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http://mathhelpforum.com/statistics/211991-please-verify-twist-birthday-problem.html | 1,524,273,705,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944848.33/warc/CC-MAIN-20180420233255-20180421013255-00380.warc.gz | 195,843,581 | 10,296 | 1. ## Please verify: a twist on the Birthday Problem
Hello,
I was assigned some optional practice on the birthday question. I got all the questions on the birthday problem itself correct, but although I got an answer for this last one (n >= 612.257), it seems VERY unintuitive to me. Can you please confirm whether it's right or what I did wrong? Here is the problem:
Given n people randomly chosen, find the smallest value of n so that the probability at least one of them were born on the same day of the year that you are born is at least 50%.
Now, here's what I thought:
P(X >= 1) >= 0.5
X~Binomial(n, 1/365)
P(X >= 1) = 1 - P(X=0) - P(X=1)
1 - P(X=0) - P(X=1) >= 0.5
0.5 >= P(X=0) + P(X=1)
0.5 >= (nCr(n,0) * (1/365)^0 * (364/365)^n) + (nCr(n,1) * (1/365) * (364/365)^(n-1))
Now, nCr(anything, 0) is always 1, and nCr(n, 1) is always n. So:
0.5 >= (364/365)^n + (n/365)*(364/365)^(n-1)
Plugging this into wolfram alpha, I get that n >= 612.257. This result seems very unintuitive to me. Is it correct?
2. ## Re: Please verify: a twist on the Birthday Problem
Hello, Gui!
Given n people randomly chosen, find the smallest value of n so that the probability that
at least one of them were born on the same day of the year that you are born is at least 50%.
We will find the probability that none of them have your borthday.
The probability that none of the $\displaystyle n$ people have your birthday is:.$\displaystyle \left(\frac{364}{365}\right)^n$
. . We want this to be less than 50%.
We have: .$\displaystyle \left(\frac{364}{365}\right)^n \,<\:0.5$
Take logs: .$\displaystyle \ln\!\left(\frac{364}{365}\right)^n \,<\;\ln(0.5) \quad\Rightarrow\quad n\!\cdot\!\ln\!\left(\frac{364}{365}\right) \,<\:\ln(0.5)$
Therefore: .$\displaystyle n \:<\:\dfrac{\ln(0.5)}{\ln\left(\frac{364}{365} \right)} \:<\:252.6519836$
. . It will take 253 people. | 614 | 1,869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-17 | latest | en | 0.876787 |
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Quantum_Physics_(Ackland)/09%3A_Indistinguishable_Particles_and_Exchange/9.05%3A_Spins_and_Exchange | 1,722,735,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00582.warc.gz | 365,712,102 | 29,480 | # 9.5: Spins and Exchange
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Now notice something strange. The exchange interaction has split the $${\bf S}=1$$ states from the $${\bf S}=0$$ states. We could write the potential as $$\hat{V} = J_{nl} − (2\hat{S} − 1)K_{nl}$$, even though the Hamiltonian does not act on the spin! This is because the sign of the exchange integral depends on the (anti)symmetry of the spatial wavefunction. Thus we can write the matrix element as
$\langle \Phi |J_{nl} − (2S − 1)K_{nl}|\Phi \rangle \nonumber$
This ‘exchange interaction’ appears to depend on the spin - the triplet states have lower energy than the singlet (this is one of Hund’s rules for determining energy levels in atoms). It is this type of exchange force which keeps spins aligned in a ferromagnet, not the magnetic interaction itself.
This page titled 9.5: Spins and Exchange is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform. | 2,021 | 5,372 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.198584 |
https://studyres.com/doc/186691/section-1.2-angle-relationships-and-similar-triangles | 1,627,850,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00371.warc.gz | 556,274,938 | 7,863 | # Download Section 1.2 Angle Relationships and Similar Triangles
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```Do Now
Find the supplement of each angle.
1.83°
2.35°
3.165°
4.73°
5.124°
Copyright © 2005 Pearson Education, Inc.
Slide 1-1
Section 1.2
Angle Relationships and Similar
Triangles
Objective:
SWBAT use geometric properties to identify similar triangles
and angle relationships.
Copyright © 2005 Pearson Education, Inc.
Vertical Angles
Vertical Angles have equal measures.
Q
R
M
N
P
The pair of angles NMP and RMQ are vertical
angles.
Do you see another pair of vertical angles?
Copyright © 2005 Pearson Education, Inc.
Slide 1-3
Parallel Lines
Parallel lines are lines that lie in the same plane
and do not intersect.
When a line q intersects two parallel lines, q, is
called a transversal. Eight angles are now
formed.
Transversal
q
m
parallel lines
n
Copyright © 2005 Pearson Education, Inc.
Slide 1-4
Angles and Relationships
q
m
n
Name
Angles
Rule
Alternate interior angles
4 and 5
3 and 6
Angles measures are equal.
Alternate exterior angles
1 and 8
2 and 7
Angle measures are equal.
Interior angles on the same
side of the transversal
4 and 6
3 and 5
Angle measures add to 180.
Corresponding angles
2 & 6, 1 & 5,
3 & 7, 4 & 8
Angle measures are equal.
Copyright © 2005 Pearson Education, Inc.
Slide 1-5
Finding Angle Measures
Find the measure of each
marked angle, given that
lines m and n are parallel.
(6x + 4)
(10x 80)
m
n
The marked angles are
alternate exterior angles,
which are equal.
Copyright © 2005 Pearson Education, Inc.
6 x 4 10 x 80
84 4 x
21 x
One angle has measure
6x + 4 = 6(21) + 4 = 130
and the other has measure
10x 80 = 10(21) 80 = 130
Slide 1-6
Finding Angle Measures
B
m<A = 58°
C
D
Z
W
Y
X
Copyright © 2005 Pearson Education, Inc.
Slide 1-7
Angle Sum of a Triangle
Take your given triangle.
Tear each corner from the triangle. (so you now have 3 pieces)
Rearrange the pieces so that the 3 pieces form a straight angle.
Convincing?!?
The sum of the measures of the angles of any
triangle is 180.
Copyright © 2005 Pearson Education, Inc.
Slide 1-8
Applying the Angle Sum
The measures of two of the
angles of a triangle are 52
and 65. Find the measure
of the third angle, x.
Solution
52 65 x 180
117 x 180
x 63
65
x
52
Copyright © 2005 Pearson Education, Inc.
Slide 1-9
Applying the Angle Sum
The measures of two of the
angles of a triangle are 48
and 61. Find the measure
of the third angle, x.
Solution:
48
x
61
Copyright © 2005 Pearson Education, Inc.
Slide 1-10
Types of Triangles: Angles
Copyright © 2005 Pearson Education, Inc.
Slide 1-11
Types of Triangles: Sides
Copyright © 2005 Pearson Education, Inc.
Slide 1-12
Homework
Page 14-16
# 4, 6, 12, 13, 16, 18, 26, 30, 34
Copyright © 2005 Pearson Education, Inc.
Slide 1-13
Do Now
Find the measures of all the angles.
(2x – 21)°
Copyright © 2005 Pearson Education, Inc.
(5x – 129)°
Slide 1-14
Section 1.2…Day 2
Angle Relationships and Similar
Triangles
Objective:
SWBAT use geometric properties to identify similar triangles
and angle relationships.
Copyright © 2005 Pearson Education, Inc.
Conditions for Similar Triangles
Similar Triangles are triangles of exactly the same
shape but not necessarily the same size.
Corresponding angles must have the same
measure.
Corresponding sides must be proportional.
(That is, their ratios must be equal.)
Copyright © 2005 Pearson Education, Inc.
Slide 1-16
Finding Angle Measures
Triangles ABC and DEF
are similar. Find the
measures of angles D and
E.
D
Since the triangles are
similar, corresponding
angles have the same
measure.
Angle D corresponds to
angle A which = 35
A
112
35
F
C
112
33
Copyright © 2005 Pearson Education, Inc.
E
Angle E corresponds to
angle B which = 33
B
Slide 1-17
Finding Side Lengths
Triangles ABC and DEF are
similar. Find the lengths of
the unknown sides in
triangle DEF.
32 64
16
x
32 x 1024
x 32
D
A
16
112
35
64
F
32
C
112
33
48
Copyright © 2005 Pearson Education, Inc.
To find side DE.
B
E
To find side FE.
32 48
16 x
32 x 768
x 24
Slide 1-18
Application
A lighthouse casts a
shadow 64 m long. At the
same time, the shadow
cast by a mailbox 3 feet
high is 4 m long. Find the
height of the lighthouse.
The two triangles are
similar, so corresponding
sides are in proportion.
3 x
4 64
4 x 192
x 48
3
4
x
The lighthouse is 48 m
high.
64
Copyright © 2005 Pearson Education, Inc.
Slide 1-19
Homework
Page 17-18
# 42-56 (evens)
Copyright © 2005 Pearson Education, Inc.
Slide 1-20
```
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• ##### Free/Non-commercial Resources:
Displaying 7 resources
• Resource Type: Website / HyperLink
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## Mathematics Common Core Unpacked: Grade Five
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• Resource Type: Website / HyperLink | 397 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-25 | latest | en | 0.748286 |
https://www.physicsforums.com/threads/i-cant-understand-this-prove-explanation-on-limit-series.286448/ | 1,723,080,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00174.warc.gz | 740,367,995 | 19,404 | # I cant understand this prove explanation on limit series
• transgalactic
In summary, the conversation discusses the proof of a continues function f(x) that is bounded on (x_0,+infinity). The proof shows that for every T, there exists a sequence X_n=+infinity such that lim [f(x_n +T) - f(x_n)]=0 as n->+infinity. The conversation also mentions the use of delta in limit proofs for finite numbers and the use of a large number M in limit proofs for approaching infinity.
transgalactic
there is a continues function f(x) and bounded on (x_0,+infinity)
proove that for every T there is a sequence
X_n=+infinity
so
lim [f(x_n +T) - f(x_n)]=0
n->+infinity
i was told:
uppose that $\lim_{x\to\infty}f(x)=a$. So we know that given any $\varepsilon>0$ there exists a $\eta>0$ such that $\eta< x\implies |f(x)-a|<\varepsilon~(1)$, and since $x_n\to \infty$ we may find a $\delta>0$ such that $\delta<n\implies \eta<x_n~(2)$. Now suppose that $T>0$ (the proof for the other cases is analgous), then choose $\delta$ such that $(2)\implies (1)$ then $\left[f\left(x+T\right)-f(x)\right|\leqslant \left|f\left(x+T\right)-a\right|+|f(x)-a|<2\varepsilon$, this implies the result.
http://www.mathhelpforum.com/math-h...e-never-learnt-well-epsilon-delta-proofs.html
i learned from the delta proofes article that when you define the delta
$\delta>0$
it needs to come with
$|x-x_3|<\delta$
in our case x_3 goes to infinity
so the inqueality that i presented not logical
but on the other hand
it how its done on the article limit proove
??
Limits in which x approaches infinity are proved differently than those for which x approaches some finite number a.
$$\lim_{x \rightarrow \infty} f(x) = L$$
means that for any $\epsilon > 0$ there exists a number M > 0, such that for any x > M, then |f(x) - L| < $\epsilon$
If I want to prove the limit above to you, you tell me how close f(x) has to be to L (you give me $\epsilon$), and I tell you a number M.
If you're not satisfied, you tell me another $\epsilon$ that's even smaller, and I have to find another M (even larger).
And so on, until you're convinced that I can force f(x) as close to L as you like, by specifiying how big x has to be.
Got it?
in your explanation M is delta>0
??
transgalactic said:
in your explanation M is delta>0
??
No. M is generally a pretty large number, while $\delta$ is usually very small. A big difference is the definition I showed doesn't try to get x within $\delta$ of infinity.
but he does use delta
for what purpose
??
If the limit is as x approaches a finite number a, you use $\delta$, since you want to make x very close to a. I.e., you want to make |x - a| < $\delta$.
If the limit is as x approaches infinity, that's a different matter, since as I explained earlier, you can't get x within $\delta$ of infinity. You can, however, make x larger than some (presumably large) number M.
I don't think I can make it any clearer than that.
## What is a limit series?
A limit series is a mathematical sequence of numbers where the terms approach a specific value, known as the limit, as the number of terms increases. It is commonly used to analyze the behavior of functions and determine their convergence or divergence.
## Why is it important to understand limit series?
Limit series are important in mathematics because they are used to prove the convergence or divergence of various mathematical functions and series. They also have applications in fields such as physics, engineering, and economics.
## How do you prove a limit series?
To prove a limit series, you must show that the terms of the series approach a specific value, known as the limit, as the number of terms increases. This can be done using various mathematical techniques such as the limit comparison test, ratio test, or root test.
## What are some common challenges in understanding limit series?
Some common challenges in understanding limit series include the complexity of the mathematical concepts involved, the use of abstract symbols, and the need for a strong foundation in calculus and algebra. It can also be challenging to apply the various tests and techniques to different types of limit series.
## What are some resources for learning more about limit series?
There are many resources available for learning about limit series, including textbooks, online tutorials, and video lectures. You can also consult with a math teacher or tutor for personalized help. Additionally, practicing problems and working through examples can greatly aid in understanding limit series.
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1K | 1,306 | 5,117 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.862172 |
https://swiftinit.org/docs/swift/swift/closedrange | 1,695,627,835,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00495.warc.gz | 627,074,506 | 7,665 | StructureSwift5.9.0
# ClosedRange
An interval from a lower bound up to, and including, an upper bound.
@frozen struct ClosedRange<Bound> where Bound : Comparable
You create a ClosedRange instance by using the closed range operator (...).
let throughFive = 0...5
A ClosedRange instance contains both its lower bound and its upper bound.
throughFive.contains(3)
// true
throughFive.contains(10)
// false
throughFive.contains(5)
// true
Because a closed range includes its upper bound, a closed range whose lower bound is equal to the upper bound contains that value. Therefore, a ClosedRange instance cannot represent an empty range.
let zeroInclusive = 0...0
zeroInclusive.contains(0)
// true
zeroInclusive.isEmpty
// false
## Using a Closed Range as a Collection of Consecutive Values
When a closed range uses integers as its lower and upper bounds, or any other type that conforms to the Strideable protocol with an integer stride, you can use that range in a for-in loop or with any sequence or collection method. The elements of the range are the consecutive values from its lower bound up to, and including, its upper bound.
for n in 3...5 {
print(n)
}
// Prints "3"
// Prints "4"
// Prints "5"
Because floating-point types such as Float and Double are their own Stride types, they cannot be used as the bounds of a countable range. If you need to iterate over consecutive floating-point values, see the stride(from:through:by:) function.
## Citizens in Swift
where Bound:Strideable, Bound.Stride:SignedInteger
## Citizens in Swift
where Bound:Comparable
## Citizens in Swift
where Bound:Comparable, Bound:Decodable
## Citizens in Swift
where Bound:Comparable, Bound:Encodable
## Citizens in Swift
where Bound:Comparable, Bound:Sendable
### Conformances
• protocol Sendable
A type whose values can safely be passed across concurrency domains by copying.
## Citizens in Swift
where Bound:Comparable, Bound:Hashable
### Conformances
• protocol Hashable
A type that can be hashed into a Hasher to produce an integer hash value. | 473 | 2,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.817759 |
https://www.discover-autism-help.com/Discover_Autism_Help-abstract-math.html | 1,611,080,729,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00478.warc.gz | 749,940,464 | 2,498 | Back to Back Issues Page Autism news: How to help our kids understand basic math September 04, 2020 Hello!How do you help a child with autism understand 2 + 4 = 6?Even if we help our kids memorize all the addition equations, do they really know what these symbols mean?We take it for granted that we know the above equation means that adding two things to four things amounts to six things.Most experts will tell you that helping students understand what they’re doing in math is important to success.So how can we help them see the meaning of what we’re teaching them?The best way I know of to overcome this diffculty is to present these concepts in a concrete form.What do I mean by that?We need to show them exactly what the math concept means by demonstrating it with real objects in real life.Videos help, and YouTube has many to choose from, which I highly recommend.But there’s nothing like presenting math concepts with objects that your child can touch, see, count, add, subtract, etc.So if you present the number 4, show your child what that means by having him count four objects. You can use four pens, four toys, four blueberries, four apples—anything you have around the house. After he counts them, ask him to write down the answer. You want to reinforce in his mind that the number 4 represents four things.For more information on how to teach your child math using concrete examples, you can check out my article at the link at the end of this message.I’ve also reposted an article I removed from my site some time ago due to a search engine issue. If you were on this newsletter list at that time, you may remember seeing it.In the updated article, I explore some reasons why our kids may be having behavior problems. It’s important to know there are causes of behavior problems for which there are remedies. If we as parents put on our detective hats, we can observe, investigate and do research to find out why our kids behave as they do.The newly updated article is the first in a three-part series I plan to be posting in the next few weeks. To see part 1 of the series on causes of behavior problems, click the link at the end of this message. Warm Regards,
Kay Donatohttps://www.discoverautismhelp.com/ Click here for more information on how to use concrete examples to teach basic math. Click here to see part 1 in the series on causes of behavior problems. Back to Back Issues Page | 519 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-04 | longest | en | 0.962691 |
https://doctroid.wordpress.com/2016/01/09/i-know-some-martian-navigators-who-can-still-do-that-today-marswalk/ | 1,500,677,276,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423812.87/warc/CC-MAIN-20170721222447-20170722002447-00494.warc.gz | 650,592,701 | 36,701 | # I know some Martian navigators who can still do that today #MarsWalk
(apologies to Tom Lewis.)
If you start up Google Earth, then go to the View menu and select Explore, then Mars, you get Mars. Very cool. The Layers include satellite images, IAU/USGS place names, guided tours, and the locations of about everything we’ve ever put on Mars, like, oh, for instance, Viking 2 and Curiosity. Curiosity’s landing site and subsequent path are both shown, though the path looks like it’s some weeks or months old.
But if you draw a path from Viking 2 to Curiosity and look at the length of that path, Google says it’s about 3059 km. Boyan’s #MarsWalk posts say it’s 2880 km. Bit of a discrepancy there.
So, first of all, are Google’s positions right?
For Curiosity’s present position see http://curiosityrover.com/rovermap1.html. Using that I made a pin in Google Earth/Mars with Curiosity’s updated position. Viking, of course, has stayed put for 40 years, and we know where it is too. The Google Earth position is accurate.
The distance of 2880 km is also given at NASA’s Mars Mileage Guide which gives a handy table of all the distances between ten landers and three geographical features. That page also states coordinates: For Viking 2, 44° N, 226° W, and for Curiosity, 4.59°S, 222.56°W. But the Viking 2 coordinates are said to be 47.97° N, 225.74° W here. In Google Earth the coordinates are 47.66° N, 134.28° W and 4.40° S, 137.22°W. Obviously a different coordinate system. If it’s just a meridian shift of 88° then Google’s reasonably consistent with the second of those pages but not the first.
Hm. If I put a marker in Google Earth at 44° N, 134° W and measure the distance to Curiosity I get about 2843 km. Close enough to 2880 to make me confident that’s the source of the discrepancy… it’s based on a wrong location for Viking 2!
I have a query in with the NASA contact for that page. Maybe I’m getting something wrong. But it looks to me as though, unfortunately for anyone walking, the actual distance is about 180 km longer than we thought.
MarsWalk spreadsheet
MarsWalk kmz file (for Google Earth — View >> Explore >> Mars)
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Nov 9 revised Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? argument below the theorem is not a proof of the theorem, just an example of some subtleties of the theorem, and the example (I hope) partially answers the question. Nov 9 answered Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? Nov 8 comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? extension. These facts are all true at least in the case when a small forcing is used. There are large forcings which will destroy certain large cardinals, large forcings which will create so-called generic large cardinals, and, if I remember correctly, forcings that will resurrect large cardinals which were previously killed. There are many articles exhibiting this type of phenomenon, though I don't have any specific references at the moment. Nov 8 comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? @JosephVanName. Your question is vague (to me at least) but now I think I get what you are after. The point of my first comment was that the consistency of an I0 gives an I1 with your embedding in a generic extension, similarly I1 suffices for an I3 with that property. These embedding already exist in the ground model and are preserved to the generic extension. This is one direction of the Levy-Solovay phenomenon. The other direction is that large cardinals are not created in the generic extension. So your desired $\lambda$, if it is not already I3, say, then it will not become I3 in the Nov 8 comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? Maybe I've misunderstood your question. Are you asking for an inaccessible above an I3 which is itself not I3, that you can collapse and find a non-trivial embedding in the generic extension? Or do you want an inaccessible (or perhaps something stronger), not necessarily above an I3 that you can singularize through forcing and introduce an embedding in the generic extension? Or maybe something else altogether? Nov 8 comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$? @JosephVanName. Don't I1 cardinals in V already imply the existence of many I3 cardinals below? If we use a small partial order, these I3 cardinals are preserved to the generic extension. More specifically, if $\lambda$ is I1, then there is a $\bar{\lambda}$ below it and a $j$ a witnessing that it is I3. If $G\subset P$ is generic and \$|P| | 657 | 2,846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-18 | latest | en | 0.89629 |
https://www.experts-exchange.com/questions/21821011/Overloading-a-function-template.html | 1,477,544,347,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721141.89/warc/CC-MAIN-20161020183841-00340-ip-10-171-6-4.ec2.internal.warc.gz | 927,757,665 | 26,130 | Solved
Posted on 2006-04-20
225 Views
Here is a function template that will accept two arguments. I must then write the function to accept three arguments and calculate the average of these arguments. I have just included my template and function- no errors anywhere else.
template <class T>
T average(T a1, T b2);
T average(T a1,T b2, T c3); //Overloaded function
template<class T>
T average(T a1, T b2);
T average(T a1, T b2, T c3);
{ if (number = 2)
average = (a1 + b2) / 2;
else if (number = 3)
average = (a1 +b2 +c3 ) / 3;
else
cout << "Try agian" <<endl;
return double average;
}
Here are the errors I am getting that is related to this code.
error C2146: syntax error : missing ';' before identifier 'average'
error C2501: 'T' : missing storage-class or type specifiers
error C2146: syntax error : missing ')' before identifier 'a1'
error C2501: 'average' : missing storage-class or type specifiers
error C2365: 'average' : redefinition; previous definition was a 'function'
error C2059: syntax error : ')'
I don't understand what is wrong with my template, it is basically written straight from the book. Please help.
0
Question by:joew08
LVL 20
Expert Comment
hi joew08,
seems you tried to declare 2 functions, try this, only the second:
template <class T>
T average(T a1,T b2, T c3); //Overloaded function
template<class T>
T average(T a1, T b2, T c3) // << remove the semicolon here
{ if (number = 2)
average = (a1 + b2) / 2;
else if (number = 3)
average = (a1 +b2 +c3 ) / 3;
else
cout << "Try agian" <<endl;
return double average;
}
hope it helps :)
ike
0
LVL 20
Expert Comment
this code should work:
template <class T> T average(T a1, T b2);
template <class T> T average(T a1,T b2, T c3);
template<class T>
T average(T a1, T b2)
{
return (a1 + b2) / 2
}
template<class T>
T average(T a1, T b2, T c3);
{
return (a1 +b2 +c3 ) / 3;
}
ike
0
LVL 20
Expert Comment
you said "basically written straight from the book". can you post the original, then we can see where you did go wrong.
in your code is a variable "number" .. where was it declared?
ike
0
Author Comment
said "basically written straight from the book". can you post the original, then we can see where you did go wrong.
in your code is a variable "number" .. where was it declared?
ike
-------------------------------------------------------------
The book example is as follows:
template<class T>
T larger(T x, T y);
template<class T>
T larger(T x, T y)
{if ........
}
-------------------------------------------------------------
So when I declare my functions and then overload one
template <class T> T average(T a1, T b2);
template <class T> T average(T a1,T b2, T c3); //overloaded function
I have to define both of them indepently? Is this correct.
