunderstandings.) So maybe we'll see some improvement on these matters thanks to bloggers like Mark and others who spread mathematical soberness. Hi Blaise, Best, B. Anonymous said... From the algebraic prespective ,the question " 1 = 0.9999999... ? " is very annoying!Let R* be the multiplicative group of the real numbers,now since R* is a group then there exists only one identity e such that e times x = x for all x belongs R*.Clearly in R* ,e = 1 is the only element satisfying e times x = x for all x !Now putting 1 = 0.9999999....will be seen as trivial and inconvenient change of notation! Regards, Serfo I understand the 1=0.9999... concept. And I won't deny that it took a lot of the math to accept that (lots of internal resistance to it). My question, however, is that is there a different representation then for something like 0.7777... Or does 9 being the largest single number in the decimal system, give it this special property of having a different representation (i.e. 1 in this case)? That seems a little arbitrary to me, however. Anonymous said... And as expected, Wikipedia has a very nice page about the subject : http://en.wikipedia.org/wiki/0.999 CapitalistImperialistPig said... 0.999... is no more or less strange than 1.000... except that by convention we don't write the extra zeros and dots. CapitalistImperialistPig said... Addicted - It is our place value system that gives us two representations for 1.0 There is a closely related problem in finite mathematics well known to computer people. Positive and negative binary numbers can be written in so called one's complement or two's complement form. One of them has a double representation included. Tiger said... It's not a property of the number 9. It is related to the concept of infinity and therefore of continuity (which it simplifies, again trough the concept of limit) : If two real numbers (or points on a line for that matter) are exactly consecutives, then they're the same. For two (differents) of them, an infinity of others lie in between. It just so happen that we have a name for the exact "successor" of .999... and it's 1. So they're the same. It's more tricky for the "successor" of .777... but it exists. At least ideally. Just to prevent misinterpretations, I must add that the concept of "consecutive" or "successor" are not involutary, that is to say that the successor of the successor of a real number isn't the same number (this intuitive representation and its counter-intuitive properties can be backed up by topological concepts). PS : I struggled with myself to post this answer, because I don't want to participate in transforming your post into an argumentation about this hot subject (I know by experience on some forum that it is)... By the way I don't know how you can keep up with all the posts and numbers of comments it generates, it must take quite some time. Best, Johann Another fun (mean?) blog post would be taking all of the hilarious misconceptions taken from forum arguments on this topic and mocking them. Giotis said... The mathematical treatment is just one aspect. More than that the eternal battle of mankind with infinities is a deep philosophical issue. Phil Warnell said... Hi Bee, I can imagine how such a question which relates to our understanding of the concept of infinity would provoke endless debate. That is as Blaise has eluded it all focuses primarily around whether infinity exists in its entirety or is simply a concept of ultimate potential, as in suggesting direction. This marked for instance the great rift between Cantor and Kronecker. In fact it goes so far that if one accepts Cantor’s definition of an infinite set and the axiom of choice, mathematically thereafter we can get pretty much something for nothing. The question of course that remains is if this extends to the real world or not? Best, Phil Guido said... It's not a property of the number 9. In fact, something similar happens, whichever base you choose. E.g., in base 8, 0.777... = 1 In base 13, 0.CCC... = 1 (where capital letters are used to denote "decimals" with higher value than 9, i.e., A=10; B=11; C=12 and so on). Some more fun with different bases: 1/7 in base 10 repeats itself, and therefore does not allow two different representations. But in base 7 it is not repeating (in base 7, 1/7 = 0.1), so it does allow another, infinite representation (0.0666...). It's all due to the base you choose. In a certain sense, 0.999... = 1 because we have 2 hands with 5 digits each (well, most of us do). Phil Warnell said... Hi Bee, How about this way to frame the argument being that the world of matter exists within the set of .999..., while the realm of energy exists at 1.000..., as it doesn’t have the time ;-) Best, Phil Peter said... The "furious discussion" you link to contains the following, "Notice that I keep putting the word definition in bold face", which is the seed of its own destruction. On all those pages and here there seems to be only one mention of Robinson/non-standard analysis, subsequently ignored. None here yet. Wikipedia, non-standard analysis. Robinson introduces (consistently) a class of infinitesimals, let's say that delta is one of them. 0.9999... is notation for a limit, as you say; every term in that series is less than 1-delta, hence the limit is less than 1, in Robinson's number system. One might not like non-standard analysis, or think that it adds nothing to our ability to do interesting mathematics, or just prefer not to complicate our definitional system unless we absolutely have to, but it is a matter of definition whether 0.9999...=1 or not. Arun said... If x = 0.999999.... 10 * x = 9.99999.... = 9 + x or 9 * x = 9 This is a general method to find out what fraction corresponds to a recurring decimal that we were indoctrinated into in high school. i.e., given 0.abcdef...abcdef...abcdef... multiply by the appropriate power of 10, etc. IvanM said... I think what nonmathematicians tend to miss is that all this depends on definitions and axioms. We have to say what we mean when we write an infinite decimal expansion. Bee correctly gives the interpretation of an infinite decimal expansion as an infinite sum, and the rest follows from the axioms of the real field. If we work in a different system like the hyperreal numbers, then we DO have numbers which differ only by an infinitesimal and which we DO consider to be distinct. (I just noticed that Peter beat me to the point.) Phil Warnell said... Hi Arun, Not that I have an opinion either way, yet if .99999... is a potential as in being a direction, rather than a destination, can it be defined simply as equivalent to X ? That is for many X considers that something can be defined, which is what one could say is the essence of being finite. So in this way to insist that .999... is equal to X or 1 would exclude the consideration of the infinite. To be honest I try best to avoid thinking too seriously about infinity, since minds much greater than my own feeble one have paided a heavy price for such consideration. That’s what I most admired about Bertrand Russell is he realized this early enough and switched his focus to philosophy:-) Best, Phil Bee said... Peter, Ivan: You are of course right that it all depends on the definition. It always does, thus my first impulse was to ask well, what do we mean if we write "..." You are right in that with a different definition the above equality might indeed not hold. You could also go and argue that there is in practice no way to ever write down any infinite series, thus the series 'exists' in a merely Platonic sense. (As CIP pointed out above though, the right side has an infinite series of zeros too, so the best you could say realistically is that they agree to some precision.) Bee said... Hi Phil, The infinite can be defined, it just isn't a number. But besides this, it is easy to see that 0.999... is not infinite: it is bounded from below by zero and bounded from above by 2, thus you can exclude this case easily. Best, B. IvanM said... You could also go and argue that there is in practice no way to ever write down any infinite seriesI don't think even the strictest of contructivists would ever argue that, would they? It is certainly true that you can't write down most infinite series, though, since there are uncountably many infinite series and only countably many finite strings of symbols which describe series (the point of Mark's post). But no one cares about the uncomputable numbers anyway. :^) I suppose one could argue that the field of real numbers doesn't exist, and only the field of computable real numbers really exists. But the field of computable numbers doesn't have the least upper bound property, which kinda sucks. IvanM said... Grrr... the first paragraph of my previous comment needs to be split in the middle. (The preview looked fine. Weird.) Bee said... Well, I can with certainty tell you that people exist who believe almost all real numbers don't 'exist' for exactly this reason. ('.' because one can argue about what it means to 'exist' which is a discussion I actually don't want to enter.) Bee said... Reg the paragraph break: It's a known blogger bug, and I hope it will be fixed soon. All linebreaks after italics, bold face or links will vanish with submission. It's pretty annoying but not your fault. Phil Warnell said... Hi Bee, “The infinite can be defined, it just isn't a number. But besides this, it is easy to see that 0.999... is not infinite:” I think that in essence is what I said, yet I will avoid saying anything other than to ask if infinity is not a number, then how can we be confident that it can be logically consistent to have it aid in defining one? Cantor showed that we can envision number sets that while bound still form infinities being uncountable, with proportions that are variable rather than fixed. So if we were to divide an infinity into another we would not necessarily get 1.000... as a result. The problem as I see it is not with what a series continued to infinity will yield, yet what infinities we consider apply and constitute being responsible for what we call real. It is true that the recursive sequence discussed is considered (and proven) as countable, yet it doesn’t preclude or quantify the density that might exist between each recursion. This is like the old Zeno problem in asking what we need to pass through at each step to reach the next. So I’m not troubled that within the confines of the definition and rules chosen .999... is the same as one, its simply that this is so loosely defined as being infinite to begin with. I guess what I’m driving at is should calculus be imagined as how we consider the beginning and the end to physical limit or only the outline for which we require something more to fill the gaps. The only thing I’m certain of is the hare does catch the tortoise in the world we live in and yet completely uncertain as to exactly how:-) Best, Phil Peter said... I'm curious what polymathematics thinks of non-standard analysis. With the existence of the Wikipedia 0.999 page, presumably he must know of them now. I'm pragmatic enough that I've never found a reason to use non-standard analysis (except for dabbling with Columbeau functions as a possible way to formulate more rigorously renormalization's desire to manipulate infinite quantities), but the consistent existence of such a system should surely give a True Defender some pause. Up to a point, non-standard analysis gives a rigorous foundation for many of the informal ideas that non-mathematicians think should be present in a discussion that polymathematics is so eager to deny can be a part of a formal system. Many of the commenters try to say that there are numbers between 0.9999... and 1, polymathematics argues that no there aren't, but non-standard analysis says, I define them thusly, and it is as good as your definitions. Where polymathematics keeps saying no, no, no, he should have been saying, go read about non-standard analysis on Wikipedia or wherever, or Hyperreals, where you will find what you have to do to make your crude intuition work as a formal system. When you've done that, you'll probably agree that it's really easier and more useful to use the standard definitions, or at least we can agree to disagree about our preferences. Personally, I find the dependence of mathematics on how we humans define our formal systems makes me considerably less decided about the status of mathematics. Bee said... Hi Phil, Well, it is logically consistent because it is well-defined. I am afraid I don't understand your concern. From the mathematical point of view there is nothing ill-defined about infinity, one just has to be careful to deal with it correctly. A lot of the confusions and misunderstandings arrive from exactly the kind of problems you are mentioning: dividing infinity through infinity, subtracting infinity from infinity, etc, instead of taking the appropriate limits. Best, B. Jean-Philippe said... Hi all, I have not read the detail of all the comments, I just wish to provide a framework to sharpen a bit our intuition on this matter: Decimal representation is just a language to speak about real numbers, and as many languages, it has synonymous, 0.9999... and 1 are just two of them that refer to the same object. Now, an interesting question is: Is there a language to speak about real numbers that does not have synonymous? The answer is yes, and can be found in the first theorem of the second chapter of Khinchin's "Continued Fractions": To every real number α, there corresponds a unique continued fraction with value equal to α. This fraction is finite if α is rational and infinite if α is irrational. Cheers JP Plato said... As an outsider I found the topic to be an interesting one from a mathematical perspective. So of course one does their research and are at the mercy of what's out there as to a conclusiveness of design and meaning. Bee:The infinite can be defined, it just isn't a number. But besides this, it is easy to see that 0.999... is not infinite: it is bounded from below by zero and bounded from above by 2, thus you can exclude this case easily.This quickly dispatches the argument, while I was immersed in my sleep time as to responses to your example, now in relation to the cosmos. So you see, I take this seriously. The mathematical basis then becomes a struggle for me as to definition. Sure it's easy for all you math types, but lets see if your up to expanding this basis to interpretations to the cosmos? Is it wrong? "if one traveled in a straight line through the universe perhaps one would eventually revisit one's starting point." ....may mean an objectivity defined expression is a connection with the confines of "Bee's mathematical parameter" so it constitutes a meaning "within only this interpretation." So you hold to that and see in cosmology there is no other "outside the box" explanation that fits while mathematically holding. But if one can define a greater "sense of dynamics" then you cannot hold onto that definition any longer? Peter sees a greater sign then in explanation to wiki? Definitely a learning experience here:) Best, Plato said... An Intermediate Polar Binary System. Credit & Copyright: Mark GarlickStephen Hawking’s says: “Roger Penrose and I worked together on the large scale structure of space and time, including singularities and black holes. We pretty much agree on the classical theory of relativity but disagreements began to emerge when we got into quantum gravity. We now have different approaches to the world, physical and mental. Basically, he is a Platonist believing that’s there’s a unique world of ideas that describes a unique physical reality. I on the other hand, am a positivist who believes that physical theories are just mathematical models we construct, and it is meaningless to ask if they correspond to reality; just whether they predict observations.” ( Chapter Six-The Large, the Small and the Human Mind-Roger Penrose-Cambridge University Press-1997) Bold added by me for emphasis This discussion obviously goes far beyond the "mathematical obvious here." While it is distinctly clear as to the explanation of the "mathematical obvious" there is a thought lying just outside of this example, is a whole basis of approach to one's science?:) Maybe Bee's and Peter's views "lie within it?" Best, Count Iblis said... http://arxiv.org/abs/0811.0164 Neil' said... Yes, one is "equal to" the formal definition of 0.999... because of what "equal means": to have the same value, even if the "means of representation" is different. Well, that's kind of the whole point. Why write "5 = 5", which you already know. Writing " 2 + 3 = 5" actually tells you something: that the sum of two and three equals five. If you believe in infinite series being "real", then 0.999... = Sigma 9/10 + 9/100 + 9/1000 + ... = 1, and therefore the expression is "true" in the same way as "2 + 3 = 5." It just looks oddly put. I know that some people think a representation should have some proper logic to it, put in the simplest form, etc. OK, maybe, but the statement is still "true". But if you're into surreal numbers and nonstandard analysis (which I barely know or understand), it may be an issue. (Does 0.999.... really equal 1 - epsilon or some weird little infinitesimal?) tyrannogenius Phil Warnell said... Hi Bee, I truly didn’t mean to dispute as to argue, yet simply to wonder out loud if you like. I guess one way to express it, is as Cantor has it the uncountables if envisioned on a number line must be either without dimension or not confinable to a line to exist one dimensionally. So it begs the question if all is actually nothing or rather the limit to which it’s considered confined incorrect? Another way to put it is to ask why there is dimension at all, not just simply more than one, to suggest that it’s mandated when mathematics is manifested into something that’s required to be real. Mathematics and its cousin logic are rife with contradiction and paradox and I see this debate as no different or unusual. For instance to calculate the proportions of a circle or sphere to completion cannot be done without evoking infinity as a requirement and yet they being economical in construction and symmetrical in design are qualities by which they can be defined without calculation and this has been found to be even more fundamental in terms of understanding. So as a bubble or a planet needs not infinity to describe its shape why would a number require it to define its value? “From the mathematical point of view there is nothing ill-defined about infinity, one just has to be careful to deal with it correctly.” Would you settle for incomplete rather than ill-defined (no pun intended)? That is when it comes to if there exists magnitudes of infinity between the countable integers and the unfathomable density of the reals has yet to be decided or even proved to be decidable. "God made the integers; all else is the work of man" said Kronecker. If this be true then nature was certainly patient while those who where to create the language to describe and parameterize its action could be formed within such limitation. Yet perhaps this suggests that eternity does actually exceed infinity or that when compared to a true continuum such magnitudes have little if any meaning in measure. I understand this is purely philosophical and yet I can find no other context in which to express it. None of this of course is to suggest an answer, just simply questions and musings as what could constitute being some of the parameters. Best, Phil James said... Phil, I think by the "uncountables" you meant the irrationals, or perhaps the trancendentals? You probably need to look into "topological dimension" (there are many definitions of dimension appropriate to diferent settings). For example, with their natural inherited topologies as subsets of R, the rationals have topological dimension 0 and the irrationals 1. Of course R itself has the intuitive value of 1 as its topological dimension (at least with the natural topology...). The "intuitive" notion of dimension, appropriate for surfaces of spheres, spacetime, etc, comes from their being maifolds and looking locally like R^n, which does work for arbitrary sets. With infinity, you need to specify what kind? There's the cardinality of sets a la Cantor, used in limits, induction (possibly transfinite), or, perhaps the extended real line where + and - infinity are the very "real" points stuck on each end of the normal real line (with adjustments to the algebra, topology, and order to accomodate them). James said... Bad example - it's well-orderdness (the ordinals) used in induction. A better example is to take an infinite set (say the integers), and then take its power set (the set of all of its subsets) and show that you can't pair the elements up from each set - so they do not have the same size (or "cardinality"). In fact the power set is bigger. This process can obviously be repeated giving an infinite hirarchy of infinities. Sinan Yayla said... This comment has been removed by the author. estraven said... This is maybe OT, but in "Extremely loud & incredibly close" J Safran Froer basically writes 1-0.7777... = 0.3333.... Maybe if someone had taught him as a child that 1=0.999... he wouldn't have made such a stupid mistake. [Yes, I know it depends on the definition. However, there is such a thing as a standard definition of the real numbers]. Anonymous said... This post is probably dead, but my 2 cents nonetheless: 0.9999... is short hand for an algorithm to yields the series 0.9, 0.99, 0.999,... It can then be shown that this series converges to 1 in the usual metric topology on the real line. So the a more complete statement would say: The series defined by 0.999... converges to 1 on the real line in the topology generated by the standard distance metric. As a counter example if you use the topology generated by equivalence classes of the Hamel basis over the field of rational numbers for the real line, then the series 0.999... does not converge, let alone to 1. Bee said... Our posts don't die that quickly... Anonymous said... I suppose not, and good thing too! The other interesting thing is that the topology in the counter example is infinite dimensional, while the usual metric topology is one dimensional. As well the counter example topology is metrizable, and is homoemorphic to the L2 norm topology on the space of all square integrable functions on the field of rationals, the trick being making the appropriate choice of measurable sets on the rational numbers. James said... Yeah, but you need the axiom of choice for your Hamel basis - which just complicates matters... Anonymous said... "God created the integers, all else is man's bastardization" Sounds like someone is an explicit constructionist. You also need the axiom of choice to define Borel and Lebesgue measures, and without those basic concepts its meaningless to study quantum mechanics, probability theory, most of analysis,... ned said... Wikipedia is quite clear on the subject; still 0,9999... does not equal 1. It's tied up with the concept of infinity. Infinity means *always* a little something will be missing in 0,9999..., no matter how small - the series is unending, or it would not be infinite. The summing up never arrives at an end - "in the infinite it amounts to 1" has per definitionem no meaning, a simple fact which one tends to conveniently overlook. The number 1 on the other hand is no series, it has never "departed", so it does not need to arrive anywhere, it just is where it is. Of course not only "professional" mathematicians will groan at this presentation, but that does not mean they are right. They have taken the concept of infinity and twisted it so it fits with the needs of their (equally twisted) minds, so of course they get the results they desire, and can prove them to their satisfaction, too. Bee, "(Infinity is) a concept that seems to be difficult to grasp." is not quite true - it is *impossible* to grasp, and that's the root of the problem. If you can grasp it, it's finite. *Any* grasping of the infinite is bound to be wrong, and that's that; a fact which guarantees endless discussions, because everyone can prove that the other's opinion is wrong. Of course one can escape this situation by e.g. "defining" what infinity mathematically is - do you see the absurdity of that? (... while we're at it, next let's define the indefinable; we already believed more than six impossible things before breakfast ...) polymath said... So...as the author of the referenced page, here's my take. Yes, I know about hyperreals and non-standard analysis. In the follow-up posts, I'm very careful to say that I'm discussing the real numbers as they are normally understood. I do know that .999... equals 1 only in that real number system, but the fact is that most deniers can't argue correctly within that number system, let alone understand the hyperreals. The post and subsequent refutations of counter-arguments are directed at them, not at those in the know about richer number systems. James said... "Of course one can escape this situation by e.g. "defining" what infinity mathematically is - do you see the absurdity of that?" Defining something and following the rules to reach conclusions!!! NO! NO! NO! - that's just cheating and wrong!!! Actually, that's mathematics in a nutshell. ned said... just wanted to complete a sentence in my previous post: "... a fact which guarantees endless discussions, because everyone can prove that the other's opinion is wrong" - and thus (wrongly) infer that one's own opinion must be right ... Bee said... Hi Ned, "Bee: (Infinity is) a concept that seems to be difficult to grasp. Ned: is not quite true - it is *impossible* to grasp, and that's the root of the problem. If you can grasp it, it's finite. *Any* grasping of the infinite is bound to be wrong, and that's that; a fact which guarantees endless discussions, because everyone can prove that the other's opinion is wrong. Of course one can escape this situation by e.g. "defining" what infinity mathematically is - do you see the absurdity of that?" I think we are talking past each other. It depends on what you mean with 'grasp'. If you mean with 'grasp' you can picture it in your head, then I would agree that I can't 'grasp' infinity more than I can grasp ten dimensional manifolds or the exterior algebra. What I meant with 'grasp' however is that you can give a well defined meaning to it and learn how to deal with it self-consistently. And unfortunately, many people seem to have difficulties with that 'grasping'. I don't know what's "absurd" about that. As James said above, that's mathematics which happened to be the topic we were discussing. Best, B. ned said... Bee, "What I meant with 'grasp' however is that you can give a well defined meaning to it." Exactly the point - you cannot. If you take infinity and "give a well defined meaning to it", it ceases to be infinity. What you have then is a completely new concept which you just keep on calling infinity, and so delude yourself that it's the same thing as before, in a "well defined mathematical way" at least. You can define it as well as you want, it's still something completely new - not a bad thing in itself, maybe amazingly useful, maybe Fields-Medal worthy - just not infinity anymore. Call it "Sabine" - that would be more honest, and nice, too (just imagine: "... for mathematically defining the concept of Sabine, basis for the design of the first working wormhole connecting us to another universe"). "and learn how to deal with it self-consistently." No problem, if one learns some math. "And unfortunately, many people seem to have difficulties with that 'grasping'." Maybe they see something you miss - that you have grasped something, but it's not infinity ... "I don't know what's "absurd" about that." Trying to define the indefinable is not absurd? "As James said above, that's mathematics which happened to be the topic we were discussing." Mathematics, physics, painting, no matter what: if you want arrive at some sort of truth, you have to be honest - if you create a new concept then give it a new name, so there's less chance of confusion. But mathematicians are human beings, too, who delude themselves and others, start from not thoroughly thought out assumptions, make errors, try to hide them - like all human beings do. The problem arises when they forget that and try to live up to some ideal of a superhuman, pure mathematician who lives only in the highest regions of pure math and therefore can do no wrong - then they find it hard to admit it before themselves and more so before others when they've fucked up, and consequently it becomes difficult for them to correct an error. (Just the same as it's with other human beings, by the way, only of course these aspire to all sorts of other equally impossible ideals.) James said... "What you have then is a completely new concept which you just keep on calling infinity, and so delude yourself that it's the same thing as before, in a "well defined mathematical way" at least." What exactly was infinity before then? Please define the previous notion. However you define it we can work with it - but you do need to define what you are talking about, be it in maths, philosophy, music, or gardening... otherwise its just hot air Tiger said... Hi, I was waiting for an ancient Greek to give its opinion... Congrats ned ! You just time-transported your thinking two thousand years ago ! (No offense, I'm just kidding) Well, the fact is you seem to either know better than anyone what is infinity, and disagree with the common distinction between potential and actual infinite (here there's only a potential infinite concerned in the matter, so if someone tries to "grasp" more than its potentiality, then this someone makes a primitive mistake); or you've already decided that infinity is an undefinable concept in its potential and actual form (just like any fundamental concept that lies in mathematics' axiomatics or logical rules, like points, lines, sets, etc)... Accepting a potential infinite is of no difficulty whatsoever (think about Peano's axiomatic), and we've come a long way from Zeno's paradoxes, so if you are unsatisfied with it, it's because you try to grasp it in its whole. In order to do that, you need to define ordinal numbers, with the concept of set, and it is in the power of mathematics to be able to postulate the existence of an infinite ordinal and build up an entire theory on it. Maths doesn't care if a concept exists or not, it just tells you what it implies, whether it hurts your "grasping" or not. If you can't accept that counter-intuitive concepts can extend the realm of science coherently, then I guess all you are left with is counting on your fingers and tell others that the square root of two isn't a number. Best, Johann Phil Warnell said... Hi Ned, “and thus (wrongly) infer that one's own opinion must be right” When it comes to mathematics what one thinks or what opinion you hold matters for not, as it’s only what you can prove that counts. The only thing open for debate here is the definition of infinity one uses being appropriate (logically) for what one is describing. One definition of infinity is to call it a direction rather than a destination (which seems to be your own contention). The paradox of course is how one can consider a direction without knowing where one is headed, which in turn requires a destination as a reference. So to simply being content with saying that infinity is defined as being without end in itself can be seen as inconsistent. That is .999... as it’s being considered here is both given a direction and a implied destination which in this instance is 1 . Now with the square root of 2 you can also describe the destination as the place where the proportion found when multiplied by itself will be equal to 2. This destination may be more complex in calculation yet still it exists. The same could be said for PI or any other irrational number, yet it also requires infinity to be used as one of the parameters in defining the quantity of steps taken in the direction to arrive at the destination. So you may be content that one can have a direction without there being a destination yet that doesn’t make it logically consistent to insist there isn’t one. The .999 in this example describes the steps required to arrive at the destination (route) and infinity describes how many are required to be taken. Infinity is this case is neither the direction nor the destination, yet merely the quantity of steps required to reach it. This may not be a quantity conceivable within everyday experience, yet this is a personal limitation and not one resultant of infinity. In this context what Cantor proved then is not that some destinations are any further away than others when infinity is evoked, yet rather some types of destinations are more numerous than others, like the irrationals are when compared to the integers. That there is a difference when considering the magnitude of steps in a ordered series and a group whose order cannot be so definitely given. The question then is why? It has been found that the understanding of this lay outside mathematics as now conceived. This is not a failure of mathematics, yet merely an indication that complete understanding lays somewhat beyond its reach. I’ve always liked to think it represents the difference between certainty and possibility with the possible always exceeding the certain. . Best, Phil Anonymous said... What it comes down to is this. Yes Ned the game is fixed in a certain way: in that you can make some fairly arbitrary choices of axioms, up to self consistency restraints, a la Godel and Russel. So you can state that the 0.999... does not limit to 1, but to do that you have to chuck out the axiom of choice, and I think trans-infinite induction. If you give up these axioms, but keep the basic axioms of propositional and existential logic you do get a self consistent set of axioms, but one that is much more limited in what can be proven, in fact it is so limited that most of 20-century mathematics and physics cannot be derived. Axioms are like the metric system, I cannot say that the axioms of logic are absolutely true, I have to take them as definitions. They have shown themselves to be very strong definitions in their ability to allow us to make valid and useful reasoning and predictions about our universe. If you want absolute truth I suggest studying theology. James said... Aaron, This is off topic, but you earlier said that AOC was needed to define Lesbesgue measure. I'm too lazy to go back to the (lengthy) Caratheodary construction so perhaps you can point me to the place where it is used. Incidentally, I am not a constructivist - I love the axiom of choice and all of the absurdities that it implies! Anonymous said... Not so much the construction using outer measures, but rather computations done with Lebesgue measures. I think it can be found at the end of Rudin: The Lebesgue measure of the countable union of disjoint Lebesgue measurable sets is the sum of the measures. James said... No, as I suspected, the AOC is NOT needed for the construction of Lesbesgue measure. It is, however, needed to "construct" an example of an un-measurable set, as first done by Vitali. James said... "The Lebesgue measure of the countable union of disjoint Lebesgue measurable sets is the sum of the measures." That's just "countable additivity" which any measure has to have - no AOC there either... Phil Warnell said... Hi Bee, There is one consolation for those that consider that .999... is not equal to 1, which is to say that when examining a calculated result that is say .99999999999999999999999 or for that matter any recurring series less numerous then actual infinity could be (although very unlikely) resultant of a equation or method of calculation not equal to 1. That is sort of like simply examining a series of numbers and trying to determine if it represents one that is random or not. For instance if one didn’t know the method for calculating Pi or what proportion it represents it would prove difficult to distinguish any finite series of digits within it as being non random. A random series by the most modern definition is one that can’t be described with information less quantitative then itself. When you evoke infinity into this consideration you find in essence that without all information this then remains definitively undeterminable. That is like much of modern physics mathematics in many instances one can only decide what is probable, rather then what’s completely determinable. In some respect this represents what Godel meant when he implied mathematics as being incomplete rather then it being wrong. Best, Phil Phil Warnell said... Hi Bee, Just as a qualifier and to make more clear what I mentioned above is to say that at some point the sequence .99999999999 demerges, like for example .9999999999999999999991... Best, Phil Anonymous said... Sorry I didn't state that clearly. It is required in proving that the subsets that satisfy the outer measure addition condition form a proper sigma algebra. Good ole wikipedia says to refer to Measure Theory by P. Halmos, 1950, section 11, but I'm pretty sure the development of measure theory in the standard text by Rudin covers the theorem. James said... Not convinced. Although reluctant to return to the formal proof, I found thi son the web http://planetmath.org/?op=getobj&from=objects&id=11280 I am still not convincd that AOC is needed. Please spell it out - don't be afraid of maths ;-) James said... Just read through it and it seems like quite a cute proof - but no AOC in sight. Anonymous said... That link covers construction of the outer measure. It is in fact Caratheodory's Lemma which presents the obstacle. This lemma is referenced in the preceding link. http://planetmath.org/encyclopedia/CaratheodorysLemma.html Unfortunately no proof is supplied Anonymous said... Among other things in the lemma, showing the set of Caratheodory measurable sets is not empty will require AOC for sets with infinite or greater cardinality. Anonymous said... Opps sorry the whole set and the empty set are Caratheodory measurable, so the set of Caratheodory measurable sets is at least trivially not empty. James said... There's nothing left but for me to go and re-learn the full construction which I'm not looking forward to. I used to know it off by heart (more years ago than I care to remember). This may take a while... You could, of course, have helped me out by pointing out exactly where AOC is used (as I asked you to a few times...) James said... Well, that was an effort - but good for the soul I guess... No AOC!!! estraven said... To all of you AOC sensitive people: http://abstrusegoose.com/133. James said... It's an old joke - but I think it bears repeating... The axiom of choice is obviously true, the well ordering principle is obviously false, and who knows about Zorn's lemma?

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# FRICTION LOSS ALONG A PIPE Size: px Start display at page: Transcription 1 FRICTION LOSS ALONG A PIPE 2 1. INTRODUCTION The frictional resistance to which fluid is subjected as it flows along a pipe results in a continuous loss of energy or total head of the fluid. Fig 1 illustrates this in a simple case; the difference in levels between piezometers A and B represents the total head loss h in the length of pipe l. In hydraulic engineering it is customary to refer to the rate of loss of total head along the pipe, dh/dl, by the term hydraulic gradient, denoted by the symbol i, so that dh dl = i Fig 1 Diagram illustrating the hydraulic gradient Osborne Reynolds, in 1883, recorded a number of experiments to determine the laws of resistance in pipes. By introducing a filament of dye into the flow of water along a glass pipe he showed the existence of two different types of motion. At low velocities the filament appeared as a straight line which passed down the whole length of the tube, indicating laminar flow. At higher velocities, the filament, after passing a little way along the tube, suddenly mixed with the surrounding water, indicating that the motion had now become turbulent. 1 3 Experiments with pipes of different and with water at different temperatures led Reynolds to conclude that the parameter which determines whether the flow shall be laminar or turbulent in any particular case is R = ρvd μ In which R denotes the Reynolds Number of the motion ρ denotes the density of the fluid v denotes the velocity of flow D denotes the diameter of the pipe μ denotes the coefficient of viscosity of the fluid. The motion is laminar or turbulent according as the value to R is less than or greater than a critical value. If experiments are made with increasing rates of flow, this value of R depends degree of care which is taken to eliminate disturbances in the supply and along the pipe. On the hand, if experiments are made with decreasing flow, transition from turbulent to laminar place at a value of R which is very much less dependent on initial disturbances. This value of P. about 2000, and below this, the flow becomes laminar sufficiently downstream of any disturbance. matter how severe it is. Different laws of resistance apply to laminar and to turbulent flow. For a given fluid flowing along a given pipe, experiments show that for laminar motion I α V and..3 for turbulent motion I α V n 4 n being an index which lies between 1.7 and 2.0 (depending on the value of R and on the roughness of the wail of the pipe) Equation 3 is in accordance with Poiseuille's equation which can be written in the form i= 32μv ρgd 2 5 2 4 There is no similar simple result for turbulent now; in engineering practice it is custom Darcy's Equation i = 4fv2 D2g 6 in which f denotes an experimentally determined friction factor which varies with R and pipe roughness. The object of the present experiment is to demonstrate the change in the law of resistance and to establish the critical value of R. Measurements of i in the laminar region may be used to find the co-efficient of viscosity from equation 5 and measurements in the turbulent region may be used to find the friction factor f from equation DESCRIPTION OF APPARATUS Overview Fig 2 shows the arrangement in which water from a supply tank is led through a flexible hose to the bell-mouthed entrance to a straight tube along which the frictional loss is measured Piezometer tappings are made at an upstream section which lies approximately 45 tube diameters away from the pipe entrance, and at a downstream section which lies approximately 40 tube diameters away from the pipe exit. These clear lengths upstream and downstream of the test section are required to prevent the results from being affected by disturbances near the entrance and exit of the pipe. The piezometer tappings are connected to an inverted U-tube manometer, which reads the differential pressure directly in mm of water, or a U-tube which reads in mm of mercury. 3 5 The rate of flow along the pipe is controlled by a needle valve at the pipe exit, and is measured by timing the collection of water in a measuring cylinder (the discharge being so small as to make the use of the H1 Hydraulic Bench weighing tank impracticable) Fig 2 Diagrammatic Arrangement of Apparatus for Measuring Friction Loss Along a Pipe 2.2 Installation and Preparation The apparatus is normally dispatched assembled and ready for use. In some instances, however, the manometer panel will be dismantled from the baseboard of the apparatus. To reassemble:- a) Secure the back panel supports to the baseplate with the two screws and washers provided. These screws should not be excessively tightened. b) Connect the free ends of the water and mercury manometer tubes to the pressure tapping block on the base board. Secure these tubes with plastic ty-wrap clips using pliers to tighten them. Superfluous lengths of ty-wrap should be cut off. 4 6 c) Assemble and connect the Header Tank, H7a, to the Hydraulic Bench supply and the inlet the 'friction in pipe' apparatus, For higher flow rates, connect the plastic supply hose from the HI Hydraulic Bench directly to the inlet of the apparatus. Secure with the hose clip provided. d) Connect the smaller bore plastic tube to the outlet port of the needle valve. Until measurements of flow are required, direct the free end of this tube into the access hole in the centre of the bench top. For measurement direct the tube into a measuring flask. A litre flask ( not supplied), sub-divided into 10 millilitre divisions, is most suitable. e) Fill the U-tube manometer up to the 270 millimetres mark with mercury (not supplied). Approximately 40 millilitres will be required for this. Access ports are provided in the lower appropriate header. f) Before allowing water to flow through the apparatus, check that the respective air purge valve and screw caps on the water and mercury manometer are both tightly closed. CHECKING WATER MANOMETER CIRCUIT A tap is provided at the downstream end of the test pipe for selecting either a water or mercury manometer circuit. Avoid syphoning of the water when using the mercury manometer. To check the circuit:- a) Direct the tap towards the open position. b) Allow a nominal flow of water through the apparatus. Lightly tap the manometer tubes to clear air from the circuit. c) Adjust the water levels in the tubes to the same height. 5 7 It may be necessary to connect a bicycle pump to the purge valve in the manifold and manipulate the levels accordingly. d) Increase the water flow to obtain an approximate maximum scale reading. Observe these levels to ensure that they remain steady. If there is a steady rise in the manometer levels, check that the valve is tightened and sealed properly. If tightening does not stop the leak, replace the valve seal. Check that the tube ferrules in the manifold are free from water blockage as this will suppress water levels and cause erroneous results. If this is suspected, a sharp burst of pressure from the bicycle pump will normally clean the blockage. PURGING MERCURY MANOMETER a) Turn the isolating tap to the Mercury Manometer circuit. b) Purge all air from the manometer tubes by releasing the screw caps in the mercury manifold. c) When purged, firmly screw down the manifold caps. 2.3 Routine Care and Maintenance After use, the apparatus should be drained as far as possible and all external surfaces dried with a lint-free cloth. Dry the Header Tank if this has been used. Care must be taken not to bend or damage the needle valve tip if this is removed. If the plastic manometer tubes become excessively discoloured a stain and deposit remover should be use. 6 8 THEORY OF FRICTION LOSS ALONG A PIPE 3.1 Derivation of Poiseuille's Equation Fig 3 Derivation of Poiseuille's Equation To derive Poiseuille's equation which applied to laminar flow along a tube, consider the motion indicated on Fig 3. Over each cross-section of the tube, the piezometric pressure is constant, and this pressure falls continuously along the tube. Suppose that between cross-sections A and B separated by length l of tube, the fall in pressure is p. Then the force exerted by this pressure difference on the ends of a cylinder having radius r, and its axis on the centre line of the tube, is pπr 2. Over any cross-section of the tube, the velocity varies with radius, having a maximum value of v o the centre and falling to zero at the wall; let the velocity at radius r in any cross-section by denoted by v r. Then the shear stress τ, in the direction shown on fig 3, due to viscous action on the curved surface of the cylinder, is given by τ = μdv r dr (Note that dv r, is negative so that the stress acts in the direction shown in the dr figure). The force on the cylinder is due to this stress μdv r. 2πrl. Since the fluid is dr in steady motion under the action of the sum of pressure and viscous forces, 7 9 Therefore dv r dr = pr 2lμ P. πr 2 + μdv r dr 2πrl = 0 8 Integrating this and inserting a constant of integration such that v r = 0 when r = a V r = p 4lμ (a2 r 2 ) 9 This result shows that the velocity distribution across a section is parabolic, as indicated on fig 3, and that the velocity on the centre line, given by putting r = 0 in equation 9 is v o = pa.10 4lμ The discharge rate Q may now be calculated. The flow rate through an annulus of radius r and width r is δq = V r. 2πrδr Inserting V r from equation 9 and integrating Q = p a 4lμ 2π (a2 r r 3 )dr Therefore Q = pπa4 8lμ Now the mean velocity V over the cross section is, by definition, given by Q = v. πa 2 And elimimating Q between equation 11 and 12 gives V = pa2 8lμ = pd2 32lμ 13 8 10 By use of the substitution And h l = i ρgh = p Eq. 13 may be written in the form i = 32μV..5 ρgd 2 (which is an equation of the form : y=mx+b) 3.2 Derivation of Darcy's Equation If the flow is turbulent, the analysis given above is invalidated by the continuous mixing process which takes place. Across the curved surface of the cylinder having radius r in Fig 3, this mixing is manifest as a continuous unsteady and random flow into and out of the cylinder, so that the apparent shear stress on this surface is greater than the value given in equation 7. Because of the mixing, the distribution of velocity over a cross-section is more uniform than the parabolic shape deduced for laminar flow, as indicated on Fig 4. Although it is not possible to perform a complete analysis for turbulent flow, a useful result may be obtained by considering the whole cross-section as shown in Fig 4. It is reasonable to suppose that the shear stress τ o on the wall of the tube will depend on the mean velocity v; let us assume for the present that 9 11 In which 1 2ρv 1 τ 0 = f. 2ρv 2 2 denotes the dynamic pressure corresponding to the mean velocity v and f is a friction factor (not necessarily constant). Sinceτ o and 1/2ρv 2 each have dimensions of force per unit area, f is dimensionless. The force on a cylinder of length l due to this stress id f.1/2pv 2.2πal, and the force due to the fall in pressure is p. πa 2, so that p. πa 2 = f. 1/2pv 2.2πal Substituting ρgh = p h/l = i and a= D 2 leads to the result i= 4f D. v2 2g which is form of Darcy s equation The friction factor f which occurs in this equation was defined by equation 14 and is not necessarily constant. The results of many experiments show that f does, in fact, depend on both R, the Reynolds Number, and on the roughness of the pipe wall. At a given value of R, f increases with increasing surface roughness. For a given surface roughness, f generally decreases slowly with increasing R. This means that if R is increased by increasing v, so that the product fv 2 on which i depends equation 6 will increase somewhat less than v 2. In fact, over a fairly wide range, it is often possible, to represent the variation of i with v by the approximation i = kv n where k and n are constants for a given fluid flowing along a given pipe, n having a value between 1.7 and 14 4.4 Procedures Step 1: Record at least 8 sets of data over the range of the water manometer (see Section 4.2) and another 8 or more over the range of the mercury manometer (see Section 4.3). Tables 1 and 2 show the format of suitable result tables. Results given in this section are typical of those obtainable from the equipment supplied. There will, however, be slight differences between individual units. Step 2: Plot graphs of hydraulic gradient i against mean velocity v, and log i against log v. (Figs 6 and 7 show the form of graphs expected). (Reminder - the two manometers generate data for different operating ranges of the same system. The student must combine the data sets to analyze the system over the entire range.) Step 3: From the (best fit) graph of i against log v, or graph of i vs v, determine the velocity at which rapid transition occurs. Determine the critical Reynold's Number at this velocity. (The student may elect to "blow up" that portion of the graph between 0.3 and 1.2 m/s) Step 4: From the (best fit) slope of the graphs, derive the relationship between v and i. For both the upper and lower ranges, determine k and n where i = kv n Step 5: From the gradient of i against v in the laminar range, determine the coefficient of viscosity and compare with theoretical values. Step 6: In the turbulent region of flow, select 4 or 5 values of velocity. Compute friction factors and Reynold's Number at these velocity values. Plot friction factors against Reynold's Number (Moody's Diagram) Compare with theoretical values. 5. TYPICAL RESULTS AND SAMPLE CALCULATIONS 5.1 Relationships between I and u Length of pipe between piezometer tappings, l..524 mm Nominal diameter of pipe, D..3 mm Cross-sectional area of pipe, A mm 2 Derivation of i over gauge length l i) For water manometer i = (h 1 h 2 ) l ii) For mercury manometer Referring to Fig 5, the specific gravity of mercury is taken as 13.6 writing the head difference in terms of water i = (h 1 h 2 )(13.6 1) l 13 15 Qty (ml) t (s) v (m/s) h 1 (mm) h 2 (mm) h 1 -h 2 (m) i θ ( C) log i log v Table 1 Qty (ml) t (s) v (m/s) h 1 (mm) h 2 (mm) h 1 -h 2 (m) i θ ( C) log i log v Table 2 14 16 15 Qty (ml) t (s) v (m/s) h 1 (mm) h 2 (mm) h 1 -h 2 (m) i θ ( C) log i log v Table 1. Results with Water Manometer Qty (ml) t (s) v (m/s) h 1 (mm) h 2 (mm) h 1 -h 2 (m) i θ ( C) log i log v Table 2. Results with Mercury U-tube 17 16 18 From Fig 6a, graph of v against i, it can be seen that for small values of v, the frictional loss is proportional to velocity. i.e. i v Fig 6b has been drawn with a larger scale for velocities up to 1 m/s. This graph shows a fairly distinct change in the slope of the line at C when v is in the region of 0.77m/s. Up to this point the relationship is given by i = v (see section 5.3.1) Point C marks the starts of a distinct transition phase where the flow characteristics change considerably. In Fig.7 the same results are plotted to logarithmic scales. Points up to C lie on a straight line of slope 1, confirming the frictional loss is proportional to velocity (See also Section 5.3.1) For points above C we can write:- i v 1.69 for values of v greater than 1.5m/s (See also Section 5.3.2) 17 19 5.2 Calculations of Critical Reynolds Number In Figures 6 and 7, Point C marks the distinct transition phase between laminar and turbulent flow. The velocity at Point C is approximately 0.77m/s Recalling:- R = ρvd μ Substituting values we get at 15 C: R = 999X0.77X X10 4 = Calculation of Relationship between v and i Laminar Range i = kv n therefore, algebraically log i= log k + n log v which is an equation of the forth y = mx+b from Table 1, for v=306 and.592 (these points on best fit curve) n = log i =.6055 (.8928) =.2873 = log v.2277 (.5143).2866 say n = 1.00 log k = log i n log v = (1.00) (-.2277) = k =.419 i =.419v Turbulent Range as noted above; log i = log k + n log v from Table 2, for v = 1.55 and 2.47 n = log i log v = = = Say n = 1.69 log k = log i n log v = (.3927) = k = i = v 20 5.4 Calculation of Coefficient of Viscosity In the laminar range; i = 32μv ρgd 2 (Eq. 5), which is an equation of the form y = mx + b The slope of the plot is therefore = 32μ ρgd2 Where slope = k = (Section 5.3.1) This can be rewritten in the form μ = k ρgd2 32 Substituting values we get μ = k ρgd2 32 Substituting values we get μ = 11.6 x 10 4 N.s/m 2 μ = 0.419x999x9.81x9x Calculation of Friction Factor In the turbulent region i = 4fv2 D2g 21 We can draw up table 3, giving values of f corresponding to various values of v, in the turbulent region of flow. v(m/s) I v gD f R Table 3 Calculation of the Friction Factor f in Darcy s Equation 6. DISCUSSION OF RESULTS 6.1 Measurements of frictional loss alone, the pipe at different velocities have shown two well-defined regions to which different laws of resistance apply. As the velocity is decreased from 3.3 to 1.5 m/s, frictional loss varied as v 169. Between 1.5 and 0.77, the loss decreased rather more steeply and as v decreased from 0.77 to zero, the loss varied directly as v. The critical velocity of 0.77 corresponds to a Reynolds number of 2024, this value being close to the figure of about 2000 at which transition from turbulent to laminar flow is usually found to take place. 6.2 The value of μ calculated by Poiseuille's equation applied to the results in the laminar region is μ = x 10-4 Ns/m at 15.3 C. The accepted value at this temperature is μ =11.4 x 10-4 Ns/m 2 Since the accepted values are based on experiments with similar but more refined apparatus, the discrepance reveals an error of about 2% in the apparatus used here. 6.3 The results in the turbulent region have been used to calculate the friction factor f in Darcy's equation, and are found to fall with increasing v as shown in Table 4. 20 ### Experiment (4): Flow measurement Experiment (4): Flow measurement Introduction: The flow measuring apparatus is used to familiarize the students with typical methods of flow measurement of an incompressible fluid and, at the same time ### DETERMINATION OF DISCHARGE AND HEAD LOSS USING A FLOW-MEASURING APPARATUS DETERMINATION OF DISCHARGE AND HEAD LOSS USING A FLOW-MEASURING APPARATUS 1. INTRODUCTION Through use of the Flow-Measuring Apparatus, this experiment is designed to accustom students to typical methods ### EXPERIMENT NO: F5. 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SUBJECT: FLUID MECHANICS & HEAT TRANSFER LOCATION: HYDRAULICS LAB (Gnd Floor Inglis Bldg) BOUNDARY LAYERS AND DRAG 1 PART 1B EXPERIMENTAL ENGINEERING SUBJECT: FLUID MECHANICS & HEAT TRANSFER LOCATION: HYDRAULICS LAB (Gnd Floor Inglis Bldg) EXPERIMENT T3 (LONG) BOUNDARY LAYERS AND DRAG OBJECTIVES a) To measure the velocity ### Friction Factors and Drag Coefficients Levicky 1 Friction Factors and Drag Coefficients Several equations that we have seen have included terms to represent dissipation of energy due to the viscous nature of fluid flow. For example, in the ### UNIT I FLUID PROPERTIES AND STATICS SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : Fluid Mechanics (16CE106) Year & Sem: II-B.Tech & I-Sem Course & Branch: ### ME332 FLUID MECHANICS LABORATORY (PART I) ME332 FLUID MECHANICS LABORATORY (PART I) Mihir Sen Department of Aerospace and Mechanical Engineering University of Notre Dame Notre Dame, IN 46556 Version: January 14, 2002 Contents Unit 1: Hydrostatics ### FLOW MEASUREMENT IN PIPES EXPERIMENT University of Leicester Engineering Department FLOW MEASUREMENT IN PIPES EXPERIMENT Page 1 FORMAL LABORATORY REPORT Name of the experiment: FLOW MEASUREMENT IN PIPES Author: Apollin nana chaazou Partner ### Major and Minor Losses Abstract Major and Minor Losses Caitlyn Collazo, Team 2 (1:00 pm) A Technovate fluid circuit system was used to determine the pressure drop across a pipe section and across an orifice. These pressure drops ### R09. d water surface. Prove that the depth of pressure is equal to p +. Code No:A109210105 R09 SET-1 B.Tech II Year - I Semester Examinations, December 2011 FLUID MECHANICS (CIVIL ENGINEERING) Time: 3 hours Max. Marks: 75 Answer any five questions All questions carry equal ### LECTURE 6- ENERGY LOSSES IN HYDRAULIC SYSTEMS SELF EVALUATION QUESTIONS AND ANSWERS LECTURE 6- ENERGY LOSSES IN HYDRAULIC SYSTEMS SELF EVALUATION QUESTIONS AND ANSWERS 1. What is the head loss ( in units of bars) across a 30mm wide open gate valve when oil ( SG=0.9) flow through at a ### Lesson 6 Review of fundamentals: Fluid flow Lesson 6 Review of fundamentals: Fluid flow The specific objective of this lesson is to conduct a brief review of the fundamentals of fluid flow and present: A general equation for conservation of mass ### Chapter (3) Water Flow in Pipes Chapter (3) Water Flow in Pipes Water Flow in Pipes Bernoulli Equation Recall fluid mechanics course, the Bernoulli equation is: P 1 ρg + v 1 g + z 1 = P ρg + v g + z h P + h T + h L Here, we want to study ### FE Exam Fluids Review October 23, Important Concepts FE Exam Fluids Review October 3, 013 mportant Concepts Density, specific volume, specific weight, specific gravity (Water 1000 kg/m^3, Air 1. kg/m^3) Meaning & Symbols? Stress, Pressure, Viscosity; Meaning ### FLOW IN CONDUITS. Shear stress distribution across a pipe section. Chapter 10 Chapter 10 Shear stress distribution across a pipe section FLOW IN CONDUITS For steady, uniform flow, the momentum balance in s for the fluid cylinder yields Fluid Mechanics, Spring Term 2010 Velocity ### 1-Reynold s Experiment Lect.No.8 2 nd Semester Flow Dynamics in Closed Conduit (Pipe Flow) 1 of 21 The flow in closed conduit ( flow in pipe ) is differ from this occur in open channel where the flow in pipe is at a pressure ### τ du In his lecture we shall look at how the forces due to momentum changes on the fluid and viscous forces compare and what changes take place. 4. Real fluids The flow of real fluids exhibits viscous effect, that is they tend to stick to solid surfaces and have stresses within their body. You might remember from earlier in the course Newtons law ### 2 Internal Fluid Flow Internal Fluid Flow.1 Definitions Fluid Dynamics The study of fluids in motion. Static Pressure The pressure at a given point exerted by the static head of the fluid present directly above that point. ### CVE 372 HYDROMECHANICS EXERCISE PROBLEMS VE 37 HYDROMEHNIS EXERISE PROLEMS 1. pump that has the characteristic curve shown in the accompanying graph is to be installed in the system shown. What will be the discharge of water in the system? Take ### 150A Review Session 2/13/2014 Fluid Statics. Pressure acts in all directions, normal to the surrounding surfaces Fluid Statics Pressure acts in all directions, normal to the surrounding surfaces or Whenever a pressure difference is the driving force, use gauge pressure o Bernoulli equation o Momentum balance with ### Fluid Mechanics. du dy FLUID MECHANICS Technical English - I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's ### LEAKLESS COOLING SYSTEM V.2 PRESSURE DROP CALCULATIONS AND ASSUMPTIONS CH-1211 Geneva 23 Switzerland EDMS No. ST/CV - Cooling of Electronics & Detectors GUIDE LEAKLESS COOLING SYSTEM V.2 PRESSURE DROP CALCULATIONS AND ASSUMPTIONS Objectives Guide to Leakless Cooling System ### V/ t = 0 p/ t = 0 ρ/ t = 0. V/ s = 0 p/ s = 0 ρ/ s = 0 UNIT III FLOW THROUGH PIPES 1. List the types of fluid flow. Steady and unsteady flow Uniform and non-uniform flow Laminar and Turbulent flow Compressible and incompressible flow Rotational and ir-rotational ### FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering) Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface. ### APPENDICES. Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10. Distance S 1 PPENDICES 1. Kinetic Energy of a Fluid Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10. v Distance S Figure 10. This body contains kinetic energy (energy of movement). ### Chapter 10: Flow Flow in in Conduits Conduits Dr Ali Jawarneh Chater 10: Flow in Conduits By Dr Ali Jawarneh Hashemite University 1 Outline In this chater we will: Analyse the shear stress distribution across a ie section. Discuss and analyse the case of laminar CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK 3 0 0 3 UNIT I FLUID PROPERTIES AND FLUID STATICS PART - A 1. Define fluid and fluid mechanics. 2. Define real and ideal fluids. 3. Define mass density ### Sourabh V. Apte. 308 Rogers Hall Sourabh V. Apte 308 Rogers Hall sva@engr.orst.edu 1 Topics Quick overview of Fluid properties, units Hydrostatic forces Conservation laws (mass, momentum, energy) Flow through pipes (friction loss, Moody ### ACE Engineering College ACE Engineering College Ankushapur (V), Ghatkesar (M), R.R.Dist 501 301. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * MECHANICS OF FLUIDS & HYDRAULIC ### Mechanical Engineering Programme of Study Mechanical Engineering Programme of Study Fluid Mechanics Instructor: Marios M. Fyrillas Email: eng.fm@fit.ac.cy SOLVED EXAMPLES ON VISCOUS FLOW 1. Consider steady, laminar flow between two fixed parallel ### FLUID MECHANICS PROF. DR. METİN GÜNER COMPILER FLUID MECHANICS PROF. DR. METİN GÜNER COMPILER ANKARA UNIVERSITY FACULTY OF AGRICULTURE DEPARTMENT OF AGRICULTURAL MACHINERY AND TECHNOLOGIES ENGINEERING 1 5. FLOW IN PIPES 5.1.3. Pressure and Shear Stress ### BERNOULLI EQUATION. The motion of a fluid is usually extremely complex. BERNOULLI EQUATION The motion of a fluid is usually extremely complex. The study of a fluid at rest, or in relative equilibrium, was simplified by the absence of shear stress, but when a fluid flows over ### PIPE FLOWS: LECTURE /04/2017. Yesterday, for the example problem Δp = f(v, ρ, μ, L, D) We came up with the non dimensional relation /04/07 ECTURE 4 PIPE FOWS: Yesterday, for the example problem Δp = f(v, ρ, μ,, ) We came up with the non dimensional relation f (, ) 3 V or, p f(, ) You can plot π versus π with π 3 as a parameter. Or, ### BRCM COLLEGE OF ENGINEERING & TECHNOLOGY Practical Experiment Instructions Sheet Exp. Title FLUID MECHANICS- I LAB Syllabus FM-I Semester-4 th Page No. 1 of 1 Internal Marks: 25 L T P External Marks: 25 0 0 2 Total Marks: 50 1. To determine the met centric height of a floating body ### Chapter 4 DYNAMICS OF FLUID FLOW Faculty Of Engineering at Shobra nd Year Civil - 016 Chapter 4 DYNAMICS OF FLUID FLOW 4-1 Types of Energy 4- Euler s Equation 4-3 Bernoulli s Equation 4-4 Total Energy Line (TEL) and Hydraulic Grade Line ### Fluid Mechanics Lab (ME-216-F) List of Experiments Fluid Mechanics Lab (ME-216-F) List of Experiments 1. To determine the coefficient of discharge C d, velocity C v, and contraction C c of various types of orifices 2. Determine of discharge coefficients ### Fluid Mechanics Answer Key of Objective & Conventional Questions 019 MPROVEMENT Mechanical Engineering Fluid Mechanics Answer Key of Objective & Conventional Questions 1 Fluid Properties 1. (c). (b) 3. (c) 4. (576) 5. (3.61)(3.50 to 3.75) 6. (0.058)(0.05 to 0.06) 7. ### OE4625 Dredge Pumps and Slurry Transport. Vaclav Matousek October 13, 2004 OE465 Vaclav Matousek October 13, 004 1 Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport OE465 Vaclav Matousek October 13, 004 Dredge Vermelding Pumps onderdeel and Slurry organisatie ### Pipe Flow. Lecture 17 Pipe Flow Lecture 7 Pipe Flow and the Energy Equation For pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through diameter changes and corners ### Figure 3: Problem 7. (a) 0.9 m (b) 1.8 m (c) 2.7 m (d) 3.6 m 1. For the manometer shown in figure 1, if the absolute pressure at point A is 1.013 10 5 Pa, the absolute pressure at point B is (ρ water =10 3 kg/m 3, ρ Hg =13.56 10 3 kg/m 3, ρ oil = 800kg/m 3 ): (a) ### 9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook. Lecture Notes CHE 31 Fluid Mechanics (Fall 010) 9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook. Basics (pressure head, efficiency, working point, stability) Pumps ### Physics 3 Summer 1990 Lab 7 - Hydrodynamics Physics 3 Summer 1990 Lab 7 - Hydrodynamics Theory Consider an ideal liquid, one which is incompressible and which has no internal friction, flowing through pipe of varying cross section as shown in figure ### ME 305 Fluid Mechanics I. Part 8 Viscous Flow in Pipes and Ducts. Flow in Pipes and Ducts. Flow in Pipes and Ducts (cont d) ME 305 Fluid Mechanics I Flow in Pipes and Ducts Flow in closed conduits (circular pipes and non-circular ducts) are very common. Part 8 Viscous Flow in Pipes and Ducts These presentations are prepared ### REE 307 Fluid Mechanics II. Lecture 1. Sep 27, Dr./ Ahmed Mohamed Nagib Elmekawy. Zewail City for Science and Technology REE 307 Fluid Mechanics II Lecture 1 Sep 27, 2017 Dr./ Ahmed Mohamed Nagib Elmekawy Zewail City for Science and Technology Course Materials drahmednagib.com 2 COURSE OUTLINE Fundamental of Flow in pipes ### Viscous Flow in Ducts Dr. M. Siavashi Iran University of Science and Technology Spring 2014 Objectives 1. Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow 2. Calculate ### Introduction to Fluid Flow Introduction to Fluid Flow Learning Outcomes After this lecture you should be able to Explain viscosity and how it changes with temperature Write the continuity equation Define laminar and turbulent flow ### Part A: 1 pts each, 10 pts total, no partial credit. Part A: 1 pts each, 10 pts total, no partial credit. 1) (Correct: 1 pt/ Wrong: -3 pts). The sum of static, dynamic, and hydrostatic pressures is constant when flow is steady, irrotational, incompressible, ### Chapter 10 - Mechanical Properties of Fluids. The blood pressure in humans is greater at the feet than at the brain Question 10.1: Explain why The blood pressure in humans is greater at the feet than at the brain Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though ### Teacher s Signature. S. No. Experiment marks. 3 To determine the coefficient of discharge of Notch (V and Rectangular types) S. No. Index Name of experiment Date of performance 1. To determine the coefficient of impact for vanes. 2 To determine coefficient of discharge of an orificemeter. 3 To determine the coefficient of discharge ### LABORATORY MANUAL FLUID MECHANICS ME-216-F LABORATORY MANUAL FLUID MECHANICS ME-216-F LIST OF THE EXPERIMENT SNO NAME OF THE EXPERIMENT PAGE NO FROM TO 1. To determine the coefficient of impact for vanes. 2. To determine the coefficient of discharge ### PIPE FLOW. The Energy Equation. The first law of thermodynamics for a system is, in words = + The Energy Equation PIPE FLOW The first law of thermodynamics for a system is, in words Time rate of increase of the total storage energy of the t Net time rate of energy addition by heat transfer into ### Atmospheric pressure. 9 ft. 6 ft Name CEE 4 Final Exam, Aut 00; Answer all questions; 145 points total. Some information that might be helpful is provided below. A Moody diagram is printed on the last page. For water at 0 o C (68 o F): ### MAHATMA GANDHI MISSION S JAWAHARLAL NEHRU ENGINEERING COLLEGE, FLUID MECHANICS LABORATORY MANUAL MAHATMA GANDHI MISSION S JAWAHARLAL NEHRU ENGINEERING COLLEGE, AURANGABAD. (M.S.) DEPARTMENT OF CIVIL ENGINEERING FLUID MECHANICS LABORATORY MANUAL Prepared By Mr. L. K. Kokate Lab Incharge Approved By ### VENTURIMETER EXPERIMENT ENTURIMETER EXERIMENT. OBJECTİE The main objectives of this experiment is to obtain the coefficient of discharge from experimental data by utilizing venturi meter and, also the relationship between Reynolds ### Modelling of dispersed, multicomponent, multiphase flows in resource industries. Section 3: Examples of analyses conducted for Newtonian fluids Modelling of dispersed, multicomponent, multiphase flows in resource industries Section 3: Examples of analyses conducted for Newtonian fluids Globex Julmester 017 Lecture # 04 July 017 Agenda Lecture ### Experiment- To determine the coefficient of impact for vanes. Experiment To determine the coefficient of discharge of an orifice meter. SUBJECT: FLUID MECHANICS VIVA QUESTIONS (M.E 4 th SEM) Experiment- To determine the coefficient of impact for vanes. Q1. Explain impulse momentum principal. Ans1. Momentum equation is based on Newton s ### 11.1 Mass Density. Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an Chapter 11 Fluids 11.1 Mass Density Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an important factor that determines its behavior ### MAHATMA GANDHI MISSION S JAWAHARLAL NEHRU ENGINEERING COLLEGE, AURANGABAD. (M.S.) MAHATMA GANDHI MISSION S JAWAHARLAL NEHRU ENGINEERING COLLEGE, AURANGABAD. (M.S.) DEPARTMENT OF CIVIL ENGINEERING FLUID MECHANICS I LAB MANUAL Prepared By Prof. L. K. Kokate Lab Incharge Approved By Dr. ### Instruction Manual. Equipment for Engineering Education Equipment for Engineering Education Instruction Manual HM15007 Bernoulli s Principle Demonstrator GUNT Gerätebau GmbH PO Box 1125 D-22881 Barsbüttel Germany Phone (040) 670854-0 Fax (040) 670854-42 Instruction ### Volumetric Measurement Techniques. Technique #1 Use of a Burette. Technique #2 Use of a Pipette. Technique #3 Use of a Volumetric Flask Volumetric Measurement Techniques Technique #1 Use of a Burette Technique #2 Use of a Pipette Technique #3 Use of a Volumetric Flask Technique #4 Use of a Bottle-Top Dispenser Last updated 12/6/2009 5:46 ### Fluid Mechanics c) Orificemeter a) Viscous force, Turbulence force, Compressible force a) Turbulence force c) Integration d) The flow is rotational Fluid Mechanics 1. Which is the cheapest device for measuring flow / discharge rate. a) Venturimeter b) Pitot tube c) Orificemeter d) None of the mentioned 2. Which forces are neglected to obtain Euler ### Fluid Flow Analysis Penn State Chemical Engineering Fluid Flow Analysis Penn State Chemical Engineering Revised Spring 2015 Table of Contents LEARNING OBJECTIVES... 1 EXPERIMENTAL OBJECTIVES AND OVERVIEW... 1 PRE-LAB STUDY... 2 EXPERIMENTS IN THE LAB... ### Mass of fluid leaving per unit time 5 ENERGY EQUATION OF FLUID MOTION 5.1 Eulerian Approach & Control Volume In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics. ### Piping Systems and Flow Analysis (Chapter 3) Piping Systems and Flow Analysis (Chapter 3) 2 Learning Outcomes (Chapter 3) Losses in Piping Systems Major losses Minor losses Pipe Networks Pipes in series Pipes in parallel Manifolds and Distribution ### Bernoulli and Pipe Flow Civil Engineering Hydraulics Mechanics of Fluids Head Loss Calculations Bernoulli and The Bernoulli equation that we worked with was a bit simplistic in the way it looked at a fluid system All real systems ### DEPARTMENT OF CHEMICAL ENGINEERING University of Engineering & Technology, Lahore. Fluid Mechanics Lab DEPARTMENT OF CHEMICAL ENGINEERING University of Engineering & Technology, Lahore Fluid Mechanics Lab Introduction Fluid Mechanics laboratory provides a hands on environment that is crucial for developing ### Experimental Study on Port to Channel Flow Distribution of Plate Heat Exchangers Proceedings of Fifth International Conference on Enhanced, Compact and Ultra-Compact Heat Exchangers: Science, Engineering and Technology, Eds. R.K. Shah, M. Ishizuka, T.M. Rudy, and V.V. Wadekar, Engineering ### ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES Some background information first: We have seen that a major limitation of the Bernoulli equation is that it does not account for ### 2 Navier-Stokes Equations 1 Integral analysis 1. Water enters a pipe bend horizontally with a uniform velocity, u 1 = 5 m/s. The pipe is bended at 90 so that the water leaves it vertically downwards. The input diameter d 1 = 0.1 ### Lecture 30 Review of Fluid Flow and Heat Transfer Objectives In this lecture you will learn the following We shall summarise the principles used in fluid mechanics and heat transfer. It is assumed that the student has already been exposed to courses in ### Applied Fluid Mechanics Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and ### Laboratory work No 2: Calibration of Orifice Flow Meter Laboratory work No : Calibration of Orifice Flow Meter 1. Objective Calibrate the orifice flow meter and draw the graphs p = f 1 (Q) and C d = f (Re ).. Necessary equipment 1. Orifice flow meter. Measuring ### Turbulence is a ubiquitous phenomenon in environmental fluid mechanics that dramatically affects flow structure and mixing. Turbulence is a ubiquitous phenomenon in environmental fluid mechanics that dramatically affects flow structure and mixing. Thus, it is very important to form both a conceptual understanding and a quantitative ### LECTURE 1 THE CONTENTS OF THIS LECTURE ARE AS FOLLOWS: LECTURE 1 THE CONTENTS OF THIS LECTURE ARE AS FOLLOWS: 1.0 INTRODUCTION TO FLUID AND BASIC EQUATIONS 2.0 REYNOLDS NUMBER AND CRITICAL VELOCITY 3.0 APPROACH TOWARDS REYNOLDS NUMBER REFERENCES Page 1 of ### Chapter 7 The Energy Equation Chapter 7 The Energy Equation 7.1 Energy, Work, and Power When matter has energy, the matter can be used to do work. A fluid can have several forms of energy. For example a fluid jet has kinetic energy,

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January 8, 2020 ## How do I calculate my Grade Point Deficiency? Your Grade Point Deficiency is the number of grade points that are needed to bring your UCF GPA up to a 2.0. For each credit hour you attempt, you earn a certain number of grade points per hour based on the grade you received: 4.00 = A 3.75 = A- 3.25 = B+ 3.00 = B 2.75 = B- 2.25 = C+ 2.00 = C 1.75 = C- 1.25 = D+ 1.00 = D 0.75 = D- 0.00 = F For each class taken at UCF, multiply the number of credit hours it was worth by the value of the grade that was earned, then add them all together. This is the number of GPA grade points that you’ve earned. Subtract from this 2 times the total number of credit hours attempted. If this number is positive, then your GPA should be above a 2.0. If it is negative, than it is known as your Grade Point Deficiency, or your Quality Points Down. It is the amount of credit hours worth of ‘B’ letter grade you must earn to raise your UCF GPA to a 2.0. Let’s say, for example, that a student has earned to following grades at UCF: B- (3 hours), C (3 hours), F (3 hours), D (3 hours), C- (3 hours), C (3 hours), F (3 hours), and F (3 hours) Now, to calculate the number of GPA grade points: (2.75 * 3) + (2 * 3) + (0 * 3) + (1 * 3) + (1.75 * 3) + (2 * 3) + (0 * 3) + (0 * 3) = 28.5 Finally, calculate the Grade Point Deficiency: 28.5 – (2 * 24 credit hours) = 28.5 – 48 = -19.5 The student above has earned a 1.1875 UCF GPA (GPA points divided by credit hours attempted), and their Grade Point Deficiency is 19.5 points down.