Joe
0
LVL 20
Accepted Solution
>> I have to define both of them indepently? Is this correct.
yes, at least you have to define both of them. you only have to declare them, if the function is used in the code, before the definition occured.
// declaration
template <class T> T average(T a1, T b2);
template <class T> T average(T a1,T b2, T c3);
/* anywhere in the code the function is used, now at least "template <class T> T average(T a1, T b2)" has to be declared above */
int a = average( b, c );
// definition
template <class T> T average(T a1, T b2)
{
// code here
}
template <class T> T average(T a1,T b2, T c3)
{
// code here
}
hope it helps :)
ike
0
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Last updated on Apr 14, 2024
## Latest Lagrangian and Hamiltonian Formalism MCQ Objective Questions
#### Lagrangian and Hamiltonian Formalism Question 1:
The Lagrangian of a particle in one dimension is L = $$\frac{m}{2}\dot{x}^2-ax^2-V_0e^{-10x}$$ where a and V0 are positive constants. The best qualitative representation of a trajectory in the phase space is
Option 2 :
#### Lagrangian and Hamiltonian Formalism Question 1 Detailed Solution
Explanation:
Corresponding Hamiltonian is given as :
$$H=\frac{P^2}{2m}+ax^2+V_0e^{−10x}$$
Now, clearly, this is not an elliptical relation between p and x. So, we can eliminate options 3 and 4.
• Potential energy $$V(x)=ax^2+V_0e^{−10x}$$ has its minimum at x>0.
• When we decrease energy, the phase space closed curve shrinks at a potential minimum or we can say stable equilibrium.
• First graph is to shrink at $$x_0<0$$.
• The second graph is to shrink at $$x_0>0$$. Hence, is the answer.
#### Lagrangian and Hamiltonian Formalism Question 2:
The Lagrangian of a system of two particles is L = $$\frac{1}{2}\dot{x}^2_1+2\dot{x}^2_2-\frac{1}{2}(x^2_1+x^2_2+x_1x_2)$$. The normal frequencies are best approximated by
1. 1.2 and 0.7
2. 1.5 and 0.5
3. 1.7 and 0.5
4. 1.0 and 0.4
Option 4 : 1.0 and 0.4
#### Lagrangian and Hamiltonian Formalism Question 2 Detailed Solution
Explanation:
The equations of motion for the system in terms of the normal coordinates are obtained by applying Euler-Lagrange equations for i = 1, and i = 2.
Euler-Lagrange equations are: $$\frac d{dt} (\frac{∂L}{∂ẋᵢ}) - \frac{∂L}{∂xᵢ} = 0$$ for i = 1, 2. So the equations of motion are $$\ddot x₁ - x₁ - 0.5x₂ = 0$$ ...(1) & $$4\ddot x₂ - 0.5x₁ - x₂ = 0$$ ....(2)
Solve the characteristic equation for eigenvalues (which correspond to the squares of the normal mode frequencies), which is in the form of $$|m - λI| = 0$$, where m is the 2x2 matrix of coefficients of terms linear in x, I is the 2x2 identity matrix and λ are the eigenvalues.
This equation gives us a quadratic equation for the eigenvalues: $$(1 - λ)(4 - λ) - (0.5)(0.5) = λ² - 5λ + 4 - 0.25 = λ² - 5λ + 3.75 = 0$$
Solving this quadratic equation for λ gives: $$λ = \frac {[5 ± \sqrt{(5)² - 15}]}{(2\times1)} = \frac {[5 ± \sqrt{(25 - 15)]}} { 2} = \frac{[5 ± \sqrt{10}]}{ 2} = \frac {[5 ± √10]} { 2 }$$
The corresponding frequencies are the square roots of these eigenvalues: $$ω = \sqrt{(λ)} = \sqrt{\frac{5 ± √10}{ 2}}$$
So we have two frequencies, ω₁ and ω₂: $$ω₁ = \sqrt{5 - √10}\times{ 2} = 0.4$$ (approximately, when rounding) $$ω₂ = \sqrt{\frac {[5 + √10]}{ 2} }= 1$$
#### Lagrangian and Hamiltonian Formalism Question 3:
Which of the following terms, when added to the Lagrangian L(x, y, $$\dot x$$, $$\dot y$$) of a system with two degrees of freedom, will not change the equations of motion?
1. $$x\ddot x - y\ddot y$$
2. $$x\ddot y - y\ddot x$$
3. $$x\dot y - y\dot x$$
4. $$y{\dot x^2} + x{\dot y^2}$$
Option 2 : $$x\ddot y - y\ddot x$$
#### Lagrangian and Hamiltonian Formalism Question 3 Detailed Solution
Concept:
The Lagranges equation of motion of a system is given by
$${d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0$$
Calculation:
The Lagrangian L depends on
L(x,y,$$̇ x$$,$$̇ y$$)
$${d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0$$
$${d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0$$
L' = L(x,y,$$\dot x$$,$$\dot{y}$$)
$${d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = {d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0$$
$$\dot{y} = c_1$$
Similarly $$\dot{x} = c_2$$
The correct answer is option (2).
## Top Lagrangian and Hamiltonian Formalism MCQ Objective Questions
#### Lagrangian and Hamiltonian Formalism Question 4
Which of the following terms, when added to the Lagrangian L(x, y, $$\dot x$$, $$\dot y$$) of a system with two degrees of freedom, will not change the equations of motion?
1. $$x\ddot x - y\ddot y$$
2. $$x\ddot y - y\ddot x$$
3. $$x\dot y - y\dot x$$
4. $$y{\dot x^2} + x{\dot y^2}$$
Option 2 : $$x\ddot y - y\ddot x$$
#### Lagrangian and Hamiltonian Formalism Question 4 Detailed Solution
Concept:
The Lagranges equation of motion of a system is given by
$${d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0$$
Calculation:
The Lagrangian L depends on
L(x,y,$$̇ x$$,$$̇ y$$)
$${d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0$$
$${d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0$$
L' = L(x,y,$$\dot x$$,$$\dot{y}$$)
$${d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = {d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0$$
$$\dot{y} = c_1$$
Similarly $$\dot{x} = c_2$$
The correct answer is option (2).
#### Lagrangian and Hamiltonian Formalism Question 5:
Which of the following terms, when added to the Lagrangian L(x, y, $$\dot x$$, $$\dot y$$) of a system with two degrees of freedom, will not change the equations of motion?
1. $$x\ddot x - y\ddot y$$
2. $$x\ddot y - y\ddot x$$
3. $$x\dot y - y\dot x$$
4. $$y{\dot x^2} + x{\dot y^2}$$
Option 2 : $$x\ddot y - y\ddot x$$
#### Lagrangian and Hamiltonian Formalism Question 5 Detailed Solution
Concept:
The Lagranges equation of motion of a system is given by
$${d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0$$
Calculation:
The Lagrangian L depends on
L(x,y,$$̇ x$$,$$̇ y$$)
$${d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0$$
$${d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0$$
L' = L(x,y,$$\dot x$$,$$\dot{y}$$)
$${d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = {d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0$$
$$\dot{y} = c_1$$
Similarly $$\dot{x} = c_2$$
The correct answer is option (2).
#### Lagrangian and Hamiltonian Formalism Question 6:
The Lagrangian of a particle in one dimension is L = $$\frac{m}{2}\dot{x}^2-ax^2-V_0e^{-10x}$$ where a and V0 are positive constants. The best qualitative representation of a trajectory in the phase space is
Option 2 :
#### Lagrangian and Hamiltonian Formalism Question 6 Detailed Solution
Explanation:
Corresponding Hamiltonian is given as :
$$H=\frac{P^2}{2m}+ax^2+V_0e^{−10x}$$
Now, clearly, this is not an elliptical relation between p and x. So, we can eliminate options 3 and 4.
• Potential energy $$V(x)=ax^2+V_0e^{−10x}$$ has its minimum at x>0.
• When we decrease energy, the phase space closed curve shrinks at a potential minimum or we can say stable equilibrium.
• First graph is to shrink at $$x_0<0$$.
• The second graph is to shrink at $$x_0>0$$. Hence, is the answer.
#### Lagrangian and Hamiltonian Formalism Question 7:
The Lagrangian of a system of two particles is L = $$\frac{1}{2}\dot{x}^2_1+2\dot{x}^2_2-\frac{1}{2}(x^2_1+x^2_2+x_1x_2)$$. The normal frequencies are best approximated by
1. 1.2 and 0.7
2. 1.5 and 0.5
3. 1.7 and 0.5
4. 1.0 and 0.4
Option 4 : 1.0 and 0.4
#### Lagrangian and Hamiltonian Formalism Question 7 Detailed Solution
Explanation:
The equations of motion for the system in terms of the normal coordinates are obtained by applying Euler-Lagrange equations for i = 1, and i = 2.
Euler-Lagrange equations are: $$\frac d{dt} (\frac{∂L}{∂ẋᵢ}) - \frac{∂L}{∂xᵢ} = 0$$ for i = 1, 2. So the equations of motion are $$\ddot x₁ - x₁ - 0.5x₂ = 0$$ ...(1) & $$4\ddot x₂ - 0.5x₁ - x₂ = 0$$ ....(2)
Solve the characteristic equation for eigenvalues (which correspond to the squares of the normal mode frequencies), which is in the form of $$|m - λI| = 0$$, where m is the 2x2 matrix of coefficients of terms linear in x, I is the 2x2 identity matrix and λ are the eigenvalues.
This equation gives us a quadratic equation for the eigenvalues: $$(1 - λ)(4 - λ) - (0.5)(0.5) = λ² - 5λ + 4 - 0.25 = λ² - 5λ + 3.75 = 0$$
Solving this quadratic equation for λ gives: $$λ = \frac {[5 ± \sqrt{(5)² - 15}]}{(2\times1)} = \frac {[5 ± \sqrt{(25 - 15)]}} { 2} = \frac{[5 ± \sqrt{10}]}{ 2} = \frac {[5 ± √10]} { 2 }$$
The corresponding frequencies are the square roots of these eigenvalues: $$ω = \sqrt{(λ)} = \sqrt{\frac{5 ± √10}{ 2}}$$
So we have two frequencies, ω₁ and ω₂: $$ω₁ = \sqrt{5 - √10}\times{ 2} = 0.4$$ (approximately, when rounding) $$ω₂ = \sqrt{\frac {[5 + √10]}{ 2} }= 1$$ | 3,142 | 8,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.79876 |
http://list.seqfan.eu/pipermail/seqfan/2005-November/006496.html | 1,716,655,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00244.warc.gz | 19,755,669 | 2,654 | # Generating functions for sum of digits
franktaw at netscape.net franktaw at netscape.net
Fri Nov 4 04:40:02 CET 2005
``` For A053735 Sum of digits of (n written in base 3), the given generating function,
(Sum_{k>=0} x^(3^k)/(1+x^(3^k)))/(1-x),
is wrong. The correct generating function is
(Sum_{k>=0} (x^(3^k)+2*x^(2*3^k))/(1+x^(3^k)+x^(2*3^k)))/(1-x).
In general, the sum of digits of (n written in base b) has generating function
(Sum_{k>=0} (Sum_{0<=i<b} i*x^(i*b^k))/(Sum_{0<=i<b} x^(i*b^k)))/(1-x);
in particular, for b=4, A053737 the generating function is
(Sum_{k>=0} (x^(4^k)+2*x^(2*4^k)+3*x^(3*4^k))/(1+x^(4^k)+x^(2*4^k)+x^(3*4^k))/(1-x).
(There is a typo in the example line of this sequence: it should end with 4, not 3.)
Also, the generating function for the number of digits equal to d in the base b representation of n (0<d<b) is
(Sum_{k>=0} x^(d*b^k)/(Sum_{0<=i<b} x^(i*b^k)))/(1-x);
in particular, for d=1, b=3, A062756 has generating function
(Sum_{k>=0} x^(3^k)/(1+x^(3^k)+x^(2*3^k)))/(1-x).
For d=0, use the above formula with d=b:
(Sum_{k>=0} x^(b^(k+1))/(Sum_{0<=i<b} x^(i*b^k)))/(1-x),
adding 1 if you consider the representation of 0 to have one zero digit. Thus for A077267, the G.F. is
(Sum_{k>=0} x^(3^(k+1))/(1+x^(3^k)+x^(2*3^k)))/(1-x),
and for A081602 it is
1+(Sum_{k>=0} x^(3^(k+1))/(1+x^(3^k)+x^(2*3^k)))/(1-x).
(And these two sequences should have cross-references indicating that they are essentially the same.)
Finally, the sequence that got me into this: A033095 (total number of 1's in all bases) has G.F.
x+(Sum_{b>=2} (Sum_{k>=0} x^(b^k)/(Sum_{0<=i<b} x^(i*b^k)))/(1-x) - x).
If the initial term of A033095 was 0, the initial "x+" would not be needed. This value is rather arbitrary; changing the "n+1" in the definition to "n" would make it 0.
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``` | 814 | 2,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-22 | latest | en | 0.805574 |
https://uglyduckling.nl/blog/software/octave/using-octave-for-asset-price-simulation/ | 1,709,650,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00790.warc.gz | 547,832,086 | 15,213 | # Using Octave For Asset Price Simulation
In this blog we want to demonstrate the power of Octave for doing simulations. Specifically we will take a look at the Black Scholes formula and how fast an option price computed using Monte Carlo simulation will converge to the actual value using the closed-form solution. The idea is to demonstrate how Octave can be used for this kind of simulations. In a previous blog we showed how to created plots, so here will will focus on the simulation only. The mantra when using Octave is "use vectors and matrices". If you can pull that off, your code will be efficient. On the other hand if you need to resort to for-loops, that will slow things down significantly.
# Black Scholes Closed-Form
Fist we need to implement the Black Scholes formula in Octave. This is pretty straight forward if you have been working with either Matlab or Octave for a while. We looked here for inspiration and modified the code to make it more compatible with this experiment.
```function [C, P]= blackScholes(S,X,r,sigma,T)
d1 = (log(S/X) + (r + 0.5*sigma^2)*T)/(sigma*sqrt(T));
d2 = d1 - sigma*sqrt(T);
N1 = 0.5*(1+erf(d1/sqrt(2)));
N2 = 0.5*(1+erf(d2/sqrt(2)));
C = S*N1-X*exp(-r*T)*N2;
N1 = 0.5*(1+erf(-d1/sqrt(2)));
N2 = 0.5*(1+erf(-d2/sqrt(2)));
P = X*exp(-r*T)*N2 - S*N1;
end```
The result from this function will give us an absolute "truth" to verify the accuracy of the simulation result.
# Geometric Brownian Motion
We use the geometric Brownian motion for the simulation of the underlying asset. To get the option price from the simulation one:
• uses the MAX function to evaluate the payoff
• discounts to get present value.
In the code below "W" is a vector of size i * step by one filled with random normal variables. This will result in W being a vector (hence the 1) of random numbers depending on i. The index i will be increased using a for-loop so we can see the convergence for each step.
Computation of the present value is done by discounting using the "exp(-r*T)"-term.
``` W = randn(i * step,1);
S_T = S * exp((mu-sigma*sigma*0.5)*T+sigma*sqrt(T)*W);
C_mc(i) = mean(max(S_T - X, 0))*exp(-r*T);
P_mc(i) = mean(max(X - S_T, 0))*exp(-r*T);```
A complete listing of the code used for filling C_mc and P_mc with call and put values for increasing number of simulations is given in the appendix below. Most of the listing contains the code used to generate the plots presented in the next chapter.
# Results
On our development machine the code runs very fast (couple of seconds). That makes it feasible to really experiment with these kinds of simulations. In practical cases where simulation needs to be done multiple times, for example when calibrating a Monte Carlo model, this is a big advantage of using Octave. As a showcase of the kind of output that we normally get please refer to the two series of 3 plots below for the call and put options. The first plot shows the convergence to the closed-form price as the number of scenarios increases in terms of the option price. The second plot zooms in on the errors as the difference between the theoretical price and the simulated price. Finally, the third plot shows the relative error between theoretical and simulated option prices in basis points as absolute number. The number as absolute so it's easier to see convergence.
# Code Listing
## Main Script
```% option simulation script
clear; close all;
%constants
TOLERENCE = 1e-12;
% C: call price
% P: put price
% S: stock price at time 0
% X: strike price
% r: risk-free interest rate
% sigma: volatility of the stock price measured as annual standard deviation
% days: numbers of days remaining in the option contract, this will be converted into unit of years.
% Black-Scholes formula:
% C = S N(d1) - Xe-rt N(d2)
% P = Xe-rt N(-d2) - S N(-d1)
T = 0.25
r = 0.02;
S = 95
X = 100;
sigma = 1.21;
% Analytical value
[C, P]= blackScholes(S,X,r,sigma,T)
% put-call parity
% https://en.wikipedia.org/wiki/Put%E2%80%93call_parity
"Check put call parity"
(C-P) - (S - X * exp(-r * T)) < TOLERENCE
% simulation random walk
n = 2.5e6;
step = 2.5e5;
mu = r;
x = transpose([step:step:n]);
C_mc =zeros(size(x));
P_mc =zeros(size(x));
for i = 1:n/step
rand('seed', i * n * pi);
W = randn(i * step,1);
S_T = S * exp((mu-sigma*sigma*0.5)*T+sigma*sqrt(T)*W);
C_mc(i) = mean(max(S_T - X, 0))*exp(-r*T);
P_mc(i) = mean(max(X - S_T, 0))*exp(-r*T);
end
scaler = 1000;
h=figure(1);
stairs(x/scaler,[C_mc, repmat(C, size(x))]);
title('Call Convergence');
xlabel('Simulated Paths');
ylabel('Price');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "comvergence call.png");
h=figure(2);
bar(x/scaler,[C_mc - repmat(C, size(x))]);
title('Call Convergence Absolute Error');
xlabel('Simulated Paths (x10000)');
ylabel('Price Error');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "error call.png");
h=figure(3);
bar(x/scaler,[abs(C_mc ./ repmat(C, size(x)) - ones(size(x))) * 10000]);
title('Call Convergence Absolute Relative Error in basis points');
xlabel('Simulated Paths (x1000)');
ylabel('Price Error');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "relative absolute error call.png");
h=figure(4);
stairs(x/scaler,[P_mc, repmat(P, size(x))]);
title('Put Convergence');
xlabel('Simulated Paths');
ylabel('Price');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "comvergence put.png");
h=figure(5);
bar(x/scaler,[P_mc - repmat(P, size(x))]);
title('Put Convergence Absolute Error');
xlabel('Simulated Paths (x1000)');
ylabel('Price Error');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "error put.png");
h=figure(6);
bar(x/scaler,[abs(P_mc ./ repmat(P, size(x)) - ones(size(x))) * 10000]);
title('Put Convergence Absolute Relative Error');
xlabel('Simulated Paths (x1000)');
ylabel('Price Error in basis points');
grid on;
xlim([step/scaler, n/scaler]);
defaultSavePlot(h, "relative absolute error put.png");
```
## BackScholes
```function [C, P]= blackScholes(S,X,r,sigma,T)
d1 = (log(S/X) + (r + 0.5*sigma^2)*T)/(sigma*sqrt(T));
d2 = d1 - sigma*sqrt(T);
N1 = 0.5*(1+erf(d1/sqrt(2)));
N2 = 0.5*(1+erf(d2/sqrt(2)));
C = S*N1-X*exp(-r*T)*N2;
N1 = 0.5*(1+erf(-d1/sqrt(2)));
N2 = 0.5*(1+erf(-d2/sqrt(2)));
P = X*exp(-r*T)*N2 - S*N1;
end```
## Default Save Plot
```function x = defaultSavePlot(plotHandle, name)
W = 8; H = 6;
set(plotHandle,'PaperUnits','inches');
set(plotHandle,'PaperOrientation','portrait');
set(plotHandle,'PaperSize',[H,W]);
set(plotHandle,'PaperPosition',[0,0,W,H]);
print(plotHandle,'-dpng','-color', name);
end```
## One thought on “Using Octave For Asset Price Simulation”
1. Hello,
Great Post. It's very Useful Information. In Future, Hope To See More Post. Thanks You For Sahring. | 1,977 | 6,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | longest | en | 0.834511 |
http://www.ashishsheth.com/ | 1,553,464,986,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203493.88/warc/CC-MAIN-20190324210143-20190324232143-00312.warc.gz | 226,409,086 | 13,398 | Jun 26 2017
Bugs error and software quality
Category: Learning | Podcast ashish sheth @ 18:24
Some notes:
• Bugs are experienced failures, failures comes from faults within the software.
• Faults are introduced in software when some process is skipped during the SDL, such as code review
• Software cannot be tested 100%.
• Test automation just makes software testing faster, it does not improve software quality.
• Test automation should be context driven and adaptive.
• People and managers should put time and money to improve skills of the people involved in software development to improve software quality.
• Early feedback from actual users is important.
• Tools and technology does not improve quality, it is how way use them affects the quality.
Tags: ,
Nov 21 2016
How to generate Random numbers
Category: Algorithm | Maths ashish sheth @ 14:58
If you want to generate a random number for some business logic you are implementing, what would you do?
You would use Random class if you use Java or C#. Most programming language has some library function or class to give you random number.
But suppose you need to produce random numbers by your own without using any library function what would you do?
There are many algorithms to use to produce random number and here I will demonstrate a very basic algorithm which uses the modulo operator (%) in C#. The goal is not to come up with a foolproof algorithm to generate random numbers, the goal is to just use simple math trick to understand how random numbers can be generated. If you really need to generate random numbers in your programs then you should use the inbuilt library functions provided by the language or framework you are using.
You know what is modulo (%) operator is, right? It gives you the remainder when you divide the left hand number by right hand number.
so doing 20 % 20 will give you 0. And 20 % 19 will give you 1 and 20 % 18 will give to 2 and so on.
20 % 20 = 0
20 % 19 = 1
20 % 18 = 2
20 % 17 = 3
20 % 16 = 4
20 % 15 = 5
20 % 14 = 6
20 % 13 = 7
20 % 12 = 8
20 % 11 = 9
20 % 10 = 0
20 % 9 = 2
20 % 8 = 4
20 % 7 = 6
20 % 6 = 2
20 % 5 = 0
20 % 4 = 0
20 % 3 = 2
20 % 2 = 0
20 % 1 = 0
You can see when you divide 20 by numbers from 1 to 20 you get number 0 to 9 as remainders. The trick is the larger the dividend, larger the range of number you get as remainders.
So let’s set dividend d to a some large number, for the purpose of this post I will choose 10000. This will give you the range of 0 to 4999 as remainders.
But as you can see, this method produces sequential numbers, not random numbers. Well, on every iteration you use the remainder to produce a new dividend and you can see that instead of sequential numbers you are getting the random numbers.
So, let’s use below equation to produce a new dividend on every iteration, which uses the current remainder as new dividend:
remainder = a * current remainder + b % divisor
Using the above equation, we get below result, when a = 100, b = 100 and divisor is set to 19:
2100 % 19 = 10
1100 % 19 = 17
1800 % 19 = 14
1500 % 19 = 18
1900 % 19 = 0
100 % 19 = 5
600 % 19 = 11
1200 % 19 = 3
400 % 19 = 1
200 % 19 = 10
1100 % 19 = 17
1800 % 19 = 14
1500 % 19 = 18
1900 % 19 = 0
100 % 19 = 5
600 % 19 = 11
1200 % 19 = 3
400 % 19 = 1
200 % 19 = 10
Notice that the above equation produces a random number on every step but it repeats after few iteration. This is because the chosen values of the a, b and divisor.
Setting the divisor to a really large value can give us random numbers which may not repeat to soon. Such as below:
100 % 12345 = 100
10100 % 12345 = 10100
1010100 % 12345 = 10155
1015600 % 12345 = 3310
331100 % 12345 = 10130
1013100 % 12345 = 810
81100 % 12345 = 7030
703100 % 12345 = 11780
1178100 % 12345 = 5325
532600 % 12345 = 1765
176600 % 12345 = 3770
377100 % 12345 = 6750
675100 % 12345 = 8470
847100 % 12345 = 7640
764100 % 12345 = 11055
1105600 % 12345 = 6895
689600 % 12345 = 10625
1062600 % 12345 = 930
93100 % 12345 = 6685
668600 % 12345 = 1970
Here, a and b is set to 100 and divisor is set to 12345, while current remainder is initialized to 0 at the start. Note that setting divisor to 12345 did not produce repeated numbers in the first 20 iterations, but it can still produce repeated numbers after few hundred iterations. But you get the idea, right?
The equation a + b * currentRemainder % divisor is called Linear Congruential Generator. You can read more about it here.
Tags: ,
Oct 20 2014
Manage your emails faster with Outlook Quick Steps
Category: Productivity ashish sheth @ 22:34
Just noticed this setion on my Outlook toolbar:
When I saw the above in my outlook toolbar I realized that it could help in a lot of productivity gain while managing email.
There are pre-built steps which you can use to manage email as mentioned in the above image. You can add your own quick steps too. Notice that “Archive” on the top-left. I setup that by myself as follows:
Click on the “down arrow” on the left-bottom part (highlighted in blue) of the above screen. A “Manage Quick Steps” dialog box will be opened as below:
Clicking on the “New” button will open up a dialog box as below, where you can setup the action that you want to take. Note that you can also setup a shortcut key which you can use for the action instead of every time selecting the action from the menu.
Tags: , ,
Sep 30 2014
Category: Quotes ashish sheth @ 20:28
Obeying best practices blindly is not a best practice.
Tags:
Sep 18 2014
Category: .Net | C# | Design Principles ashish sheth @ 00:13
Frequently you will see following type of code in your code base.
```if(result != 1)
{
if(someOtherResult == 101)
{
if(anotherValue == 500)
{
// do something
}
}
else
{
// do some other thing
}
}
return;```
Here the code forms a shape of an arrow-head, as below:
```if
if
if
if
do something
end
end
end
end```
If you see, the main logic is deep down into the nested condition, it increases the cyclomatic complexity of the code.
A better version of the same code could be as below:
```if(result == 1)
{
return;
}
if(someOtherResult == 101 && anotherValue == 500)
{
// do something
return;
}
// do some other thing
return;```
The above code does a number of things to flatten the code and make it better:
1. Validations are performed first and it returns at the first opportunity.
2. Multiple nested conditions are combine into one (with “&&”(logical And) operator. If there are multiple expressions forming one such condition, they can be moved to a separate method returning boolean. That method can then be use in the if condition as below:
```if(IsValidResult(someOtherResult, anotherResult)
{
// do something
return;
}
bool IsValidResult(int someOtherResult, int anotherResult)
{
if(someOtherResult == 101 && anotherValue == 500)
{
return true;
}
return false;
}```
Sep 16 2014
Podcast that I listen
Category: .Net | Learning | miscellaneous ashish sheth @ 00:45
In the big sea of web, you will find many people telling you the podcast that every developers should listen. Here is my list, and yes, I too think these are the podcast every developer(.Net developer) should listen.
1. .Net Rocks: This is more than a decade old talk show focusing on Microsoft .Net Technologies. Earlier it used to be a weekly show, but now it has three show a week. Apart from covering anything and everything related to .Net, it also covers mobile app development on Android and iOS.
2. Hanselminutes: This is the show by Scott Hanselman, which focuses on technology including Microsoft .Net Technology.
3. HerdingCode: This is a show where discussion goes into nitty-gritty of development tasks. As with the above two, not just the .Net technology, but other aspects of Software development practises are also discussed.
4. Software Engineering Radio: This is a monthly podcast targeted towards not just .Net developers, but to all software developers in general. So it covers topics which include Software Engineering practises, architecture, design patterns, etc. It also has a rather academic feel and many a times the guests include leading researchers in the field of Computer Science.
5. MS Dev show: This is relatively a new podcast for Microsoft developers focussing fully to Microsoft technology including topics such as Azure/Cloud, Windows, Windows Phone, .Net etc.
6. IEEE Software’s On Computing: In this podcast Grady Booch reads from his column “On Computing” in the IEEE Software magazine. In this, he talks software’s impact on humanity.
7. IEEE Software’s Tools of the Trade: This is again the podcast of an on going column in IEEE Software with the same name. In every episode, it covers software tools to be used within a certain aspect of Software Development, such as documentation, debugging, testing, etc.
8. HBR IdeaCast: This is a weekly podcast from HBR.org and it is mostly about analysis and advice from leading minds in management.
Let me know what you listen to.
Tags: ,
Oct 17 2013
Domain Specific Languages - MindMap
Category: ashish sheth @ 06:07
Here is the MindMap that I prepared for Domain Specific Languages after listening to the podcast on the same topic by Martin Fowler on SE Radio.
Jun 26 2013
History of Computing
Category: miscellaneous ashish sheth @ 23:34
An interesting infographic on history of computing.
Tags:
Jun 17 2011
.Net Guidance Map
Category: .Net | .Net Framework | asp.net ashish sheth @ 18:05
I thought these would be very helpful to those who are learning .Net. J. D. Meier has put up Development Guidance Map for various .Net technologies. This material includes links to technical articles, videos, how-tos, blogs, tutorials, trainings, code samples on various .Net technologies and about everything you might need to know about .net.
Happy Learning.
Tags: ,
Sep 16 2010
Use Properties to Encapsulate the Hidden Fields or ViewState in Asp.Net
Category: .Net | asp.net | C# | VS2008 ashish sheth @ 05:16
In asp.net you might be using lot of hidden variables and ViewState to maintain state between page postbacks. This can make your code look cluttered. You can use properties to encapsulate the hidden fields or ViewState.
If you are writing lot of code like this:
```if(myHidden.Value == string.Empty)
{
myHidden.Value = "someValue";
}
someVariable = myHidden.Value
```
You can use properties to encapsulate the access to myHidden field. For example:
```public string MyHiddenFieldValue
{
get
{
if(myHidden.Value == string.Empty)
{
myHidden.Value = "someValue";
}
return myHidden.Value
}
set
{
myHidden.Value = value;
}
}
```
Then you can access the myHidden field just by the property
```someVariable = MyHiddenFieldValue;
```
Similarly if you are storing some custom values in the ViewState of the page the you can create property for the ViewState also.
```public string MyCustomViewState
{
get
{
if(ViewState["MyViewState"] == null)
{
ViewState["MyViewState"] = "someValue";
}
return ViewState["MyViewState"].ToString();
}
set
{
ViewState["MyViewState"] = value;
}
}
```
Tags: , , , | 2,871 | 11,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-13 | latest | en | 0.916926 |
https://www.beatthegmat.com/problem-solving-f6.html?sid=a52546e2669b3be1170535e100e809e8 | 1,670,127,256,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00629.warc.gz | 698,738,619 | 288,915 | Problem Solving
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Sun Sep 25, 2022 10:28 pm | 3,665 | 12,365 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-49 | latest | en | 0.903969 |
http://math.stackexchange.com/questions/91553/is-the-cross-product-of-two-displacement-vectors-orthogonal-to-both-of-them | 1,469,343,245,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823963.50/warc/CC-MAIN-20160723071023-00045-ip-10-185-27-174.ec2.internal.warc.gz | 156,712,617 | 18,524 | # Is the cross-product of two displacement vectors orthogonal to both of them?
I have the point P,Q and R given. I calculate the displacement vectors PQ and PR. If I then compute their cross product I get a vector orthogonal to the plane they're in. But the value of the cross product vector is not orthogonal to PQ and PR (it is not 0). Shouldn't it be 0?
Thank you
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Let $u = PQ$, $v = PR$, and $w = u \times v$. Then $w$ is orthogonal to $u$ and $v$ (and normal to the plane which contains the points P, Q, and R). So the dot product of $w$ with both $u$ and $v$ should be zero. If it isn't, you've made a mistake. – Bill Cook Dec 14 '11 at 20:41
The cross product of two vectors is always orthogonal to both vectors. In three dimensions, a vector does not have to be zero to be orthogonal to two other vectors; as you said, the result is orthogonal to the plane that $P$, $Q$, and $R$ are in, so it is also orthogonal to both $PQ$ and $PR$.
Here's a slightly more technical discussion.
In general, a vector is only necessarily zero if it is orthogonal to everything. The dimension of a subspace and the dimension of its 'orthogonal complement' (which just means everything that is orthogonal to everything in the subspace) must add to the dimension of the whole space. For example, in 3 dimensional space, two vectors define a 2 dimensional subspace (as long as they're linearly independent, but you don't have to worry about what that means if you don't know). So everything orthogonal to that whole 2 dimensional subspace, or its orthogonal complement, is a 1 dimensional subspace.
This also explains why the cross product can only be used in 3 dimensional space. Two vectors only determine a 1 dimensional orthogonal complement, and therefore a unique cross product, if the original space is three dimensional. It turns out there isn't a really useful generalization of this to higher dimensions.
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it is only in $R^2$ 0 when it's orthogonal? How can I know then in $R^3$? – Andrew Dec 14 '11 at 20:50
@Andrew does this expanded discussion answer your question? – smackcrane Dec 15 '11 at 20:58
yes it does, thank you – Andrew Dec 18 '11 at 9:32
The Cross Product $v \times w$ is always orthogonal to both $v$ and $w$. This is easy to see by a direct calculation:
Write $v = \langle v_1, v_2 , v_3\rangle$ and $w = \langle w_1, w_2, w_3\rangle$. Then
$$v \times w = \langle v_2 w_3 - v_3 w_2, v_3w_1 - v_1w_3, v_1w_2 - v_2 w_1 \rangle$$
which implies
\begin{align} v \cdot (v \times w) &= v_1 (v_2 w_3 - v_3 w_2) + v_2 (v_3 w_1 - v_1 w_3) + v_3 (v_1 w_2 - v_2 w_1) \\ &= v_1 v_2 w_3 - v_1 v_3 w_2 + v_2 v_3 w_1 - v_1 v_2 w_3 + v_1 v_3 w_2 - v_2 v_3 w_1 = 0. \end{align} The same calculation works to show $w \cdot (v \times w) = 0$.
- | 837 | 2,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-30 | latest | en | 0.944158 |
https://stacks.math.columbia.edu/tag/06M2 | 1,685,719,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00754.warc.gz | 584,752,673 | 6,123 | Lemma 99.3.5. Let $P$ be a property of morphisms of algebraic spaces as above. Let $\tau \in \{ {\acute{e}tale}, smooth, syntomic, fppf\}$. Let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks representable by algebraic spaces. Assume
1. $\mathcal{X} \to \mathcal{Y}$ is surjective and étale, smooth, syntomic, or flat and locally of finite presentation,
2. the composition has $P$, and
3. $P$ is local on the source in the $\tau$ topology.
Then $\mathcal{Y} \to \mathcal{Z}$ has property $P$.
Proof. Let $Z$ be a scheme and let $Z \to \mathcal{Z}$ be a morphism. Set $X = \mathcal{X} \times _\mathcal {Z} Z$, $Y = \mathcal{Y} \times _\mathcal {Z} Z$. By (1) $\{ X \to Y\}$ is a $\tau$ covering of algebraic spaces and by (2) $X \to Z$ has property $P$. By (3) this implies that $Y \to Z$ has property $P$ and we win. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 356 | 1,077 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.729964 |
https://www.gamedev.net/forums/topic/606804-guessing-and-finding-point-coordinates-with-screenshot/ | 1,527,040,752,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865023.41/warc/CC-MAIN-20180523004548-20180523024548-00633.warc.gz | 748,259,550 | 21,321 | # Guessing and finding point coordinates (with screenshot)
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Hi, given a triangle with known P1,P2,P3 coordinates, I'd like to be able to guess n points that lie inside of it, and their coordinates. See attached image for a better explanation.
Thanks
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What do you mean by guess? Do you want to generate n random points inside the triangle? Do you want a uniform distribution?
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Not a uniform distribution: just say you want to guess 100 points in sparse order which lie inside the triangle...
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You can use barycentric coordinates like in the following pseudocode:
[source]a = rand(0,1)
b = rand(0,1 - a)
c = 1 - a - b
P = a*P1 + b*P2 + c*P3[/source]
EDIT: Corrected a typo
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You can use barycentric coordinates like in the following pseudocode:
[source]a = rand(0,1)
b = rand(0,1 - a)
c = 1 - a - c
P = a*P1 + b*P2 + c*P3[/source]
Thank you very much, I'm going to try this !
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• 14 | 327 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-22 | latest | en | 0.878971 |
https://cheatsheeting.com/show.html?sheet=lbp-to-dop-conversions | 1,610,954,044,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00173.warc.gz | 280,730,639 | 10,647 | Home > Conversions (Currency) > Conversion tables from/to Lebanese Pound > LBP to DOP Conversion Cheat Sheet (Interactive)
From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 You could also enter the values to convert and print directly on the table
[Formula: DOP = LBP x 0.0384785619504] [Printer friendly] [Dominican Pesos to Lebanese Pounds]
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# Lebanese Pounds to Dominican PesosExchange Rate and Conversion Table
LBP = 0.0384785619504 DOP rate as of 2021-01-18 05:00:00 GMT (Source)
# How to convert from Lebanese Pounds to Dominican Pesos
Since 1 Lebanese Pound is equal to 0.0384785619504 Dominican Pesos, we could say that n Lebanese Pounds are equal to 0.0384785619504 times n Dominican Pesos. In other words, we could use the following formula:
Dominican Pesos = Lebanese Pounds x 0.0384785619504
For example, let's say that we want to convert 2 Lebanese Pounds to Dominican Pesos. Then, we just replace Lebanese Pounds in the abovementioned formula with 2:
Dominican Pesos = 2 x 0.0384785619504
That is, 2 Lebanese Pounds are equal to 0.0769571239008 Dominican Pesos. | 666 | 1,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | latest | en | 0.262507 |
https://www.bartleby.com/questions-and-answers/chapter-7-section-7.2-question-058-a-travel-mug-of-90c-coffee-is-left-on-the-roof-of-a-parked-car-on/4922bb0d-ca17-47aa-87a3-8df8846eef69 | 1,579,801,833,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00277.warc.gz | 773,652,559 | 22,055 | # Chapter 7, Section 7.2, Question 058A travel mug of 90°C coffee is left on the roof of a parked car on a 0 C winter day. The temperature of the coffee after t minutes isgiven by H 90(0.6)5. When will the coffee be only lukewarm (30C)?Round your answer to the nearest integerThe coffee will be only lukewarm after aboutminutes.the absolute tolerance is +/-1
Question
Asked Nov 25, 2019
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https://math.stackexchange.com/questions/tagged/positive-semidefinite?sort=frequent&pageSize=50 | 1,555,641,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526966.26/warc/CC-MAIN-20190419021416-20190419043416-00177.warc.gz | 480,968,624 | 27,706 | # Questions tagged [positive-semidefinite]
Relating to a symmetric $n\times n$ real matrix $(M)$ such that the scalar $x^TMx\ge 0\ \forall x\in \Bbb{R}^n\backslash \{0\}$
31 questions
48k views
### Is the product of symmetric positive semidefinite matrices positive definite?
I see on Wikipedia that the product of two commuting symmetric positive definite matrices is also positive definite. Does the same result hold for the product of two positive semidefinite matrices? ...
30k views
### Prove that every positive semidefinite matrix has nonnegative eigenvalues
There is a theorem which states that every positive semidefinite matrix only has eigenvalues $\ge0$ How can I prove this theorem?
898 views
193 views
### Definiteness of a general partitioned matrix $\mathbf M=\left[\begin{matrix}\bf A & \bf B\\\bf B^\top & \bf D \\\end{matrix}\right]$
If $\mathbf M=\left[\begin{matrix}\bf A & \bf b\\\bf b^\top & \bf d \\\end{matrix}\right]$ such that $\bf A$ is positive definite, under what conditions is $\bf M$ positive definite, positive ...
1k views
### Positive semidefinite cone is generated by all rank-$1$ matrices.
The positive semidefinite cone is generated by all rank-$1$ matrices $xx^T$, which form the extreme rays of the cone. Positive definite matrices lie in the interior of the cone. Positive semidefinite ...
3k views
### The product of two symmetric, positive semidefinite matrices has non-negative eigenvalues
How can I prove the following? If $A$ and $B$ are two symmetric, positive semidefinite matrices then all eigenvalues of $AB$ are non-negative.
16k views
### Checking if a matrix is positive semidefinite
Determine whether the following $2 \times 2$ matrix is positive semidefinite (PSD) $$\begin{bmatrix}\frac{2}{x} & \frac{-2y}{x^2} \\\frac{-2y}{x^2} & \frac{2y^2}{x^3}\end{bmatrix}$$ ...
1k views
### Are positive definite matrices robust to “small changes”?
Let $A$ be a positive-definite matrix and let $B$ be some other symmetric matrix. Consider the matrix $$C=A+\varepsilon B.$$ for some $\varepsilon>0$. Is it true that for $\varepsilon$ small ...
5k views | 564 | 2,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-18 | latest | en | 0.75597 |
http://icodeguru.com/vc/10book/books/book10/9803o/9803o.htm | 1,716,558,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00867.warc.gz | 13,846,760 | 7,037 | # Exponential Smoothing
Dr. Dobb's Journal March 1998
### By William Stallings
William is the author of High-Speed Networks: TCP/IP and ATM Design Principles (Prentice Hall, 1998). He can be reached at http://www.shore.net/~ws/ or ws@shore.net.
Whenever your software interfaces with hardware or other software, it's hard to know what to expect. How much memory will be required? How long will it take? The most robust systems measure past performance, then use that to predict future response. In this way, your software can smoothly adapt to varying situations.
In "Improving Kermit Performance" (DDJ, February 1996; available electronically, see "Resource Center," page 3), I used a "decaying average" to predict future delays, then used those predictions to optimize error handling. This month, William explores this technique -- also known as "exponential smoothing" -- in more detail and explains a variety of uses.
-- Tim Kientzle
You often need to guess the next value in a time series based on previous values in the time series. For example, the nonpreemptive scheduling policy known as "shortest process next" (SPN) selects the process with the shortest expected processing time (before it is blocked by an I/O or system call) to run next. For this purpose, the operating system keeps a running average of all of the previous processing bursts for each process, and uses these averages to estimate the next bursts. Running averages for estimating future values also show up in many areas of communications protocol design.
### Smoothing Techniques
Suppose you have a series of measured values V(1), V(2), V(3),..., where V(i) is the value observed at the ith observation. Assume you want to predict (estimate) the value V(K+1) based on the value up through V(K). One approach would be simply to take the average of observed values, as in Figure 1(a), where AV(K) is the average of the first K values. The estimate for the next value in the series, V(K+1), is equal to AV(K).
Figure 1(b) shows the expression reformulated so that it is not necessary to recalculate the entire summation each time.
Each term in the summation is given equal weight; that is, each term is multiplied by the same constant 1/K. Typically, you would like to give greater weight to more recent instances because they are more likely to reflect future behavior. A common technique for predicting the next value on the basis of a series of past values is known as "exponential averaging," or "exponential smoothing." Figure 2(a) shows how you compute SV(K), the "smoothed estimate."
Compare Figure 2(a) with Figure 1(b). By using a constant value of (0<<1), independent of the number of past observations, you have a circumstance in which all past values are considered, but more distant ones have less weight. To see this more clearly, consider Figure 2(b). Since both and (1-) are less than 1, each successive term in the preceding equation is smaller. For example, for a=0.8, the expansion is Figure 2(c). The older the observation, the less it is counted in the average.
Figure 3 shows the size of the coefficient as a function of its position in the expansion. The smaller the value of , the greater the weight given to the more-recent observations. For a=0.5, virtually all of the weight is given to the four or five most recent observations, whereas for =0.875, the averaging is effectively spread out over the ten or so most recent observations. The advantage of using a small value of is that the estimate will quickly reflect a rapid change in the observed quantity. The disadvantage is that brief surges in the value of the observed quantity will result in jerky changes in the smoothed average.
Figure 4 compares simple averaging with exponential averaging (for two different values of ). In Figure 4(a), the observed value begins at 1, grows gradually to a value of 10, and then stays there. In Figure 4(b), the observed value begins at 20, declines gradually to 10, and then stays there. In both cases, you start out with an estimate of SV(1)=0. Exponential averaging tracks changes in process behavior faster than does simple averaging and the smaller value of a results in a more rapid reaction to the change in the observed value.
### TCP Retransmission Timer
In the Transmission Control Protocol (TCP), two TCP entities exchange segments over a TCP connection. Each side retains a copy of each segment that it sends and, if an acknowledgment is not received within a retransmission timeout (RTO) interval, resends the segment. Thus, if a segment is lost in transit, it will automatically be resent. If RTO is set too small, then a TCP entity may retransmit a segment for which an acknowledgment is on its way but delayed because of network congestion; such unnecessary retransmissions actually worsen the congestion. On the other hand, if RTO is too large, the protocol will be sluggish in responding to a lost segment.
Initially, TCP used a smoothed average to estimate the round-trip time. However, this didn't work well if the round-trip time varied significantly. To cope with this, Van Jacobson proposed a refinement in the standard algorithm that has now been officially adopted for TCP. Figure 5 summarizes this algorithm.
The Van Jacobson algorithm provides an improvement by estimating both the round-trip time (RTT) and the standard deviation of the RTT. The RTO value is then set to the estimated value of RTT plus a constant multiplied by the estimated standard deviation. This enables the use of more reasonable values of the retransmission timer. Standard implementations of TCP use the parameters g=1/8=0.125, h=1/4=0.25, and f=4. Experience has shown that Van Jacobson's algorithm can significantly improve TCP performance.
### ATM ABR Service
Support for bursty data traffic on ATM networks, such as traffic generated by TCP/IP-based applications, has traditionally been carried on the so-called Unspecified Bit Rate (UBR) service. UBR is designed to make use of available capacity on the ATM network not consumed by time-sensitive traffic such as voice and video. All of this unused capacity could be made available for the UBR service. This service is suitable for applications that can tolerate variable delays and some cell losses, which is typically true of TCP-based traffic. With UBR, cells are forwarded on a first-in-first-out (FIFO) basis using the capacity not consumed by other services; both delays and variable losses are possible. No initial commitment is made to a UBR source and no feedback concerning congestion is provided; this is referred to as a "best-effort service."
To improve the service provided to bursty sources that would otherwise use UBR, the Available Bit Rate (ABR) service has been defined. An application using ABR specifies a peak cell rate (PCR) that it will use and a minimum cell rate (MCR) that it requires. The network allocates resources so that all ABR applications receive at least their MCR capacity. Any unused capacity is then shared in a fair and controlled fashion among all ABR sources. The ABR mechanism uses explicit feedback to sources to assure that capacity is fairly allocated. Any capacity not used by ABR sources remains available for UBR traffic.
To determine how much capacity to allow to each connection, each ATM switch must monitor the traffic through itself. One of the techniques recommended by the ATM Forum for this purpose uses the equation MACR(I)=(1-)×MACR(I-1) +×CCR(I) for each virtual connection through the switch, where CCR is a measure of the current cell rate on a given connection, and MACR (mean allowed cell rate) is an estimate of the cell rate for the connection in the next sampling period. Typically, =1/16, so that more weight is given to past values of CCR than to the current value. Based on the estimates for each connection, if a switch experiences congestion, it restricts the cell rate of the connections passing through it in a manner that is proportional to the estimated demand from each connection.
### Real-Time Transport Protocol (RTP)
Real-time transport protocol (RTP) is designed to support both point-to-point and multicast real-time applications. RTP overcomes three deficiencies in TCP for real-time applications:
1. TCP is a point-to-point protocol that sets up a connection between two end points. Therefore, it is not suitable for multicast distribution.
2. TCP includes mechanisms for retransmission of lost segments, which then arrive out of order. Such segments are not usable in most real-time applications.
3. TCP contains no convenient mechanism for associating timing information with segments, which is another real-time requirement.