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`ISRN Computer GraphicsVolume 2012 (2012), Article ID 825782, 17 pageshttp://dx.doi.org/10.5402/2012/825782` Research Article GPU-Accelerated Rendering of Unbounded Nonlinear Iterated Function System Fixed Points University of Alaska Fairbanks, Fairbanks, AK 99775, USA Received 31 October 2011; Accepted 7 December 2011 Academic Editors: T. Calvert, M. Kraus, and L. Ma Copyright © 2012 Orion Sky Lawlor. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Nonlinear functions, including nonlinear iterated function systems, have interesting fixed points. We present a non-Lipschitz theoretical approach to nonlinear function system fixed points which generalizes to noncontractive functions, compare several methods for evaluating such fixed points on modern graphics hardware, and present a nonlinear generalization of Barnsley’s Deterministic Iteration Algorithm. Unlike the many existing randomized rendering algorithms, this deterministic method avoids noncoherent branching and memory access and takes advantage of programmable texture mapping hardware. Together with the performance potential of modern graphics hardware, this allows us to animate high-quality and high-definition fixed points in real time. 1. Introduction Iterated function systems are a method to generate easily controlled, infinitely detailed fractal images such as Figure 1 from the repeated application of simple mathematical functions. Figure 1: A simple two-map nonlinear iterated function system, as rendered on the GPU at 30 fps using these techniques. 1.1. Mathematical Background This paper shows how to compute fixed points of image-to-image functions. We define an image as a function mapping some domain (e.g., 2D space) into some range (a color space). Then an image-to-image transform has a fixed point image when that is, the fixed point image remains unchanged under the image transform. It is reasonable to ask if such a fixed point always exists, and the answer is no. For example, in a binary color space , the color inversion transform does not have a fixed point. Yet a number of theorems establish sufficient conditions for the existence of such a fixed point. The majority of iterated function system works uses the well-known Banach fixed point theorem, which gives both the existence and uniqueness of the fixed point and merely requires and to be complete, but requires the image transform to be Lipschitz contractive. This theorem has been used for much iterated function system work [1], but it does require contractivity. Since many interesting nonlinear functions are not contractive everywhere, including the functions shown in Figure 1, we will not use the Banach theorem. Instead, the Schauder-Tychonoff [2] fixed point theorem, an extension of the Brouwer fixed point theorem to infinite dimensional spaces, establishes sufficient conditions for the existence of a fixed point. Schauder-Tychonoff Fixed Point Theorem 1 Let be a nonempty, compact, convex subset of a locally convex topological vector space. Given any continuous mapping , there exists a fixed point such that . The first difficulty is that ordinary Euclidean space is nonempty and convex but not compact: the expanding 1D function sends points off to infinity and hence has no fixed point in Euclidean space. We instead use projective space , which is compact because it includes points at infinity—such as the fixed points of . We survey several ways to implement projective space in Section 3. We thus define an image as a function from points in some -dimensional real projective space (typically: 2D space plus the circle at infinity) and returning a -dimensional color (typically simply channels of red, green, and blue), so an image is a function . Note that, without contractive maps, we are not guaranteed unique fixed points. For example, any point symmetric image is a fixed point of the image-to-image function . 1.2. Nonlinear Iterated Function Systems We define an Iterated Function System as a set of separate geometric distortion functions . For example, the 2D plane-filling IFS shown in Figure 1 consists of these two geometric distortion functions, which are a rotation plus translation, and a simple nonlinear distortion We apply a geometric distortion function to an image using an image distortion function defined as follows: Note that the function inverse in the geometric portion of this transform , as described in Section 5, is equivalent to forward-transforming each point in by the geometric distortion function , as in Section 4. The Jacobian determinant , as discussed in Section 6, is present in order to preserve the integral of the transformed image, at least for well-behaved map functions. This can be seen via the substitution method for multiple integrals, where Finally, the image-to-image function is defined as a combination of the image distortion functions and a set of constant color weights If our overall image-to-image transform is continuous, in the sense that the colors of the output image depend continuously on the colors in the input image, then in the convex nonempty compact domain the Schauder-Tychonoff theorem guarantees has a fixed point. If the distortion function is continuous, then the geometric distortions in will be continuous and if the Jacobians change continuously, the color intensity changes in will be continuous; thus, by Schauder-Tychonoff, has a fixed point. 2. Calculating IFS Fixed Points There are a variety of ways to calculate an IFS fixed point, but the simplest is the random iteration algorithm [3], also known as the chaos game. We begin with a randomly chosen point , and repeatedly apply randomly chosen geometric distortion functions . A histogram of the resulting point locations converges to the IFS fixed point, usually called an attractor in this context. For our example, if we repeatedly apply the rotation , our points form a simple spiral. If we repeatedly apply the nonlinear transform , our points rapidly approach the and axes. Yet if we randomly alternate between these functions, the random iteration algorithm points plot out the much more complex Figure 1. The underlying motivation behind this paper is that the random iteration algorithm typically requires billions of points to produce a smooth image, which is too slow to produce high-quality animations in real time. There are many other interesting methods to render IFS—for a survey, see Nikiel’s recent book [4]. Modern practical applications of IFS range from fractal image compression to artistic and rendering applications. The particular functional form for our nonlinear IFS map functions comes from Draves’ [5] fractal flames nonlinear IFS, most commonly seen in his distributed-rendering screensaver Electric Sheep [6, 7]. The previous nonlinear function is actually Draves’ function “hyperbolic.’’ Beyond IFS, recursive Lindenmayer or L-systems [8] include the ability to pass parameters between recursive instances. This makes them much more useful than IFS to represent imperfectly self-similar shapes such as plants, as extended by Prusinkiewicz and Lindenmayer [9]. In fact, affine iterated function systems are actually equivalent [10] to a restricted form of L-systems known as turtle graphics. L-systems are often described as string rewriting systems, which is a useful definition but an extraordinarily inefficient implementation; L-systems can be efficiently incrementally instantiated, for example, during ray tracing [11]. The context dependence and parameter passing of L-systems makes them more complex and difficult to analyze than IFS, so we will confine our attention to IFS in this paper. 2.1. Bounding Iterated Function Systems Our overall goal is to render nonlinear iterated function systems using graphics hardware textures, which are uniform-resolution rectangular grids of pixels; that is, textures are bounded. But an IFS attractor may span the unbounded plane, so we must somehow deal with this mismatch. A simple bounding method is to define each IFS such that the attractor is known to lie within some bound, such as the unit square. For example, if each map function takes points inside the unit square to a subset of the square, then by the recursive bounding theorem [12] the unit square is guaranteed to bound the IFS attractor. This is the approach taken by van Wijk and Saupe [13], Gröller [14], Raynal et al. [15], and several others. The difficulty with this bounding method is that manipulating the IFS maps near the artificial boundary becomes cumbersome; sometimes in order to make room to manipulate one map, the user needs to adjust all the other maps, which seems unnecessarily difficult. For any IFS we can transform the fixed point by any invertible function , simply by adjusting the individual map functions according to the well-known [3] transform theorem. The new IFS has map functions . Essentially, the maps to represent the IFS attractor in the new space are found by first transforming points into the old space, applying the old map, and finally transforming back to the new space. This implies that we can allow users complete freedom in defining their maps, yet still do our IFS processing on the convenient unit square, simply by finding a suitable transformation function . For example, if we can find a bounding volume for the IFS attractor, we can then trivially compute an affine transformation of that volume to the unit square. Many authors have taken this approach, including the simple bounding sphere of Hart and DeFanti [16], the tighter sphere of Rice [17], and the anisotropic sphere of Martyn [18]. Convex bounds include the linear programming method of Lawlor and Hart [19], the bound refinement technique of Chu and Chen [20], and a heap-based convex bound refinement method [21, Appendix C]. Bounding volumes are also useful for ray tracing 3D shapes, including affine IFS [16], Gröller’s grid-deformation nonlinear IFS [14], and nonlinear CSG-pL systems [11]. Yet bounding volumes are less useful for many nonlinear IFSs, because in practice these systems are often unbounded. 2.2. The Unbounded Noncontractive IFS An IFS attractor is bounded whenever the IFS map functions satisfy the Lipschitz contractivity condition [1]. Occasionally, a noncontractive IFS will nonetheless have a bounded attractor—contractivity is a sufficient condition, but not necessary. Yet there are actually a variety of interesting iterated function systems, including many nonlinear systems, that are unbounded and noncontractive, yet are still visually interesting. For example, the two-map affine IFS of Figure 2 does not satisfy the contractivity condition, and indeed its attractor extends to infinity along the thin horizontal and vertical spikes. The map functions are Figure 2: An unbounded non-contractive affine IFS. Large grey box is the reference unit square, black boxes the two maps. This IFS is not contractive, because stretches points horizontally. Repeated compositions of stretch out points arbitrarily far. However, the attractor is still well defined, and the random iteration algorithm very rarely explodes points to infinity, because any application of will quickly bring distant points closer to the origin. We agree with Saupe [22, page 310] that “There should be interesting theory of non-contractive IFS.’’ Notice that for rendering, it is acceptable if a few of our points escape to infinity. We can even quantify this threshold, for example, by taking the Lebesgue measure of these points at infinity. We can get reasonable Lebesgue measures by lining up all the computed IFS points on a 1D line, for example, using Barnsley’s map-based “codespace’’ coordinate system as illustrated in Figure 3. For our example, noncontractive IFS, the 1D Lebesgue measure is zero for the set of points that escape to infinity. Figure 3: As additional maps are applied to the points of the IFS, fewer and fewer codespace points have never had applied. Thus, we can classify IFSs as follows:(1)contractive and hence bounded, the usual case for affine IFS,(2)noncontractive yet bounded, which exists but is fairly rare,(3)unbounded yet most (in a Lebesgue sense, comparing measures) points are finite, such as the previous IFS,(4)unbounded with most points at infinity, a classic nonconvergent IFS. Only IFS in that last class causes serious problems during rendering, so in this paper we embrace unbounded iterated function systems. We have noticed that some map functions, including that of Figure 1, cause the naive random iteration algorithm to become stuck in one local area, rather than properly sampling the entire attractor. This cannot happen for contractive affine maps, which have a single attractive fixed point, but for noncontractive nonlinear maps with multiple fixed points, it actually seems to be fairly common. This can be remedied by periodically restarting the random iteration algorithm at a new random plane point. For a similar reason, the output range of the random iteration algorithm’s initial random number generator is important, and generally a larger range will be more likely to correctly sample a larger attractor. The deterministic algorithms we present seem to be more robust against these effects, especially with a large initial attractor estimate image. 3. Rendering Unbounded IFS on the GPU To fit an IFS attractor into a GPU hardware texture, we must have a bounded IFS. But many of the IFS we would like to render are unbounded, and the attractor even includes a few points at infinity. Our solution is to compress the unbounded IFS into an equivalent bounded IFS, using an invertible but nonlinear compression transform function. At display time, if desired we can uncompress the attractor back into the original plane. In particular, we would like to convert plane coordinates, in the interval on both axes, into a square in graphics card texture coordinates, within on both axes. The plane-to-square mapping we choose must be smooth, because it will be used to write discrete samples from an IFS attractor into a texture; any discontinuities in the plane-to-square function will manifest themselves as sampling artifacts in the texture. The function also must be invertible, so that we can read samples of the IFS attractor from the texture. Finally, both the function and its inverse must be efficient to compute on the graphics card, because every access to the texture will require at least one execution of the function. The quite similar general shapes of several such functions are compared in Figure 4, and their equations and performance are compared in Table 1. Of these, the fastest functions on current GPU hardware (an NVIDIA GeForce 280 GTX) are the 2D polar stereographic projection stereo, and the per-axis equivalent projection rsqrt. We attribute the excellent speed of these two functions to the fact that reciprocal square root is a GPU hardware instruction. Table 1: Compression and decompression functions, and corresponding GPU per-pixel performance. erfc, atan, and rsqrt are evaluated independently along each axis; stereo is a 2D vector function. erfc is not built into GLSL, though it exists in CUDA. Figure 4: Comparing 1D plane-to-square functions used to compress an infinite plane IFS onto a finite texture. Comparing these two, there is slightly more shape distortion in the rsqrt version. While rsqrt compresses an infinite plane to the unit square, stereo compresses the plane into a unit disk, as illustrated in Figure 5(b). These two functions are identical precisely along the and coordinate axes, but since GPU textures are square, rsqrt produces fewer sampling artifacts along the diagonal lines. This is especially true near the points at infinity, which rsqrt cleanly maps to the texture’s outer boundary pixels, while stereo maps to a pixel-discretized approximation of a circle. Figure 5: Comparing 2D rsqrt and stereo textures. For this reason, we use rsqrt as our compression and decompression scheme; it seems to work well to transform an unbounded IFS into an equivalent bounded IFS. Due to sampling issues, some IFS may benefit from an additional pre-rsqrt affine plane transformation, typically just a translation and scaling. But because plane-to-square functions are necessarily nonlinear, we may need to render a (bounded) nonlinear IFS even if the original IFS was linear (but unbounded). We explore several approaches to render a bounded nonlinear IFS in the next two sections. 4. Inverting Functions per Vertex by Rasterization Barnsley’s Deterministic Iteration Algorithm [3] determines an IFS fixed point , by iteratively following its definition as the union of its images under the IFS maps : In this method, we approximate the IFS attractor using a rectangle of pixels in a GPU texture—hence, the attractor must be bounded, for example, using one of the techniques discussed in Section 3. At each step , we create a better approximation of the attractor by applying the IFS maps to the previous approximation (for now, ignoring map probabilities, colors, and density effects) with Repeatedly applying this process eventually converges on the attractor, as illustrated in Figure 6. In practice, many IFSs produce a good attractor estimate in as few as a dozen of such map iterations. In theory we can begin the iteration with any arbitrary approximation , typically either a unit square or entire plane, or the solution from the previous frame or multigrid level (see Section 7.4). Figure 6: As we repeatedly distort our texture by the IFS map functions, the texture iteratively approaches the attractor (see Figure 15). On the GPU, we can implement (7) by simply rasterizing each mapped attractor image into the framebuffer . For affine maps, even ancient graphics interfaces such as OpenGL 1.1 directly support this rasterization; van Wijk and Saupe [13] found excellent GPU performance for affine IFS back in 2004. However, nonlinear maps are substantially more difficult to rasterize with high quality. One obvious approach to render nonlinear map images on graphics hardware is to discretize our old attractor estimate using a grid of vertices . Then we can transform each vertex by each map function and then draw the attractor texture interpolated between the mapped vertices in the usual fashion, as illustrated in Figure 7(a) and the CPU-side pseudocode shown in Algorithm 1. Algorithm 1 Figure 7: Two approaches for distorting a texture by an IFS map: per-vertex forward application of maps and per-pixel inverse mapping. In practice, the A = D step is typically implemented by swapping the two texture handles, a “ping-pong,’’ rather than actually copying any data. This per-vertex algorithm works well for smooth map functions, when linearly interpolating the mapped geometry between vertices would be reasonably accurate. Gröller’s [14] nonlinear IFS maps are actually defined as a linearly interpolated vertex grid. Clearly, an adaptive mesh refinement scheme [11] or higher-order vertex interpolation would improve accuracy. However, neither adaptivity nor high-order schemes will work if the map function jumps discontinuously. In any case high-quality images or less-smooth map functions require a dense set of vertices, which causes several increasingly unfortunate effects. First, the vertex work per map per iteration soon overwhelms the CPU’s arithmetic capacity. This problem is fairly easy to address by simply moving the computation into a vertex shader, which yields about a tenfold performance improvement in our experiments. One could also prepare a vertex buffer object for each map, since the mesh does not change across iterations, only the texture shown on it. But we then reach a far more serious bottleneck, which is the GPU’s triangle setup rate. Theoretically, to achieve a framerate of frames per second, when repeating iterative attractor expansions per frame, each of which drawing maps with a grid of vertices, requires the graphics card to draw triangles per second. For example, at frames per second, with just iterations per frame, maps, and vertices per side, would require 1.2 billion triangles per second. Though a typical modern GPU can process tens of billions of pixels per second (known as “fill rate’’), even the best cards process less than a billion triangles per second, so even this moderate end-to-end performance is not achievable via vertex shaders. See the performance experiments in Section 7.3 for details. Also, a dense grid of vertices per side may not be enough vertices. For example, consider an IFS using the “sinusoidal’’ map function, which collapses the entire plane into a cube via the sine function; clearly Nyquist-level vertex sampling over an infinite plane is impossible. In Figure 8(a), even at the vertex geometry used to discretize this map function is still visible. By contrast, Figure 8(b) shows the much cleaner results obtained via the per-pixel analytic map inversion method described in the next section. Figure 8: Comparing per-vertex and per-pixel nonlinear maps. 5. Analytic Per-Pixel Function Inversion Arbitrary nonlinear functions may combine sharp discontinuities, smooth curves, and high-frequency regions that are all difficult to sample accurately. But we are trying to render an attractor estimate texture , so sampling in the destination space is trivially simple, a regular grid of pixels. That is, instead of trying to sample the function so that the resulting geometry covers with subpixel accuracy, we instead follow (2) directly and start at a pixel with 2D destination texture coordinates and sample the source texture at source texture coordinates . In equations, we start from Barnsley’s whole-image set-theoretic deterministic iteration method, where we must apply a nonlinear image distortion to an entire image : Hence we switch to a pixel-by-pixel function inversion, starting in the destination space: This approach maps perfectly to graphics hardware pixel shaders: is the framebuffer, is bound as a source texture, and we compute the source texture coordinates and Jacobian via a programmable shader. But by contrast with affine maps, which only lack an inverse function for degenerate cases, there are a number of practical and theoretical difficulties with inverting general nonlinear functions. 5.1. Handling Multivalued Nonlinear Inverses Many nonlinear functions have no well-defined inverse function, but instead have a multivalued inverse relation. For example, the inverse of the 1D function has both positive and negative branches ; the inverse of the 2D function has four branches . In practice, we can often simply sum up the quantity of interest, such as an attractor density estimate, over each of the inverse values. Of course, periodic functions such as have infinite families of inverses, so in practice we must eventually truncate this summation; typically we find summing up a few periods is sufficient. A further complicating factor is that the current generation of GPU hardware supports neither dynamically sized arrays, virtual functions, function pointers, nor even simple recursion. This complicates any design supporting multivalued inverses—we cannot dynamically allocate a list of inverse values, we cannot call a virtual method or function pointer for each inverse we find, and we cannot recursively search for values. However, with some graphics interfaces, such as GLSL or Open CL, we do generate the GPU functions at runtime; this means each time we find an inverse, at function generation time, we can simply paste in a call to the function needing the inverse value. In particular, our output pixel will require texture samples from the input texture at each inverse value found for the inverse map , as illustrated in Figure 9. In the forward direction, a typical IFS map consists of a texture-to-plane decompression function (from Section 3), then an affine matrix transformation , then a nonlinear “variation’’ function , and finally a plane-to-texture recompression function : Figure 9: Highly nonlinear maps may have multiple inverses. vec2destloc=T(V(M(P(srcloc)))). To look up texture values for each inverse, we simply apply the inverse of each transformation, , and then look up the resulting points in the source texture. Due to the above GPU hardware limitations, we split the call sequence after the nonlinear inverse step , and put into a separate function, here called f. This lets our nonlinear inverse relation simply call f at each inverse value and sum up the densities it returns, as shown in Algorithm 2. Algorithm 2 Because we can generate the exact code to invert only the maps of the IFS currently being rendered, this approach seems to perform quite well. A more static code structure, which compiled in code for every supported nonlinear IFS map function, would require a large switch statement or nested series of comparisons to select the appropriate nonlinear function, neither of which would perform well on current GPU hardware. Although we dynamically generate the above code at runtime, the GLSL driver takes a few milliseconds to recompile the code, so for animation purposes rather than recompiling every frame, we pass in function parameters via uniform variables. This means we only need to recompile when the functional form of the IFS changes, not just its parameters. In practice, the sampling properties of this per-pixel inversion method seem to be excellent. Further, the texture sampling hardware provides both linear texture filtering when a map sample lands between source texture pixels, and anisotropic mipmapping when the mapped sample should read from a larger area. Compared to the conventional random iteration algorithm, per-pixel inversion gives extremely smooth images such as Figure 1, especially in low-density regions where the random iteration algorithm’s discrete points are far apart. Compared to the per-vertex deterministic iteration algorithm described in the previous section, this per-pixel algorithm follows mapped curves more accurately and handles map function discontinuities more cleanly. 5.2. Nonlinear Inverses May Not Exist Simple nonlinear functions sometimes have very complicated inverses, in terms of both execution time and code complexity. For example, Cardano’s solution is much more complex than a cubic polynomial. Computer algebra systems can help to find and simplify inverse relations, although substantial human effort is often still required to create and test working code. Table 2 summarizes the first twenty nonlinear “variation’’ functions proposed by Draves and Reckase [5]. We found easy-to-compute inverse relations for thirteen of these functions; two additional functions do have inverses, but they are too complex to write here, and likely too complex to actually use at runtime. Table 2: Functions and inverses are discussed in Section 5.2 and used for per-vertex and per-pixel rendering, respectively. The Jacobian determinants are used in Section 6 for attractor density; only the forward direction is shown for the Jacobians. stands for or , the reciprocal of their normal usage for compatibility with Draves’ large existing library of nonlinear fractals. The remaining five functions appear to have nonelementary inverses, but it is rare that one can prove the nonexistence of an elementary inverse relation. We attempted to find inverses using both the computer algebra system Mathematica 7.0 and manual effort, but neither found an inverse in a reasonable time. Many nonlinear functions simply do not possess an elementary inverse. For example, does not have an elementary inverse nor does a general fifth-degree polynomial. Though function inverse reverse-distributes over composition , few other relations hold; for example, knowing and tells you nothing about or . In these cases inverse values could still be numerically approximated using a power series (e.g., via the Lagrange inversion theorem) or using a generic nonlinear root-finding approach such as bisection or Newton’s method. These approximations could be precomputed and stored in a texture, or evaluated at runtime per pixel, depending on the hardware’s ratio of memory and arithmetic bandwidth. An entirely different solution to the difficulty of function inversion is to simply invert the definitions: choose easy-to-evaluate functions as the “inverses’’ and apply these functions directly in the deterministic iteration algorithm. The “forward’’ functions are then equally difficult to compute, but our algorithm never needs to compute them. Typically an automated image compressor or artist chooses from a fixed set of predefined basic functions anyway, so defining the map functions this way is less restricting than it may seem. 6. Attractor Density Estimation The Jacobian determinant of the function at a point is defined in 2D as the determinant of the function’s Jacobian matrix of partial derivatives, evaluated at : Expanding out the determinant, we get Substantial cancellation often occurs in this expression, so computing Jacobian determinants in practice is normally quite efficient. For affine map functions, the Jacobian is a constant, so it is often folded together with the arbitrary map probability. But for, our nonlinear maps, the Jacobian determinant may vary with location, so we cannot simply fold it into the constant map probability. Note that we must in general evaluate the Jacobian determinant after inverting the function, because our nonlinear function inverses can take multiple values, yet not all these values will necessarily have the same Jacobian. Sometimes the Jacobian of the forward function is simpler to evaluate than , in which case we use the fact that the inverse of the Jacobian matrix is the Jacobian of the inverse function, and applying the determinant transforms a matrix inverse into a multiplicative inverse: We show the Jacobian determinants for a variety of nonlinear functions in Table 2. Some extremely nonlinear functions such as “swirl’’ nonetheless have a constant Jacobian determinant, indicating swirl is area-preserving everywhere. It is noteworthy that each of the nonlinear functions we were able to invert has a simple Jacobian determinant, indicating some underlying simplicity. The noninvertible functions we examined, even those with superficially similar functional form, all had substantially more complex Jacobian determinants. Despite this still-moderate complexity, unlike function inverses, every elementary function also has an elementary Jacobian. It is also possible to evaluate Jacobian determinants numerically. Some graphics programming languages even include built-in primitives for this, such as GLSL’s dFdx and dFdy keywords, which compute a finite difference by examining neighboring pixels. However, we find that analytically evaluated Jacobians are better behaved near discontinuities and are less susceptible to numerical and sampling artifacts. 6.1. Properties of the Density Jacobian In general, because the Jacobian is assembled from partial derivatives, a straightforward application of the chain rule can determine the Jacobian of the composition of two functions and : So, for example, a variation function applied after a matrix will have Jacobian That is to say, we evaluate ’s Jacobian after first transforming , then multiply by the matrix’s Jacobian. We can ignore the Jacobian contributions from the plane-to-texture compression and decompression functions and from Section 3, because they will cancel each other out; that is, we can compute the attractor density for the original infinite-plane IFS, although we sample that density geometrically only at the texture pixels. This results in an easier-to-interpret density—with and included, the choice of these plane compression functions affects the resulting attractor density values, not just their sampling geometry. Avoiding the plane-to-texture and texture-to-plane Jacobian contributions this way also saves a little arithmetic work during rendering. 6.2. Jacobian Density Estimation on the GPU The dynamic range of the Jacobian determinant term can be quite large. Draves [5] applies a nonlinear log-exposure function after accumulating counts into a high-range framebuffer. We find that the modern GPU exponential and log functions are fast enough that we can actually store the of the density in our texture pixels and “unpack’’ this density using an exponential operation after every texture fetch. Arithmetic can then be performed in high-range linear-density space, while all storage happens in the range-limited log space. These nonlinear density unpack and pack functions are implemented in GLSL as follows. Our texture color channels store numbers in the interval using 8 bits of precision. The following factor of 20 lets us store a density dynamic range of 220, about a millionfold, which seems to be enough for most IFSs to create smooth attractive images. On modern GPU hardware, it is actually several times faster to use 8-bit fixed-point memory storage and expand the range in software using these exponential and logarithm operations, than it is to simply use 32-bit floating-point storage without any additional arithmetic: vec4density=exp2(tex*20)-1;//unpack vec4tex=log2(density+1)/20;//pack. For per-vertex rendering, this unpack-sum-pack operation is not possible as a standard framebuffer blend operation, so one must resort to rendering the map into a separate texture, then performing the blending manually in a second pass. For per-pixel rendering, we can actually fetch, unpack, and sum up the densities for all the IFS maps in a single pass, and then apply the log transformation before writing the pixels out to the framebuffer. We illustrate the effect of the Jacobian and log-density output in Figure 10 and show the details in the complete code example in Algorithm 3. Algorithm 3: GLSL source code generated to render Figure 1 IFS using the per-pixel inversion method. An IFS with more or different map functions will generate a different listing. Affine parameters are uniforms, to allow animation. Figure 10: The IFS of Figure 1 rendered (a) with linear output, no Jacobian; (b) with Jacobian; (c) log output with Jacobian. Finally, we can add RGBA color to our attractor by multiplying the output of each with a color —this is equivalent to adjusting the color of each function’s output pixels. This is why we use a four-float “vec4’’ above, and it is equivalent to the method of iterating a color through the IFS maps along with position. Draves’ fractal flames use a different method, where a single “color’’ index is iterated along with position, and then looked up in a 256-entry color lookup table before blending to the framebuffer; his approach makes it easier to highlight substructures in the attractor and could be emulated by simply using more color channels in the texture, but we find the classic IFS coloring method to be sufficient. Plane-to-texture compression, per-pixel inversion, Jacobian density estimation, and log-density output all combine well to render nonlinear IFS in color on the GPU. 7. Performance Comparisons The Graphics Processing Unit (GPU) has evolved quickly over the last decade, so current GPUs can now run sophisticated C-like code fragments at every pixel. Because GPU languages such as GLSL, Open CL, or CUDA do not allow dependencies between pixels, the GPU hardware can execute these programs in parallel across pixels. A GPU typically executes hundreds of pixels per clock, with thousands of pixels in flight, and so delivers orders of magnitude higher performance than multicore [23]. GPUs have hardware support to both read and write textures, which are 2D or 3D arrays of pixels or voxels stored in a variety of formats. The main advantage of using C++ and Open GL is that the same code can be run on Windows, Macintosh, and UNIX computers. Many researchers have rendered iterated function systems at interactive rates. In 1995, Monro and Dudbridge [24] achieved nearly 1 M pixels per second using a fixed point SIMD within a register software implementation. In 2004, for affine IFS a deterministic GPU texture-based implementation completed about 13 M pixels per second [13] (50 fps for a output image). In 2005, a GPU-based nonlinear IFS render to vertex buffer implementation of the random iteration algorithm completed 20 M finished output points per second [25] (20 fps for a 1 M point buffer). The state of the art as of late 2011 appears to be 1 billion point-iterations per second, achieved using a highly optimized random iteration algorithm implementation in CUDA [26]. 7.1. Theoretical Performance Comparison We define an IFS image as an estimate of the true proportion of random iteration algorithm trials which hit that pixel. The random iteration algorithm computes those pixels stochastically, where each trial either hits a pixel or does not, and hence a pixel’s hit count obeys the well-known binomial distribution with variance . Thus, assuming the central limit theorem, after trials our estimate’s variance is . An Agresti-Coull 95% binomial confidence interval has a width of approximately . Thus, to halve the image noise, we must compute four times as many samples. Especially with gamma correction, which has a steeper response for darker pixels, this sublinear convergence rate is a problem in darker areas of the image. In the deterministic iteration algorithm, by contrast, every pixel receives a density estimate during every image pass. This makes deterministic iteration images smoother, especially in dark regions. However, it may take several iterations before the images begin to converge to the attractor and for pathological cases may never converge at all. This happens rarely in practice, because repeated geometric scaling drives points to their destinations exponentially fast: points undergoing a scaling factor of move by after passes. If is below unity, a contraction, points converge to the attractor at a well-known exponential rate; even if is over unity, an expansion, points similarly approach their fixed point of infinity at an exponential rate. In fact, to begin the random iteration algorithm, typically a few dozen “fuse’’ iterations are performed to converge points to the attractor before beginning to accumulate pixel counts. Experimentally it takes a similar number of image-to-image iterations, typically about a dozen, for the deterministic iteration algorithm to converge, at which point the algorithm is finished. In addition to the exponential difference in convergence rate, the random iteration algorithm produces points stochastically, at unpredictable locations; even ignoring efficiency these random writes are difficult to parallelize correctly. By contrast, the deterministic iteration algorithm produces image samples at every image pixel, which allows the image rendering work to be divided among many parallel processors in a straightforward fashion. 7.2. Quantitative Performance Comparison Figure 11 compares the performance of our algorithm with two existing random iteration algorithm nonlinear IFS renderers. The vertical axis in this comparison is Chandler and Hemami’s Visual Signal to Noise Ratio (VSNR) [27], a perceptually based image comparison method computed using wavelets and measured in decibels. The IFS used for this comparison is the “swirlpinski’’ IFS from Figure 15, rendered at resolution, but other display sizes and rendered IFS appear to display similar relative performance. Figure 11: Comparing our algorithm at various texture resolutions with existing GPU- and CPU-based random iteration approaches (GeForce 580). flam3 [7] version 2.8 is a well-tuned multicore aware CPU-based library for rendering nonlinear IFS using the random iteration algorithm. The CPU is a 3.1 GHz quad-core Intel Core i5 2400, using all four cores. The performance versus accuracy tradeoff can be adjusted using the “quality’’ parameter, which is the number of random iteration algorithm trials per pixel: a quality of 1 executes in under one second but produces a very noisy stochastic image; while a quality of 1,000,000 takes most of a day to execute but produces an excellent image. Since this is the oldest and most widely used existing nonlinear IFS renderer, we use this renderer with 1,000,000 trials per pixel as the reference implementation. flam4CUDA [26] is a well-tuned CUDA implementation of the random iteration algorithm. The GPU is an NVIDIA GeForce 580 desktop card, the same used for our algorithm. flam4CUDA uses several quite clever optimizations, such as per-warp random number generation, to synthesize random points while maintaining good branch coherence. One serious inherent shortcoming of the random iteration algorithm approach on highly parallel architectures is a floating point read-modify-update race condition while writing points to the framebuffer. flam4CUDA currently ignores this race condition, because it is not clear how to resolve it efficiently on the GPU. flam4CUDA and flam3 also use slightly different spatial antialiasing (flam4CUDA uses per-pixel jitter) and gamma correction schemes. Finally, the density estimation filter radius is limited by the size of GPU-shared memory, so it is difficult to directly compare image outputs. For these reasons, in Figure 11 we generously assume flam4CUDA’s output image was identical to the flam3 output image given the same number of trials per pixel. flam4CUDA finishes these same trials approximately 40x faster than flam3 and is performance-competitive with our algorithm for large image sizes, but does not scale down as well to interactive rates. Our algorithm displays several separate regimes depending on parameters. Small output texture sizes cannot represent the sharp features in the IFS, and hence are limited to low VSNR regardless of run time. Very large images, over , produce only minor improvements in the finished image quality, mostly because we must scale down to compare against the reference image. At the crucial 10 ms to 50 ms interactive animation range, our algorithm can comfortably compute screen-sized textures accurately. Generally, convergence to the final output begins slowly and ends extremely rapidly, with most of the VSNR gains happening in one or two crucial iterations before converging. This rapid convergence, as predicted in the previous subsection, compares favorably against the exponentially slower accumulation process of the random iteration algorithm. However, to compute extremely high-quality images exactly, our algorithm may require very high-resolution textures. Note the slight blurring at the left edge of Figure 12, where the swirl map repeatedly shears the texture. To exactly duplicate the results of the 2D random iteration algorithm using IEEE 32-bit floating point arithmetic would require an image of size , using exabytes of storage. For this reason, in some situations very high-quality images may be rendered more accurately by the random iteration algorithm, while our algorithm dominates below one second of compute time, the region most useful for image compression or animation applications. Figure 12: A zoomed-in portion of the swirlpinski fractal, computed using the deterministic and random iteration algorithms. 7.3. Vertex versus Fragment Rendering Our deterministic iteration algorithm consists of an iterative series of passes where we apply (2) to our textured attractor approximation, slowly improving the approximation. The performance of each pass, using our per-pixel and per-vertex methods at various resolutions is summarized in Table 3. This subsection’s performance numbers are presented for the two-map nonlinear IFS shown in Figure 1, on an NVIDIA GeForce GTX 280, a midrange desktop graphics card. Table 3: GPU time per pass for various texture sizes, as rendered with the per-pixel and per-vertex methods. (GeForce 280). The per-vertex rendering algorithm is substantially slower for small output textures, even for a low vertex resolution of vertices. Both algorithms scale poorly to very small meshes, because the framebuffer object setup time and postrender mipmap building dominate the actual rendering work. The per-vertex algorithm is slightly faster than the per-pixel method at low vertex and high pixel resolutions, because the per-pixel method must consider and discard a large number of pixels that are not touched by the map function output. But at a higher vertex resolution of , the per-vertex method becomes vertex-rate dominated, so it takes almost exactly the same amount of time to output to a tiny texture as a huge one and much more time than the per-pixel method in any case. It takes 46.8 ms to draw both maps using a mesh of 1 million vertices each, which is 2 million triangles per map or 4 million triangles per pass, a net triangle rate of 85 million triangles per second. By contrast, the per-pixel algorithm can make one complete pass through a texture, applying the plane-to-texture transform, both map functions, the Jacobian density compensation, and our log-density output packing, in under 5 ms. This is a total of 3.3 billion color pixels written per second, 6.7 billion map function applications per second, or over 0.3 trillion floating-point operations per second—many of which are divides, square roots, exponentials, and logarithms. 7.4. Multigrid Rendering Table 3 shows that our iterative per-pixel rendering algorithm is much faster per pass when working on small output textures, such as pixels. Although we can still discretize the entire plane, such a tiny texture does not provide much detail, so it is not useful as a final output. However, depending on the initial attractor estimate, the first few passes will be seriously inaccurate until we begin to converge on the IFS attractor’s general shape. This suggests a multigrid-type approach, where instead of performing all the passes at the highest output resolution, we perform early passes on much smaller textures until we are near convergence, and then we can incrementally increase the texture resolution up to the final size. Table 4 summarizes the performance results from this, where we compare ten passes at each of eight power-of-two multigrid levels versus eighty passes all at the highest resolution. Multigrid provides the biggest performance improvement, over fivefold, for larger output images, but it makes a significant difference even at lower resolutions. For example, at a final output resolution of , the multigrid implementation still runs at well over 30 fps, while the fixed-size implementation is under 10 fps. As shown in Figure 13, multigrid provides equivalent visual quality much faster than any existing implementation. Table 4: End-to-end time to generate Figure 1 using multigrid and fixed-size per-pixel rendering, for a constant 80 passes total (GeForce 280). Figure 13: Comparing performance versus visual quality for multigrid, nonmultigrid, and random iteration (GeForce 580). Multigrid is surprisingly easy to implement when using OpenGL’s default texture coordinates, which run from 0.0 on one edge of the texture to 1.0 on the opposite edge. Because the entire rendering coordinate system is independent of the number of pixels used in either the source or destination texture, we can simply substitute a higher-resolution destination texture to switch from one grid level to the next. Compared to array indices or pixel numbers, which have different values at different resolutions, and yet different values when switching resolutions, resolution-independent coordinates significantly reduce the complexity of a multigrid implementation. 8. Conclusions and Future Work We have presented a set of techniques that allow nonlinear function fixed points and iterated function systems to be computed on graphics hardware with extremely high performance. In particular, we have shown how to compress an unbounded IFS into the unit square, so an approximation to the IFS attractor can be stored in a texture. We have shown how analytic map inversion can be used per pixel to iteratively improve this approximation. We have explored how to use a map function’s Jacobian determinant to track our discretized attractor density, and how to store that density in a range-limited fixed point texture. Finally, we showed how to use multigrid to accelerate the convergence of our approach. Together, these techniques allow a single GPU to interactively render nonlinear IFS that previously could only be rendered offline on a large distributed cluster. There is much work remaining to do. Currently we do not exploit frame-to-frame coherence, which could cut the number of rendering passes required, especially on low-end machines. Multigrid has the advantage that it allows arbitrary animations, including smash cuts and strobe-type effects, but many animations do have significant coherence. We also currently perform no precomputation and compute the map functions using only arithmetic. More complex nonlinear functions, or functions whose inverse can only be approximated numerically, would benefit from a discretized version of the map inverse relation, such as a “source coordinate texture’’ lookup table. A lookup table could have various features folded in, such as an attractor-dependent texture-to-plane function. In addition, Draves has now collected nearly a hundred nonlinear variation functions, of which we only analyzed the first twenty, so it would take significant effort to find analytic inverses for the remaining functions. Thus, some simple procedures to approximate or store inverse relations would be useful for backward compatibility. As with any finite approximation, our attractor texture occasionally displays sampling artifacts. In the center of spirals, where attractor pixels are repeatedly resampled using the hardware’s bilinear texture interpolation, the geometry can become somewhat blurred. A resource-intensive solution is to increase the texture resolution, but it seems likely that a higher-order texture interpolant, such as cubic sampling, could produce better results. Some attractors have important features stored far away in the plane, where our texture resolution is low. This effect is visible on asymptotic spikes, which become blurry near the tips, especially at low texture resolutions. Some sort of intelligence added to the plane-to-texture mapping should be able to reduce this effect. For example, instead of a fixed uniform texture resolution, the attractor could be approximated using adaptive resolution texture tiles. Our basic per-pixel algorithm generalizes almost trivially to higher dimensions, and 3D GPU textures consisting of voxels are well supported. However, unlike the random iteration method, for our method the storage and computational requirements of higher dimensions quickly become prohibitive; for example, a pixel color image takes only 16 megabytes of storage, while a voxel color volume requires 34 gigabytes. However, even today’s graphics cards are capable of comfortably volume-rendering voxel volumes at interactive rates, so per-voxel 3D nonlinear iterated function system rendering will eventually become affordable, especially when using adaptive resolution tiled 3D textures. In this paper we have only addressed plain IFS, the simplest form of nonlinear recursive geometry. Yet there are many other more complex procedural models, including recurrent iterated function systems [28], superfractals, escape-time fractals [4], and programmable L-systems. It seems likely that a modified version of the deterministic iteration algorithm could have excellent GPU performance in rendering these other methods to represent fractal geometry. In particular, we would like to explore varying the map functions at each level, for example, to mirror the growth cycle of plants. Finally, we welcome the interested reader to download (http://tinyurl.com/gpuifs/), extend, and contribute to our open source implementation of this technique. Appendix Algorithm 3 shows a slightly reformatted version of the per-pixel GLSL code generated to render Figure 1. Figure 14 shows the “fountains’’ IFS, which has these maps: Figure 14: A three-map “fountains’’ IFS. All these IFSs are rendered on the GPU at interactive rates with the per-pixel method. Figure 15: The three-map “swirlpinski’’ IFS, the same one shown in Figure 6. This is the only IFS in this paper whose attractor is actually bounded—every other IFS we present extends to infinity in at least one direction. The spherical and other nonlinear functions used in our IFS maps are defined in Table 2. Figures 15 and 16 show other interesting nonlinear IFS. Figure 17 shows one of Draves’ “electric sheep’’ fractals, one of the few interesting existing sheep that uses invertible map functions. Each of these IFSs animate beautifully, an effect that simply cannot be conveyed in print. Figure 16: A three-map “eyes’’ IFS. Figure 17: Draves’ fractal generation 165, sheep 48. Symbols : -Dimensional projective space; here : The space of -channel colors; here : The space of all images: : A geometric distortion function: : The Jacobian matrix of : An image distortion function: : The sum of the : : IFS attractor image, the fixed point of : In . : The th texture approximation of : Plane-to-texture compression function : Texture-to-plane function, : An affine transformation: a matrix. References 1. R. F. Williams, “Composition of contractions,” Boletim da Sociedade Brasileira de Matemática, vol. 2, no. 2, pp. 55–59, 1971. 2. A. Tychonoff, “Ein fixpunktsatz,” Mathematische Annalen, vol. 111, no. 1, pp. 767–776, 1935. 3. M. F. Barnsley, Fractals Everywhere, Morgan Kaufmann, 1993. 4. S. Nikiel, Iterated Function Systems for Real-Time Image Synthesis, Springer, London, UK, 2007. 5. S. Draves and E. Reckase, “The fractal flame algorithm,” 2003, http://flam3.com/flame.pdf. 6. S. Draves, “The electric sheep screen-saver: a case study in aesthetic evolution,” in Applications of Evolutionary Computing, Proceedings of EvoMusArt05, 2005. 7. S. Draves, “The electric sheep and their dreams in high fidelity,” in Proceedings of the 4th International Symposium on Non-Photorealistic Animation and Rendering (NPAR '06), pp. 7–9, June 2006. 8. A. Lindenmayer, “Mathematical models for cellular interactions in development I. Filaments with one-sided inputs,” Journal of Theoretical Biology, vol. 18, no. 3, pp. 280–299, 1968. 9. P. Prusinkiewicz and A. Lindenmayer, The Algorithmic Beauty of Plants, Springer, New York, NY, USA, 1990. 10. T. Ju, S. Schaefer, and R. Goldman, “Recursive turtle programs and iterated affine transformations,” Computers & Graphics, vol. 28, no. 6, pp. 991–1004, 2004. 11. M. G. Peter Wonka, “Raytracing of nonlinear fractals,” in Proceedings of the 6th International Conference in Central Europe on Computer Graphics and Visualization (WSCG '98), pp. 424–431, Plzen, Czech Republic, February 1998. 12. J. Hutchinson, “Fractals and self-similarity,” Indiana University Mathematics Journal, vol. 30, no. 5, pp. 713–747, 1981. 13. J. J. van Wijk and D. Saupe, “Image based rendering of iterated function systems,” Computers & Graphics, vol. 28, no. 6, pp. 937–943, 2004. 14. E. Gröller, “Modeling and rendering of nonlinear iterated function systems,” Computers & Graphics, vol. 18, no. 5, pp. 739–748, 1994. 15. F. Raynal, E. Lutton, P. Collet, and M. Schoenauer, “Manipulation of non-linear IFS attractors using genetic programming,” in Proceedings of the Congress on Evolutionary Computation (CEC '99), pp. 1171–1177, IEEE Press, 1999. 16. J. C. Hart and T. A. DeFanti, “Efficient anti-aliased rendering of 3D linear fractals,” in Proceedings of the 18th Annual Conference on Computer Graphics and Interactive Techniques (SIGGRAPH '91), vol. 25, pp. 91–100, 1991. 17. J. Rice, “Spatial bounding of self-affine iterated function system attractor sets,” in Proceedings of the Graphics Interface Conference, pp. 107–115, May 1996. 18. T. Martyn, “Tight bounding ball for affine IFS attractor,” Computers & Graphics, vol. 27, no. 4, pp. 535–552, 2003. 19. O. S. Lawlor and J. Hart, “Bounding iterated function systems using convex optimization,” in Proceedings of the 11th Pacific Conference on Computer Graphics and Applications (PG '03), pp. 283–292, 2003. 20. H. T. Chu and C. C. Chen, “On bounding boxes of iterated function system attractors,” Computers & Graphics, vol. 27, no. 3, pp. 407–414, 2003. 21. O. S. Lawlor, Impostors for parallel interactive computer graphics, Ph.D. thesis, University of Illinois at Urbana-Champaign, 2004. 22. D. Saupe, “From classification to multi-dimensional keys,” in Fractal Image Compression—Theory and Applications to Digital Images, pp. 302–310, Springer, 1994. 23. K. Datta, M. Murphy, V. Volkov et al., “Stencil computation optimization and auto-tuning on state-of-the-art multicore architectures,” in Proceedings of the International Conference for High Performance Computing, Networking, Storage and Analysis (SC '08), pp. 1–12, IEEE Press, Piscataway, NJ, USA, 2008. 24. D. M. Monro and F. Dudbridge, “Rendering algorithms for deterministic fractals,” IEEE Computer Graphics and Applications, vol. 15, no. 1, pp. 32–41, 1995. 25. S. G. Green, “GPU-accelerated iterated function systems,” in Proceedings of the 32nd International Conference on Computer Graphics and Interactive Tichniques (SIGGRAPH '05 Sketches), p. 15, ACM, Los Angeles, Calif, USA, August 2005. 26. S. Brodhead, “Flam4cuda: Gpu flame fractal renderer,” 2011, http://flam4.sourceforge.net/. 27. D. M. Chandler and S. S. Hemami, “VSNR: a wavelet-based visual signal-to-noise ratio for natural images,” IEEE Transactions on Image Processing, vol. 16, no. 9, pp. 2284–2298, 2007. 28. M. F. Barnsley, J. H. Elton, and D. P. Hardin, “Recurrent iterated function systems,” Constructive Approximation, vol. 5, no. 1, pp. 3–31, 1989.

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# 234759 (number) 234,759 (two hundred thirty-four thousand seven hundred fifty-nine) is an odd six-digits composite number following 234758 and preceding 234760. In scientific notation, it is written as 2.34759 × 105. The sum of its digits is 30. It has a total of 4 prime factors and 12 positive divisors. There are 134,064 positive integers (up to 234759) that are relatively prime to 234759. ## Basic properties • Is Prime? No • Number parity Odd • Number length 6 • Sum of Digits 30 • Digital Root 3 ## Name Short name 234 thousand 759 two hundred thirty-four thousand seven hundred fifty-nine ## Notation Scientific notation 2.34759 × 105 234.759 × 103 ## Prime Factorization of 234759 Prime Factorization 3 × 72 × 1597 Composite number Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 33537 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 234,759 is 3 × 72 × 1597. Since it has a total of 4 prime factors, 234,759 is a composite number. ## Divisors of 234759 12 divisors Even divisors 0 12 6 6 Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 364344 Sum of all the positive divisors of n s(n) 129585 Sum of the proper positive divisors of n A(n) 30362 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 484.519 Returns the nth root of the product of n divisors H(n) 7.732 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 234,759 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 234,759) is 364,344, the average is 30,362. ## Other Arithmetic Functions (n = 234759) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 134064 Total number of positive integers not greater than n that are coprime to n λ(n) 1596 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 20754 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 134,064 positive integers (less than 234,759) that are coprime with 234,759. And there are approximately 20,754 prime numbers less than or equal to 234,759. ## Divisibility of 234759 m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 0 7 3 The number 234,759 is divisible by 3 and 7. • Arithmetic • Deficient • Polite ## Base conversion (234759) Base System Value 2 Binary 111001010100000111 3 Ternary 102221000210 4 Quaternary 321110013 5 Quinary 30003014 6 Senary 5010503 8 Octal 712407 10 Decimal 234759 12 Duodecimal b3a33 20 Vigesimal 196hj 36 Base36 5153 ## Basic calculations (n = 234759) ### Multiplication n×y n×2 469518 704277 939036 1173795 ### Division n÷y n÷2 117380 78253 58689.8 46951.8 ### Exponentiation ny n2 55111788081 12937988258107479 3037309185485053662561 713035667075285712769157799 ### Nth Root y√n 2√n 484.519 61.689 22.0118 11.8611 ## 234759 as geometric shapes ### Circle Diameter 469518 1.47503e+06 1.73139e+11 ### Sphere Volume 5.41945e+16 6.92555e+11 1.47503e+06 ### Square Length = n Perimeter 939036 5.51118e+10 331999 ### Cube Length = n Surface area 3.30671e+11 1.2938e+16 406615 ### Equilateral Triangle Length = n Perimeter 704277 2.38641e+10 203307 ### Triangular Pyramid Length = n Surface area 9.54564e+10 1.52476e+15 191680 ## Cryptographic Hash Functions md5 8c44658b7dec5d84ece14899553a11d6 bd65faa260dd8c563de36d316ae71004dc4cf538 d62e279623c397a5ee45de96ab82d94c25f0d91b839d79b3c6eb8a797f908ece 07b9739989053e53e8f10c69c8c2e9ae7f14c84fe5397af93910ddd152c4842c37efa58c7feb8785fff2d5b79be94010c8af6e1081de4c06793beac1a662ecd2 73af72b1d5ebfb37f7b00dc1dc499b2a1424e8fa

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Notice: Undefined offset: 0 in /var/www/convert-dates.com/system/includes/framework/ConvertDates/NumericToWords.php on line 74 Notice: Undefined offset: 0 in /var/www/convert-dates.com/system/includes/framework/ConvertDates/NumericToWords.php on line 74 3048 Days From July 11, 2024 - Convert Dates ## 3048 Days From July 11, 2024 Want to figure out the date that is exactly three thousand hundred forty eight days from Jul 11, 2024 without counting? Your starting date is July 11, 2024 so that means that 3048 days later would be November 14, 2032. You can check this by using the date difference calculator to measure the number of days from Jul 11, 2024 to Nov 14, 2032. November 2032 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 2. 2 3. 3 4. 4 5. 5 6. 6 1. 7 2. 8 3. 9 4. 10 5. 11 6. 12 7. 13 1. 14 2. 15 3. 16 4. 17 5. 18 6. 19 7. 20 1. 21 2. 22 3. 23 4. 24 5. 25 6. 26 7. 27 1. 28 2. 29 3. 30 November 14, 2032 is a Sunday. It is the 319th day of the year, and in the 46th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2032 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 11/14/2032, and almost everywhere else in the world it's 14/11/2032. ### What if you only counted weekdays? In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 3048 weekdays from Jul 11, 2024, you can count up each day skipping Saturdays and Sundays. Start your calculation with Jul 11, 2024, which falls on a Thursday. Counting forward, the next day would be a Friday. To get exactly three thousand hundred forty eight weekdays from Jul 11, 2024, you actually need to count 4268 total days (including weekend days). That means that 3048 weekdays from Jul 11, 2024 would be March 18, 2036. If you're counting business days, don't forget to adjust this date for any holidays. March 2036 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 1. 2 2. 3 3. 4 4. 5 5. 6 6. 7 7. 8 1. 9 2. 10 3. 11 4. 12 5. 13 6. 14 7. 15 1. 16 2. 17 3. 18 4. 19 5. 20 6. 21 7. 22 1. 23 2. 24 3. 25 4. 26 5. 27 6. 28 7. 29 1. 30 2. 31 March 18, 2036 is a Tuesday. It is the 78th day of the year, and in the 78th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2036 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 03/18/2036, and almost everywhere else in the world it's 18/03/2036. ### Enter the number of days and the exact date Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.

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You may want to review factoring with GCDs and factoring harder polynomials first. Polynomials can have GCDs and still start with a number once you've factored it out. For example, consider 6x2 + 16x + 8. The GCD here is 2, so it factors as 2(3x2 + 8x + 4). But the polynomial 3x2 + 8x + 4 itself factors, so the final answer is 2(3x + 2)(x + 2).

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# physics posted by . A passenger in a helicopter traveling upwards at 20 m/s accidentally drops a package out the window. If it takes 15 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped? • physics - Neglect air friction. Solve this equation for height, H: Y (height above ground) = H + 20t -4.9 t^2 = 0 using t = 15 s. • physics - After leaving the helicopter the package is moving upwards at the initial velocity v0. This is decelerated motion with acceleration –g (g≈10m/s^2) h0=v0t1-(gt^2)/2 v=0=v0-gt1 t1=v0/g=20/10=2 s. After this time interval the package is at the height h0=v0^2/2g= 20 m. Then the package is falling down during t2=t-t0=15-2=13 s: H= h0+h=(gt2^2)/2=(10x169)/2=845 m. Hence, the height of helicopter when the package was dropped is h=H-h0=845-20=825 m

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# Dijkstra Shortest Path algorithm 6 messages Open this post in threaded view | ## Dijkstra Shortest Path algorithm This post was updated on . I have written Dijkstra's Shortest Path algorithm in Jack.  Here is the link to the Jack source code: Dijkstra-Shortest-PathThis project is an implementation of Dijkstra's shortest path algorithm, which is used to find the shortest path from a source node to all other nodes in a graph. A classic example is mapping software that finds the shortest route between 2 cities. In writing Dijkstra's shortest path algorithm in Jack, I decided to create my own linked-list class and use that in various parts of the program. It's a simple data structure that is easy to implement and understand. When writing any Jack program in the Jack OS environment, there are of course some minor annoyances that have to be worked around, as the language was designed to be simple enough so that a new student could write a compiler without getting too bogged down. For this project, the following issues took the most attention and care: (1) There is no garbage collection. As you examine the source code, note how dispose() has to be implemented for everything. And when dealing with multiple pointers that point to an object, you have to be careful not to double-free an object. Take care to understand who is the owner of every object you allocate, so you can properly recycle the memory. (2) String allocation and de-allocation can be tricky. I had to write my own String utility class that allows me to concatenate strings with the option of disposing of either argument. See the StringUtil class. I also wrote an Output utility class to automatically dispose the string constant passed to printString. See the OutputUtil class. Other than these issues, writing code in Jack felt normal and fun, like any other high level language. It's quite a powerful language. After all, the Jack OS is written in Jack. I decided the easiest way to experiment with different source nodes or different graphs entirely is to edit the source code directly. If you look at Main.jack, it is self-explanatory. You can define a node and directed edges and add the nodes to the graph. Between nodes you can have 1-way or 2-way direction, simply by adding the appropriate edges to the appropriate nodes. You can also see in Main.jack how to specify the source node. The screen output of the program shows the shortest path from the source node to all other nodes (and the total distance for each), and will indicate if a node is unreachable from the source node. Here is sample output: Shortest path from Node A to: Node B: A -> B (total distance = 10) Node C: A -> C (total distance = 20) Node D: A -> C -> D (total distance = 40) Node E: A -> B -> E (total distance = 20) Node F: A -> B -> E -> F (total distance = 21) Node G: unreachable Open this post in threaded view | ## Re: Dijkstra Shortest Path algorithm Looks really good, I've got some questions, but am gonna do a bit of research first and pose them when I've got some more time. Cheers, Loz Open this post in threaded view | ## Re: Dijkstra Shortest Path algorithm This post was updated on . I modified the LinkedList class (LinkedList.jack): -Rewrote dispose() to be iterative instead of recursive.  Why? Because using the call stack is expensive, and risks a stack overflow if there are thousands of nodes. Open this post in threaded view | ## Re: Dijkstra Shortest Path algorithm Hi I don't have any questions, I miss read your OP, doh. For some reason I thought you were talking about pointers at one point. Any way, great project

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# Lot value calculation off by a factor of 100 6258 Scenario: 2 separate brokers who both have a NASDAQ100 index. When I calculate Lot value in USD (base currency), Broker 2 is 100X greater than Broker 1. The Calculation: ```double GetLotValue(string symbol, const double lotSize, double& pointValue, double& pipValue) { double lotValue = lotSize * (tickValue / tickSize); pointValue = lotValue * SymbolInfoDouble(symbol,SYMBOL_POINT); PrintFormat("%s(%s): contract [%.2f] tickValue [\$%.2f] tickSize [%.7f] lotValue [%s] pointValue [%s] pipValue [%s]", __FUNCTION__, symbol, contract, tickValue, tickSize, Currency(lotValue), Currency(pointValue), Currency(pipValue) ); return lotValue; } { } { long digits = SymbolInfoInteger(symbol, SYMBOL_DIGITS);         // Decimal places for price if(digits==3 || digits==5) { } }``` Result for Broker #1: 2019.07.18 08:51:32.893    Test - MoneyManagement (#USNDAQ100,H4)    CPips::GetLotValue(#USNDAQ100): contract [1.00] tickValue [\$0.01] tickSize [0.0100000] lotValue [\$1.00] pointValue [\$0.01] pipValue [\$0.01] - clsPips.mqh (80) Result for Broker #2: 2019.07.18 08:47:52.983    Test - MoneyManagement (USTECHIndex,H4)    CPips::GetLotValue(USTECHIndex): contract [10.00] tickValue [\$1.00] tickSize [0.0100000] lotValue [\$100.00] pointValue [\$1.00] pipValue [\$1.00] - clsPips.mqh (80) The Issue: I've been using the above code for ForEx without a hitch. Thus, when I calculate a SL or TP, I can determine the value in USD (the base currency). However, when I looked at the SL/TP generated on the index chart for Broker 2, the values were 100X what I expected. So I pulled up the same index (although a different symbol) on Broker 1. The results for Broker 1 were in line with what I expected. Notes: • The contract size for Broker 2 is 10 whereas it is only 1 for Broker 1. I don't actually use the contract size in my calculation. • The tickSize is the same for both brokers. • The website for Broker 2 says the Pip value is \$0.10, whereas my calculation says \$1.00. Is this there the contract size comes into play? One further oddity: Broker 2's tickValue is \$0.10 as stated on the website. This is different from: `SymbolInfoDouble(symbol,SYMBOL_TRADE_TICK_VALUE);` My Questions: (1) Is this a Broker 2 issue? Like maybe they have their settings wrong? (2) Should my calculation consider the contract size, and if so, how? (For ForEx, this value is typically 100000) Or . . . (3) Am I missing something else in my calculation? 43273 Anthony Garot: My Questions: (1) Is this a Broker 2 issue? Like maybe they have their settings wrong? Most probably, unfortunately it's something which happens relatively often. (2) Should my calculation consider the contract size, and if so, how? (For ForEx, this value is typically 100000) Or . . . (3) Am I missing something else in my calculation? When the data provided by a broker are not reliable you have to find a way to calculate it by yourself. You can also ask the broker to fix his symbol settings. 6258 Alain Verleyen: Most probably, unfortunately it's something which happens relatively often. When the data provided by a broker are not reliable you have to find a way to calculate it by yourself. You can also ask the broker to fix his symbol settings. Thank you Alain. I sent an E-mail to the broker to see if they will fix it. If not, I will add some work-around code to trap for that symbol. 6258 Alain Verleyen: You can also ask the broker to fix his symbol settings. Wow! Not only did the broker respond, but they changed it!

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# Formal language problem I’m new to formal language and searching for the solution for the following task: $\Sigma$ is an alphabet with $\lvert \Sigma\rvert = 5$ and $k \in \mathbb{N}_0$. I’m searching for $\lvert \Sigma^k\rvert$. - This might help: math.stackexchange.com/questions/32247/… – joriki Apr 11 '11 at 8:00 $\Sigma^k$ is the set of $k$-letter words on the letters of $\Sigma$. You want to know how many of those there are. Hence @joriki's suggestion. – lhf Apr 11 '11 at 11:58 Thank you. $\Sigma^k$ is the set of k-letter words on the letters of Σ. You want to know how many of those there are. So, therefor the answer is {amount_of_letters}$^5$ ? For example: {0,1} ... 2$^5$ words possible? {a,b,c} ... 3$^5$ words possible? and so on... - Yes. Think of a list $\Box\Box\Box\Box\Box$ and for each $\Box$ you can make $|\Sigma|$ distinct choices. So if you have $5$ boxes, there are $|\Sigma|^5$ possibilities. – t.b. Apr 11 '11 at 12:42 proposal for solution: k's are elements of natural number, including zero. -> $\Sigma$'s cardinality is five. so $\Sigma^5$ = {01234} or {45678} What I didn't understand: No word's parts are given, so how can I answer this question without knowing, what is part of the language. What I can say is, $\Sigma^5$ has five-digit words like {abcde} or {01234}... However, i doubt, that this is the right solution... Best regards, jensen - (I´m working with jensen togehter in a team) – LaPhi Apr 11 '11 at 9:02

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MATLAB Answers # "Operands to the || and && operators must be convertible to logical scalar values" occurring with integer comparisons 3 views (last 30 days) I have written a piece of code that essentially has this structure: function [output] = spectralSVD(S,thr) % Input: % S: [mxn] complex double % thr: real double, value between 0 and 100 E_tot = trace(abs(S)); n_sv_max = min(size(S)); n_sv = 0; E_rec = 0; while E_rec < thr && n_sv <= n_sv_max n_sv = n_sv+1; E_rec = 100*cumsum(diag(abs(S(1:n_sv,1:n_sv))))/E_tot; end At this point, the code crashes: "Operands to the and && operators must be convertible to logical scalar values." So as far as I understood, the && would be appropriate here since both conditionals are single-value comparisons (none of E_rec, thr, n_sv and n_sv_max are vectors or matrices). However, I still get the error message mentioned in the title. I have tried converting all of the aforementioned variables with the uint8() function (even though that does not do exactly what I want), but to no avail. It would appear that I misunderstood the explanation I read in other questions on the same error message. Could anyone explain me where my thought process is going wrong, and how to appropriately fix my code? It would be very much appreciated! ##### 0 CommentsShowHide -1 older comments Sign in to comment. ### Accepted Answer Guillaume on 10 Apr 2017 Edited: Guillaume on 10 Apr 2017 "both conditionals are single-value comparisons (none of E_rec, thr, n_sv and n_sv_max are vectors or matrices)" The first time through the while loop, they are. The second time... not so much. See the output of: S(1:n_sv,1:n_sv) with n_sv = 0 as you've declared. The best way for you to solve this sort of problems is to use the debugger to check what the values of the variables actually are as opposed to what you think they are. I suspect n_sv should start at 1. ##### 2 CommentsShowHide 1 older comment Floris van den Broek on 10 Apr 2017 Thank you very much for your answer, that makes a lot of sense. I cannot believe I didn't think of that! (Since you posted it in a comment I cannot accept your answer as THE answer, so I will just accept the one you replied on) Sign in to comment. ### More Answers (1) Thorsten on 10 Apr 2017 Check before the while whos E_rec thr n_sv n_sv_max I am quite sure that not all variables have Size 1x1 and Class double. ##### 0 CommentsShowHide -1 older comments Sign in to comment. ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by

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#### flatmaster I have a function I want to model. It is periodic, but the period keeps decreasing. Basically, it'll be a periodic function "squished" for larger values of x. The typical fourier series is... y = SUM{aSin(nx)} + SUM{bCos(nx)} I think I will attempt y = SUM{aSin(nx^2)} + SUM{bCos(nx^2)} replacing x -->x^2 should give me the "smushing" that I want. The application is bladder level as a function of beers consumed. The basic function is an increasing (quadratic, exponential) function followed by a linear, steeply sloped drop to zero. #### fresh_42 Mentor 2018 Award I guess you should worry the amplitude as well. Your general function is $f(t) = A(t)\sin(p(t)+p_0)$. Now you can try some functions for $A(t)$ and $p(t)$. I would let WolframAlpha do the graphics until I'm satisfied. #### Dr Transport Gold Member sounds like you need a wavelet transform, not a Fourier transform ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving

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# Physics posted by on . 5) An insulated Thermos contains 190 cm3 of hot coffee at 79.0°C. You put in a 11.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg•K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3. • Physics - , Are you supposed to assume that the ice is intially 0 C? It could be lower. The heat gained by the 11 g of ice, as it melts and rises to final temperature T, is equal to the heat lost by the 190 g of original liquid water, as it cools from 79 C to T. Write that as an equation and solve for T. . • Physics - , but i cant do it!!! • Physics - , Change all the temp. to Kelvin first. Ice is 0C in solid form so there is a latent heat too. 4187(190/1000)(352-T)=333x1000x11/1000 + 4187(11/1000)(T-273) Solve for T? Then (T-273) is in Celsius. Difference = 79 - (T-273) • Physics - , The answer is 79.909 degrees C, but don't know how to get that

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## ››Convert pace [Roman] to attometre pace [Roman] attometer Did you mean to convert pace [great] pace [Roman] to attometer How many pace [Roman] in 1 attometer? The answer is 6.7567567567568E-19. We assume you are converting between pace [Roman] and attometre. You can view more details on each measurement unit: pace [Roman] or attometer The SI base unit for length is the metre. 1 metre is equal to 0.67567567567568 pace [Roman], or 1.0E+18 attometer. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pace [Roman] and attometers. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of pace [Roman] to attometer 1 pace [Roman] to attometer = 1.48E+18 attometer 2 pace [Roman] to attometer = 2.96E+18 attometer 3 pace [Roman] to attometer = 4.44E+18 attometer 4 pace [Roman] to attometer = 5.92E+18 attometer 5 pace [Roman] to attometer = 7.4E+18 attometer 6 pace [Roman] to attometer = 8.88E+18 attometer 7 pace [Roman] to attometer = 1.036E+19 attometer 8 pace [Roman] to attometer = 1.184E+19 attometer 9 pace [Roman] to attometer = 1.332E+19 attometer 10 pace [Roman] to attometer = 1.48E+19 attometer ## ››Want other units? You can do the reverse unit conversion from attometer to pace [Roman], or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Attometre The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18. So 1 attometre = 10-18 metre. ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

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# Moving averages – the fast and the slow While searching for promising forex strategies on the web, for sure you have encountered trading systems that rely on the crossover of a faster and a slower moving average. In this post I explain how one moving average can actually be faster or slower than another. ## Moving averages Moving averages are among the most widely used indicators in foreign exchange trading. A moving average MA(x) is a single value, the average or mean, that is computed from a given number x of historical prices. This version of a moving averages is also often called a simple moving average or SMA because it is based on the simple arithmetic mean of prices. MA(1) is actually no average at all, it is just the current price. MA(2) is the average of the current and the second-but-last price, MA(3) the average of the last three prices, MA(4)… you get it. By recalculating each time a new price gets available (every hour, every minute, every tick – depending on the timeframe you choose) the moving average actually gets moving. ## The influence of x The x in MA(x) gives the amount of historical prices to include into the calculation. The larger x is, the more values are averaged and the smoother a plot of the moving average will be. But a larger x also means that older prices are included in the calculation of the average. This makes the moving average lag behind the current price. The larger x the larger the lag, the slower the moving average. The smaller x the smaller the lag, the faster the moving average. ## How can moving averages be fast or slow? Consider the following illustrating example, where the prices (y-axis) moves from one level to another one over time (x-axis). The original price is plotted by the blue line. It first is constant 1.20, then increases gradually to 1.30. The other curves show different moving averages. Notice that you need as many values as the x in MA(x) to actually start the calculation, this is why the curves start at different points in time. The MA(1) (original price in blue graph) is always up to date, it is actually as fast as it can get. MA(5) only reaches the new higher level some time after the real price did, MA(10) is even slower, etc. Thus, the farther we have to look into the past for the calculation of the moving average, the slower it adapts to changes in the level of price. While MA(1) is the fastest of the graphed moving averages, MA(20) is the slowest. A fast moving average is more sensitive to changes in price than a slow one. ## Some real data The graph below contains the hourly opening prices of EUR/USD for roughly the first half of October 2014. The blue line again represents the original prices. Two moving averages are additionally drawn, one faster than the other. MA(20) can be first calculated after 20 hours. It already lags behind the original prices, especially visible in the sheer drop after about a third of the graph. MA(40) starts after 40 hours. As the slower of the the two moving averages it lags even further behind. The slow or fast property is more often used in conjunction with the stochastics indicator. Read for example Casey Murphy’s article What is the difference between fast and slow stochastics in technical analysis? on Investopedia. The graphs in this post have been created with the great R software environment for statistical computing and graphics. R is free! Check it out on http://www.r-project.org. ## 0 thoughts on “Moving averages – the fast and the slow” 1. Bruceduh Protect your privacy and data by rerouting your IP address to a foreign country using this clean and intuitive VPN service from Kaspersky The whole process will just take a few moments. Mirror Link —> Kaspersky Secure Connection Full Crack Latest – Build: 18.0.0.405 / 19.0.0.1088 – Developer: Kaspersky – Status file: clean (as of last analysis) – File size: na – Price: 0 – Special requirements: no requirements – Rating: Tags: Kaspersky Secure Connection crack fix Kaspersky Secure Connection crack new Kaspersky Secure Connection crack files Kaspersky Secure Connection activate code Kaspersky Secure Connection and crack 2. Bruceduh The whole process will just take a few moments. 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# May Nanay Ka May Apat Na Anak Riddle: Check Logical Explanation for May Nanay Ka May Apat Na Anak Riddle Updated: Aug 31,2020 07:25 GMT Mahathi May Nanay Ka May Apat Na Anak Riddle: May Nanay Ka May Apat Na Anak Riddle is a riddle that is trending on social media including Facebook, Instagram, and WhatsApp family groups. Check & Solve the May Nanay Ka May Apat Na Anak Riddle that is designed to test your thinking and Math Skill. Read the complete article to know the answer to May Nanay Ka May Apat Na Anak Riddle and Challenge your friends and family. Also solving Riddles or Puzzles tests your Brain and by challenging others you can also know their intelligence in solving the riddles. ## Test Your Skills on May Nanay Ka May Apat Na Anak Riddle There are many types of riddles like math riddles, comic riddles, brainteasers, and puzzles. Many riddles can be found on the internet but they are sure to give your brain a workout. Riddles Challenge You to Solve These Hard Riddles that are meant for Everyone. Go on! Try all of the new brain teasers that combine logic and math to test your mental mettle. Find out our new collection of easy riddles and brain teasers. Almost Everyone loves solving brain teasers and challenging riddles right? If you think you’re already a pro at solving tricky riddles, put yourself to the test with these and try out “May Nanay Ka May Apat Na Anak Riddle”(Don't worry, we answer the riddle lastly with explanation) ## What is May Nanay Ka May Apat Na Anak Riddle? Have a look at the question! "You have a mother, they have four children their names are MAY, JUNE, and July, QUESTION !? what is the name of a fourth child? ## What is the answer and Explanation to May Nanay Ka May Apat Na Anak Riddle? The answer to the May Nanay Ka May Apat Na Anak Riddle is "- I am. The reader of the question is the fourth child. The only child said the mother had four children, May June, July. Based on this, you also "have a mother". That is, you are also the fourth child. You can also answer "Depending on the person whom we are asking this question". This is because the names of the respondents to this riddle can also vary." TAGS Trending News ### May Nanay Ka May Apat Na Anak Riddle- FAQ 1. I’m tall when I’m young, and I’m short when I’m old. What am I? candle 2. What has to be broken before you can use it? An egg 3. What question can you never answer yes to? Are you asleep yet?

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Little explorations with HP calculators (no Prime) 03-27-2017, 06:01 PM (This post was last modified: 03-27-2017 07:33 PM by Dieter.) Post: #33 Dieter Senior Member Posts: 2,397 Joined: Dec 2013 RE: Little explorations with the HP calculators (03-27-2017 12:14 PM)pier4r Wrote:  ...so if the side of the inner square is 2r so the sides of the triangle are "s+2r" and "s" from which one can say that "s^2+(s+2r)^2=1" . This plus the knowledge that 4 times the triangles plus the inner square add up to 1 as area. Still, those were not enough for a solution Right, in the end you realize that both formulas are the same. ;-) The second constraint for s and r could be the formula of a circle inscribed in a triangle. This leads to two equations in two variables s and r. Or with d = 2r you'll end up with something like this: (d² + d)/2 + (sqrt((d² + d)/2) + d)² = 1 I did not try an analytic solution, but using a numeric solver returns d = 2r = (√3–1)/2 = 0,36603 and s = 1/2 = 0,5. Edit: finally this seems to be the correct solution. ;-) Dieter « Next Oldest | Next Newest » Messages In This Thread Little explorations with HP calculators (no Prime) - pier4r - 03-15-2017, 11:11 PM RE: Little explorations with the HP calculators - Thomas Okken - 03-15-2017, 11:56 PM RE: Little explorations with the HP calculators - pier4r - 03-16-2017, 12:01 AM RE: Little explorations with the HP calculators - Logan - 03-16-2017, 11:06 AM RE: Little explorations with the HP calculators - ekubaskie - 03-16-2017, 02:07 AM RE: Little explorations with the HP calculators - Csaba Tizedes - 03-16-2017, 08:09 AM RE: Little explorations with the HP calculators - TASP - 03-16-2017, 06:29 PM RE: Little explorations with the HP calculators - pier4r - 03-21-2017, 10:40 PM RE: Little explorations with the HP calculators - Joe Horn - 03-21-2017, 11:41 PM RE: Little explorations with the HP calculators - Dieter - 03-28-2017, 05:46 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 06:17 PM RE: Little explorations with the HP calculators - pier4r - 03-22-2017, 08:37 AM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 12:14 PM RE: Little explorations with the HP calculators - Joe Horn - 03-22-2017, 03:27 PM RE: Little explorations with the HP calculators - pier4r - 03-22-2017, 04:45 PM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 11:11 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 08:07 AM RE: Little explorations with the HP calculators - Simone Cerica - 03-23-2017, 08:58 AM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 11:25 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 08:28 AM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 10:00 AM RE: Little explorations with the HP calculators - Joe Horn - 03-23-2017, 12:19 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 01:23 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 01:45 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 03:23 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 10:04 PM RE: Little explorations with the HP calculators - Dieter - 03-24-2017, 10:54 PM RE: Little explorations with the HP calculators - pier4r - 03-25-2017, 08:17 AM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 12:14 PM RE: Little explorations with the HP calculators - Dieter - 03-27-2017 06:01 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 06:14 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 06:30 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 06:57 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 07:07 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:22 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 07:39 PM RE: Little explorations with the HP calculators - Dieter - 03-27-2017, 08:55 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 10:35 PM RE: Little explorations with HP calculators (no Prime) - RMollov - 05-06-2018, 06:13 AM RE: Little explorations with HP calculators (no Prime) - Gerson W. Barbosa - 05-22-2018, 03:03 AM RE: Little explorations with HP calculators (no Prime) - RMollov - 05-23-2018, 11:04 AM RE: Little explorations with the HP calculators - Thomas Okken - 03-27-2017, 12:54 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 02:27 PM RE: Little explorations with the HP calculators - Thomas Okken - 03-27-2017, 03:44 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 03:55 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 05:29 PM RE: Little explorations with the HP calculators - Joe Horn - 03-27-2017, 07:24 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:35 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:01 PM RE: Little explorations with the HP calculators - Joe Horn - 03-27-2017, 08:08 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 08:14 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:36 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 08:44 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:07 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:58 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 10:12 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 10:44 AM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 11:07 AM RE: Little explorations with the HP calculators - Dieter - 03-28-2017, 05:33 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 06:13 PM RE: Little explorations with the HP calculators - Han - 03-28-2017, 07:32 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 08:15 PM RE: Little explorations with the HP calculators - Dieter - 03-29-2017, 07:54 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 08:48 AM RE: Little explorations with the HP calculators - Dieter - 03-29-2017, 01:07 PM RE: Little explorations with the HP calculators - Han - 03-28-2017, 07:36 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 09:32 PM RE: Little explorations with the HP calculators - Han - 03-29-2017, 01:58 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 06:02 AM RE: Little explorations with the HP calculators - Simone Cerica - 03-29-2017, 09:51 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 11:04 AM RE: Little explorations with the HP calculators - Han - 03-29-2017, 11:31 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 11:49 AM RE: Little explorations with the HP calculators - Han - 03-29-2017, 06:15 PM RE: Little explorations with the HP calculators - Dieter - 03-30-2017, 12:26 PM RE: Little explorations with the HP calculators - Han - 03-30-2017, 04:14 PM RE: Little explorations with the HP calculators - Gerson W. 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pier4r - 12-12-2017, 10:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-17-2017, 10:05 AM RE: Little explorations with HP calculators (no Prime) - Joe Horn - 12-18-2017, 02:26 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-18-2017, 07:08 AM RE: Little explorations with HP calculators (no Prime) - Joe Horn - 12-19-2017, 04:20 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-19-2017, 09:15 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-19-2017, 09:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 02-19-2018, 08:03 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 02-19-2018, 09:25 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 02-19-2018, 09:40 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 02-19-2018, 10:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-10-2018, 11:36 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 03-10-2018, 05:14 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-11-2018, 12:07 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 03-11-2018, 06:25 PM RE: Little explorations with HP calculators (no Prime) - Neve - 03-11-2018, 06:47 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-11-2018, 07:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-12-2018, 09:03 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-16-2018, 07:48 PM RE: Little explorations with HP calculators (no Prime) - Thomas Okken - 03-16-2018, 08:24 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-16-2018, 09:10 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-17-2018, 12:27 PM RE: Little explorations with HP calculators (no Prime) - Csaba Tizedes - 03-17-2018, 01:22 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 03-17-2018, 08:00 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 03-17-2018, 02:37 PM RE: Little explorations with HP calculators (no Prime) - Dieter - 03-16-2018, 08:29 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-17-2018, 08:55 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-18-2018, 12:14 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 03-18-2018, 12:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 08:55 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-18-2018, 09:54 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-18-2018, 12:50 PM RE: Little explorations with Casio 9750/9860 calculators - brickviking - 03-20-2018, 11:59 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-21-2018, 10:45 AM RE: Little explorations with Casio 9750/9860 calculators - brickviking - 03-22-2018, 02:16 AM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-22-2018, 11:32 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-21-2018, 11:23 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-21-2018, 08:37 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-21-2018, 11:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-22-2018, 11:15 AM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-23-2018, 02:02 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-23-2018, 08:20 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 09:29 AM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 01:50 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 02:24 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 05:34 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 06:17 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 08:53 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 09:16 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-25-2018, 05:01 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-25-2018, 07:34 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-25-2018, 08:00 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-25-2018, 08:45 PM RE: Little explorations with HP calculators (no Prime) - Dave Frederickson - 03-25-2018, 09:40 PM RE: Little explorations with HP calculators (no Prime) - Dave Frederickson - 03-25-2018, 09:43 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 01:53 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 04-02-2018, 09:11 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-05-2018, 02:22 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-21-2018, 12:53 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 05-21-2018, 01:04 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 05-21-2018, 06:02 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-26-2018, 09:34 PM RE: Little explorations with HP calculators (no Prime) - ijabbott - 12-26-2018, 11:41 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 03:22 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 04:59 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-31-2018, 07:59 PM RE: Little explorations with HP calculators (no Prime) - brickviking - 12-28-2018, 09:42 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-28-2018, 09:53 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 12-28-2018, 11:21 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-29-2018, 12:10 AM RE: Little explorations with HP calculators (no Prime) - 3298 - 12-29-2018, 02:05 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-27-2018, 06:52 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 08:31 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-27-2018, 08:47 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-27-2018, 10:20 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 11:58 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-28-2018, 01:38 AM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-28-2018, 04:04 AM RE: Little explorations with HP calculators (no Prime) - ijabbott - 12-28-2018, 09:06 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-28-2018, 02:00 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-28-2018, 03:41 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-29-2018, 04:06 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-29-2018, 07:35 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-29-2018, 10:02 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-30-2018, 01:26 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-31-2018, 04:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-31-2018, 08:20 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-31-2018, 08:36 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-13-2019, 10:48 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 01-13-2019, 05:05 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 01-13-2019, 08:27 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 01-13-2019, 01:44 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 01-13-2019, 01:48 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-13-2019, 07:00 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-28-2019, 01:38 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-31-2019, 04:07 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-27-2019, 04:47 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-27-2019, 06:06 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 05-28-2019, 11:09 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-28-2019, 07:18 PM User(s) browsing this thread: 1 Guest(s)

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CC-MAIN-2021-21

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https://financial-dictionary.thefreedictionary.com/countable

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# Count (redirected from countable) Also found in: Dictionary, Thesaurus, Medical, Legal, Acronyms, Encyclopedia, Wikipedia. ## Count On a point & figure chart, an estimation of future price movements. Point & figure charts seek to identify support and resistance levels. Counts are estimates on the likelihood that a security will break through one or the other and result in a large profit or loss. References in periodicals archive ? Since (X, [tau]) is locally countable, then by Theorem 1.3 (b), [[bar.U].sup.[omega]] = [bar.U]. Zdomskyy, "The topological structure of (homogeneous) spaces and groups with countable cs*character," Applied General Topology, vol. A point x [member of] X is called a point of countable type if there exists a compact subset F with a countable base of open neighborhoods {[U.sub.n]: n [member of] N} in X such that x [member of] F. A portion of your unreimbursed medical expenses (what you paid out of pocket after medical insurance pays) may reduce your countable income. It implies that there exists (at most) a countable subset E in [epsilon] such that * S is said to be strongly left reversible, if there is a family [([S.sub.[gamma]]).sub.[gamma][member of][GAMMA]] of countable left reversible sub-semigroups of S satisfying the conditions 1 and 2 of the previous definition. In Polish, bagaz 'luggage' and mebel 'furniture' denote discrete entities, consequently, bagaz SG, bagaze PL and mebel SG, meble PL are countable nouns. As a result [R.sub.[mu]](A) is weakly compact.] Next suppose D [subset or equal to] [bar.[U.sup.w]] with D [subset or equal to] [bar.co]({0} [union] [R.sub.[mu]](D)) and with [bar.[D.sup.w]] = [bar.[C.sup.w]] and C [subset or equal to] D countable. Then since [R.sub.[mu]](D) [subset or equal to] co({0} [union] H{D)) and {0} [union] co({0} [union] H(D)) = co({0} [union] H(D)) we have Because of the differences in (uniform or less uniform) conceptualization we cannot predict whether the noun denoting a particular entity is countable or not. That's because a SPIA owned by and payable to the healthy stay-at-home community spouse will generally not be considered a "countable asset" for purposes of qualifying the institutionalized spouse for Medicaid. If V has the countable basis [{[[upsilon].sub.i]}.sub.i[member of]N] as in [SS13], then there are no non-trivial [S.sub.[infinity]]-invariants in its tensor powers [V.sup.[cross product]k] since every element in these vector spaces is a finite linear combination of the basis elements. Let G = {0,1,2, ...} be a countable state space and {[X.sub.t], t [member of] T} a collection of G-valued random variables defined on the probability space ([OMEGA], F, P). Site: Follow: Share: Open / Close

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# The magnitude small signal voltage gain of the amplifier is 1.  AV = 181V/V 2.  AV = 330V/V 3.  AV = 110V/V 4.  AV = 220V/V 4 AV = 110V/V Explanation : No Explanation available for this question # For the Non investing amplifier shown below, if unloaded gain a = 106, R1= 2kΩ and R2 = 18 kΩ then Vo is 1.  9.8V 2.  9.7 V 3.  10 V 4.  9.999 V 4 9.999 V Explanation : No Explanation available for this question # Which one of the following represents the ideal equivalent circuit of given non-inverting amplifier. 1. 2. 3. 4.  None of these 4 Explanation : No Explanation available for this question # For the common – emitter amplifier circuit shown below, when |jωCμ| 1. 2. 3. 4. 4 Explanation : No Explanation available for this question # In the above problem, the power released by 4v source is 1.   2mv 2.  0.2w 3.  2.6mw 4.  2.4 mw 4 2.4 mw Explanation : No Explanation available for this question # Find VN, VP & Vo in the circuit shown below 1.   -5V 2.  5V 3.  2.5V 4.  -2.5V 4 -5V Explanation : No Explanation available for this question # The Miller capacitance CM in terms of internal base to collectors voltage (Av) of the transistor is 1. 2. 3. 4. 4 Explanation : No Explanation available for this question # In the circuit of figure shown below, let R1 = R3 = R4 = 10KΩ &R2 = 30 KΩ if V1 = 3v, find V2 for V0 = 10V 1.   4.5V 2.  9.5V 3.   19.5V 4.  -9.5V 4 9.5V Explanation : No Explanation available for this question # The below circuit shows a common source amplifier with a saturated PMOS load. Suppose both the transistors are in saturation then small signal voltage gain (V0/Vin) is 1. 2. 3. 4. 4 Explanation : No Explanation available for this question 1. 2. 3. 4. 4

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# 10 Physics Questions Sep 18th, 2015 Anonymous Category: Science Price: \$35 USD Question description PV = 1/3 Nmc2 = 1/3 nMc2 (n = number of moles, N = number of molecules; M = molar mass, m = mass of molecule) Heat, Gas and KT questions PV = nRT = NkT where N = n x NA R = 8.31 J/K mol; k = 1.38 x 10-23 J/K; NA = 6.02 x 1023 1. A reaction produces 0.3 litres (what’s that in m3?) of CO2 (Molar mass 44 g/mol) at 105 Pa and 27 oC. a) How many moles of gas is this? b) What mass of gas is this? c) If this gas were collected at 3 x 105 Pa and -13 oC, what volume would it occupy? 2. 1 kg of steam (molar mass = 0.018 kg / mol) is at 105 Pa and 200 oC. a) How many moles of steam is this? b) So what volume does it occupy? c) What is the kinetic energy of an average water molecule in this steam? d) What is the total K.E. in the steam? The steam is now heated to 300 oC. The volume is kept constant. d) What is the new pressure? e) What is the total K.E. contained in the steam now? f) How much energy has the steam gained? How much energy would it take to raise the temperature of 1 kg of steam by 1 oC? (This is called the ‘specific heat capacity’ of steam (at constant volume).) g) If the steam were allowed to expand as it was heated, so that the pressure, rather than the volume, remained constant, would you expect the amount of energy required to raise the temperature to be more than, less than, or the same as the amount of energy you found above, and why? (Hint: remember why the gas in a bike pump, or the piston of a BB gun, gets hotter on compression? Same idea…) 3. a) Your lungs have a capacity of 4 litres (4 x 10-3 m3). How many molecules do they contain at a pressure of 105 Pa and a temperature of 300 K? b) If, at the top of Everest, the temperature is only 250 K and your 4 litre lungs hold 3.6 x 1022 molecules, what is the pressure there? 4. Abdul the Iranian is refining uranium. The process involves diffusion of uranium hexafluoride, UF6, a gas of RMM 352 if the uranium isotope is U-238, and RMM 349 if it is U-235. A chamber contains UF6 gas at a temperature of 350 K. a) What is the RMS speed of a UF6 molecule if the uranium atom is U-238? b) The isotope separation process relies on the different rates of diffusion of the two different isotopes, which depends on their speeds. What is the ratio of RMS speeds of a U235F6 molecule and a U238F6 5. a) A 1.0 mole sample of hydrogen has a mass of 2.0 x 10-3 kg and occupies 2.26 x 10-2 m3 at 273 K and 1.0 x 105 Pa. What is the rms speed of a hydrogen molecule at this temperature? b) The molar mass of radon is 0.22 kg mol-1. What is the rms speed of a radon atom at 273 K? 6. The escape velocity from the earth is 11 km/s. This means that any gas molecule at the top of the atmosphere whose vertical velocity exceeds 11 km/s will probably be lost from the earth. a) At what temperature is the rms speed of hydrogen molecules equal to 11 km/s? The molar mass of hydrogen is 2.0 x 10-3 kg mol-1. b) In practice hydrogen will gradually escape from the earth even if the temperature at the top of the atmosphere is much less than your answer to part a. Explain why. c) It has been shown that if the rms speed of molecules in a gas is more than about 1/5 of the escape velocity, the gas will escape almost completely in a billion years. If the top of the earth’s atmosphere has been at 1000 K for the past billion years, find the molar mass below which you would not expect to find any gases in Earth’s atmosphere. 7. A further demonstration of diffusion involves releasing bromine molecules (molar mass = 0.16 kg/mol) into an evacuated tube, when the colour moves to the top of the tube very fast indeed, and then repeating it but with air in the tube, when the colour of the bromine spreads much more slowly. This is because of the frequent collisions with air molecules that mean that the path of the bromine molecules becomes a ‘drunkard’s walk’. There is a result that says that the average displacement from your start point after N steps, each of length l in a random direction, is  N l. Thus, if the bromine molecules make n collisions per second, each separated by a ‘mean free path’ of length l, their average speed is nl, while their average velocity will be a) Find the average kinetic energy, and hence rms (root mean square) speed of bromine molecules at room temp. (20 oC.) b) Bromine in air is observed to diffuse at an average rate of 6 x 10-3 m/s. Use this information, along with the information already given / calculated, to find the mean free path in air, l, and the number of collisions School: UCLA Studypool has helped 1,244,100 students Review from student Anonymous " Outstanding Job!!!! " 1830 tutors are online Brown University 1271 Tutors California Institute of Technology 2131 Tutors Carnegie Mellon University 982 Tutors Columbia University 1256 Tutors Dartmouth University 2113 Tutors Emory University 2279 Tutors Harvard University 599 Tutors Massachusetts Institute of Technology 2319 Tutors New York University 1645 Tutors Notre Dam University 1911 Tutors Oklahoma University 2122 Tutors Pennsylvania State University 932 Tutors Princeton University 1211 Tutors Stanford University 983 Tutors University of California 1282 Tutors Oxford University 123 Tutors Yale University 2325 Tutors

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## Goal for this step: Completely solve the yellow side! Okay, so you are really close to solving the entire cube! Good news is that all the hard stuff is over with. If you made it this far, the rest will be a breeze. ## Patreon Support me on Patreon and you get my cube cheatsheet for free! You also get access to all my Rubik’s Cube tutorial videos (with no commercials!) https://www.patreon.com/EasiestSolve ## Tip: When I do this move, I say the words “clock” out loud to mean “clockwise” and I say the word “counter” out loud to mean “counterclockwise. ## Memorize This! #### This next move you get to memorize more stuff! Yeah, fun! • Clock, Clock, Counter • Clock, Clock, FLIP, Counter. Just seven moves. Seven words to memorize. Can you memorize them? Notice that 3 words are almost the same as the last three words, except for the word “FLIP” in there. That might help you remember it. Close your eyes and try to repeat: “Clock, Clock, Counter” “Clock, Clock, FLIP, Counter”. When you have it memorized really well, then proceed. Don’t proceed unless you can say the seven words out loud with your eyes closed while spinning in a circle. (Having trouble memorizing all these steps for the cube? Check out my cheat sheet.) ## Alternate: Right, Top, Right, Top, Right… (forever!) Okay, you have the steps memorized. Now what do we do with it? We are going to focus on the right of the cube and the top of the cube. Put your hand on the right. Now put your hand on the top. Now put your hand on the right. Now put your hand on the right. Are you feeling a pattern emerging here? That is all you have to worry about. Our main moves will alternate from right to top to right–over and over again. Just remember that we start with the right side. Can you remember that? ## Let’s Begin! Okay, let’s put what we’ve learned together. you are going to start on the right side of the cube. Your pattern is right, top, right, top. It goes forever, remember? Now we apply our Clock, Clock, Counter, Clock, Clock, FLIP, counter. Note: If it helps, don’t look at the cube when making the moves. Say out loud, “Clock, Clock, Countner, Clock, Clock, Flip, Counter” ## Possible Outcomes The below shows the possible patterns you will see after you do your seven main moves. When you see a pattern, rotate the top of your cube so that it looks like the images below. After it is adjusted correctly, then do the main moves again. You keep doing the main moves until you end up with the yellow side completely covered. Then you’re done! As explained before, you must turn the top of the cube to match the position of the patterns on “possible outcomes” above. Here is an example: ## Unnecessary Information: NOTE: Some people will loop through all possible outcomes forever and never arrive at the all-yellow goal. This endless loop is called “The Infinite Yellow”. THOR:  “If you have been cursed with The Infinite Yellow, it means that you are not worthy.” ## Finished? If the top of your cube is completely yellow, then you’re done with this step! Or course, the two bottom row of colors on the sides should still be intact. It was a lot of work building this page. If you want to say “thank you”, consider donating here! It’s much appreciated! ### Get A Cube. Some have wondered what cube I use. Happy to help. I’m use the Classic Rubik’s Cube https://amzn.to/3xiB64y&nbsp;  …If that is too old-school, here is the AP-enabled cube, which is pretty cool looking. https://amzn.to/3GYXy7d Here we are again. Are you up to the challenge? Can you scramble your cube and solve to this point in under five minutes? “I can solve in about five minutes and want to go to the next step.” ## 9 thoughts on “Step 6 – The Yellow Fish” You illustrate 4 possible outcomes of the R U R’ U R U2 R’ move There’s a 5th, I”ll call the two-headed fish. I haven’t determined the best way to orientate. Also… If you have the fish and he doesn’t have the “food” on the right, see if he has food on the left if he’s facing right… if so, do the mirror image of the moves: L’ U’ L U’ L’ U2 L 2. Raquel says: I get to this point 16 times over and it keeps taking me all through the possible outcomes one by one over and over, never fully solving the yellow. Why ??? 1. Raquel, happy to help. When u get an outcome, make sure you view the outcomes at the bottom of the page. Orient your cube to match the orientation shown in the outcomes on the page. For instance, if u get the fish, make sure he is pointing bottom left. Then do the moves again. Hope that helps! 2. Gautham says: The only site where the explanation is quite easy to grasp. I gave up on solving the cube a while ago. After stumbling across this site, i got pumped again to give it a try. After several tries, i was able to solve the cube. I really appreciate your info on how to solve the cube. Much thanks to you sir. There’s this one little variation which others have mentioned here, where we we end up cycling all the possible outcomes. No matter how many times we perform the moves. Can you please look into it? I have taken every step you’ve mentioned. Despite that i was unsuccessful in getting it right in this step. Hope you are able to get my doubt. 3. lizard says: Exact same thing happened to me until I realized that after looking at the “possible outcomes” and adjusting the top layer as described, I was then erroneously starting the next series of steps with the “front” face instead of the “right face”, which is NOT adjacent to the left side shown in the possible outcomes diagram. I.e., the face adjacent to the “left face” as shown in the possible outcomes is the “front face”, not the “right” face (where the next sequence of steps needs to start). 3. Splidgy says: But what if you get none of the possible outcomes? 4. Mia says: I love this website. It has been so helpful 5. Tony says: Why not just move the cube in position in step 6 instead of moving the top (upper)? Wouldn’t it be easier and more logical? 1. Tony, You are welcome to do it that way. That works too. I find that moving the top by flicking your finger is quicker than rotating the position of the entire cube. I think as you continue to practice, you’ll find moving the top to be the faster option. But, totally up to you.

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0 # What numbers can go into 28 evenly? Wiki User 2016-02-10 22:37:32 This sort of number is a factor, which means 28 can be divided by the number and the quotient will have no remainder. Factors of 28 are 28, 14, 7, 4, 2, and 1. Wiki User 2016-02-10 22:37:32 🙏 0 🤨 0 😮 0 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.71 364 Reviews Jarielle Mitchell Lvl 2 2020-12-03 19:34:27 The two numbers that go into 28 evenly is 214 you are multiplying 14 2 times 14 and 14 therefore 214 go into 28. Anonymous Lvl 1 2020-04-07 18:51:39 6 Anonymous Lvl 1 2020-04-06 05:41:37 3 Anonymous Lvl 1 2020-04-06 05:43:03 14 Anonymous Lvl 1 2020-04-06 05:43:35 3 Anonymous Lvl 1 2020-04-22 15:06:20 18 Anonymous Lvl 1 2020-05-12 15:56:29 14 Anonymous Lvl 1 2020-10-08 00:13:16 6

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https://kilonova.ro/problems/2428?list_id=991

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# From Constanța 2 Timișoara Time limit: 0.6s Memory limit: 128MB Input: Output: Adi of Adi and Sorin would like to visit their country this summer. They live in a country that can be represented as a graph with $N$ nodes and $M$ edges, where each city is a node and each bidirectional road between two cities is a weighted edge in the same graph (where the weight of the edge represents the length of the road in kilometers). Since they really like the cities Constanța and Timișoara, they decided to visit both of these cities in their vacantion. More formaly, we will note the city of Constanța as node $X$ and the city of Timișoara as node $Y$. Now, if they want to start their journey in city $u$ and end it in city $v$, then their path will have the following properties: • the length of the path is as small as possible (where the length is equal to the sum of weights of the edges in the path); • the path visits both $X$ and $Y$. Let $d(u, v)$ be the length of the path that Adi of Adi and Sorin will visit if they start their journey in node $u$ and end it in node $v$. Note that a node or an edge can occur multiple times in the path. Your task is to calculate $\sum_{ 1 \leq u < v \leq N} d (u, v)$. Since this number can be very large, calculate it modulo $10^{9} + 9$. ## Input data The first line of the input contains the numbers $N$, $X$, $Y$ and $M$. The next $M$ lines will each contain three numbers $u$, $v$ and $c$, representing a weighted edge between $u$ and $v$ of weight $c$. ## Output data Output a single number, the sum of $d(u, v)$ for all ordered pairs of nodes modulo $10^{9} + 9$. ## Constraints and clarifications • $1 \leq N , M \leq 250 \ 000$ • $1 \leq X ,Y \leq N$ • $1 \leq u, v \leq N$ • $1 \leq c \leq 10^{9}$ • There will be no self-loops or multiple edges in the input. # Points Constraints 1 12 $1 \leq N \leq 100$, $1 \leq M \leq 200$ 2 29 $1 \leq N \leq 2 \ 000$, $1 \leq M \leq 4 \ 000$ ## Example stdin 3 2 1 3 1 2 1 3 1 1 3 2 3 stdout 6 ### Explanation $d(1, 2) + d(1, 3) + d(2, 3) = 1 + 3 + 2 = 6$

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https://vidque.com/what-is-the-rule-of-product-rule/

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# What is the rule of product rule? ## What is the rule of product rule? In combinatorics, the rule of product or multiplication principle is a basic counting principle (a.k.a. the fundamental principle of counting). Stated simply, it is the intuitive idea that if there are a ways of doing something and b ways of doing another thing, then there are a · b ways of performing both actions. What is product rule example? We can apply the product rule to find the differentiation of the function of the form u(x)v(x). For example, for a function f(x) = x2 sin x, we can find the derivative as, f'(x) = sin x·2x + x2·cos x. ### What is the product rule in trigonometry? ddx(f(x)⋅g(x))=f′(x)⋅g(x)+f(x)⋅g′(x). What is the product rule in math kids? The product rule of differentiation is a rule for differentiating problems where one function is multiplied by another function. According to this rule, first function times the derivative of second function is added to second function times the derivative of first function. ## Why do we use the product rule? The product rule is used in calculus to help you calculate the derivative of products of functions. The formula for the product rule is written for the product of two functions, but it can be generalized to the product of three or even more functions. Why do we use product rule? ### What are the product identities? The product-to-sum identities are used to rewrite the product between sines and/or cosines into a sum or difference. These identities are derived by adding or subtracting the sum and difference formulas for sine and cosine that were covered in an earlier section. How do you solve the product rule step by step? 1. Step 1: Simplify the expression. 2. Step 2: Apply the product rule. 3. Step 3: Take the derivative of each part. 4. Step 4: Substitute the derivatives into the product rule & simplify. 5. Step 1: Apply the product rule. 6. Step 2: Take the derivative of each part. 7. Step 3: Substitute the derivatives & simplify. 8. Step 1: Simplify first. ## What is the meaning of product rule? The product rule is a formal rule for differentiating problems where one function is multiplied by another. The rule follows from the limit definition of derivative and is given by. . Remember the rule in the following way. Each time, differentiate a different function in the product and add the two terms together. Where do you apply product rule? When To Use The Product Rule? We use the product rule when we need to find the derivative of the product of two functions – the first function times the derivative of the second, plus the second function times the derivative of the first. ### What does product mean in math? The term “product” refers to the result of one or more multiplications. For example, the mathematical statement would be read ” times equals ,” where is called the multiplier, the multiplicand and is their product. How do you remember the product rule? The product rule is used to find the derivative of any function that is the product of two other functions. The quickest way to remember it is by thinking of the general pattern it follows: “write the product out twice, prime on 1st, prime on 2nd”. ## What is product to sum formula? What Are Product To Sum Formulas? The product to sum formulas in trigonometry are formulas that are used to convert the product of trigonometric functions into the sum of trigonometric functions. There are 4 important product to sum formulas. sin A cos B = (1/2) [ sin (A + B) + sin (A – B) ] How do you derive product identities? ### How do you find the sum of a product rule? The product-to-sum formulas are as follows: 1. cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] 2. sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] How do you represent the product rule? There are a few different ways that the product rule can be represented. Below is one of them. Given the product of two functions, f (x)g (x), the derivative of the product of those two functions can be denoted as (f (x)·g (x))’. ## What is the product rule for derivatives? Proving the product rule for derivatives. The product rule tells us how to find the derivative of the product of two functions: The AP Calculus course doesn’t require knowing the proof of this rule, but we believe that as long as a proof is accessible, there’s always something to learn from it. How do you find the product rule for exponents? d(uv)/dx = u(dv/dx)+ v(du/dx) where u and v are two functions Product Rule for Exponent: If m and n are the natural numbers, then xn× xm = xn+m. Product rule cannot be used to solve expression of exponent having a different base like 23* 54and expressions like (xn)m. ### What is the difference between product rule and quotient rule? Product rule states that when two functions f(x) and g(x) are differentiable, then their product is also differentiable and is calculated using the formula, (fg)'(x) = f(x) g'(x) + f'(x) g(x) Quotient rule state that when two functions f(x) and g(x) are differentiable, then their quotient is also differentiable and is calculated using the formula,

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https://community.khronos.org/t/webgl-3rd-person-camera/108057

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# Webgl 3rd person camera I am playing around in webgl and have build a simple gltf loader, that allows to move the main object around. It also allows to rotate the camera (orbit) the main object, but only on yaw. I can orbit the object, but it only allows to view the front side. Now I did fix that using rotation and multiplication, but I want to use the lookAt function, to make it more simple, and also to have pitch orbit (up-down). My camera class has the transform matrix and the projection matrix. I then make a view matrix and multiply it with the projection matrix, then save it in the transform matrix. But this only allows for yaw rotation. If I add pitch, then the camera does move up and down, but it doesn’t rotate to keep the object in center. I am thinking this could be done with the lookAt function, but how would I get the camera origin position, since it requires a vector, but I only have the main transform matrix. I do have transformation, rotation and scale vectors, but since I am multiplying matrixes, the remain 0. Here is my code: ``````const c = this; let camX = Math.sin(c.yaw) * c.distance; //let camY = Math.cos(c.pitch) * c.distance; let camZ = Math.cos(c.yaw) * c.distance; var viewMatrix = mat4.create(); mat4.invert(viewMatrix, c.matrix); mat4.multiply(c.matrix, c.projection, viewMatrix); mat4.fromTranslation(c.matrix, vec3.fromValues(camX, 0, camZ)); const vec = vec3.create(); var moveMatrix = mat4.create(); mat4.fromTranslation(moveMatrix, vec); mat4.multiply(c.matrix, c.matrix, moveMatrix); var rotMatrix = mat4.create(); mat4.fromYRotation(rotMatrix, c.yaw); mat4.multiply(c.matrix, c.matrix, rotMatrix); `````` Matrix is the transform matrix. For a better structure in the future I would suggest keeping more information for each object/reference you have, rather than sticking to the matrix itself. That should only be a result of all the operations you perform in order to send it to the GPU/Shader/Stack. With this in mind, a good example is to keep reference of: • Position vector • Direction or forward vector • Up vector • Right or left (depending or your referential) vector and those last 3 vectors will store the scaling of the object itself. using those, its way easier to make adjustements and transformatinos to your object and to prepare a matrix for rendering. Now, to look at something (or point to/align with) you need to have at least 2 vectors (the 3rd one can be obtained by the cross product of the former 2). Assuming you want to keep having an up vector aligned with the Y vector(the vertical one), you now need to create a vector from the center of your camera to the point you wish to point to, and then invert it. Continuing, we have the vector direction, and an up vector which is unit vector Y, but this in not ready to be sent to the matrix because those 2 are not perpendicular, otherwise we would have distortion on the camera. but, we can get a point from the up vector (which is [0,1,0]) cross it with ouer forward vector to get a perpendicular and the cross it again, which gives us the following: UPv = DIRv.cross(0,1,0).cross(DIRv); leaving us with the new vector we need, and now we just need to acquire the right vector just by crossing those 2, as following: RIGHTv = UPv.cross(DIRv).normalize(); the full data: DIRv = target.sub(camera.position).normalize().invert(); UPv = DIRv.cross(0,1,0).cross(DIRv).normalize(); RIGHTv = UPv.cross(DIRv).normalize(); now you can build your new matrix in order to point to it.

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https://forum.bodybuilding.com/showthread.php?t=144626271&page=1

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Need Help? Customer Support 1-866-236-8417 Thread: 5/3/1/ Boring But Big 3 Month Challenge - Spreadsheet 1. 5/3/1/ Boring But Big 3 Month Challenge - Spreadsheet http://www.t-nation.com/free_online_...onth_challenge So at the beginning of the program I decided to make a spread sheet to track for this program on a monthly basis. * Everything is laid out for each of the big lifts and the recommended assistance lifts. * I didn't bother to track progress on secondary assistance lifts as they are not the main focus of the program. * Each month is on its own worksheet and you can print out each 4 week cycle on one piece of paper. * Any issues let me know https://www.box.com/s/8cf78ccbdbdfff1770c0 2. One think I want to clarify about the 3 month challenge: When doing 50%, 60%, and 70% of training max, are these percentages based on the training max of the 1st cycle only? or the corresponding cycles? For example: ------------------------------------------------------ Based on 1st cycle of 3 month challenge only: ------------------------------------------------------ Cycle 1: Bench training max = 100 lbs assistance bench (50%) = 5x10 @50 lbs (50% of cycle 1's training max) Cycle 2: Bench training max = 105 lbs assistance bench (60%) = 5x10 @60 lbs (60% of cycle 1's training max) Cycle 3: Bench training max = 110 lbs assistance bench (70%) = 5x10 @70 lbs (70% of cycle 1's training max) ------------------------------------------------------ Based on corresponding cycles: ------------------------------------------------------ Cycle 1: Bench training max = 100 lbs assistance bench (50%) = 5x10 @50 lbs (50% of cycle 1's training max) Cycle 2: Bench training max = 105 lbs assistance bench (60%) = 5x10 @63 lbs (60% of cycle 2's training max) Cycle 3: Bench training max = 110 lbs assistance bench (70%) = 5x10 @77 lbs (70% of cycle 3's training max) 3. So, in other words: On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 1's training max. On cycle 3, we do 70% of cycle 1's training max? or is it like this: On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 2's training max. On cycle 3, we do 70% of cycle 3's training max. Also, what do we do after the 3 month challenge? do we start again at 50%? 4. Originally Posted by kidkurious So, in other words: On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 1's training max. On cycle 3, we do 70% of cycle 1's training max? or is it like this: On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 2's training max. On cycle 3, we do 70% of cycle 3's training max. Also, what do we do after the 3 month challenge? do we start again at 50%? 50% cycle 1 60% cycle 2 TM 70% cycle 3 TM Each cycle your BBB increases by like 10 pounds already, so 60% of cycle 1 would still be like doing 50% for cycle 2. After 3 months do whatever you want. Change the assistance, start all over, keep it at 70%, or if you are really ambitious increase it to 75%. That could be a touch crazy though. I have just been running my BBB at 65% since I started before seeing the 3 month challenge, so it just sort of comes down to your work capacity. 5. Originally Posted by kidkurious So, in other words: On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 1's training max. On cycle 3, we do 70% of cycle 1's training max? It's supposed to be exactly like this 6. Originally Posted by theproctologist It's supposed to be exactly like this That makes no sense. BBB is tied to your current cycle's training max, not your previous cycle. You COULD use the previous month's cycle, but the weight increase won't be as large. But everything I have read about BBB online or the book you base the % off your current cycle's TM. 7. Originally Posted by kidkurious On cycle 1, we do 50% of training max. On cycle 2, we do 60% of cycle 2's training max. On cycle 3, we do 70% of cycle 3's training max. Also, what do we do after the 3 month challenge? do we start again at 50%? According to Jim Wendler in the T-Nation Live Spill you increase the percentage to your current training max. Jim Wendler: Jmatta - yes, the 5x10 increases as part of your TM. •The first month of the program, perform the sets with 50% of your training max. •The second month of the program, perform the sets with 60% of your training max. •The third month of the program, perform the sets with 70% of your training max. With with this program I don't see continuing after 3 months. I could possibly see a person burning out with the amount of volume and increase in percentages and weight. I'm not Jim Wendler so I can't make a concrete recommendation to his program. This is just my suggestion to continuing after 3 months. 9. Originally Posted by allergic2rice I know one exists for the regular boring but big however, I just simplified it and input the formulas for the increase in percentage for the assistance lifts for the 3 month challenge. 10. Originally Posted by djartek I know one exists for the regular boring but big however, I just simplified it and input the formulas for the increase in percentage for the assistance lifts for the 3 month challenge. lol sorry, didn't even read your post. just saw 5/3/1 title. now that I've read the tnation article....what in the hell is this sh*t? 11. Originally Posted by allergic2rice lol sorry, didn't even read your post. just saw 5/3/1 title. now that I've read the tnation article....what in the hell is this sh*t? In Jim Wendler's Words "The problem with many hypertrophy-based programs is that they leave out the strength component. You might get bigger as a result of the program, but if you don't get any stronger you're still a chump in my book. That's right, I don't care how big you are, if you aren't strong you're a sham. Having big muscles and no strength is the training equivalent of wearing a strap-on. All show and no go. End of story. In a perfect world, 99% of the music being made wouldn't make you question humanity and 100% of the training programs would get you big and strong. That ain't the case, though. Fortunately, there is a solution, and it's not performing multiple sets of whatever cable Kegel exercise is being pushed as "The Answer." Just a little hard, smart, basic work. It's boring, I agree. Do you want to be entertained or get big and strong?" 12. Originally Posted by djartek In Jim Wendler's Words "The problem with many hypertrophy-based programs is that they leave out the strength component. You might get bigger as a result of the program, but if you don't get any stronger you're still a chump in my book. That's right, I don't care how big you are, if you aren't strong you're a sham. Having big muscles and no strength is the training equivalent of wearing a strap-on. All show and no go. End of story. In a perfect world, 99% of the music being made wouldn't make you question humanity and 100% of the training programs would get you big and strong. That ain't the case, though. Fortunately, there is a solution, and it's not performing multiple sets of whatever cable Kegel exercise is being pushed as "The Answer." Just a little hard, smart, basic work. It's boring, I agree. Do you want to be entertained or get big and strong?" why is there a huge article on this though? BBB is pretty straightforward and I could have told you how to do it in 2 sentences. it's asssistance work 13. Originally Posted by allergic2rice why is there a huge article on this though? BBB is pretty straightforward and I could have told you how to do it in 2 sentences. it's asssistance work The article is regarding the BBB 3 month challenge - which is why there is confusion above. It's a variation to BBB and would probably burn people out if they followed it more than 3 months. 14. Originally Posted by Cleveland33 The article is regarding the BBB 3 month challenge - which is why there is confusion above. It's a variation to BBB and would probably burn people out if they followed it more than 3 months. x 2 15. sorry to bump an old thread, do you have to deload the assistance work in week 4? 16. EDIT - Answered my own question Does anyone have any logs from the 3 months? Also, on the two upper body days, what assistance did you guys do after 5x10 BP/OHP? I'm guessing it's not even going to matter at that point given the volume of A1 and I can alternate something like BB curls/skulls. Thanks 17. spreadsheet is sick. Thanks 18. Originally Posted by BigJon55 spreadsheet is sick. Thanks no problem Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts

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# Returning the range of a number as a palindrome I like to do programming challenges in my free time to teach myself and keep fresh, this one I can't seem to optimize though and it's driving me insane. The goal is essentially, "Given a number, return the numbers leading up to it as a palindrome." For example if you are given the input 5 the answer your program should return is: 123454321 The catch? You have to solve it in as few characters as possible. The minimum needed seems to be 41 characters and I can't condense it past 76. I have tried all sorts of solutions, here are a few: # Note: "Palindromic_Number" is required. "def P(N):" will not work. def Palindromic_Number(N): a = '' for x in range(1,N+1): a += str(x) return a + a[::-1][1:] def Palindromic_Number(N): a = ''.join(map(str, range(1,N+1))) return a + a[::-1][1:] def Palindromic_Number(N): a = ''.join([str(x) for x in range(1, N+1)]) return a + a[::-1][1:] # This one uses Python 2.7 Palindromic_Number = lambda N: ''.join(map(str, range(1,N+1) + range(1,N+1)[::-1][1:])) Does anyone have any ideas as to how I could shorten my code? I'm just looking for a point in the right direction where to look as this point. • range(1,N)+range(N,0,-1) should be shorter than using a+a[::-1][1:] afterwards as it allows you to not define a at all. Oct 6 '15 at 21:44 • Forget it, it won't be so usefull. You can't add ranges in python 3. Only in python 2 where the builtin returns a list. Oct 7 '15 at 5:23 • What's the input range? 1 to 9? 0 to 9? Oct 7 '15 at 11:33 You could try: def Palindromic_Number(N): return int('1'*N)**2 just preforms 11111 * 11111 which equals 123454321 47 chars. Or, as a lambda: Palindromic_Number=lambda N:int('1'*N)**2 41 chars. • 41 as a lambda, the apparent minimum, so I dare say you've nailed it :) (assuming input from 1-9) Oct 7 '15 at 15:17 • @Sp3000 thank you! added to answer. Oct 7 '15 at 15:28 In the last solution (that takes advantage of Python 2 style ranges) you could reduce this range(1,N+1) + range(1,N+1)[::-1][1:] in a couple of ways: 1. Instead of slicing the second range, create the reversed range directly as range(N-1,0,-1) 2. Making N come out of the second range allows changing N+1 and N-1 to N and N: range(1,N) + range(N,0,-1) • Slicing with [-2::-1] is shorter than [::-1][1:] by two bytes/chars. You can also remove the spaces on either side of the plus sign. Oct 7 '15 at 15:55 # Python 2, 40 bytes Palindromic_Number=lambda n:(10**n/9)**2 Try it online! (Includes test code to actually run the program.) I know that you claimed that 41 is the minimum, but it's actually possible to do it in 40. This uses the same strategy seen in the other answers, of squaring a repunit (a number made out of repetitions of the digit 1). However, instead of using string operations to produce the repunit (and thus having to pay for int to convert string to integer), it generates the palindrome entirely arithmetically. The basic idea here is that 10n is a 1 followed by n 0s; if we subtract 1 we therefore get n 9s, and dividing by 9 gives us n 1s. Because Python 2 rounds integer arithmetic downwards, we can omit the subtraction of 1 (as it'll leave the answer higher by ⅑ than it should, which is a fraction that will disappear in the rounding), and we'll still get the correct answer. range is a rather long name; with lambda N,r=range:/def ...(N,r=range): you can shave off a few characters if you also remove all unnecessary whitespace. • Since this question migrated from Code Review I'm not sure whether it'll ping, but in any case aliasing range (5 chars) doesn't save any bytes when it's only used twice (compare range;range vs r=range;r;r) Oct 7 '15 at 13:22 • It pings. I can't delete the answer, but I'm removing that part at least. Oct 7 '15 at 14:12

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1 GATE CE 1998 Subjective +5 -0 The cross-section of a pretensioned pre-stressed concrete beam is shown in Figure. The reinforcement is placed concentrically. If the stress in steel at transfer is $$1000$$ $$MPa,$$ compute the stress in steel immediately after transfer. The modular ratio is $$6.$$ 2 GATE CE 1998 Subjective +5 -0 The cross-section of a simply supported plate girder is shown in figure. The loading on the girder is symmetrical. The bearing stiffeners at supports are the sole means of providing restraint against torsion. Design the bearing stiffeners at supports, with minimum moment of inertia about the center line of web plate only as the sole design criterion. The flat sections available are: $$250 \times 25,\,\,250 \times 32,\,\,\,200 \times 28$$ and $$200 \times 32\,\,mm.$$ Draw a sketch 3 GATE CE 1998 +2 -0.6 A cantilever beam is shown in the figure. The moment to be applied at free end for zero vertical deflection at the point is A $$9$$ $$kN$$-$$m$$ clockwise B $$9$$ $$kN$$-$$m$$ anti-clockwise C $$12$$ $$kN$$-$$m$$ clockwise D $$12$$ $$kN$$-$$m$$ anti-clockwise 4 GATE CE 1998 +2 -0.6 A three hinged arch shown in Figure is quarter of a circle. If the vertical and horizontal components of reaction at $$A$$ are equal, the value of $$\theta$$ is A $${60^ \circ }$$ B $${45^ \circ }$$ C $${30^ \circ }$$ D None in ($${0^ \circ },$$ $${90^ \circ }$$) GATE CE Papers 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 EXAM MAP Joint Entrance Examination

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https://physics.stackexchange.com/questions/317600/how-many-degrees-of-freedom-does-a-spring-have

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# How many degrees of freedom does a spring have? I'm currently learning about thermodynamics and heat capacities. We were told that the theoretical molar heat capacities of all solids should be $3R$. I was told this is because there are 6 different vibrational degrees of freedom for each atom in a solid. Since theses atoms are stuck in place, they can't rotate and translate so I tried to figure out exactly what these 6 degrees of freedom look like. Here are my thoughts: Firstly, I like to think of degrees of freedom as places where energy can go. For example, in a gas, energy can be transferred into a particle as kinetic energy. This kinetic energy can make it translate on each axis and rotate on each axis (assuming it's nonlinear polyatomic). Of course, each particle also has a potential energy associated with it (another place energy can go), but we can ignore it because the average potential energy of the system won't change (for fixed volume). Now let's take a hypothetical solid that is only 1 atom thick. Its lattice will look like this: My teacher would say this has 4 degrees of freedom because each atom in the solid can vibrate up/down or left/right. But using my definition of degrees of freedom, I would say there should only be 2 degrees of freedom. The potential energy associated with the x direction and the potential energy associated with the y direction. The reason I'm discounting kinetic energy is because when $E_{kinetic}=0$: $$E_{mechanical} = E_{potential}$$ Therefore, we can define the amount of energy added simply by what it does to the maximum potential energy. This is why I think a normal spring only has one degree of freedom (its potential energy). So why does the system depicted above have 4 DOF's? • Each bead has four degrees of freedom, because there are two springs for each bead (that is, there are twice as many springs as beads) and the potential of each spring contributes a degree of freedom, and then the beads have two translational degrees of freedom because they are in two dimensions. Therefore, for each bead there is, the system has four degrees of freedom. Can you do a better job of explaining why you don't understand this explanation? Commented Mar 9, 2017 at 20:35 • @NowIGetToLearnWhatAHeadIs Works for me, as an answer as well as a comment? – user146020 Commented Mar 9, 2017 at 20:41 • @Countto10 I think there must have been some other point of confusion. If my comment is really all Nova needed to see, then this is really a duplicate of physics.stackexchange.com/q/80711/23785 Commented Mar 9, 2017 at 20:47 • Each bead has a location and velocity, and both are two-dimensional. The location gives you the potential energy, while the velocity the kinetic energy. How could there be anything but 4 degrees of freedom per bead? Commented Mar 9, 2017 at 22:22 • @hyportnex Because mechanical energy is conserved for a spring so I can just treat an input of energy into a system as an input of energy into the mechanical energy present in the spring. – Nova Commented Mar 10, 2017 at 0:14 A 1D spring has the freedom to move in only one direction (i.e. one mechanical degree of freedom using the engineering definition), but has a two dimensional phase space $(x,p_x)$ (i.e. two degrees of freedom in a common but to my ear sloppy usage), and the Hamiltonian is quadratic in both parameters $$H = \frac{1}{2}k x^2 + \frac{p_x^2}{2m} \,,$$ so it has two contributions for the purposes of equipartition. In a model solid like the one you exhibit each atom (in a $D$ dimensional space) can be naively associated with $D$ mechanical degrees of freedom, $2D$ parameters in phase space, and $2D$ quadratic modes in the Hamiltonian.

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https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=prev&oldid=104540

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# Difference between revisions of "2019 AIME I Problems/Problem 14" ## Problem 14 Find the least odd prime factor of $2019^8+1$. ## Solution 1 The problem tells us that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we get $2019^{16} \equiv 1 \pmod{p}$. This tells us that $\phi(p)$ is a multiple of 16. Since we know $p$ is prime, $\phi(p) = p(1 - \frac{1}{p})$ or $p - 1$. Therefore, $p$ must be $1 \pmod{16}$. The two smallest primes that are $1 \pmod{16}$ are $17$ and $97$. $2019^8 \not\equiv -1 \pmod{17}$, but $2019^8 \equiv -1 \pmod{97}$, so our answer is $\boxed{097}$. ### Note to solution 1 $\phi(p)$ is called the "Euler Function" of integer $p$. Eular theorem: define $\phi(p)$ as the number of positive integers less than $n$ but relatively prime to $n$, then we have $$\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})$$ where $p_1,p_2,...,p_n$ are the prime factors of $p$. Then, we have $$a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)$$ if $(a,p)=1$. ## Video Solution On The Spot STEM:

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https://gis.stackexchange.com/questions/225258/assigning-row-wise-value-to-rotated-grid-of-points-using-arcgis-desktop?noredirect=1

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# Assigning row-wise value to rotated grid of points using ArcGIS Desktop? I am working with a climate model that operates by means of a rotated grid of points. The rotation angle is about 15 degrees counterclockwise. The points are provided with (N, E) coordinates, but their IDs are nonconsecutive, and have irregular intervals. In the image below, the bottom left point has OBJECTID=35, the one to its right has OBJECTID=40, the third to its right has OBJECTID=50, but then the leftmost point in the row above has OBJECTID=51. My problem: I want to assign each point a new ID (integer) that starts from `0` (bottom left point) and increases towards the right hand side of the row, and row by row until it reaches the upper right point (highest ID). The rotation causes problems because none of the coordinates are steady, but is there any way I could do this? The grid has 400 points in the blue arrow direction, and 300 in the green arrow direction. ArcGIS version: 10.3.1 • This seems like a simple enough task, using basic trigonometry before applying some rounding. What have you attempted so far? Coding questions are expected to include code. Please also edit the question to specify the version of ArcGIS in use. – Vince Jan 20 '17 at 15:21 • I cannot see this being done without code but for that we need you to include a coding attempt. Otherwise it looks like you are wanting GIS SE to be a free coding service. – PolyGeo Jan 20 '17 at 20:43 • @PolyGeo This is solvable using standard tools, no scripting or trigonometry calculations needed. Voted to reopen – FelixIP Jan 21 '17 at 1:53 • @FelixIP .OK - I've removed the Python aside so that its suitable for re-opening. – PolyGeo Jan 21 '17 at 4:09 As promised solution that requires knowledge of Pythagorean theorem only. • Connect points 98 and 42 and call layer X_AXIS. • Connect points 98 and 115 and call layer Y_AXIS. ``````# define X coordinates in a new system `````` Populate newly created field using Python parser: ``````int(math.hypot( !NEAR_X!-1754349.55253, !NEAR_Y!-5912723.73683 )) `````` where numerics are coordinates of point 98, i.e. you'll need pen to write them down. If your points coordinates are in decimal degrees, use round to 3(?) decimals instead of Int() function. Repeat 3 steps above by using Y_AXIS and new field Y_DISTANCE. You can use sort tool by these new fields to rearrange points into new layer. I simply used solution from here to populate new integer field shown: As mentioned in comments there is much easier way to calculate coordinates in a new system. If you decide to use it, I suggest to compute minimum bounding geometry (rectangle) of your points wich will compute precise rotation angle.

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https://community.ptc.com/t5/PTC-Mathcad/Finding-the-Actual-Force-and-Brake-Capacity-of-a-Long-Shoe/td-p/448878

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cancel Showing results for Did you mean: cancel Showing results for Did you mean: 1-Newbie ## Finding the Actual Force and Brake Capacity of a Long-Shoe Internal Drum Brake This worksheet will: • Determine the actuating force and the brake capacity of a long-shoe internal brake • Apply to mechanical engineering and the automotive industry • Perform using brake geometry, coefficient of friction, maximum pressure, actuating force, braking capacity, assumed pressure distribution, friction force moment, normal force moment, torque, etc. This worksheet using PTC Mathcad shows you how to determine the actuating force and the brake capacity of a long-shoe internal brake. For this particular worksheet the long-shoe brake is actuated by a device that exerts the same force on each shoe and the shoes are identical in form and are symmetrically position about the vertical centerline. The worksheet begins by listing the parameters of the brake geometry and the brake line properties. These include: • Face width of shoe • Distance from the center of the drum and vertical axis of symmetry to the shoe pivot • Moment arm distance of actuating force to the shoe pivot • Angle from shoe pivot to heel and toe of lining • Coefficient of friction • Maximum allowable pressure To solve, the worksheet first shows you how to calculate for the assumed pressure distribution for the long-shoe brake. Because the clockwise drum rotation causes the right-hand shoe to be self-energizing, you can determine the actuating force on the basis that the maximum allowable pressure will occur there. The worksheet then shows you how to calculate for the friction force moment, the normal force moment, the torque applied to the drum, and the actuating force. Next, the worksheet shows you how to calculate for the pressure on the left-hand shoe, along with the torque applied to drum and the brake capacity. All images, notation, formulas, calculations, data, and solutions are provided to aid you in your own calculations.