One key requirement for receivers of real-time traffic is to estimate the amount of delay variation, or jitter, experienced on a connection in order to determine buffering requirements. In essence, the receiver would like to pass incoming traffic through a delay buffer that will smooth out delay variability so that the received data has the same timing characteristics as the transmitted data. There is no simple way to measure this quantity at the receiver, but it is possible to estimate the average jitter in the following way. At a particular receiver, you define the following parameters for a given source:
• S(I) = Timestamp from RTP data packet I.
• R(I) = Time of arrival for RTP data packet I, expressed in RTP timestamp units. The receiver must use the same clock frequency (increment interval) as the source, but need not synchronize time values with the source.
• D(I) = The difference between the interarrival time at the receiver and the spacing between adjacent RTP data packets leaving the source.
• J(I) = Estimated average interarrival jitter up to the receipt of RTP data packet I.
The value of D(I) is calculated as D(I)=(R(I)-R(I-1))-(S(I)-S(I-1)). Thus, D(I) measures how much the spacing between arriving packets differs from the spacing between transmitted packets. In the absence of jitter, the spacings will be the same and D(I) will have a value of 0. The interarrival jitter J(I) is calculated continuously as each data packet I is received, according to the formula in Figure 6. J(I) is calculated as an exponential average of observed values of D(I). Only a small weight is given to the most recent observation, so that temporary fluctuations do not invalidate the estimate. The estimated value of jitter is reported to other participants in the connection. The jitter measure may provide a warning of increasing congestion before it leads to packet loss.
### Internet Congestion Control
When routers in a network become congested to the point of buffer saturation, it becomes necessary to discard incoming or buffered packets. It may be desirable for routers to begin to discard packets before total saturation occurs. In this way, TCP connections receive early warning of increased congestion as the result of a few lost packets. The TCP connections can then back off the volume that they generate to avoid more catastrophic congestion. The most important example of a proactive packet discard scheme is Random Early Detection (RED), introduced by Sally Floyd. RED has been implemented by a number of router vendors.
In essence, RED monitors the queue length for an output port on the router. Associated with each output buffer are two thresholds THmax and THmin. When the queue length is less than THmin, no discard occurs; when it is greater than THmax, any arriving packet is discarded. Between these two thresholds, an incoming packet has an increasing probability of being discarded as the queue length grows from the lower to the higher threshold. The use of two thresholds and a sliding probability scale affords the opportunity to fine-tune the algorithm, and experience has shown that it can be effective in anticipating congestion without wasteful discard.
However, rather than use the actual queue length for the calculation, RED uses a smoothed average queue length SQ(K)=(1-W)×SQ(K-1)+W×Q(K), where Q is the actual queue length, SQ is the smoothed average, and W is a weighting factor.
You might wonder why an average queue size is used when it would be simpler to use the actual queue size. The purpose of using an average queue size is to filter out transient congestion at the router. The weight W determines how rapidly SQ changes in response to changes in actual queue size. Floyd recommends a quite small value of 0.002. As a result, SQ lags considerably behind changes in actual queue size. The use of this small weight prevents the algorithm from reacting to short bursts of congestion.
### Summary
Exponential averaging is used in a surprisingly wide variety of protocols and congestion-control algorithms. By varying the weighting factor used in the average, more or less weight can be given to recent observations compared to more distant ones. Anyone needing a simple technique for making estimates based on past observations should be aware of this versatile tool.
### References
The algorithm used by TCP to adjust its retransmission timeout was described by Van Jacobson in "Congestion Avoidance and Control." Proceedings, SIGCOMM '88, Computer Communication Review, August 1988. The article was reprinted in Computer Communication Review, January 1995. A revised version is available at ftp.ee.lbl.gov/papers/congavoid.ps.Z.
The RED technique was described by Sally Floyd and Van Jacobson in "Random Early Detection Gateways for Congestion Avoidance," IEEE/ACM Transactions on Networking, August 1993.
DDJ | 2,840 | 13,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.911087 |
https://mathematica.stackexchange.com/questions/8977/automatically-return-output-using-decimals-instead-of-fractions | 1,723,628,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00106.warc.gz | 300,890,570 | 41,671 | # Automatically return output using decimals instead of fractions
I have a mathematica file with an algorithm that feeds another program with data. my problem is that some times (randomly) the output is in fraction instead of decimal. The other program (java) does not read fractions, so I have to feed it with decimal numbers. One solution is to use the N[.] function before it output, but this is really painful procedure. Is there any setting in mathematica forcing it to return as output only decimals?
• "...but this is really painful procedure..." - why? Commented Aug 1, 2012 at 13:28
• By the way, welcome to Mathematica.SE! Please consider registering your account, so that any votes you might get for this question are added to those received for future questions. As you gain reputation points, you will be able to do more things like participate in the chat room. Commented Aug 1, 2012 at 13:42
• I have several algorithms, pretty big ones, and I was seeking for a general way to overcome my problem before getting inside the code and start messing with it. Thanx for the hint, i will do register. Commented Aug 1, 2012 at 13:54
• You only need to use N at the very last stage before sending the output to Java. Commented Aug 1, 2012 at 14:06
• Take @Verbeia 's advice on using real inputs seriously, specially if you have a long heavy weight algorithm. The difference in space and time it can make is enormous, and only generated by inputing, e.g, 2 instead of 2. or 2
– Rojo
Commented Aug 1, 2012 at 19:02
1. Make sure all your input is in floating point (Real) form, not Integers or Rational. You might find this tutorial helpful. I am willing to bet that the cases where you get a fraction are not really random.
2. If wrapping your output in N is unacceptable to you for some reason, you could instead use a rule once you have your output: data /. x_Rational :> N@x. This will almost certainly be slower than N.
For example:
{1/2, 256/225, 0.6} /. x_Rational :> N@x
(* output is {0.5, 1.13778, 0.6}*)
N is Listable, so the following also works:
N[{1/2, 256/225, 0.6}]
• first of all thank you for your reply! I will check the tutorial u sent me. I was overexaggerating when i said randomly, but frankly, i give an initial input of 5 variables in my algorithm, and i get hundreds of outputs, and the percentage of getting fractions instead of decimal numbers is not greater than 1% based on my hundreds and hundreds of tests. nevertheless, once i get fraction, the second program is useless (java). The second solution you suggest seems to be similar to the N[.] function, with respect to the editing i have to do in my code. i was wondering if there was a setting hidden Commented Aug 1, 2012 at 13:59
• As long as 1 value is MachinePrecision the output will be that too, isn't it? Commented Aug 1, 2012 at 14:53
You can use $Post: If you set $Post to N, N is applied to the output of every evaluation:
Cos[Pi/4]
$Post = N; Cos[Pi/4] (* 0.707107 *) To restore $Post to its default value, use
\$Post =.
` | 793 | 3,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.918265 |
https://resourcecenter.byupathway.org/math/m01-11 | 1,685,658,322,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00212.warc.gz | 543,887,598 | 30,410 | Back
Percentages in Excel
> ... Math > Excel > Percentages in Excel
Percentages in Excel
There are many ways to format a cell as a percentage in Excel. This lesson goes over the following five ways.
• Number and % symbol
• Format Cell
• % Style Button
• Format Several Cells At Once
## Number and % symbol
Type a number into the cell.
At the end of the number, type the % symbol.
Press Enter.
This will automatically tell Excel that the number you typed is a percentage, and it will format the cell as a percentage cell. Any number you put in the cell thereafter will appear with a % symbol.
## Format Cell
Right click on the cell you want to display as a percentage.
A menu will appear. Select the Format Cell item on the list.
This will bring up the Format Cell window.
Under the Numbers tab, you will see a category list of all the number formats you can assign to a cell. Select the Percentage category.
Notice that you can change the number of decimal places displayed in the Decimal places box. You can use this box to adjust the decimal places no matter what approach you originally used to format the cell as a percentage.
In the Home Toolbar, you will find the number format menu box. The default in the box says General.
Select the arrow to the right of the box to view the other options. Select the Percentage item on the list.
This will automatically format the cell as a percentage with two places after the decimal point. You can adjust the places after the decimal point using the format cell technique explained above.
## % Style Button
Select the cell you want to format by clicking within it.
Select the Percent Style button in the Home Toolbar. It is a button with a % symbol on it.
This will format the cell to display as a percentage.
## Format Several Cells At Once
Highlight all the cells you want to format as a percentage.
Example image:
Once the cells are selected, use either the Format Cell, Number Format Menu, or Percent Style button technique to format the cells as percentages.
## Need More Help?
1. Study other Math Lessons in the Resource Center.
2. Visit the Online Tutoring Resources in the Resource Center. | 454 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.823902 |
http://pinboard.in/t:lifts-projections/ | 1,553,584,210,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204857.82/warc/CC-MAIN-20190326054828-20190326080828-00116.warc.gz | 164,294,566 | 11,629 | lifts-projections 14
Section 10 Chi-squared goodness-of-fit test.
- pf that chi-squared statistic for Pearson's test (multinomial goodness-of-fit) actually has chi-squared distribution asymptotically
- the gotcha: terms Z_j in sum aren't independent
- solution:
- compute the covariance matrix of the terms to be E[Z_iZ_j] = -sqrt(p_ip_j)
- note that an equivalent way of sampling the Z_j is to take a random standard Gaussian and project onto the plane orthogonal to (sqrt(p_1), sqrt(p_2), ..., sqrt(p_r))
- that is equivalent to just sampling a Gaussian w/ 1 less dimension (hence df=r-1)
QED
pdf nibble lecture-notes mit stats hypothesis-testing acm probability methodology proofs iidness distribution limits identity direction lifts-projections
october 2017 by nhaliday
Covering space - Wikipedia
A covering space of X is a topological space C together with a continuous surjective map p: C -> X such that for every x ∈ X, there exists an open neighborhood U of x, such that p^−1(U) (the inverse image of U under p) is a union of disjoint open sets in C, each of which is mapped homeomorphically onto U by p.
concept math topology arrows lifts-projections wiki reference fiber math.AT nibble preimage
january 2017 by nhaliday
Answer to What is it like to understand advanced mathematics? - Quora
thinking like a mathematician
some of the points:
- small # of tricks (echoes Rota)
- web of concepts and modularization (zooming out) allow quick reasoning
- comfort w/ ambiguity and lack of understanding, study high-dimensional objects via projections
- above is essential for research (and often what distinguishes research mathematicians from people who were good at math, or majored in math)
math reflection thinking intuition expert synthesis wormholes insight q-n-a 🎓 metabuch tricks scholar problem-solving aphorism instinct heuristic lens qra soft-question curiosity meta:math ground-up cartoons analytical-holistic lifts-projections hi-order-bits scholar-pack nibble giants the-trenches innovation novelty zooming tricki virtu humility metameta wisdom abstraction skeleton s:*** knowledge expert-experience
may 2016 by nhaliday
Copy this bookmark:
description:
tags: | 569 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | latest | en | 0.730768 |
https://www.mrexcel.com/board/threads/compounded-rate-of-return-on-land.275873/ | 1,679,888,719,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00613.warc.gz | 1,002,170,118 | 16,645 | # Compounded rate of return on land?
#### gidsy
##### Board Regular
Hi There,
I am trying to work out the best way to return the Compounded rate of return for the increase in land value.
So for a property purchased for 64,000 35 years ago is now worth 1,900,000. That's an increase of 52,457 pa. What is the Compounded rate of return?
There must be a function, I just can't find it.
### Excel Facts
If Excel says you have links but you can't find them, go to Formulas, Name Manager. Look for old links to dead workbooks & delete.
#### acw
##### MrExcel MVP
Hi
Compound Annual Growth Rate (CAGR)
(Vtn/Vt0)^(1/tn)-1
Where Vtn is the value at time n (1900000)
Vt0 is the starting value (64000)
tn is the number of years (35)
I get 10.17%
HTH
Tony
#### gidsy
##### Board Regular
Thanks for that....Is that a standard function...I do not have that one...I'm on version 2002
#### acw
##### MrExcel MVP
Hi
I just gave you the formula required to calculate from the data given. If you look up compound annual growth rate in Excel help, it will give you the XIRR function, but I don't think it will work for you example data.
Tony
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Go back | 643 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.871224 |
http://forum.allaboutcircuits.com/threads/full-bridge-rectifier-transfer-function.60849/ | 1,484,763,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280310.48/warc/CC-MAIN-20170116095120-00328-ip-10-171-10-70.ec2.internal.warc.gz | 108,473,368 | 16,047 | # full bridge rectifier transfer function
Discussion in 'Homework Help' started by suzuki, Oct 18, 2011.
1. ### suzuki Thread Starter Member
Aug 10, 2011
119
0
Hi,
having some difficulty putting this altogether so i thought i would ask here.
i want to find the transfer function of a full wave rectifier (four diodes in bridge configuration). Mathematically speaking, i believe this is the equivalent of passing a signal through an absolute value, so i tried to model it as such in simulation.
However, since the absolute value is non-linear, i'm not sure if i can use the frequency domain, bode plots etc for the purpose of analysis. So my question is, what other ways can i model my full bridge rectifier?
tia
2. ### steveb Senior Member
Jul 3, 2008
2,433
469
This is a tricky subject.
Are you trying to find the transfer function of input voltage to load current to determine the power factor or loading conditions that a rectifier puts on an AC source? Or, are you interested in the transfer function from input voltage to output voltage?
Either way, it is nonlinear, as you said. You are asking the right questions because you have to be very careful in trying to define a transfer function and if you want to transform to the frequency domain in a meaningful way.
A few issues come up. For one, you might be interested in a large signal transfer function, and this might depend on amplitude. Or, you might be interested in what small AC signals might do when riding on top of the main large signal.
I'm guessing you want to find the voltage transfer function for large signals. If so, the first issue is to define what you mean by "transfer function". One possible definition is to look at the Fourier components of the output voltage and compare only the fundamental frequency of the output to the input. Is this definition useful to you? If not, you have to use another definition.
3. ### suzuki Thread Starter Member
Aug 10, 2011
119
0
hi,
thanks for your response. just to clarify, my overall goal of this task is to determine the transfer function (of the plant) such that i can design a control loop to regulate the plant output voltage. i want to be able to determine and achieve specific crossover frequencies, as well as phase margins.
my input is an ac signal, which passes through a resonant network and is then rectified by the diode bridge to get a dc output. i also wanted to do the rectification at this stage, since i could more easily compare a dc setpoint value with my output voltage (to get my error signal).
As for the transfer function, i *think* what i want is the effects small ac signal riding on top of the dc signal, since this is usually how we obtain the bode plot, and this should give us a relationship between vout and vin. My other reason for thinking i need the small signal and not the large signal method as you suggested, is that i dont think my output voltage would have a fundamental frequency as it is a dc signal. Or, unless you meant "output voltage" as the the voltage before entering the rectifier, but im not sure if that would be useful in my case (since i want my control loop to compare a dc setpoint as stated above).
I guess the thing that is still confusing me is that, any time we look at the dc output voltage, we must already passed through the non-linear rectifier, but i cannot treat it simply as just doing a "absolute value" of my transfer function.
hopefully this gives some clarification on my overall objective, please let me know if you need any more information, as your help is most welcome and appreciated.
4. ### tgotwalt1158 Member
Feb 28, 2011
111
18
Hi Suzuki!
I have drawn full rectifier with input and out put waveforms. Please have a look at the attached drawing. Obviously the input is sine wave and the out put is composed of +ve half cycle plus absolute value of the negative half cycle. If you could transform both the waves in mathematical form, you will get the transfer function, which is a tricky part, at least for me for the time being. I hope this will help up to some extent.
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5. ### steveb Senior Member
Jul 3, 2008
2,433
469
Yes, most definitely it does. This is still a tricky task, but I think you can do it with some thought. I don't have the direct answer off the top of my head, but I would use the typical linearization approach used with nonlinear circuits.
Basically, you want to think of the large signal as creating a type of fixed bias point at any point in time. Think of that point as the operating Q-point similar to when you bias a transistor circuit with DC. Even though this so-called bias point is not fixed, it can still behave as a fixed point for high frequency small signals that might get through your system and cause instability/oscillations. Once you have the bias point, you can think of the forward biased diodes as having a dynamic resistance at that operating point, and then you can make an equivalent AC linearized circuit for small signals. The AC circuit will have parameters that vary in time, depending on where you are biased on your large AC signal that is driving the rectifier.
One issue is whether you need a more advanced high frequency equivalent circuit for a diode. This depends on the diode type and on how high a frequency you are interested.
Thinking quickly, one thing I'm wondering about is whether your small AC signals will do a 180 degree phase shift on the negative half-cycles. My intuition says that it might do that, but I guess I'd have to work out the math to be sure.
Hopefully you can work this out. This week, I'm actually working in a remote location on a project for my job. This means my internet access is very limited and my time is limited too because we are working 16 hours per day. If you have any issues, hopefully others can guide you if you think this approach might be useful to you, and I'll also check in here periodically.
Last edited: Oct 21, 2011
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https://docs.itascacg.com/itasca900/flac3d/docproject/source/options/dynamic/models/review.html | 1,716,726,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00482.warc.gz | 185,092,860 | 9,901 | Comprehensive Liquefaction Constitutive Models
The state of practice for seismic analysis involving liquefiable materials is currently experiencing a shift from using empirical schemes (first developed in the 1970s) to simulate liquefaction, to time-marching numerical methods incorporating liquefaction constitutive models currently at various stages of development. There are several ways liquefaction behavior is included in numerical methods, ranging from total-stress empirical schemes to estimate liquefaction conditions (e.g., the UBCTOT model—see Beaty and Byrne 2000), to simple effective-stress shear-volume coupling schemes (e.g., the Finn model, and the “Roth” model—see Dawson et al. 2001), to more comprehensive constitutive models (e.g., the UBCSAND model—see Byrne et al. 1995; and bounding surface models such as the model described in Wang 1990, and the model described in Papadimitriou et al. 2001) that address cyclic shearing via kinematic hardening.
To help practicing engineers choose a procedure best suited to their needs, a selection of approaches is outlined below, ranging from simple to elaborate in terms of complexity and model parameter determination. This is not a comprehensive list, but is a selection that illustrates the different types of liquefaction models that have been developed and used in two-dimensional FLAC, and could be used in FLAC3D. [1] One important factor to keep in mind is that, in engineering practice, the use of a very complicated model for liquefaction analyses is often hardly justified, considering the many uncertainties with respect to soil properties and earthquake motions, and the numerous approximations that must be made (see Dawson et al. 2001). Before describing the individual models, it will be helpful to review the current state of practice for liquefaction analysis. (See Byrne et al. 2006 for further discussion on state-of-practice analysis.)
State of Practice — The standard practice approach for liquefaction analysis of earthquake loading is based on a total-stress analysis in which it is assumed that the liquefiable soil remains undrained at the in-situ void ratio (Byrne and Wijewickreme 2006). Typically, this analysis approach is divided into three steps:
1. Triggering Evaluation: Typically, an equivalent-linear elastic, dynamic analysis (such as SHAKE) using strain-compatible moduli and damping is conducted for the design earthquake. The cyclic stress ratio (CSR) [2] is evaluated from the numerical simulation and compared to the value of the cyclic stress resistance that the soil has because of its density (CRR) [3], derived from empirical curves. A factor of safety against triggering liquefaction is evaluated using the ratio of CRR and CSR (e.g., see Byrne and Anderson 1991, and Youd et al. 2001).
2. Flow Slide Assessment: Post-liquefaction (undrained) strengths are assigned in zones predicted to liquefy from the triggering evaluation analysis, and a standard limit-equilibrium analysis is carried out to evaluate the factor of safety against a flow slide. Post liquefaction strengths may be derived from penetration resistance (corrected blow count $$(N_1)_{60}$$) using empirical charts (e.g., see Seed and Harder 1990, and Olson and Stark 2002).
3. Seismic Displacements: Displacements are evaluated using the Newmark approach (see Newmark 1965). In this step, the potential sliding block of soil is modeled as a rigid mass resting on an inclined plane. The design time history of acceleration is applied at the base, and the equation of motion is solved to obtain the displacement of the mass caused by the shaking.
The main shortcomings of the standard practice approach are that the three aspects of liquefaction (triggering, flow slide, and deformation) are treated sequentially, when in reality they may interact locally in various zones of the soil structure and affect the overall behavior of the soil mass. Also, no direct account is made of excess pore-pressure redistribution and dissipation.
Total-Stress Synthesized Procedure
The synthesized procedure of Beaty and Byrne (2000) uses FLAC and the UBCTOT constitutive model to combine the three steps of the standard practice approach (triggering, flow slide, and estimate of liquefaction-induced displacements) into one single analysis. The procedure, which assumes undrained behavior, uses a (two-dimensional) total-stress approach to liquefaction analysis and relies on adjustment of liquefied element properties (stiffness and strength) at the instant of triggering of liquefaction. The main features of the UBCTOT model are summarized below.
A seismic analysis using the synthesized procedure starts from a static state of equilibrium for the FLAC model. The seismic analysis is conducted in total-stress space. UBCTOT uses Mohr-Coulomb elasto-plastic logic with zero friction and a value of cohesion equal to the undrained shear strength, in combination with Rayleigh damping. The elastic shear modulus is assigned a value of $$G_{max}$$ multiplied by a modulus reduction factor (MRF). Unlike equivalent-linear methods, this approach is not iterative, and appropriate values of MRF and damping are selected at the start of the seismic analysis. Triggering of liquefaction is based on changes of shear stress on the horizontal plane, $$\tau_{xy}$$. The irregular shear-stress history caused by the earthquake is interpreted in each FLAC zone as a succession of half-cycles with the contribution to triggering determined by the maximum value of $$\tau_{cyc}$$, defined as the difference between $$\tau_{xy}$$ and the initial horizontal shear-stress prior to earthquake loading (i.e., the static bias). A cumulative damage approach is used to combine the effects of each half-cycle. The approach converts the nonuniform history into an equivalent series of uniform stress cycles with amplitude equal to $$\tau_{15}$$ (i.e., the value of $$\tau_{cyc}$$ required to cause liquefaction in 15 cycles, which is approximately the number of cycles in a magnitude 7.5 earthquake). This is done using an empirical chart giving the cyclic stress ratio (CSR) versus cycles to liquefaction. Several property changes are imposed when liquefaction is detected in a FLAC zone: a residual shear strength is assigned; a reduced loading stiffness is used; and unloading uses a stiffer modulus than loading, according to a bilinear stiffness model. Also, a hydrostatic stress state is imposed when a zone experiences a shear-stress reversal. Finally, reduced viscous damping is assigned in a liquefied zone. The constitutive model also has logic to account for anisotropy in stiffness and strength. See Beaty (2001) for additional information.
The UBCTOT model removes some of the limitations associated with the sequential approach to problem solving used in the state-of-practice procedure, while relying on similar empirical charts for triggering of liquefaction and residual strength.
The following are some of the drawbacks of the model:
• the use of equivalent modulus ratio that may not capture the pre-liquefaction phase well;
• the cyclic shear stresses are accounted for on the horizontal plane only;
• the simplified manner in which the undrained shear strength is specified;
• pore pressure is not taken into account explicitly; and
Loosely Coupled Effective-Stress Procedure
The Roth model is a loosely coupled effective-stress constitutive model to generate pore pressure from shear stress cycles using the Seed cyclic stress approach (Seed and Idriss 1971). This model is built around the standard FLAC Mohr-Coulomb model. The (two-dimensional) model counts shear stress cycles by tracking the shear stress acting on horizontal planes ($$\tau_{xy}$$) and looking for stress reversals. The cyclic stress ratio (CSR) of each cycle is measured, and this is used to compute the incremental “damage” that is then translated into an increment of excess pore pressure. The procedure is “loosely coupled” because pore pressures are only computed after each 1/2 cycle of strain or stress as the analysis proceeds. The model incorporates residual strength (a critical parameter for seismic stability analyses) by using a two-segment failure envelope consisting of a residual cohesion value and zero friction angle that is extended to meet with the traditional Mohr-Coulomb failure envelope.