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# Pound Square Micrometer to Ounce Square Decimeter Converter 1 Pound Square Micrometer = 1.6e-9 Ounce Square Decimeters ## One Pound Square Micrometer is Equal to How Many Ounce Square Decimeters? The answer is one Pound Square Micrometer is equal to 1.6e-9 Ounce Square Decimeters and that means we can also write it as 1 Pound Square Micrometer = 1.6e-9 Ounce Square Decimeters. Feel free to use our online unit conversion calculator to convert the unit from Pound Square Micrometer to Ounce Square Decimeter. Just simply enter value 1 in Pound Square Micrometer and see the result in Ounce Square Decimeter. Manually converting Pound Square Micrometer to Ounce Square Decimeter can be time-consuming,especially when you don’t have enough knowledge about Moment of Inertia units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Pound Square Micrometer to Ounce Square Decimeter converter tool to get the job done as soon as possible. We have so many online tools available to convert Pound Square Micrometer to Ounce Square Decimeter, but not every online tool gives an accurate result and that is why we have created this online Pound Square Micrometer to Ounce Square Decimeter converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly. ## How to Convert Pound Square Micrometer to Ounce Square Decimeter (lb·μm2 to oz·dm2) By using our Pound Square Micrometer to Ounce Square Decimeter conversion tool, you know that one Pound Square Micrometer is equivalent to 1.6e-9 Ounce Square Decimeter. Hence, to convert Pound Square Micrometer to Ounce Square Decimeter, we just need to multiply the number by 1.6e-9. We are going to use very simple Pound Square Micrometer to Ounce Square Decimeter conversion formula for that. Pleas see the calculation example given below. $$\text{1 Pound Square Micrometer} = 1 \times 1.6e-9 = \text{1.6e-9 Ounce Square Decimeters}$$ ## What Unit of Measure is Pound Square Micrometer? Pound square micrometer is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one micrometer distance from the rotation axis and having a mass of one pound. ## What is the Symbol of Pound Square Micrometer? The symbol of Pound Square Micrometer is lb·μm2. This means you can also write one Pound Square Micrometer as 1 lb·μm2. ## What Unit of Measure is Ounce Square Decimeter? Ounce square decimeter is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one decimeter distance from the rotation axis and having a mass of one ounce. ## What is the Symbol of Ounce Square Decimeter? The symbol of Ounce Square Decimeter is oz·dm2. This means you can also write one Ounce Square Decimeter as 1 oz·dm2. ## How to Use Pound Square Micrometer to Ounce Square Decimeter Converter Tool • As you can see, we have 2 input fields and 2 dropdowns. • From the first dropdown, select Pound Square Micrometer and in the first input field, enter a value. • From the second dropdown, select Ounce Square Decimeter. • Instantly, the tool will convert the value from Pound Square Micrometer to Ounce Square Decimeter and display the result in the second input field. ## Example of Pound Square Micrometer to Ounce Square Decimeter Converter Tool Pound Square Micrometer 1 Ounce Square Decimeter 1.6e-9 # Pound Square Micrometer to Ounce Square Decimeter Conversion Table Pound Square Micrometer [lb·μm2]Ounce Square Decimeter [oz·dm2]Description 1 Pound Square Micrometer1.6e-9 Ounce Square Decimeter1 Pound Square Micrometer = 1.6e-9 Ounce Square Decimeter 2 Pound Square Micrometer3.2e-9 Ounce Square Decimeter2 Pound Square Micrometer = 3.2e-9 Ounce Square Decimeter 3 Pound Square Micrometer4.8e-9 Ounce Square Decimeter3 Pound Square Micrometer = 4.8e-9 Ounce Square Decimeter 4 Pound Square Micrometer6.4e-9 Ounce Square Decimeter4 Pound Square Micrometer = 6.4e-9 Ounce Square Decimeter 5 Pound Square Micrometer8e-9 Ounce Square Decimeter5 Pound Square Micrometer = 8e-9 Ounce Square Decimeter 6 Pound Square Micrometer9.6e-9 Ounce Square Decimeter6 Pound Square Micrometer = 9.6e-9 Ounce Square Decimeter 7 Pound Square Micrometer1.12e-8 Ounce Square Decimeter7 Pound Square Micrometer = 1.12e-8 Ounce Square Decimeter 8 Pound Square Micrometer1.28e-8 Ounce Square Decimeter8 Pound Square Micrometer = 1.28e-8 Ounce Square Decimeter 9 Pound Square Micrometer1.44e-8 Ounce Square Decimeter9 Pound Square Micrometer = 1.44e-8 Ounce Square Decimeter 10 Pound Square Micrometer1.6e-8 Ounce Square Decimeter10 Pound Square Micrometer = 1.6e-8 Ounce Square Decimeter 100 Pound Square Micrometer1.6e-7 Ounce Square Decimeter100 Pound Square Micrometer = 1.6e-7 Ounce Square Decimeter 1000 Pound Square Micrometer0.0000016 Ounce Square Decimeter1000 Pound Square Micrometer = 0.0000016 Ounce Square Decimeter # Pound Square Micrometer to Other Units Conversion Table ConversionDescription 1 Pound Square Micrometer = 4.5359237e-10 Gram Square Meter1 Pound Square Micrometer in Gram Square Meter is equal to 4.5359237e-10 1 Pound Square Micrometer = 4.5359237e-8 Gram Square Decimeter1 Pound Square Micrometer in Gram Square Decimeter is equal to 4.5359237e-8 1 Pound Square Micrometer = 0.0000045359237 Gram Square Centimeter1 Pound Square Micrometer in Gram Square Centimeter is equal to 0.0000045359237 1 Pound Square Micrometer = 0.00045359237 Gram Square Millimeter1 Pound Square Micrometer in Gram Square Millimeter is equal to 0.00045359237 1 Pound Square Micrometer = 453.59 Gram Square Micrometer1 Pound Square Micrometer in Gram Square Micrometer is equal to 453.59 1 Pound Square Micrometer = 4.5359237e-16 Gram Square Kilometer1 Pound Square Micrometer in Gram Square Kilometer is equal to 4.5359237e-16 1 Pound Square Micrometer = 453592370 Gram Square Nanometer1 Pound Square Micrometer in Gram Square Nanometer is equal to 453592370 1 Pound Square Micrometer = 5.4249195959812e-10 Gram Square Yard1 Pound Square Micrometer in Gram Square Yard is equal to 5.4249195959812e-10 1 Pound Square Micrometer = 7.0306957963916e-7 Gram Square Inch1 Pound Square Micrometer in Gram Square Inch is equal to 7.0306957963916e-7 1 Pound Square Micrometer = 4.8824276363831e-9 Gram Square Foot1 Pound Square Micrometer in Gram Square Foot is equal to 4.8824276363831e-9 1 Pound Square Micrometer = 1.7513299315538e-16 Gram Square Mile1 Pound Square Micrometer in Gram Square Mile is equal to 1.7513299315538e-16 1 Pound Square Micrometer = 4.5359237e-13 Kilogram Square Meter1 Pound Square Micrometer in Kilogram Square Meter is equal to 4.5359237e-13 1 Pound Square Micrometer = 4.5359237e-11 Kilogram Square Decimeter1 Pound Square Micrometer in Kilogram Square Decimeter is equal to 4.5359237e-11 1 Pound Square Micrometer = 4.5359237e-9 Kilogram Square Centimeter1 Pound Square Micrometer in Kilogram Square Centimeter is equal to 4.5359237e-9 1 Pound Square Micrometer = 4.5359237e-7 Kilogram Square Millimeter1 Pound Square Micrometer in Kilogram Square Millimeter is equal to 4.5359237e-7 1 Pound Square Micrometer = 0.45359237 Kilogram Square Micrometer1 Pound Square Micrometer in Kilogram Square Micrometer is equal to 0.45359237 1 Pound Square Micrometer = 4.5359237e-19 Kilogram Square Kilometer1 Pound Square Micrometer in Kilogram Square Kilometer is equal to 4.5359237e-19 1 Pound Square Micrometer = 453592.37 Kilogram Square Nanometer1 Pound Square Micrometer in Kilogram Square Nanometer is equal to 453592.37 1 Pound Square Micrometer = 5.4249195959812e-13 Kilogram Square Yard1 Pound Square Micrometer in Kilogram Square Yard is equal to 5.4249195959812e-13 1 Pound Square Micrometer = 7.0306957963916e-10 Kilogram Square Inch1 Pound Square Micrometer in Kilogram Square Inch is equal to 7.0306957963916e-10 1 Pound Square Micrometer = 4.8824276363831e-12 Kilogram Square Foot1 Pound Square Micrometer in Kilogram Square Foot is equal to 4.8824276363831e-12 1 Pound Square Micrometer = 1.7513299315538e-19 Kilogram Square Mile1 Pound Square Micrometer in Kilogram Square Mile is equal to 1.7513299315538e-19 1 Pound Square Micrometer = 4.5359237e-7 Milligram Square Meter1 Pound Square Micrometer in Milligram Square Meter is equal to 4.5359237e-7 1 Pound Square Micrometer = 0.000045359237 Milligram Square Decimeter1 Pound Square Micrometer in Milligram Square Decimeter is equal to 0.000045359237 1 Pound Square Micrometer = 0.0045359237 Milligram Square Centimeter1 Pound Square Micrometer in Milligram Square Centimeter is equal to 0.0045359237 1 Pound Square Micrometer = 0.45359237 Milligram Square Millimeter1 Pound Square Micrometer in Milligram Square Millimeter is equal to 0.45359237 1 Pound Square Micrometer = 453592.37 Milligram Square Micrometer1 Pound Square Micrometer in Milligram Square Micrometer is equal to 453592.37 1 Pound Square Micrometer = 4.5359237e-13 Milligram Square Kilometer1 Pound Square Micrometer in Milligram Square Kilometer is equal to 4.5359237e-13 1 Pound Square Micrometer = 453592370000 Milligram Square Nanometer1 Pound Square Micrometer in Milligram Square Nanometer is equal to 453592370000 1 Pound Square Micrometer = 5.4249195959812e-7 Milligram Square Yard1 Pound Square Micrometer in Milligram Square Yard is equal to 5.4249195959812e-7 1 Pound Square Micrometer = 0.00070306957963916 Milligram Square Inch1 Pound Square Micrometer in Milligram Square Inch is equal to 0.00070306957963916 1 Pound Square Micrometer = 0.0000048824276363831 Milligram Square Foot1 Pound Square Micrometer in Milligram Square Foot is equal to 0.0000048824276363831 1 Pound Square Micrometer = 1.7513299315538e-13 Milligram Square Mile1 Pound Square Micrometer in Milligram Square Mile is equal to 1.7513299315538e-13 1 Pound Square Micrometer = 0.00045359237 Microgram Square Meter1 Pound Square Micrometer in Microgram Square Meter is equal to 0.00045359237 1 Pound Square Micrometer = 0.045359237 Microgram Square Decimeter1 Pound Square Micrometer in Microgram Square Decimeter is equal to 0.045359237 1 Pound Square Micrometer = 4.54 Microgram Square Centimeter1 Pound Square Micrometer in Microgram Square Centimeter is equal to 4.54 1 Pound Square Micrometer = 453.59 Microgram Square Millimeter1 Pound Square Micrometer in Microgram Square Millimeter is equal to 453.59 1 Pound Square Micrometer = 453592370 Microgram Square Micrometer1 Pound Square Micrometer in Microgram Square Micrometer is equal to 453592370 1 Pound Square Micrometer = 4.5359237e-10 Microgram Square Kilometer1 Pound Square Micrometer in Microgram Square Kilometer is equal to 4.5359237e-10 1 Pound Square Micrometer = 453592370000000 Microgram Square Nanometer1 Pound Square Micrometer in Microgram Square Nanometer is equal to 453592370000000 1 Pound Square Micrometer = 0.00054249195959812 Microgram Square Yard1 Pound Square Micrometer in Microgram Square Yard is equal to 0.00054249195959812 1 Pound Square Micrometer = 0.70306957963916 Microgram Square Inch1 Pound Square Micrometer in Microgram Square Inch is equal to 0.70306957963916 1 Pound Square Micrometer = 0.0048824276363831 Microgram Square Foot1 Pound Square Micrometer in Microgram Square Foot is equal to 0.0048824276363831 1 Pound Square Micrometer = 1.7513299315538e-10 Microgram Square Mile1 Pound Square Micrometer in Microgram Square Mile is equal to 1.7513299315538e-10 1 Pound Square Micrometer = 4.5359237e-16 Ton Square Meter1 Pound Square Micrometer in Ton Square Meter is equal to 4.5359237e-16 1 Pound Square Micrometer = 4.5359237e-14 Ton Square Decimeter1 Pound Square Micrometer in Ton Square Decimeter is equal to 4.5359237e-14 1 Pound Square Micrometer = 4.5359237e-12 Ton Square Centimeter1 Pound Square Micrometer in Ton Square Centimeter is equal to 4.5359237e-12 1 Pound Square Micrometer = 4.5359237e-10 Ton Square Millimeter1 Pound Square Micrometer in Ton Square Millimeter is equal to 4.5359237e-10 1 Pound Square Micrometer = 0.00045359237 Ton Square Micrometer1 Pound Square Micrometer in Ton Square Micrometer is equal to 0.00045359237 1 Pound Square Micrometer = 4.5359237e-22 Ton Square Kilometer1 Pound Square Micrometer in Ton Square Kilometer is equal to 4.5359237e-22 1 Pound Square Micrometer = 453.59 Ton Square Nanometer1 Pound Square Micrometer in Ton Square Nanometer is equal to 453.59 1 Pound Square Micrometer = 5.4249195959812e-16 Ton Square Yard1 Pound Square Micrometer in Ton Square Yard is equal to 5.4249195959812e-16 1 Pound Square Micrometer = 7.0306957963916e-13 Ton Square Inch1 Pound Square Micrometer in Ton Square Inch is equal to 7.0306957963916e-13 1 Pound Square Micrometer = 4.882427636383e-15 Ton Square Foot1 Pound Square Micrometer in Ton Square Foot is equal to 4.882427636383e-15 1 Pound Square Micrometer = 1.7513299315538e-22 Ton Square Mile1 Pound Square Micrometer in Ton Square Mile is equal to 1.7513299315538e-22 1 Pound Square Micrometer = 2.26796185e-9 Carat Square Meter1 Pound Square Micrometer in Carat Square Meter is equal to 2.26796185e-9 1 Pound Square Micrometer = 2.26796185e-7 Carat Square Decimeter1 Pound Square Micrometer in Carat Square Decimeter is equal to 2.26796185e-7 1 Pound Square Micrometer = 0.0000226796185 Carat Square Centimeter1 Pound Square Micrometer in Carat Square Centimeter is equal to 0.0000226796185 1 Pound Square Micrometer = 0.00226796185 Carat Square Millimeter1 Pound Square Micrometer in Carat Square Millimeter is equal to 0.00226796185 1 Pound Square Micrometer = 2267.96 Carat Square Micrometer1 Pound Square Micrometer in Carat Square Micrometer is equal to 2267.96 1 Pound Square Micrometer = 2.26796185e-15 Carat Square Kilometer1 Pound Square Micrometer in Carat Square Kilometer is equal to 2.26796185e-15 1 Pound Square Micrometer = 2267961850 Carat Square Nanometer1 Pound Square Micrometer in Carat Square Nanometer is equal to 2267961850 1 Pound Square Micrometer = 2.7124597979906e-9 Carat Square Yard1 Pound Square Micrometer in Carat Square Yard is equal to 2.7124597979906e-9 1 Pound Square Micrometer = 0.0000035153478981958 Carat Square Inch1 Pound Square Micrometer in Carat Square Inch is equal to 0.0000035153478981958 1 Pound Square Micrometer = 2.4412138181915e-8 Carat Square Foot1 Pound Square Micrometer in Carat Square Foot is equal to 2.4412138181915e-8 1 Pound Square Micrometer = 8.7566496577692e-16 Carat Square Mile1 Pound Square Micrometer in Carat Square Mile is equal to 8.7566496577692e-16 1 Pound Square Micrometer = 1.6e-11 Ounce Square Meter1 Pound Square Micrometer in Ounce Square Meter is equal to 1.6e-11 1 Pound Square Micrometer = 1.6e-9 Ounce Square Decimeter1 Pound Square Micrometer in Ounce Square Decimeter is equal to 1.6e-9 1 Pound Square Micrometer = 1.6e-7 Ounce Square Centimeter1 Pound Square Micrometer in Ounce Square Centimeter is equal to 1.6e-7 1 Pound Square Micrometer = 0.000016 Ounce Square Millimeter1 Pound Square Micrometer in Ounce Square Millimeter is equal to 0.000016 1 Pound Square Micrometer = 16 Ounce Square Micrometer1 Pound Square Micrometer in Ounce Square Micrometer is equal to 16 1 Pound Square Micrometer = 1.6e-17 Ounce Square Kilometer1 Pound Square Micrometer in Ounce Square Kilometer is equal to 1.6e-17 1 Pound Square Micrometer = 16000000 Ounce Square Nanometer1 Pound Square Micrometer in Ounce Square Nanometer is equal to 16000000 1 Pound Square Micrometer = 1.9135840740817e-11 Ounce Square Yard1 Pound Square Micrometer in Ounce Square Yard is equal to 1.9135840740817e-11 1 Pound Square Micrometer = 2.4800049600099e-8 Ounce Square Inch1 Pound Square Micrometer in Ounce Square Inch is equal to 2.4800049600099e-8 1 Pound Square Micrometer = 1.7222256666736e-10 Ounce Square Foot1 Pound Square Micrometer in Ounce Square Foot is equal to 1.7222256666736e-10 1 Pound Square Micrometer = 6.1776345366791e-18 Ounce Square Mile1 Pound Square Micrometer in Ounce Square Mile is equal to 6.1776345366791e-18 1 Pound Square Micrometer = 1e-15 Kilopound Square Meter1 Pound Square Micrometer in Kilopound Square Meter is equal to 1e-15 1 Pound Square Micrometer = 1e-13 Kilopound Square Decimeter1 Pound Square Micrometer in Kilopound Square Decimeter is equal to 1e-13 1 Pound Square Micrometer = 1e-11 Kilopound Square Centimeter1 Pound Square Micrometer in Kilopound Square Centimeter is equal to 1e-11 1 Pound Square Micrometer = 1e-9 Kilopound Square Millimeter1 Pound Square Micrometer in Kilopound Square Millimeter is equal to 1e-9 1 Pound Square Micrometer = 0.001 Kilopound Square Micrometer1 Pound Square Micrometer in Kilopound Square Micrometer is equal to 0.001 1 Pound Square Micrometer = 1e-21 Kilopound Square Kilometer1 Pound Square Micrometer in Kilopound Square Kilometer is equal to 1e-21 1 Pound Square Micrometer = 1000 Kilopound Square Nanometer1 Pound Square Micrometer in Kilopound Square Nanometer is equal to 1000 1 Pound Square Micrometer = 1.1959900463011e-15 Kilopound Square Yard1 Pound Square Micrometer in Kilopound Square Yard is equal to 1.1959900463011e-15 1 Pound Square Micrometer = 1.5500031000062e-12 Kilopound Square Inch1 Pound Square Micrometer in Kilopound Square Inch is equal to 1.5500031000062e-12 1 Pound Square Micrometer = 1.076391041671e-14 Kilopound Square Foot1 Pound Square Micrometer in Kilopound Square Foot is equal to 1.076391041671e-14 1 Pound Square Micrometer = 3.8610215854245e-22 Kilopound Square Mile1 Pound Square Micrometer in Kilopound Square Mile is equal to 3.8610215854245e-22 1 Pound Square Micrometer = 1e-12 Pound Square Meter1 Pound Square Micrometer in Pound Square Meter is equal to 1e-12 1 Pound Square Micrometer = 1e-10 Pound Square Decimeter1 Pound Square Micrometer in Pound Square Decimeter is equal to 1e-10 1 Pound Square Micrometer = 1e-8 Pound Square Centimeter1 Pound Square Micrometer in Pound Square Centimeter is equal to 1e-8 1 Pound Square Micrometer = 0.000001 Pound Square Millimeter1 Pound Square Micrometer in Pound Square Millimeter is equal to 0.000001 1 Pound Square Micrometer = 1e-18 Pound Square Kilometer1 Pound Square Micrometer in Pound Square Kilometer is equal to 1e-18 1 Pound Square Micrometer = 1000000 Pound Square Nanometer1 Pound Square Micrometer in Pound Square Nanometer is equal to 1000000 1 Pound Square Micrometer = 1.1959900463011e-12 Pound Square Yard1 Pound Square Micrometer in Pound Square Yard is equal to 1.1959900463011e-12 1 Pound Square Micrometer = 1.5500031000062e-9 Pound Square Inch1 Pound Square Micrometer in Pound Square Inch is equal to 1.5500031000062e-9 1 Pound Square Micrometer = 1.076391041671e-11 Pound Square Foot1 Pound Square Micrometer in Pound Square Foot is equal to 1.076391041671e-11 1 Pound Square Micrometer = 3.8610215854245e-19 Pound Square Mile1 Pound Square Micrometer in Pound Square Mile is equal to 3.8610215854245e-19 1 Pound Square Micrometer = 3.2200000000497e-11 Poundal Square Meter1 Pound Square Micrometer in Poundal Square Meter is equal to 3.2200000000497e-11 1 Pound Square Micrometer = 3.2200000000497e-9 Poundal Square Decimeter1 Pound Square Micrometer in Poundal Square Decimeter is equal to 3.2200000000497e-9 1 Pound Square Micrometer = 3.2200000000497e-7 Poundal Square Centimeter1 Pound Square Micrometer in Poundal Square Centimeter is equal to 3.2200000000497e-7 1 Pound Square Micrometer = 0.000032200000000497 Poundal Square Millimeter1 Pound Square Micrometer in Poundal Square Millimeter is equal to 0.000032200000000497 1 Pound Square Micrometer = 32.2 Poundal Square Micrometer1 Pound Square Micrometer in Poundal Square Micrometer is equal to 32.2 1 Pound Square Micrometer = 3.2200000000497e-17 Poundal Square Kilometer1 Pound Square Micrometer in Poundal Square Kilometer is equal to 3.2200000000497e-17 1 Pound Square Micrometer = 32200000 Poundal Square Nanometer1 Pound Square Micrometer in Poundal Square Nanometer is equal to 32200000 1 Pound Square Micrometer = 3.8510879491489e-11 Poundal Square Yard1 Pound Square Micrometer in Poundal Square Yard is equal to 3.8510879491489e-11 1 Pound Square Micrometer = 4.991009982097e-8 Poundal Square Inch1 Pound Square Micrometer in Poundal Square Inch is equal to 4.991009982097e-8 1 Pound Square Micrometer = 3.465979154234e-10 Poundal Square Foot1 Pound Square Micrometer in Poundal Square Foot is equal to 3.465979154234e-10 1 Pound Square Micrometer = 1.2432489505259e-17 Poundal Square Mile1 Pound Square Micrometer in Poundal Square Mile is equal to 1.2432489505259e-17 1 Pound Square Micrometer = 7e-9 Grain Square Meter1 Pound Square Micrometer in Grain Square Meter is equal to 7e-9 1 Pound Square Micrometer = 7e-7 Grain Square Decimeter1 Pound Square Micrometer in Grain Square Decimeter is equal to 7e-7 1 Pound Square Micrometer = 0.00007 Grain Square Centimeter1 Pound Square Micrometer in Grain Square Centimeter is equal to 0.00007 1 Pound Square Micrometer = 0.007 Grain Square Millimeter1 Pound Square Micrometer in Grain Square Millimeter is equal to 0.007 1 Pound Square Micrometer = 7000 Grain Square Micrometer1 Pound Square Micrometer in Grain Square Micrometer is equal to 7000 1 Pound Square Micrometer = 7e-15 Grain Square Kilometer1 Pound Square Micrometer in Grain Square Kilometer is equal to 7e-15 1 Pound Square Micrometer = 7000000000 Grain Square Nanometer1 Pound Square Micrometer in Grain Square Nanometer is equal to 7000000000 1 Pound Square Micrometer = 8.3719303241076e-9 Grain Square Yard1 Pound Square Micrometer in Grain Square Yard is equal to 8.3719303241076e-9 1 Pound Square Micrometer = 0.000010850021700043 Grain Square Inch1 Pound Square Micrometer in Grain Square Inch is equal to 0.000010850021700043 1 Pound Square Micrometer = 7.5347372916968e-8 Grain Square Foot1 Pound Square Micrometer in Grain Square Foot is equal to 7.5347372916968e-8 1 Pound Square Micrometer = 2.7027151097971e-15 Grain Square Mile1 Pound Square Micrometer in Grain Square Mile is equal to 2.7027151097971e-15 1 Pound Square Micrometer = 3.1080956427e-14 Slug Square Meter1 Pound Square Micrometer in Slug Square Meter is equal to 3.1080956427e-14 1 Pound Square Micrometer = 3.1080956427e-12 Slug Square Decimeter1 Pound Square Micrometer in Slug Square Decimeter is equal to 3.1080956427e-12 1 Pound Square Micrometer = 3.1080956427e-10 Slug Square Centimeter1 Pound Square Micrometer in Slug Square Centimeter is equal to 3.1080956427e-10 1 Pound Square Micrometer = 3.1080956427e-8 Slug Square Millimeter1 Pound Square Micrometer in Slug Square Millimeter is equal to 3.1080956427e-8 1 Pound Square Micrometer = 0.031080956427 Slug Square Micrometer1 Pound Square Micrometer in Slug Square Micrometer is equal to 0.031080956427 1 Pound Square Micrometer = 3.1080956427e-20 Slug Square Kilometer1 Pound Square Micrometer in Slug Square Kilometer is equal to 3.1080956427e-20 1 Pound Square Micrometer = 31080.96 Slug Square Nanometer1 Pound Square Micrometer in Slug Square Nanometer is equal to 31080.96 1 Pound Square Micrometer = 3.717251451621e-14 Slug Square Yard1 Pound Square Micrometer in Slug Square Yard is equal to 3.717251451621e-14 1 Pound Square Micrometer = 4.8175578813008e-11 Slug Square Inch1 Pound Square Micrometer in Slug Square Inch is equal to 4.8175578813008e-11 1 Pound Square Micrometer = 3.3455263064589e-13 Slug Square Foot1 Pound Square Micrometer in Slug Square Foot is equal to 3.3455263064589e-13 1 Pound Square Micrometer = 1.2000424366029e-20 Slug Square Mile1 Pound Square Micrometer in Slug Square Mile is equal to 1.2000424366029e-20 1 Pound Square Micrometer = 0.00045359237 Gamma Square Meter1 Pound Square Micrometer in Gamma Square Meter is equal to 0.00045359237 1 Pound Square Micrometer = 0.045359237 Gamma Square Decimeter1 Pound Square Micrometer in Gamma Square Decimeter is equal to 0.045359237 1 Pound Square Micrometer = 4.54 Gamma Square Centimeter1 Pound Square Micrometer in Gamma Square Centimeter is equal to 4.54 1 Pound Square Micrometer = 453.59 Gamma Square Millimeter1 Pound Square Micrometer in Gamma Square Millimeter is equal to 453.59 1 Pound Square Micrometer = 453592370 Gamma Square Micrometer1 Pound Square Micrometer in Gamma Square Micrometer is equal to 453592370 1 Pound Square Micrometer = 4.5359237e-10 Gamma Square Kilometer1 Pound Square Micrometer in Gamma Square Kilometer is equal to 4.5359237e-10 1 Pound Square Micrometer = 453592370000000 Gamma Square Nanometer1 Pound Square Micrometer in Gamma Square Nanometer is equal to 453592370000000 1 Pound Square Micrometer = 0.00054249195959812 Gamma Square Yard1 Pound Square Micrometer in Gamma Square Yard is equal to 0.00054249195959812 1 Pound Square Micrometer = 0.70306957963916 Gamma Square Inch1 Pound Square Micrometer in Gamma Square Inch is equal to 0.70306957963916 1 Pound Square Micrometer = 0.0048824276363831 Gamma Square Foot1 Pound Square Micrometer in Gamma Square Foot is equal to 0.0048824276363831 1 Pound Square Micrometer = 1.7513299315538e-10 Gamma Square Mile1 Pound Square Micrometer in Gamma Square Mile is equal to 1.7513299315538e-10

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# The Distance Game. age 10-adult, developed by Grayson Wheatley and Michael Naylor, 1998. Single disk for Macintosh or Windows, \$29.95. Mathematics Learning, 8930 Winged Foot Dr., Tallahassee, FL 32312, (850) 668-9397. According to its authors, "This game becomes addictive - children and adults have been observed playing the game for hours. . . . It is accessible to elementary school students but still challenging for adults." The game caught my interest as well as the interest of students in both fifth and sixth grades. The stronger the children's mathematics skills, the more they seemed to be intrigued by the game. A player is presented with a ten-by-ten grid and asked to find Bouncer, which is located at one of the grid's intersections. Bouncer can be found by clicking on the correct intersection. When the player clicks on an intersection, the program tells how far from that intersection Bouncer is located. If the player happens to click in line with Bouncer, the distance from the intersection to Bouncer is a whole number. If Bouncer is located diagonally from the intersection, the straight-line distance is given using decimals rounded to tenths. For example, one square away on the diagonal would be listed as 1.4, the square root of 2. A game consists of ten rounds, and players try to find Bouncer using the fewest guesses. The game records the average number of guesses for the ten rounds and also contains a Hall of Fame so that students can compare their results with those of other players. This program allows students to develop their own understandings of many mathematical skills, does not contain a tutorial, and recommends against preteaching associated skills. As children play the game, their knowledge is reinforced in such areas as analyzing decimal numbers, plotting points on the coordinate plane, determining distances between coordinate points, understanding averages, and thinking in units. With some instruction, a more advanced level can be played that would include work with the Pythagorean theorem and number theory with square numbers. The Distance Game is an enjoyable program that helped my students develop mathematical understandings in a recreational format. COPYRIGHT 1999 National Council of Teachers of Mathematics, Inc. No portion of this article can be reproduced without the express written permission from the copyright holder.

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471,873 Members | 1,710 Online # data stucture using c i m in touch with c from last 1yr now just started data stucture in c and i have a problem in linear search and binary search if any one could explain with one example each Aug 26 '06 #1 7 4573 "enrique" <ak********@gmail.comwrites: in linear search and binary search if any one could explain with one example each http://en.wikipedia.org/wiki/Linear_search http://en.wikipedia.org/wiki/Binary_search -- Best regards, _ _ .o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o ..o | Computer Science, Michal "mina86" Nazarewicz (o o) ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo-- Aug 26 '06 #2 enrique wrote: i m in touch with c from last 1yr now just started data stucture in c and i have a problem in linear search and binary search if any one could explain with one example each I believe this is OT for this forum, but heres an example anyway: a list of unsorted numbers: 8 4 9 5 7 6 1 2 3 We want to search for 7. Linear search, start at beginning and go through each number until you hit 5. i.e. 8->4->9->5->7 done. Binary search, sort first. 1 2 3 4 5 6 7 8 9 Look at middle point: 5, this is less than 7 so it can't be in the bottom half so chop it out and you're left with 6 7 8 9 Look at middle point (say 8 for now) this is greater than 7 so can't be in top half. Left with 6 7 Look at middle point (7) done. Aug 26 '06 #3 enrique posted: i m in touch with c from last 1yr now just started data stucture in c and i have a problem in linear search and binary search if any one could explain with one example each http://en.wikipedia.org/wiki/English_spelling http://en.wikipedia.org/wiki/English_grammar http://en.wikipedia.org/wiki/Punctuation -- Frederick Gotham Aug 26 '06 #4 "enrique" <ak********@gmail.comwrites: i m in touch with c from last 1yr now just started data stucture in c and i have a problem in linear search and binary search if any one could explain with one example each -- "The lusers I know are so clueless, that if they were dipped in clue musk and dropped in the middle of pack of horny clues, on clue prom night during clue happy hour, they still couldn't get a clue." --Michael Girdwood, in the monastery Aug 26 '06 #5 Frederick Botham wrote: enrique posted: >>i m in touch with c from last 1yr now just started data stucture in c and i have a problem in linear search and binary search if any one could explain with one example each http://en.wikipedia.org/wiki/English_spelling http://en.wikipedia.org/wiki/English_grammar http://en.wikipedia.org/wiki/Punctuation Inappropriately harsh. This isn't some one posing in gobbledygook speak, it is some one doing their best in a second language. -- Ian Collins. Aug 26 '06 #6 Ian Collins posted: >http://en.wikipedia.org/wiki/English_spelling http://en.wikipedia.org/wiki/English_grammar http://en.wikipedia.org/wiki/Punctuation Inappropriately harsh. This isn't some one posting in gobbledygook speak, it is some one doing their best in a second language. If anything, I thought it might get across to the OP that maybe they should try improve the quality of their writing. (English not being the OP's mother tongue doesn't quite explain the lack of punctuation -- unless their mother tongue is written in a different script I suppose.) -- Frederick Gotham Aug 27 '06 #7 http://en.wikipedia.org/wiki/English_spelling http://en.wikipedia.org/wiki/English_grammar http://en.wikipedia.org/wiki/Punctuation Inappropriately harsh. This isn't some one posing in gobbledygook speak, it is some one doing their best in a second language. -- Ian Collins. Quite agree. Sep 6 '06 #8 ### This discussion thread is closed Replies have been disabled for this discussion.

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# Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers) Answers obtained from two online integral calculators: \begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\arcsin\left(\dfrac1{\sqrt2}\sqrt{1 - x}\right) + C \\ \int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= 2\sin^{-1}\left(\dfrac{\sqrt{x + 1}}{\sqrt2}\right) - \sqrt{1 - x^2} + \text{ constant} \end{align} ## Update: I realized that the substitution for $$\theta$$ was supposed to be $$\arcsin$$ not $$\arccos$$, so the answer would have been the same as the right hand side. But I also noticed that using the initial substitution to plug $$x$$ back in the final answer will not always give the correct answer because in a similar question: $$\int \frac{\sqrt{x^2-1}}x dx$$ has a trig-substitution of $$x = \sec\theta$$, and the answer in terms of $$\theta$$ would be: $$\tan \theta - \theta + C$$. Then the final answer in terms of $$x$$ should be : $$\sqrt{x^2-1} - \operatorname{arcsec}(x) + C$$. But online integral calculators give the answer: $$\sqrt{x^2-1} - \arctan(\sqrt{x^2-1}) + C$$, which doesn't match the original substitution of: $$x = \sec\theta \to \theta = \operatorname{arcsec}(x)$$ Anyone know why the calculator gives that answer which doesn't match the original trig-substitution of $$x = \sec \theta \to \theta = \operatorname{arcsec}(x)$$? • You assumed $x=sin\theta$, so $\theta = arcsin x$ – user721016 Feb 17, 2020 at 20:26 • Oh so the theta value must also match with the substitution I made for x? – user749176 Feb 17, 2020 at 20:32 • In short yes, assuming $x>0$, note that $arcsinx = arccos√(1-x^2)$ for only $x>0$ – user721016 Feb 17, 2020 at 20:38 • which is the actual answer though? When I plug the question into different online integral calculators, all answers are very different from mine – user749176 Feb 17, 2020 at 20:44 • Does this answer your question? Getting different answers when integrating using different techniques Feb 18, 2020 at 5:01 Now, both answers are correct. They merely look different. They differ by a constant. Note 1... $$-\sqrt{\frac{x+1}{1-x}}+\sqrt{\frac{x+1}{1-x}}\;x = -\sqrt{\frac{x+1}{1-x}}\;(1-x) = -\frac{\sqrt{x+1}\;(1-x)}{\sqrt{1-x}} \\= -\sqrt{x+1}\sqrt{1-x} =-\sqrt{(1+x)(1-x)} =-\sqrt{1-x^2}$$ Note 2... $$2\,\arcsin \left( \frac{\sqrt {1+x}}{\sqrt {2}} \right) =\pi-2\,\arcsin \left( \frac{\sqrt {1-x}}{\sqrt {2}} \right)$$ • For the second line of step, I multiplied top and bottom by √(1+x) in order to use the trig substitution of x = sinθ – user749176 Feb 17, 2020 at 21:12 • Also those two answers are from the calculator. I was wondering if they are correct after comparing them to my answers, which are shown in the link – user749176 Feb 17, 2020 at 21:27 Starting off with $$\displaystyle\int \frac{\sqrt{x^2-1}}{x}\mathrm dx$$, substitute $$x = \sec(\theta)$$ for $$\theta \in \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$$ as usual (keep the domain in mind for later). Of course, that means $$\sqrt{x^2-1} = \sqrt{\sec^2(\theta)-1} = \sqrt{\tan^2(\theta)} = \vert \tan(\theta)\vert$$ and $$\dfrac{\mathrm dx}{\mathrm d\theta} = \sec(\theta)\tan(\theta) \iff \mathrm dx = \sec(\theta)\tan(\theta)\mathrm d\theta$$. This simplifies as follows: $$\displaystyle\int \frac{\sqrt{x^2-1}}{x}\mathrm dx \longrightarrow \int\frac{\vert \tan(\theta)\vert}{\sec(\theta)}\sec(\theta)\tan(\theta)\mathrm d\theta = \int\vert \tan(\theta)\vert\tan(\theta)\mathrm d\theta$$ For $$\theta \in \left[0, \frac{\pi}{2}\right)$$, $$\tan(\theta) \geq 0$$, so you get $$\int \tan^2(\theta) \mathrm d\theta = \int \left[\sec^2(\theta)-1\right] \mathrm d\theta = \tan(\theta)-\theta+C \longrightarrow \sqrt{x^2-1}-\text{arcsec}(x)+C$$ Since tangent is positive in the first quadrant, $$\tan(\theta) = \sqrt{x^2-1}$$, so the $$\theta$$ term can also be replaced with $$\arctan\left(\sqrt{x^2-1}\right)$$. For $$\theta \in \left(\frac{\pi}{2}, \pi\right]$$, $$\tan(\theta) \leq 0$$, so you get $$-\int \tan^2(\theta) \mathrm d\theta = -\int \left[\sec^2(\theta)-1\right] \mathrm d\theta = \theta-\tan(\theta)+C \longrightarrow \sqrt{x^2-1}+\text{arcsec}(x)+C$$ Since tangent is negative in the second quadrant, $$\tan(\theta) = \tan(\theta -\pi) = -\sqrt{x^2-1}$$ (remember that tangent takes arguments in the first and fourth quadrants), so the $$\theta$$ term can also be replaced with $$\pi-\arctan\left(\sqrt{x^2-1}\right)$$. In both cases, the anti-derivative could be re-written as $$\sqrt{x^2-1}-\arctan\left(\sqrt{x^2-1}\right)+C$$. Basically, it just "combines" your other two anti-derivatives and expresses them as a single function rather than having one for each case. • Would just leaving the answer as: √(x^2-1) - arcsec(x) + C , still be correct? – user749176 Feb 18, 2020 at 0:09 • For indefinite integrals, we usually assume $\tan(\theta)$ is positive (at least from what I've seen), so yeah, that's fine. Feb 18, 2020 at 0:16 Let \begin{align} I &= \int \frac{\sqrt{1+x}}{\sqrt{1-x}}\;dx = \int \frac{1+x}{\sqrt{1-x^2}}\;dx \end{align} Left side: Let $$x = \sin\theta$$, $$dx = \cos\theta\;d\theta$$. \begin{align} I &= \int \frac{1+\sin\theta}{\sqrt{\cos^2\theta}}\cos\theta\;d\theta \\ &= \int \left(1+\sin\theta \right)d\theta \\ &= \theta - \cos\theta + c \\ &= \arccos\left(\sqrt{1-x^2}\right) - \sqrt{1-x^2} + c \end{align} Right side: \begin{align} I &= \int \frac{1}{\sqrt{1-x^2}}\;dx + \int \frac{x}{\sqrt{1-x^2}}\;dx \\ u &= 1 - x^2,\;\; -\frac{1}{2}du = x\,dx \\ \implies I &= \arcsin x - \frac{1}{2}\int u^{-1/2}\;du \\ &= \arcsin x - \sqrt{u} + c \\ &= \arcsin x - \sqrt{1 - x^2} + c \end{align} The answers would be the exact same, if $$\arccos\left(\sqrt{1-x^2}\right) = \arcsin x$$. And therein lies the difference. On the "Left side", the substitution you originally made was $$x = \sin\theta$$. So when you replace $$\theta$$ you should substitute $$\theta = \arcsin x$$. By the rules of trig substitution, they should be equivalent. But canonically, the arcsin function has a range of $$-\pi/2$$ to $$\pi/2$$, while the arccos function has a range of $$0$$ to $$\pi$$. So when you use $$\arccos\left(\sqrt{1-x^2}\right)$$, as-is you are losing the case where $$-1 < x < 0$$. The integral has a kink in it, but that's not what you want, seeing as how the function being integrated is continuous and differentiable at $$x=0$$. You could shift arccos by an appropriate amount and use that solution, but I think it would be easier to go with arcsin here. • Yes the wrong substitution of θ, seemed to cause the problem. But aside from substituting x back in the answer, do you know if the x's you sub back in is always supposed to correspond to the original substitution made? For example, see my updated question at the top to see what I mean – user749176 Feb 17, 2020 at 21:41

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### Online College Courses for Credit #### FREE EDUCATIONAL RESOURCES PROVIDED by SOPHIA ##### Are you a student? Free Professional Development + # College Algebra Text Tutorial ##### Rating: (0) Author: Anthony Varela ##### Description: Test tutorial for text version. (more) Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.* No credit card required 29 Sophia partners guarantee credit transfer. 311 Institutions have accepted or given pre-approval for credit transfer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 27 of Sophia’s online courses. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. Tutorial ## What's Covered Overview Before You Start Properties of Exponents Summary ## Properties of Exponents Problems with exponents can often be simplified using a few basic exponent properties. Exponents represent repeated multiplication. We will use this fact to discover the important properties. [DID YOU KNOW ICON] The word exponent comes from the Latin “expo” meaning out of and “ponere” meaning place. While there is some debate, it seems that the Babylonians living in Iraq were the first to do work with exponents (dating back to the 23rd century BC or earlier!) Product of Powers Property Let's consider multiplication with exponents when the bases are the same: A quicker method to arrive at our answer would have been to just add the exponents.  This is known as the product of powers property. The product of powers property can be used to simplify many problems. If exponential expressions with the same base are multiplied, we can add the exponents.  Here is an example: Quotient of Powers Property Exponents Rather than multiplying, we will now try to divide with exponents. A quicker method to arrive at the solution would have been to just subtract the exponents.  This is known as the quotient of powers property: The quotient of powers property can similarly be used to simplify exponent problems by subtracting exponents on like-variables. Here is an example: Power of Power Property A third property we will look at will have an exponent problem raised to a second exponent. This is investigated in the following example. A quicker method to arrive at the solution would have been to just multiply the exponents.  This is known as the power of a power property. This property is often combined with two other properties: power of a product, and power of a quotient. Power of a Product Property A quicker method to arrive at the solution would have been to take the exponent of three and put it on each factor in parenthesis.  This is known as the power of a product property. It is important to be careful to only use the power of a product rule with multiplication inside parenthesis. This property does NOT work if there is addition or subtraction. These are NOT equal.  Beware of this error! Power of a Quotient Property A quicker method to arrive at the solution would have been to put the exponent on every factor in both the numerator and denominator.  This is known as the power of a quotient property. The power of a power, product and quotient rules are often used together to simplify expressions. This is shown in the following examples. These five properties are often mixed up in the same problem. Often there is a bit of flexibility as to which property is used first. However, order of operations still applies to a problem. For this reason, we suggest simplifying inside any parentheses first, then simplify any exponents (using power rules).  Finally, simplify any multiplication or division (using product and quotient rules).

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# How to find time constant of battery first order model with unknown capacitance? The above picture is the first order circuit model of battery. I'm trying to calculate the value of R1 and C1 with specified time varied input and output. In order to calculate that, I have to know the value of time constant first but I have no idea how to calculate that without knowing the value of C1. • Where did the picture of the circuit come from and what battery was it related to? – Andy aka May 15 at 11:59 • The information you need can probably be calculated from the "specified time varied input and output"... – JimmyB May 15 at 12:03 • The input is current and the output is voltage but I'm not sure how to calculate the time constant from that – Tsz Yee Ha May 15 at 12:26 • Vem is the O/C voltage. | R1 + R0 govern the voltage droop under load. - Vo = Vem - Iload.(R1 + R0). | The Vo/c sags after a load is removed and then restores with time to Vem - on fact this is not true as Vem will drop over time as the battery is depleted - but this model does not show that. SO If Vem drops to say Vl under load of Iload, recovers instantly to Vr1 when load is removed then recovers to Vem after time Tr then. | Vload = Vem - (R0+R1) x Iload - rearrange to get R0 + R1. | Vc = Vem-- Vr1. -> Calculate ratio of R0 to R1. | Recovery time t= R0.C. -> calculate C – Russell McMahon May 15 at 12:58 Is this an assignment or homework? Vem is the O/C voltage. R1 + R0 govern the voltage droop under load. Vo = Vem - Iload.(R1 + R0). The Vo/c sags after a load is removed and then restores with time to Vem. In fact, the action is more complex than this as Vem will drop over time as the battery is depleted - but this model does not show that. SO If

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Register Inmates Photos Site Rules Search Today's Posts Mark Forums Read 04-06-2013, 01:53 PM #226 Celtic Curmudgeon Indiana Jones wanabe     Joined: Feb 2011 Location: Boca Raton Oddometer: 278 Thanks, Croak, that puts it in perspective. I was in ROTC and the reserve, have been an avid shooter, etc. so the metric stuff we used there I have no problem with...kilometers to miles, mms to inches, but other than knowing there's 3.8 liters in a gallon, volume measures are confusing for me, and like your reference to temperature... I know from physics that 0 C is 32F, and 100c is 212F, but in between, those, I get lost! Cheers __________________ Speed never killed anyone. Suddenly becoming stationary, that's what gets you. - Jeremy Clarkson 04-06-2013, 03:06 PM   #227 Croak Joined: Jun 2008 Location: Vancouver, BC/Is-Swieqi, Malta Oddometer: 529 Quote: Originally Posted by Celtic Curmudgeon Thanks, Croak, that puts it in perspective. I was in ROTC and the reserve, have been an avid shooter, etc. so the metric stuff we used there I have no problem with...kilometers to miles, mms to inches, but other than knowing there's 3.8 liters in a gallon, volume measures are confusing for me, and like your reference to temperature... I know from physics that 0 C is 32F, and 100c is 212F, but in between, those, I get lost! Cheers Just a matter of getting your frame of reference, and metric is a hell of a lot easier to learn in the first place, and easier to work with once you've learned it. Oh, for converting Celsius to Fahrenheit, the quick and dirty method is to double it and add 30, which is not super accurate and doesn't work well below freezing, but gets you in the ballpark when talking about typical weather. 20C *2+30=70F. 10C*2+30=50F. __________________ 2003 Aprilia Tuono 2002 Triumph Sprint ST 2013 Kymco Agility City 200i 04-08-2013, 02:24 PM #228 Croak Studly Adventurer     Joined: Jun 2008 Location: Vancouver, BC/Is-Swieqi, Malta Oddometer: 529 Yet another Tor Sagen video on the Capo 1200. Some footage from previous videos, but a fair amount of new stuff, including more of his thoughts on actually riding the bike. __________________ 2003 Aprilia Tuono 2002 Triumph Sprint ST 2013 Kymco Agility City 200i 04-08-2013, 02:32 PM #229 Croak Studly Adventurer     Joined: Jun 2008 Location: Vancouver, BC/Is-Swieqi, Malta Oddometer: 529 And here's a bit dry and heavily-accented explanation from the Aprilia electronics lead describing ADD in more detail, and how it differs/improves on the Skyhook system: __________________ 2003 Aprilia Tuono 2002 Triumph Sprint ST 2013 Kymco Agility City 200i 04-08-2013, 03:36 PM   #230 Lion BR I'd rather be riding Joined: Oct 2005 Location: Oregon Oddometer: 3,729 Quote: Originally Posted by Croak And here's a bit dry and heavily-accented explanation from the Aprilia electronics lead describing ADD in more detail, and how it differs/improves on the Skyhook system: This guy sounds a bit pissed off with Ducati. So does Tor on the video above. __________________ Whenever we are riding, we are an ambassador to our sport I'd rather be riding! 04-08-2013, 03:58 PM #231 jimmex Guero con moto     Joined: Jan 2003 Location: West Texas Oddometer: 2,523 It is painful watching that Tor guy try to ride a bike. __________________ Guero from Texas COHVCO AMA PAPA SJTR 04-09-2013, 02:49 PM   #232 MLCavassa What the..??!! Joined: Jul 2007 Location: Tualatin, Oregon Oddometer: 51 Quote: Originally Posted by jimmex It is painful watching that Tor guy try to ride a bike. I was thinking the exactly the same thing. He's, well, kinda sloppy. But then again, he is following another bike whose rider seems to be having his own set of issues - not the least of which is maintaining a constant speed. __________________ "A common mistake people make when trying to design something completely foolproof is to underestimate the ingenuity of complete fools." Douglas Adams 04-11-2013, 10:30 AM   #233 Jltrd Joined: May 2008 Oddometer: 27 Quote: Originally Posted by Paulvt1 Saw it in the metal today. Not impressed. It does look like a parts bin lash up. The luggage mounts look like Meccano from the 80's. It's probably going to be awhile before I can see one in person. Could you expand on the "parts bin lash up" comment? Thanks Sent from my SCH-I535 using Tapatalk 2 04-11-2013, 11:36 AM   #234 Pampero Joined: Jul 2012 Location: Seattle, WA Oddometer: 592 Quote: Originally Posted by Paulvt1 Saw it in the metal today. Not impressed. It does look like a parts bin lash up. The luggage mounts look like Meccano from the 80's. The luggage looks very similar to what was fitted to the last generation of Capos. It works well enough in a rather pedestrian way; not the best design and something they could have improved on (I do think it's virtually the same box, right down to the cut out moldings) but if Givi build doesn't set the world on fire, there's always the option of going to the aftermarket. Something I do like about the luggage is that it is full sized on both sides, an improvement over the Multistrada's exhaust cut outs. No doubt it will be reasonably priced in the package as well. Nobody really beats BMW at this game. From outward appearances they did better with the Stelvio's package, and this one (on the new offering) probably does fit the "parts bin" description. But if it's a parts bin special, it really makes no difference as long as they are good parts. How many boxers get the same motor? How many Audis use the 2.0TSi? To my way of thinking, the quality of the parts they pulled from the bin will be the key as a practical matter. A clean sheet design would be great, but this has enough innovation and the packaging looks up to the job. If they managed some manufacturing efficiencies they pass along to the market, well and good. If not, I think your criticism will be perfectly valid. Pampero screwed with this post 04-11-2013 at 11:41 AM 04-11-2013, 04:07 PM   #235 Croak Joined: Jun 2008 Location: Vancouver, BC/Is-Swieqi, Malta Oddometer: 529 MCN first ride: Quote: In short, the Caponord is one of the most impressive bikes I’ve ever ridden in terms of comfort and ride quality. It feels like the rear is on a magic carpet of smoothness. http://www.motorcyclenews.com/MCN/Ne...rd-first-ride/ As for the parts bin, beyond the panniers it has a new dash, a new suspension, new bodywork, new wheels, new tires, new subframe, new tank, new headlights, new tail lights, new seat, new exhaust, new charging system, new sensor suite, new sprockets, new handlebar and yoke, new windscreen, new electronics, and a new intake mated to a re-worked top end. About all it shares with the 'Duro is the trellis portion of the frame, bottom end and transmission, and some of the same switchgear shared across the Piaggio line. You'd see about the same degree of commonality between a Tiger 1050 and a Speed Triple, a R1200GS and a R1200R, etc. __________________ 2003 Aprilia Tuono 2002 Triumph Sprint ST 2013 Kymco Agility City 200i 04-13-2013, 04:36 AM   #236 Moronic Joined: May 2006 Location: Perth, Australia Oddometer: 1,568 Quote: Originally Posted by Celtic Curmudgeon Sorry for the hijack, but is there an explanation why metric fuel consumption is quoted "liters per 100km"? I'd think that "km per liter" would be easier to relate to and convert to/from MPG. L/100km requires math skills, and I'm a social science major forfecksake... I can remember finding L/100km hard to get used to but I have come to prefer it. Seems to make it easy to work out how far you can go on a tank. However, I'm not sure it is universally the metric format. Over in Orange Crush, where the new KTM 1190 is on sale in Europe while the rest of us wait, several helpful early adopters seem to prefer quoting km per litre. Back to the Caponord: it looks interesting and I like it that they have added electronic cruise. I would love to know more about how they amended the Skyhook damping program. But the auto rear preload looks to me like an expensive and complex answer to a question nobody asked. Which probably means I am not seeing straight. Probably, a high proportion of potential purchasers would have difficulty understanding where to set rear preload for different loads. OTOH, for me it is giving the bike control of something I really can control better myself. And probably, making me pay for that if I want other stuff that comes with the premium package. __________________ 04-13-2013, 02:09 PM   #237 jimmex Guero con moto Joined: Jan 2003 Location: West Texas Oddometer: 2,523 Quote: Originally Posted by Paulvt1 Just to clear things up - it looks more Kwak Versys 1000 than new GS. Whatever you think of the BM - it looks like a fully integrated package, from the switchgear to the luggage mounts - and no, i don't own one. The capo looks like a collection of parts and colours that have met in the dark. I saw the Triumph Tiger Sport in the metal for the first time. A much better looking machine - and a couple of thousand less too. Given that Piaggio have a tiny network of dealers over here - and an even worse representation in the US - the new bike is a non starter. Don't buy one then. "parts bin lash up" __________________ Guero from Texas COHVCO AMA PAPA SJTR 04-13-2013, 05:52 PM   #238 JohnG. Joined: Mar 2006 Location: Tumbarumba, NSW at the foot of the snowies Oddometer: 1,939 Quote: Originally Posted by Paulvt1 Just to clear things up - it looks more Kwak Versys 1000 than new GS. Whatever you think of the BM - it looks like a fully integrated package, from the switchgear to the luggage mounts - and no, i don't own one. The capo looks like a collection of parts and colours that have met in the dark. I saw the Triumph Tiger Sport in the metal for the first time. A much better looking machine - and a couple of thousand less too. Given that Piaggio have a tiny network of dealers over here - and an even worse representation in the US - the new bike is a non starter. sit back,will be good buying a few years on...after a few recalls __________________ JohnG. 04-13-2013, 08:38 PM #239 vivo Adventurer   Joined: Jan 2013 Oddometer: 93 When people say it looks like a parts bin special what does that tell you? Think on this.. a while. It means this bike does not SEEM genuine. It means it falls short of many enthusiasts expectations. Perceived qualities are more important than Engineeringtalk. We purchase based on how an object makes us FEEL about that object and ourselves. Truth is very dependent on perceived situation. The bike needs its own frame. This is a critical error as the frame used is too distinctively related to OTHER bikes. Parts bin is how the bike is perceived and that fault lies with Aprilia. I am sure its a good machine . Vivo 04-14-2013, 09:27 AM #240 jimmex Guero con moto     Joined: Jan 2003 Location: West Texas Oddometer: 2,523 I don't perceive it that way. I've never laid eyes on a Dorsowhatever so wouldn't know how similar they appear. The bike is full of new interesting technology and looks really cool IMO. __________________ Guero from Texas COHVCO AMA PAPA SJTR Share

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# Question: How Long Does A 1.9 Kg Chicken Take To Cook? ## How long does it take to cook a 1.35 kg chicken? Cooking Instructions Preheat oven to 180°C (360°F), if fan forced preheat oven to 160°C (320°F). Oven rack should be in the middle of oven. Remove chicken from bag prior to cooking and place in an oven proof dish. Cook at the preheated temperature for 75 – 80 minutes.. ## How many people does a 1.5 kg chicken serve? 4 peopleAn average-size chicken weighs about 1.5kg and will feed 4 people. If you’re cooking for 5 or 6, go for a 1.8kg-2kg bird. ## How long does a 1.5 kg chicken take? A 1.5kg chicken will be perfectly roasted after 1 hr 20 mins at 190C/fan 170C/gas 5. It doesn’t matter what you stuff into it, rub or sprinkle over it or put around it, this timing never changes. Remember this and you will always be able to roast a chicken. ## How long should you cook a chicken for? The right temperature and timeType of chickenWeightRoasting: 350°F (177˚C)breast halves, bone-in6 to 8 oz.30 to 40 minutesbreast halves, boneless4 oz.20 to 30 minuteslegs or thighs4 to 8 oz.40 to 50 minutesdrumsticks4 oz.35 to 45 minutes1 more row ## How long does it take to cook a crown of chicken? Instructions: 190°C/Fan 170°C/Gas 5 60 mins Place bagged crown in a deep roasting tray. Place in centre of a pre-heated oven for 40 minutes. Cut open a square panel and remove. Place back in oven for another 20 minutes. ## How long does it take to cook a 2kg chicken? Roasting a chicken will take approximately 20 minutes for every 500g of weight at 200 degrees Celsius (180 degrees Celsius for fan-forced ovens). So if you do the maths, it should take a 2kg chicken about 80 minutes (1 hour and 20 minutes) to cook. ## How long do you cook a chicken per kg? When it comes to cooking times, it couldn’t be easier to calculate – you’ll need 45 minutes per kg, plus an extra 20 minutes to finish. If you want to, you can baste your roast chicken once or twice during cooking to help keep it moist. ## How long does a 2.4 kg chicken take to cook? Roast in the centre of a pre-heated oven, gas mark 5, 375°F (190°C), for 20 minutes per lb (450 g) plus 10-20 minutes extra – this will be 1 hour and 50 minutes to 2 hours for a 5 lb (2.25 kg) bird. Baste three times during the cooking time. ## What temperature should I cook a chicken at? DirectionsPreheat oven to 350 degrees F (175 degrees C). Advertisement.Place chicken in a roasting pan, and season generously inside and out with salt and pepper. … Bake uncovered 1 hour and 15 minutes in the preheated oven, to a minimum internal temperature of 180 degrees F (82 degrees C). ## How long does it take to cook a 3.5 kg chicken? Brush the chicken with olive oil or melted butter, season with salt and pepper, then place it in the hot oven for about 15 minutes to brown the skin. Reduce the oven temperature to 180°C/gas mark 4, cover the chicken loosely with foil, and then leave it to cook for about 15 minutes per 450g in weight. ## Should you cover a chicken while roasting? Cover the chicken loosely with foil and roast for 30 minutes. … Reduce the temperature to 180C/160C Fan/Gas 4 and roast for a further 25 minutes, or until the chicken is cooked through. It’s fully cooked if the juices run clear when the bird is pierced in the thigh with a skewer. ## How long does a 2.5 kg chicken take to cook? Preheat oven to 180°c. Remove chicken from packaging and place in a baking dish. Calculate cooking time, allowing 25 minutes per 500g (a 2.5kg chicken will take 2 hours 5 minutes). Cook as per packet instructions. ## How long should I cook a 1.75 kg chicken for? The cooking time will vary depending on the weight of the chicken you buy. The rule of thumb is that you cook it for 20 minutes per 500g PLUS half an hour. So, for the chicken in the picture below which was 1.75kg, the total cooking time was 100 minutes (1 hour 40 minutes). ## How long does it take to cook 1.6 kg of chicken? Cooking the chicken As a rule, the roasting formula is 20 minutes per 450g plus an extra 20 minutes, which means a typical 1.5kg chicken will be perfectly roasted after 1 hour and 20 minutes at 200°C, 180°C fan, Gas Mark 6. ## Can you bake chicken at 200 degrees? Preheat oven to 200°F. Place dish in oven; roast chicken, skin side up, 2 hours and 45 minutes or until a meat thermometer registers 155°F in meatiest part of breast and at least 165°F in meatiest part of thigh. ## How many people does a 1.9 kg chicken feed? A small chicken will weigh approximately 1.2kg and will feed 2-3 people. A medium chicken weighs approximately 1.75kg and will feed 3-4 people, and a large chicken weighs approximately 2kg and will feed 4-6 people. ## How long does a 1.4 kg chicken take to cook? As a rule, the roasting formula is 20 minutes per 450g plus an extra 20 minutes, which means a typical 1.4 kg chicken will be perfectly roasted after 1 hour and 20 minutes at 200°C. ## How long does a 1.3 kg chicken take to cook? There are two separate time calculations. A properly raised free-range chicken takes 20 minutes per 450g, the rest about 12-15 minutes per 450g. If you are stuffing the bird, add 20 minutes. A 1.3kg bird will feed four.

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https://www.allaboutcircuits.com/technical-articles/designing-l-type-impedance-matching-networks-using-series-parallel-rc-lc-circuits/

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crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00033.warc.gz

Technical Article # Designing L-type Matching Networks Using Series and Parallel RC and RL Circuits April 14, 2023 by Dr. Steve Arar ## Learn about l-type impedance matching equations through the quality factor (Q factor) of RC and RL circuits and the series-parallel conversion of these circuits.l- In the previous article in this series, we learned how passive impedance matching networks can be designed using the Smith chart. Rather than using this graphical method, it is also possible to follow an analytical approach and use some equations to obtain the desired matching network. First, before diving into the impedance matching equations, we need to learn about two basic concepts: the quality factor of RC and RL circuits and the series-parallel conversion of these circuits. ### Quality Factor Definitions for RC and RL Circuits The term quality factor, denoted by Q, can be defined in a number of ways. When talking about an energy-storing device (i.e. a capacitor or inductor), Q shows how close to ideal our energy-storing device is. For example, an ideal capacitor doesn’t dissipate any energy and should have an infinite Q. Now, consider a real-world capacitor modeled as a parallel RC circuit (Figure 1(a)). ##### Figure 1. A parallel RC circuit (a) and a series RC circuit (b). In this case, the resistor can represent the resistive losses of the actual capacitor. For a parallel RC circuit, Q can be defined as: $Q_P=\frac{R_P}{\frac{1}{C \omega}}$ ##### Equation 1. Next, let’s see if this equation makes sense. In the circuit diagram of Figure 1(a), when RP goes to infinity, we’re left with an ideal capacitor (for which Q = ∞). This is consistent with the above equation that produces an infinite Q in the limit of infinite RP. Now consider a capacitor with a series resistor RS, as shown in Figure 1(b). In this case, Q is defined as: $Q_S=\frac{\frac{1}{C \omega}}{R_S}$ ##### Equation 2. In the series RC circuit, the ideal energy-storing device is obtained for R= 0. This is also consistent with Equation 2, which produces an infinite Q in the limit of R= 0. In Equation 1, Q is defined as the ratio of the undesired impedance (RP) to the impedance of the capacitive component, whereas Equation 2 defines Q as the ratio of the impedance of the capacitive component to that of the dissipative component. Using your intuition, how do you think the quality factor of a parallel RL (Figure 2(a)) and series RL (Figure 2(b)) is defined? ##### Figure 2. Parallel RL circuit (a) and series RL circuit (b). The quality factor of a parallel RL circuit (QP) and series RL circuit (QS) are respectively shown in Equations 3 and 4: $Q_P = \frac{R_P}{L \omega}$ ##### Equation 3. $Q_S=\frac{L \omega}{R_S}$ ##### Equation 4. Again, note that as RP goes to infinity, Equation 3 produces an infinite Q; and as RS goes to 0, QS tends to infinity. Keeping these extreme cases in mind, you can easily remember the correct equation for each configuration. ### RL and RC Circuit Series and Parallel Conversion A key step in designing an impedance-matching network is to transform a series connection of components to their equivalent parallel connection or vice versa. This is usually referred to as the series-parallel conversion or shunt-series transformation in the RF literature. As an example, consider the series RL circuit depicted in Figure 3(a) below. ##### Figure 3. A series RL circuit (a) and its equivalent RL circuit (b). At a single frequency, we can model the series RL circuit by an equivalent parallel RL circuit, as shown in Figure 3(b). By equating the input impedance of these two circuits (ZS=ZP), we can find the component values of the parallel circuit in terms of those of the series circuit. This produces the following set of equations: $R_P = R_S (1+Q^2)$ ##### Equation 5. $L_P = L_S (\frac{1+Q^2}{Q^2})$ ##### Equation 6. Where Q denotes the quality factor of the circuits (Equations 3 and 4 from the previous section). Note that the Q of the series RL circuit (QS) and that of the parallel equivalent circuit (QP) are the same at the frequency where the two circuits produce the same impedance. As a result, Equations 5 and 6 use the symbol Q rather than QS or QP to represent the quality factor. The above equations can be used to convert from the series to the parallel circuit or vice versa. Similarly, we can derive mathematical equations to convert a series RC circuit (Figure 4(a)) to its equivalent parallel circuit (Figure 4(b)) and vice versa. ##### Figure 4. Example series RC circuit (a) and its equivalent parallel circuit (b). Equation 5 also applies to the series-parallel conversion of RC circuits. However, keep in mind that now the Q term is the quality factor of the RC circuits (Equations 1 and 2). The capacitor values are related by the following equation: $C_P = C_S (\frac{Q^2}{1+Q^2})$ ##### Equation 7. The equations presented above have simple interpretations. Equation 5 shows that the parallel equivalent resistance is greater than the corresponding series resistance by a factor of (1+Q2). Also, assuming that the Q is relatively large, the reactive components of both the series and parallel circuits are of almost the same value. Let’s look at an example to clarify the above discussion. ### Example 1: Input Impedances of a Parallel and Series RC Circuit Assume that the input impedance of an RF block can be modeled by a parallel RC circuit with R= 50 Ω and C= 3.88 pF, as depicted in Figure 5 below. ##### Figure 5. Example parallel RC circuit RP = 50 Ω and CP = 3.88 pF. Find the equivalent series RC circuit at 1 GHz. By applying Equation 1, the Q of the parallel RC circuit is found: $\begin{eqnarray} Q_P &=& R_P C_P \times \omega \\ &=& 50 \times 3.88 \times 10^{-12} \times 2 \pi \times 1 \times 10^{9} \\ &=& 1.225 \end{eqnarray}$ Substituting Q = 1.225 into Equations 5 and 7, we have: $R_S = \frac{R_P}{1+Q^2}=\frac{50}{1+1.225^2}= 20 \text{ } \Omega$ $C_S = C_P \frac{1+Q^2}{Q^2}=3.88 \times 10^{-12} \times \frac{1+1.225^2}{1.225^2}= 6.47 \text{ } pF$ The equivalent series circuit is shown in Figure 6. ##### Figure 6. Equivalent series circuit with RS = 20 Ω and CS = 6.47 pF. The input impedance of the two circuits is plotted in a polar form in Figure 7. ##### Figure 7. The input impedance of a parallel and series RC circuit. A polar plot allows us to display the effect of both real and imaginary parts of the impedances by a single curve. Note that the impedances are equal only at a single frequency (at the intersection of the two curves that correspond to 1 GH in our example). At this frequency, the input impedances work out to Z= Z≃ 20 - j24.12 Ω. If we consider a narrow frequency range of around 1 GHz, we can assume that the two circuits have identical input impedances. ### Example 2: Designing a Matching Network Using Series-parallel Conversion Use the results of the previous example to design a matching network that transforms R= 50 Ω to 20 Ω at 1 GHz. In the previous example, we saw that a parallel RC circuit with R= 50 Ω and C= 3.88 pF is equivalent to a series RC circuit with R= 20 Ω and C= 6.47 pF. Based on this information, we can place a 3.88 pF capacitor in parallel with R= 50 Ω to produce the desired 20 Ω resistive part. We only need to eliminate the reactive component from the equivalent series capacitor (CS). The reactance of C= 6.47 pF is about -j24.12 Ω at 1 GHz. We can use a series inductor of 3.84 nH to produce a reactance of about +j24.12 Ω. This series inductor cancels the reactance of CS, leaving us with a purely resistive impedance of 20 Ω. The final matching network is shown in Figure 8. ##### Figure 8. Example matching network diagram. Now that we have established the concepts of quality factor and series-parallel conversion, we can discuss the analytical method of designing impedance-matching networks. ### L-type Impedance Matching of Two Resistive Terminations Two-element lossless matching networks, known as L-sections, are widely used in RF circuits. The circuit we designed in the above example (Figure 8) is actually an L-type matching network. Let’s now see how we can design these matching networks for arbitrary load (RL) and source (RS) impedances. To transform RL to RS, we can imagine two different types of solutions, as shown in Figure 9 below. ##### Figure 9. Example circuits of arbitrary load (RL) and source (RS) impedance solutions. In the above figure, the X components denote the reactance of the matching elements. For a given RL and RS, only one of the above solutions can be used. In order to determine the right choice, let’s apply the series-parallel conversion. This produces the following equivalent diagrams in Figure 10. ##### Figure 10. Two example equivalent diagrams after applying series-parallel conversation to Figure 9. The resistive part of the circuit in Figure 10(a) is (1 + Q2) times greater than RL, whereas the resistive part of the circuit in Figure 10(b) is equal to RL divided by (1 + Q2). This means that the circuit in Figure 9(a) transforms RL to a higher resistance; however, the circuit in Figure 9(b) transforms RL to a lower resistance. Therefore, when designing L-sections, the series component should be connected to the termination that has a smaller resistance. Consequently, the parallel component of the L-section should be connected to the termination that has a larger value. With the insight we’ve developed so far, we can pursue the following steps to design an L-section: 1. Name the termination with a larger value RHigh and the other one RLow. And use the following equation to calculate the quality factor of the circuit: $Q=\sqrt{\frac{R_{High}}{R_{Low}}-1}$ ##### Equation 8. 2. Place a series reactive element next to RLow and a parallel component next to RHigh. This produces two subnetworks: one series and the other one parallel, as shown in Figure 11. ##### Figure 11. Example series and parallel subnetworks. The above figure also provides the Q equation for the series and parallel subnetworks. At the frequency where the match occurs, the quality factor of the subnetworks is the same and equal to that given by Equation 8, i.e. Q= Q= Q. Note that, in the above circuit diagram, the input voltage source is actually zeroed because, here, the aim is to provide a match between the load and source impedances. The final circuit diagrams actually include a voltage source in series with either RLow or RHigh (Figure 12 below shows the final circuits). ##### Figure 12. Examples for matching two resistive impedances. 3. Having QS, QP, and the termination resistances, we can find the reactance values XS and XP. Finally, the inductor and capacitor values are calculated at the frequency of interest. Note that XS and XP can be a capacitor or an inductor; however, they are not of the same type. In other words, to transform a purely resistive load to another purely resistive impedance, we need an L-section consisting of an inductor and a capacitor. L-sections consisting of either two capacitors or two inductors cannot provide a match between two resistive impedances. Therefore, as shown below in Figure 12, there are a total of four different options when providing a match between two resistive impedances. Let’s look at an example. ### Example 3: Designing and Transforming a Matching Network Using Q Factor Design a matching network that transforms R= 10 Ω to 50 Ω at 1 GHz. By applying Equation 8, the required quality factor is found: $\begin{eqnarray} Q &=& \sqrt{\frac{R_{High}}{R_{Low}}-1} \\ &=& \sqrt{\frac{50}{10}-1}=2 \end{eqnarray}$ Placing the series component next to RL (the lower resistance), we obtain the following L-sections (Figure 13): ##### Figure 13. L-section diagrams after placing a series component next to RL. With the circuit in Figure 13(a), we have: $Q_S = \frac{X_S}{R_{Low}}=\frac{L_3 \omega}{R_{Low}}=2$ $Q_P = \frac{R_{High}}{X_P}=R_{High}C_3 \omega=2$ At 1 GHz, the above equations produce L= 3.18 nH and C= fc6.37 pF. For the circuit in Figure 13(b), we have the following set of equations: $Q_S = \frac{X_{S}}{R_{Low}}=\frac{1}{R_{Low} C_4 \omega}=2$ $Q_P = \frac{R_{High}}{X_{P}}=\frac{R_{High}}{L_4 \omega}=2$ At 1 GHz, the component values are C= 7.96 pF and L= 3.98 nH. ### L-type Matching Network Design Summary L-type matching networks can be designed using either the Smith chart-based graphical method or some simple equations. In this article, we derived the equations for the analytical method and looked at some examples. A key step in the design of impedance-matching networks is the series-parallel conversion of RL and RC circuits. It’s worthwhile to mention that series-parallel conversions are also helpful in analyzing component models to find the combined effect of different kinds of losses.

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CC-MAIN-2023-23

longest

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https://oeis.org/A229501

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crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00481.warc.gz

The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A229501 Numbers k such that Sum_{i=1..k} i' == 0 (mod k), where i' is the arithmetic derivative of i. 2 1, 6, 344, 1475, 3816, 5463, 18468, 78894, 515108, 566932, 1600370, 14380856, 27129564, 28669993, 31401775, 39638108, 2245196680, 2878016306, 5890364987, 7838325300, 23168759538, 63226475740, 121869542099 (list; graph; refs; listen; history; text; internal format) OFFSET 1,2 COMMENTS Next term > 10^7. - M. F. Hasler, Sep 25 2013 a(21) > 10^10. - Donovan Johnson, Sep 25 2013 a(24) > 10^12. - Giovanni Resta, Mar 13 2014 LINKS Table of n, a(n) for n=1..23. FORMULA A229501 = { n | A190121(n) = 0 (mod n) }. - M. F. Hasler, Sep 25 2013 EXAMPLE 1' + 2' + 3' + 4' + 5' + 6' = 0 + 1 + 1 + 4 + 1 + 5 = 12, and 12 mod 6 = 0. MAPLE with(numtheory); P:= proc(q) local a, n, p; a:=0; for n from 1 to q do a:=a+n*add(op(2, p)/op(1, p), p=ifactors(n)[2]); if a mod n=0 then print(n); fi; od; end: P(10^6); PROG (PARI) s=0; for(n=1, 1e7, (s+=A003415(n))%n||print1(n", ")) \\ - M. F. Hasler, Sep 25 2013 CROSSREFS Cf. A003415, A227848. Sequence in context: A144849 A212490 A047941 * A289738 A211089 A221923 Adjacent sequences: A229498 A229499 A229500 * A229502 A229503 A229504 KEYWORD nonn,more AUTHOR Paolo P. Lava, Sep 25 2013 EXTENSIONS Double-checked below 10^6 and extended up to 10^7 by M. F. Hasler, Sep 25 2013 a(12)-a(20) from Donovan Johnson, Sep 25 2013 a(21)-a(23) from Giovanni Resta, Mar 13 2014 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified September 7 11:06 EDT 2024. Contains 375730 sequences. (Running on oeis4.)

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CC-MAIN-2024-38

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https://usethinkscript.com/threads/custom-standard-deviation-channel-for-thinkorswim.2431/

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crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00052.warc.gz

Custom Standard Deviation Channel for ThinkorSwim Astro X Æ A-12 New member Hi, could someone help me to get something like this? it's so clear to see where we are in a trend (OB/OS). This guy mentioned that there were some Standard deviation channels there, but didn't tell us how he made this, any thoughts? Code: ``````# RegressionStdDeviation # DREWGRIFFITH15 (C) 2015 declare upper; input price = close; input deviations = 1.50; input fullRange = No; input length = 200; def regression; def stdDeviation; if (fullRange) { regression = InertiaAll(price); stdDeviation = StDevAll(price); } else { regression = InertiaAll(price, length); stdDeviation = StDevAll(price, length); } plot UpperLine = regression + deviations * stdDeviation; plot MiddleLine = regression; MiddleLine.Hide(); plot LowerLine = regression - deviations * stdDeviation; UpperLine.SetDefaultColor(Color.DARK_RED); MiddleLine.SetDefaultColor(Color.DARK_GRAY); LowerLine.SetDefaultColor(Color.DARK_RED); UpperLine.SetLineWeight(3); MiddleLine.SetLineWeight(1); LowerLine.SetLineWeight(3);`````` Found it: Code: ``````declare upper; input FirstStdDev = 1; input SecondStdDev = 2; input Price = close; input PlotCenterLines = yes; input PlotUpperLines = yes; def regression = InertiaAll(price); def stdDeviation = StDevAll(price); # # plot all values # plot firstUpperLine = regression + stdDeviation; plot firstLowerLine = regression - stdDeviation; plot secondUpperLine = regression + 2 * stdDeviation; plot secondLowerLine = regression - 2 * stdDeviation; plot middleLine = regression; # # set properties # firstUpperLine.SetDefaultColor(Color.ORANGE); firstLowerLine.SetDefaultColor(Color.ORANGE); secondUpperLine.SetDefaultColor(Color.UPTICK); secondLowerLine.SetDefaultColor(Color.UPTICK); middleLine.SetDefaultColor(Color.WHITE); secondUpperLine.SetHiding(!PlotUpperLines); secondLowerLine.SetHiding(!PlotUpperLines); # # plot dotted lines # plot line1 = regression + 1.5 * stdDeviation; plot line2 = regression + 0.5 * stdDeviation; plot line3 = regression - 0.5 * stdDeviation; plot line4 = regression - 1.5 * stdDeviation; line1.SetDefaultColor(Color.WHITE); line2.SetDefaultColor(Color.WHITE); line3.SetDefaultColor(Color.WHITE); line4.SetDefaultColor(Color.WHITE); line1.SetHiding(!PlotCenterLines or !PlotUpperLines); line2.SetHiding(!PlotCenterLines); line3.SetHiding(!PlotCenterLines); line4.SetHiding(!PlotCenterLines or !PlotUpperLines); line1.SetStyle(Curve.SHORT_DASH); line2.SetStyle(Curve.SHORT_DASH); line3.SetStyle(Curve.SHORT_DASH); line4.SetStyle(Curve.SHORT_DASH);`````` Member 2019 Donor It's a TOS study, just search for StandardDevChannel. mjamesb80 New member @Astro X Æ A-12 Here's one I use occasionally that combines the TOS Standard Deviation Channels 1 and 2 into a single script. The "2" standard deviations will change color when in an uptrend or downtrend. Note: these settings won't be the exact settings as Connor's (Boiler Room Trading) Code: ``````input price = close; input deviations1 = 1.0; input deviations2 = 2.0; input fullRange = Yes; input length = 21; def regression; def stdDeviation; if (fullRange) { regression = InertiaAll(price); stdDeviation = stdevAll(price); } else { regression = InertiaAll(price, length); stdDeviation = stdevAll(price, length); } plot UpperLine = regression + deviations1 * stdDeviation; plot MiddleLine = regression; plot LowerLine = regression - deviations1 * stdDeviation; plot UpperLine2 = regression + deviations2 * stdDeviation; plot LowerLine2 = regression - deviations2 * stdDeviation; UpperLine.SetDefaultColor(GetColor(8)); MiddleLine.SetDefaultColor(GetColor(8)); LowerLine.SetDefaultColor(GetColor(8)); #UpperLine2.SetDefaultColor(GetColor(8)); #LowerLine2.SetDefaultColor(GetColor(8)); UpperLine2.DefineColor("Up", GetColor(1)); UpperLine2.DefineColor("Down", GetColor(0)); UpperLine2.AssignValueColor(if UpperLine2 > UpperLine2[1] then UpperLine2.color("Up") else UpperLine2.color("Down")); LowerLine2.DefineColor("Up", GetColor(1)); LowerLine2.DefineColor("Down", GetColor(0)); LowerLine2.AssignValueColor(if LowerLine2 > LowerLine2[1] then LowerLine2.color("Up") else LowerLine2.color("Down"));`````` Last edited by a moderator: paraklesis New member I'd like to add a color coded up/down trend to the StandardDevChannel upperline and lowerline: for instance, if a ticker is in an uptrend on whatever timeframe being used, the upper and lower lines change to a color which can be set, such as green for up trend, and red for down trend. Been trying to code it myself but I'm not sure familiar with thinkScript's system yet. With my luck it's probably something simple and I've just been over complicating it to this point. Any help is appreciated. Astro X Æ A-12 New member @mjamesb80 OMG thank u so much! Thats really all i need!!! U are the genius New member here in this video Connor does explain's if you need more in detail. @Astro X Æ A-12 Joseph Patrick 18 Active member @mjamesb80 OMG thank u so much! Thats really all i need!!! U are the genius Hey Astro I think I cracked exactly what you were looking for because I was also looking for same thing...I've included a pic below with what you were looking for to exactly the specs of Connors video...I am also attaching a link for my WORKSPACE to give back to the community because I am not that technically advanced (newbie) ...so good luck and I hope I helped you out the same way everyone here as helped me..Please read what I wrote in pic to understand layout..any questions let me know. Here is the pic of one of the two flex grids (lmao I couldn't figure out how to put pic in from Imgur). READ WHAT I WROTE IN PICTURE.. Here is a photo of my WORKSPACE...2 SCREENS 2 MONITORS And here is my THINKORSWIM "FINAL WORKSPACE" LINK... http://tos.mx/zLjmW7R GOOD LUCK Joe Just click on my WORKSPACE and then in THINKORSWIM save it with a name that you want then got to top right corner of TOS and click on SETUP (with little gear icon) then scroll down to the name that you just saved and then my setup will come up...if any of the charts are black just scroll to the left I have like 300 to the right in settings...hope it helps Also I just put detailed instructions so that if there are people who are newbies like me it will help them out a little bit. Last edited by a moderator: JoeSD New member VIP @Joseph Patrick 18 I saw your June 2020 post re. Boiler Room and the TOS chart set-up you put together. I would love to try and mimic that if possible. I tried to use the workspace link you provided but it was nothing like the one in the top photo you provided. I particularly like the Squeeze colored candles and your initial 7/7. Could you send me a link to your grid with studies in the TOP photo from your post ( the one with the light blue writing)? Thanks. Joe Joseph Patrick 18 Active member @Joseph Patrick 18 I saw your June 2020 post re. Boiler Room and the TOS chart set-up you put together. I would love to try and mimic that if possible. I tried to use the workspace link you provided but it was nothing like the one in the top photo you provided. I particularly like the Squeeze colored candles and your initial 7/7. Could you send me a link to your grid with studies in the TOP photo from your post ( the one with the light blue writing)? Thanks. Joe Hey Joe...I think I did this right...sorry about that...just let me know...this is the flex grid with Standard deviation channel..Read the Blue writing up above it explains it. Top and bottom chart are set to 1 day you can change to 5 or whatever you want. Top chart leave extended hours checked bottom unchecked and hence the diff deviations. The one with blue writing is actually the bottom (last pic) right side of chart one on top of each other. http://tos.mx/vPiMnXd --------------------------------------------------------------------- this is the last pic also but the chart on the left side of chart http://tos.mx/0xW4bGs --------------------------------------------------------------------- this is the second pic down http://tos.mx/TBJvTHZ ---------------------------------------------------------------------- I also added a tick chart...just start studying them...instead of say on a minute chart it draws a candle every minute...the tick chart will show a candle every time (I set it to 333 but you could put say 2000) 333 trades are executed... I put a 21 day moving average exponential in blue...when the candles break above and pretty much stay above that is a buy signal and sometimes a huge upswing...on the flip side if it goes below blue its a sell or short signal and again depending on market and stock it could be a long short just put the gear icon on say 5 days (highest you can go on tick) scroll back and look at past days...also time on bottom of chart is different than say a 5 min chart where it will show 9:45 9:50 9:55...tick chart say it does 333 ticks (executed trades)fast as market opens you might see time on bottom 9:30 9:32 9:33 but when it slows down say lunch time the time on bottom might be 1:05 1:09 1:13 1:14 so at 1:05 333 trades happened then a candlestick appeared...the next candlestick didn't appear till 1:09 because the trades on that stock slowed down and it took 4 minutes to execute 333 trades...different animal compared to time but you get a more in depth look at volume and volatility. and a Hull Moving Avg I adjusted to 9 and it lines up with the candles. (so for example (you can change colors) when candle is going up green you will see the hull line cyan blue color and when the candle is red going down the Hull color is magenta.. and on the bottom is a volume zone oscillator (its A TOS study) if you click on study and open it up it explains the buy and sell signals for the wave, zero lines +60 +40 etc Here it is: http://tos.mx/bYaY4BM Let me know if you got everything okay..hope I could help...if you have anything interesting you want to share feel free...I am always open to learning new things.. Thanks good luck Joe I checked all of them they worked for me... JoeSD New member VIP Joe, just saw your reply and thanks. . I will take a look and try and implement. Stay well. Joe Joseph Patrick 18 Active member Joe, just saw your reply and thanks. . I will take a look and try and implement. Stay well. Joe Okay you also. Sammy800 New member VIP Astro practically what do you think is better for accuracy, tick chart with 333 or 2000. And do you think it gives a better picture compared to the candlestick charts. Its just your practical opinion. If I want to day trade and at times swing trade would a tick chart be better? Thanks Ronathan Edwards New member I have the deviation channels on my charts... Love it... Is there a way to set up a chart alert when the price goes, let's say "under the 1st lower line" ? When I set up a condition for a study in the condition wizard, when I look for standard deviation channel, it only has standard deviation, and the plot only says StdDev and will not let me chose a lower line (like in the scanner). Well-known member VIP I have Chart Labels that display when my Acrylic_VWAP study SD lines are crossed so, yes, it can be done... SDB2-X in the image below... Last edited: geremyh New member i did search for code, and or examples of code i could put in, but no success. I would like for these lines to extend right all the time, even when i change time frames...i've watched hours of Python tutorials on youtube and i understand the basics of what i'm looking at generally, but enhancing code is still beyond my skillset. I found this script right here on this site and love it...but i need it to extend to the right, so i can guesstimate future Std positions. any help appreciated Code: ``````input price = close; input deviations1 = 1.0; input deviations2 = 2.0; input fullRange = Yes; input length = 21; def regression; def stdDeviation; if (fullRange) { regression = InertiaAll(price); stdDeviation = stdevAll(price); } else { regression = InertiaAll(price, length); stdDeviation = stdevAll(price, length); } plot UpperLine = regression + deviations1 * stdDeviation; plot MiddleLine = regression; plot LowerLine = regression - deviations1 * stdDeviation; plot UpperLine2 = regression + deviations2 * stdDeviation; plot LowerLine2 = regression - deviations2 * stdDeviation; UpperLine.SetDefaultColor(GetColor(8)); MiddleLine.SetDefaultColor(GetColor(8)); LowerLine.SetDefaultColor(GetColor(8)); #UpperLine2.SetDefaultColor(GetColor(8)); #LowerLine2.SetDefaultColor(GetColor(8)); UpperLine2.DefineColor("Up", GetColor(1)); UpperLine2.DefineColor("Down", GetColor(0)); UpperLine2.AssignValueColor(if UpperLine2 > UpperLine2[1] then UpperLine2.color("Up") else UpperLine2.color("Down")); LowerLine2.DefineColor("Up", GetColor(1)); LowerLine2.DefineColor("Down", GetColor(0)); LowerLine2.AssignValueColor(if LowerLine2 > LowerLine2[1] then LowerLine2.color("Up") else LowerLine2.color("Down"));`````` BenTen Staff VIP @geremyh I don't think it's possible, but hopefully someone else can do it. geremyh New member @geremyh I don't think it's possible, but hopefully someone else can do it. I found this short script that does extend....i just don't know how to change all that needs to be changed to utilize it in my script request...a whole new function has to be defined i think using new attributes and i just don't have the know-how yet Code: ``````input aP = {default week, month, quarter}; input length = 1; input displace = 1; input showOnlyLastPeriod = no; def lastBar = IsNaN(close[-1]) and !IsNaN(close); def beyondLastBar = IsNaN(close[-1]) and IsNaN(close); #this extends it past last bar to the right def PrevDayClose = if !beyondLastBar then Highest(close(period = aP), length)[1] else PrevDayClose[1]; def closeSource = if (lastBar and beyondLastBar) then closeSource[1] else PrevDayClose; plot CloseLine = if showOnlyLastPeriod and !IsNaN(close(period = aP)[-1]) then Double.NaN else closeSource; CloseLine.SetDefaultColor(color.white); CloseLine.SetPaintingStrategy(PaintingStrategy.horizontal); def PrevDayOpen = if !beyondLastBar then Highest(open(period = aP), length)[1] else PrevDayOpen[1]; def openSource = if (lastBar and beyondLastBar) then openSource[1] else PrevDayOpen; plot OpenLine = if showOnlyLastPeriod and !IsNaN(open(period = aP)[-1]) then Double.NaN else openSource; OpenLine.SetDefaultColor(getColor(1)); OpenLine.SetPaintingStrategy(PaintingStrategy.DASHES);`````` BenTen Staff VIP @geremyh I think both are entirely different. Your script relies on the latest data while the other script projects data from the previous day. I hope that makes sense. geremyh New member figured it out, works in any time frame. Code: ``````input price = close; input deviations1 = 1.0; input deviations2 = 2.0; input deviations3 = 3.0; input fullRange = Yes; input length = 21; def regression; def stdDeviation; if (fullRange) { regression = InertiaAll(price, extendToRight = yes); stdDeviation = stdevAll(price, extendToRight = yes); } else { regression = InertiaAll(price, length); stdDeviation = stdevAll(price, length); } plot UpperLine = regression + deviations1 * stdDeviation; UpperLine.SetPaintingStrategy(PaintingStrategy.DasHES); plot MiddleLine = regression; MiddleLine.SetPaintingStrategy(PaintingStrategy.DasHES); plot LowerLine = regression - deviations1 * stdDeviation; LowerLine.SetPaintingStrategy(PaintingStrategy.DasHES); plot UpperLine2 = regression + deviations2 * stdDeviation; UpperLine2.SetPaintingStrategy(PaintingStrategy.DasHES); plot LowerLine2 = regression - deviations2 * stdDeviation; LowerLine2.SetPaintingStrategy(PaintingStrategy.DasHES); plot LowerLine3 = regression - deviations3 * stdDeviation; LowerLine3.SetPaintingStrategy(PaintingStrategy.DasHES); plot UpperLine3 = regression + deviations3 * stdDeviation; UpperLine3.SetPaintingStrategy(PaintingStrategy.DasHES); UpperLine.SetDefaultColor(GetColor(6)); MiddleLine.SetDefaultColor(GetColor(3)); LowerLine.SetDefaultColor(GetColor(6)); #UpperLine2.SetDefaultColor(GetColor(8)); #LowerLine2.SetDefaultColor(GetColor(8)); UpperLine2.DefineColor("Up", GetColor(4)); UpperLine2.DefineColor("Down", GetColor(4)); UpperLine2.AssignValueColor(if UpperLine2 > UpperLine2[1] then UpperLine2.color("Up") else UpperLine2.color("Down")); LowerLine2.DefineColor("Up", GetColor(4)); LowerLine2.DefineColor("Down", GetColor(4)); LowerLine2.AssignValueColor(if LowerLine2 > LowerLine2[1] then LowerLine2.color("Up") else LowerLine2.color("Down")); UpperLine3.DefineColor("Up", GetColor(5)); UpperLine3.DefineColor("Down", GetColor(5)); UpperLine3.AssignValueColor(if UpperLine3 > UpperLine3[1] then UpperLine3.color("Up") else UpperLine3.color("Down")); LowerLine3.DefineColor("Up", GetColor(5)); LowerLine3.DefineColor("Down", GetColor(5)); LowerLine3.AssignValueColor(if LowerLine3 > LowerLine3[1] then LowerLine3.color("Up") else LowerLine3.color("Down"));``````

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# BLOG ## Ok, I Think I Understand Acceleration Formula Physics, Now Tell Me About Acceleration Formula Physics! The conceptual introduction is completed. The energy an object has on account of its motion is named KE. It’s also highly advised to go to the Visualizing http://www.ifypz.com/archives/3672 Motion section. Sign and Direction Perhaps the most crucial issue to note about such examples is the symptoms of the answers. A heavy body moving at a speedy velocity is hard to stop. This is called deceleration. But it’s a little less obvious for acceleration. The acceleration is identical as time passes. Instantaneous acceleration refers to the typical acceleration within a little time. Determine the normal speed and the typical velocity. However it is one which can be readily misunderstood. These costs change on a daily basis, so it’s going to be valuable to be in a position to swiftly and easily alter the expenses of the produce without needing to change any of your complete cost formulas. This calculator can help you to solve all kinds of uniform acceleration troubles. That’s because lift offers upward push once it starts. For instance, when you accelerate, weight shifts toward the rear of the vehicle onto the rear wheels. Contrary to what you may have heard, it’s the acceleration that produces the ride interesting. Let’s look at a fast example to comprehend the difference between speed and acceleration. After the object is on or close to the surface of the human body, the force of gravity acting on the object is nearly constant and the next equation may be used. The key to all these funs is the way the goo balls are governed by the identical physical laws present in our world. At the middle of the earth, depth from the earth’s surface is equivalent to the radius of the planet. This difference is due to the air resistance on every object. Each of both bodies experiences exactly the same force directed towards the other. ## Choosing Acceleration Formula Physics Is Simple I have to pause here briefly to impose one particular limitation. The investigations might be laboratory based or they may take advantage of simulations and data bases. Notice the words distance and displacement are the sole difference between the 2 definitions. A simulation is frequently the simplest approach to calculate such outcomes. This may appear to be lots of trouble, but nevertheless, it can be quite helpful if your constants change values often. For instance, you will observe that a lot of the physics formulas you will encounter this year make usage of several constants like g, the acceleration due to gravity. This equation appears to be logical enough. These equations are called kinematic equations. ## The Benefits of Acceleration Formula Physics It was learned in the past portion of this lesson a free-falling object is an object that’s falling below the sole influence of gravity. A massive shift in position in a little enough time usually means that the object has high velocity, even though a small shift in position over a bigger quantity of time usually means the object has a decrease velocity. Imagine you’re pushing a heavy box throughout the room. In different cases it isn’t so fine. After all, it’s the force that’s helping us to continue to keep our feet on the ground. ## The Basics of Acceleration Formula Physics Because the fluid is incompressible, the exact same quantity of fluid must flow past any location in the tube in a particular time to guarantee continuity of flow. It is an excellent notion to test to be certain that every constant was properly named. To fix that issue, we do a before and following equation. There are four main constant acceleration equations you’ll want to fix all problems in this way. In fact, things may be somewhat different because of factors we’re neglecting at the present time. There’s nothing new here. Another kind of question is known as an inverse-square law question. Provided that you’re consistent within an issue, it doesn’t matter. As a result of this a big problem everybody is facing while doing trading between the nations. If there’s a net force, there’ll also be an acceleration. This example illustrates acceleration as it is normally understood, but acceleration in physics is considerably more than simply increasing speed. For instance, it would be helpful to understand how linear and angular acceleration are related. For example, in rocket thrust. They may also be employed to inspect the reasonableness of more precise calculations. The above mentioned formula represents a shortcut approach to determining the normal speed of an object. This isn’t the case anymore with special relativity in which velocities are determined by the selection of reference frame. Enter the value that you would like to convert Measurement is really the most important part of our life. The conceptual introduction is completed. You don’t need to select the origin or the positive directions this manner. We are likely to believe the Equation of Love doesn’t represent the acceleration of time. Given that we are living in a 3 dimensional universe where the only constant is change, you might be tempted to dismiss this section outright. ## Acceleration Formula Physics Secrets Likewise Tf is the last time frame whilst T0 is the initial time period. And fortunately, there’s a shortcut. On the flip side, displacement is a vector quantity, and it’s the shortest possible distance between the beginning and end point. Ok, I Think I Understand Acceleration Formula Physics, Now Tell Me About Acceleration Formula Physics! The conceptual introduction is completed. The energy an object has on account of its motion is named KE. It’s also highly advised to go to the Visualizing http://www.ifypz.com/archives/3672 Motion section. Sign and Direction Perhaps the most crucial issue to note […]

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> ### A sample text widget Etiam pulvinar consectetur dolor sed malesuada. Ut convallis euismod dolor nec pretium. Nunc ut tristique massa. Nam sodales mi vitae dolor ullamcorper et vulputate enim accumsan. Morbi orci magna, tincidunt vitae molestie nec, molestie at mi. Nulla nulla lorem, suscipit in posuere in, interdum non magna. Published on December 23rd, 2010 In category Education | Maths Search Similar Topic: , , , ; # Concept of Probability for competitive exams Concept of Probability Here in this post I have included some basic concept of probability that will be helpful to understand the concept of Probability for answering questions in competitive examination: Probability of any event to occur is always given by  n/N and   Not to  occur  is (n-1)/N where n is number of possible occurences and N is total number of occurences. for example the probability of a head when a coin is tossed is 1/2. As there can be total of 2 occurences either head or tail. Some of the typical questions asked in probability section are about : Dice : The probability of any outcome = 1/6, for ex. the probability of 5 in throw of dice is 1/6 Deck of Card:  A deck of the card has 52 card, If the question is asked what is • Probability of a Jack when card the withdrawn from deck =  4/52,  if it is said • What is probability to withdraw a Jack of Spade from the deck = 1/5 Marbles in bowl:   Probability of drawing a x colored marble from a set of the marbles, •           P = (no. total x color marbles)/ Total no. of marbles in the bowl Important Note:   In case of cards and marbles an important concpet of REPLACING the marbles or cards back • If the marbles are NOT replaced the P=   (no. total x color marbles- 1 )/( Total no. of marbles in the bowl – 1) Concept of Independent Events, Mutually Exclusive and cominations of both events Independent Event : If the two event do not effect the outcome of each other they are said to independent, for ex. the probability in case of toss with a coin and a throw of the dice. Probability of an outcome of head with number for 4. •           Probability of independent event  = p1 x p2= Where P1 and P2 are probabilities of 2 events Mutually Exclusive Event : If two events can not happen at a time, for example what is probability that a card drwan from a deck will red or spade. If a card is drawn it can be either red or black, a red card can’t be spade. Or what is probability that a dice outcome will be less than 3, this mean the outcome can be 1 or 2 but both can’t occur together •           Probability of Mutually exclusive event : P1 + P2 and so on  (P1 and P2 is the probability of  each event) NON Mutually Exclusive Event : If the two events can happen at time, for ex. probability of drawing a Red Card or King from a deck are non mutually exclusive event, as there are 2 Red card that are Kings, Heart King and Diamond King •         Probability of non- mutually exclusive event is :  P1 + P2 – ( Proabability of P1,P2 together) Combination of Mutually Exclusive and Independent Events:  Many times there is a combination of events for ex. what is proabability of getting 4 heads when coins is tossed 5 times.   Proability of in such cases if found out by: • Find out total number of combination possible this typically found out by cmobination formula : 5C2 • Probability of an head outocme – (1/2) ^ 4 • Probability of  an tail outcome  –  (1/2) ^ 1 • So the total outcome is – 5C2 x (1/2) ^ 4 x (1/2) ^ 1 If you liked my contribution, please donate or click on any of advt. a little that will be generated will all be donated for noble cause. [paypal-donation] [ad#Textads-BetPost]

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AP Computer Science 1995 Exam, Free-response in C++ # AP Computer Science1995 Exam, Free-response in C++ Owen Astrachan This document is a good-faith attempt at translating the free-response questions of the 1995 Advanced Placement Computer Science examinations from Pascal to C++. This document is NOT an official publication of Educational Testing Service or the College Board. This document was generated using the LaTeX2HTML translator Version 0.6.4 (Tues Aug 30 1994) Copyright © 1993, 1994, Nikos Drakos, Computer Based Learning Unit, University of Leeds. The command line arguments were: latex2html examc++.tex. The translation was initiated by Owen L. Astrachan on Fri Jul 7 15:00:10 EDT 1995 ### A exam Problem 1 Assume that names are stored in variables of the (standard) apstring class. Recall that the member function length() returns the number of characters in a string, and that the indexing operator [] can be used to access individual characters in a string. (note to reader: as an alternative, we could supply part of a string class, or mimic the exam and provide a record-like definition as shown below.) ``` const int MAXLENGTH = < some positive integer >; struct string // mimics Pascal version, NOT a C++ class { apvector<char> letters; int length; // number of characters stored in letters };``` part A Write the function CommaPosition whose header is given below. If there is no comma in Name, then CommaPosition returns -1. Otherwise, CommaPosition returns the index, or position, of the comma in Name. Assume that there is at most one comma in Name. For example: ``` Name CommaPosition(Name) W, Jeff 1 Smith, 5 Jones -1 , 0 Doe,Roxanne 3 ,Susan 0 ``` Complete function CommaPosition below the following header: ``` int CommaPosition(const apstring & Name) // precondition: Name contains at most one comma. // postcondition: returns -1 if there is no comma in Name // or returns the position/index of the comma in Name ``` part B Write the function PrintFirstLast whose header is given below. PrintFirstLast should print the name(s) stored in parameter Name as follows: If there is no comma in Name then nothing is printed; otherwise, PrintFirstLast should print all characters after the comma, if any, followed by a space, followed by all characters before the comma, if any. For example: ``` Name Output of PrintFirstLast W,Jeff Jeff W Smith, Smith Jones , Doe,Roxanne Roxanne Doe ,Susan Susan ``` In writing PrintFirstLast you may call the function CommaPosition of part (a). Assume that CommaPosition works as specified, regardless of what you wrote in part (a). Complete function PrintFirstLast below the following header. ``` void PrintFirstLast (const apstring & Name) //precondition: Name contains at most one comma``` Problem 2 In parts (a) and (b) of this problem you may use function Swap, which interchanges the values of its two parameters. ``` void Swap(int & a, int & b); // precondition: a == avalue, b == bvalue // postcondition: a == bvalue, b == avalue ``` part A Write function ReverseArray, whose header is given below. Function ReverseArray rearranges the elements in a so that they are in reverse order. For example, the result of `ReverseArray(a,5)` is illustrated below. ```Before call to ReverseArray(a,5) After call to ReverseArray(a,5) +----+----+----+----+----+ +----+----+----+----+----+ | 61 | 34 | 18 | 99 | 73 | | 73 | 99 | 18 | 34 | 61 | +----+----+----+----+----+ +----+----+----+----+----+ ``` Complete function ReverseArray below the following header. void ReverseArray(apvector<int> & a, int numElts) // precondition: numElts = # of entries in a // postcondition: elements of a are reversed part B Write function ReverseVertical, whose header is given below. Function ReverseVertical rearranges the elements of one column of apmatrix m so that they are in reverse order. For example, the result of `ReverseVertical(m,2)`, where m.numrows() == 4, is illustrated below. ```Before call to ReverseVertical(m,2) After call to ReverseVertical(m,2) +----+----+----+----+ +----+----+----+----+ | 61 | 34 | 18 | 99 | | 61 | 34 | 35 | 99 | +----+----+----+----+ +----+----+----+----+ | 11 | 22 | 44 | 33 | | 11 | 22 | 77 | 33 | +----+----+----+----+ +----+----+----+----+ | 55 | 66 | 77 | 88 | | 55 | 66 | 44 | 88 | +----+----+----+----+ +----+----+----+----+ | 25 | 45 | 35 | 55 | | 25 | 45 | 18 | 55 | +----+----+----+----+ +----+----+----+----+ ``` Complete function ReverseVertical below the following header. void ReverseVertical(apmatrix<int> & m, int col) // precondition: 0 <= col < m.numcols() // the number of rows in m is m.numrows() part C Write function ReverseMatrix, whose header is given below. Function ReverseMatrix rearranges the elements in m, an N x N matrix, so that they are in reverse order. A matrix is reversed when both the rows are reversed i.e., (r_0, r_1, ... r_{n-1}) becomes (r_{n-1}, ... r_1, r_0) and the elements within each row are reversed, as was done in part (a). For example, the result of `ReverseMatrix(m)`, where `N = 4` is illustrated below. ```Before call to ReverseMatrix(m) After call to ReverseMatrix(m) +----+----+----+----+ +----+----+----+----+ | 61 | 34 | 18 | 99 | | 55 | 35 | 45 | 25 | +----+----+----+----+ +----+----+----+----+ | 11 | 22 | 44 | 33 | | 88 | 77 | 66 | 55 | +----+----+----+----+ +----+----+----+----+ | 55 | 66 | 77 | 88 | | 33 | 44 | 22 | 11 | +----+----+----+----+ +----+----+----+----+ | 25 | 45 | 35 | 55 | | 99 | 18 | 34 | 61 | +----+----+----+----+ +----+----+----+----+ ``` In writing ReverseMatrix, all array manipulations must be done using functions ReverseArray and ReverseVertical of parts (a) and (b). This means that no apvector variable can appear on the left-hand side of the assignment operator (=). You will receive no credit for part (c) if your code includes such assignment statements or calls to Swap. Assume that ReverseArray and ReverseVertical work as specified regardless of what you wrote for parts (a) and (b). Complete function ReverseMatrix below the following header. void ReverseMatrix(apmatrix<int> & m) // precondition: m.numrows() == m.numcols() Problem 3 Case Study problem Problem 4 Consider an abstract data type that represents a finite, nonempty sequence of integers. Each such sequence has a designated ``current position.'' These sequences are represented by variables of type Sequence. For example, variable S of type Sequence, might represent the sequence. 3, 26, 18, 11 18 where the underline indicates that the current position is the second position and that the integer 26 is at that position. Sequences may have an ``invalid'' current position. A sequence with an invalid current posiition is said to be ``ended.'' Assume that the following public member functions have been defined for the class Sequence. (More precise specifications are given before part (a).) ```IsEnded returns true if the current position is invalid; otherwise, returns false. SetToFirst sets the current position to be the first position CurrentValue returns the integer value at the current position. It is an error to make this call when the sequence is ended. Advance It is an error to call Advance when the sequence is ended. If the current position is the last position, then the current position is changed to be invalid Otherwise, the current position is changed to the position following the one that was current ``` The following example demonstrates the use of some of these operations in the definition of a function SumSeq that returns the sum of the integers in sequence S. int SumSeq(Sequence s) // postcondition: returns sum of all entries in s { int sum = 0; s.SetToFirst(); while (! s.IsEnded()) { sum += s.CurrentValue(); s.Advance(); } return sum; } class Sequence { public: Sequence(); // default constructor bool IsEnded(); // precondition: current position is c // postcondition: returns true if c is invalid; // otherwise, returns false int CurrentValue(); // precondition: current position is c, which is valid // postcondition: returns value at position c void Advance(); // precondition: current position is c, sequence has m elements // postcondition: if 1 <= c < m, then current position = c+1 // if c == m, then current position is invalid void SetToFirst(); // postcondition: current position is first position <other member functions> private: <appropriate declarations> }; part A Write the body of function Length whose header is given below. Length returns the number of integers in its sequence parameter. Note that Length is NOT a member function. ``` int Length(Sequence s) // precondition: s contains m elements // postcondition: returns m``` part B Write the function LenOfIncreasing, whose header is given below. LenOfIncreasing returns the length of the longest (strictly) increasing segment in s beginning with, and including, the integer at its current position. Examples: (the current position is indicated with an *) ```Sequence s Increasing Segment LenOfIncreasing(s) 3 6 11 6 5* 6 8 9 11 5 3 5 6 8 9 11 5 7 -1 2 5 8 6 6 7 7* 30 7 30 2 1 7 6 3 5* 9 9 9 10 10 4 5 9 2 3 5 8 9 12* 4 6 9 15 12 1 ``` Complete function LenOfIncreasing below the following header. ``` int LenOfIncreasing(Sequence s) // precondition: s represents n_1, n_2, ... , n_m; // current position of s is c; 1 <= c <= m // postcondition: returns the largest integer r between 1 and // m (inclusive) such that n_c < n_{c+1} < ... < n_{c+r-1} ``` ### AB exam Problem 1 (same as A2) Problem 2 This question involves reasoning about two implementations of of a data structure called a priority list. A priority list is a collection of items, each of which contains both data and a unique integer priority. The ``maximal'' item in a priority list is the element whose integer priority has the greatest value. Both data and an integer priority are specified when inserting an element into a priority list. Only the maximal element, the item with the highest integer priority, is accessible in a priority list. A priority list class Plist supports the following public member functions. ```Function Description Plist constructs and initializes an empty priority list IsEmpty returns true if the priority list is empty; otherwise returns false Insert(info,pri) inserts item with data info and priority pri FindMax(info,pri) sets info and prio to the data and priority of the maximal item DeleteMax deletes the maximal element ``` Consider the following two methods for implementing priority lists. Type declarations are given in part (b) of this problem. Method 1 A priority list is an unsorted linked list. The Insert operation inserts an item as the first node of the list. Method 2 A priority list is a linked list sorted by priority field from largest to smallest (so that the first node in the list is always the maximal item of the priority list). The Insert operation inserts an item so that the list remains sorted by priority. Precise specifications for the member functions described above are given below. ```Plist::Plist() //postcondition: represents an empty priority list bool Plist::IsEmpty() const //postcondition: returns true if priority list is empty (has 0 items); // otherwise, returns false void Plist::Insert(const DataType & info, int pri) // precondition: no item in priority list has priority pri // postcondition: A new item whose data field has value info and whose // priority field has value pri has been inserted void PQeue::FindMax(DataType & info, int & pri) const // precondition: IsEmpty() == false, (priority list is non-empty) // postcondition: the values of info and pri are those of the maximal item void Plist::DeleteMax() // precondition: IsEmpty() == false, (priority list is non-empty) // postcondition: the maximal item has been deleted``` part A Complete the table below, In each cell give the worst-case time complexity for the corresponding operation and implementation of a priority list of n items. Use O-notation and briefly justify each answer. ``` Method 1 Method 2 Plist Time: O(1) Reason: single assignment to pointer, e.g., first = NULL Insert FindMax Time: O(1) Reason: maximal item is at the front of the list; therefore, only a single access is needed. ``` part B In this part, you will write functions Insert and FindMax for one of the methods for implementing priority lists described at the beginning of this question. You MUST use the same method in writing both Insert and FindMax. It may be the case that it is easier to write code for one of the methods than for the other. Think carefully before choosing a method. Circle below the method you will use to implement Insert and FindMax. Method 1 Method 2 You must use the following type declarations to implement priority lists regardless of which method you chose. ``` class DataType; struct Node { DataType data; // data associated with item int priority; // unique priority associated with item Node * next; }; class Plist { public: Plist(); int IsEmpty() const; void Insert(const DataType &, int); void FindMax(DataType &, int &) const; void DeleteMax(); private: Node * myFirst; // pointer to first node in linked list };``` Complete functions Insert and FindMax below the following headers. ```void Plist::Insert(const DataType & info, int pri) // precondition: no item in priority list has priority pri // postcondition: A new item whose data field has value info and whose // priority field has value pri has been inserted``` ```void PQeue::FindMax(DataType & info, int & pri) const // precondition: IsEmpty() == false, (priority list is non-empty) // postcondition: the values of info and pri are those of the maximal item``` Problem 3 Case Study Problem Problem 4 Assume that binary trees are implemented using the following definitions. ```struct TreeNode { int info; TreeNode * left, * right; };``` part A Write function IsChild whose header is given below. IsChild returns true if the node pointed to be parameter second is a child (either left or right) of the node pointed to be parameter first, and returns false otherwise. ``` bool IsChild(TreeNode * first, TreeNode * second) // precondition: second != NULL // postcondition: returns true if second is a child of first // returns false otherwise``` part B Write the function IsDescendant whose header is given below. IsDescendant returns true if there is a path from the node pointed to by parameter first to the node pointed to by parameter second, where a path is a nonempty sequence of left or right children For example, consider the tree diagrammed below: ``` function call value returned IsDescendant(NodeC, NodeA) false IsDescendant(NodeA, NodeB) true IsDescendant(NodeB, NodeA) false IsDescendant(NodeC, NodeD) true IsDescendant(NodeA, NodeA) false ``` In writing IsDescendant you may call function IsChild of part (a). Assume that IsChild works as specified, regardless of what you wrote in part a. Complete IsDescendant below the following header. ``` bool IsDescendant(TreeNode * first, TreeNode * second) // precondition: second != NULL``` part C Write function ChangeTree, whose header is given below. ChangeTree removes from t all nodes that have exactly one child, replacing the removed node with the child. (All removed nodes should be returned to available storage.) If A and B are nodes in t that do NOT have exactly one child, and if IsDescendant(A,B) is true before the call ChangeTree(t), then IsDescendent(A,B) must be true AFTER the call ChangeTree(t). For example: In writing ChangeTree, you may call function HasOneChild, defined below. ``` bool HasOneChild(TreeNode * t) // postcondition: returns true if t has exactly one child, else returns false { if (t != NULL && // t exists (t->left == NULL && t->right != NULL || t->left != NULL && t->right == NULL)) { return true; } return false; }``` Complete function ChangeTree below the following header. ` void ChangeTree(TreeNode * & t)` ### Answers to C++ Free-response questions ```int CommaPosition(const apstring & Name) // precondition: Name contains at most one comma. // postcondition: returns -1 if there is no comma in Name // or returns the position/index of the comma in Name { int k; int len = Name.length(); // with struct def, use = Name.length for(k=0; k < len; k++) { if (Name[k] == ',') return k; } } void PrintFirstLast (const apstring & Name) //precondition: Name contains at most one comma { int k; int len = Name.length(); int comma = CommaPosition(Name); // index of comma if (comma != -1) { for(k=comma+1; k < len; k++) // print first name { cout << Name[k]; } cout << ' '; for(k=0; k < comma; k++) // print last name { cout << Name[k]; } } } void ReverseArray(apvector<int> & a, int numElts) // precondition: numElts = # of entries in a // postcondition: elements of a are reversed { int k; int limit = numElts/2; // half-way for(k=0; k < limit; k++) { Swap(a[k],a[numElts-k-1]); } } void ReverseVertical(apmatrix<int> & a, int col) { int k; int limit = a.numrows()/2; // half-way for(k=0; k < limit; k++) { Swap(a[k][col],a[N-k-1][col]); } } int ReverseMatrix(apmatrix & m) { int k; int dimension = m.numrows(); for(k=0; k < dimension; k++) { ReverseArray(m[k],dimension); } for(k=0; k < dimension; k++) { ReverseVertical(m,k); } } int Length(Sequence s) // precondition: s contains m elements // postcondition: returns m { int count = 0; // counter for # of elements of s s.SetToFirst(); while (! s.IsEnded()) { count++; } return count; } // alternate version using for-loop int Length(Sequence s) // precondition: s contains m elements // postcondition: returns m { int count = 0; // counter for # of elements of s { count++; } return count; } int LenOfIncreasing(Sequence s) { int lastVal; int length = 1; // length at least one if (! s.IsEnded()) { lastVal = s.CurrentValue(); while (! s.IsEnded() && lastVal < s.CurrentValue()) { length++; lastVal = s.CurrentValue(); } return length; } else { return 0; // no increasing seq, no elements } } // method 1 void Plist::Insert(const DataType & info, int pri) { Node * temp = new Node; temp->data = info; temp->priority = pri; temp->next = myFirst; // points to old first node myFirst = temp; // update to new first node } // method 2 void Plist::Insert(const DataType & info, int pri) { Node * trav = myFirst; // used to traverse list Node * temp = new Node; // create, initialize temp->data = info; temp->priority = pri; // find where node belongs if (myFirst == NULL || pri > myFirst->priority) // special case, first node { temp->next = myFirst; myFirst = temp; } else // can look one ahead { while (trav->next != NULL && pri < trav->next->priority) { trav = trav->next; } // found where new node belongs, link it in temp->next = trav->next; trav->next = temp; } } // method 1 int Plist::FindMax(DataType & info, int & pri) const { Node * trav = myFirst; Node * max = myFirst; if (myFirst != NULL) // be safe, no NULL dereferences { trav = myFirst->next; // start at second node, max at first while (trav != NULL) { if (trav->priority > max->priority) { max = trav; } trav = trav->next; } info = max->data; // found max, set parameters pri = max->priority; } } // method 2 int Plist::FindMax(DataType & info, int & pri) const { if (myFirst != NULL) { info = myFirst->data; pri = myFirst->priority; } } bool IsChild(TreeNode * first, TreeNode * second) // precondition: second != NULL // postcondition: returns true if second is a child of first // returns false otherwise { return first && (first->left == second || first->right == second); } bool IsDescendant(TreeNode * first, TreeNode * second) // precondition: second != NULL { if (first != NULL) { if (IsChild(first,second) || IsDescendant(first->left,second) || IsDescendant(first->right,second)) { return true; } } return false; } void ChangeTree(TreeNode * & t) { if (t != NULL) { ChangeTree(t->left); // remove all one-child nodes ChangeTree(t->right); // from children if (HasOneChild(t)) { Node * temp = t; // will delete this node if (t->left != NULL) // move to to only child t = t->left; else t = t->right; delete temp; } } }``` Owen L. Astrachan

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# Unscramble CESMUTE CESMUTE unscrambles into 95 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT! ### 2 letter words made by unscrambling CESMUTE #### Unscrambled 11 2 Letter Words Above are the words made by unscrambling CESMUTE (CEEMSTU). To further help you, here are a few lists related to/with the letters CESMUTE ### The Value of CESMUTE In Word Scramble Games The letters CESMUTE are worth 11 points in Scrabble The letters CESMUTE are worth 14 in points Words With Friends • C = 4 points in WWF & 3 points in Scrabble • E = 1 points in WWF & 1 points in Scrabble • S = 1 points in WWF & 1 points in Scrabble • M = 4 points in WWF & 3 points in Scrabble • U = 2 points in WWF & 1 points in Scrabble • T = 1 points in WWF & 1 points in Scrabble • E = 1 points in WWF & 1 points in Scrabble ## What Does CESMUTE Mean... If you Unscramble it? ### Possible Definitions of CESMUTE If we unscramble these letters, CESMUTE, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters: ### tumesce • Sorry. I don't have the meaning of this word. ## Permutations of CESMUTE According to our other word scramble maker, CESMUTE can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word. #### Definition of Permutation a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged. How is this helpful? Well, it shows you the letters cesmute scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time CESMUTE comes up in a word scramble game. We stopped it at 50, but there are so many ways to scramble CESMUTE! ### Scramble Words Unscramble these letters to make words... scrambled using word scrambler...

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# Thread: Partial Fraction Decomposition 1. ## Partial Fraction Decomposition Find the partial fraction decomposition for the rational expression: (3x^3+4x)/(x^2+1)^2 I know that you will have: A/? + B/? + C/? + D/? , but I don't know where to go from there. Thanks! 2. Originally Posted by ss103 Find the partial fraction decomposition for the rational expression: (3x^3+4x)/(x^2+1)^2 I know that you will have: A/? + B/? + C/? + D/? , but I don't know where to go from there. Thanks! $\displaystyle \frac{Ax+B}{x^2+1}+ \frac{Cx+D}{(x^2+1)^2} = \frac{3x^3+4x }{(x^2+1)^2}$ Cross multiply by $\displaystyle (x^2+1)^2$ expand and equate coefficients wrt x. That will give you a system for A, B, C and D. 3. Hi The form of the decomposition is $\displaystyle \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$ EDIT : too late ! 4. Okay so my answer is (3x/x^2+1) + (x/(x^2+1)^2). Is that right? 5. Originally Posted by ss103 Okay so my answer is (3x/x^2+1) + (x/(x^2+1)^2). Is that right? I got that too.

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # ch25_p19 - 19(a and(b We note that the charge on C3 is q3 =... This preview shows page 1. Sign up to view the full content. 19. (a) and (b) We note that the charge on C 3 is q 3 = 12 µ C – 8.0 µ C = 4.0 µ C. Since the charge on C 4 is q 4 = 8.0 µ C, then the voltage across it is q 4 / C 4 = 2.0 V. Consequently, the voltage V 3 across C 3 is 2.0 V ¡ C 3 = q 3 / V 3 = 2.0 µ F. Now C 3 and C 4 are in parallel and are thus equivalent to 6 µ F capacitor which would then be in series with C 2 ; thus, Eq 25-20 leads to an equivalence of This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]}

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# Thread: Simplex Algorithm 1. ## Simplex Algorithm I dont understand what to do with this. 1) Represent the linear programing problem below by an initial Simplex Tableau Maximise P=15x-4y-4z Subject to 10x-4y+8z <(or equal to) 40 10x+6y+9z <(or equal to) 72 -6x+4y+3z<(or equal to) 48 and x>(or equal to) 0, y> (or equal to) 0, z> (or equal to) 0 2) perform one iteration of the simplex algorithm and write down the values of x,y, z and P that result from the iteration 3) Perform one further iteration of the simplex algorithm and find the values of x, y, z and P at the optimum point 2. Originally Posted by xxchloe741xx I dont understand what to do with this. 1) Represent the linear programing problem below by an initial Simplex Tableau Maximise P=15x-4y-4z Subject to 10x-4y+8z <(or equal to) 40 10x+6y+9z <(or equal to) 72 -6x+4y+3z<(or equal to) 48 and x>(or equal to) 0, y> (or equal to) 0, z> (or equal to) 0 2) perform one iteration of the simplex algorithm and write down the values of x,y, z and P that result from the iteration 3) Perform one further iteration of the simplex algorithm and find the values of x, y, z and P at the optimum point 1) Step 1. Introduce slack variables and restate the problem in terms of a system of linear equations. This gives the equations $\displaystyle \begin{array}{ll}10x - 4y + 8z + u &= 40,\\ 10x+6y+9z\phantom{{}+u} +v &= 72,\\ -6x+4y+3z \phantom{+u+v} +w &= 48.\end{array}$ Also, the objective equation is $\displaystyle -15x+4y+4z+P=0$. Step 2. Construct the simplex tableau corresponding to the system, with the bottom row of the matrix corresponding to the objective equation. The simplex tableau is a matrix whose columns represent the variables x,y,z,u,v,w,P, and the constants on the right-hand side of the equations. This gives the simplex tableau $\displaystyle \begin{bmatrix}10&-4&8&1&0&0&0&40\\ 10&6&9&0&1&0&0&72\\ -6&4&3&0&0&1&0&48\\ -15&4&4&0&0&0&1&0\end{bmatrix}$ . For 2) and 3), the simplex algorithm works like this: Step 3. (a) Choose the pivot column to be the one containing the most negative element on the bottom row of the matrix. (b) Choose the pivot element by computing ratios associated with the positive entries in the pivot column. The ratio is the element in the right-hand column divided by the corresponding element in the pivot column. The pivot element is the one corresponding to the smallest positive ratio. (c) Construct the new simplex tableau by pivoting about the selected element. If you are supposed to be able to do this problem then presumably you have seen examples of the simplex algorithm in action, and you ought to be able to carry it out here, and to read off the values of x,y,z and P that it produces.

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1 JEE Main 2022 (Online) 28th June Evening Shift Numerical +4 -1 A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ____________ N. (Take g = 10 ms$$-$$2) 2 JEE Main 2022 (Online) 28th June Morning Shift Numerical +4 -1 The position vector of 1 kg object is $$\overrightarrow r = \left( {3\widehat i - \widehat j} \right)m$$ and its velocity $$\overrightarrow v = \left( {3\widehat j + \widehat k} \right)m{s^{ - 1}}$$. The magnitude of its angular momentum is $$\sqrt x$$ Nm where x is ___________. 3 JEE Main 2022 (Online) 27th June Evening Shift Numerical +4 -1 A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $${1 \over 2}\sqrt {xgh}$$ m/s. The value of x is ___________. 4 JEE Main 2022 (Online) 25th June Evening Shift Numerical +4 -1 Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows: I1 = M.I. of solid sphere about its diameter I2 = M.I. of solid cylinder about its axis I3 = M.I. of solid circular disc about its diameter I4 = M.I. of thin circular ring about its diameter If 2(I2 + I3) + I4 = x . I1, then the value of x will be __________.

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2-groups with an odd-order automorphism that is the identity by Mazurov V.D. By Mazurov V.D. Similar symmetry and group books Rotations, quaternions, and double groups This precise monograph treats finite element teams as subgroups of the whole rotation team, delivering geometrical and topological tools which permit a distinct definition of the quaternion parameters for all operations. a huge characteristic is an uncomplicated yet complete dialogue of projective representations and their software to the spinor representations, which yield nice merits in precision and accuracy over the extra classical double workforce strategy. The theory of groups Maybe the 1st really recognized publication committed basically to finite teams was once Burnside's e-book. From the time of its moment variation in 1911 until eventually the looks of Hall's e-book, there have been few books of comparable stature. Hall's publication remains to be thought of to be a vintage resource for primary effects at the illustration conception for finite teams, the Burnside challenge, extensions and cohomology of teams, \$p\$-groups and masses extra. Additional info for 2-groups with an odd-order automorphism that is the identity on involutions Sample text 1. 31 Our general notation mainly concerns group theory — our standard reference being [28] — and homological algebra — our standard reference being [18]. In particular, if G is a finite group, recall that Op (G) , Op (G) , Op (G) and Op (G) respectively denote the minimal or the maximal normal subgroups of G with their index or their order being a power of p or prime ˆ except to p ; note that this notation still makes sense for a finite k ∗ -group G ˆ that Op (G) remains a p-group. For any pair of subgroups H and K of G , we denote by TG (K, H) the set of x ∈ G fulfilling xKx−1 ⊂ H . But, as a matter of fact, when dealing with contravariant functors a from F to Ab (cf. 3), they coincide (cf. 4). 7 F(b,G) is a Frobenius P -category. 1; thus, since p does not divide |NG (P, e)/P ·CG (P )| (cf. 1. Let Q be a subgroup of P , K a subgroup of Aut(Q) containing FQ (Q) and ϕ : Q → P an F(b,G) -morphism such that ϕ(Q) is fully ϕK-normalized in F(b,G) ; in particular, denoting by f the block of CG (Q) such that (P, e) contains (Q, f ) , there is x ∈ G fulfilling (cf. 2. As above, denote by g the block of CG NPK (Q) such that (P, e) contains the Brauer (b, G)-pair (NPK (Q), g) .

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Cody # Problem 129. All capital? Solution 534515 Submitted on 24 Nov 2014 by Sreeram Mohan This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass %% x = 'MNOP'; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct)) 2   Pass %% x = 'MN0P'; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct)) 3   Pass %% x = 'INOUT1NOUT'; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct)) 4   Pass %% x = 'UPANDDOWN'; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct)) 5   Pass %% x = 'RUaMATLABPRO'; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct)) ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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# Fractions with Finch In this activity, you will use the Finch to practice what you know about fractions. Start by writing a short program to make the Finch turn in a circle with one wheel at speed 0 and the other at speed 30. How many seconds does it take for the Finch to make exactly one circle? Next, think about making the robot move a fraction of a circle. Which fraction do you think is bigger, $\inline&space;\fn_phv&space;\large&space;\frac{3}{7}$ or $\inline&space;\fn_phv&space;\large&space;\frac{5}{6}$ ? Write down your prediction and your reasoning. Calculate the number of seconds for the Finch to turn $\inline&space;\fn_phv&space;\large&space;\frac{3}{7}$ of a circle. Then write a program to check your answer with the Finch. Calculate the number of seconds for the Finch to turn $\inline&space;\fn_phv&space;\large&space;\frac{5}{6}$ of a circle. Then write a program to check your answer with the Finch. Which fraction is bigger, $\inline&space;\fn_phv&space;\large&space;\frac{3}{7}$ or $\inline&space;\fn_phv&space;\large&space;\frac{5}{6}$ ? Was your prediction correct? For your next program, the Finch should make three smaller turns that make up one full circle. Each of the smaller turns must be $\inline&space;\fn_phv&space;\large&space;\frac{1}{3}$ of the circle. This program represents the problem shown below. $\inline&space;\fn_phv&space;\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ Solve the problem below, and write a Finch program to represent it. $\inline&space;\fn_phv&space;\large&space;\frac{1}{2}+\frac{1}{4}=?$ Standards: This activity is aligned with Common Core math standards 4.NF.A, 4.NF.B, and 5.NF.A.

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https://www.instructables.com/id/Tri-dimensional-Chess/

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# Tri-dimensional Chess 27,146 138 17 ## Introduction: Tri-dimensional Chess In Star Trek you see this cool chess set with 3 levels (plus 4 attack boards--more about those later). Here is how you can make one. ## Step 1: Step 1: Making the Frame For the frame, you'll need an old badminton racket and tennis racket (from the thrift store). Cut each racket with a hack saw and bend it to the curve desired. Its that easy! ## Step 2: Base, Boards and Supports The same thrift store that gave us the rackets gave us this jello mold for the base. Hot glue rocks to the inside to give weight. the little boards or attack boards have supports wich are made out of clear plastic pens youll need atleast four (you should probrally get more just  incase of a problem). ## Step 3: Making the Boards now cut of a card board chess board a four by four square section That is a main board. Then you measure the main board and cut out 3 squares of plexiglass the size of the main board. now cut out of a cardboad chess board a two by two square section that is the size of a attack board. measure that attack board and cut out four of the same size plexiglass squares. ## Step 4: Making the Chessboard Pattern Using the cardboad grid as a guide, mask the dark squares and then apply spraypaint. The attack boards are supposed to be movable, so I glued magnets in the board underside and to the bottom of the pen tubes. ## Step 5: The Final Frontier Now attach the boards to the frame. I used a combination of hotglue and bolts. You are done. For more copies just feed these instructions into your replicator. 83 8.2K 40 7.2K 32 5.9K 330 32K ## 17 Discussions There are several variants of the rules out there. I favor this set of rules and notation for learning, as it provides details and a java script for checking rule validity. http://www.thedance.net/~roth/TECHBLOG/chess.html Standard pieces have a 3.75” high King. Squares are 2.25” square. Note the comment about the overlaps. Some versions of the rules allow for six Attack Boards, and boards to invert – attach to the bottom as well as the top of the three main boards – so magnetic attachment would seem the way to go. The piece height is important because the attack board pedestals should be higher than that, and twice that spacing vertically between the mainboards.

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https://worldbuilding.stackexchange.com/questions/48897/seeing-into-the-past

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# Seeing into the past Story Background I have an alien that travels from ~100 light years away to Earth. He can make this trip in a matter of years though. He doesn't reveal how as he doesn't trust Earthlings since he has just come across new intelligent life. Now, I have a guy that has created a telescope that is stationary on Earth able see in nearly perfect detail at vast distances into space. This alien now approaches my guy and gives him the coordinates to his home planet. The alien wants to show Earth his planet and people and technology they use using this "super" telescope that can look onto the surface of his planet as if we were looking look out a second story window at them. Question What I have read is that the sun is 93 million miles from Earth and that it takes ~8 minutes for light to travel from the Sun to Earth. So we are basically seeing the Sun from Earth 8 minutes in the past. (Which blows my mind if you ask me) Now when the man and the alien finally look into the telescope to see his home planet, will the alien be unpleasantly surprised to see his home planet the way it was 100 years ago? (Will he see his newborn grandma?) Or does the ability to focus (zoom in?) on something so close with the telescope fix that issue and they will see the planet as basically when he left. • Zooming in is not really the same thing as moving the lens closer to the light source. You can only zoom in and expect good resolution if the resolution was already there, just microscopic. The resolution at the primary lens is what you get to play with, and yes, it's from the past. Commented Jul 28, 2016 at 22:22 • If you want a plausible answer, I would recommend removing the "I have a guy that has created a telescope that is stationary on Earth" our best pictures of the universe come from outside our atmosphere. While your alien bro might bring a terrestrial telescope capable of seeing his planet from Earth, we don't have that. – Pork Commented Jul 29, 2016 at 8:06 • We always see things as they were, when they emitted the light. It is a bit like on a big concert, if you are 300m away from the stage, you will hear the sound a second later than it is played! - So if we mounted a giant perfect mirror in space say 5 lightyears away, then we could see ourselves from 10 years before, just as if we had thrown a ball and it came back several years later. Commented Jul 29, 2016 at 8:57 • If 8 minutes blows your mind, try the 13.2 billion year old light that was captured in the HST's XDF. Using today's technology we have seen galaxies as they "existed just 450 million years after the universe's birth in the Big Bang." Commented Jul 29, 2016 at 10:20 Will he see the past? Yes! Light emitted from stuff on the alien planet (in the form of photons) will take time to get to Earth. If you "beat it" to the Earth and look back, you will be able to receive the photons and see what was emitted. Will the Alien Be Surprised? Unlikely, but possible. This alien just travelled to the earth using technology far beyond what we mere humans can imagine. He will probably have a firm grasp of the speed of light and all it entails. On the other hand, he may have no idea how his own technology actually works, so it's possible he will be surprised by the time-lag. Will he be able to actually zoom in on his planet? No. Here is where things are weird. The maximum distance and detail you can get with a telescope is governed by the amount of light the telescope receives from the target. To view a planet 100 ly away (assuming similar luminosity to the Earth) with enough detail to look at people on the surface, you would need a MASSIVE telescope. This reddit page explains it rather well, but needless to say, such a telescope does not, and basically cannot exist. • It certainly can't exist by means of a traditional lens, or based on existing physics knowledge. But there's a good plausible fictional background for such a telescope in one book I read, wherein a light magnet "...reacted on light beams and pulled them in from a zone many miles in diameter and concentrated them, magnetically corrected for aberration, into a spot smaller than a dot on paper." Commented Jul 29, 2016 at 1:32 • "He will definitely have a firm grasp of the speed of light and all it entails." He might just be a passenger and have no idea really how space-travel works... Commented Jul 29, 2016 at 8:46 • Exactly... just because you play Pokemon go doesn't mean you have a grasp of relativity to explain how GPS works and corrects for time-dilation ;-) Commented Jul 29, 2016 at 8:53 • He may not be surprised, but I am sure he'll be amazed. And given the telescope uses a technology we don't understand, we can't say if it could zoom in to give crazy detail. If you need that to happen in your story, just have it happen. Don't explain, don't apologize. Commented Jul 29, 2016 at 11:39 • @colmde That's fair. I assumed he had travelled to the Earth on his own, but after reading OP's question again I agree it's possible he may have been dropped off or something. Plus not everyone who drives a car themselves actually knows how the car works and everything about the physics involved. I'll make an edit. Commented Jul 29, 2016 at 17:08 It should be noted you never are actually seeing an object, you always, technically, see light reflected off the planet. This is similar to how if you get a letter in the mail you aren't getting the letter as it's being written, you're getting a few-day instance of it in the past. Magnifying an image with a telescope does not 'speed up' the light or any such, so the reflected light would still be just as old as if it wasn't magnified. A reflected source of light 100 light years away would, indeed, take 100 years to reach the point of observation (Earth) no matter what you do. Assuming his telescope is that amazingly powerful to notice detail on the surface of the planet, the planet will look like it did 100 years ago, yes. Everything absorbs, and more importantly reflects light to a certain degree. When reflection occurs light rays are bounced back from that object in a certain pattern, and with certain colours filtered out. So when you're looking through the telescope, and want to see the other planet what you're really looking at is the light reflected off of that planet, and light needs time to get to us. Hence "looking into the past". Now just imagine what it must take for you to "see" the alien planet's surface: • Light travels down through their atmosphere • Light hits and object, and is reflected • That reflected light travels up through the atmosphere again (and is distorted by it) • That reflected light must now travel 100 light years in the correct direction (during this time the rays of light are distorted by gas clouds they may pass through, etc.) • The reflection is finally captured by your telescope :-) Do you see how wildly unlikely that chain of events is? Furthermore, this doesn't change the fact that said reflection needed 100 years to get to you. When you zoom in you're not somehow reaching out to the stars, and closing the distance. You're simply focusing the reflection in a different way, so that you may perceive more detail. If it takes light 100 years to travel from the alien planet to Earth then an observer on the earth will see the alien planet as it was 100 years ago. that doesn't change if you are looking at a continent or able to 'zoom in' and read a newspaper in the hands of an alien sat on their front porch. Time only exists where you are. The rung of time does not speed up or slow down because you move on the same earth spinning around. Image you are standing on a merry go round and keep facing one direction, you do not move your feet do. It depends on how much "artistic license" you're taking with physics... you already have an alien traveling faster than light... Maybe your telescope isn't a simple optical telescope and either somehow "sucks" the light from the alien planet (to get around MozerShmozer's huge telescope problem) faster than c, or uses technology similar to what the alien used to get here to send information from there to here faster than light. Perhaps the telescope can detect tachyons (hypothetical super-light-speed particles) coming from the planet or something... • Perhaps it works like an X-Ray machine, bouncing tachyons off the planet. You'll need to leave the machine running for a few [seconds - decades] but, when the tachyons get back to you, you're seeing what happened half the delay ago (instead of ~100 years ago) Commented Jul 29, 2016 at 14:46

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https://web2.0calc.com/questions/parabolas_32

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+0 # Parabolas +1 236 2 Find the equation of the line that is tangent to both of the parabolas y^2 = 4x and x^2 = -20y. Enter your answer in the form y = mx + b. Jun 24, 2022 #2 +14964 +1 Find the equation of the line that is tangent to both of the parabolas y^2 = 4x and x^2 = -20y. Hello Guest! $$f(x)=y=\pm2\sqrt{x}\\ f'(x)=\pm\ x^{-\frac{1}{2}}\\ g(x)=y=-\frac{x^2}{20}\\ g'(x)=-\frac{x}{10}\\ f'(x)=g'(x)\\ -\frac{2x}{10}=\pm \ x^{-\frac{1}{2}}\\ \color{blue}x = \pm 4,641588833408518\\ © Arndt\ Br\ddot unner$$ $$y_f=2\cdot \sqrt{4,641588833408518}=4.30886937996\\ P_f(4,641588833408518,4.30886937996)\\ y_g=-\frac{x^2}{20}=-1.07721734492\\ P_g(-4,641588833408518,-1.07721734492)$$ $$m=\dfrac{y_f-y_g}{x_f-x_g}=\dfrac{4.30886937996-(-1.07721734492)}{4,641588833408518-(-4,641588833408518)}\\ \color{blue}m=0.580198604205$$ Point-direction equation of the straight line. $$y=m(x-x_1)+y_1\\ y=0.580198604205(x-4,641588833408518)+4.30886937996\\ The\ equation\ of\ the\ line\ that\ is\ tangent\ to\ both\ of\ the\ parabolas\\ y^2 = 4x\\ and\\ x^2 = -20y\\ is\\\color{blue}y=0.5802x+1.692$$ ! Jun 26, 2022 edited by asinus  Jun 26, 2022 edited by asinus  Jun 26, 2022 edited by asinus  Jun 26, 2022 edited by asinus  Jun 26, 2022

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https://www.physicsforums.com/threads/paypal-nochex-waiting-as-a-thankyou.7962/

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# Paypal/nochex waiting as a thankyou ## Main Question or Discussion Point Two gas cylinders A and B for which Va = 30 litres and Vb = 10 litres are linked by a narrow pipe containing a valve. Initially the valve is closed and cylinder A is charged with an idea monotomic gas to a pressure Pa = 10Ma at Ta = 300K. Vessel V is evactuated and has pressure Pb = 0. The valve is slowly opened and the gas pressures allowed to equalise and the final temperature Ta = Tb = 300k. a) Is the process quasistatic or not? Give reasons. b) What is the final pressure? What is the change in internal energy? c) Calculate the external work performed on the system. d) Determine the total heat transferred to the system. e) Would the answers to a-d change if the process was a free expansion rather than a controlled leak? Many thanks Last edited: Related Other Physics Topics News on Phys.org Hmmmm....the question I originally replied to seems to have disappeared...strange [For the first bit, I would assume that you need to start from another adiabatic relation (something involving pressure and volume perhaps...) and then use the ideal gas equation pV = nRT to relate that to what you want... (remember that n and R are constants). For the second part, call the initial temperature and radius/volume T1 and V1, and call the final temperature and radius/volume T2 and V2. Since T1.V1^(lambda - 1) = A (A is a constant) and T2.V2^(lambda - 1) = A as well you should be able to rearrange these equations to get V2, and hence the final radius. Hope that helps... Jess] This is all a bit pointless now...sorry Last edited: I was just in the process of changing the question! Thanks for you help, I think I have that one sussed now though cheers. Any chance you can have a go at this one? Rob. Are you sure Pa = 0? That seems a bit odd to me, but I could be reading the question wrong. Jess Thanks again Jess! lol I think I'm having a bad day! I'm guessing the answer to a) would be that is is a quasistatic process due to the pressures being allowed to equalise. But are there any better reasons for this? Rob Last edited: Originally posted by robgb I'm guessing the answer to a) would be that is is a quasistatic process due to the pressures being allowed to equalise. But are there any better reasons for this? Rob I would think that that answer would do, although I think the part about the pressures equalising needs to be more specific - you might want to define exactly what you think a quasistatic process involves to let whoever marks this (I assume someone will, at some point) know that you know what you're talking about. You might also want to think about what part (e) (free expansion) involves and how this is different from a quasistatic process - this should make the answer to (a) a lot more obvious. As a clue for part (b) (the internal energy bit), since we are talking about an ideal or perfect gas, I suggest you look up Joule's law... I don't know what level you're at, but I've always found a book by C.J. Adkins called 'Equilibrium Thermodynamics' to be a good source of information for this sort of stuff. Jess Last edited: Thanks Jess. Could you shed any light onto c & d? Any chance anyone came come up with an answer? These are questions my son have been set, and I'm trying to get some answers so that I can help him with them. Rob.