The model is simple, robust, and practice-oriented; it is based on the widely accepted cyclic-stress approach with input parameters readily obtainable from routine field investigations. (Note that liquefaction due to monotonic loading is not considered.) A disadvantage of the model is that liquefaction-induced consolidation settlements are not captured, because the actual physical mechanism of liquefaction, whereby pore pressure is generated through contraction of the soil skeleton, is bypassed. The model is applicable to problems where slope movements due to reduced shear strength are the main concern (such as seismic stability of dams and waterfront retaining structures), while shaking-induced consolidation settlements are of secondary importance. See Roth et al. (1991), Inel et al. (1993), Roth et al. (1993), and Perlea et al. (2008) for some field applications.
The Roth model is similar to the built-in Finn model, which is also considered a loosely coupled effective-stress model. The primary difference is that in the Finn model, the volumetric strains induced by cyclic loading are evaluated based on an experimental curve of irrecoverable volumetric strain versus number of constant amplitude cycles. Pore pressures are then generated from these volumetric strains, as well as from contraction of the soil skeleton. Also, the Finn model in FLAC (and FLAC3D ), at present, does not include a post-liquefaction residual strength.
Fully Coupled Effective-Stress Procedure
The UBCSAND model is a fully coupled (kinematic hardening) effective-stress constitutive model to predict seismic response and liquefaction of cohesionless soils in plane strain problems. The model uses an elasto-plastic formulation based on an assumed hyperbolic relation between stress ratio and plastic shear strain, similar to the Duncan and Chang (1970) formulation. It is applicable for monotonic as well as cyclic loading (e.g., see Byrne et al. 2003, 2006).
The model implementation is a modified form of the built-in Mohr-Coulomb model in FLAC that accounts for a strain-hardening frictional behavior, neglects cohesion, and applies to plane strain conditions. The hardening law is a hyperbolic function of plastic shear strain. Unloading is assumed to be nonlinear elastic, with bulk and shear modulus as functions of mean (in-plane) effective stress. Stress reversal is detected by a change of sign in horizontal shear stress, $$\tau_{xy}$$. Reloading is elasto-plastic, with the yield locus reset to the value at the reversal point. Plastic flow is nonassociated; the logic is based on a variation of Rowe’s stress-dilatancy theory. According to this theory, there is a constant-volume stress ratio, $$\phi_{cv}$$, below which the material contracts (i.e., for mobilized friction, $$\phi_m$$, smaller than $$\phi_{cv}$$), while for higher stress ratios (i.e., for $$\phi_m > \phi_{cv}$$), the material dilates. The effect of relative density is addressed through the choice of material properties. Most properties are calibrated to field experience as well as centrifuge tests and are conveniently related to blow count, $$(N_1)_{60}$$.
The model is able to capture the stiff pre-liquefaction stage, the onset of liquefaction at the appropriate number of cycles, and the much softer post-liquefaction response observed in cyclic, simple shear-constant volume tests.
The coupled effective-stress approach corrects many drawbacks of the previous approaches. Although most parameters are related to blow count and rely on a growing body of data and experience, it is always good practice to check on model parameters for each layer (using numerical simulation of a simple shear test) to verify that if it has to liquefy in $$N$$ cycles according to the field data during dynamic loading, it will. Also, because comparison with standard procedures may not be straightforward, it is recommended that the model be used under the supervision of an experienced practitioner. With time, and with the increase of its usage, the model should become more prominent and be used for problems ranging from simple to complex with little effort.
The primary disadvantages are: 1) the logic for detection of stress reversal is based on horizontal shear stress only; and 2) the present formulation applies only to two-dimensional analysis.
Fully Coupled Effective-Stress Bounding-Surface Procedure
Bounding surface plasticity provides a framework to account for cyclic stress reversal in two and three dimensions (see Dafalias 1986, and Wang 1990). The models developed by Wang (1990) (herein named the WANG model) and Papadimitriou et al. (2001) (herein named the PAPADIMITRIOU model) are two (kinematic-hardening) constitutive models that have been implemented in FLAC based on that logic.
WANG Model — The WANG model is an effective stress, bounding-surface hypoplasticity model for (cohesionless) soil that is capable of reproducing, in detail, typical monotonic and cyclic, drained and undrained, hardening and softening behavior observed in classical laboratory tests on initially dense and loose soils (Wang 1990). (The term hypoplasticity characterizes the dependence of loading and plastic strain-rate directions on stress-rate direction.)
The model formulation includes a noncircular pyramidal failure (bounding) surface, a loading surface, a surface of phase transformation (at which contractive behavior changes to dilative during shearing), and a critical state surface (defining an ultimate state in which the sand deforms at constant volume under constant stress). The three-dimensional effective-stress model requires the specification of 15 constants for a given sand (eight parameters are required for two-dimensional analysis). One disadvantage is that model calibration is an arduous task because most model constants are not related to properties with which the practitioner is familiar, and the body of available parameter data is not yet sufficiently well-developed. Also, comparison to state-of-practice analysis is not straightforward. The relation between cyclic stress ratio, number of cycles to liquefaction and normalized blow count can be used to calibrate the model (Wang et al. 2001), but there is no direct relation between empirical rules used in standard practice and theoretical laws used in the model theory. The WANG model is a sophisticated research tool for laboratory-scale experiments. Application of the model to study boundary-value problems at the field scale should probably not be attempted without the advice of the model developer, whose assistance may be required for model calibration, interpretation of results, and support on possible issues with numerical implementation.
PAPADIMITRIOU Model — The PAPADIMITRIOU model is an effective stress, bounding-surface model for loose and dense sand that is based on critical state elasto-plasticity (Papadimitriou et al. 2001, Papadimitriou and Bouckovalas 2002). The model applies to monotonic as well as cyclic loading (in two and three dimensions) of non-cohesive soils under small and large strains. The model uses a kinematic hardening noncircular cone as the (loading) yield surface. In addition to bounding and dilatancy (marking the transition between contractive and dilatants behavior) surfaces, the model also contains a critical-state surface (defining an ultimate state in which the sand deforms at constant volume under a constant shear and confining stress). The model contains a total of 14 parameters: 11 of the parameters can be derived from in-situ and laboratory tests, while the remaining three must be derived indirectly via trial-and-error simulations of drained and undrained laboratory tests. (Note that each parameter set is independent of initial and drainage conditions, as well as cyclic shear-strain amplitude.)
The model has the capability to reproduce, qualitatively, the characteristic behavior observed in cyclic experiments, including the degradation of shear modulus and increase of hysteretic damping with cyclic shear strain amplitude, the shear and volumetric strain accumulation at a decreasing rate with increasing number of cycles, and the increase in liquefaction resistance with density. Comparisons with centrifuge experiments have been made (Andrianopoulos et al. 2006a), and the ability of the model to study a practical problem of geotechnical earthquake engineering has been demonstrated (Andrianopoulos et al. 2006b). The disadvantages are the model calibration, which is a rather tedious procedure and requires a test database not readily available in most cases, and the long computational time required for the solution of practical problems. Also, comparison to standard practice is not straightforward. Application of the model to study boundary-value problems at the field scale is not recommended at this time without the assistance of the model developer for model calibration, interpretation of results, and eventual support for issues related to numerical implementation. | 3,534 | 17,386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.920002 |
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# GATE | GATE 2017 MOCK II | Question 18
• Last Updated : 28 Jun, 2021
Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. Let X – Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.
(A) 4
(B) 8
(C) 16
(D) 32
Explanation: Number of simple graphs possible with n labeled vertices is 2^(n(n-1)/2).
Number of simple graphs possible with n unlabeled vertices is n+1.
Number of spanning tree possible with n vertices complete graph n^(n-2)
X =8
Y = 4
X-Y=4 | 168 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-39 | latest | en | 0.763152 |
https://www.physicsforums.com/threads/mathematica-simple-problem-with-the-plot-function.690661/ | 1,686,048,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00430.warc.gz | 1,008,994,260 | 15,517 | # [Mathematica] Simple Problem with the plot function
• Mathematica
• PEZenfuego
#### PEZenfuego
I have a formula for the fibonacci sequence (with 1 being the first) and I noticed that the 12th fibonacci number was 144. I thought that was a neat coincidence, so I I headed over to mathematica to see if this (and 1) were the only numbers that had this property. I was almost certain that it was. So I plotted the two graphs, but only the x^2 graphed showed up. Here was my input. My question is about what I did wrong.
phi := (1 + Sqrt[5])/2
phih := (1 - Sqrt[5])/2
f:={(phi^n)-(phih^n)}/{Sqrt[5]}
y:=n^2
Plot[{f, y}, {n, -20, 20}]
Now, this works when I remove the phih term, which works well enough that I can clearly see that only 1 and 12 are solutions. On the other hand, this is not plotting something that it should be plotting...
Anyway, thank you in advance for the help.
The problem is that plot tries to make a continuous plot. It then runs into the problem that phih^x is not a real-valued function for non-integer x, and refuses to plot it altogether. To just plot the values at integer values of x, you can use DiscretePlot:
DiscretePlot[{f, y}, {n, -20, 20, 1}]
The problem is that plot tries to make a continuous plot. It then runs into the problem that phih^x is not a real-valued function for non-integer x, and refuses to plot it altogether. To just plot the values at integer values of x, you can use DiscretePlot:
DiscretePlot[{f, y}, {n, -20, 20, 1}]
That's really neat and useful. I was so accustomed to just glazing over (1-Sqrt[5])/2, that I forgot that it was negative. If you raise it to the power of 1/2 for example, then the answer is imaginary. That's why it is not a real-valued function, correct? Thank you, sir.
That's really neat and useful. I was so accustomed to just glazing over (1-Sqrt[5])/2, that I forgot that it was negative. If you raise it to the power of 1/2 for example, then the answer is imaginary. That's why it is not a real-valued function, correct? Thank you, sir.
You are welcome. And, yeah, if the function has a non-zero imaginary part for at least one value of x, then clearly it's not real-valued on the whole domain (i.e. the plot range). However, Mathematica is able to handle cases where the function is piecewise real-valued, it just doesn't plot the part where the values are comples, see eg.
Plot[(1 + I*HeavisideTheta[x - 5]*HeavisideTheta[6 - x]), {x, 0, 10}]
(I would prefer it to give an error message or a warning, but it doesn't seem to do that.)
Your function is a bit worse though as it is only real-valued on integer values of x. The proof just uses the general definition of the power of a number a (negative, complex etc.), which makes use of the complex logarithm:
$a^b=e^{b log a}$.
If a is real and negative, then this simplifies to
$a^b=e^{bLog(|a|)+ib\pi}=|a|^b e^{ib\pi}$,
which is only real for integer values of b. | 809 | 2,914 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-23 | latest | en | 0.961818 |
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# 20 Periodic Table Puns Worksheet Answers
Periodic Table Scavenger Hunt Worksheet Answer Key for periodic table zirconium, periodic table podcast, periodic table xenon, periodic table legend, periodic table i, via: pinterest.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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https://www.physicsforums.com/threads/doppler-effect.71186/ | 1,480,876,551,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541361.65/warc/CC-MAIN-20161202170901-00136-ip-10-31-129-80.ec2.internal.warc.gz | 997,144,113 | 16,604 | # Doppler effect
1. Apr 12, 2005
### stunner5000pt
A stationary light source S wit ha natural frequency Fo is viewed in a mirror M by a stationary observer O. The mirror moves away from the observer wit ha velocty of Vrel << c
a) what frequency of light is recorded by a detector attached to the moving mirror
because Vrel << c classical may be used
let F1 be this observed frequency observed
then $$f_{1} = f_{0} (1 - \frac{v_{rel}}{c})$$
is this correct??
b) what frequency in terms of fo will the stationary observer measure for the light reflected off the mirror?
the mirror will now emit the f1 from above wouldn't it ??
sine this mirror is moving away wouldnt the doppler shift be $$f_{2} = f_{1} \frac{c}{c+v_{rel}}$$
which would be $$f_{2} = f_{0} (1 - \frac{v_{rel}}{c}) \frac{c}{c+v_{rel}} = f_{0} \frac{c-v}{c+v}$$
but i got the second part wrong! Whats wrong with it??
Also when asked for the lowest average speed of atom at some temperature T given some molar mass M
which formula should be used??
is it $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ or $$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$
Last edited: Apr 12, 2005
2. Apr 12, 2005
### Nylex
Surely if v_rel << c, the second term in that bracket would go to 0?
3. Apr 12, 2005
### stunner5000pt
are you talking about part a) or b)? I did get the first part correct by the way
4. Apr 12, 2005
### Nylex
Either! Assuming v_rel is << c in both cases..
5. Apr 12, 2005
### jdavel
stunner,
Can you find a polynomial expression that approximates (c-v)/(c+v) when v<<c?
6. Apr 12, 2005
### stunner5000pt
k first of all the first one isnt wrong because i wasnt marked wrong thae fact that v<<c doesnt mean that the result in null so get off that!
Of course my approximation is lousy but im trying to answer my prof's question properly according to him, at least
$$\frac{c-v}{c+v} = \frac{1-\frac{v}{c}}{1+\frac{v}{c}} = \frac{1-\beta}{1+\beta} = 1 - \beta + \frac{\beta^2}{2} + ...$$
something like that? Doesnt that give the same answer as a) though??
Last edited: Apr 12, 2005
7. Apr 12, 2005
### jdavel
stunner,
Try this. Divide the numerator and denominator of (c-v)/(c+v) by c. Then define x = v/c. The new denominator will be 1+x. Can you find a power series for 1/(1+x)?
8. Apr 12, 2005
### jdavel
stunner,
I can't keep up with you!
You're very close to the right answer, but you're guessing on the power series. Figure it out. Write 1/(1+x) as (1+x)^-1, then it's easy to see all the derivatives: -(1+x)^-2, 2(1+x)^-3....
9. Apr 12, 2005
### stunner5000pt
what you're saying is put v/c = x whihc would give
$$\frac{1 + x}{1 - x}$$ whiuch is clearly not $$\frac{1}{1-x}$$ and thus u cannt ot expand it out like the latter
10. Apr 12, 2005
### stunner5000pt
and power series for
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ...$$ for abs (x) < 1 where abs means absolute value
11. Apr 12, 2005
### jdavel
stunner,
But (1-x)/(1+x) = (1-x)*1/(1+x). And 1-x is already a power series. So just get the series for 1/1+x and you'll see the answer.
12. Apr 12, 2005
### stunner5000pt
are you sure that can be done??
13. Apr 12, 2005
### jdavel
stunner,
A power series for 1/(1+x)? Why not?
y = (1+x)^-1 >> y(0) = 1
y' = -(1+x)^-2 >> y'(0) =-1
y'' = 2(1+x)^-3 >> y''(0) =2
etc....
So, y = 1/(1+x) = 1 - x + x^2....
So, what's 1/(1+x) when x << 1? | 1,152 | 3,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-50 | longest | en | 0.901179 |
https://math.stackexchange.com/questions/2647560/numbers-defined-with-matrices | 1,581,989,869,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00346.warc.gz | 468,569,100 | 40,340 | # Numbers defined with matrices
We know that a complex number, written as $c=(a,b)$, can be expressed with the help of a matrix as $$\begin{bmatrix}a & -b\\ b & a\end{bmatrix}$$ and operations on such matrices resemble operations on complex numbers.
However with $2 \times 2$ matrices we could imagine a definition of another type "number" $x=(a,b)$, for example $$(a,b) \longleftrightarrow \begin{bmatrix}a & b\\b & a\end{bmatrix}.$$
Here the operations are quite well defined - multiplication and addition are commutative - the only difference to the complex numbers it seems is that not all numbers have their inverses - for example for $(a,a)$ or $(a,-a)$ it's hard to say what is its inverse.
Why don't we use such "numbers"? Are they numbers at all? When can we say that a given matrix represents number?
The same is true for $4 \times 4$ matrices ... it seems only one way of defining numbers - known as quaternions - has found its way into the numbers world... (even though the number of possible ways for constructing matrices with $4$ values when every value is repeated in the matrix $4$ times is much greater).
• Hi, complex noob here, can you point me to a link where I can read about matrix representation of complex numbers? Thanks! – Gaurang Tandon Feb 12 '18 at 15:51
• @GaurangTandon Ok. See en.wikipedia.org/wiki/… – Widawensen Feb 12 '18 at 15:52
• It is not really clear what you are asking. The usual matrix representation of a complex number is not arbitrary, it is invertible iff $(a,b) \neq 0$. – copper.hat Feb 12 '18 at 15:54
• The point is not calling them numbers or not, but the properties they satisfy or fail to satisfy. Many choices are vector spaces over the field of coefficients, some are associative algebras too using the matrix multiplication, fewer are division algebras. – user530511 Feb 12 '18 at 15:55
• What is "number"? – Dietrich Burde Feb 12 '18 at 15:56
This is to some extent a question of representation theory. So we do use these things.
For example, suppose you want to extend the field of rational numbers to include some weird number $\xi$ which satisfies $\xi^2 - N = 0$. You can represent multiplication by a number $a+b\xi$ with the 2x2 matrix $\pmatrix{a & Nb \\ b & a}$. Then yours is just a special case of $N = -1$, which (if the underlying elements $a$ and $b$ are in $\mathbb{R}$) is one way to represent complex numbers. If $N = 0$ then we have the dual numbers, and if $N=1$ then we have the split-complex numbers.
Note that extensions like this can cause problems and we may lose field properties, for instance there is no way to divide a real number by a pure dual number since eliminating the dual from the denominator constitutes division by zero. So always check that the basic rules of arithmetic are preserved, or if we need additional constraints. Just because we lose field properties doesn't mean the algebraic structure isn't interesting or useful.
This kind of extension can continue. If your underlying field is $\mathbb{Q}$ and you extend it to include $\xi_N$ (as above) and then want to extend it again to include another $\xi_M$ then you have a 4x4 matrix representation.
• Thank you for the explanation and in-depth links, however I'm not sure when we can exclude a given form of matrix from consideration it as a number...perhaps I'm to weak in abstract algebra.. – Widawensen Feb 12 '18 at 16:12
• @Widawensen I don't believe there's a simple answer to that other than working out the arithmetic to see which properties are preserved. If you understand matrix arithmetic then it is straightforward (but often tedious) to check. For instance moving from $\mathbb{R}$ to $\mathbb{C}$ is straightforward but then moving from $\mathbb{C}$ to $\mathbb{H}$ you lose some arithmetic properties (in particular you lose the commutative property). – law-of-fives Feb 12 '18 at 16:17
• +1 out of curiosity, are multicomplex numbers also special cases of this? – user541686 Feb 12 '18 at 22:57
• @Mehrdad It is straightforward. If you picture extending your base field by the quadratic element $\xi_i$ giving a 2x2 matrix, then these 2x2 matrices are your numbers, and you can extend quadratically again getting a 4x4 matrix. Something like $a \binom{\sqrt{-1}}{\rightarrow} \pmatrix{a & -b\\ b & a} \binom{\sqrt{-1}}{\rightarrow} \pmatrix{A & -B \\ B & A}$. But solving "the same" equation twice causes other problems as your link shows. – law-of-fives Feb 13 '18 at 0:19
• If I understand you correctly, you are interested in $\mathbb R$-algebras that are subalgebras of $Mat(n \times n, \mathbb R)$ for some $n \in \mathbb N$. Perhaps surprisingly, the only $\mathbb R$-algebras that can be construct this way, which also happen to be Skew-fields (i.e, every element is invertible, but multiplication is not necessarily commutative) are isomorphic to either $\mathbb R$, $\mathbb C$ or $\mathbb H$ (This is known as Frobenius theorem). – H1ghfiv3 Feb 13 '18 at 11:33
To put what you've done in algebraic context: you've identified a subset of the ring of 2x2 matrices (with real number entries I assume, since those are the type we use to represent the complex numbers) and showed that within this subset, multiplication is commutative, and the subset is closed under addition and multiplication, so it's a commutative subring. In quid's answer to this related question, quid shows that your subring is isomorphic to the direct product of two copies of $\mathbb{R}$. I refer you there to the algebraic proof: I'll show it for your example in a more concrete way.
Associate to each matrix $\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ the pair $(a+b,a-b)$. It's clear that, conversely, given such a pair, you can recover the matrix. Let's check how matrix addition and multiplication affect the pair. $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} + \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} a+a' & b+b' \\ b+b' & a+a' \end{bmatrix}$$, so $$(a+b,a-b)+(a'+b',a'-b') = (a+a'+b+b',a+a'-b-b')$$. And $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} \cdot \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} aa'+bb' & ab'+ba' \\ ab'+ba' & aa'+bb'' \end{bmatrix}$$, so $$(a+b,a-b)\cdot(a'+b',a'-b') = (aa'+bb'+ab'+ba',aa'+bb'-ab'-ba')= ((a+b)(a'+b'),(a-b)(a'-b'))$$. Note that the matrix addition and multiplication on the matrices induce regular addition and multiplication on the pairs, performed independently on the first and second entries.
So it turns out these "numbers" are equivalent to pairs of numbers on which addition and multiplication are performed separately.
• Very interesting approach.. so these pairs of numbers can be treated here separately, but in the case of complex representation I suppose not, is a good conclusion? – Widawensen Feb 12 '18 at 16:24
• @Widawensen Yes. To say that the complex numbers can't be turned into pairs treated separately we'd formally say that $\mathbb{C}$ is not isomorphic to $\mathbb{R} \times \mathbb{R}$. This can be proven from the observation (which you made) that all complex numbers are invertible, while not all these pairs are invertible, e.g. the pair $(2,0)$ cannot multiply by anything to give $(1,1)$. – BallBoy Feb 12 '18 at 16:30
• Such observations as you've made are clarifying the situation, that's what I'm looking for .... for essential differences in both cases... it's interesting that all explanations given here are lightening the problem as if from different points of view.. – Widawensen Feb 12 '18 at 16:39
The point is that we want to have a field. At least, this is what is suggested here by speaking of "numbers". Now in general, subalgebras of $M_2(K)$ need not be a field. In the first case, however, we obtain a field, namey the field of complex numbers $\mathbb{C}$ . In the second example, we do not obtain a field. For a field $(K,+,\cdot)$, we need that $(K,+)$ and $(K^*,\cdot)$ are both abelian groups. This is not the case here.
• Aha, connections with abstract algebra are essential here, that's when I'm weak. But I memorize these "abelian groups" and I will rethink them.. – Widawensen Feb 12 '18 at 16:04
• I'm also interested how the construct of field is referenced to the quaternions? – Widawensen Feb 12 '18 at 16:17
• You have been right .. maybe the "field" is the central construct here ...according to Wikipedia necessary is condition of existence "multiplicative inverse $b^−1$ for every nonzero element $b$", so "numbers" are only numbers when are appearing on a field.. quaternions have also always (except 0) inverse elements therefore they are the numbers.. – Widawensen Feb 12 '18 at 17:05
• Hamilton's real quaternions do not form not a field, but they are still close to it. They are sometimes called "a number system that extends the complex numbers", see here. So you are right in a way. – Dietrich Burde Feb 12 '18 at 19:50
The matrix representation of the complex numbers shows that there exists an isomorphism between the complex numbers and that particular subset of matrices in $\mathbb R^{2 \times 2}$.
The point is, they started with the complex numbers and then searched for a matrix representation.
I don't want to get all Zen on you***, but what is a number?
The integers, whole number, natural numbers, real numbers, complex numbers, quaternions, and so on all have different properties. Yet we call them all numbers.
What about differential forms? What about $m \times n$ arrays of real numbers? What about the set of all permutations of the set $\{1,2,3 \dots \}$? Are they numbers?
You wonder, since there exists a particular subset of $\mathbb R^{2 \times 2}$ that behaves exactly like the field of complex numbers, can that process be turned around to create new $\text{$$numbers$"$}$?
Sure it can. I suppose that you should expect at least for some form of closure to happen but everything else is just a matter of what you discover to be true and decide is important.
The big question is, when you show what you have discovered to the rest of the world, will they call them numbers too? That is to say, what is a number?