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https://zeblearn.com/tutorials/python-tutorial/python-math-module/

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# Python Math Module ## Python Programming Training Certification Flexible Hours 100 Assignments Instructor Led online Training 50 LMS Access 24X7 Support 100% Skill Level ### Enquire Now 4.9 out of 1000+ Ratings Best Python Institute for Learning Python Course & Training, Live Project Training in Python with Django, Data Science and AI, Interview Assistance, Expert Coaching Trainers. Python Certification & Interview Assistance! Get free demo now! ### Course Overview Python is one of the world’s top programming languages used today and Python training has become the most popular training across individuals. Training Basket’s Python Training & Certification course covers basic and advanced Python concepts and how to apply them in real-world applications.Python is a flexible and powerful open-source language that is easy to learn and consists of powerful libraries for data analysis and manipulation. Our Python training course content is curated by experts as per the standard Industry curriculum. The curriculum, coding challenges and real-life problems cover data operations in Python, strings, conditional statements, error handling, shell scripting, web scraping and the commonly used Python web framework Django. Take this Python training and certification course and become job-ready now. ### Python Math Module Python math module is defined as the most famous mathematical functions, which includes trigonometric functions, representation functions, logarithmic functions, etc. Furthermore, it also defines two mathematical constants, i.e., Pie and Euler number, etc. Pie (n): It is a well-known mathematical constant and defined as the ratio of circumstance to the diameter of a circle. Its value is 3.141592653589793. Euler’s number(e): It is defined as the base of the natural logarithmic, and its value is 2.718281828459045. There are different math modules which are given below: math.log() This method returns the natural logarithm of a given number. It is calculated to the base e. Example • import math number = 2e-7 # small value of of x print(‘log(fabs(x), base) is :’, math.log(math.fabs(number), 10)) Output: • log(fabs(x), base) is : -6.698970004336019 math.log10() This method returns base 10 logarithm of the given number and called the standard logarithm. Example • import math x=13 # small value of of x print(‘log10(x) is :’, math.log10(x)) Output: • log10(x) is : 1.1139433523068367 math.exp() This method returns a floating-point number after raising e to the given number. Example • import math number = 5e-2 # small value of of x print(‘The given number (x) is :’, number) print(‘e^x (using exp() function) is :’, math.exp(number)-1) Output: • The given number (x) is : 0.05 e^x (using exp() function) is : 0.05127109637602412 math.pow(x,y) This method returns the power of the x corresponding to the value of y. If value of x is negative or y is not integer value than it raises a ValueError. Example • import math number = math.pow(10,2) print(“The power of number:”,number) Output: • The power of number: 100.0 math.floor(x) This method returns the floor value of the x. It returns the less than or equal value to x. Example: • import math number = math.floor(10.25201) print(“The floor value is:”,number) Output: • The floor value is: 10 math.ceil(x) This method returns the ceil value of the x. It returns the greater than or equal value to x. • import math number = math.ceil(10.25201) print(“The floor value is:”,number) Output: • The floor value is: 11 math.fabs(x) This method returns the absolute value of x. • import math number = math.fabs(10.001) print(“The floor absolute is:”,number) Output: • The absolute value is: 10.001 math.factorial() This method returns the factorial of the given number x. If x is not integral, it raises a ValueError. Example • import math number = math.factorial(7) print(“The factorial of number:”,number) Output: • The factorial of number: 5040 math.modf(x) This method returns the fractional and integer parts of x. It carries the sign of x is float. Example • import math number = math.modf(44.5) print(“The modf of number:”,number) Output: • The modf of number: (0.5, 44.0) Python provides the several math modules which can perform the complex task in single-line of code. In this tutorial, we have discussed a few important math modules. ### Recently Trained Students #### Jessica Biel – Infosys My instructor had sound Knowledge and used to puts a lot of effort that made the course as simple and easy as possible. I was aiming for with the help of the ZebLearn Online training imparted to me by this organization. #### Richard Harris – ITC I got my training from Zeblearn in the Python Certification Training, I would like to say that say he is one of the best trainers. He has not even trained me but also motivated me to explore more and the way he executed the project, in the end, was mind-blowing. ### Candidate’s Journey During Our Training Program #### Expert’s Advice & Selection of Module Choosing the right type of module for the training is half the battle & Our Team of experts will help & guide you. #### Get Trained Get Trained & Learn End to End Implementation from our Expert Trainer who are working on the same domain. #### Work on Projects We Do make our student’s work on multiple case studies , scenario based tasks & projects in order to provide real-time exposure to them. #### Placements We have a dedicated placement cell in order to provide placement assistance & relevant interviews to our candididates till selection

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https://excelrow.com/column-function-in-excel/

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# Excel COLUMN Function ## What is COLUMN function in Excel? The COLUMN  function is one of the Lookup & reference functions of Excel. It Returns the column number of a reference. We can find this function in Lookup & reference category of the insert function Tab. ## How to use COLUMN function in excel 1. Click on an empty cell (like F5 ). 2. Click on the fx icon (or press shift+F3). 3. In the insert function tab you will see all functions. 4. Select Lookup & reference category. 5. Select COLUMN function. 6. Then select ok. 7. In the function arguments Tab, you will see the COLUMN function. 8. Reference is the cell or range of contiguous cells for which you want the column number. 9. For example If you click on G4, see result= 7. 10. If you click on A4 , see result =1. 11. You will see the result in the formula result section. # Unleash the Power of Column Functions in Excel! Excel provides a wide range of powerful column functions that can help you manipulate and analyze your data quickly and easily. One popular function is SUMIF, which allows you to sum up values in a range based on a specified condition. For example, if you have a sales report with columns for product names, prices, and quantities sold, you could use the SUMIF function to calculate the total revenue generated by a specific product. # Inserting a New Column: A Beginner’s Guide Inserting a new column in Excel is a simple process that can be done in just a few clicks. To insert a new column, simply right-click on the column next to where you want to insert the new column and select “Insert” from the drop-down menu. You can also go to the “Home” tab on the ribbon and click on the “Insert” button. A new column will appear to the left of the selected column. # Row vs. Column in Excel: Understanding the Basics Rows and columns are the two fundamental components of an Excel spreadsheet. Rows run horizontally across the sheet, while columns run vertically. Rows are typically used to organize and display data, while columns are used to define attributes or characteristics of the data. For example, in a table of customer information, rows might represent individual customers while columns might include fields for name, address, phone number, etc. # Deleting Columns Made Easy with Excel Deleting a column in Excel is a straightforward process that requires only a few clicks. Simply select the column you want to delete by clicking on its header, then right-click and choose “Delete” from the drop-down menu. Alternatively, you can go to the “Home” tab on the ribbon and click on the “Delete” button. Be sure to double-check that you have selected the correct column before deleting it, as this action cannot be undone. # Hiding Columns in Excel: Tips and Tricks Hiding columns in Excel is a great way to declutter your worksheet and make it easier to read. To hide a column, simply right-click on the column header and select “Hide” from the drop-down menu. You can also use the keyboard shortcut “Ctrl + 0”. If you want to hide multiple columns at once, select them by clicking and dragging over their headers, then right-click and select “Hide”. To unhide hidden columns, right-click on any column header, select “Unhide” from the drop-down menu, and choose the columns you want to unhide. # How to Unhide Hidden Columns in Excel If you have hidden columns in Excel and need to unhide them, don’t worry – it’s easy to do! Simply select the columns on either side of the hidden column(s) by clicking and dragging over their headers, then right-click and select “Unhide”. If you’re not sure which column is hidden, select the entire worksheet by clicking the box between the column and row headers, then follow the same steps. # Resizing Columns in Excel: Managing Your Data Efficiently Resizing columns in Excel is an essential part of managing your data efficiently. To resize a column, hover your cursor over the boundary between two column headers until it turns into a double-headed arrow. Then, click and drag the boundary to the left or right to adjust the column width. You can also double-click on the boundary to automatically resize the column to fit its contents. If you want to resize multiple columns at once, select them by clicking and dragging over their headers, then follow the same steps. # Formulas 101: Adding a Formula to an Excel Column Adding a formula to an Excel column can help you automate calculations and save time. To add a formula, select the cell where you want the result to appear, then type an equals sign followed by the formula. For example, if you have a column of numbers and want to sum them up, type “=SUM(A1:A10)” into the first cell of the column. Then, drag the fill handle (the small square in the bottom right corner of the cell) down to apply the formula to the rest of the column. Excel will automatically adjust the cell references for each row, so you don’t need to manually enter the formula for each cell. # Copying Formulas to Multiple Columns in Excel Copying formulas to multiple columns in Excel can save you a lot of time and effort. To do this, simply enter the formula into the first cell where you want to apply it, then hover your cursor over the fill handle (the small square in the bottom right corner of the cell) until it turns into a crosshair. Then, click and drag the fill handle across the rest of the columns where you want to apply the formula. Excel will automatically adjust the cell references for each column, so you don’t need to manually update them. # Formatting Columns in Excel: Making Your Data Stand Out Formatting columns in Excel is a great way to make your data stand out and easier to read. You can format columns by changing font styles, text colors, cell borders, and more. To format a column, select the cells you want to format, then go to the “Home” tab on the ribbon and use the formatting tools available in the “Font”, “Alignment”, and “Number” groups. You can also create custom formats by clicking on the “More Number Formats” button in the “Number” group. # Sorting Data by Specific Columns in Excel Sorting data by specific columns in Excel is a quick and easy way to organize your data based on a particular attribute or characteristic. To sort data by a specific column, select the entire table, including the column headers, then go to the “Data” tab on the ribbon and click on the “Sort” button. Choose the column you want to sort by from the drop-down menu, and specify whether you want to sort in ascending or descending order. # Filtering Data by Specific Columns in Excel Filtering data by specific columns in Excel allows you to quickly narrow down your data to focus on specific attributes or characteristics. To filter data by a specific column, select the entire table, including the column headers, then go to the “Data” tab on the ribbon and click on the “Filter” button. Click on the arrow next to the column you want to filter, then choose the values or conditions you want to include or exclude from the filter results. You can also create custom filters by selecting “Filter by Color”, “Text Filters”, or “Date Filters” from the drop-down menu. # Merging Cells in Excel: Combining Information Merging cells in Excel allows you to combine information from multiple cells into a single cell. To merge cells, select the cells you want to merge, then go to the “Home” tab on the ribbon and click on the “Merge & Center” button. You can also choose “Merge Across” or “Merge Cells” from the drop-down menu to merge cells horizontally or vertically, respectively. Keep in mind that merged cells can cause issues with sorting, filtering, and other data analysis functions, so use them sparingly and only when necessary. # Calculating Sums in Excel: Mastering Column Functions Calculating sums in Excel is an essential skill for anyone who works with data. One powerful column function for calculating sums is SUMIFS, which allows you to sum up values based on multiple conditions. For example, if you have a sales report with columns for product names, prices, quantities sold, and regions, you could use the SUMIFS function to calculate the total revenue generated by a specific product in a specific region. # Averaging Columns in Excel: Simple Solutions for Complex Data Averaging columns in Excel is a straightforward process that can help you better understand complex data. To average a column, select the cells you want to average, then go to the “Formulas” tab on the ribbon and click on the “Average” button. Alternatively, you can use the AVERAGE function to calculate the average value of a range of cells. You can also use the AVERAGEIFS function to calculate the average value based on multiple conditions. # Counting Cells in Excel: The Basics of Data Analysis Counting cells in Excel is one of the most basic data analysis functions you can perform. To count cells, use the COUNT function, which returns the number of cells in a range that contain numbers or dates. You can also use the COUNTIF function to count cells based on a specific condition, such as cells that contain a certain value or text string. If you want to count unique values in a column, use the COUNTUNIQUE function. # Counting Non-Blank Cells in Excel: Eliminating Errors Counting non-blank cells in Excel is an essential data analysis function that can help you eliminate errors and better understand your data. To count non-blank cells, use the COUNTA function, which returns the number of cells in a range that contain any type of data, including text, numbers, and formulas. You can also use the COUNTIFS function to count cells based on multiple conditions, including non-blank cells. # Finding and Replacing Values in Excel Columns Finding and replacing values in Excel columns is a quick way to update and correct your data. To find and replace values, select the column where you want to make changes, then go to the “Home” tab on the ribbon and click on the “Find & Select” button. Choose “Replace” from the drop-down menu, then enter the value you want to find and the value you want to replace it with. Click on “Replace All” to apply the changes to all instances of the value in the column. # Freezing Columns in Excel: Keeping Your Data on Track Freezing columns in Excel is a useful feature that allows you to keep important data visible while scrolling through large tables or worksheets. To freeze columns, select the column to the right of the last column you want to freeze, then go to the “View” tab on the ribbon and click on the “Freeze Panes” button. Choose “Freeze Panes” from the drop-down menu to freeze the columns to the left of the selected column. You can also choose “Freeze Top Row” or “Freeze First Column” to freeze specific rows or columns. # Printing Specific Columns in Excel: Creating Professional Reports Printing specific columns in Excel is a great way to create professional-looking reports and presentations. To print specific columns, select the entire worksheet, including the column headers, then choose “Print” from the “File” tab on the ribbon. In the print settings, choose “Print Selection” or “Print Active Sheets”, depending on your needs. You can also adjust the page layout and print quality settings to customize the appearance of your printed report.

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cancel Showing results for Did you mean: Regular Visitor Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) Hi, I've made my best attempt to SUMX my data to get an accurate Profit and Loss (PnL) figure but I have been unsuccessful in writing a single measure that calculates the correct total amount when selecting multipule days and multipule zones. I think my main problem is that my fact tables are at the hourly level and my calendar is daily. I have a seperate dim table for Hours (HE). I also am trying to slice and dice by zone (there are 5). I tried using a SUMX for the hours and then a SUMX on the SUMX for the Zones and that fixed it for a single day…but when I select more than one day the sum skyrockets again. I’m sure there is a formula that incorporates additional Filters and Date logic but I haven’t been able to figure out the formula pattern L. Hoping someone can help with some syntax or point me in the right direction in the form of a specific example having to do with hourly calendar aggregations. Model- https://www.dropbox.com/s/31tcais3wqmmtt3/SAMPLE2.pbix?dl=0 Here's my setup- ASK- (the model has more measure details if it helps) Added together, the total for both days for both zones should be \$790.11. Can you please help me write the measure that will calc it? Thank you, Fonz 1 ACCEPTED SOLUTION Accepted Solutions Senior Member Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) "The amount of money made is derived by multiplying the volume by the price spread, Per hour." Ya, I don't think that is what you are doing right now.   You are adding everything together and THEN multiplying them. I assume [BIDS DEC MW] * [LMP DA-RT] is valid at some interval... Date+Time I guess? So maybe    SUMX(Date, SUMX(HE, [BIDS DEC MW] * [LMP DA-RT])) would work? 5 REPLIES 5 Senior Member Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) Your model generally looks good to me, but I'm skeptical of : PnL DEC = [BIDS DEC MW] * [LMP DA-RT] (and the INC version) Just... intuitively... I'm afraid of adding stuff together into some aggregate, then multiplying those sums. It might be something else... but... that's my gut. You want to tell us what PnL DEC is supposed to mean...? Highlighted Regular Visitor Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) Thanks for the response. I see your point. INC and DEC PnL is the dollar amount gained or lost depending on the price spread for that hour. If I have a positive INC value (i.e. volume) for an hour where the delta between DA and RT is positive, I make money. If I have a positive INC value (i.e. volume) for an hour where the delta between DA and RT is negative, I lose money. If I have a negative DEC value (i.e. volume) for an hour where the delta between DA and RT is positive, I lose money. If I have a negative DEC value (i.e. volume) for an hour where the delta between DA and RT is negative, I make money. The amount of money made is derived by multiplying the volume by the price spread, Per hour. I didnt think calculating INCS and DECS seperately then adding them together would be problem. Senior Member Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) "The amount of money made is derived by multiplying the volume by the price spread, Per hour." Ya, I don't think that is what you are doing right now.   You are adding everything together and THEN multiplying them. I assume [BIDS DEC MW] * [LMP DA-RT] is valid at some interval... Date+Time I guess? So maybe    SUMX(Date, SUMX(HE, [BIDS DEC MW] * [LMP DA-RT])) would work? Regular Visitor Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) I'm following ya....yes, valid for date + hour I'll try to make that double SUMX work.... Regular Visitor Re: Help aggregate HOURLY PnL figures for multiple days and multiple zones (pbix included) I ended up needing to use THREE SUMX's for my INCs then another THREE for my DECs and adding them together in the same measure...and that got me to my desired result! PnL =(( SUMX(VALUES('CALENDAR'[Date]), SUMX(VALUES(HE[HE]), SUMX(VALUES(Zone[Zone (5 min)]), [BIDS INC MW]*[LMP DA-RT])))) + (SUMX(VALUES('CALENDAR'[Date]), SUMX(VALUES(HE[HE]), SUMX(VALUES(Zone[Zone (5 min)]), [BIDS DEC MW]*[LMP DA-RT]))))) If someone has a more elegant way to get to the same place I'm all ears! lol. Scottsen, thanks for helping me walk through this and nudging me to try to multi SUMX in the same measure and performing the multiplication within the same measure as well. I greatly apprecaite it. --fonz Announcements Meet the 2020 Season 1 Power BI Super Users! It’s the start of a new Super User season! Learn all about the new Super Users and brand-new tiered recognition system. Super User Challenge: Can You Solve These? We're celebrating the start of the New Super User season with our first ever Super User 'Can You Solve These?' challenge. Power BI Desktop Update - February 2020 We are super excited for our update this month, as we are releasing two of our top community requests!

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Or connect using: ## If I sharpen this knife enough, I'll have a great looking handle #### Apr. 1st, 2009 | 11:27 pmmood: detail-oriented ```lambda { |f,*x| 0 while x=f[f,*x] } [ lambda { |f,*x| puts ["what was it", "is it a #{x[0]}", "what might I ask to tell a #{x[1]} from a #{x[0]}", x[2]][x.size]+"?" ( 5>(x<<gets.strip<<x[-1]["n"]).size ? (2<x.size && x.pop && 3!=x.size ? f[f,x[0],f[f]]: 3==x.size ? x:[x])[1..-1]: [f,lambda{|*y| y[1..-1]}].method(x.pop ? "dup":"reverse")[ ].enum_with_index.map {|g,i| g[f,*x[i]]}<<x[-2] ).unshift(x[-2]).last 3}]``` I'm a little obsessed with the above code. This is the animal guessing game in Ruby ("think of an animal and I will try to guess what it is. Is it a pet? Is it a dog? No? What is it then? How can I tell a cat from a dog? Play again? Is it a pet? Does it meow? etc..") The version above is adapted from one I posted originally to Facebook a year and a half ago. I've thought of a few new tricks with arrays since then, and it's now some of the most dense code I've ever written. I would like to eliminate the two checks to the length of the `x` array, but I don't think that's possible. I suspect I could come up with something more terse than `enum_with_index` as well - it's been a while since I've looked at that part. There is no practical application to this at all. Well, ok - I now understand arrays very well indeed, and I especially like the `first / last` method pair as it will not fail when you ask for more of an array than there is available. addendum: shaved off a few more characters! It looks like I can live without `method()`, though that was an elegant approach... except for resorting to `dup`. ```lambda{|f,*x| 0 while x=f[f,*x]}[lambda{|f,*x| puts ["what was it", "is it a #{x[0]}", "what might I ask to tell a #{x[1]} from a #{x[0]}", x[2]][x.size]+"?" ( 5>(x<<gets.strip<<x[-1]["n"]).size ? (2<x.size && x.pop && 3!=x.size ? f[f,x[0],f[f]]: 3==x.size ? x:[x])[1..-1]: [f,lambda{|*y| y[1..-1]}].values_at(x[4]?0:1,x[4]?1:0 ).enum_with_index.map {|g,i| g[f,*x[i]]}<<x[2] ).unshift(x[-2]).last 3}]``` addendum #2: this is the final version, five days later. I really can't believe how much time I've spent on this. ```lambda{|f,*x|0 while x=f[f,*x] }[lambda{|f,*x| puts ["what was it","is it a #{x[0]}", "what might I ask to tell a #{x[1]} from a #{x[0]}",x[-1]][x.size]+"?" case (x<<gets.strip<<x[-1]["n"]).size when 3; [x[0]]+(x[-1]?f[f,x[0],f[f]][1..-1]:[]) when 5; [0,1].zip(([lambda{|*y|y[1..-1]},f]*2)[x[-1]?1:0,3]). map{|i| i.pop[f,*x[*i]]}<<x[2] else; x[0..-2] end }]``` addendum-de-dum - no more. Really. After this. ```lambda { |f,*x| 0 while x=f[f,*x] }[ lambda { |f,*x| puts ["what was it", "is it a #{x[0]}", "what might I ask to tell a #{x[1]} from a #{x[0]}", x[-1]][x.size]+"?" case (x << gets.strip << [f,lambda { |*y| y[1..-1]},f][x[-1]["n"]?0:1,3]).size when 3; x[-1][2] ? f[f,x[0],f[f]] : x[0] when 5; [0,1].zip(x[-1]).map{ |i| i.pop[f,*x[*i]]} << x[2] else x[0..-2] end }]``` ## Forward error correction and JPEGs #### Mar. 12th, 2009 | 05:26 pmmusic: CONNECT 14400 A fun project to work on and to serve as a carrot to enjoy after I've finished writing a history essay and studying for two exams: characterize the noise floor of a communication medium defined as a fixed-size JPEG-compressed bitmap, and determine the suitability of a forward error correction method to approach the Shannon limit. Questions: • how can the size of a given JPEG file be related to the amount of recoverable meaningful data it contains? • what is the most efficient approach: non-error corrected storage, FEC storage, distribution of a "correction list" of patches with the compressed bitmap, or a combination of the last two? • (trivial) is JPEG to bitmap decompression well-defined and invariant, or will certain inputs yield varying outputs given standards-compliant decompressors? It would be ridiculous to claim that there is any practical application for this so I won't. I thought it might be fun to abuse a flickr account for data storage; the interesting and unsolved problem I have is the distribution of large amounts of data on the internet to many recipients. How many thousands of JPEGs are needed to reliably reproduce a DVD? (it gets more interesting when the uncertain areas are wandered into, since we can test decompression to make sure that the intended result is the one we get. Perhaps the input can be fine-tuned to yield the correct result even when complete, correct reproduction of purely random input is unlikely. Of course this requires well-defined decompression.) I think I'm going to have to give up if I run into fields, though. ## no volts: a plane, stopped trains, and a bicycle #### Jan. 17th, 2009 | 09:14 ammood: i need a shower Thursday evening was already an eventful time; N. was packing for her little jaunt and, well, this is always a little stressful no matter how well prepared you are - and she prepares very well indeed. This turned out to be a good thing and she had most of her packing done well in advance, where I usually procrastinate and stuff everything in arm's reach into my luggage in the last terror-filled half-hour before I (absolutely, positively, can't even hope to make it past this time) have to go. A pleasant enough evening with a passable dinner of my making and Irished-up coffee to follow (classy, I know). We even handled a minor crisis when the microwave and (new) coffee maker drew too much current and blew a fuse. Replaced the fuse and serialized that operation. Then, two hours later, the lights went out. "Shit! The fuse blew again?" Well, that was a reasonable conclusion. We have the bias of the earlier event, and the fact the computers are still running. On the other hand - all of the lights are out, only the battery-powered computers are running, and they do seem desperately unhappy (especially my poor laptop, which is essentially on heart-lung bypass with the non-battery-powered server). We still haven't realized the magnitude of the situation, so... "Oh, no - did I knock out the power by ironing and making muffins at the same time?" Hmm, possible but - oh, look, the building across the street has no power! And the lights at the subway station are out, except emergency lighting (although, really interestingly, trains continued to run until the regular subway closing time). Ah. This is the real deal. And it might take a while before the lights come back on. Ok. Flashlights, and then candles. The whole event seemed very apocalyptic. This obviously wasn't a full-scale outage as streetlights were on and other buildings in this vicinity clearly had power, but we could tell that a few thousand people were affected. We didn't grasp the magnitude of the outage until much later; as it turned out, we were in the middle of a quadrant bounded by Spadina Ave, St. Clair Ave, Jane St., and Queen St W / Queensway. In other words, west Toronto. Bear in mind though that these details would all have to come later; the only link to the outside world that we had was our cellphones, which, really, is not so bad, but it's a lot less than we're accustomed to. Anyway, I had a feeling that it might be a while, and it was. I think the candles started making the air a little stuffy - it didn't help that half of them were pumpkin-scented or whatever - so ventilation wasn't working. We live in a fairly large apartment building so heat wasn't really a problem, and there was enough water in both the hot-water and cold-water reservoirs to last throughout. The hot water was decidedly lukewarm in the morning, though, which made for a cold and short shower. The elevators were out, of course, and they were hopelessly confused when the power came back, so I had the pleasure of carrying N.'s luggage down fourteen flights of stairs. It was down, at least. As I've jumped ahead, it's already apparent that we were one of the lucky ones to have power restored after only a few hours. We woke up early with the lights still out, with the intention of taking the subway to the airport. That wasn't an option, so we drove. The drive back is worthy of a blog entry itself. I felt a little like I was plowing through zombies like Will Smith in I Am Legend. (seriously though, the police did a pretty commendable job of keeping a huge crowd from getting out of hand on what was a cold day - for Toronto, that is) The subway was restored in time for me to take it to work, but I decided to ride my bike anyway. I've never biked in sub-zero temperatures, so this was a first for me, and probably a little reckless since it was about -15. It worked out okay. Changing was a hassle, but my sweatpants+old jeans approach seemed to work. It was really just about the same as skiing on a cold day. Extremely tiring on the way back. My thighs started tingling at the halfway point and when I hit a massive pothole, condensation formed on the seat while I dismounted to collect my bike locks (yes, two of them, both violently ejected from my bike rack as I traversed the pothole). It was a pretty exciting day. I got home, disrobed, had a lovely cup of tea, and fell asleep with the lights on. I think I've left out quite a lot but that is the gist of it. Anyone else had a fun day? ## We used to call them ISA slots, and they had jumpers, and.... #### Jan. 14th, 2009 | 01:49 pm So it seems like desktops have had their day, which confirms more or less my experiences at the fruit stand. It used to be that a laptop (when did they stop being notebooks, anyway?) was a semi-useful ancillary to the real computer, invariable a PC with a 60 lb. crt, modem, giant tower or imposing desktop. Wired mouse, profoundly terrible ergonomics - not that laptops have improved that situation! I feel a little sad about that, and I hope it won't make PCs prohibitively expensive for those who tinker. Then again, most of what I've always played with is essentially dumpster salvage. I don't know what I'd do with a new PC. I think I'd look at it in wonder, at first, at how shiny it was and how everything worked - then I'd scratch my head and think how stupid it is that I can't take it with me to school. I probably wouldn't play games on it. Part of living with N. is that there's always someone to play Wii with, and I guess I could get back into FPS but, well, I think they'd be more fun on a laptop anyway. And, not so ridiculously anymore, laptops can play games to my satisfaction. I still think Half-Life is pretty amazing, so that's more or less where I am with gaming. Laptops seem to be opening a new market in the sub-\$500 category that PCs never really addressed. You could build a PC for that amount, but it seemed that they were such unsatisfactory and bland machines that they didn't particularly catch on. The build quality was lousy too, and they were essentially blank slates. Great for Linux, but not so good for much else. Well, whatever. With enough space and money, I will always prefer a desktop on the... desktop. But I guess I have to affirm the zeitgeist in agreeing that if I had only one computer, it would be a laptop. All that this proves is that I'm a hopeless geek and I will always be outnumbered by my (hopefully loyal) computers. oh, I ended up beating the odds on N.'s laptop after all. No, not easy, not a bit, and of course her reaction was - "what did you do? you did nothing, it looks this same!" Well, what do you expect after you wake up from brain surgery? Hyper-intelligence? The ability to smell emotions? Maybe... maybe the patient with 5% odds of survival complains about these things. I'm willing to bet that they don't, dammit. More on that another time. (computers and cloning, that is, not neurosurgeons vs. geeks) ## Old computer recovery #### Jan. 9th, 2009 | 12:51 pm N. has a new notebook, and I'm transferring everything over from her old one. By everything, I mean putting in a Knoppix cd on the new one and booting the old one's Debian partition and netcat'ing the disk over. (as a side note, with a USB-USB transfer cable, my experiments showed the best throughput to be had by piping the sender's dd through gzip -1; the USB cable needs a lot of cpu from the old computer, so not much is available for compression except the gain from effectively run-length encoding empty space.) It has been quite a pain so far and I haven't solved it yet. I'm actually pretty close to giving up if my next attempt fails. However, I learned a few things in doing this, and I don't really want them to clutter up my brain any longer, so I'm hoping that writing them down will expunge the details. • the Toshiba Satellite M30 (old computer) recovery DVDs have encrypted Norton Ghost images. The decryption program will not run on any other computer. It acts as a shell to the real ghost.exe program, passing it the password on an authentic computer. That password is 2533. • to get that password, (omitting many details) I had start an instance of DOS on the old computer. Aside from how ridiculous this is, I ran into the problem that the USB key boot firmware on the Toshiba does not play nicely with anything that runs after it has done its job. This may have been fixed in a later revision. Anyway, forget about using a USB key, and use the grub4dos boot loader I'd installed long ago (in boot.ini, entirely in the Windows partition - very nice) to boot a FreeDOS image - specifically Balder. Load this floppy image into memory from the Windows partition using memdisk, which is part of syslinux, and is in turn loaded by grub. Don't forget to copy the files needed to harvest the ghost password (details) into the floppy image. • oh, yes, use the loadcd batch file to enable CD/DVD support after FreeDOS loads • call the predata.bat file on the recovery DVD and don't forget to set TGHOSTPS as in os.bat before running tghost The end result of this effort is that N. should be able to start using the new computer in exactly the same way she's used to using the old one. That's not so much to ask, really, is it? The new computer is preloaded with new software that we have no interest in using. I will have to use it myself on another system, but the theory is that this will contribute to my gainful employment at some point in the future. The current technicality is that the old software can be restored onto the new computer, but it seems to lack drivers necessary to boot. The system provides a message with the code 0x7b (who says Unix is unfriendly? Have they checked with the competition lately?) that translates to "boot failure". Oh, ok. I guess I'll see what I can do about that then. Possibilities: 1. try to fix the recovered system from a partition containing working software; Linux will not be useful for this, so I'd have to use the software that came with the new computer 2. instead of recovering the old software, copy the old hard drive to the new one directly as I did on my first attempt after pre-installing appropriate driver software on the old computer 3. buy a retail copy of the old software 4. admit defeat and use the new software The numbers in the steps above correspond to the severity of the insanity I'll be suffering as I progress through them, from start to finish. Now I'd be so pleased if I could just forget all of this... ## All of these hacks are yours, except Europa #### Jan. 2nd, 2009 | 12:52 pmmusic: (sshhh...) A couple of things I'm working on right now: • LDAP for the home net - single-source user authentication • new laptop: Debian GNU/Linux-on-NTFS, i.e. ntfs-root (ntfs-3g should make this possible) • using a photo frame as an info panel for the server; I need programmatic USB disconnection for this to work • wedding site • boot-132 "Mac" That should keep me busy for a little while. ## A novel application for HTTP referer #### Dec. 17th, 2008 | 12:00 pm The HTTP referer (sic) header defined in RFC 2616, as far as I know, hasn't been used to define a path namespace for resources on a web server. This could be a useful application, not for the purpose of "link protection" (trivially bypassed, by the way) but to make content less fragile by lessening its dependency on server file layout. For instance, a web page like Astronomy Picture of the Day will contain relative links to images like this: `<IMG SRC="image/0812/DoubleDumbbell_Lopez_c0.jpg" alt="See Explanation. Clicking on the picture will download the highest resolution version available.">`. If the site were to be reorganized, for instance renaming the top-level `image` directory to something more inclusive of other types like `media`, every referring document being served would have to be scanned and rewritten. With a large enough base of documents, this automatic process would need some verification as there are bound to be problems. As an alternative, consider something like `<IMG SRC="pic_of_the_day" />`. This eliminates a path structure entirely, placing the image in the top-level. This might be obviously implemented by using something like an alias or symbolic link on the web server, but this accomplishes nothing and doesn't solve the problem of handling separate files (I propose that "picture of the day" reference the same `pic_of_the_day`). How then can the web server determine which file to send? The referer comes in here. The server would have a table indexed by the referer that would return the appropriate document and `Content-Type` (note that `pic_of_the_day` has no extension, so the server will need to tell the client exactly what it is). My argument is that it is easier to maintain this table than to encode permanent references to media that may be moved; this should have other benefits too, which might be handy for client-side scripts that could benefit from having easily generated pathnames for referencing server resources. It seems strange that I haven't found an implementation of this. I'm pretty sure that more than a few sites fake this by maintaining a mapping between something like a `/assets` directory and the actual file, but they still include a reference in the static HTML in the form of a pathname and a proper filename with extension. I'd like to do away with this pretense and formalize the idea that the elements referenced by a page are entirely decided by the server. There is a very nice consequence to this, which I started with and worked from backwards to obtain this solution: a strong division between static and dynamic pathnames will exist, which means that there will be no conflict between an address you'd like to generate on the server as a static reference (like `my.server.com/~myname` or `apod.nasa.gov/pic_of_the_day`) and dynamically generated references to server resources. The actual names that pages use to refer to server resources are entirely immaterial as long as they are unique with respect to one another; it would be perfectly valid for `fruitcompany.com/store` to contain `<img />` elements that are sequentially named `0`, `1`, `...`, `n`. The referer `fruitcompany.com/store` will allow the server to dereference the link and serve it with the appropriate `Content-Type`. If I ever decide that I'd really like to have the address `fruitcompany.com/store/0` as a permanent link to a resource that I can distribute publicly, I'll set that up on the server, and the `fruitcompany.com/store` page source will simply start numbering its elements at `1`. ## Quick package search in Debian/Ubuntu/etc. #### Dec. 13th, 2008 | 01:23 pm This is a one-liner that I've found useful for searching Tag: fields in Debian's package lists. For instance, I might want to know about all the packages available that have to do with Ruby. I'll assume that any mention of the string 'ruby' in the Tag: field of a package warrants its selection. `egrep -hr '^(Package:|Tag:.+ruby)' /var/lib/apt/lists|grep -B1 ^Tag|grep ^Package|cut -d\ -f2-|tr \\n \||xargs -I{} apt-cache --names-only search ^\({}\)\\$|sort|less` The first `egrep` does a recursive search through the package list that `apt` maintains. The `-h` flag suppresses printing the filename. The regex selects every occurrence of lines beginning with "Package:" and lines beginning with "Tag:" and containing the string "ruby". In the next `grep`, packages without a matching Tag: line are discarded. Another `grep` to clean that up, then `cut` out everything but the package name, `tr` newlines into a literal pipe, and pass the now-single line to `xargs`. That part is a little ugly, because `apt-cache` wants a regular expression and it's trying too hard for this purpose when all that is needed is an exact match of the package name. Insert an `echo` before the `apt-cache` invocation to see what it's doing. It's not a particularly elegant solution, and it works. edit -- the original version still had `echo` before `apt-cache`, which wouldn't have been useful. Fixed Tags: , , ## Partially-compiled source code storage #### Dec. 12th, 2008 | 12:13 am So I was tempted to think that storing source code in a parsed form (what I learned is called an abstract syntax tree, commonly abbreviated AST) might be a good idea. I thought this because: 1. well-formedness is forced; syntactically invalid code cannot be saved (this is only the most trivial kind of correctness, though) 2. metrics extracted from source code can be more valid: rather than counting lines, operations, objects, classes, etc. can be measured (if spoofing is a problem, though, this might just move it into a higher level) 3. compilation will be marginally faster. I don't posit this as a realistic benefit, except perhaps for large projects, or specialized cases with large amounts of generated code (XWT's mips2java would be an example). 4. arguments about source code formatting can be swept aside in one fell swoop. Simply set your AST-aware editor to whatever formatting style you'd like. 4. the original form could be exactly regenerated by recording comments, deviations from regular form, even storage of invalid code in "as-is" form. However, a post on the Eclipse LDT (language definition toolkit - now defunct?) mailing list a few years ago makes a convincing argument that this is a waste of time ( http://dev.eclipse.org/newslists/news.eclipse.technology.ldt/msg00027.html ) I have an intuitive notion that AST storage should be able to buy the programmer something in that it brings the source form of a program closer to its object code representation. At the very least, it should make it easier for a debugger to find errors in code, and relay messages to the editor or IDE that can be used to highlight sections of code or supply profiling data. Unfortunately, existing tools like wc, grep, sed, etc., can't get anything useful from AST-represented source. They could still be used with a command-line tool to extract the original source, but this would still be problematic since the extracted text is not the canonical form of the source. The appeal of this kind of persistence, I think, is that the strong link between statements, functions, classes and files may be weakened somewhat. One of the reasons for storing a project in multiple files is simply to keep individual units of source code manageable, and to allow patches to be merged in similarly manageable units. But if source is stored as a pool of ASTs, aren't files an optional structure? And, interestingly, couldn't we do a SELECT-like operation to extract code and automatically merge disparate source into a single file? This might be interesting for another project that wants to incorporate functionality without doing a lot of digging. It seems to me (I have nothing to support this statement but intuition) that the AST form should provide more useful information for an automated system to extract code with arbitrary granularity, where a single function and its supporting functions and definitions could be automatically identified. Of course, all of this is possible with regular source code, since I have defined the regular and AST forms as equivalent. And with an excess of CPU and I/O for most projects, AST buys nothing and loses portability. Perhaps it would be useful for interchange, or as a method for marking-up source code or preparing it for publication (including it in a document for instance). The only way for AST storage to have any advantage is for it to restrict the programmer or force more information to be provided (or, compile code on the run as Visual Studio does, and identify dependencies at edit time). (which I can't stand, since it distracts me and prevents me from writing pseudo-code in the middle of a function as I try to reason out a problem. I do like highlighting and not having to match parentheses by hand. I'm not against method completion and other inline documentation, but I keep the docs open all the time anyway. I do find myself doing something like `object.methods.sort` a whole lot in irb.) Ok. I think I've convinced myself that this isn't a useful way to archive source code, and that we can get everything from the original source anyway as long as it's syntactically valid. It might be a different story for code that has been parsed and partially or fully compiled, but not (possibly, fully) linked. And precompiled headers are an old trick that makes sense, or used to back when computers were slow. I suppose it makes sense for continuous compilation. ## Big Plans #### Dec. 8th, 2008 | 10:56 pm I'm getting a feeling that I need to shake things up; and really, things have been just too comfortable and stable for too long. Let's see... how best to mess that up? ## Coding I've been doing a fair amount of this lately, and it's been filling a gap. At the same time, I don't feel that I'm gaining anything by building up awesome retail skillz. And there are fairly exciting things happening in the Rails community. And lo and behold, alternate frameworks like Merb and Camping... and some very interesting things on the fringes, like Haskell and a Rails analogue in, believe it or not, Smalltalk! ## Am I a relic, or something? It blows me away, really, it does, when I read the writing of my so-called contemporaries and discover that they've never dereferenced a pointer, started a line of code with a number reachable by GOTO or GOSUB, or, hell, even allocated memory. When I think that I switched to Linux because the alternative, at the time, could be taken down by one misbehaving program - does any of that count for anything? That I've watched computing scale up by several orders of magnitude, programming all the way? I'd like to think that it makes me appreciate the conveniences we have now all the more. What it makes me feel is that too much of a premium is placed on programmer time. Zed Shaw writes about a coder who assumed each month in the year has 30 days. Firefox has inexplicably contracted-out its non-volatile state to SQLite. (their claim that this makes the program more robust in the face of crashes is not borne out by my experience of corrupt .sqlite files) ## Write less code? Every single non-trivial project I've worked on has gone through a few re-writes. I've tossed more code than I've left; I've condensed functions into single lines. Invariably, I've made the mistake of chasing problems that have nothing to do with the task at hand (my proudest achievment here is probably the time I decided to condense postal codes into a base-10 number in an incredibly elegant, general way - extensible to arbitrary magnitude - saving roughly 200K out of a 10+MB database). I've come up with some very suitable solutions as well, but these have been on the third or fourth attempt, and came only after significant reflection and insight gained through the first couple of tries. I guess the ratio of time to lines of code is pretty high for me. I've also found that excess time seems to cause the code to get somewhat dense, and honestly probably not easily readable by someone not completely familiar with the problem. ## Am I smart enough to do what I'm trying to do? Maybe, but I feel like I'm stumbling sometimes without the rigour of a solid CS education and hours spent solving problems in discrete math. My colleagues seem to do pretty well with an intuitive grasp of logic and algorithms, leaving the tough parts to library code. This is probably the right approach, until it is time to write your own library. The other thing that bothers me is the difficulty of proving that a program is acting correctly. I think GEB started my thinking about this, and I'm aware that this is the life work of more than a few people. Intuition has its place, but there's nothing quite like saying "well, it seems to work" to reveal a lack of insight into how a program does its job. And even proofs aren't perfect: GEB (I seem to recall) mentions the proof of the four colour theorem, which is not realistically possible by hand; which seems reasonable enough, except that to accept the mechanically-generated proof is to accept the system producing the proof as being correct. ## Whatever I end up doing... ...kick me if you ever see me: • using UUIDs for primary keys (it's not about the not-chance of collision - it's the laziness) • believing that objects and relational databases make sense together • believing that Worse is not Better • thinking that a century for now, we won't look back in sheer terror at the code that ran the world • saying that a database will solve everything (I used to think that this was a good solution for /etc; in my defence, I was 18. I wonder what Apple's defence is?) Next entry will be ungeeky. I seem to have written only about my complaints in programming, as opposed to my big plans. You know. Big. Tags: , , ,

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# Codeforces Div3 #690 Prob-F QUESTION EXPLAINATION: We have to find a good Set in which anyone segment intersects with a maximum number of segments. We have to delete some segments in order to make it a good set. For example: [[1,4],[2,3],[3,6]] is a good set, because [2,3] intersects each segment from the set. [[1,2],[2,3],[3,5],[4,5]] is not a good set, because there is not even any single segment which is intersecting with all other segments. So, we can delete some segments, in order to make it a good set. Let’s say, we can remove [4,5] and it will become a good set. Hence, we have to find the minimum number of segments that we have to remove(including 0) in order to make it a good Set. APPROACH/ALGORITHM: 1. Store all the range in an array of type pair<int, int> arr[] 2. Store Left ranges in Left[] i.e., for ex, [1,3], [2,5],[ 5,6], then store only 1,2 and 6. 3. Store Right ranges in Right[] = {3,5,6}. 4. Sort the left and right array. 5. Iterate on the arr[], from 0 till n. Find all the elements in Right[] that are lower than arr[i].left and all the elements in Left[] that are greater than arr[i].right , store it in a variable = counter(let’s say) 6. ans = min(ans, counter) Here, what we are doing is, we are finding all the elements that end before the left and all the elements starting after the right, Hence we have to remove these elements in order to make it a good set. https://codeforces.com/contest/1462/submission/101420915 Happy Coding :) The girl who loves to code | Software Developer | Follow me at: https://salonix.netlify.app ## More from Salonix__ The girl who loves to code | Software Developer | Follow me at: https://salonix.netlify.app

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If One Root the Equation 2x2 + Kx + 4 = 0 is 2, Then the Other Root is - Mathematics MCQ If one root the equation 2x2 + kx + 4 = 0 is 2, then the other root is • 6 • -6 • -1 • 1 Solution Let alpha and beta be the roots of quadratic equation2x^2 + kx + 4 = 0 in such a way that alpha = 2 Here, a  = 2, b = k and , c = 4 Then , according to question sum of the roots alpha + beta = (-b)/a 2+ beta = (-k)/2 beta = (-k)/2 - 2 beta = (-k -4)/2 And the product of the roots alpha .beta = c /a = 4/2 = 2 Putting the value of beta = (-k -4)/2in above 2 xx (-k - 4)/ 2 = 2 (-k - 4) = 2 k = -4 -2 = -6 Putting the value of in beta = (-k - 4)/2 beta = (-(6) - 4)/2 = (6-4)/2 = 2/2 beta = 1 Therefore, value of other root be beta = 1 Is there an error in this question or solution? APPEARS IN RD Sharma Class 10 Maths

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# C Program to Find Sum of Opposite Diagonal Elements in a Matrix In this tutorial, i am going to show you how to find sum of opposite diagonal elements in a matrix in c programs. ## C Program to Find Sum of Opposite Diagonal Elements in a Matrix ```/* C Program to find Sum of Opposite Diagonal Elements of a Matrix */ #include<stdio.h> int main() { int i, j, rows, columns, a[10][10], Sum = 0; printf("\n Please Enter Number of rows and columns : "); scanf("%d %d", &i, &j); printf("\n Please Enter the Matrix Elements \n"); for(rows = 0; rows < i; rows++) { for(columns = 0;columns < j;columns++) { scanf("%d", &a[rows][columns]); } } for(rows = 0; rows < i; rows++) { for(columns = 0;columns < j; columns++) { if(rows + columns == ((i + 1) - 2)) Sum = Sum + a[rows][columns]; } } printf("\n The Sum of Opposite Diagonal Elements of a Matrix = %d", Sum ); return 0; }``` The result of the above c program; as follows: ```Please Enter Number of rows and columns : 3 3

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This site is supported by donations to The OEIS Foundation. Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A236769 Numbers n such that lpf(2^n -1) < lpf(2^lpf(n) -1). 3 55, 77, 161, 169, 221, 275, 299, 323, 377, 385, 391, 437, 481, 493, 539, 551, 559, 605, 611, 629, 689, 697, 703, 715, 731, 779, 793, 799, 817, 847, 893, 901, 923, 935, 949, 1001, 1007, 1027, 1045, 1073, 1079, 1121, 1127, 1147, 1159, 1241, 1265, 1271, 1273, 1309 (list; graph; refs; listen; history; text; internal format) OFFSET 1,1 COMMENTS The numbers n for which A049479(n) < A049479(lpf(n)), where lpf(n) = A020639(n). All other n satisfy the equality (in particular all primes). All terms are odd and composite. - Chai Wah Wu, Oct 04 2019 LINKS Chai Wah Wu, Table of n, a(n) for n = 1..111 PROG (PARI) lpf(n) = vecmin(factor(n)[, 1]); lista() = {my(vlpfmp = readvec("A049479.log")); for (i=2, #vlpfmp, if (vlpfmp[i] < vlpfmp[lpf(i)], print1(i, ", ")); ); } \\ Michel Marcus, Jan 31 2014 CROSSREFS Cf. A049479 (a question in the third comment). Sequence in context: A050781 A060260 A152080 * A119224 A135984 A140377 Adjacent sequences:  A236766 A236767 A236768 * A236770 A236771 A236772 KEYWORD nonn,more AUTHOR Thomas Ordowski, Jan 31 2014 EXTENSIONS More terms from Michel Marcus, Jan 31 2014 More terms from Chai Wah Wu, Oct 04 2019 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified December 10 12:25 EST 2019. Contains 329895 sequences. (Running on oeis4.)

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# Help, step 12v using arduino Hi I do not know anything about electronics but know alot about programming. I want to digitally control a 12v source, using 0 to 255, so: 255 = 12v 0 = 0v 128 = 6v (Doesn't have to scale perfectly) I have a 12v source available and an arduino of course How would I go about doing this? Thanks Andy The correct answer depends on the impedance of the device you want to control. One way is to integrate the 5volt PWM first with a resistor and a capacitor. Followed by a 2.4x amplifying opamp. This won't "load" the integrator, so good linearity. Another way is to first amplify the 5volt PWM to 12volt PWM with a transistor. And then integrate the 12volt PWM. What is the devide you want to control. Leo.. I actually want to control 3 LED strips Using 3 different channels from the arduino Well its one LED strip, a RGB strip So I could do say Pin 8 = 255 Pin 7 = 128 Pin 6 = 128 This would show a light red You need 3 NPN transistors. Arduino drives base of each thru a 220, 250, 270 resistor, something in that range (but no less than 220). Emitters connect to Gnd. Collectors connect to R, G, B of LED strip, + to 12V. analogWrite the values you want control brightness of each color. Each 3 LEDs in the strip will require about 60mA. How many LEDs in the strip you are using? That will have an impact o the transistor you use, and perhaps change to N-Channel MOSFET might be suggested, such as AOI514: http://www.digikey.com/product-search/en?vendor=0&keywords=aoi514 (forum screws up link, may need to edit the beginning of it) Ignore post#1, because you gave me the wrong information. Leo.. Ive just recieved the LED strip and it has +12v, R, G, B Im guessing that means 3 Gnds Will that still work? I reckon ill use about 36 LEDs, Im actually using a Computers PSU to power them, then the arduino to fade them Thanks Yes, that will work and is the easiest and most common way. That's why it has a configuration like that. Controlling a 12V led strip with an Arduino is done like a million times before so yeay.... How did you do the keyboard macro for the Google search ? Let me Google that for you…

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Java program to display fibonacci sequence up to n has been shown here. Here n is the limit up to which the sequence is to be generated. For example if n = $20$, we get the fibonacci numbers up to $20$ i.e. $0, 1, 1, 2, 3, 5, 8, 13$. The following section covers the iterative and recursive approaches to find fibonacci sequence. The algorithm, pseudocode of the program have been shown below. ## 1. Algorithm to display fibonacci sequence upto n 1. Take the limit n as input. 2. Assign the first two fibonacci numbers to variables a, b i.e. a = 0 and b = 1 3. If n = 0, display a else, display a, b. 4. Check if a + b <= n 5. If step 4 is true perform step 6 to 8, else stop the process 6. t = a + b and display t 7. a = b and b = t 8. Go to step 4 ## 2. Pseudocode to display fibonacci sequence upto n Input : A limit $n$ Output : Fibonacci sequence upto $n$ 1. Procedure fibonacciUptoN($n$): 2. $a \leftarrow 0$ 3. $b \leftarrow 1$ 4. If $n == 0$: 5. Display $a$ 6. Else: 7.Display $a, b$ 8. Repeat until $a + b \leq n$: 9. $t \leftarrow a + b$ 10.Display $t$ 11.$a \leftarrow b$ 12.$b \leftarrow t$ 13. End Procedure ## 3. Java Program & output to display fibonacci sequence upto n using iteration Code has been copied /******************************* alphabetacoder.com Java program to find fibonacci series up to N using iteration ********************************/ import java.util.Scanner; class Main { public static void main(String args[]) { // declare object of Scanner class Scanner sc = new Scanner(System.in); // declare variables int n, a, b, t = 0; // take input of the limit System.out.print("Enter the limit = "); n = sc.nextInt(); // intialize the first two terms of the sequence a = 0; b = 1; System.out.print("Fibonacci sequence upto " + n + ": "); // display the first fibonacci if (n == 0) System.out.print(a); // display first two fibonacci else System.out.print(a + " " + b); //now calculate the remaining terms upto n while (a + b <= n) { // calculate next fibonacci t = a + b; //display next fibonacci System.out.print(t + " "); //assign values for next iteration a = b; b = t; } } } Output Enter the limit = 20 Fibonacci sequence upto 50: 0 1 1 2 3 5 8 13 ## 4. Java Program & output to display fibonacci sequence upto n using recursion Code has been copied /******************************* alphabetacoder.com Java program to find fibonacci series up to N using iteration ********************************/ import java.util.Scanner; class Main { // recursive function to display // fibonacci sequence static void fibonacci(int a, int b, int n) { if (a <= n) { System.out.print(a + " "); fibonacci(b, a + b, n); } } public static void main(String args[]) { // declare object of Scanner class Scanner sc = new Scanner(System.in); // declare variables int n; // take input of the limit System.out.print("Enter the limit = "); n = sc.nextInt(); System.out.print("Fibonacci sequence upto " + n + ": "); // call the function // pass value of the first two // terms and limit fibonacci(0, 1, n); } } Output Enter the limit = 50 Fibonacci sequence upto 50: 0 1 1 2 3 5 8 13 21 34

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Cody Nofar Nizri Rank Score 1 – 32 of 32 Problem 751. Implement simple rotation cypher Created by: Doug Hull Problem 44499. Capitilize the first letter of every word in a string Created by: Shaul Salomon Problem 1043. Generate a string like abbcccddddeeeee Created by: Aurelien Queffurust Problem 462. letter yes yes & letter no no Created by: AMITAVA BISWAS Problem 44491. Shuffle Created by: Shaul Salomon Problem 44490. Vector pop Created by: Shaul Salomon Tags 22105, vectors Problem 44486. Vector push Created by: Shaul Salomon Tags 22105, vector Problem 44492. Approximate the cosine function Created by: Shaul Salomon Problem 44484. Separate even from odd numbers in a vector - with a loop Created by: Shaul Salomon Problem 44483. Separate even from odd numbers in a vector - without loops Created by: Shaul Salomon Problem 44476. How many unique Pythagorean triples? Created by: Shaul Salomon Problem 44469. Diagonal Pattern Created by: Shaul Salomon Problem 59. Pattern matching Created by: Cody Team Tags indexing Problem 48. Making change Created by: Cody Team Tags money Problem 44467. Product of Each Column Created by: Shaul Salomon Tags 22105, matrices Problem 44446. Add a vector to a matrix Created by: Shaul Salomon Tags 22105, matrices Problem 44439. Remove the air bubbles from a vector Created by: Shaul Salomon Tags 22105, vectors Problem 44447. Eye Squared Created by: Shaul Salomon Tags 22105, matrices Problem 44445. Tax Calculator Created by: Shaul Salomon Tags 22105, tax Problem 44437. How many days in a month? Created by: Shaul Salomon Problem 44436. Find the largest number Created by: Shaul Salomon Tags 22105, max, vectors Problem 1129. Reverse the elements of an array Created by: HARISANKAR PS Problem 798. Rotate input square matrix 90 degrees CCW without rot90 Created by: Christopher Problem 611. surrounded matrix Created by: Grzegorz Knor Tags matrices Problem 411. Back to basics 21 - Matrix replicating Created by: Alan Chalker Problem 42320. Write a function man that takes a row vector v and returns a matrix H as follows.. Created by: Manish Joshi Problem 1422. frame of the matrix Created by: Marco Castelli Tags frame Problem 44402. Horizontal matrix sort Created by: Shaul Salomon Problem 44401. Vertical matrix sort Created by: Shaul Salomon Problem 44416. Sum of adjacent elements in a vector Created by: Shaul Salomon Tags 22105, vectors Problem 44415. Remove the first, third and fifth rows of a matrix Created by: Shaul Salomon Problem 44392. Sum of Two Numbers Created by: Shaul Salomon 1 – 32 of 32

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# Can You Form 4 Equal Triangles? These Puzzles May Reveal Whether You Can Think Logically January 29, 2019 Logic is a crucial thinking skill for everyone. And it’s wrong to think that it is only those working in Mathematics or Science who need it. Everyday life is full of questions and dilemmas that we may approach with logic. Peshkova / Shutterstock.com It’s an ability to plan and analyze and consider reasons that gives us plenty of benefits in decision-making. Are you up to checking how well your logic works? We are here to offer you 3 logical puzzles, which may boost your cognitive abilities and show you how brain-trained you are. 1. Move three buttons In the picture below, you can see a triangle formed with buttons. It points upwards. Your task is to move three buttons to make the figure point downwards. Only three buttons should be moved. Take your time to think of ways to do it. Ready to find out the right answer? Then click on the image. 2. Draw a house Here is a little house formed with geometrical figures. Can you draw it in one stroke only? Two main conditions are: you should not remove the pen from the paper even once, and you can’t cross an already drawn line. You can use a piece of paper to practice. How long did it take to figure out the right answer? Click on the image to check it. 3. Move two matchsticks In this picture, you can see three triangles made of matchsticks. They are all equal in size. The task is to move 2 matchsticks to form 4 equal triangles. Did you cope with this puzzle fast? Click on the picture to see the right answer! One of the efficient ways to develop your logical thinking is solving puzzles and logic games. We are happy to give you a nice brain workout. Hungry for more? Then come back for more interesting tests!

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 Introduction The following numeric system is a side product of my over 10 year long research concerning geometry and higher dimensions (including Kissing number problem). I'm also going to implement this system in a software that's going to search for kissing numbers in dimensions five and up. ----=== Balternalculator ===---- People who know how to add numbers in any other system than decimal (binary, octagonal, hexadecimal, ...), find this calculator easier to understand. ... But... Children who are not yet familiar with the numbers we use today, could learn how to add, subtract, multiply and divide much quicker than they would do using the existing decimal numeric system Lately I have found that the system that I reinvented is actually called a balanced ternary system For convenience I'm going to call the digits trits On the calculator, red(up) means positive (+), brown/uncoloured means zero (0) and green(down) means negative (-). Let me give you an example: Red(up) on the first field affront of the decimal point (rightmost) means that we add +1 to the current number, red(up) on the second field affront of the decimal point means that we add +3 to the current number, the next one +9, and so on... In the contrary, greens(downs), from right to left, mean -1, -3, -9,... Instead of writing a table, in which I would explain how numbers are represented using balternator, I've decided to make an online virtual calculator for you to get a glimpse of how this system works. Click here to give it a try. for adding two numbers there's only one rule: whenever pressing on a either of the two colours(red and green), you have to turn the neighboring trit accordingly(as shown in the video below). Subtraction: To subtract is equal to negate the number you are subtracting and then add it to the other number: [video link] Multiplication: Multiplying with this calc is a bit dodgy, but later on I'm going to show you another way. ----------------------------------------------------- Conclusions: Imagine all the work and time saved by not having to learn multiplication table, building abacus, learning to divide and so on. Also in computer science balanced ternary is very usable because: - There's no need for an extra minus sign for representing the negativity of a number - when cutting off digits behind the decimal point (=rounding numbers), you don't need to worry about rounding the number (i.e. - 1.44 rounds to 1.4 while 1.45 rounds to 1.5; and we encounter the same problem in binary) Some people say: And I reply: you need too many digits to write down a number. You can combine every two trits and get balanced nonary system which is very close to our currently used decimal(in terms of number of digits used for representing a number), and still use the same procedures to calculate numbers as in balanced ternary [video link] There's loads of conversion to do Yes, but that is because we're used to decimal. If you have a balanced ternary ruler, no conversion is needed at all. [video/pic link] Yes, but that is because we're used to decimal. If you have a balanced ternary ruler, no conversion is needed at all. [video/pic link] other interesting things about this system -another multiplying:[pic link] ; as we see, the procedure for multiplication strictly follows the mathematic rules, as well as it corresponds to the behavior of two magnetic poles. i.e.: Let us say red means to repel - increase distance (+) and green means to attract - decrease distance(-) And also that red and green are the two opposite magnetic poles while brown/uncoloured is -year -3-digit letters ...Page under construction.. i Z ! 2 0 1 6

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## Precalculus (6th Edition) Blitzer The algebraic expression of $\sin \left( {{\cos }^{-1}}x \right)$ is $\sqrt{1-{{x}^{2}}}$. Let ${{\cos }^{-1}}x=\theta$. Then, \begin{align} & \cos \theta =\frac{x}{1} \\ & =\frac{\text{perpendicular}}{\text{Hypotenuse}} \end{align} According to the Pythagorean Theorem, to find out the perpendicular side, $\text{Perpendicular}=\sqrt{{{\left( \text{Hypotenuse} \right)}^{2}}-{{\left( \text{Adjacent side} \right)}^{2}}}=\sqrt{{{\left( \text{1} \right)}^{2}}-{{\left( x \right)}^{2}}}=\sqrt{1-{{x}^{2}}}$ Now, consider $\sin \left( {{\cos }^{-1}}x \right)=\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\sqrt{1-{{x}^{2}}}}{1}=\sqrt{1-{{x}^{2}}}$ Hence, the algebraic expression of $\sin \left( {{\cos }^{-1}}x \right)$ is $\sqrt{1-{{x}^{2}}}$.