*** Whenever someone says "I don't want to...", they want to.
• Hmm, this gives a freedom in the decision what the number is.. really we have so much freedom? – Widawensen Feb 12 '18 at 16:43
• @Widawensen No, you don't. There was a time when people argued whether or not $0$ and $1$ were numbers. Look at the names of some numbers, they all have distainful connotations: negative, complex, imaginary, surd (as in a number that is not surd must be an absurd number), irrational. Ultimately, the decision whether a particular set is a set of numbers is a matter of (mathematical) public opinion based mostly on utility. – steven gregory Feb 12 '18 at 17:03
• ..but what with the proposition of Dietrich that numbers are the numbers only on - so called - field ? Maybe this is what is widely accepted nowadays? – Widawensen Feb 12 '18 at 17:08
• That's his definition. How well has it been received? – steven gregory Feb 12 '18 at 17:49
Not sure if I fully understand your question, but I'll give it a try. The reason that we can represent a complex number $a+bi$ as $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}=a\begin{bmatrix}1&0\\0&1\end{bmatrix}+b\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ is because they behave the same way. For instance $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}^2=(-1)\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ behaves just like $i$, since $i^2=-1$.
You can also do this with the real numbers by identifying $a\in\mathbb{R}$ with the matrix $$\begin{bmatrix}a&0\\0&a\end{bmatrix}=a\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ but the crucial point is that the "numbers" and the "matrices" have to behave the same way (this is called an isomorphism).
• Thank you for the answer, I understand correspondence between complex numbers and these matrices because we have rules of translation one representation to the other. Any way I appreciate your answer... – Widawensen Feb 12 '18 at 16:08
Just to maybe help widen the number concept.. =)
Any function from a finite subset of $\mathbb Z$ onto itself can be represented as a binary matrix with column sum = 1. For example square modulo 4: $\cases{0\to 0\\1\to 1\\2\to 0\\3\to 1}$
$$\left[\begin{array}{cccc}1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0\end{array}\right] \text{ assuming column vectorization of variable:}\left[\begin{array}{cccc}mod=0\\mod=1\\mod=2\\mod=3\end{array}\right]$$
Now matrix multiplication represents function concatenation so we can actually treat functions as numbers. Cool huh?
• Functions as numbers? Is it not going too far ?? – Widawensen Feb 13 '18 at 11:34
• @Widawensen the really cool stuff does not happen until you find ways to do the same for subsets of $\mathbb R$, then you can do some real magic. ;) – mathreadler Feb 13 '18 at 22:07
You can use whatever "numbers" you like! At a bare minimum, you probably want to be able to add, subtract, and multiply your "numbers". Division might also be nice too. It turns out that many of these "numbers" can be thought of as matrices.
Here is an interesting example: the "dual numbers". A dual number $a + b \varepsilon$ has a real part $a$, and an infinitesimal part $b\varepsilon$. The interesting thing about the dual numbers is that $\varepsilon^2 = 0$ (think of $\varepsilon$ as being really really small, so when it gets squared it goes away). In general, the multiplication rule is $$(a + b \varepsilon)(c + d \varepsilon) = ac + bc\varepsilon + ad\varepsilon + bd\varepsilon^2 = ac + (bc + ad)\varepsilon$$
We can embed the dual numbers into the $2 \times 2$ matrices, via $$a + b \varepsilon = \begin{bmatrix}a & b \\ 0 & a \end{bmatrix}$$ and double-check that this multiplication works: $$(a + b \varepsilon)(c + d \varepsilon) = \begin{bmatrix}a & b \\ 0 & a \end{bmatrix}\begin{bmatrix}c & d \\ 0 & c \end{bmatrix}=\begin{bmatrix}ac & ad+bc \\ 0 & ac \end{bmatrix}$$
One super-cool thing about the dual numbers is that they somehow know how to differentiate: We know via calculus that if $f(x) = x^2 + 5x - 1$, then $f'(x) = 2x + 5$. But the dual numbers just know this anyway: $$f(x + \varepsilon) = x^2 + 2x\varepsilon + \varepsilon^2 + 5x + 5\varepsilon - 1 = (x^2 + 5x - 1) + (2x + 5)\varepsilon$$
• Presented connection with differentation is really cool .. it seems that everything can be "replaced" with matrices :) – Widawensen Feb 13 '18 at 11:45 | 4,000 | 14,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-10 | latest | en | 0.901699 |
https://prepinsta.com/harmonic-progression/how-to-solve-quickly/ | 1,713,575,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00670.warc.gz | 416,867,007 | 36,835 | # How To Solve Harmonic Progression Questions Quickly
## How to Solve Harmonic Progression Problems Quickly
### How to Solve HP Questions Quickly
A harmonic progression is a sequence of real numbers formed by taking the reciprocals of an Arithmetic progression.
An HP is represented in the form $\frac{1}{a_{1}}, \frac{1}{a_{2}} ,\frac{1}{a_{3}}……\frac{1}{a_{n}}$
where,
$\frac{1}{a_{1}}$= the first term,
d = the common difference taken from AP = d
Here are some of the best way for how to solve harmonic progression questions effectively.
### Type 1: nth term of an HP : $a_{n} = \frac{1}{a+(n-1)d}$
Question 1. If the sum of reciprocals of first 11 terms of an HP series is 110, find the 6th term.
Options:
A. 10
B. $\frac{1}{10}$
C. $\frac{1}{6}$
D. $\frac{1}{5}$
Solution Reciprocals of first 11 terms of an HP will be AP
Therefore, $s_{n} =\frac{n}{2}[2a+(n-1)\times d]$
$S_{n}$ = 110
n = 11
$110 = \frac{11}{2}[2a + (11 − 1) d]$
$110 = \frac{11}{2} \times [2a + 10d]$
220 = 22a + 110d
22a + 110d = 220
a + 5d = 10
which is the 6th term of the AP series
Therefore, the 6th term in HP = $\frac{1}{10}$
Correct option: B
Question 2 Find the infinite sum of series $\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$ …….
Options:
A. 0.45
B. 0.69
C. 0.87
D. 2
Solution In the given series,
$\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$
$\frac{1}{(3^{2}-2^{2})} + \frac{1}{(4^{2}-2^{2})} + \frac{1}{(5^{2}-2^{2})}$
We know, that a2-b2 = (a-b) (a+b)
S = $\frac{1}{(3-2)(3+2)} + \frac{1}{(4-2)(4+2)} +\frac{1}{(5-2)(5+2)}$ …..
S = $\frac{1}{4}(1-\frac{1}{5}) + \frac{1}{4}(\frac{1}{2} – \frac{1}{6}) + \frac{1}{4}(\frac{1}{3} – \frac{1}{7})$
S = $\frac{1}{4}\times 1.833$
S = 0.45
Correct option: A
Question 3. Identify the 4th and 8th term of the series 6, 4, 3….
Options:
A. $\frac{12}{5}$ and $\frac{13}{9}$
B. $\frac{12}{9}$ and $\frac{13}{5}$
C. $\frac{12}{5}$ and $\frac{12}{9}$
D. $\frac{12}{5}$ and $\frac{12}{7}$
Solution Consider the series in the form of $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$
We know that $T_{2} – T_{1} = T_{3} – T_{2} = \frac{1}{12}$
As $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$ is an AP, Thus 4th term of the AP = a +3d = $\frac{1}{ 6} + 3 \times \frac{1}{12} = \frac{5}{12}$
Similarly the 8th term of the series = a +7d = $\frac{1} {6} + 7 \times \frac{1}{12} = \frac{9}{12}$
Thus the 4th term of an AP = $\frac{12}{5}$ and 8th term = $\frac{12}{9}$ respectively.
Correct option: C
### Type 2: Harmonic mean of the series: HM = $\mathbf{\frac{n} {\frac{1}{a_{1}} + \frac{1}{a_{2}} ……\frac{1}{a_{n}}}}$
Question 1. Aarti walked first one-third of the distance at a speed of 2 km/hr. The next one-third of the distance was covered by running at the speed of 3km/hr. The last one-third of the distance was covered by cycling at the speed of 6 km/hr. Find the average speed for the whole journey covered by Aarti?
Options:
A. 5 km/hr
B. 6 km/hr
C. 4 km/hr
D. 3 km/hr
Solution According to the question, the distance covered is same in all the three cases.
Therefore the average speed = HM of 2, 3, and 6
Average speed = $\frac{3}{ (\frac{1}{2} + \frac{1}{3} + \frac{1}{6})}$
Average speed = $\frac{3}{1}$
Average speed = 3 km/hr
Correct option: D
Question 2. Find the Harmonic mean of 5, 10 ,15
Options:
A. 4.5
B. 6.5
C. 8.33
D. 6.4
Solution We know that,
HM = $\frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}}…\frac{1}{a_{n}}}$
HM = $\frac{3}{\frac{1}{5}+\frac{1}{10}+\frac{1}{15}}$
HM = $\frac{3}{0.36}$
HM = 8.33
Correct option: C
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Android Question
# Calculate angle of touched point and rotate it in Android
Math has defeated me once again. This is such a simple task, but I can't manage to get it done.
Scenario: I draw on a SurfaceView a round image. The user touches a point on image border and starts to drag it adround. I need to rotate the circle image according to user movement.
I have two important piece of information, the image center X,Y coordinates and the touched points coordinates.
As you can see in the image, the user touched a point, according to my draw the touched point angle should be around 40. I can't manage to calculate it properly.
I tried using this formula:
``````angle = Math.atan2(touchedY - centerY, touchedX - centerX) * 180 / Math.PI
``````
I can't manage to understand how I should calculate the angle, as it is now, it doesn't work properly and values are not good. For instance, in the case of the image, the angle calculate is -50.
LE: Actually I did a mistake I think, as mentioned below. Should the circle be like:
Let's reformulate the problem: You want to find the angle between two vectors. The first vector is the upvector going straigt up from your center-point (u), and the second vector is the vector from the center point to the touch point (v).
Now we can recall (or google) that
cos a = uv / (|u|*|v|)
Where a is the angle between the vectors and |u| is the length of a vector. The upvector, u, is (0, 1) and has length 1.
Multiplying the vectors by hand cancels the x-term and gives us something like this.
``````double tx = touch_x - center_x, ty = touch_y - center_y;
double t_length = Math.sqrt(tx*tx + ty*ty);
double a = Math.acos(ty / t_length);
``````
Note how the v vector is obtained by subtracting the center point from the touch point. Remember to convert to degrees if needed.
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 476 | 1,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-30 | latest | en | 0.925746 |
https://encyclopediaofmath.org/index.php?title=Riemann-Hilbert_problem&oldid=48540 | 1,721,421,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00347.warc.gz | 196,431,951 | 8,978 | # Riemann-Hilbert problem
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
linear conjugation problem, Riemann problem, Hilbert problem, Riemann boundary value problem, Riemann–Privalov problem, Hilbert–Privalov problem
One of the main boundary value problems of analytic function theory. It can be stated in the simplest case as follows. Let $L$ be a simple smooth closed contour that splits the plane into a bounded interior domain $D ^ {+}$ and the domain $D ^ {-}$ complementary to it, containing the point at infinity. Let two functions $G ( t)$ and $g ( t)$ be given on $L$, satisfying a Hölder condition ( $H$- condition), with $G ( t) \neq 0$ everywhere on $L$. It is required to find two functions $\Phi ^ \pm ( z)$, analytic in $D ^ \pm$, respectively, continuous up to the contour except for finitely many points $t _ {k}$, where they may have discontinuities satisfying
$$| \Phi ( z) | < \ \frac{A}{| z - t _ {k} | ^ {\alpha _ {k} } } ,$$
and satisfying on $L$ the boundary condition
$$\tag{* } \Phi ^ {+} ( t) = G ( t) \Phi ^ {-} ( t) + g ( t) ,$$
where the function $G ( t)$ is called the coefficient of the problem. The integer
$$\kappa = \mathop{\rm ind} G ( t) = \frac{1}{2 \pi } \int\limits _ { L } d \mathop{\rm arg} G ( t) = \ \frac{1}{2 \pi i } \int\limits d \mathop{\rm ln} G ( t)$$
is called the index of the coefficient $G ( t)$ and at the same time the index of the Riemann–Hilbert problem. Suppose that one is looking for a solution satisfying the condition $\Phi ^ {-} ( \infty ) = 0$. Then for $\kappa > 0$, the homogeneous Riemann–Hilbert problem (that is, when $g ( t) \equiv 0$) and the non-homogeneous Riemann–Hilbert problem are unconditionally solvable; the solution depends linearly on $\kappa$ arbitrary constants and are expressed linearly in terms of a polynomial of degree $\kappa - 1$ with arbitrary coefficients. If $\kappa = 0$, the homogeneous Riemann–Hilbert problem has only the trivial zero solution, and the non-homogeneous problem is solvable unconditionally and uniquely. If $\kappa < 0$, the homogeneous problem has only the trivial solution, and the non-homogeneous problem has a unique solution only if $| \kappa |$ solvability conditions are satisfied, which can be expressed linearly in terms of a polynomial with variable coefficients.
The solution of the Riemann–Hilbert problem in all cases is represented in closed form by quadratures in terms of integrals of Cauchy type (cf. Cauchy integral). The boundary values of the unknown functions necessarily satisfy the $H$- condition on the contour. The result listed above can be carried over unchanged to the case of a multiply-connected domain $D ^ {+}$ bounded by finitely many simple mutually non-intersecting closed curves. The case of a contour consisting of open curves presents the peculiarity that at the ends of the curves of the contour the solutions of the Riemann–Hilbert problem, depending on the chosen class of solutions, may become infinite or remain bounded. The index depends on the chosen class of solutions. The index in the class of solutions with an admissible infinity at the end of order less than one (an integrable infinity) is equal to one plus the index in the class of solutions bounded at this end. In accordance with this, the number of linearly independent solutions increases by one or the number of solvability conditions decreases by one. A similar situation occurs in the case when the coefficient $G ( t)$ has discontinuities of the first kind; the solution at the points of discontinuity behaves in the same way as at the ends of the contour.
In the general case, when the contour consists of finitely many arbitrarily situated closed and open curves, the Riemann–Hilbert problem is solved by the same methods as the simple cases mentioned above, and similar results hold. Some difficulties are presented by the investigation of the solution at the points where several curves of the contour meet.
In the case when $G ( t)$ and $g ( t)$ are only continuous, but do not satisfy an $H$- condition, the results stated above remain valid, except that here the boundary values of the solutions exist only as the contour is approached along non-tangential paths, and they are not continuous, but $\Phi ^ \pm ( t) \in L _ {p}$ for any $p > 0$; if $G ( t)$ is continuous and $g ( t) \in L _ {p}$, then $\Phi ^ \pm ( t) \in L _ {p}$. The most general assumption for the coefficient $G ( t)$ under which the Riemann–Hilbert problem has been solved is that it belongs to the class of measurable functions with an additional condition on the value of the jump of the argument; here also $g ( t) \in L _ {p}$.
Riemann–Hilbert problems with infinite index have been considered, in which simple smooth curves have been chosen for the contours with one or both ends going to infinity. The following cases have been investigated: 1) a polynomial order of growth, when as $| t | \rightarrow \infty$ the asymptotic equalities
$$\mathop{\rm Ind} G ( t) \sim \pm | t | ^ \rho$$
are satisfied ( $0 < \rho < \infty$ for the case one infinite end, $0 < \rho < 1$ for both ends infinite); and 2) a logarithmic order of growth, when as $| t | \rightarrow \infty$,
$$\mathop{\rm Ind} G ( t) \sim \pm \mathop{\rm ln} ^ \alpha | t | ,\ \ 0 < \alpha < \infty .$$
At both ends with a positively infinite index the number of linearly independent solutions is infinite and is expressed in terms of an entire function whose form depends on the order of the indices. For a negative infinite index the homogeneous problems has no non-trivial solutions, and the non-homogeneous problem is solvable only if an infinite set of solvability conditions is satisfied. The main difficulty lies in distinguishing the finite solutions.
The solution of the Riemann–Hilbert problem on a Riemann surface, and the equivalent problem on the fundamental domain of an automorphic function belonging to a group of permutations, has been investigated for automorphic functions of this class. The number of solutions or solvability conditions depends on the index, and in certain (singular) cases also on the genus of the surface or on the fundamental domain.
If in condition (*) $G$ is a matrix and $\Phi ^ \pm$ and $g$ are ( $n$- dimensional) vectors, then there arises the Riemann–Hilbert problem for a componentwise-analytic vector. This is significantly more complicated than the scalar case $( n = 1 )$ considered above. The investigation is carried out by reduction to a system of integral equations. The number of linearly independent solutions or solvability conditions depends on $n$ unknown quantities, called partial indices, the dependence of which on the coefficient matrix has not been established explicitly. An important role is played by the index of the determinant of $G$, called the total index of the problem.
The Riemann–Hilbert problem (for a componentwise-analytic vector) occurred first with B. Riemann (see [1]) in connection with the solution of the problem of constructing a linear differential equation from a given group of permutations (monodromy group). However, in the approximate form stated above the Riemann–Hilbert problem was first considered by D. Hilbert in 1905 (see [2]) under less general conditions; the first results (of an alternative character) were obtained by reduction to an integral equation. In 1908 J. Plemelj (see [3]), using for the first time integrals of Cauchy type, reduced the Riemann–Hilbert vector problem to an integral equation and completely solved the problem of differential equations posed by Riemann. The non-homogeneous Riemann–Hilbert problem (in a somewhat different formulation) was first considered by I.I. Privalov [4]; his results are concerned mainly with function-theoretical generalizations.
The Riemann–Hilbert problem has many applications. The main ones are in the theory of singular integral equations. Generalizations have been given in various directions: the Riemann–Hilbert problem with a shift or with conjugation, with differentials for generalized analytic functions, and others. The theory of the Riemann–Hilbert problem and its generalizations and applications are more fully reflected in [5] and [6].
#### References
[1] B. Riemann, "Collected works" , Dover, reprint (1953) [2] D. Hilbert, "Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen" , Chelsea, reprint (1953) [3] J. Plemelj, "Riemannsche Funktionenscharen mit gegebenen Monodromiegruppe" Monatsh. Math. Phys. , 19 (1908) pp. 211–245 [4] I.I. Privalov, "On a boundary problem in analytic function theory" Mat. Sb. , 41 : 4 (1934) pp. 519–526 (In Russian) (French abstract) [5] N.I. Muskhelishvili, "Singular integral equations" , Wolters-Noordhoff (1972) pp. Chapt. 2 (Translated from Russian) [6] F.D. Gakhov, "Boundary value problems" , Pergamon (1966) (Translated from Russian) | 2,244 | 8,889 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.836394 |
https://www.jiskha.com/display.cgi?id=1526483171 | 1,527,046,797,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00059.warc.gz | 775,118,713 | 4,111 | # Science
11. Create a scale for your model thermometer. Divide the distance between the two
marks into 5-mm intervals. Starting with the lowest point, label the intervals on the
straw 0, 1, 2, 3, and so on. Describe your scale below. PLEASE HELPPPP IM super behind and school is almost over someone please be nice enough to help me :)
1. bobpursley
I need help tho @bobpursley
3. Ms. Sue
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Waves are coming towards your boat at a speed of 4 m/s, you are standing on the fish scale and notice that the reading on the scale changes up and down such that the lowest scale reading is ½ of the maximum scale reading. If the total … | 532 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-22 | latest | en | 0.912043 |
http://www.mayhematics.com/d/d2.htm | 1,508,450,974,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00545.warc.gz | 499,477,937 | 4,425 | ç Index to Chessboard Dissection Problems
# 2-Square Pieces: Dominising the Chessboard
The subject of enumerating the dissections of a rectangular board m×n into two-celled pieces (dominoes) was briefly discussed in Chessics (1986, vol.1, #28, p.138) where (in different notation) the following recurrence relation was given.
T(m,n) = S(s=1 ... n) of V(m,s)·T(m,n–s)............................................................................(1)
where V(m,s) is the number of m×s dissections without vertical fault lines, and T(m,0) = 1.
Proof of the formula: V(m.s) is the number of ways of dissecting the part of the rectangle before the first fault line (or the whole rectangle if there is no fault line). T(m,n–s) is the number of ways of dissecting the rest of the rectangle, becoming T(m,0) = 1 when there are no fault lines. Multiplying these together gives the number of ways of dissecting the rectangle for each position s of the first fault line. Summing over all possible positions of the first fault line gives the required total.
This method is practical for calculating the totals when m </= 4 since the values of V(m,s) are then easily found by direct calculation, mostly being zero (in particular they are zero when m and s are both odd, since dominoes always cover an even number of cells).
The nonzero cases are illustrated here:
From these diagrams we find the values:
V(1,2) = 1, V(2,1) = 1, V(2,2) = 1, (V,3,2) = 3, V(3,2h) = 2, V(4,1) = 1, V(4,2) = 4, V(4,2h–1) = 2, V(4,2h) = 3,
and using the above summation we can deduce the following recurrence relations to calculate T(2,n), T(3,n) and T(4,n).
T(2,n) = T(2,n–1) + T(2,n–2)...................................................................................................(2)
T(3,2k) = 4·T(3,2k–2) – T(3,2k–4)..........................................................................................(3)
T(4,n) = T(4,n–1) + 5·T(4,n–2) + T(4,n–3) – T(4,n–4).............................................................(4)
Equation (2) was shown by W. L. Patten in American Mathematical Monthly 1961. This recurrence relation is of course the one that generates the Fibonacci sequence. Equation (3) was given by Chris Holt in Mathematical Spectrum (1994/5, volume 27, #3, p.62). Equation (4) was supplied by me in a letter to Mathematical Spectrum (1995/6, volume 28, #2, p.44).
The subject was revisited in The Games and Puzzles Journal (volume 2, #13, 1996, pp204-5) in an article on 'Dominizing the Chessboard', based mainly on new information in a letter to me (24 Jan 1996) from Robin J. Chapman, Department of Mathematics, University of Exeter. He pointed out that the problem, on boards of all sizes, was completely solved as long ago as 1961!
The relevant paper by P. W. Kasteleyn, Shell-Laboratorium, Amsterdam, has the title 'The statistics of dimers on a lattice, I: The number of dimer arrangements on a quadratic lattice', and appeared in the journal Physica (volume 27, 1961, pp1209–25). This reference seems to have escaped the notice of the recreational mathematics fraternity, presumably because it was published in a physics journal and its title makes no mention of dominoes, dissections or chessboards; also it gives no explicit numerical results. The formula for the number T(m,n) of dissections m by n can be expressed in the form:
T(m,n) = P(j=1...m) P(k=1...n) [4·cos2{jp/(m+1)} + 4·cos2{kp/(n+1)}]1/4...............................(5)
[I'm not sure how cos and pi get into the act, but am assured that this is just a bit of mathematical flim-flam and that they cancel out each other's effect (rather like the White Knight's plan to dye his whiskers green and always carry so large a fan that they could not be seen).]
An unpublished paper by James Propp, 'Dimers and dominoes', Massachusetts Institute of Technology, dated 24/ix/1992, based on Kasteleyn, notes that the number of domino tilings of an 8×8 board is 12,988,816 = 3604².
Robin Chapman kindly evaluated the formula for all cases up to 20×20 using the computer algebra system MAPLE. Here are the results up to 72 cells:
2×2=2, 2×3=3, 2×4=5, 2×5=8, 2×6=13, 2×7=21, 2×8=34, 2×9=55, 2×10=89, 2×11=144, 2×12=233, 2×13=377, 2×14=610, 2×15=987, 2×16=1597, 2×17=2584, 2×18=4181, 2×19=6765, 2×20=10946, 2×21=17711, 2×22=28657, 2×23=46368, 2×24=75025, 2×25=121393, 2×26=196418, 2×27=317811, 2×28=514229, 2×29=832040, 2×30=1346269, 2×31=2178309, 2×32=3524578, 2×33=5702887, 2×34=9227465, 2×35=14930352, 2×36=24157817. 3×4=11, 3×6=41, 3×8=153, 3×10=571, 3×12=2131, 3×14=7953, 3×16=29681, 3×18=110771, 3×20=413403, 3×22=1542841, 3×24=5757961. 4×4=36, 4×5=95, 4×6=281, 4×7=781, 4×8=2245, 4×9=6336, 4×10=18061, 4×11=51205, 4×12=145601, 4×13=413351, 4×14=1174500, 4×15=3335651, 4×16=9475901, 4×17=26915305, 4×18=76455961. 5×6=1183, 5×8=14824, 5×10=185921, 5×12=2332097, 5×14=29253160. 6×6=6728, 6×7=31529, 6×8=167089, 6×9=817991, 6×10=4213133, 6×11=21001799, 6×12=106912793. 7×8=1292697, 7×10=53175517. 8×8=12988816. 8×9=108435745
Propp also notes that the total is a square on square boards of side 4k, and twice a square on square boards of side 4k+2.
For 6×6 we have: 6728 = 2×58². The 10×10 gives: 258584046368 = 2×359572²
and 12×12 gives 53060477521960000 = 230348600²
Chapman notes that Kasteleyn's formula also gives the number of tilings of an m×n rectangle with a given number of horizontal tiles. He has computed these for an 8×8 square. The results are listed below. If A(n) is the number of tilings with n horizontal tiles (n even) then A(n) = A(32–n).
A(0) = A(32) = 1, A(2) = A(30) = 70, A(4) = A(28) = 1785, A(6) = A(26) = 21656, A(8) = A(24) = 144092, A(10) = A(22) = 580620, A(12) = A(20) = 1511368, A(14) = A(18) = 2644858, A(16) = 3179916.
The problem remains of how to calculate V(m,n) for larger values of m, and of determining the number F(m,n) of fault-free dissections (with neither vertical nor horizontal fault lines).
Problem 150 in Chessics (volume 2, #23, 1985) was: Dissect a chessboard into dominoes so that the number of 2×2s (dominoes with long sides matching) is minimised. The unique solution (apart from rotation) is diagrammed here.
This is a case of a general result on dominoes that I found: Every domino dissection of a rectangle must contain at least one 2×2 inset, i.e. two dominoes side by side; and a domino dissection that has exactly one 2×2 inset must be either square (2k by 2k) or near-square (2k by 2k–1), with the 2×2 centrally placed and the other dominoes in 'rings' around it. To prove the first part of this, start at the a1 corner and try to construct a dissection without a 2×2. We are forced to place dominoes in a 'herringbone' pattern diagonally from the corner, and this forces the formation of a 2×2 where it meets the opposite edge. | 2,307 | 6,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-43 | latest | en | 0.814493 |
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# Integral inverse DMO
DMO is a method of transformation of finite-offset data to zero-offset data. Let the normal moveout corrected input data be denoted and the zero-offset desired output denoted . Assume known relationships between the coordinates of the general form
(14)
The DMO operator can be defined in the zero-offset frequency and midpoint wavenumber as Liner (1988)
(15) (16)
whereas its inverse can be defined as
(17)
where
(18)
A detailed derivation of J2 is given by Liner 1988. The method is based on a general formalism Beylkin (1985); Cohen and Hagin (1985) for inverting integral equations such as dmo.eq. It involves inserting dmo.eq into dmoinv.eq and expanding the resulting amplitude and phase as a Taylor series and making a change of variables according to Beylkin 1985. The solution provides an asymptotic inverse for dmo.eq, where the weights are given by
(19)
In this expression, is the Jacobian of the change of variables in the forward DMO given by
(20)
which reduces to , assuming the general coordinate relationships coord.relat where is independent of t2, leading to a zero lower left element in the determinant matrix above.
The quantity is the inverse of the Beylkin determinant, H, and is given by
(21)
If we recognize that is independent of , then the lower element of H-1 is zero and Beylkin_inv reduces to
(22)
where and are, respectively,
(23)
(24)
Notice that and depend on the coordinate relationships coord.relat. Therefore, the Beylkin determinant, H, varies according to the DMO operator but is constant for kinematically equivalent operators.
Next: Hale DMO and its Up: Amplitude-preserving AMO Previous: Chaining DMO and inverse
Stanford Exploration Project
1/18/2001 | 430 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.882017 |
https://nethercraft.net/if-5-0-moles-of-c3h8-react-how-many-molecules-of-water-are-formed/ | 1,701,392,289,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00229.warc.gz | 477,486,589 | 12,379 | # If 5.0 moles of C3H8 react, how many molecules of water are formed?
0
• The reaction of propane with oxygen is C3H8 + 5 O2 —-> 3 CO2 + 4 H2O
Therefore for every atom of C3H8, four atoms of water are produced.
Therefore 5 moles of C3H8, 5 x 4 moles of water are produced
On the basis of Avogadro’s constant, 1 mole of water contains 6.02214129 x 10^23 molecules
Therefore 5 x 4 = 20 moles
20 moles of water contains 20 x 6.02214129 x 10^23 molecules
= 120.4428158 x 10^23 = 1.204428158 x 10^25 molecules of water
• Minors in something do now no longer qualify you to alter into specialists in those fields. Mathematicians, physicists, and chemists pick PhDs, and engineers choose a Bachelors at minimum. Sorry to break it to you, yet English majors are in many circumstances ineffective very final, alongside with all diverse liberal arts majors, at the same time because it contains finding worthwhile jobs. in case you pick to alter perfect right into a mathematician, chemist , physicist, or an engineer, bypass to grad college and get a PhD (a Masters indoors the case of engineering). Or significant in a single among those matters, and minor in English. or you would be waiting to choose to purpose stepping perfect right into a joint Masters-PhD application in English. in any diverse case, choose like hell which you get into regulation college.
Also Check This What is the nutritional advantage of eating range-fed buffalo? | 371 | 1,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.923267 |
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### Powerpoint Subtraction Of Decimals L2
1. 1. Adult numeracy Level 2 Subtraction of Decimals
2. 2. Decimals at Level 2 <ul><li>At level 2 we should be able to add and subtract decimals up to three decimal places </li></ul><ul><li>A Decimal place is the position of the digit to the right of the decimal point, you should recognise these values as tenths, hundredths, and thousandths </li></ul>
3. 3. Decimals at Level 2 <ul><li>For example; Let’s consider a piece of cloth that measures 42cm long, in metres this is 0.42m. </li></ul><ul><li>If we were then to cut off a length of 228mm from the cloth, to calculate the remaining length we need to perform the following calculation </li></ul><ul><li>0.42-0.228=? </li></ul>
4. 4. <ul><li>0.420 </li></ul><ul><li>0.228 </li></ul>We start by setting out the calculation in the usual way, keeping the decimals in line Notice that we have added a zero after the 0.42 to keep the place values correct
5. 5. <ul><li>0.420 </li></ul><ul><li>0.228 </li></ul>We cannot subtract the eight from the two (Thousandths) so we borrow one of the hundredths from the next column Note; one hundredth is the same as ten thousandths
6. 6. <ul><li>0.420 </li></ul><ul><li>0.228 </li></ul>We cannot subtract the two from the one(hundredths) so we borrow one of the tenths from the next column Note; one tenth is the same as ten hundreths
7. 7. <ul><li>0.420 </li></ul><ul><li>0.228 </li></ul>It now just remains for us to subtract the two tenths from the remaining three tenths Answer; the remaining cloth is 0.192m long | 593 | 1,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-07 | latest | en | 0.755902 |
https://convertoctopus.com/604-decimeters-to-yards | 1,611,225,837,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00304.warc.gz | 269,226,087 | 7,611 | ## Conversion formula
The conversion factor from decimeters to yards is 0.10936132983377, which means that 1 decimeter is equal to 0.10936132983377 yards:
1 dm = 0.10936132983377 yd
To convert 604 decimeters into yards we have to multiply 604 by the conversion factor in order to get the length amount from decimeters to yards. We can also form a simple proportion to calculate the result:
1 dm → 0.10936132983377 yd
604 dm → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 604 dm × 0.10936132983377 yd
L(yd) = 66.054243219598 yd
The final result is:
604 dm → 66.054243219598 yd
We conclude that 604 decimeters is equivalent to 66.054243219598 yards:
604 decimeters = 66.054243219598 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.015139072847682 × 604 decimeters.
Another way is saying that 604 decimeters is equal to 1 ÷ 0.015139072847682 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that six hundred four decimeters is approximately sixty-six point zero five four yards:
604 dm ≅ 66.054 yd
An alternative is also that one yard is approximately zero point zero one five times six hundred four decimeters.
## Conversion table
### decimeters to yards chart
For quick reference purposes, below is the conversion table you can use to convert from decimeters to yards
decimeters (dm) yards (yd)
605 decimeters 66.164 yards
606 decimeters 66.273 yards
607 decimeters 66.382 yards
608 decimeters 66.492 yards
609 decimeters 66.601 yards
610 decimeters 66.71 yards
611 decimeters 66.82 yards
612 decimeters 66.929 yards
613 decimeters 67.038 yards
614 decimeters 67.148 yards | 487 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | latest | en | 0.872037 |
http://www.goodmath.org/blog/2006/12/08/pathological-stack-hell-underload/ | 1,719,086,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00093.warc.gz | 39,327,175 | 21,652 | Our pathological language this week is [Underload][underload]. Underload is, in some ways, similar to Muriel, only it’s a much more sensible language. In fact, there are actually serious
practical languages called *concatenative languages* based on the same idea as Underload: [Joy][joy] and [Factor][factor] are two examples.
[muriel]: http://scienceblogs.com/goodmath/2006/11/friday_pathological_programmin_5.php
[joy]: http://www.latrobe.edu.au/philosophy/phimvt/joy.html
[factor]: http://www.factorcode.org/
Underload is a remarkably simple language. It’s stack based, like Forth, so all of the data is stored on the stack. Its only means of control is constructing a program on the stack and executing it; and the only data that it can manipulate is lists of characters.
* `~` – swap the top two elements on the stack.
* `:` – duplicate the top element on the stack.
* `!` – discard the top element on the stack.
* `*` – concatenate the two elements on top of the stack into a single list.
* `a` – take the element on top of the stack, and wrap it in “()”s.
* `(…)` – push the contents of the parens on the stack as a single stack element.
* `S` – print the element on top of the stack.
* `^` – take the element on top of the stack, and append it to the currently executing program
string.
(Hello world!)S
Simple, right?
Suppose we wanted to add two numbers. The easiest way to handle numbers in Underload is unary format. So suppose want to add 3 + 5. First, we need to put 3 and 5 on the stack. We can represent three as a list `xxx` and 5 as a list `xxxxx`. To push those onto the stack, we need
to wrap them in parens; and t add them, we want to just concatenate the two lists. So the program to add 3 + 5 and print the result is:
(xxx)(xxxxx)*S
As you’d probably expect, an Underload quine is extremely simple:
(:aSS):aSS
I’ll walk through that just to make it clear. First, the list `”:aSS”` is pushed onto the stack, so writing the stack as a list of comma separated elements, the stack is “`[:aSS]`”. Then we execute “:”, which duplicates the top of the stack, leaving “`[:aSS,:aSS]`”. Then we execute “a”, which wraps the element on top of the stack in parens, giving us “`[(:aSS),:aSS]`”. Now there are two “S” commands, which output the two top stack elements; so the out is `(:aSS):aSS`.
A program to generate the fibonacci series is also pretty simple:
(()(*))(~:^:S*a~^a~!~*~:(/)S^):^
It looks like a nighmare, but it’s really not bad. It starts with “()” (0) and “(*)” (1) on the stack. And the rest of the program basically copies the larger number, adds the smaller and larger (giving the next element of the sequence), leaving two fibonacci numbers of the stack. It duplicates the larger, prints it, and then goes on to the next iteration. The body of the program is `(~:^:S*a~^a~!~*~:(/)S^)`, which at the very beginning `(~:^)` duplicates itself.
There’s an online interpreter for [Underload][interp] which does single-stepping. I recommend popping over there, and watching the fibonacci series program execute. It’s much more interesting to watch it in action than it would be to read my detailed description of it!
Now, is this Turing complete? It’s a little hard to see. But the author of Underload took care of that question, by showing how to compile [Unlambda][unlambda] into Underload.
* Unlambda “`s`” translates to Underload “`((:)~*(~)*a(~*(~^)*))`”
* Unlambda “`k`” translates to Underload “`(a(!)~*)`”
* Unlambda “`i`” translates to Underload “`()`”.
* Unlambda “““” translates to Underload “~^”.
* Unlambda “`.x`” translates to Underload “`((x)S)`”.
[unlambda]: http://scienceblogs.com/goodmath/2006/08/friday_pathological_programmin_3.php
0 thoughts on “Pathological Stack Hell: Underload”
1. Oliver
(…) – push the contents of the parents on the stack as a single stack element.
I think that should be “parens”. Amazing how that t makes it so much more confusing.
2. Jonathan Vos Post
I loved my parens, Mom and Dad. Now I’m a paren, also.
There are some steps missing in the Turing Completeness of Underload, via Unlambda.
I hope that you, or one of your active readers, will complete the proof.
And is there a Categorical apprach to that proof, with Currying and so forth?
3. John Armstrong
JVP: of course this whole thing is very categorical. Let X be an object in a category with finite products.
Here’s the category theory interpretations for a few of the operations
~ – twist map from X2 to itself
: – canonical comultiplication from X to X2
! – canonical counit from X to T
* – multiplication from X2 to X
The others I’m not entirely sure about, but a few guesses follow
(…) – defines an arrow from T to X. This is the categorical notion of an “element of X”, familiar from topoi.
a – sends X to hom(T,X) (related to currying?)
S – categorically extraneous
^ – no clue
Of course, if I put more time into it I could probably find a good interpretation for the other terms. Actually, I think this probably manages to be a very close (if terse) parallel to a hypothetical language growing out of Baez’ seminar.
Pathological maybe, but categorically beautiful.
4. Mark C. Chu-Carroll
John:
Thanks for that! It is beautiful in a way, isn’t it? That’s really what I look for in the languages for my friday posts – languages that are bizzare, and yet have a kernel of beauty laying underneath the insanity.
Underload is also *very* fun to program it.
5. David
Hah, what a marvellous language! Now I wish all programming languages involved writing a program that writes dozens of little programs onto the stack that does some foo, then calls the next little program down the line in a lovely cascade. Nest as necessary.
To hell with plodding old single-step iteration! Let recursion disappear up its own bumhole! This is what control flow should be about! It’s not the end result that counts, it’s the getting there!
My magnum opus so far is a program that takes the factorial of however many colons are in the first set of parents. I feel I should share, in case anyone out there has a pressing need for exactly 5060 colons:
(:::::::):(:((^:()~((:)*~^)a~*^!!()~^))~*()~^^)~(^a(*~^)*a~*()~^!()~^)a~**^!!^S | 1,639 | 6,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.88296 |
https://pennylane.readthedocs.io/en/stable/code/api/pennylane.utils.decompose_hamiltonian.html | 1,660,155,967,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00430.warc.gz | 418,650,518 | 9,153 | # qml.utils.decompose_hamiltonian¶
decompose_hamiltonian(H, hide_identity=False, wire_order=None)[source]
Decomposes a Hermitian matrix into a linear combination of Pauli operators.
Parameters
• H (array[complex]) – a Hermitian matrix of dimension $$2^n\times 2^n$$
• hide_identity (bool) – does not include the Identity observable within the tensor products of the decomposition if True
Returns
a list of coefficients and a list of corresponding tensor products of Pauli observables that decompose the Hamiltonian.
Return type
tuple[list[float], list[Observable]]
Example:
We can use this function to compute the Pauli operator decomposition of an arbitrary Hermitian matrix:
>>> A = np.array(
... [[-2, -2+1j, -2, -2], [-2-1j, 0, 0, -1], [-2, 0, -2, -1], [-2, -1, -1, 0]])
>>> coeffs, obs_list = decompose_hamiltonian(A)
>>> coeffs
[-1.0, -1.5, -0.5, -1.0, -1.5, -1.0, -0.5, 1.0, -0.5, -0.5]
We can use the output coefficients and tensor Pauli terms to construct a Hamiltonian:
>>> H = qml.Hamiltonian(coeffs, obs_list)
>>> print(H)
(-1.0) [I0 I1]
+ (-1.5) [X1]
+ (-0.5) [Y1]
+ (-1.0) [Z1]
+ (-1.5) [X0]
+ (-1.0) [X0 X1]
+ (-0.5) [X0 Z1]
+ (1.0) [Y0 Y1]
+ (-0.5) [Z0 X1]
+ (-0.5) [Z0 Y1]
This Hamiltonian can then be used in defining VQE problems using ExpvalCost.
Using PennyLane
Development
API | 459 | 1,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | latest | en | 0.647477 |
https://slack-chats.kotlinlang.org/t/502097/warning-day-2-solution-thread-here-be-spoilers-warning | 1,696,285,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00216.warc.gz | 561,883,815 | 64,909 | n
Nir
12/02/2020, 5:24 PM
⚠️ Day 2 Solution Thread - here be spoilers ⚠️
Copy code
``````package day2
import utils.*
val (rangeStart, rangeEnd, char, password) = split(": ", " ", "-")
assert(char.length == 1)
}
val charPresent = { i: Int -> password.elementAtOrNull(i-1)?.let { it == char} ?: false }
return charPresent(range.first) xor charPresent(range.endInclusive)
}
fun countValid(validator: RuledPassword.() -> Boolean) = (aocDataDir / "day2.txt").useLines { it.count {
}
}
a
12/02/2020, 5:35 PM
Copy code
``````private val input = readInput("input2.txt")
fun main() {
var part1 = 0
var part2 = 0
input.split("\n").forEach { line ->
val (rangeT, lT, password) = line.split(" ")
val l = lT.removeSuffix(":")
val ranges = rangeT.split("-").map { it.toInt() }
val low = ranges[0]
val high = ranges[1]
val num = password.filter { it.toString() == l }.length
if (num in low..high) part1++
if ((password[low - 1].toString() == l) xor (password[high - 1].toString() == l)) part2++
}
println("Part 1: \$part1")
println("Part 2: \$part2")
}``````
j
Joris PZ
12/02/2020, 6:26 PM
May I suggest making the daily solution threads stand out a bit more with some markup? At my company we name them ⚠️ Day N: XYZ - here be spoilers ⚠️ which makes them easy to spot and find
👍 1
n
Nir
12/02/2020, 6:29 PM
Hope you don't mind my blatant copy paste 🙂
👍 1
😄 1
e
ephemient
12/02/2020, 6:41 PM
r
renatomrcosta
12/02/2020, 7:58 PM
https://github.com/renatomrcosta/adventofcode/blob/main/src/main/kotlin/aoc2020/day2.kt I think I’m massively overcomplicating this whole thing
😂 2
n
Nir
12/02/2020, 8:05 PM
i... cannot tell if this is intentional satire or not
j
Joris PZ
12/02/2020, 8:08 PM
You seem to be very focused on accurately naming things, and closely modeling your domain objects to align with 'the business'. Both important in general, but for AoC I'd say the story/domain can be safely ignored in favor of focusing on the algorithms and minimal data structures required for performance. In later puzzles especially you'll find that the enterprisey OO models just don't work well for these types of problems
n
Nir
12/02/2020, 8:16 PM
I still want to know, without being mean, whether it's satire or not... I can totally see it going either way
👆 1
r
renatomrcosta
12/02/2020, 10:07 PM
no, not satire, sorry
Another person accurately described this as “Defensive Programming”, which I guess it is true. I’ll try getting into the idea of streamlining the solutions from here on out. (not that writing this one was an undertaking or anything, but I did go overboard on modelling domain and being prepared for the eventual changes)
n
Nir
12/02/2020, 10:20 PM
Yeah, I mean this reminds me a lot of the very over the top OO Java code from the late 90's
The problem is basically you write your codebase entirely in this ultra defensive style, now you have quadrupled your whole codebase's size and made things very hard to read because of all the indirection. And it comes in useful maybe 10% of the time, or even less.
And, even this so called "defensive" programming doesn't work great because the new code is still fragile to some changing requirements. It's robust against some changes yes but fragile against others, so often when you need to make changes, it won't save you from significant refactorings.
r
renatomrcosta
12/02/2020, 10:30 PM
It did work pretty well for the jump between parts 1 and 2, and it is easy to test in pieces, Participating here made it clearer to me the correct strat for these sort of events / competitions. Dropping the abstractions and following the requirements to the letter probably will make me faster in the next days of the event
j
Joris PZ
12/02/2020, 11:01 PM
For me, a big part of the fun of aoc is to compare all kinds of solutions, try to understand what people's thought processes were, and have some discussion about pros and cons. So just write whatever you like and let them howl 😎
n
Nir
12/02/2020, 11:02 PM
I'm curious mostly for when the algorithms needed will really become non-trivial to get good enough performance
j
Joris PZ
12/02/2020, 11:17 PM
Yeah that's a fair point. I know in previous years my coding style starts getting into performance problems with puzzles where the choosing the proper data structure and algorithm becomes crucial for reaching acceptable performance.
n
Nir
12/02/2020, 11:18 PM
even the first two days for me have been a big reminder for me of why you need mutable data structures
I was looking at using groupBy and things like that and they return Map, but in the next step I'd often have a need for a MutableMap to do something more efficiently
e
ephemient
12/03/2020, 2:31 AM
``.groupByTo(mutableMapOf()) { ... }``
has the same result as
``.groupBy { ... }.toMutableMap()``
but with one fewer intermediate data structure
n
Nir
12/03/2020, 4:12 AM
@ephemient that's still pretty inefficient if you just want the counts of each group. Ultimately just counting occurrences into a MutableMap<T, Int>, it's actually simpler to just write the code yourself than try to use the standard library and wonder if it's efficient
e
ephemient
12/03/2020, 2:38 PM
if you wanted counts, you'd use
``.groupingBy { ... }.eachCount()``
or
``.groupingBy { ... }.eachCountTo(...)``
when I run this with JMH,
Copy code
``````Benchmark Mode Cnt Score Error Units
MapBench.manual sample 12505 0.399 ± 0.003 ms/op
MapBench.stdlib sample 12769 0.391 ± 0.002 ms/op``````
it shows stdlib marginally outperforming the obvious hand-written loop
n
Nir
12/03/2020, 3:35 PM
Eh I'd still prefer the hand written version in this case 🙂
1
4 Views | 1,629 | 5,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.683489 |
https://www.airmilescalculator.com/distance/dtw-to-sfb/ | 1,660,394,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00081.warc.gz | 571,411,709 | 47,285 | # Distance between Detroit, MI (DTW) and Orlando, FL (SFB)
Flight distance from Detroit to Orlando (Detroit Metropolitan Airport – Orlando Sanford International Airport) is 934 miles / 1503 kilometers / 811 nautical miles. Estimated flight time is 2 hours 16 minutes.
Driving distance from Detroit (DTW) to Orlando (SFB) is 1116 miles / 1796 kilometers and travel time by car is about 21 hours 23 minutes.
934
Miles
1503
Kilometers
811
Nautical miles
2 h 16 min
146 kg
## How far is Orlando from Detroit?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 933.784 miles
• 1502.779 kilometers
• 811.436 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 935.768 miles
• 1505.973 kilometers
• 813.160 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Detroit to Orlando?
Estimated flight time from Detroit Metropolitan Airport to Orlando Sanford International Airport is 2 hours 16 minutes.
## What is the time difference between Detroit and Orlando?
There is no time difference between Detroit and Orlando.
## Flight carbon footprint between Detroit Metropolitan Airport (DTW) and Orlando Sanford International Airport (SFB)
On average flying from Detroit to Orlando generates about 146 kg of CO2 per passenger, 146 kilograms is equal to 322 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Detroit to Orlando
Shortest flight path between Detroit Metropolitan Airport (DTW) and Orlando Sanford International Airport (SFB).