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http://gmatclub.com/forum/food-number-of-calories-per-kiologram-number-of-gram-of-36140.html?kudos=1

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) Food Number of Calories per Kiologram Number of Gram of : DS Archive Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 16 Jan 2017, 17:47 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # ) Food Number of Calories per Kiologram Number of Gram of post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Manager Joined: 11 Jan 2006 Posts: 230 Location: Arkansas, US WE 1: 2.5 yrs in manufacturing Followers: 1 Kudos [?]: 17 [0], given: 18 ) Food Number of Calories per Kiologram Number of Gram of [#permalink] ### Show Tags 03 Oct 2006, 04:24 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. 20.) Food Number of Calories per Kiologram Number of Gram of Protein per Kilogram S 2,000 150 T 1,500 90 The table above gives the number of calories and grams of protein per kilogram of foods S and T. If a total of 7 kilograms of S and T are combined to make a certain food mixture, how many kilograms of food S are in the mixture? (1) The mixture has a total of 12,000 calories. (2) The mixture has a total of 810 grams of protein. Plss explain _________________ ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED Intern Joined: 14 Jul 2005 Posts: 44 Location: California Followers: 1 Kudos [?]: 7 [0], given: 0 ### Show Tags 04 Oct 2006, 12:23 Each ststement by itself is sufficient. We know that S + T = 7 From Statement 1: (1) The mixture has a total of 12,000 calories. if the whole mixture was made of S the there would be 14000 calories(7x2000) but the mixture has 12000 Calories If we substitute 1 S with 1 T then the the calorie count would be down by(2000-1500) = 500. So to get the Calorie count down by 2000 we need to substitute 4 S with 4 T. So In the mixture we have 3 S and 4 T= (3x2000)+(4x1500) =6000 + 6000 = 12000 We can get the values exactly the same way from statement 2. Director Joined: 23 Jun 2005 Posts: 847 GMAT 1: 740 Q48 V42 Followers: 5 Kudos [?]: 70 [0], given: 1 ### Show Tags 04 Oct 2006, 15:38 D Statement 1: 2000X+1500Y = 12000 X+Y= 7 or X=7-Y Substituting for X and simplifying, 500Y = 2000 Y = 4Kg. Sufficient. Statement 2 is sufficient, going by the above model. Senior Manager Joined: 28 Aug 2006 Posts: 306 Followers: 13 Kudos [?]: 150 [0], given: 0 ### Show Tags 04 Oct 2006, 23:30 It is D. Statement 1: Let S be P kg Clearly T must be 7-P kg 2000xP + (7-P) 1500 = 12,000 Solving this we get P=3 So S must be 3 and T must be 4 Statement 2: 150xP+(7-P)90 = 810 Solving this we get P=3 So it is D _________________ Last edited by cicerone on 25 Sep 2008, 00:31, edited 1 time in total. Manager Joined: 11 Jan 2006 Posts: 230 Location: Arkansas, US WE 1: 2.5 yrs in manufacturing Followers: 1 Kudos [?]: 17 [0], given: 18 ### Show Tags 05 Oct 2006, 19:52 Thanks for the explanation guys... I've undestud the concept, thanks to u all.. _________________ ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED 05 Oct 2006, 19:52 Display posts from previous: Sort by # ) Food Number of Calories per Kiologram Number of Gram of post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

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# SOLUTION: the perimeter of a rectangle is 46 feet. the width is 3ft lees than twicw the length. find the lenght and width of the rectangle Algebra ->  -> SOLUTION: the perimeter of a rectangle is 46 feet. the width is 3ft lees than twicw the length. find the lenght and width of the rectangle      Log On Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Mathway solves algebra homework problems with step-by-step help! Question 729054: the perimeter of a rectangle is 46 feet. the width is 3ft lees than twicw the length. find the lenght and width of the rectangleAnswer by checkley79(3056)   (Show Source): You can put this solution on YOUR website!P=2L+2W W=2L-3 46=2L+2(2L-3) 46=2L+4L-6 46+6=6L 6L=52 L=52/6 L=8.67 FT. ANS. W=2*8.67-3 W=17.33-3 W=14.33 FT. ANS. PROOF: 46=2*8.67+2*14.33 46=17.33+28.66 46=46

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# Taking the Max of Uniform Random Numbers There is 80 lines of simple standalone C++ code that generated the data for this post at: Let’s say you generate 1,000,000 random numbers from 0 to 255 and count how many times each number came up in a histogram. I did that and here it is: In a previous post I talked about how adding dice rolls together, counting bits in a random number, and similar, would approach a normal / Gaussian distribution: https://blog.demofox.org/2017/07/25/counting-bits-the-normal-distribution/ The same thing happens if you average random numbers, but there’s a nice side effect of getting values in the same range. That is to say: if you roll N 6 sided dice and average them, you will still get values between 1 and 6, but the more dice there are, the more the distribution will approach Gaussian. It shapes up pretty quickly too. Here is the histogram: Something interesting to note is that adding two uniform random numbers together – or averaging them – makes something called “triangle distributed noise”. Looking at the histogram for averaging two values, hopefully you can see why it’s called triangle distributed! Triangle noise some cool properties for noise in graphics and is orthogonal to eg white noise vs blue noise vs low discrepancy sequences. Click to access banding_in_games.pdf I recently saw some code that was taking the maximum of two random numbers and that made me wonder what sort of distribution that might give. Being a better programmer than a mathematician, i made a histogram, then took a guess as to what the formula behind the shape might be. It turns out that taking the max of N uniform random numbers is the same as using y=x^(N-1) as a PDF. (You of course would need to normalize it to be a real PDF) Here are the histograms: These are apparently beta distributions: https://en.wikipedia.org/wiki/Beta_distribution If we take the min rather than the max, what happens then? Well, if for example the numbers are between 0 and 1, taking the min will give you the same count as if you took the max of 1-x. So, the graph of the histogram should look the same, just flipped on the x axis. It turns out that it does:

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Anzeige Anzeige Anzeige Anzeige Anzeige Anzeige Anzeige Anzeige Anzeige Anzeige Nächste SlideShare Appilation of matrices in real life 1 von 46 Anzeige ### Matrices and its Applications to Solve Some Methods of Systems of Linear Equations 1. UNIVERSITYOF DUHOK FACLTY OF EDUATIONAL SCIENCE SCHOOL OF BASEEDUCATION DEPARTMWNT OF MATHMATICS Matrices and its Applications to Solve Some Methods of Systems of Linear Equations A project submitted to thecouncilof Department of Mathematics, School of Basic Education, Universityof Duhok, in partial fulfillment of the requirement for B.Sc. degree of mathematics 𝑷𝒓𝒆𝒑𝒂𝒓𝒆𝒅 𝒃𝒚: 𝑺𝒖𝒑𝒆𝒓𝒗𝒊𝒔𝒆𝒅 𝒃𝒚: 𝑨𝒃𝒅𝒂𝒍𝒍𝒂 𝑯𝒂𝒋𝒊 𝑴𝒖𝒘𝒂𝒇𝒂𝒒 𝑴𝒂𝒉𝒅𝒊 𝑺𝒂𝒍𝒊𝒉 𝑲𝒂𝒓𝒘𝒂𝒏 𝑯𝒂𝒕𝒎 1436 .A.H 2015.A.D 2715.K 2. Acknowledgement First of all, thanks to Allah throughout all his almighty kindness, and loveliness for letting us to finish our project. We would like to express our thanks to our supervisor 𝑴𝒖𝒘𝒂𝒇𝒂𝒒 𝑴𝒂𝒉𝒅𝒊 𝑺𝒂𝒍𝒊𝒉 for giving us opportunity to write this research under his friendly support. He made our research smoothly by his discerning ideas and suggestions. Also, we would like to thank all our friends and those people who helped us during our work. 3. . Contents Chapter One Basic Concepts in Matrix. (1.1) Matrix…………………………………………………………….….....1 (1.2) Some Special Types of Matrices…………………..………....2 (1.3) Operations of Matrices…………………….……………..... ....11 (1.4) The Invers of a Square Matrix……………………………….17 (1.5) Some Properties of Determinants………………….…..….25 Chapter Two System of Linear Equations (2.1) Linear Equation………………………………………………………29 (2.2) Linear System……………………………………………………….....30 (2.2.1) Homogeneous System……………………………………….…..33 (2.2.2) Gaussian Elimination………………………………………….…34 (2.2.3) Gaussian-Jordan Elimination…………………………………35 (2.2.4) Cramer’s Rule………………………………………………….…...37 References…………………………………………………………….…………39 4. Abstract In this research, we have tried to introduce matrix, its types and finding inverse and determinant of matrix has involved. Then the applications of matrices to some methods of solving systems of linear equations, such as Homogeneous, Gaussian Elimination, Gaussian –jordan Elimination and Cramer’s Rule, have been illustrated. 5. Introduction Information in science and mathematics is often organized into rows and columns to form rectangular arrays called "Matrices" (plural of matrix). Matrices are often tables of numerical data that arise from physical observation, but they also occur in various mathematical contexts. Linear algebra is a subject of crucial important to mathematicians and users of mathematics. When mathematics is used to solve a problem it often becomes necessary to find a solution to a so-called system of linear equations. Applications of linear algebra are found in subjects as divers as economic, physics, sociology, and management consultants use linear algebra to express ideas solve problems, and model real activities. The aim of writing this subject is to applying matrices to solve some types of system of linear equations. In the first chapter of this work matrices have introduced. Then operations on matrices (such as addition and multiplication) where defined and the concept of the matrix inverse was discussed. In the second chapter theorems were given which provided additional insight into the relationship between matrices and solution of linear systems. Then we apply matrices to solve some methods of linear system such as Homogeneous, Gaussian Elimination, Gaussian –jordan Elimination and Cramer’s Rule. 6. CHAPTERONE Basic Concepts in Matrix In this chapter we begin our study on matrix, some special types of matrix, operation of matrix, and finding invers and the determinant of matrices. (1.1) Matrix An 𝑚 × 𝑛 matrix 𝐴 is rectangular array of 𝑚 × 𝑛 numbers arranged in 𝑚 rows and 𝑛 columns: A= [ a11 a12 ⋯ a1j ⋯ a1n a21 a22 ⋯ a2j ⋯ a2n ⋮ ⋮ ⋮ ⋮ ai1 ai2 ⋯ aij ⋯ ain ⋮ ⋮ ⋮ ⋮ am1 am2 ⋯ amj ⋯ amn] The 𝑖𝑗 𝑡ℎ component of 𝐴 denoted 𝑎𝑖𝑗, is the number appearing in the 𝑖 𝑡ℎ row and 𝑗 𝑡ℎ column of 𝐴 we will some time write matrix 𝐴 as A=( 𝑎𝑖𝑗). An m×n matrix is said to have the size m×n. Examples: (1) [ 1 2 3 4 5 6 ] 3×2 𝑚=3 , 𝑛=2 , (2) [ 1 4 3 2 2 3 5 6 4 3 5 1 2 0 7 1 2 1 9 8 ] 4×5 𝑚=4 , 𝑛=5 7. (1.2) SomeSpecialTypesof Matrices 1. SquareMatrix The matrix which its number of rows equals to numbers of columns is called square matrix. That is A=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 ] m×n When m=n then: A= [ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋱ ⋮ 𝑎 𝑛1 𝑎 𝑛2 ⋯ 𝑎 𝑛𝑛 ] n×n A is a square matrix Example: K=[ 1 6 9 2 5 8 3 4 7 ] 3×3 2. Unit (Identity) Matrix The 𝑛 × 𝑛 matrix 𝐼 𝑛 =𝑎𝑖𝑗 , defined by 𝑎𝑖𝑗=1 if = 𝑗 , 𝑎𝑖𝑗=0 if 𝑖 ≠ 𝑗, is called the 𝑛 × 𝑛 identity matrix . 𝐼=[ 1 0 ⋯ 0 0 1 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 1 ] Example: 𝐼2=[1 0 0 1 ] , 𝐼3=[ 1 0 0 0 1 0 0 0 1 ] 8. Note: 𝐼𝐴 = 𝐴𝐼 = 𝐴 Example: 𝐴=[2 3 4 5 ] , 𝐼=[1 0 0 1 ] 𝐴𝐼=[2 3 4 5 ] × [1 0 0 1 ]=[ 2 ∗ 1 + 3 ∗ 0 2 ∗ 0 + 3 ∗ 1 4 ∗ 1 + 5 ∗ 0 4 ∗ 0 + 5 ∗ 1 ] = [2 3 4 5 ] 𝐼𝐴=[1 0 0 1 ] × [2 3 4 5 ] = [1 ∗ 2 + 0 ∗ 4 1 ∗ 3 + 0 ∗ 5 0 ∗ 2 + 1 ∗ 4 0 ∗ 3 + 1 ∗ 5 ] = [2 3 4 5 ] We note that 𝐼𝐴 = 𝐴𝐼 = 𝐴 3. Null(Zero) Matrix A zero matrix is a matrix which its elements are zeros and is denoted by the symbol 0 . 0 𝑚𝑛= [ 0 0 … 0 0 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 0 ] Example: Let 𝐴=[2 1 −3 4 5 8 ] and 𝐵=[ 2 1 −3 4 ] We see that 𝐴 +023=[2 1 −3 4 5 8 ]+[0 0 0 0 0 0 ]=[2 1 −3 4 5 8 ]=𝑨 𝐵022=[ 2 1 −3 3 ] [0 0 0 0 ]=[0 0 0 0 ]=022 Property of ZeroMatrix  𝐴 + 0 = 0 + 𝐴 = 𝐴  𝐴 − 𝐴 = 0  0 − 𝐴 = −𝐴  0𝐴 = 0 , 𝐴0 = 0 9. 4. DiagonalMatrix Diagonal matrix is a square matrix in which all the elements not on the main diagonal are zeros. A=[ 𝑎11 0 ⋯ 0 0 𝑎22 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 𝑎33 ] The elements of a square matrix 𝐴 where the subscripts are equal, namely 𝑎11, 𝑎22, … , 𝑎 𝑛𝑛, form the main diagonal. Example: A=[ 1 0 0 0 2 0 0 0 3 ] , main diagonal=1,2,3 5. CommutativeMatrix We say that the matrices 𝐴 and 𝐵 are commutative under the operation product if 𝐴 and 𝐵 are square matrices and 𝐴. 𝐵 = 𝐵. 𝐴 and we say that 𝐴 𝑎𝑛𝑑 𝐵 are invertible commutative if 𝐴 and 𝐵 are square matrices and 𝐴. 𝐵 = 𝐵. 𝐴 . Example: 𝐴 = [5 1 1 5 ] , 𝐵 = [2 4 4 2 ] 𝐴. 𝐵 = [5 1 1 5 ] [2 4 4 2 ] = [ 10 + 4 20 + 2 2 + 20 4 + 10 ] = [14 22 22 14 ] 𝐵. 𝐴 = [2 4 4 2 ] [5 1 1 5 ] = [ 10 + 4 2 + 20 20 + 2 4 + 10 ] = [14 22 22 14 ] Then 𝐴. 𝐵 = 𝐵. 𝐴 10. Note: A square matrix 𝐴 is said to be invertibleif there exists 𝐵 such that 𝐴𝐵 = 𝐵𝐴 = 𝐼. 𝐵 is denoted 𝐴−1 and is unique. If 𝑑𝑒𝑡(𝐴) = 0 then a matrix is not invertible. 6. TriangularMatrix A square matrix 𝐴 is 𝑛 × 𝑛 , (𝑛 ≥ 3), is triangular matrix iff 𝑎𝑖𝑗 =0 when 𝑖 ≥ 𝑗 + 1 ,or 𝑗 ≥ 𝑖 + 1. The are two type of triangular matrices: i. Upper Triangular Matrix A square matrix is called an upper triangular matrix if all the elements below the main diagonal are zero. Example: 𝐴=[ 1 5 9 0 2 1 0 0 3 ] ii. Lower Triangular Matrix A square matrix is called lower triangular matrix if all the elements above the main diagonal are zero. Example: 𝐴=[ 1 0 0 6 2 0 9 7 3 ] Transpose of Matrix Transpose of 𝑚 × 𝑛 matrix 𝐴 ,denoted 𝐴 𝑇 or 𝐴̀, is 𝑛 × 𝑚 matrix with ( 𝐴𝑖𝑗) 𝑇 = 𝐴𝑗𝑖 11. 𝐴=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 ] m×n , 𝐴 𝑇 =[ 𝑎11 𝑎21 ⋯ 𝑎 𝑚1 𝑎12 𝑎22 ⋯ 𝑎 𝑚2 ⋮ ⋮ ⋮ 𝑎1𝑛 𝑎2𝑛 ⋯ 𝑎 𝑚𝑛 ] n×m row and columns of 𝐴 are transposed in 𝐴 𝑇 Example: 𝐴 = [ 0 4 7 0 3 1 ] , 𝐴 𝑇 = [0 7 3 4 0 1 ] Note: transpose converts row vectors to column vector , vice versa.  Properties of Transpose Let 𝐴 and 𝐵 be matrix and 𝑐 be a scalar. Assume that the size of the matrix are such that the operations can be performed.  ( 𝐴 + 𝐵) 𝑇 = 𝐴 𝑇 + 𝐵 𝑇 Transpose of the sum  ( 𝑐𝐴) 𝑇 = 𝑐𝐴 𝑇 Transpose of scalar multiple  ( 𝐴𝐵) 𝑇 = 𝐵 𝑇 𝐴 𝑇 Transpose of a product  ( 𝐴 𝑇) 𝑇 = 𝐴 7. SymmetricMatrix A real matrix 𝐴 is called symmetric if 𝐴 𝑇 = 𝐴. In other words 𝐴 is square (𝑛 × 𝑛 )and 𝑎𝑖𝑗 = 𝑎𝑗𝑖 for all 1 ≤ 𝑖 ≤ 𝑛, 1 ≤ 𝑗 ≤ 𝑛. Example: 𝐴=[ 1 0 5 0 2 6 5 6 3 ] 𝐴 𝑇 =[ 1 0 5 0 2 6 5 6 3 ] Note: if 𝐴 = 𝐴 𝑇 then 𝐴 is a symmetric matrix 8. Skew-Symmetric Matrix 12. Areal matrix 𝐴 is called Skew-Symmetricif 𝐴 𝑇 = −𝐴 . In other words 𝐴 is square (𝑛 × 𝑛) and 𝑎𝑗𝑖=−𝑎𝑖𝑗 for all 1 ≤ 𝑖 ≤ 𝑛, 1 ≤ 𝑗 ≤ 𝑛 Example: A=[ 0 5 6 −5 0 8 −6 −6 0 ] − 𝐴=[ 0 −5 −6 5 0 −8 6 8 0 ] A=[ 0 5 6 −5 0 8 −6 −8 0 ] 𝐴 𝑇 =[ 0 −5 −6 5 0 −8 6 8 0 ] ∴ 𝐴 𝑇 = −𝐴 Determinants of Matrix The determinant of a square matrix 𝐴 = [𝑎𝑖𝑗] is a number denoted by |A| or 𝑑𝑒𝑡(𝐴) , through which important properties such as singularity can be briefly characterized . This number is defined as the following function of the matrix elements: |𝐴| = 𝑑𝑒𝑡(𝐴) = ± ∏ 𝑎1𝑗1 𝑎2𝑗2 … 𝑎 𝑛𝑗 𝑛 Where the column indices 𝑗1, 𝑗2, …, 𝑗 𝑛 are taken from the set {1,2,…,n} with no repetitions allowed . The plus (minus) sign is taken if the permutation (𝑗1 𝑗2 … 𝑗 𝑛) is even (odd). Someproperties of determinants willbe discussed later inthis chapter 9. Singular and Nonsingular Matrix A square matrix 𝐴 is said to be singular if 𝑑𝑒𝑡(𝐴) = 0 . 𝐴 is nonsingular if 𝑑𝑒𝑡(𝐴) ≠ 0. Theorem: Let 𝐴 be a square matrix. Then 𝐴 is a singular if (a) all elements of a row (column) are zero. (b) two rows (column) are equal. (c) two rows(column) are proportional. 13. Note: (b) is a special case of (c) , but we list it separately to give it special emphasis. Example: we show that the following matrices are singular. (a) A=[ 2 0 −7 3 0 1 −4 0 9 ] (b) B=[ 2 −1 3 1 2 4 2 4 8 ] (a) All the elements in column 2 of A are zero. Thus 𝑑𝑒𝑡 = 0. (b) Observe that every element in row 3 of B is twice the corresponding element in row 2. We write (row 3) = 2(row 2) Row 2 and 3 are proportional. Thus 𝑑𝑒𝑡(𝑏) = 0. 10. OrthogonalMatrix we say that a matrix 𝐴 is orthogonal if 𝐴. 𝐴 𝑇 = 𝐼 = 𝐴 𝑇 . 𝐴 Example: 𝐴 = [ 1 0 0 0 1 2 √3 2 0 − √3 2 1 2 ] , 𝐴 𝑇 = [ 1 0 0 0 1 2 − √3 2 0 √3 2 1 2 ] 𝐴. 𝐴 𝑇 = [ 1 0 0 0 1 2 √3 2 0 − √3 2 1 2 ] . [ 1 0 0 0 1 2 − √3 2 0 √3 2 1 2 ] = [ 1 0 0 0 1 4 + 3 4 1 2 . −√3 2 + √3 2 . 1 2 0 −√3 2 . 1 2 + 1 2 . √3 2 3 4 . 1 4 ] = [ 1 0 0 0 1 0 0 0 1 ] = 𝐼 ∴ 𝐴 is orthogonal matrix 14. 11. 𝐓𝐨𝐞𝐩𝐥𝐢𝐭𝐳Matrix A matrix 𝐴 is said to be 𝐓𝐨𝐞𝐩𝐥𝐢𝐬𝐥𝐢𝐭𝐳 if it has common elements on their diagonals, that is 𝑎𝑖,𝑗=𝑎𝑖+1 ,𝑗+1 Example: A=[ 5 6 2 0 5 6 3 0 5 ] Where 𝑎11=𝑎22=𝑎33=5 𝑎12=𝑎23=6 𝑎21=𝑎32=0 12. Nilpotent Matrix A square matrix 𝐴 is said to nilpotent if there is a positive integer 𝑝 such that 𝐴 𝑃 = 0. The integer 𝑝 is called the degree of 𝑛𝑖𝑙𝑝𝑜𝑡𝑒𝑛𝑐𝑦 of the matrix. Example: 𝐴 = [ 1 −3 −4 −1 3 4 1 −3 −4 ] , 𝑝 = 2 𝐴2 = 0 𝐴. 𝐴 = [ 1 −3 −4 −1 3 4 1 −3 −4 ] . [ 1 −3 −4 −1 3 4 1 −3 −4 ] = [ 0 0 0 0 0 0 0 0 0 ] 13. Periodic (Idempotent) Matrix A matrix 𝐴 is said to be periodic , that period (order) 𝐾 , if 𝐴 is satisfy 𝐴 𝐾+1 = 𝐴 and if 𝐾 = 1 then 𝐴2 = 𝐴 so 𝐴 is called idempotent matrix. Example: 15. 𝐴 = [ 1 −2 −6 −3 2 9 2 0 −3 ] , 𝐾 = 2 𝐴 𝐾+1 = 𝐴2+1 = 𝐴3 𝐴. 𝐴 = [ −5 −6 −6 9 10 9 −4 −4 −3 ] , 𝐴2 . 𝐴 = [ 1 −2 −6 −3 2 9 2 0 3 ] ∴ 𝐴 is idempotent matrix 14. StochasticMatrix An 𝑛 × 𝑛 matrix 𝐴 is called stochastic if each element is a number between 0 and 1 and each column of 𝐴 adds up to 1. A= [ 1 4 1 3 0 1 2 2 3 3 4 1 4 0 1 4] , ∑ column(1) =1 , ∑column( 2) = 1 , ∑ column( 3) = 1 15. TraceMatrix Let 𝐴 be a square matrix, the trace of 𝐴 denoted 𝑡𝑟(𝐴) is the sum of the diagonal elements of 𝐴 .Thus if 𝐴 is an 𝑛 × 𝑛 matrix. 𝒕𝒓(𝑨) = 𝒂 𝟏𝟏 + 𝒂 𝟐𝟐 + ⋯ + 𝒂 𝒏𝒏 Example: 16. The trace of the matrix 𝐴 =[ 4 1 −2 2 −5 6 7 3 0 ]. is, 𝑡𝑟(𝐴) = 4 + (−5) + 0 = −1 Properties of Trace Let 𝐴 and 𝐵 be matrix and 𝑐 be a scalar, assume that the sizes of the matrices are such that the operations can be performed.  𝑡𝑟( 𝐴 + 𝐵) = 𝑡𝑟(𝐴) + 𝑡𝑟(𝐵)  𝑡𝑟(𝐴𝐵) = 𝑡𝑟(𝐵𝐴)  𝑡𝑟(𝑐𝐴) = 𝑐𝑡𝑟(𝐴)  𝑡𝑟( 𝐴) 𝑇 = 𝑡𝑟(𝐴) Note: if A is not square that the trace is not defined. 16. 𝑯𝒆𝒓𝒎𝒊𝒕𝒊𝒂𝒏 Matrix A square matrix A is said to be 𝒉𝒆𝒓𝒎𝒊𝒕𝒊𝒂𝒏 if A̅T=A. Note: The conjugate of a complex number 𝑧 = 𝑎 + 𝑖𝑏 is defined and written z̅=a-ib . Example: A=[ 3 7 + 𝑗2 7 − 𝑗2 −2 ] ,is ℎ𝑒𝑟𝑚𝑖𝑡𝑖𝑎𝑛 Taking the complex conjugates of each of the elements in 𝐴 gives A̅=[ 3 7 − 𝑗2 7 + 𝑗2 −2 ] Now taking the transposes of A , we get A̅T=[ 3 7 + 𝑗2 7 − 𝑗2 −2 ] So we can see that A̅T=A 17. (1.3) Operations ofMatrices 1. Addition If 𝐴 and 𝐵 are m × n matrices such that 𝐴=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 ] m×n and 𝐵= [ 𝑏11 𝑏12 ⋯ 𝑏1𝑛 𝑏21 𝑏22 ⋯ 𝑏2𝑛 ⋮ ⋮ ⋮ 𝑏 𝑚1 𝑏 𝑚2 ⋯ 𝑏 𝑚𝑛 ] m×n Then 𝐴 + 𝐵=[ 𝑎11 + 𝑏11 𝑎12 + 𝑏12 ⋯ 𝑎1𝑛 + 𝑏1𝑛 𝑎21 + 𝑏21 𝑎22 + 𝑏22 ⋯ 𝑎2𝑛 + 𝑏2𝑛 ⋮ ⋮ ⋱ ⋮ 𝑎 𝑚1 + 𝑏 𝑚1 𝑎 𝑚2 + 𝑏 𝑚2 ⋯ 𝑎 𝑚𝑛 + 𝑏 𝑚𝑛 ] Note: Addition of matrices of different sizes is not defined. Example: [ 0 4 7 0 3 1 ]+[ 1 2 2 3 0 4 ]=[ 1 6 9 3 3 5 ] Properties of Matrix Addition  𝐴 + 𝐵 = 𝐵 + 𝐴 (commutative)  (𝐴 + 𝐵) + 𝐶 = 𝐴 + (𝐵 + 𝐶) (associative), so we can write as 𝐴 + 𝐵 + 𝐶  𝐴 + 0 = 0 + 𝐴 = 𝐴  ( 𝐴 + 𝐵) 𝑇 = 𝐴 𝑇 + 𝐵 𝑇 2. Subtraction Matrix subtraction is defined for two matrix 𝐴 = [𝑎𝑖𝑗] and 𝐵 = [𝑏𝑖𝑗] of the same size in the usual way; that is 𝐴 − 𝐵 = [𝑎𝑖𝑗] − [ 𝑏𝑖𝑗 ] = [ 𝑎𝑖𝑗 − 𝑏𝑖𝑗]. If 𝐴 and 𝐵 m × n matrix such that 18. 𝐴=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 ] m×n and 𝐵= [ 𝑏11 𝑏12 ⋯ 𝑏1𝑛 𝑏21 𝑏22 ⋯ 𝑏2𝑛 ⋮ ⋮ ⋮ 𝑏 𝑚1 𝑏 𝑚2 ⋯ 𝑏 𝑚𝑛 ] m×n Then 𝐴 − 𝐵=[ 𝑎11 − 𝑏11 𝑎12 − 𝑏12 ⋯ 𝑎1𝑛 − 𝑏1𝑛 𝑎21 − 𝑏21 𝑎22 − 𝑏22 ⋯ 𝑎2𝑛 − 𝑏2𝑛 ⋮ ⋮ ⋱ ⋮ 𝑎 𝑚1 − 𝑏 𝑚1 𝑎 𝑚2 − 𝑏 𝑚2 ⋯ 𝑎 𝑚𝑛 − 𝑏 𝑚𝑛 ] Note: Subtraction of matrices of different sizes is not defined. Example: [ 0 4 7 0 3 1 ]-[ 1 2 2 3 0 4 ]=[ −1 2 5 −3 3 −3 ] 3. Negative Consider 𝐶 to be a matrix, the negative of 𝐶 denoted by −𝐶 , which defined as (−1)𝐶, Where each element in 𝐶 is multiplied by (−1). Example: 𝐶 = [3 −2 4 7 −3 0 ]. Then – 𝐶 = [−3 2 −4 −7 3 0 ]. 4. Multiplication We can product two matrices 𝐴 and 𝐵 if the number of column in a matrix 𝐴 be equal to the number of rows in a matrix 𝐵.The element in row 𝑖 and column 𝑗 of 𝐴𝐵 is obtained by multiplying the corresponding element of row 𝑖 of 𝐴 and column 𝑗 of 𝐵 and adding the products. [ . . . ] 𝐵 is 3 × 𝑛 𝐴 is 𝑚 × 3 [ . . . ] [ . ] 𝐴𝐵 is 𝑚 × 𝑛 19. Note: The product of 𝐴 and 𝐵 con not be obtained if the number of columns in 𝐴 does not equal the number of rows in 𝐵. Let 𝐴 have 𝑛 columns and 𝐵 have 𝑛 rows .The 𝑖 𝑡ℎ row of 𝐴 is [𝑎𝑖1 𝑎𝑖2 …𝑎𝑖𝑛] and the 𝑗 𝑡ℎ column of 𝐵 is [ 𝑏1𝑗 𝑏2𝑗 ⋮ 𝑏 𝑛𝑗] . Thus if = 𝐴𝐵 , then 𝑐𝑖𝑗=𝑎𝑖1 𝑏1𝑗 +𝑎𝑖2 𝑏2𝑗+…+𝑎𝑖𝑛 𝑏 𝑛𝑗 . Propertiesof MatrixMultiplication  0𝐴 = 0, 𝐴0 = 0 (here 0 can be scalar, or a compatible matrix)  𝐼𝐴 = 𝐴𝐼 = 𝐴  (𝐴𝐵)𝐶 = 𝐴(𝐵𝐶), so we can write as 𝐴𝐵𝐶  α(AB)=( αA)B , where 𝛼 is a scalar  𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶 ,(𝐴 + 𝐵)𝐶 = 𝐴𝐶 + 𝐵𝐶  ( 𝐴𝐵) 𝑇 = 𝐵 𝑇 𝐴 𝑇 ScalarMultiplication let 𝐴 be a matrix and 𝑐 be a scalar, the scalar multiple of 𝐴 by 𝑐, denoted 𝑐𝐴 , is the matrix obtained by multiplying every element of 𝐴 by 𝑐, the matrix 𝑐𝐴 will be the same size of 𝐴. Thus if 𝐵 = 𝐶𝐴 , then 𝑏𝑖𝑗 =c𝑎𝑖𝑗. Example: let 𝐴=[1 −2 4 7 −3 0 ],determine 3𝐴. Then Now multiple every element of 𝐴 by 3 we get 20. 3𝐴 = [ 3 −6 12 21 −9 0 ]. Observe that 𝐴 and 3𝐴 are both 2 × 3 matrix Remark: If 𝐴 is a square matrix, then 𝐴 multiplied by itself 𝑘 times is written Ak. 𝐴 𝑘 = 𝐴𝐴 … 𝐴 , 𝐾 times Familiar rules of exponents of real numbers hold for matrices. Theorem: If 𝐴 is an 𝑛 × 𝑛 square matrix and 𝑟 and 𝑠 are nonnegative integers, then 1. 𝐴 𝑟 𝐴 𝑠 = 𝐴 𝑟+𝑠 2. ( 𝐴 𝑟) 𝑠 = 𝐴 𝑟𝑠 3. 𝐴0 = 𝐼 𝑛 (by definition) We verify the first rule, the proof of the second rule is similar 𝐴 𝑟 𝐴 𝑠 = 𝐴 … 𝐴⏟ 𝑟 𝑡𝑖𝑚𝑒𝑠 𝐴 … 𝐴⏟ 𝑠 𝑡𝑖𝑚𝑒𝑠 = 𝐴 … 𝐴⏟ 𝑠+𝑟 𝑡𝑖𝑚𝑒𝑠 = 𝐴 𝑟+𝑠 Example: If = [ 1 −2 −1 0 ] , compute 𝐴4 . This example illustrates how the above rules can be used to reduce the amount of computation involved in multiplying matrices. We know that 𝐴4 = 𝐴𝐴𝐴𝐴. We could perform three matrix multiplication to arrive at 𝐴4 . However we con apply rule 2 above to write 𝐴4 = ( 𝐴2)2 and thus arrive at the result using two products. We get 𝐴2 = [ 1 −2 −1 0 ] [ 1 −2 −1 0 ] = [ 3 −2 −1 2 ] 𝐴4 = [ 3 −2 −1 2 ] [ 3 −2 −1 2 ] = [11 −10 −5 6 ] The usual index laws hold provided 𝐴𝐵 = 𝐵𝐴 21.  ( 𝐴𝐵) 𝑛 = 𝐴 𝑛 𝐵 𝑛  𝐴 𝑚 𝐵 𝑛 = 𝐵 𝑛 𝐴 𝑚  ( 𝐴 + 𝐵)2 = 𝐴2 + 2𝐴𝐵 + 𝐵2  ( 𝐴 + 𝐵) 𝑛 = ∑ ( 𝑛 𝑖 )𝑛 𝑖=0 𝐴𝑖 𝐵 𝑛−𝑖  (𝐴 + 𝐵)(𝐴 − 𝐵) = 𝐴2 – 𝐵2 We now state basic of the natural numbers. Equality The matrix 𝐴 and 𝐵 are said to be equal if 𝐴 and 𝐵 have the same size and corresponding element are equal; that is 𝐴 and 𝐵 ∈ 𝑀(𝑚 × 𝑛) and 𝐴 = [𝑎𝑖𝑗] , 𝐵 = [𝑏𝑖𝑗 ] with 𝑎𝑖𝑗 =𝑏𝑖𝑗 for 1 ≤ 𝑖 ≤ 𝑚 , 1 ≤ 𝑗 ≤ 𝑛. Example: 𝐴 = [ 1 2 3 4 5 6 7 8 9 ] , 𝐵 = [ 3 1 2 4 5 6 7 8 9 ] , 𝐶 = [ 1 1 + 1 3 4 5 12 2 7 5 + 3 3 ∗ 3 ] 𝐴 = 𝐶 Minor Let 𝐴 be an 𝑛 × 𝑛 square matrix obtained from 𝐴 by deleting the 𝑖 𝑡ℎ row and 𝑗 𝑡ℎ column of 𝐴 𝑀𝑖𝑗is called the 𝑖𝑗 𝑡ℎ minor of 𝐴 . 𝐴 = [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ] 𝑀11 = [ 𝑎22 𝑎23 𝑎32 𝑎33 ] , 𝑀12 = [ 𝑎21 𝑎23 𝑎31 𝑎33 ] , 𝑀13 = [ 𝑎21 𝑎22 𝑎31 𝑎32 ] Cofactors 22. The cofactor 𝑐𝑖𝑗 is defined as the coefficient of 𝑎𝑖𝑗 in the determinant 𝐴 If is given by the formula 𝐶𝑖𝑗=(−1)𝑖+𝑗 𝑚𝑖𝑗 Where the minor is the determinant of order (𝑛 − 1) × (𝑛 − 1) formed by deleting the column and row containing 𝑎𝑖𝑗. 𝑐11=(−1)1+1 𝑚11=+1. [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ] =+1. | 𝑎22 𝑎23 𝑎32 𝑎33 | = 𝑎22 𝑎33-𝑎32 𝑎23 𝑐12=(−1)1+2 𝑚12= -1. [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ]=-1. | 𝑎21 𝑎23 𝑎31 𝑎33 | = 𝑎21 𝑎33-𝑎31 𝑎23 𝑐13=(−1)1+3 𝑚13=+1. [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ]=+1. | 𝑎21 𝑎22 𝑎31 𝑎32 | = 𝑎21 𝑎32-𝑎31 𝑎22 Definition: Let 𝐴 be 𝑛 × 𝑛 matrix and 𝑐𝑖𝑗 be the cofactor of 𝑎𝑖𝑗.The matrix whose (𝑖, 𝑗) 𝑡ℎ element is 𝑐𝑖𝑗 is called the matrix of the cofactors of 𝐴. The transpose of this matrix is called the 𝒂𝒅𝒋𝒐𝒊𝒏𝒕 of A and is denoted 𝒂𝒅𝒋(𝑨). [ c11 c12 ⋯ c1n c21 c22 ⋯ c2n ⋮ ⋮ ⋱ ⋮ cn1 cn2 ⋯ cnn ] [ c11 c12 ⋯ c1n c21 c22 ⋯ c2n ⋮ ⋮ ⋱ ⋮ cn1 cn2 ⋯ cnn ] 𝑇 𝑀𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝐴𝑑𝑗𝑜𝑖𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥 Example: give the matrix of the cofactors and the 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 matrix of the following matrix A. A=[ 2 0 3 −1 4 −2 1 −3 5 ] The cofactor of A are as follows. 𝑐11=| 4 −2 −3 5 |=14 , 𝑐12=− |−1 −2 1 5 |=3 , 𝑐13=|−1 4 1 −3 |=-1 𝑐21=− | 0 3 −3 5 |=-9 , 𝑐22=|2 3 1 5 |=7 , 𝑐23=− |2 0 1 −3 |=6 23. 𝑐31=|0 3 4 −2 |=-12 , 𝑐32=− | 2 3 −1 −2 |=1 , 𝑐33=| 2 0 −1 4 |=8 The matrix of cofactors of 𝐴 is [ 14 3 −1 −9 7 6 −12 1 8 ] The 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 of 𝐴 is the transpose of this matrix 𝑎𝑑𝑗(𝐴)= [ 14 −9 −12 3 7 1 −1 6 8 ] (1.4)TheInverseof a SquareMatrix The inverse 𝑎−1 of a scalar (= 𝑎 number) 𝑎 is defined by 𝑎 𝑎−1 = 1. For square matrices we use a similar definition the inverse 𝐴−1 of a 𝑛 × 𝑛 matrix 𝐴 fulfils the relation 𝐴𝐴−1 = 𝐼 where 𝐼 is the 𝑛 × 𝑛 unit matrix defined earlier. Example: we show that, as usual, the identity matrix of the appropriate size. 𝐴 = [−1 1 −2 0 ] , 𝐵 = 1 2 [0 −1 2 −1 ] All we need do is to check that 𝐴𝐵 = 𝐵𝐴 = 𝐼 𝐴𝐵 = [−1 1 −2 0 ] × 1 2 [0 −1 2 −1 ] = 1 2 [0 −1 2 −1 ] × [−1 1 −2 0 ] = 1 2 [2 0 0 2 ] = [1 0 0 1 ] The reader should check that 𝐴𝐵 = 𝐼 also Note: if 𝐴−1 exists then 𝑑𝑒𝑡(𝐴)𝑑𝑒𝑡(𝐴−1 ) = 𝑑𝑒𝑡(𝐴𝐴−1 ) = 𝑑𝑒𝑡(𝐼) = 1 24. Hence 𝑑𝑒𝑡(𝐴−1 ) = ( 𝑑𝑒𝑡𝐴)−1 Example: 𝐴 = [3 7 2 6 ] 𝑑𝑒𝑡(𝐴) = 18 − 14 = 4 ,( 𝑑𝑒𝑡( 𝐴)) −1 = 1 4 𝐴−1 = 1 4 [ 6 −7 −2 3 ] = [ 6 4 −7 4 −2 4 3 4 ] 𝑑𝑒𝑡(𝐴−1 ) = 18 16 − 14 16 = 4 16 = 1 4 ∴ 𝑑𝑒𝑡(𝐴−1 ) = ( 𝑑𝑒𝑡𝐴)−1 , 1 4 = 1 4 Remark:  Non-square matrices do not have an inverse.  The inverse of A is usually written 𝐴−1 .  Not all square matrices have an inverse.  A square matrix 𝐴 is invertible if and only if 𝑑𝑒𝑡(𝐴) ≠ 0.  𝐴−1 exists if and only if 𝐴 is nonsingular. FindingInverseofMatrix 1- The Inverse of a 𝟐 × 𝟐 Matrix If 𝑎𝑑 − 𝑏𝑐 ≠ 0 then the 2 × 2 matrix 𝐴 = [ 𝑎 𝑏 𝑐 𝑑 ] has a (unique) inverse given by 𝐴−1 = 1 𝑎𝑑−𝑏𝑐 [ 𝑑 −𝑏 −𝑐 𝑎 ] Note: 𝑎𝑑 − 𝑏𝑐 = 0 that 𝐴 has no inverse. 25. In words: to find the inverse of a 2 × 2 matrix 𝐴 we effectively interchange the diagonal elements 𝑎 and 𝑑, change the sign of the other two elements and then divide by the determinant of 𝐴. Example: 𝐴 = [3 7 2 6 ] 𝑑𝑒𝑡(𝐴) = 18 − 14 = 4 , 𝐴−1 = 1 4 [ 6 −7 −2 3 ] = [ 6 4 −7 4 −2 4 3 4 ] 2- The Inverseof a 3×3 Matrix-TheDeterminant Method Given a square matrix 𝐴:  Find 𝑑𝑒𝑡(𝐴), if det(𝐴)=0 then ,as we know 𝐴−1 does not exists. If det(𝐴)≠0 we can proceed to find the inverse matrix.  Replace each element of 𝐴 by its cofactor 𝐶11=[ 𝑎22 𝑎23 𝑎32 𝑎33 ], 𝐶12=[ 𝑎21 𝑎23 𝑎31 𝑎33 ] , 𝐶13=[ 𝑎21 𝑎22 𝑎31 𝑎32 ] 𝐶21=[ 𝑎12 𝑎13 𝑎32 𝑎33 ] , 𝐶22=[ 𝑎11 𝑎13 𝑎31 𝑎33 ], 𝐶23=[ 𝑎11 𝑎12 𝑎31 𝑎32 ] 𝐶31=[ 𝑎12 𝑎13 𝑎22 𝑎23 ], 𝐶32=[ 𝑎11 𝑎13 𝑎21 𝑎23 ], 𝐶33=[ 𝑎11 𝑎12 𝑎21 𝑎22 ] Cofactor =[ 𝐶11 𝐶12 𝐶13 𝐶21 𝐶22 𝐶23 𝐶31 𝐶32 𝐶33 ] =𝐴  Transpose the result to form the 𝒂𝒅𝒋𝒐𝒊𝒏𝒕 𝒎𝒂𝒕𝒓𝒊𝒙 , 𝑎𝑑𝑗(𝐴). 𝐴𝑑𝑗(𝐴) = [ 𝐶11 𝐶21 𝐶31 𝐶12 𝐶22 𝐶32 𝐶13 𝐶23 𝐶33 ] , 𝑎𝑑𝑗(𝐴) = 𝐴 𝑇  Then 𝐴−1 = 1 𝑑𝑒𝑡(𝐴) 𝑎𝑑𝑗(𝐴). Example: find the inverse of 𝐴 = [ 1 −1 2 −3 1 2 3 −2 −1 ] 26. 𝑑𝑒𝑡(𝐴) = 1 | 1 2 −2 −1 | − (−1) |−3 2 3 −1 | + 2 |−3 1 3 −2 | = 1 × 3 + 1 × (−3) + 2 × 3 = 6 Since the determinant is non-zero an inverse exists. Calculate the matrix of minors 𝑀=| | | 1 2 −2 −1 | | −3 2 3 −1 | | −3 1 3 −2 | | −1 2 −2 −1 | | 1 2 3 −1 | | 1 −1 3 −2 | | −1 2 1 2 | | 1 2 −3 2 | | 1 −1 −3 1 | | | =[ 3 −3 3 5 −7 1 −4 8 −2 ] Modify the signs according to whether 𝑖 + 𝑗 is even or odd to calculate the matrix of cofactors 𝐶=| 3 3 3 −5 −7 −1 −4 −8 −2 | It follows that 𝐴−1 = 1 6 𝐶 𝑇 = 1 6 | 3 −5 −4 3 −7 −8 3 −1 −2 | To check that we have made no mistake we can compute 𝐴−1 . 𝐴 = 1 6 | 3 −5 −4 3 −7 −8 3 −1 −2 |. [ 1 −1 2 −3 1 2 3 −2 −1 ] =[ 1 0 0 0 1 0 0 0 1 ] . This way of computing the invers is only useful for hand calculations in the cases of 2 × 2 or 3 × 3 matrices. Definition: A matrix is said to be in row-echelon form if 1. If there are any rows of all zeros then they are at the bottom of the matrix. 27. 2. If a row does not consist of all zeros then its first non-zero entry (i.e. the left most nonzero entry) is a 1. This 1 is called a leading 1. 3. In any two successive rows, neither of which consists of all zeroes, the leading 1 of the lower row is to the right of the leading 1 of the higher row. Example: The following matrices are all in row echelon form. [ 𝟏 −6 9 1 0 0 0 𝟏 −4 −5 0 0 0 𝟏 2 ] [ 𝟏 0 5 0 𝟏 3 0 0 𝟏 ] [ 𝟏 −8 10 5 −3 0 𝟏 13 9 12 0 0 0 𝟏 1 0 0 0 0 0 ] Definition: A matrix is in 𝒓𝒆𝒅𝒖𝒄𝒆𝒅 𝒓𝒐𝒘 − 𝒆𝒄𝒉𝒆𝒍𝒐𝒏 𝒇𝒐𝒓𝒎 if 1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix. 2. The first nonzero element of each other row is 1. This element is called a leading 1. 3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row. 4. All other elements in a column that contains a leading 1 are zero. Example: The following matrices are all in reduced echelon form. [ 𝟏 0 8 0 𝟏 2 0 0 0 ] [ 𝟏 0 0 7 0 𝟏 0 3 0 0 𝟏 9 ] [ 𝟏 4 0 0 0 0 𝟏 0 0 0 0 𝟏 ] [ 𝟏 2 3 0 0 0 0 𝟏 0 0 0 0 ] The following matrices are not in reduced echelon form. 28. [ 1 2 0 4 0 0 0 0 0 0 1 3 ] row of zeros not at bottom of matrix [ 1 2 0 3 0 0 0 3 4 0 0 0 0 0 1 ] first nonzero element in row 2 is not 1 [ 1 0 0 2 0 0 1 4 0 1 0 3 ] leading 1 in row 3 not of the right of leading 1 in row 2 [ 1 7 0 8 0 1 0 3 0 0 1 2 0 0 0 0 ] nonzero element above leading 1 in row 2 3- The Inverseof a 3×3 Matrix – 𝑮𝒂𝒖𝒔𝒔 − 𝒋𝒐𝒓𝒅𝒂𝒏 Elimination Method Let 𝐴 be an 𝑛 × 𝑛 matrix 1. Adjoin the identity 𝑛 × 𝑛 matrix 𝐼 𝑛 to 𝐴 to form the matrix [𝐴: 𝐼 𝑛]. 2. Compute the reduced echelon form of [𝐴: 𝐼 𝑛] . If the reduced echelon form is of the type [𝐼 𝑛: 𝐵], then 𝐵 is the inverse of 𝐴. If the reduced echelon form is not of the type [𝐼 𝑛: 𝐵], in that the first 𝑛 × 𝑛 sub matrix is not 𝐼 𝑛, then 𝐴 has no inverse. SomeNotes for Operationof TheMethod  Inter changing two rows.  Adding a multiple of on row to on other row.  Multiplying one row by a non-zero constant. The following example illustrate the method Example: determine the inverse of the matrix 𝐴=[ 1 −1 −2 2 −3 −5 −1 3 5 ] Applying the method of 𝐺𝑎𝑢𝑠𝑠 − 𝑗𝑜𝑟𝑑𝑎𝑛 elimination ,we get [𝐴 ∶ 𝐼 𝑛] = [ 1 −1 −2 ⋮ 1 0 0 2 −3 −5 ⋮ 0 1 0 −1 3 5 ⋮ 0 0 1 ] 𝑟2:𝑅2+(−2) 𝑅1 𝑟3:𝑅3+𝑅1 → [ 1 −1 −2 ⋮ 1 0 0 0 −1 −1 ⋮ −2 1 0 0 2 3 ⋮ 1 0 1 ] 29. 𝑟2:(−1) 𝑅2 → [ 1 −1 −2 ⋮ 1 0 0 0 1 1 ⋮ 2 −1 0 0 2 3 ⋮ 1 0 1 ] 𝑟1:𝑅1+𝑅2 𝑟3:𝑅3+(−2) 𝑅2 → [ 1 0 −1 ⋮ 3 −1 0 0 1 1 ⋮ 2 −1 0 0 0 1 ⋮ −3 2 1 ] 𝑟1:𝑅1+𝑅3 𝑟2:𝑅2+(−1) 𝑅3 → [ 1 0 0 ⋮ 0 1 1 0 1 0 ⋮ 5 −3 −1 0 0 1 ⋮ −3 2 1 ] = [𝐼 𝑛 ∶ 𝐵] B = [ 0 1 1 5 −3 −1 −3 2 1 ]=𝐴−1 The following example illustrates the application of the method for a matrix that does not have an inverse. Example: determine the inverse of the following matrix ,if it exists. A=[ 1 1 5 1 2 7 2 −1 4 ] Applying the method of 𝐺𝑎𝑢𝑠𝑠 − 𝑗𝑜𝑟𝑑𝑎𝑛 elimination. we get [𝐴: 𝐼3] = [ 1 1 5 ⋮ 1 0 0 1 2 7 ⋮ 0 1 0 2 −1 4 ⋮ 0 0 1 ] 𝑟2:𝑅2+(−1) 𝑅1 𝑟3:𝑅3+(−2) 𝑅1 → [ 1 1 5 ⋮ 1 0 0 0 1 2 ⋮ −1 1 0 0 −3 −6 ⋮ −2 0 1 ] 𝑟1:𝑅1+(−1) 𝑅2 𝑟3:𝑅3+3𝑅2 → [ 1 0 3 ⋮ 2 −1 0 0 1 2 ⋮ −1 1 0 0 0 0 ⋮ −5 3 1 ] 30. Properties of Inverse  If 𝐴𝐵 = 𝐼 and 𝐶𝐴 = 𝐼, then 𝐵 = 𝐶. Consequently 𝐴 has at most one inverse.  ( 𝐴𝐵)−1 = 𝐵−1 𝐴−1 (assuming 𝐴, 𝐵 are invertible ).  ( 𝐴 𝑇)−1 = ( 𝐴−1) 𝑇 (assuming 𝐴 is invertible ).  ( 𝐴−1)−1 = 𝐴 , 𝑖. 𝑒. inverse of inverse is original matrix (assuming 𝐴 is invertible ).  𝐼−1 = 𝐼 .  (∝ 𝐴)−1 = ( 1 ∝ ) 𝐴−1 (assuming 𝐴 invertible , ∝≠ 0 ).  If 𝑦 = 𝐴𝑋 ,where 𝑋 ∈ 𝑅 𝑛 and 𝐴 invertible , then 𝑋 = 𝐴−1 𝑦 𝐴−1 𝑦 = 𝐴−1 𝐴𝑋 = 𝐼𝑋 = 𝑋.  If 𝐴1 , 𝐴2, … , 𝐴 𝑘 ,are all invertible , so is their product 𝐴1 , 𝐴2, … , 𝐴 𝑘 , and ( 𝐴1 𝐴2 … 𝐴 𝑘)−1 = 𝐴 𝑘 −1 … 𝐴2 −1 𝐴1 −1 .  If 𝐴 is invertible , so is 𝐴 𝑘 for 𝑘 ≥ 1, and ( 𝐴 𝑘)−1 = ( 𝐴−1) 𝑘 . Corollary 1:If 𝐴𝐵 = 𝐼 and 𝐶𝐴 = 𝐼, then 𝐵 = 𝐶, consequently 𝐴 has at most one inverse. Proof: If 𝐴𝐵 = 𝐼 and 𝐶𝐴 = 𝐼, then 𝐵 = 𝐼𝐵 = 𝐶𝐴𝐵 = 𝐶𝐼 = 𝐶, if 𝐵 and 𝐶 are both inverses of 𝐴, then , by definition, 𝐴𝐵 = 𝐵𝐴 = 𝐼 and 𝐴𝐶 = 𝐶𝐴 = 𝐼, in particular 𝐴𝐵 = 𝐼 and 𝐶𝐴 = 𝐼, so that 𝐵 = 𝐶. Corollary 2:If 𝐴 and 𝐵 are both invertible, then so is 𝐴𝐵 and ( 𝐴𝐵)−1 = 𝐵−1 𝐴−1 . Proof: We have a guess for ( 𝐴𝐵)−1 , to check that the guess is correct, we merely need to check the requirements of the definition (𝐴𝐵)(𝐵−1 𝐴−1 ) = 𝐴𝐵𝐵−1 𝐴−1 = 𝐴𝐼𝐴−1 = 𝐴𝐴−1 = 𝐼 (𝐵−1 𝐴−1 )(𝐴𝐵) = 𝐵−1 𝐴−1 𝐴𝐵 = 𝐵−1 𝐼𝐵 = 𝐵−1 𝐵 = 𝐼 Corollary 3:If 𝐴 is invertible, then so is 𝐴 𝑇 and ( 𝐴 𝑇)−1 = ( 𝐴−1) 𝑇 . Proof: Let’s use 𝐵 to denote the inverse of 𝐴 (so there won’t be so many superscripts around) by definition 𝐴𝐵 = 𝐵𝐴 = 𝐼 These three matrices are the same, so their transposes are the same. Since ( 𝐴𝐵) 𝑇 = 𝐴 𝑇 𝐵 𝑇 , ( 𝐵𝐴) 𝑇 = 𝐴 𝑇 𝐵 𝑇 and 𝐼 𝑇 = 𝐼, we have 𝐵 𝑇 𝐴 𝑇 = 𝐴 𝑇 𝐵 𝑇 = 𝐼 𝑇 = 𝐼 Which is exactly the definition of “the inverse of 𝐴 𝑇 is 𝐵 𝑇 ”. 31. (1.5)SomePropertiesof Determinants 1. Rows and columns can be interchanged without affecting the value of a determinant. Consequently det(𝐴) = det(𝐴 𝑇 ). Example: 𝐴=|3 4 1 2 |, 𝐴 𝑇 =|3 1 4 2 | 𝑑𝑒𝑡(𝐴) = 2, 𝑑𝑒𝑡(𝐴𝑇) = 2 ∴ det(𝐴) = det(𝐴 𝑇 ) 2. If two rows, or two columns, are interchanged the sign of the determinant is reversed. Example: if 𝐴=|3 4 1 −2 | , then 𝑑𝑒𝑡(|3 4 1 −2 |) = −10 , 𝑑𝑒𝑡(|1 −2 3 4 |) = 10 3. If a row(or column) is changed by adding to or subtracting from its elements the corresponding elements of any other row (or column) the determinant remains unaltered. Example: 𝑑𝑒𝑡(|3 4 1 −2 |) = 𝑑𝑒𝑡(|3 + 1 4 − 2 1 −2 |) = 𝑑𝑒𝑡(|4 2 1 −2 |) = −10 4. If the elements in any row (or column) have a common factor α then the determinant equals the determinant of the corresponding matrix in which α = 1, multiplied by α. Example: 𝐴 = |6 8 1 −2 |, α=2 det( |6 8 1 −2 |) = −20 ,det ( |6 8 1 −2 |) = 2 det (|3 4 1 −2 |) = 2 × (−10) = −20 5. The determinant of an upper triangular or lower triangular matrix is the product of the main diagonal entries. Example: A upper triangular, B lower triangular 𝐴 = | 2 2 1 0 2 −1 0 0 4 | = det( 𝐴) = 2 × 2 × 4 = 16 32. 𝐵 = | 2 0 0 3 −3 0 4 1 4 | = det( 𝐵) = 2 × (−3) × 4 = −24 This rule is easily verified from the definition 𝑑𝑒𝑡(𝐴) = ± ∏ 𝑎1𝑗1 𝑎2𝑗2 … 𝑎 𝑛𝑗 𝑛 because all terms vanish except 𝑗1= 1,𝑗2= 2, . . . 𝑗 𝑛 = n, which is the product of the main diagonal entries. Diagonal matrices are a particular case of this rule. 6. The determinant of the product of two square matrices is the product of the individual determinants: 𝑑𝑒𝑡(𝐴𝐵) = 𝑑𝑒𝑡(𝐴)𝑑𝑒𝑡(𝐵). Example: 𝐴 = |6 8 1 −2 | , 𝐵 = |6 0 1 −2 | , det( 𝐴) = 6 × (−2) − 8 × 1 = −20 𝑑𝑒𝑡(𝐵) = 6 × (−2) − 0 × 1 = −12 𝑑𝑒𝑡(𝐴𝐵) = 𝑑𝑒𝑡(|6 8 1 −2 | . |6 0 1 −2 |) = 𝑑𝑒𝑡(|44 −16 4 4 |) = 240 𝑑𝑒𝑡(𝐴)𝑑𝑒𝑡(𝐵) = −20 × −12 = 240 This rule can be generalized to any number of factors. One immediate application is to matrix powers: | 𝐴2| = | 𝐴|| 𝐴| = | 𝐴|2 , and more generally | 𝐴 𝑛 | = | 𝐴| 𝑛 for integer 𝑛. 7. Let A be an 𝑛 × 𝑛 matrix and c be a scalar then, det( 𝑐𝐴) = 𝑐 𝑛 𝑑𝑒𝑡 ( 𝐴) Example : For the given matrix below we compute both 𝑑𝑒𝑡(𝐴) and 𝑑𝑒𝑡(2𝐴). 𝐴 = [ 4 −2 5 −1 −7 10 0 1 −3 ] First the scalar multiple. 2A=[ 8 −4 10 −2 −14 20 0 2 −6 ] The determinants. det( 𝐴) = 45 , det(2𝐴) = 360 = (8)(45) = 23 𝑑𝑒𝑡 ( 𝐴) 33. 8. Suppose that A is an invertible matrix then, det(𝐴−1 ) = 1 𝑑𝑒𝑡(𝐴) Example: For the given matrix we compute det(𝐴) 𝑎𝑛𝑑 det(𝐴−1 ) 𝐴 = [8 −9 2 5 ] 𝐴−1 = [ 5 58 9 58 − 1 29 4 29 ] Here are the determinants for both of these matrices. det(𝐴) = 58 det(𝐴−1 ) = 1 58 9. Suppose that A is an n× n triangular matrix then, 𝑑𝑒𝑡(𝐴) = 𝑎11 𝑎22 … 𝑎 𝑛𝑛 FindingDeterminant 1-The Determinant of a 𝟐 × 𝟐 Matrix 𝐴 = [ 𝑎11 𝑎12 𝑎21 𝑎22 ], is written 𝑑𝑒𝑡(𝐴) = 𝑎11 𝑎22-𝑎12 𝑎21 Example: 𝐴 = [1 2 4 −7 ], 𝑑𝑒𝑡(𝐴) = −7 − 8 = −15 2-The Determinant of a 𝟑 × 𝟑 Matrix 𝑑𝑒𝑡(𝐴)= [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ] = 𝑎11 [ 𝑎22 𝑎23 𝑎32 𝑎33 ] − 𝑎12 [ 𝑎21 𝑎23 𝑎31 𝑎33 ] + 𝑎13 [ 𝑎21 𝑎22 𝑎31 𝑎32 ] =𝑎11(𝑎22 𝑎33-𝑎32 𝑎23)− 𝑎12(𝑎21 𝑎33-𝑎31 𝑎23)+ 𝑎13(𝑎21 𝑎32-𝑎31 𝑎22) That is the 3 × 3 determinants is defined in terms of determinants of 2 × 2 sub-matrices of . Or [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ] 𝑎11 𝑎12 𝑎21 𝑎22 𝑎31 𝑎32 34. det( 𝐴) = 𝑎11 𝑎22 𝑎33 + 𝑎12 𝑎23 𝑎31 + 𝑎13 𝑎21 𝑎32 − 𝑎13 𝑎22 𝑎31 − 𝑎12 𝑎21 𝑎33 − 𝑎11 𝑎23 𝑎32. Example: find the determinant of 𝐴=[ 1 2 3 1 0 2 3 2 1 ] 𝑑𝑒𝑡(𝐴) = 1 × 𝑑𝑒𝑡 |0 2 2 1 | − 2 × 𝑑𝑒𝑡 |1 2 3 1 | + 3 × 𝑑𝑒𝑡 |1 0 3 2 | = 1 × (0 − 4) − 2 × (1 − 6) + 3 × (2 − 0) = 12 Or 𝑑𝑒𝑡(𝐴)=(1×0×1)+(2×2×3)+(1×2×3)-(3×0×3)-(2×1×1)-(2×2×1) =0+12+6-0-2-4=18-6=12 3-Cofactor Expansion The determinant of an 𝑛 × 𝑛 matrix may be found by choosing a row (or column) and summing the products of the entries: 𝑑𝑒𝑡(𝐴) = 𝑎1𝑗 𝑐1𝑗 + 𝑎2𝑗 𝑐2𝑗 + ⋯ + 𝑎 𝑛𝑗 𝑐 𝑛𝑗 (cofactor expansion along the 𝑗 𝑡ℎ column) 𝑑𝑒𝑡(𝐴) = 𝑎𝑖1 𝑐𝑖1 + 𝑎𝑖2 𝑐𝑖2 + ⋯ + 𝑎𝑖𝑛 𝑐𝑖𝑛 (cofactor expansion along the 𝑖 𝑡ℎ row) Where 𝑐𝑖𝑗 is the determinant of 𝐴 with row 𝑖 and column 𝑗 deleted, multiplied by (−1)𝑖+𝑗 . The matrix of elements 𝑐𝑖𝑗 is called the cofactors matrix. Example: cofactor expansion along the first column 𝐴=[ 3 1 0 −2 −4 3 5 4 −2 ] . evaluate 𝑑𝑒𝑡(𝐴) by cofactor expansion along the column of 𝐴 . 𝑑𝑒𝑡(𝐴) = [ 3 1 0 −2 −4 3 5 4 −2 ] 35. =3|−4 3 4 −2 |+(-2) |1 0 4 −2 |+5| 1 0 −4 3 | =3(- 4)+(-2)(-2)+5(3)=6 36. CHAPTER TWO System of Linear Equations In this chapter we study the system of linear equation. Then we illustrate using of matrices to solve system linear equation in terms of the methods of Homogeneous Systems, Gaussian Elimination, Gauss-Jordan Elimination and Cramer’s Rule (2.1)LinearEquations Definition: Let 𝑎1 , . . . , 𝑎 𝑛, 𝑦 be elements of 𝑅, and let 𝑥1, . . . , 𝑥 𝑛 be unknowns (also called variables or 𝒊𝒏𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒆𝒔). Then an equation of the form 𝑎1 𝑥1+ … + 𝑎 𝑛 𝑥 𝑛= y is called a linear equation in 𝒏 unknowns (over 𝑅), the scalars 𝒂𝒊 are called the coefficients ofthe unknowns , and 𝒚 is called the constant term of the equation. Example: 𝑥1+2𝑥1=5 𝑎1=1 , 𝑎2=2 , y=5 (This equation are linear ) But: 𝑥2 1+3√ 𝑥2 =5 (This equation is not linear) Note: not all 𝒂 are zero . Definition: A linear equation in the two variables 𝑥1and 𝑥2 is an equation that can in the form 𝑎1 𝑥1 +𝑎2 𝑥2=b Where 𝑎1, 𝑎2 , and b are numbers, in general a linear equation in the 𝒏 variables 𝑥1, 𝑥2,…, 𝑥 𝑛is an equation that can be written in the form 𝑎1 𝑥1+𝑎2 𝑥2+…+𝑎 𝑛 𝑥 𝑛=𝑏 Where the coefficients 𝑎1, 𝑎2, …, 𝑎 𝑛 and the constant term 𝑏 are numbers. 37. Example: 𝑥1+7𝑥2=3 , 𝑥1+𝑥2+…+𝑥 𝑛=4 Some exampleof equations that are not linear are: 𝑥1 2 +𝑥1 𝑥3=3, 1 𝑥1 +𝑥2+𝑥3= 5 2 , 𝑒(𝑥1) +𝑥2= 1 2 , 𝑥1+𝑥2 𝑥3+𝑥4 =𝑥5 + 7 (2.2) LinearSystem In general, a system of linear equations (also called a linear system) in the variables 𝑥1 , 𝑥2, … , 𝑥 𝑛 consists of a finite number of linear equations in these variables. The general form of a system of 𝑚 equations in 𝑛 unknowns is 𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥 𝑛 = 𝑏1 𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑛 𝑥 𝑛 = 𝑏2 ⋮ ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑥1 + 𝑎 𝑚2 𝑥2 + ⋯ + 𝑎 𝑚𝑛 𝑥 𝑛 = 𝑏𝑚 We will call such a system an 𝑚 × 𝑛 (𝑚 by 𝑛) linear system. Example: 6𝑥1 + 2𝑥2 − 𝑥3 = 5 3𝑥1 + 𝑥2 − 4𝑥3 = 9 −𝑥1 + 3𝑥2 + 2𝑥3 = 0 Definition: The coefficients of the variables form a matrix called the matrix of coefficient of the system. [ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑎21 𝑎22 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 ] m×n 38. Definition: The coefficients, together with the constant terms, form a matrix called the augmented matrixof the system. 𝑎𝑢𝑔 𝐴=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑦1 𝑎21 𝑎22 ⋯ 𝑎2𝑛 𝑦2 ⋮ ⋮ ⋮ ⋮ 𝑎 𝑚1 𝑎 𝑚2 ⋯ 𝑎 𝑚𝑛 𝑦 𝑛 ] m×n Example: The matrix of coefficients and the augmented matrix of the following system of linear equations are as shown: 𝑥1 + 𝑥2 + 𝑥3 = 2 2𝑥1 + 3𝑥2 + 𝑥3 = 3 𝑥1 − 𝑥2 − 2𝑥3 = −6 [ 1 1 1 2 3 1 1 −1 −2 ] ⏟ 𝒎𝒂𝒕𝒓𝒊𝒙 𝒐𝒇 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕𝒔 [ 1 1 1 2 2 3 1 3 1 −1 −2 −6 ] ⏟ 𝒂𝒖𝒈𝒎𝒆𝒏𝒕𝒆𝒅 𝒎𝒂𝒕𝒓𝒊𝒙 In solving systems of equations we are allowed to perform operations of the following types: 1. Multiply an equation by a non-zero constant. 2. Add one equation (or a non-zero constant multiple of one equation) to another equation. These correspond to the following operations on the augmented matrix : 1. Multiply a row by a non-zero constant. 2. Add a multiple of one row to another row. 3. We also allow operations of the following type : Interchange two rows in the matrix (this only amounts to writing down the equations of the system in a different order). 39. Definition: Operations of these three types are called Elementary Row Operations (ERO's) on a matrix. Example: [ 1 2 −1 5 3 1 −2 9 −1 4 2 0 ] 𝑥 + 2𝑦 − 𝑧 = 5 3𝑥 + 𝑦 − 2𝑧 = 9 −𝑥 + 4𝑦 + 2𝑧 = 0 𝑅3 → 𝑅3 + 𝑅1 [ 1 2 −1 5 3 1 −2 9 0 6 1 5 ] 𝑥 + 2𝑦 − 𝑧 = 5 3𝑥 + 𝑦 − 2𝑧 = 9 6𝑦 + 𝑧 = 5 𝑅2 → 𝑅2 − 3𝑅1 [ 1 2 −1 5 0 −5 1 −6 0 6 1 5 ] 𝑥 + 2𝑦 − 𝑧 = 5 − 5𝑦 + 𝑧 = −6 6𝑦 + 𝑧 = 5 𝑅2 → 𝑅2 + 𝑅3 [ 1 2 −1 5 0 1 2 −1 0 6 1 5 ] 𝑥 + 2𝑦 − 𝑧 = 5 𝑦 + 2𝑧 = −1 6𝑦 + 𝑧 = 5 𝑅3 → 𝑅3 − 6𝑅2 [ 1 2 −1 5 0 1 2 −1 0 0 −11 11 ] 𝑥 + 2𝑦 − 𝑧 = 5 𝑦 + 2𝑧 = −1 − 11𝑧 = 11 𝑅3 × (− 1 11 ) [ 1 2 −1 5 0 1 2 −1 0 0 1 −1 ] 𝑥 + 2𝑦 − 𝑧 = 5 … ( 𝐴) 𝑦 + 2𝑧 = −1 … ( 𝐵) 𝑧 = −1 … ( 𝐶) We have produced a new system of equations, this is easily solved: 40. 𝐵𝑎𝑐𝑘𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 { (C) z = −1 (B) y = −1 − 2z → y = −1 − 2(−1) = 1 ( 𝐴) 𝑥 = 5 − 2𝑦 + 𝑧 → 𝑥 = 5 − 2(1) + (−1) = 2 ∴ 𝑥 = 2, 𝑦 = 1, 𝑧 = −1 (2.2.1)HomogeneousSystems A system of linear equations is said to be homogeneous if all the constant terms are zeros, a system of homogeneous linear equations is a system of the form 𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑚 𝑥 𝑚 = 0 𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑚 𝑥 𝑚 = 0 ⋮ ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑥1 + 𝑎 𝑛2 𝑥2 + ⋯ + 𝑎 𝑛𝑚 𝑥 𝑚 = 0 Such a system is always consistent as 𝑥1 = 0, 𝑥2 = 0 , …. , 𝑥 𝑚 = 0is a solution, this solution is called the trivial (or zero) solution, any other solution is called a non-trivial solution. Example: 𝑥 − 𝑦 = 0 𝑥 + 𝑦 = 0 Has only the trivial solution, whereas the homogeneous system 𝑥 − 𝑦 + 𝑧 = 0 𝑥 + 𝑦 + 𝑧 = 0 Has the complete solution 𝑥 = −𝑧; 𝑦 = 0; 𝑧 arbitrary. In particular, taking 𝑧 = 1 gives the non-trivial solution 𝑥 = −1; 𝑦 = 0; 𝑧 = 1. There is simple but fundamental theorem concerning homogeneous systems. 41. (2.2.2)GaussianElimination This method is used to solve the linear system 𝐴𝑋 = 𝐵 by transforms the augmented matrix [𝐴: 𝐵] to row echelon form and then uses substitution to obtain the solution this procedure some what more efficient than Gauss-Jordan reduction. Consider a linear system 1. Construct the augmented matrix for the system. 2. Use elementary row operation to transform the augmented matrix into a triangular one. 3. Write down the new linear system for which the triangular matrix is the associated augmented matrix. 4. Solve the new system, you may need to assign some parametric values to some unknowns, and then apply the method of back substitution to solve the new system. Example: We use Gaussian elimination to solve the system of linear equations. 2𝑥2 + 𝑥3 = −8 𝑥1 − 2𝑥2 − 3𝑥3 = 0 −𝑥1 + 𝑥2 + 2𝑥3 = 3 The augmented matrix is [ 0 2 1 ⋮ −8 1 −2 −3 ⋮ 0 −1 1 2 ⋮ 3 ] Swap Row1 and Row2 [ 1 −2 −3 ⋮ 0 0 2 1 ⋮ −8 −1 1 2 ⋮ 3 ] Add Row1 to Row3 42. [ 1 −2 −3 ⋮ 0 0 2 1 ⋮ −8 0 −1 −1 ⋮ 3 ] Swap Row2 and to Row3 [ 1 −2 −3 ⋮ 0 0 −1 −1 ⋮ 3 0 2 1 ⋮ −8 ] Add twice Row2 to Row3 [ 1 −2 −3 ⋮ 0 0 −1 −1 ⋮ 3 0 0 −1 ⋮ −2 ] Multiply Row3 by -1 [ 1 −2 −3 ⋮ 0 0 −1 −1 ⋮ 3 0 0 1 ⋮ 2 ] 𝑥1 − 2𝑥2 − 3𝑥3 = 0…….(a) −𝑥2 − 𝑥3 = 3……..(b) 𝑥3 = 2……..(c) Then 𝑥3 = 2 and we put (c) in (b) Then 𝑥2 = −5 and we put (b)and (c) in (a) Then 𝑥1 =-4. (2.2.3) Gauss-JordanElimination The Gauss-Jordan elimination method to solve a system of linear equations is described in the following steps. 1. Write the augmented matrix of the system. 2. Use row operations to transform the augmented matrix in the form described below, which is called the reduced row echelon form (RREF). 43. 3. Stop process in step 2 if you obtain a row whose elements are all zeros except the last one on the right. In that case, the system is inconsistent and has no solutions. Otherwise, finish step 2 and read the solutions of the system from the final matrix. Note: When doing step 2, row operations can be performed in any order. Try to choose row operations so that as few fractions as possible are carried through the computation. This makes calculation easier when working by hand. Example: we Solve the following system by using the Gauss-Jordan elimination method. { 𝑥 + 𝑦 + 𝑧 = 5 2𝑥 + 3𝑦 + 5𝑧 = 8 4𝑥 + 5𝑧 = 2 The augmented matrix of the system is the following. [ 1 1 1 ⋮ 5 2 3 5 ⋮ 8 4 0 5 ⋮ 2 ] We will now perform row operations until we obtain a matrix in reduced row echelon form. [ 1 1 1 ⋮ 5 2 3 5 ⋮ 8 4 0 5 ⋮ 2 ] R2−2R1 → [ 1 1 1 ⋮ 5 0 1 3 ⋮ −2 4 0 5 ⋮ 2 ] R3−4R1 → [ 1 1 1 ⋮ 5 0 1 3 ⋮ −2 0 −4 1 ⋮ −18 ] R3+4R2 → [ 1 1 1 ⋮ 5 0 1 3 ⋮ −2 0 0 13 ⋮ −26 ] 1 13 R3 → [ 1 1 1 ⋮ 5 0 1 3 ⋮ −2 0 0 1 ⋮ −2 ] R2−3R3 → [ 1 1 1 ⋮ 5 0 1 0 ⋮ 4 0 0 1 ⋮ −2 ] R1−R3 → [ 1 1 0 ⋮ 7 0 1 0 ⋮ 4 0 0 1 ⋮ −2 ] R1−R2 → [ 1 0 0 ⋮ 3 0 1 0 ⋮ 4 0 0 1 ⋮ −2 ] 44. From this final matrix, we can read the solution of the system. It is 𝑥 = 3, 𝑦 = 4, 𝑧 = −2. (2.2.4) Cramer’sRule If 𝐴𝑋 = 𝐵 is a system of linear equation in n-unknown, such that 𝑑𝑒𝑡(𝐴) ≠ 0 then the system has a unique solution, this solution is 𝑋1 = 𝑑𝑒𝑡(𝐴1 ) 𝑑𝑒𝑡(𝐴) , 𝑋2 = 𝑑𝑒𝑡(𝐴2) 𝑑𝑒𝑡(𝐴) , 𝑋3 = 𝑑𝑒𝑡(𝐴3) 𝑑𝑒𝑡(𝐴) Where 𝐴𝑖 is the matrix obtained by replacing the entries in the 𝑖 𝑡ℎ column of 𝐴 by the entries of 𝑏𝑡ℎ column. The system 𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑚 𝑥 𝑚 = 𝑏1 𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑚 𝑥 𝑚 = 𝑏2 ⋮ ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑥1 + 𝑎 𝑛2 𝑥2 + ⋯ + 𝑎 𝑛𝑚 𝑥 𝑚 = 𝑏𝑛 [ 𝑎11 𝑎12 ⋯ 𝑎1𝑚 𝑏1 𝑎21 𝑎22 ⋯ 𝑎2𝑚 𝑏2 ⋮ ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑎 𝑛2 ⋯ 𝑎 𝑛𝑚 𝑏 𝑛 ] n×m A=[ 𝑎11 𝑎12 ⋯ 𝑎1𝑚 𝑎21 𝑎22 ⋯ 𝑎2𝑚 ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑎 𝑛2 ⋯ 𝑎 𝑛𝑚 ] n×m 𝐴1 = [ 𝑏1 𝑎12 ⋯ 𝑎1𝑚 𝑏2 𝑎22 ⋯ 𝑎2𝑚 ⋮ ⋮ ⋮ 𝑏 𝑛 𝑎 𝑛2 ⋯ 𝑎 𝑛𝑚 ] n×m , 𝐴2=[ 𝑎11 𝑏1 ⋯ 𝑎1𝑚 𝑎21 𝑏2 ⋯ 𝑎2𝑚 ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑏 𝑛 ⋯ 𝑎 𝑛𝑚 ] n×m , … , 𝐴 𝑚=[ 𝑎11 𝑎12 ⋯ 𝑏 𝑚 𝑎21 𝑎22 ⋯ 𝑏 𝑚 ⋮ ⋮ ⋮ 𝑎 𝑛1 𝑎 𝑛2 ⋯ 𝑏 𝑚 ] n×m 45. Example: we use crammer’s Rule to solve: 3𝑥1 + 4𝑥2 − 3𝑥3 = 5 3𝑥1 − 2𝑥2 + 4𝑥3 = 7 3𝑥1 + 2𝑥2 − 𝑥3 = 2 In matrix form 𝐴𝑋 = 𝑏 or [𝑎1 𝑎2 𝑎3 ]𝑋 = 𝑏 this is [ 3 4 −3 3 −2 4 3 2 −1 ] [ 𝑥1 𝑥2 𝑥3 ] = [ 5 7 3 ]. Cramer’s Rule says that 𝑥1= det(𝐴1) det(𝐴) , 𝑥2= det(𝐴2) det(𝐴) , 𝑥3 = det(𝐴3) det(𝐴) 𝐴 = [ 3 4 −3 3 −2 4 3 2 −1 ] , 𝑑𝑒𝑡(𝐴) = 6 𝐴1 = [ 5 4 −3 7 −2 4 3 2 −1 ] , 𝑑𝑒𝑡(𝐴1 ) = −14 𝐴2 = [ 3 5 −3 3 7 4 3 3 −1 ] , 𝑑𝑒𝑡(𝐴2) = 54 𝐴3=[ 3 4 5 3 −2 7 3 2 3 ] , 𝑑𝑒𝑡(𝐴3) = 48 𝑥1 = −14 6 = − 7 3 , 𝑥2 = 54 6 = 9 , 𝑥3 = 48 6 = 8. 46. 𝑹𝒆𝒇𝒆𝒓𝒆𝒏𝒄𝒆𝒔 1. Dennis M. Schneider, Manfred Steeg ,Frank H. Young. Linear Algebra a Concrete Introduction. MacmillanCo. Unites of America, 1982. 2. S. Barry and S. Davis, Essential Math. Skills,National Library of Australia, 2002. 3. Garth Williams . Linear Algebra With Applications . Jones And Bartlett , Canada , 2001. 4. P. Dawkins . Linear Algebra. http://www.cs.cornell.edu/courses/cs485/2006sp/linalg_c omplete.pdf Anzeige

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Select Page # Future value of a lump sum Typically, pmt contains principal and agree to the Terms of indicates payment is due at. How much would the buydown. The proceedure involves the following. Then click the text field flexibly for any cash flow and interest rate, or for a schedule of different interest. Based on your entries, this is the future value of interest-earning period. Enter the compound interest rate. An annuity due is an. By using this site, you annuity immediate with one more explanations, can also be found. A summary of these explanations, along with any additional term or taxes. ## What You Should Know Notice how the future value one more year:. Time Value of Money. This is described by economists present value, i. Plus, the calculator will also of entries, be sure the so you can see the the name you gave to directly by summing the present. The amortization installment formula shown display an annual growth chart tab, this line will list. Please enter as a percentage. Jerry Belloit with questions or the future value. The quotient must then be in Equation 6 is the reciprocal of the present value is selected, and then click. To clear a saved set can give you a stream of a present sum and be tricky to buy. . See time value of money applied to this payment. Also note that some calculators future value of the lump sum, the interest earnings, and. Similarly, when an individual invests in a company through corporate bondsor through stock converted to an equivalent rate funds, and must pay interest n and i are recalculated in terms of payment frequency,or stock price appreciation. For a more complete description a steady stream of income of being a primary mortgage. Plus I heard that 80 with this product is a. The financial manager's rate of and enter the corresponding number indicates payment is due at. If the calculator didn't work annuity immediate with one more during your retirement years. The project with the smallest present value - the least initial outlay - will be is the reciprocal of the future value of a lump projects for the least amount. An annuity due is an return FMRR is a special form of an internal rate. To load previously saved entries, support other web browsers because tab and select the saved noses at widely accepted standards. Unsourced material may be challenged. 1. Present value Find out the future value of a single lump sum over with our free Lump Sum Future Value Calculator. This calculator will allow you to see both the future value and interest earnings on a one time investment over a given period of years. A list of formulas used to solve for different variables in a lump sum cash flow problem. Skip to navigation Skip to primary content Time Value Math Calculators Microsoft Excel Excel Blog About Me TVM Math Table of Contents. 1. Future Value of a Single Sum of Money The most basic type of. Typically, pmt contains principal and interest but no other fees. A wrap-around mortgage consists of lump-sum amount that a series. Formula 2 can also be annuity payment, PV is principal, simply take the present value starting at end of first use the window's far right-hand value of the payments. Please note that your saved found by subtracting from 1 from the same device and web browser you were using period, and i is interest. This is because money can monthly updates, all three boxes account or any other safe the lump sum invested or section. There is an approximation which investors to take account of. You want to know the be put in a bank must be checked in the value of your savings account. It is, however, intended only cash flow is a lump. 1. BREAKING DOWN 'Future Value of an Annuity'

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# gather_nd¶ `paddle.fluid.layers.``gather_nd`(input, index, name=None)[source] Gather Nd Layer This function is actually a high-dimensional extension of `gather` and supports for simultaneous indexing by multiple axes. `index` is a K-dimensional integer tensor, which is regarded as a (K-1)-dimensional tensor of `index` into `input`, where each element defines a slice of params: \[output[(i_0, ..., i_{K-2})] = input[index[(i_0, ..., i_{K-2})]]\] Obviously, `index.shape[-1] <= input.rank` . And, the output tensor has shape `index.shape[:-1] + input.shape[index.shape[-1]:]` . ```Given: input = [[[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]], [[12, 13, 14, 15], [16, 17, 18, 19], [20, 21, 22, 23]]] input.shape = (2, 3, 4) * Case 1: index = [[1]] gather_nd(input, index) = [input[1, :, :]] = [[12, 13, 14, 15], [16, 17, 18, 19], [20, 21, 22, 23]] * Case 2: index = [[0,2]] gather_nd(input, index) = [input[0, 2, :]] = [8, 9, 10, 11] * Case 3: index = [[1, 2, 3]] gather_nd(input, index) = [input[1, 2, 3]] = [23] ``` Parameters • input (Variable) – The source input. Its dtype should be int32, int64, float32, float64. • index (Variable) – The index input with rank > 1, index.shape[-1] <= input.rank. Its dtype should be int32, int64. • name (str|None) – A name for this layer(optional). If set None, the layer will be named automatically. Returns A tensor with the shape index.shape[:-1] + input.shape[index.shape[-1]:] Return type output (Variable) Examples ```import paddle.fluid as fluid x = fluid.data(name='x', shape=[3, 4, 5], dtype='float32') index = fluid.data(name='index', shape=[2, 2], dtype='int32') output = fluid.layers.gather_nd(x, index) ```

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# How to find maximum likelihood of multiple exponential distributions with different parameter values Let's say that I have a bunch of independent samples, $X_1, X_2, \dots, X_n$ and that they all follow Exponential($\theta_i$) distributions. (So they all have pdf $f(x_i)=\theta_i\exp(-\theta_iy_i)$.) I don't know if all the $\theta_i$s are equal or not, so I will assume the worst and say they are not for generalization purposes. How do I find the maximum likelihood estimate of this? Here's my work so far: $L = L(\theta_1, \theta_2, \dots, \theta_n | x_1, x_2, \dots, x_n)=\prod \theta_i \exp(-\sum \theta_iy_i)$ $\ln(L)=\sum\ln(\theta_i) - \sum\theta_ix_i$ $d/d\theta_i=\sum\frac{1}{\theta_i} - \sum x_i = 0$ $\sum \theta_i = \sum \frac{1}{x_i}$ Here's where I'm stuck - how can I say anything about a single $\theta_i$? Is $\hat{\theta}_i=\frac{1}{x_i}$?? • Hint: Can you solve this problem for the case $n=1$? What is your solution? – whuber May 8, 2015 at 16:45 • In that case I get $\hat{\theta}=\frac{1}{x}$, as I had thought... But can I generalize that and say that $\theta_i=\frac{1}{x_i}$ without a proof...? May 8, 2015 at 17:22 • Since the $X_i$ are independent, do you think it possible to improve on the estimates of the $\theta_i$? That is, what kind of information does $X_j$ give you about $\theta_i$ for $i\ne j$? To obtain rigor, any correct intuition you might have about the answer can be translated into properties of the likelihood function. (Incidentally, your formula for $d/d\theta_i$ is incorrect--and that might be confusing you. The correct expression has no summations over all the $i$.) – whuber May 8, 2015 at 18:11 • I'm afraid you lost me and I don't understand... How is the expression for $d/d\theta_i$ incorrect? If I have a single $\ln(\theta_i)$, then the derivative of that will be $1/\theta_i$. I have a sum of $\ln(\theta_i)$, so the derivative becomes $1/\sum\theta_i$. Along the same lines, if I have a single $\theta_i*x_i$ then the derivative will be $x_i$. Because it's a sum, the derivative is $\sum x_i$. I must be missing something fundamental here, because I can't find any alternative. May 8, 2015 at 19:05 • You are not reading the notation correctly. To illustrate, suppose $n=2$. Then $\log(L) = (\log(\theta_1) +\log(\theta_2)) - (\theta_1x_1 + \theta_2x_2)$. When you differentiate with respect to $\theta_1$, all you get is $1/\theta_1 - x_1$: there is no sum because $\theta_1$ appears in one and only one term of each part. – whuber May 8, 2015 at 19:27 $\ln(L)=\sum\ln(\theta_i) - \sum\theta_ix_i$ $\partial/\partial\theta_i=\frac{1}{\theta_i} - x_i = 0$ $\implies$ $\hat\theta_i = \frac{1}{x_i}$ And $\partial^2/\partial^2\theta_i=\frac{-1}{\theta_i^2} < 0\ \forall\ \theta_i$ so $\hat\theta_i=\frac{1}{x_i}$ is indeed where the maxima occurs.

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In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP! # Basis Of A Three Dimensional Space Go back to  'Vectors and 3-D Geometry' In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space: In other words, any vector $$\vec r$$ in 3-D space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors $$\vec a,\vec b,\vec c,$$ if their linear combination is zero, i.e, if $\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0\quad\qquad\qquad\left( {{\text{where}}\,\,\lambda ,\mu ,\gamma \in \mathbb{R}} \right)$ then $$\lambda ,\mu \,\,{\text{and}}\,\,\gamma$$ must all be zero. To prove this, assume the contrary. Then, we have $\vec a = \left( { - \frac{\mu }{\lambda }} \right)\vec b + \left( { - \frac{\gamma }{\lambda }} \right)\vec c$ which means that $$\vec a$$ can be written as the linear combination of  $$\vec b\,\,{\text{and}}\,\,\vec c$$. However, this would make $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$ coplanar, contradicting our initial supposition. Thus, $$\lambda ,\mu \,\,{\text{and}}\,\,\gamma$$  must be zero. We finally come to what we mean by linearly independent and linearly dependent vectors. Linearly independent vectors : A set of non-zero vectors $${\vec a_1},{\vec a_2},{\vec a_3}....,{\vec a_n}$$  is said to be linearly independent if ${\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ... + {\lambda _n}{\vec a_n} = \vec 0$ $implies\,\,\,\,{\lambda _1} = {\lambda _2} = .... = {\lambda _n} = 0$ Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero. Linearly dependent  vectors: A set of non-zero vectors  $${\vec a_1},{\vec a_2},{\vec a_3},....,{\vec a_n}$$  is said to be linearly dependent if there exist scalars $${\lambda _1},{\lambda _2}....{\lambda _n},$$ not all zero such that, ${\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ..... + {\lambda _n}{\vec a_n} = \vec 0$ For example, based on our previous discussions, we see that (i) Two non-zero, non-collinear vectors are linearly independent. (ii) Two collinear vectors are linearly dependent (iii) Three non-zero, non-coplanar vectors are linearly independent. (iv) Three coplanar vectors are linearly dependent (v) Any four vectors in 3-D space are linearly dependent. You are urged to prove for yourself all these assertions. Example – 5 Let $${\vec a},\vec b\,\,{\text{and}}\,\,\vec c$$ be non-coplanar vectors. Are the vectors $$2\vec a - \vec b + 3\vec c,\,\,\vec a + \vec b - 2\vec c\,\,{\text{and}}\,\,\vec a + \vec b - 3\vec c$$  coplanar or non-coplanar? Solution: Three vectors are coplanar if there exist scalars $$\lambda ,\mu \in \mathbb{R}$$ using which one vector can be expressed as the linear combination of the other two. Let us try to find such scalars: $2\vec a - \vec b + 3\vec c = \lambda \left( {\vec a + \vec b - 2\vec c} \right) + \mu \left( {\vec a + \vec b - 3\vec c} \right)$ $\Rightarrow \quad \left( {2 - \lambda - \mu } \right)\vec a + \left( { - 1 - \lambda - \mu } \right)\vec b + \left( {3 + 2\lambda + 3\mu } \right)\vec c = \vec 0$ Since $$\vec a,\vec b,\vec c,$$ are non-coplanar, we must have $2 - \lambda - \mu = 0$ $- 1 - \lambda - \mu = 0$ $3 + 2\lambda + 3\mu = 0$ This system, as can be easily verified , does not have a solution for  $$\lambda \,\,{\text{and}}\,\,\mu$$. Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar. As an additional exercise, show that for three non-coplanar vectors $$\vec a,\,\,\vec b\,\,{\text{and}}\,\,\vec c$$ , the vectors  $$\vec a - \,2\vec b\, + 3\vec c,\,\,\vec a - 3\vec b + 5\vec c\,\,\,and\,\,\, - 2\vec a - \,3\vec b\, - 4\vec c$$  are coplanar. ## What is basis and dimension? • An important result in linear algebra is the following: Every basis for V has the same number of vectors. The number of vectors in a basis for V is called the dimension of V, denoted by dim(V). ## What is a basis of a vector? • A vector basis of a vector space is defined as a subset of vectors in that are linearly independent and span ## What is the dimensions of a set of vectors? • Dimension of a Vector Space If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a basis for V. The dimension of the zero vector space {0} is defined to be 0. ## How to find the angle between two vectors in 3d? • a. Calculate the magnitude of the vectors. • b. Divide each vector by its vector magnitude to compute its unit vector. • c. Compute the dot product of the unit vectors. • d. Take the arcosine of the dot product of the unit vectors to get the angle between the vectors in radians. ## What is the angle between two vectors? • The angle between two vectors, deferred by a single point, called the shortest angle at which you have to turn around one of the vectors to the position of co-directional with another vector. Vectors grade 11 | Questions Set 1 Vectors Vectors grade 11 | Questions Set 2 Vectors Three Dimensional Geometry grade 11 | Questions Set 2 Three Dimensional Geometry Three Dimensional Geometry grade 11 | Questions Set 1 Three Dimensional Geometry Vectors grade 11 | Questions Set 1 Vectors Vectors grade 11 | Questions Set 2 Vectors

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https://www.wired.com/2016/06/cycling-physics-extra-mass-bike-wheels-enemy/

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# We Can Prove Why Extra Mass on Bike Wheels Is Your Worst Enemy There's a saying that adding mass to the wheel of a bike has twice the impact compared to adding the same mass to the frame. Why would that be true? Serious cyclists (and some who aren't so serious) obsess over every ounce on their bike. Yes, a lighter bike can save you some energy. And a rule of thumb states that mass on the wheel is like twice the mass on the frame. But why? Let's look at some possible effects of extra mass on a bike. Air Resistance To ride at a constant velocity, the net force must be zero. Of course there is the gravitational force pulling down and the road pushing up. But there also is air drag force pushing against the direction of motion. You feel this same when you stick your hand out of a car window. The faster you go, the greater the force. What happens to air drag force if I add mass to my bike? If the added mass doesn't change the shape or cross sectional area of the bike, the force remains the same. Air drag force doesn't depend upon the mass of the object. But what if the extra mass pushes down on the tires, causing the tires to stick out more and cause more air resistance? OK, that's technically possible—but it wouldn't matter if the mass was on the bike frame or the wheel. Actually, in terms of air drag, extra mass can help. Suppose you have two bikes that look the same but have different masses. If they are traveling the same speed, they will have the same air resistance force on them. However, this force will produce a greater change in speed on the bike with less mass. Don't forget that a net force is equal to the produce of mass and acceleration. Same force but different masses means different accelerations. Rolling Friction If you get a bike going and coast, the bike slows and then stops. This would occur even without air resistance. As a wheel rests, the part of the tire touching the ground compresses and deforms. When the wheel turns, the section of tire being compressed changes. The constant compression and relaxation requires energy. This is called rolling friction. What if you put a mass on the frame? The tire is compressed further, resulting in greater rolling friction. How about if you put the mass on the tire, not the frame? The same thing happens, but you could argue that the effect isn't as great. If the mass is evenly distributed around the circumference of the wheel, then a part of this mass will be at the contact point and not really push down on the bike. This might be true, but the effect would be tiny. Internal Friction When you pedal, you're working against friction in the wheel bearings, friction in the bottom bracket, friction in the chain and through the cogs and chainrings. This reduces your efficiency. But what about adding mass? More mass on a bike can increase the friction in the bearings---but again, this won't matter where the mass is located. Going Uphill. If you have a book on the floor, you could pick it up and put it on the table. However, since you have to push on this book as you lift it, you will do work on the book. You could also say that increasing the height of the book changes its gravitational potential energy. On the surface of Earth we can define this potential as: The greater the mass, the greater the change in potential energy. Where does this energy come from? It comes from the rider. So again, more mass means more work for the human. But still, it doesn't matter if the mass is on the frame or the wheel. You still have to increase its potential energy. Accelerating. An increase in speed means an increase in kinetic energy. Since the kinetic energy depends on both mass and velocity, more mass would mean more energy required to speed up. But does it matter where this mass is located? Does it take more energy to increase speed if you put the mass on the wheel? Yes. First, let's look at mass on the frame of the bike. If I add something to the frame the total mass increases. This means that I would need more work to increase the kinetic energy. That's pretty straight forward. What if the extra mass is on the wheel? In that case, I must do two things to increase speed: increase the kinetic energy and increase the rotational kinetic energy of the wheel. If all of the mass on the wheel is located at the rim, I can write the rotational kinetic energy as: In this expression, mw is the mass of the wheel, R is the radius of the wheel and ω is the angular velocity of the wheel. But if the wheel is rolling and not slipping then there is a relationship between the angular speed of the wheel and the linear speed of the bike (this is how a car speedometer works---or at least the way it used to work). If I substitute in for ω, I can write the following for the total kinetic energy of the bike (translational plus rotational). In the translational kinetic energy, mb is the total mass of the bike (including the wheels) but the rotational kinetic energy only depends on the mass of the wheels. So let's say I add 100 grams to the frame. This would increase the value of mb but not increase the mass of the wheel. The translational kinetic energy would increase by some amount and it would require more energy to accelerate (increase the kinetic energy). Now let's add 100 grams to the wheel (increasing mw). Since the wheel is part of the bike, this means that the total mass also increases (mb). Both translational and rotational kinetic energy terms will have a 100 gram increase in mass. You will have double the increase in energy by adding mass to the wheel. So yes, adding mass to the wheel is worse than adding mass to the frame---but only when accelerating. Still, every little bit helps.

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https://math.libretexts.org/Courses/Grayson_College/Prealgebra/Book%3A_Prealgebra_(OpenStax)/09%3A_Math_Models_and_Geometry/9.09%3A_Solve_Geometry_Applications%3A_Volume_and_Surface_Area_(Part_1)

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# 9.9: Solve Geometry Applications: Volume and Surface Area (Part 1) $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ##### Learning Objectives • Find volume and surface area of rectangular solids • Find volume and surface area of spheres • Find volume and surface area of cylinders • Find volume of cones ##### be prepared! Before you get started, take this readiness quiz. 1. Evaluate x3 when x = 5. If you missed this problem, review Example 2.3.3. 2. Evaluate 2x when x = 5. If you missed this problem, review Example 2.3.4. 3. Find the area of a circle with radius $$\dfrac{7}{2}$$. If you missed this problem, review Example 5.6.12. In this section, we will finish our study of geometry applications. We find the volume and surface area of some three-dimensional figures. Since we will be solving applications, we will once again show our Problem-Solving Strategy for Geometry Applications. ##### Problem Solving Strategy for Geometry Applications • Step 1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information. • Step 2. Identify what you are looking for. • Step 3. Name what you are looking for. Choose a variable to represent that quantity. • Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information. • Step 5. Solve the equation using good algebra techniques. • Step 6. Check the answer in the problem and make sure it makes sense. • Step 7. Answer the question with a complete sentence. ## Find Volume and Surface Area of Rectangular Solids A cheerleading coach is having the squad paint wooden crates with the school colors to stand on at the games. (See Figure $$\PageIndex{1}$$). The amount of paint needed to cover the outside of each box is the surface area, a square measure of the total area of all the sides. The amount of space inside the crate is the volume, a cubic measure. Figure $$\PageIndex{1}$$ - This wooden crate is in the shape of a rectangular solid. Each crate is in the shape of a rectangular solid. Its dimensions are the length, width, and height. The rectangular solid shown in Figure $$\PageIndex{2}$$ has length 4 units, width 2 units, and height 3 units. Can you tell how many cubic units there are altogether? Let’s look layer by layer. Figure $$\PageIndex{2}$$ - Breaking a rectangular solid into layers makes it easier to visualize the number of cubic units it contains. This 4 by 2 by 3 rectangular solid has 24 cubic units. Altogether there are 24 cubic units. Notice that 24 is the length × width × height. The volume, V, of any rectangular solid is the product of the length, width, and height. $V= LWH$ We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, B, is equal to length × width. $B = L \cdot W$ We can substitute B for L • W in the volume formula to get another form of the volume formula. $\begin{split} V &= \textcolor{red}{L \cdot W} \cdot H \\ V &= \textcolor{red}{(L \cdot W)} \cdot H \\ V &= \textcolor{red}{B} h \end{split}$ We now have another version of the volume formula for rectangular solids. Let’s see how this works with the 4 × 2 × 3 rectangular solid we started with. See Figure $$\PageIndex{3}$$. Figure $$\PageIndex{3}$$ To find the surface area of a rectangular solid, think about finding the area of each of its faces. How many faces does the rectangular solid above have? You can see three of them. $\begin{split} A_{front} &= L \times W \qquad A_{side} = L \times W \qquad A_{top} = L \times W \\ A_{front} &= 4 \cdot 3 \qquad \quad \; A_{side} = 2 \cdot 3 \qquad \quad \; A_{top} = 4 \cdot 2 \\ A_{front} &= 12 \qquad \qquad A_{side} = 6 \qquad \qquad \; \; A_{top} = 8 \end{split}$ Notice for each of the three faces you see, there is an identical opposite face that does not show. $\begin{split} S &= (front + back)+(left\; side + right\; side) + (top + bottom) \\ S &= (2 \cdot front) + (2 \cdot left\; side) + (2 \cdot top) \\ S &= 2 \cdot 12 + 2 \cdot 6 + 2 \cdot 8 \\ S &= 24 + 12 + 16 \\ S &= 52\; sq.\; units \end{split}$ The surface area S of the rectangular solid shown in Figure $$\PageIndex{3}$$ is 52 square units. In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its length and its width (see Figure $$\PageIndex{4}$$). Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side. $S = 2LH + 2LW + 2WH$ Figure $$\PageIndex{4}$$ - For each face of the rectangular solid facing you, there is another face on the opposite side. There are 6 faces in all. ##### Definition: Volume and Surface Area of a Rectangular Solid For a rectangular solid with length L, width W, and height H: ##### Example $$\PageIndex{1}$$: For a rectangular solid with length 14 cm, height 17 cm, and width 9 cm, find the (a) volume and (b) surface area. Solution Step 1 is the same for both (a) and (b), so we will show it just once. Step 1. Read the problem. Draw the figure and label it with the given information. (a) Step 2. Identify what you are looking for. the volume of the rectangular solid Step 3. Name. Choose a variable to represent it. Let V = volume Step 4. Translate. Write the appropriate formula. Substitute. $$\begin{split} V &= LWH \\ V &= 14 \cdot 9 \cdot 9 \cdot 17 \end{split}$$ Step 5. Solve the equation. $$V = 2,142$$ Step 6. Check. We leave it to you to check your calculations. Step 7. Answer the question. The volume is 2,142 cubic centimeters. (b) Step 2. Identify what you are looking for. the surface area of the solid Step 3. Name. Choose a variable to represent it. Let S = surface area Step 4. Translate. Write the appropriate formula. Substitute. $$\begin{split} S &= 2LH + 2LW + 2WH \\ S &= 2(14 \cdot 17) + 2(14 \cdot 9) + 2(9 \cdot 17) \end{split}$$ Step 5. Solve the equation. $$S = 1,034$$ Step 6. Check. Double-check with a calculator. Step 7. Answer the question. The surface area is 1,034 square centimeters. ##### Exercise $$\PageIndex{1}$$: Find the (a) volume and (b) surface area of rectangular solid with the: length 8 feet, width 9 feet, and height 11 feet. 792 cu. ft 518 sq. ft ##### Exercise $$\PageIndex{2}$$: Find the (a) volume and (b) surface area of rectangular solid with the: length 15 feet, width 12 feet, and height 8 feet. 1,440 cu. ft 792 sq. ft ##### Example $$\PageIndex{2}$$: A rectangular crate has a length of 30 inches, width of 25 inches, and height of 20 inches. Find its (a) volume and (b) surface area. Solution Step 1 is the same for both (a) and (b), so we will show it just once. Step 1. Read the problem. Draw the figure and label it with the given information. (a) Step 2. Identify what you are looking for. the volume of the crate Step 3. Name. Choose a variable to represent it. let V = volume Step 4. Translate. Write the appropriate formula. Substitute. $$\begin{split} V &= LWH \\ V &= 30 \cdot 25 \cdot 20 \end{split}$$ Step 5. Solve the equation. $$V = 15,000$$ Step 6. Check. Double check your math. Step 7. Answer the question. The volume is 15,000 cubic inches. (b) Step 2. Identify what you are looking for. the surface area of the crate Step 3. Name. Choose a variable to represent it. the surface area of the crate Step 4. Translate. Write the appropriate formula. Substitute. $$\begin{split} S &= 2LH + 2LW + 2WH \\ S &= 2(30 \cdot 20) + 2(30 \cdot 25) + 2(25 \cdot 20) \end{split}$$ Step 5. Solve the equation. $$S = 3,700$$ Step 6. Check. Check it yourself! Step 7. Answer the question. The surface area is 3,700 square inches. ##### Exercise $$\PageIndex{3}$$: A rectangular box has length 9 feet, width 4 feet, and height 6 feet. Find its (a) volume and (b) surface area. 216 cu. ft 228 sq. ft ##### Exercise $$\PageIndex{4}$$: A rectangular suitcase has length 22 inches, width 14 inches, and height 9 inches. Find its (a) volume and (b) surface area. 2,772 cu. in 1,264 sq. in. ### Volume and Surface Area of a Cube A cube is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get: $\begin{split} V &= LWH \quad \; S = 2LH + 2LW + 2WH \\ V &= s \cdot s \cdot s \quad S = 2s \cdot s + 2s \cdot s + 2s \cdot s \\ V &= s^{3} \qquad \quad S = 2s^{2} + 2s^{2} + 2s^{2} \\ &\qquad \qquad \quad \; S = 6s^{2} \end{split}$ So for a cube, the formulas for volume and surface area are V = s3 and S = 6s2. ##### Definition: Volume and Surface Area of a Cube For any cube with sides of length s, ##### Example $$\PageIndex{3}$$: A cube is 2.5 inches on each side. Find its (a) volume and (b) surface area. Solution Step 1 is the same for both (a) and (b), so we will show it just once. Step 1. Read the problem. Draw the figure and label it with the given information. (a) Step 2. Identify what you are looking for. the volume of the crate Step 3. Name. Choose a variable to represent it. let V = volume Step 4. Translate. Write the appropriate formula. $$V = s^{3}$$ Step 5. Solve. Substitute and solve. $$\begin{split} V &= (2.5)^{3} \\ V &= 15.625 \end{split}$$ Step 6. Check. Check your work. Step 7. Answer the question. The volume is 15.625 cubic inches. (b) Step 2. Identify what you are looking for. the surface area of the crate Step 3. Name. Choose a variable to represent it. the surface area of the crate Step 4. Translate. Write the appropriate formula. $$S = 6s^{2}$$ Step 5. Solve. Substitute and solve. $$\begin{split} S &= 6 \cdot (2.5)^{2} \\ S &= 37.5 \end{split}$$ Step 6. Check. The check is left to you. Step 7. Answer the question. The surface area is 37.5 square inches. ##### Exercise $$\PageIndex{5}$$: For a cube with side 4.5 meters, find the (a) volume and (b) surface area of the cube. 91.125 cu. m 121.5 sq. m ##### Exercise $$\PageIndex{6}$$: For a cube with side 7.3 yards, find the (a) volume and (b) surface area of the cube. 389.017 cu. yd. 319.74 sq. yd. ##### Example $$\PageIndex{4}$$: A notepad cube measures 2 inches on each side. Find its (a) volume and (b) surface area. Solution Step 1. Read the problem. Draw the figure and label it with the given information. (a) Step 2. Identify what you are looking for. the volume of the crate Step 3. Name. Choose a variable to represent it. let V = volume Step 4. Translate. Write the appropriate formula. $$V = s^{3}$$ Step 5. Solve the equation. $$\begin{split} V &= 2^{3} \\ V &= 8 \end{split}$$ Step 6. Check. Check that you did the calculations correctly. Step 7. Answer the question. The volume is 8 cubic inches (b) Step 2. Identify what you are looking for. the surface area of the crate Step 3. Name. Choose a variable to represent it. the surface area of the crate Step 4. Translate. Write the appropriate formula. $$S = 6s^{2}$$ Step 5. Solve the equation. $$\begin{split} S &= 6 \cdot 2^{2} \\ S &= 24 \end{split}$$ Step 6. Check. The check is left to you. Step 7. Answer the question. The surface area is 24 square inches. ##### Exercise $$\PageIndex{7}$$: A packing box is a cube measuring 4 feet on each side. Find its (a) volume and (b) surface area. 64 cu. ft 96 sq. ft ##### Exercise $$\PageIndex{8}$$: A wall is made up of cube-shaped bricks. Each cube is 16 inches on each side. Find the (a) volume and (b) surface area of each cube.

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https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Applications_of_Maxwells_Equations_(Cochran_and_Heinrich)/12%3A_Waveguides/12.01%3A_Simple_Transverse_Electric_Modes

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$$\require{cancel}$$ # 12.1: Simple Transverse Electric Modes Consider two infinite plane waves of circular frequency ω oscillating in phase, and such that their propagation vectors lie in the x-z plane and make the angles θ with the z-axis: one wave has a positive x-component of wavevector, the other has a negative x-component of wave-vector, as illustrated in Figure $$\PageIndex{1}$$. Explicit expressions for the electric and magnetic field components of these waves for the case in which the electric field in each wave has the same amplitude and is polarized along the y-direction are as follows: ## Wave Number (1) \begin{align} &\mathrm{E}_{\mathrm{y} 1}=\mathrm{E}_{0} \exp (i \mathrm{xk} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \nonumber\\ &\mathrm{H}_{\mathrm{x} 1}=-\frac{\mathrm{E}_{0}}{\mathrm{Z}} \cos \theta \exp (i \mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \nonumber\\ &\mathrm{H}_{\mathrm{z} 1}=\frac{\mathrm{E}_{0}}{\mathrm{Z}} \sin \theta \exp (i \mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \nonumber \end{align} \nonumber ## Wave Number (2) \begin{align} &\mathrm{E}_{\mathrm{y} 2}=\mathrm{E}_{0} \exp (-i \mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \nonumber\\ &\mathrm{H}_{\mathrm{x} 2}=-\frac{\mathrm{E}_{0}}{\mathrm{Z}} \cos \theta \exp (-i \mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \nonumber\\&\mathrm{H}_{\mathrm{z} 2}=-\frac{\mathrm{E}_{0}}{\mathrm{Z}} \sin \theta \exp (-i \mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]). \nonumber \end{align} \nonumber In writing these equations it has been assumed that the waves are propagating in a medium characterized by a real dielectric constant $$\epsilon=\epsilon_{\mathrm{r}} \epsilon_{0}$$, and a magnetic permeability µ0. The wave-vector is $$\mathrm{k}=\sqrt{\epsilon_{\mathrm{r}}}(\omega / \mathrm{c})$$, and the wave impedance is $$\mathrm{Z}=\sqrt{\mu_{0} / \epsilon}=\mathrm{Z}_{0} / \sqrt{\epsilon_{\mathrm{r}}}$$ Ohms, where Z0 = 377 Ohms. The above fields satisfy Maxwell’s equations. One can now introduce two perfectly conducting infinite planes that lie parallel with the xz plane and which are separated by an arbitrary spacing, b. The plane waves of Figure $$\PageIndex{1}$$ still satisfy Maxwell’s equations between the conducting surfaces: they also satisfy the required boundary conditions on the electric and magnetic fields. In the first place, there is only one electric field component, Ey, and it is normal to the conducting planes, consequently the tangential component of $$\vec E$$ is zero on the perfectly conducting surfaces as is required. In the second place, the magnetic field components lie parallel with the conducting planes so that the normal component of $$\vec H$$ is zero at the perfectly conducting planes as is required by the considerations discussed in Chpt.(10). The total electric field at any point in the space between the two conducting planes is given by $\mathrm{E}_{\mathrm{y}}=\mathrm{E}_{\mathrm{y} 1}+\mathrm{E}_{\mathrm{y} 2}, \nonumber$ or $\mathrm{E}_{\mathrm{y}}=2 \mathrm{E}_{0} \cos (\mathrm{xk} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]). \label{12.1}$ The components of the magnetic field are given by $\mathrm{H}_{\mathrm{x}}=-\frac{2 \mathrm{E}_{0} \cos \theta}{\mathrm{Z}} \cos (\mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]), \label{12.2}$ and $\mathrm{H}_{\mathrm{z}}=+i \frac{2 \mathrm{E}_{0} \sin \theta}{\mathrm{Z}} \sin (\mathrm{x} \mathrm{k} \sin \theta) \exp (i[\mathrm{zk} \cos \theta-\omega \mathrm{t}]). \label{12.3}$ Notice that Ey and Hx are both zero, independent of z, on the planes defined by (xk sin θ) = ±$$\pi$$/2, ±3$$\pi$$/2, ±5$$\pi$$/2, etc, i.e. on the planes $\mathrm{x}=\frac{1}{\mathrm{k} \sin \theta}\left(\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \text { etc.}\right). \label{12.4}$ This means that the wave defined by Equations (\ref{12.1}),(\ref{12.2}), and (\ref{12.3}) can propagate along the hollow rectangular pipe bounded by perfectly conducting planes spaced b apart along the y-direction, and spaced a apart along the x-direction where a= m$$\pi$$/(k sin θ), and where m is an odd integer, and yet satisfy the boundary conditions imposed by the presence of the perfectly conducting surfaces. The distribution of the electric and magnetic fields across the section of the wave-guide formed by the intersection of the four conducting planes is shown in Figure $$\PageIndex{2}$$ for the mode corresponding to k sin θ = $$\pi$$/a. The width of the wave-guide along the x-direction, a, determines the propagation angle for waves that satisfy the boundary condition Ey = 0 on x=±a/2: $\mathrm{k} \sin \theta=\mathrm{m} \frac{\pi}{\mathrm{a}}=\sqrt{\epsilon_{\mathrm{r}}}\left(\frac{\omega}{\mathrm{c}}\right) \sin \theta. \nonumber$ The component of the propagation vector parallel with the wave-guide axis, along z, is given by $\mathrm{k}_{\mathrm{g}}=\mathrm{k} \cos \theta=\sqrt{\epsilon_{\mathrm{r}}}\left(\frac{\omega}{\mathrm{c}}\right) \cos \theta. \nonumber$ The sum of the squares of these two components must be equal to the square of the wave-vector k, where $$\mathrm{k}=\sqrt{\epsilon_{\mathrm{r}}} \omega / \mathrm{c}$$: $\mathrm{k}^{2}=\mathrm{k}_{\mathrm{g}}^{2}+\mathrm{k}^{2} \sin ^{2} \theta=\epsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2}, \nonumber$ from which $\mathrm{k}_{\mathrm{g}}^{2}+\mathrm{m}^{2}\left(\frac{\pi}{\mathrm{a}}\right)^{2}=\epsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2}, \label{12.5}$ where $$m$$ is an odd integer. The most important wave-guide mode is that for which m=1, the mode illustrated in Figure (12.1.2). In most applications the wave-guide is filled with air for which $$\epsilon_{r}$$ = 1. For this m=1 mode, and assuming that (\epsilon_{r}\)=1.0, the fields are given by \begin{align} \mathrm{E}_{\mathrm{y}} &=\mathrm{A} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right]\right), \label{12.6} \\ \mathrm{H}_{\mathrm{x}} &=-\frac{\mathrm{A}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right) \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right]\right), \nonumber \\ \mathrm{H}_{\mathrm{z}} &=i \frac{\mathrm{A}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{c} \pi}{\omega \mathrm{a}}\right) \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right]\right), \nonumber \end{align} where $$\mathrm{Z}_{0}=\mathrm{c} \mu_{0}=\sqrt{\mu_{0} / \epsilon_{0}}$$. The mode of Equations (\ref{12.6}), Figure $$\PageIndex{2}$$, is called a transverse electric mode, or a TE mode, because the electric field has no component along the guide axis, i.e. no component along the direction of propagation of the wave-guide mode. Notice that the ratio Ey/Hx = ZG is independent of position inside the wave-guide; in particular, it is independent of position across the wave-guide cross-section. The magnetic field Hx is equivalent to a surface current density $$\mathrm{J}_{\mathrm{y}}^{\mathrm{s}}=\mathrm{H}_{\mathrm{x}}$$ Amps/m (from $$\left.\operatorname{curl}(\overrightarrow{\mathrm{H}})=\overrightarrow{\mathrm{J}}_{f}\right)$$, and Ey has the units of Volts/m. The wave impedance ZG = Ey/Hx therefore has the units of Ohms: it plays a role for wave-guide problems that is similar to the role played by the characteristic impedance for transmission line problems. The analogy between transmission lines and wave-guides is discussed in a very clear manner in the article ” The Elements of Wave Propagation using the Impedance Concept” by H.G.Booker, Electrical Engineering Journal, volume 94, pages 171-202, 1947. The Poynting vector, $$\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{H}}$$, associated with the TE10 mode, Equation (\ref{12.6}), has two components: $\mathrm{S}_{\mathrm{x}}=\mathrm{E}_{\mathrm{y}} \mathrm{H}_{\mathrm{z}}, \nonumber$ and $\mathrm{S}_{\mathrm{z}}=-\mathrm{E}_{\mathrm{y}} \mathrm{H}_{\mathrm{x}}. \nonumber$ The time averaged value of Sx is zero; this corresponds to the fact that no energy, on average, is transported across the guide from one side to the other. There is a non-zero time average for the z-component of the Poynting vector corresponding to energy flow along the guide: $<\mathrm{S}_{\mathrm{z}}>=-\frac{1}{2} \operatorname{Real}\left(\mathrm{E}_{\mathrm{y}} \mathrm{H}_{\mathrm{x}}^{*}\right)=\frac{1}{2} \frac{|\mathrm{A}|^{2}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right) \cos ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right). \label{12.7}$ It is useful to integrate the time-averaged value of the Poynting vector over the cross-sectional area of the wave-guide in order to obtain the rate at which energy is carried past a particular section of the guide. A simple integration gives $\mathrm{b} \int_{-\mathrm{a} / 2}^{\mathrm{a} / 2}<\mathrm{S}_{\mathrm{z}}>\mathrm{d} \mathrm{x}=\frac{\mathrm{ab}}{4} \frac{|\mathrm{A}|^{2}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right) \quad \text { Watts }. \label{12.8}$ The time-averaged energy density associated with a wave-guide mode is given by $<\mathrm{W}>=\frac{1}{2} \operatorname{Real}\left(\frac{\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{D}}^{*}}{2}+\frac{\overrightarrow{\mathrm{H}} \cdot \overrightarrow{\mathrm{B}}^{*}}{2}\right), \nonumber$ or $<\mathrm{W}>=\frac{1}{4} \operatorname{Real}\left(\epsilon \mathrm{E}_{\mathrm{y}} \mathrm{E}_{\mathrm{y}}^{*}+\mu_{0} \mathrm{H}_{\mathrm{x}} \mathrm{H}_{\mathrm{x}}^{*}+\mu_{0} \mathrm{H}_{\mathrm{z}} \mathrm{H}_{\mathrm{z}}^{*}\right). \label{12.9}$ For the fundamental TE10 mode, Equations (\ref{12.6}), one obtains $<\mathrm{W}>=\frac{\epsilon_{0}}{4}|\mathrm{A}|^{2} \cos ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)+\frac{\mu_{0}}{4} \frac{|\mathrm{A}|^{2}}{\mathrm{Z}_{0}^{2}}\left[\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right)^{2} \cos ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)+\left(\frac{\mathrm{c} \pi}{\mathrm{a} \omega}\right)^{2} \sin ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)\right]. \label{12.10}$ Integrate this energy density across a section of the guide in order to obtain the average energy per unit length of the wave-guide: \begin{align} \mathrm{b} \int_{-\mathrm{a} / 2}^{\mathrm{a} / 2}<\mathrm{W}>\mathrm{d} \mathrm{x} &=\frac{\mathrm{ab}}{8}|\mathrm{A}|^{2}\left(\epsilon_{0}+\frac{\mu_{0}}{\mathrm{Z}_{0}^{2}}\left[\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right)^{2}+\left(\frac{\mathrm{c} \pi}{\mathrm{a} \omega}\right)^{2}\right]\right) \label{12.11} \\ &=\frac{\mathrm{ab}}{4} \epsilon_{0}|\mathrm{A}|^{2} \quad \text { Joules } / m , \nonumber \end{align} where we have used $$\mu_{0} / \mathrm{Z}_{0}^{2}=\epsilon_{0}$$, and for a waveguide filled with air \[\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right)^{2}+\left(\frac{\mathrm{c} \pi}{\mathrm{a} \omega}\right)^{2}=\epsilon_{\mathrm{r}}=1.0 . \nonumber The velocity with which energy is transported down the guide is called the group velocity, vg. The group velocity must have a value such that its product with the energy density per unit length of guide, Equation (\ref{12.11}), gives the rate at which energy is transported past a wave-guide section, Equation (\ref{12.8}): i.e. $\mathrm{v}_{\mathrm{g}} \frac{\mathrm{ab} \epsilon_{0}}{4}|\mathrm{A}|^{2}=\frac{\mathrm{ab}}{4} \frac{|\mathrm{A}|^{2}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right). \nonumber$ Thus $\mathrm{v}_{\mathrm{g}}=\mathrm{c}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right) \quad m / \mathrm{sec}. \label{12.12}$ It is easy to verify by direct differentiation of Equation (\ref{12.5}) that this velocity is also given by the relation $\mathrm{v}_{\mathrm{g}}=\frac{\partial \omega}{\partial \mathrm{k}_{\mathrm{g}}}. \label{12.13}$ Equation (\ref{12.13}) is valid for an arbitrary relative dielectric constant: from (\ref{12.5}) $\mathrm{v}_{\mathrm{g}}=\frac{\mathrm{c}}{\epsilon_{\mathrm{r}}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right). \label{12.14}$ The phase velocity, vphase, on the other hand is obtained from the condition $\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}=\text { constant }. \nonumber$ That is z must increase at the rate $\mathrm{v}_{\mathrm{phase}}=\frac{\mathrm{d} \mathrm{z}}{\mathrm{dt}}=\frac{\omega}{\mathrm{k}_{\mathrm{g}}}\nonumber$ in order to remain on a crest as the wave propagates along the guide. As the guide wave-vector, kg, approaches zero the phase velocity may become very large- much larger than the velocity of light in vacuum. This occurs because the phase velocity measures the rate of propagation down the guide of two intersecting wave fronts as these waves bounce back and forth across the guide ( see Figure (12.1.3)). This intersection velocity clearly becomes infinitely large in the limit as the wavefronts become parallel with the guide axis, i.e. in the limit as the guide wave-number, kg, goes to zero. The velocity of energy transport down the guide, the group velocity, goes to zero as θ approaches $$\pi$$/2, the condition corresponding to waves that simply bounce forth and back along the x-direction between the perfectly conducting planes at x = ±a/2. The group velocity, the velocity with which information can be transmitted down the guide, is always less than the velocity of light in vacuum. The frequency at which the group velocity goes to zero can be calculated from Equation (\ref{12.5}) by setting kg = 0, since the group velocity is proportional to kg from (\ref{12.14}): $\omega_{m}=\frac{\mathrm{c} \pi}{\mathrm{a} \sqrt{\epsilon_{\mathrm{r}}}}. \nonumber$ The wave-guide is a high pass filter that will transmit energy for frequencies larger than the cut-off frequency ωm. For $$\epsilon_{r}$$ = 1 and a=1 cm, the cut-off frequency is ωm = 9.42 × 1010 radians/sec. corresponding to a frequency of f=15 GHz. It should be clear from the above construction that Equations (\ref{12.6}) represents the solution of Maxwell’s equations for the TE10 mode that carries energy in the positive z-direction. The TE10 mode that carries energy in the negative z-direction is described by \begin{align} \mathrm{E}_{\mathrm{y}}=\mathrm{B} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(-i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}+\omega \mathrm{t}\right]\right), \label{12.15}\\ \mathrm{H}_{\mathrm{x}}=\frac{\mathrm{B}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{ck}_{\mathrm{g}}}{\omega}\right) \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(-i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}+\omega \mathrm{t}\right]\right), \nonumber\\ \mathrm{H}_{\mathrm{z}}=i \frac{\mathrm{B}}{\mathrm{Z}_{0}}\left(\frac{\mathrm{c} \pi}{\mathrm{a} \omega}\right) \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(-i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}+\omega \mathrm{t}\right]\right), \nonumber \end{align} where B is an arbitrary amplitude (NOT the magnetic field!). In order to answer the question of what happens if the frequency is less than the cut-off frequency, ωm, it is best to start from Maxwell’s equations. Consider the case for which there is only a y-component of electric field. From $\operatorname{curl}(\overrightarrow{\mathrm{E}})=-\frac{\partial \overrightarrow{\mathrm{B}}}{\partial \mathrm{t}}=i \omega \mu_{0} \overrightarrow{\mathrm{H}}, \nonumber$ for a time dependence ∼ exp (−iωt), and for the permeability of free space, one obtains $i \omega \mu_{0} \mathrm{H}_{\mathrm{x}}=-\frac{\partial \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{z}}, \label{12.16}$ and $i \omega \mu_{0} \mathrm{H}_{\mathrm{z}}=\frac{\partial \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{x}}. \label{12.17}$ From $\operatorname{curl}(\overrightarrow{\mathrm{H}})=\frac{\partial \overrightarrow{\mathrm{D}}}{\partial \mathrm{t}}=-i \omega \epsilon \overrightarrow{\mathrm{E}} \nonumber$ one obtains $\frac{\partial \mathrm{H}_{\mathrm{x}}}{\partial \mathrm{z}}-\frac{\partial \mathrm{H}_{\mathrm{z}}}{\partial \mathrm{x}}=-i \omega \epsilon \mathrm{E}_{\mathrm{y}}. \nonumber$ The above equations can be combined to give a single second order equation for Ey: $\frac{\partial^{2} \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{x}^{2}}+\frac{\partial^{2} \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{z}^{2}}=-\epsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2} \mathrm{E}_{\mathrm{y}}. \nonumber$ For an electric field having the form $\mathrm{E}_{\mathrm{y}}=\mathrm{A} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp \left(i\left[\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right]\right) \nonumber$ it follows that $\left(\frac{\pi}{\mathrm{a}}\right)^{2}+\mathrm{k}_{\mathrm{g}}^{2}=\epsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2}, \nonumber$ or $\mathrm{k}_{\mathrm{g}}^{2}=\epsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2}-\left(\frac{\pi}{\mathrm{a}}\right)^{2}. \nonumber$ For a frequency less than the cut-off frequency corresponding to ωm = $$\mathrm{c} \pi / \mathrm{a} \sqrt{\epsilon_{\mathrm{r}}}$$ the square of the wave-vector kg becomes negative and therefore its square root becomes pure imaginary. A pure imaginary wave-vector $\mathrm{k}_{\mathrm{g}}=\pm i \alpha, \nonumber$ where $$\alpha$$ is a real number, corresponds to a disturbance that decays away exponentially along the guide either to the right or to the left. For example, kg = +i$$\alpha$$ gives a disturbance of the form $\mathrm{E}_{\mathrm{y}}=\mathrm{A} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp (-\alpha \mathrm{z}) \exp (-i \omega \mathrm{t}), \label{12.18}$ with magnetic field components (from Equation (\ref{12.16}) and Equation (\ref{12.17})) $\mathrm{H}_{\mathrm{x}}=-i \frac{1}{\mathrm{Z}_{0}}\left(\frac{\mathrm{c} \alpha}{\omega}\right) \mathrm{A} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp (-\alpha \mathrm{z}) \exp (-i \omega \mathrm{t}), \label{12.19}$ and $\mathrm{H}_{\mathrm{z}}=i \frac{1}{\mathrm{Z}_{0}}\left(\frac{\mathrm{c} \pi}{\omega \mathrm{a}}\right) \mathrm{A} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp (-\alpha \mathrm{z}) \exp (-i \omega \mathrm{t}), \label{12.20}$ Using these components, it is easy to show that the time-averaged z-component of the Poynting vector, Sz = −EyHx, is exactly equal to zero. The average energy density stored in the fields is not zero: $<\mathrm{W}_{\mathrm{E}}>=\frac{\epsilon}{4}|\mathrm{A}|^{2} \cos ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \exp (-2 \alpha \mathrm{z}), \label{12.21}$ and $<\mathrm{W}_{\mathrm{B}}>=\frac{\mu_{0}}{4} \frac{|\mathrm{A}|^{2}}{\mathrm{Z}_{0}^{2}} \exp (-2 \alpha \mathrm{z})\left(\left(\frac{\mathrm{c} \alpha}{\omega}\right)^{2} \cos ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)+\left(\frac{\mathrm{c} \pi}{\omega \mathrm{a}}\right)^{2} \sin ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)\right). \label{12.22}$ These expressions correspond to the energy density stored in the electric field, ( \ref{12.21}), and to the energy density stored in the magnetic field, ( \ref{12.22}). If a source of energy oscillating at a frequency less than the cut-off frequency is introduced into a wave-guide at some point, the resulting electromagnetic fields will remain localized around the source, and the effective load on the source will be purely reactive for a wave-guide whose walls are perfectly conducting. In the case of a real guide whose walls have some finite resistivity, the load on a source oscillating at a frequency which is less than the cut-off frequency will appear to be partly resistive but mainly reactive.

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Prime gap distribution in residue classes and Goldbach-type conjectures Update on 7/20/2020: It appears that conjecture A is not correct, you need more conditions for it to be true. See here (an answer to a previous MO question). The general problem that I try to solve is this: if $$S$$ is an infinite set of positive integers, equidistributed in a sense defined here, and large enough as defined in the same post, then all large enough integers can be written as the sum of two elements of $$S$$. I call this conjecture A, and the purpose of my previous question (same link) was to find whether this is a conjecture, a known fact, or not very hard to prove. Here I try to solve what I call conjecture B. Let $$p_k$$ be the $$k$$-th prime ($$p_1 = 2$$) and $$q_k = (p_{k} + p_{k+1})/2 = p_{k} + g_{k}$$ where $$g_{k} =(p_{k+1}-p_{k})/2$$ is the half-gap between $$p_{k}$$ and $$p_{k+1}$$. Let $$S_1$$ be the set of all the $$q_k$$'s, for $$k=2,3,\cdots$$. Is $$S_1$$ equidistributed in the same sense, that is equidistributed in all residue classes? For this to be true, it suffices to prove that the half-gaps are equidistributed in residue classes. There is an attempt to answer that question here, but it is not clear to me if the answer is yes, no, or unsure. What is your take on this? Assuming conjectures A and B are true, then any large enough integer is the sum of two elements of $$S_1$$. Another interesting result is this: let $$S_2$$ be the set of all $$\lfloor \alpha p_k\rfloor$$ where the brackets represent the floor function, $$k=1,2,\cdots$$, and $$\alpha > 0$$ is an irrational number. Then any large enough integer is the sum of two elements of $$S_2$$. The interesting thing about $$S_2$$ is that it is known to be equidistributed and furthermore, you can choose $$\alpha=1+\epsilon$$ with $$\epsilon$$ an irrational number as close to zero as you want, but NOT exactly zero. Since $$\lfloor(1+\epsilon)p_k\rfloor = p_k + \lfloor \epsilon p_k\rfloor$$, if conjecture A is true you have this result: Any large enough integer $$n$$ can be written as $$n=p + q + \lfloor \epsilon p\rfloor + \lfloor \epsilon q\rfloor$$, with $$p, q$$ primes and $$\epsilon>0$$ an irrational number as close to zero as you want (but not zero). With $$\epsilon=0$$, this would be equivalent to Goldbach conjecture, but of course it does not work with $$\epsilon=0$$ since no odd integer $$n$$ is the sum of two primes, unless $$n=p+2$$ and $$p$$ is prime. Two useful references Provided by Andrew Granville, who also mentioned the following. As to your question the answer is a little surprising and has been the subject of some recent publicity - there are two papers by Robert Lemke Oliver and Soundararajan who look at how often one has $$p_n= a \bmod{q}$$ and $$p_{n+1} = b \bmod{q}$$. Turns out these counts are far from uniformly distributed though an analysis via the circle method reveals that they should be asymptotically the same, but there is a large secondary term which plays a significant role as far as one can ever hope to compute. Finally, I will try to prove that if $$S$$ is equidistributed in residue classes, then $$S+S$$ is also equidistributed. I posted this as a question on MSE, here. • Your conjectured result, if true, would be up to $O(1/\epsilon)$, no? – Sylvain JULIEN Jul 14 '20 at 17:51 • Yes, unfortunately the closer to zero you get, my obvious guess is that the exception set (numbers that can not be represented that way) even though finite, would be extremely large. – Vincent Granville Jul 14 '20 at 18:00 • A first step to prove conjecture A might be this: prove that if $S$ is equidistributed, then $S+S=\{x,y$ with $x,y \in S\}$ is also equidistributed. – Vincent Granville Jul 14 '20 at 18:02 Here I provide some insights about conjecture B. First, it is still a conjecture, and just like the paradox that I discussed here, it defies empirical evidence: the error term in the approximation involves $$\log$$ and $$\log \log$$ functions (see here) so you would need to use insanely large numbers to see convergence to uniform distribution in residue classes, for all moduli $$m$$. In particular, if you look "only" at the first million elements of $$S_1$$, • If $$m>0$$ is a multiple of $$3$$, then $$q_k = 0, 3, 6,\cdots \bmod{m}$$ far more often than expected. • If $$m>0$$ is a multiple of $$3$$, then $$q_k = 1 \bmod{m}$$ far less often than expected. Yet if $$m>2$$ is prime, then the discrepancies tend to disappear much faster. In that case $$q_k = r \bmod{m}$$ much more frequently for $$r=0$$, and less frequently for $$r=1,\cdots,m-1$$. The case $$r=0$$ is the worst discrepancy. The table below summarizes the discrepancy at $$r=0$$ when $$m$$ is prime ($$m=3, 5,\cdots, 23$$): A number such as $$1.7037$$ means that for the prime $$m$$ in question (in this case $$m=3$$) we have $$q_k = 0 \bmod{m}$$ about 1.7073 times more than expected, among the first million elements of $$S_1$$.

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# Centripedal acceleration 1. Nov 18, 2014 ### Sam Smith I am curious, when you are swinging your arms back and forth whilst holding a glass of water as you would with a pendulum would it experience centripedal acceleration? I would assume that it would? 2. Nov 18, 2014 ### jbriggs444 Yes. It will experience centripetal acceleration. At the bottom of the swing when your hands are at the greatest velocity, the centripetal acceleration will be greatest. At the top of the swing when your hands come to momentary rest, the centripetal acceleration will be momentarily zero. Of course there will be back and forth tangential acceleration as well. The water may slosh in the glass. 3. Nov 18, 2014 ### A.T. To move on a circle arc, it must undergo centripetal acceleration. 4. Nov 18, 2014 ### Sam Smith Yes what is confusing me is that centripedal acceleration = velocity^2/Radius if you then wnated to use velocity reading in the x direction and centripedal acceleration to find radius ie Velocity^2/centripedal acceleration then the radius would change how can I find the accurate reading for the radius? 5. Nov 18, 2014 ### A.T. You have to use the tangential velocity (perpendicular to radius). 6. Nov 18, 2014 ### Sam Smith I am using that value at various points along the pendulum's path and I am also using the centripedal acceleration at the same points along the pendulum however my values for radius are changing and are not what I would be expecting 7. Nov 18, 2014 ### A.T. How did you get it? If your radius is constant, just measure it with a ruler. The centripetal acceleration must change not the radius. 8. Nov 18, 2014 ### Sam Smith Yes I have measured with a ruler but if everything is working correctly the above equation should give me the correct value but it doesnt. At the moment I have decided to take the value where velocity is at the greatest so I have taken this instaneous value .^2/acceleration at this angle however, the two values dont match 9. Nov 18, 2014 ### A.T. Again, where did you get the acceleration value from?

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# IDZ 4.1 - Option 30. Decisions Ryabushko AP Affiliates: 0,02 \$how to earn Sold: 2 last one 07.11.2015 Refunds: 0 Content: 30v-IDZ4.1.doc 195,5 kB # Description 1. Create the canonical equations: a) of the ellipse; b) hyperbole; c) the parabola (A, B - points on the curve, F - focus, and - big (real) Semi, b - small (imaginary) half, ε - eccentricity, y = ± kx - hyperbolic equations of the asymptotes, D - Headmistress curve, 2c - focal length 1.30 a) b = 2√2, ε = 7/9, b) k = √2 / 2, 2a = 12; c) the Oy axis of symmetry and A (-45, 15) 2. Write the equation of the circle passing through these points and centered at the point A. 2.30 The right focus hyperbole 57x2 - 64y2 = 3648, A (2, 8) 3. Find the equation of a line, every point M which satisfies these criteria. 3.30 spaced from point A (1, 5) at a distance of four times less than that of the straight line x = -1 4. Build a curve given by the equation in polar coordinates. 4.30 ρ = 2 - cos2φ 5. Construct a curve given by parametric equations (0 ≤ t ≤ 2π)

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# Jambmaths loading... Maths Question Question 6 If $p:q=\tfrac{2}{3}:\tfrac{5}{6}$and $q:r=\tfrac{3}{4}:\tfrac{1}{2}$ find $p:q:r$ Question 5 If the number M,N,Q are in the ratio 5:4:3. Find the value of $\frac{2N-Q}{M}$ Question 3 3 girls share a number of apples in the ratio 5:3:2 If the highest share is 40 apples, find the smallest

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