## Airport information
Origin Detroit Metropolitan Airport
City: Detroit, MI
Country: United States
IATA Code: DTW
ICAO Code: KDTW
Coordinates: 42°12′44″N, 83°21′12″W
Destination Orlando Sanford International Airport
City: Orlando, FL
Country: United States
IATA Code: SFB
ICAO Code: KSFB
Coordinates: 28°46′39″N, 81°14′15″W | 533 | 2,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.855082 |
http://mathhelpforum.com/advanced-statistics/162062-conditional-probability-please-check-my-soln.html | 1,529,800,307,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865438.16/warc/CC-MAIN-20180623225824-20180624005824-00491.warc.gz | 214,589,035 | 9,796 | 1. ## Conditional Probability [please check my soln]
The number of flowers N appearing on an apple tree, is a random variable with distribution
$\displaystyle \mathbb{P}(N=n)=(1-p)p^n, (n=0,1,2,...)$ for some $\displaystyle p \in (0,1)$. Each flower turns into a fruit with probability $\displaystyle \alpha$ independently of other flowers on the tree. Given that there are $\displaystyle r$ apples on the tree what is the probability that there were $\displaystyle n$ flowers originally on the tree?
Here's my solution (I am not sure, so please correct me if I'm wrong):
Bayes' theorem says:
$\displaystyle \mathbb {P}(A|B) = {\mathbb{P}(A \cap B)}/{\mathbb{P}(B)}$
In this example A is the event: 'originally there were $\displaystyle n$ flowers on the tree'. B is the event: 'there are $\displaystyle r$ apples on the tree.
So in the numerator of the Bayes' formula we get:
$\displaystyle (1-p)p^n \alpha ^r (1 - \alpha)^{n-r}$
and for the denominator we get:
$\displaystyle {\frac {n!}_{(n-r)! r!}} \alpha$
since there are r choose n ways in total and each has probability $\displaystyle \alpha$. Is this correct? Please reply.....
2. Doesn't look correct to me.
let i = number of flowers
r = number of fruits
P(B) = summation from i = r to infinity {P(r|i).P(i)}
P(A.B) = P(B|A).P(A)
P(A) = Probabiltiy there were 'n' flowers - which is straight fwd
P(B|A) = nCr$\displaystyle \alpha ^r (1 - \alpha)^{n-r}$
Try to put these values and simplify | 422 | 1,458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-26 | latest | en | 0.884988 |
http://mymathforum.com/number-theory/345365-inert-maximal-real-subfield-cyclotomic-field.html | 1,566,653,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00142.warc.gz | 131,831,324 | 8,735 | My Math Forum Inert in the Maximal Real Subfield of Cyclotomic Field
Number Theory Number Theory Math Forum
November 24th, 2018, 09:57 PM #1 Newbie Joined: Jul 2018 From: morocco Posts: 26 Thanks: 0 Math Focus: algebraic number theory Inert in the Maximal Real Subfield of Cyclotomic Field Hello Let $Q_n$ be the Maximal Real Subfield of of $\mathbb Q(\zeta_{2^{n+1}})$. for example $Q_2=\mathbb{\sqrt 2}$ is the maximal real subfield of $\zeta_8$. How to show that if a rational prime $p$ inert in $Q_2$ then it is inert in $Q_n$. Thank you
December 9th, 2018, 07:25 PM #2
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Joined: Dec 2018
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Quote:
Originally Posted by Chems Hello Let $Q_n$ be the Maximal Real Subfield of of $\mathbb Q(\zeta_{2^{n+1}})$. for example $Q_2=\mathbb{\sqrt 2}$ is the maximal real subfield of $\zeta_8$. How to show that if a rational prime $p$ inert in $Q_2$ then it is inert in $Q_n$. Thank you
The primes that split in $Q_n$ are those congruent to $±1 \pmod {2^{n+1}}$. If $n>3$, then these primes are also congruent to $±1 \pmod 8$, which are also the primes that split in $Q_2$. The inert primes in $Q_2$ are congruent to $±3 \pmod 8$.
Tags cyclotomic, field, inert, maximal, real, subfield
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In backgammon the doubling cube is utilized to enhance the stakes through the activity. The doubling dice is a comparatively new addition to backgammon but it elevates the game to a fresh degree with regard to tactic. It is vital that you realize the principle and system features connected to the doubling dice mainly because it may very well be your critical to wonderful accomplishment.
Utilizing
You Typically Participate in backgammon in Match play, i.e. the winner will be the player who very first reaches a predetermined variety of details. Each individual match is worthy of one particular place initially of the game, so in a traditional acquire the winner receives just one point.
At first Just about every game is value 1 position. On his change prior to a player rolls the dice he may well decide to provide the doubling dice to your opponent. When the opponent accepts the dice it is turned with range two struggling with up and the opponent takes the dice into possession, indicating that only he can initiate another doubling. But now that the doubling cube has been employed once the game is truly worth two details. Must the doubling dice utilised a next time along with the opponent would accept the sport would now be worth 4 points.
If a player to whom the doubling was supplied doesn’t want to just accept the doubling he can resign. In that scenario the sport is concluded as well as the winner receives 롤듀오 as quite a few details as the sport was truly worth before the doubling was offered.
The doubling cube is a standard die While using the figures 2, four, 8, sixteen, 32 and 64 on it. Every single amount signifies a multiplier, which could alwas be doubled. Hence, In case the doubling dice continues to be applied 4 situations an individual straight get can be value 16 points. Theoretically doubling can go on forever but In fact the doubling doesn’t go beyond four.
Optional doubling dice relevant policies
Beavering is usually accustomed to continue to keep players on their toes when doubling. If a participant beavers, it signifies that he was made available the doubling cube but simply accepting it he re-doubles to another quantity! In addition he also retains control of the doubling cube. So, if the player initiating the doubling misjudged the sport the opponent might grab the specific situation and by beavering make a unpleasant problem for him by beavering and a bit later when he is in obvious lead possibly once more doubling and forcing the opponent to resign.
The Crawford rule has actually been released to limit the use of the doubling cube in essential scenarios. It can be an optional rule but a sensible 1. It states that if one player has come in just a single point of winning the match, the sport that follows is played with no doubling cube. In case the participant who is shedding wins this match the doubling dice is all over again being used. Think about a situation four-three inside of a 5 position match. Without the Crawford rule the getting rid of participant could blindly double on his initially transform becuase he has nothing to unfastened anyway. The Crawford rule makes sure that no Bizarre doubling cube motion is taking place in backgammon.
Scoring with
As mentioned earlier mentioned Just about every match is worth 1 point at first and the value of the sport may possibly enhance Together with the doubling dice. So, if the doubling cube is employed 2 times and selection 4 is struggling with up only one earn will give the winner four factors. On the other hand, If your player wins which has a gammon (truly worth two points), the value of the game is multiplied by two and in a backgammon win it can be multiplied by a few. As an example, the player received with a gammon with the doubling cube displaying 4 he scores four x 2 = 8 factors. | 790 | 3,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.96699 |
http://stats.stackexchange.com/questions/12762/measure-of-spread-of-a-multivariate-normal-distribution/13116 | 1,386,832,257,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164567400/warc/CC-MAIN-20131204134247-00080-ip-10-33-133-15.ec2.internal.warc.gz | 169,551,635 | 14,565 | Measure of spread of a multivariate normal distribution
What is a good measure of spread for a multivariate normal distribution?
I was thinking about using an average of the component standard deviations; perhaps the trace of the covariance matrix divided by the number of dimensions, or a version of that. Is that any good?
Thanks
-
as such, the spread of multivariate gaussian doesn't make sense. However, depending on your needs, there might exists approaches to answer your question. Trace of the matrix is one of the many ways, but you would be ignoring correlations, which may make a huge difference. Eigen values, PCA, etc. might be much better. Therefore, could you please elaborate on your needs? – suncoolsu Jul 7 '11 at 10:41
As such, I want an analog of the standard deviation to a multi-dimensional space. Yes, the trace would ignore the correlations, which is what I fear. Having said that, this does not need to be mathematically exact. Basically, a good indication of spread would be the hypervolume size of the hyperellipse defined by 1 std. deviation from the mean. But a nice, handy formula without deriving the exact volume would be much appreciated. – Kristian D'Amato Jul 7 '11 at 10:48
Seems like PCA could answer your question. – suncoolsu Jul 7 '11 at 10:51
What about the determinant of the sample variance-covariance matrix: a measure of the squared volume enclosed by the matrix within the space of dimension of the measurement vector. Also, an often used scale invariant version of that measure is the determinant of the sample correlation matrix: the volume of the space occupied within the dimensions of the measurement vector.
-
+1 Yes, the determinants are directly related to the "hypervolume...of the ellipse defined by 1 sd from the mean." – whuber Jul 7 '11 at 13:53
So that's the determinant of the covariance matrix, right? – Kristian D'Amato Jul 7 '11 at 14:29
@Kristian The square root of the determinant of the covariance matrix tells you the hypervolume, incorporating both shape (correlation) and size (standard deviation) information. It is the product of the standard deviations of the principal components. The determinant of the correlation matrix is basically a shape factor only, ranging from 0 for degenerate distributions up to 1 when all components are uncorrelated. – whuber Jul 7 '11 at 18:42
I would go with either trace or determinant with a preference towards trace depending on the application. They're both good in that they're invariant to representation and have clear geometric meanings.
I think there is a good argument to be made for Trace over Determinant.
The determinant effectively measures the volume of the uncertainty ellipsoid. If there is any redundancy in your system however then the covariance will be near-singular (the ellipsoid is very thin in one direction) and then the determinant/volume will be near-zero even if there is a lot of uncertainty/spread in the other directions. In a moderate to high-dimensional setting this occurs very frequently
The trace is geometrically the sum of the lengths of the axes and is more robust to this sort of situation. It will have a non-zero value even if some of the directions are certain.
Additionally, the trace is generally much easier to compute.
-
+1 Good points. This gets me thinking: any symmetric function of the $n$ eigenvalues would qualify as "good." All such polynomial functions are polynomials in the $n$ elementary symmetric functions, which include the determinant and the trace. – whuber Jul 15 '11 at 20:53
Yes, the sum (trace) isn't necessarily the best way to go. You're right that you could imagine lots of mixtures here depending on the application. I wonder if there is some standard family of functions that would be good here.... – MRocklin Jul 15 '11 at 20:56
@MR I'm not aware of anybody attempting to use a single statistic to compute the spread of a multivariate normal distribution (except, of course, when independence of all components is assumed). This leads me to believe there may be no such standard family. – whuber Jul 18 '11 at 12:44
$\frac{1}{2} \log |(2\pi e)\Lambda|$
where $\Lambda$ is the covariance matrix. The advantage of this choice is that it can be compared to the "spread" of points under other (e.g., non-Gaussian) distributions. | 969 | 4,324 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2013-48 | latest | en | 0.936846 |
https://infinitylearn.com/surge/maths/class-7/algebraic-expression/counting-terms/ | 1,669,649,975,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00392.warc.gz | 344,098,181 | 13,415 | Counting Terms
# Counting Terms
• Coefficient of a Term
• Counting Terms
• Summary
• What’s Next?
In the previous segment of Class 8 Maths, we learnt what a Term is. In this segment, we will learn how to count terms.
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## What is the Coefficient of a term?
The constant multiplied to a variable or variables in a term is known as the Coefficient of that term.
For example,
• In the term 4x, 4 is the coefficient of x.
• In the term 15?3, 15 is the coefficient of ?3.
• In the term 2xy, 2 is the coefficient of xy.
When there are no constants multiplied to a variable in a term, then the coefficient is considered as 1.
For example,
• In the term xy, the coefficient is 1.
• In the term ?, the coefficient is 1.
?
## How to count terms?
If an expression consists of multiple terms, then these terms will always be separated by a plus (+) or a minus (-) sign.
For example,
• Consider the term 13x + 6.
The terms 13x and 6 are separated by a plus sign. Hence, there are two terms: 13x and 6.
• Consider the term 19?2 + 16? − 3?.
## Related content
Numbers Factors Equations Terms Equations Algebraic Expressions 04 Algebraic Expressions 05A Algebraic Expressions 10 Algebraic Expressions 15 Algebraic Expressions 14
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Verify OTP Code (required) | 366 | 1,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-49 | latest | en | 0.857864 |
http://www.mashpedia.com/Rice_distribution | 1,369,197,211,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701314683/warc/CC-MAIN-20130516104834-00040-ip-10-60-113-184.ec2.internal.warc.gz | 581,066,920 | 22,775 | Parameters Probability density function Cumulative distribution function ν ≥ 0 — distance between the reference point and the center of the bivariate distribution, σ ≥ 0 — scale x ∈ [0, +∞) $\frac{x}{\sigma^2}\exp\left(\frac{-(x^2+\nu^2)} {2\sigma^2}\right)I_0\left(\frac{x\nu}{\sigma^2}\right)$ $1-Q_1\left(\frac{\nu}{\sigma },\frac{x}{\sigma }\right)$ where Q1 is the Marcum Q-function $\sigma \sqrt{\pi/2}\,\,L_{1/2}(-\nu^2/2\sigma^2)$ $2\sigma^2+\nu^2-\frac{\pi\sigma^2}{2}L_{1/2}^2\left(\frac{-\nu^2}{2\sigma^2}\right)$ (complicated) (complicated)
In probability theory, the Rice distribution or Rician distribution is the probability distribution of the magnitude of a circular bivariate normal random variable with potentially non-zero mean. It was named after Stephen O. Rice.
## Characterization
The probability density function is
$f(x\mid\nu,\sigma) = \frac{x}{\sigma^2}\exp\left(\frac{-(x^2+\nu^2)} {2\sigma^2}\right)I_0\left(\frac{x\nu}{\sigma^2}\right),$
where I0(z) is the modified Bessel function of the first kind with order zero.
\begin{align} &\chi_X(t\mid\nu,\sigma) \\ & \quad = \exp \left( -\frac{\nu^2}{2\sigma^2} \right) \left[ \Psi_2 \left( 1; 1, \frac{1}{2}; \frac{\nu^2}{2\sigma^2}, -\frac{1}{2} \sigma^2 t^2 \right) \right. \\[8pt] & \left. {} \qquad + i \sqrt{2} \sigma t \Psi_2 \left( \frac{3}{2}; 1, \frac{3}{2}; \frac{\nu^2}{2\sigma^2}, -\frac{1}{2} \sigma^2 t^2 \right) \right], \end{align}
where $\Psi_2 \left( \alpha; \gamma, \gamma'; x, y \right)$ is one of Horn's confluent hypergeometric functions with two variables and convergent for all finite values of $x$ and $y$. It is given by:[3][4]
$\Psi_2 \left( \alpha; \gamma, \gamma'; x, y \right) = \sum_{n=0}^{\infty}\sum_{m=0}^\infty \frac{(\alpha)_{m+n}}{(\gamma)_m(\gamma')_n} \frac{x^m y^n}{m!n!},$
where
$(x)_n = x(x+1)\cdots(x+n-1) = \frac{\Gamma(x+n)}{\Gamma(x)}$
is the rising factorial.
## Properties
### Moments
The first few raw moments are:
$\mu_1^'= \sigma \sqrt{\pi/2}\,\,L_{1/2}(-\nu^2/2\sigma^2)$
$\mu_2^'= 2\sigma^2+\nu^2\,$
$\mu_3^'= 3\sigma^3\sqrt{\pi/2}\,\,L_{3/2}(-\nu^2/2\sigma^2)$
$\mu_4^'= 8\sigma^4+8\sigma^2\nu^2+\nu^4\,$
$\mu_5^'=15\sigma^5\sqrt{\pi/2}\,\,L_{5/2}(-\nu^2/2\sigma^2)$
$\mu_6^'=48\sigma^6+72\sigma^4\nu^2+18\sigma^2\nu^4+\nu^6\,$
and, in general, the raw moments are given by
$\mu_k^'=\sigma^k2^{k/2}\,\Gamma(1\!+\!k/2)\,L_{k/2}(-\nu^2/2\sigma^2). \,$
Here Lq(x) denotes a Laguerre polynomial:
$L_q(x)=L_q^{(0)}(x)=M(-q,1,x)=\,_1F_1(-q;1;x)$
where $M(a,b,z) = _1F_1(a;b;z)$ is the confluent hypergeometric function of the first kind. When k is even, the raw moments become simple polynomials in σ and ν, as in the examples above.
For the case q = 1/2:
\begin{align} L_{1/2}(x) &=\,_1F_1\left( -\frac{1}{2};1;x\right) \\ &= e^{x/2} \left[\left(1-x\right)I_0\left(\frac{-x}{2}\right) -xI_1\left(\frac{-x}{2}\right) \right]. \end{align}
The second central moment, the variance, is
$\mu_2= 2\sigma^2+\nu^2-(\pi\sigma^2/2)\,L^2_{1/2}(-\nu^2/2\sigma^2) .$
Note that $L^2_{1/2}(\cdot)$ indicates the square of the Laguerre polynomial $L_{1/2}(\cdot)$, not the generalized Laguerre polynomial $L^{(2)}_{1/2}(\cdot).$
## Related distributions
• $R \sim \mathrm{Rice}\left(\nu,\sigma\right)$ has a Rice distribution if $R = \sqrt{X^2 + Y^2}$ where $X \sim N\left(\nu\cos\theta,\sigma^2\right)$ and $Y \sim N\left(\nu \sin\theta,\sigma^2\right)$ are statistically independent normal random variables and $\theta$ is any real number.
• Another case where $R \sim \mathrm{Rice}\left(\nu,\sigma\right)$ comes from the following steps:
1. Generate $P$ having a Poisson distribution with parameter (also mean, for a Poisson) $\lambda = \frac{\nu^2}{2\sigma^2}.$
2. Generate $X$ having a chi-squared distribution with 2P + 2 degrees of freedom.
3. Set $R = \sigma\sqrt{X}.$
• If $R \sim \text{Rice}\left(\nu,1\right)$ then $R^2$ has a noncentral chi-squared distribution with two degrees of freedom and noncentrality parameter $\nu^2$.
• If $R \sim \text{Rice}\left(\nu,1\right)$ then $R$ has a noncentral chi distribution with two degrees of freedom and noncentrality parameter $\nu$.
• If $R \sim \text{Rice}\left(0,\sigma\right)$ then $R \sim \text{Rayleigh}\left(\sigma\right)$, i.e., for the special case of the Rice distribution given by ν = 0, the distribution becomes the Rayleigh distribution, for which the variance is $\mu_2= \frac{4-\pi}{2}\sigma^2$.
• If $R \sim \text{Rice}\left(0,\sigma\right)$ then $R^2$ has an exponential distribution.[5]
## Limiting cases
For large values of the argument, the Laguerre polynomial becomes[6]
$\lim_{x\rightarrow -\infty}L_\nu(x)=\frac{|x|^\nu}{\Gamma(1+\nu)}.$
It is seen that as ν becomes large or σ becomes small the mean becomes ν and the variance becomes σ2.
## Parameter estimation (the Koay inversion technique)
There are three different methods for estimating the parameters of the Rice distribution, (1) method of moments,[7][8][9] (2) method of maximum likelihood,[7][8][9] and (3) method of least squares.[citation needed] In the first two methods the interest is in estimating the parameters of the distribution, ν and σ, from a sample of data. This can be done using the method of moments, e.g., the sample mean and the sample standard deviation. The sample mean is an estimate of μ1' and the sample standard deviation is an estimate of μ21/2.
The following is an efficient method, known as the "Koay inversion technique".[10] for solving the estimating equations, based on the sample mean and the sample standard deviation, simultaneously . This inversion technique is also known as the fixed point formula of SNR. Earlier works[7][11] on the method of moments usually use a root-finding method to solve the problem, which is not efficient.
First, the ratio of the sample mean to the sample standard deviation is defined as r, i.e., $r=\mu^{'}_1/\mu^{1/2}_2$. The fixed point formula of SNR is expressed as
$g(\theta) = \sqrt{ \xi{(\theta)} \left[ 1+r^2\right] - 2},$
where $\theta$ is the ratio of the parameters, i.e., $\theta = \frac{\nu}{\sigma}$, and $\xi{\left(\theta\right)}$ is given by:
$\xi{\left(\theta\right)} = 2 + \theta^2 - \frac{\pi}{8} \exp{(-\theta^2/2)}\left[ (2+\theta^2) I_0 (\theta^2/4) + \theta^2 I_1(\theta^{2}/4)\right]^2,$
where $I_0$ and $I_1$ are modified Bessel functions of the first kind.
Note that $\xi{\left(\theta\right)}$ is a scaling factor of $\sigma$ and is related to $\mu_{2}$ by:
$\mu_2 = \xi{\left(\theta\right)} \sigma^2.\,$
To find the fixed point, $\theta^{*}$, of $g$, an initial solution is selected, ${\theta}_{0}$, that is greater than the lower bound, which is ${\theta}_{\mathrm{lower bound}} = 0$ and occurs when $r = \sqrt{\pi/(4-\pi)}$[10] (Notice that this is the $r=\mu^{'}_1/\mu^{1/2}_2$ of a Rayleigh distribution). This provides a starting point for the iteration, which uses functional composition,[clarification needed] and this continues until $\left|g^{i}\left(\theta_{0}\right)-\theta_{i-1}\right|$ is less than some small positive value. Here, $g^{i}$ denotes the composition of the same function, $g$, $i$-th times. In practice, we associate the final $\theta_{n}$ for some integer $n$ as the fixed point, $\theta^{*}$, i.e., $\theta^{*} = g\left(\theta^{*}\right)$.
Once the fixed point is found, the estimates $\nu$ and $\sigma$ are found through the scaling function, $\xi{\left(\theta\right)}$, as follows:
$\sigma = \frac{\mu^{1/2}_2}{\sqrt{\xi\left(\theta^{*}\right)}},$
and
$\nu = \sqrt{\left( \mu^{'~2}_1 + \left(\xi\left(\theta^{*}\right) - 2\right)\sigma^2 \right)}.$
To speed up the iteration even more, one can use the Newton's method of root-finding.[10] This particular approach is highly efficient. | 2,656 | 7,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 78, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2013-20 | latest | en | 0.544569 |
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IMPORTANT CONCEPTS Banker’s Discount: Suppose a merchant A buys goods worth, say Rs. 10,000 from another merchant B at a credit of say 5 months.…
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Stock Capital:The total amount of money needed to run the company is called the stock capital. Shares or Stock:The whole capital is divided into small units,…
## Tips and Tricks to simplify algebra problems
(a + b)(a – b) = (a2 – b2) (a + b)2 = (a2 + b2 + 2ab) (a – b)2 = (a2 + b2 – 2ab) (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (a3 + b3) = (a + b)(a2 – ab + b2) (a3 – b3) = (a – b)(a2 + ab + b2) (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac) When a + b + c =…
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Solving systems of equations in two variables A system of a linear equation comprises two or more equations and one seeks a common solution to…
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Downstream/Upstream:In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream. If the speed of a boat in still…
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## बोर्ड परीक्षा के लिए महत्वपूर्ण नोट्स – कार्बन एवं उसके यौगिक
बोर्ड परीक्षा में अच्छे अंक लाने के लिए पाठ को अच्छे से समझना बहुत ज़रूरी है | इस article में कार्बन एवं उसके यौगिक पाठ…
## बोर्ड परीक्षा के लिए महत्वपूर्ण नोट्स – धातु एवं अधातु
बोर्ड परीक्षा में अच्छे अंक लाने के लिए पाठ को अच्छे से समझना बहुत ज़रूरी है | इस article में धातु एवं अधातु पाठ से… | 712 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-31 | latest | en | 0.498655 |
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Unformatted text preview: Muraj, Hamza Homework 27 Due: Apr 5 2006, 4:00 am Inst: Florin 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 0 points A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass of 84 g and a length 5 cm. A piece of clay with mass 34 g and velocity 1 . 7 m / s hits the very top of the rod, gets stuck and causes the clay- rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is 1 . 7 m / s O 34 g 84 g (a) (b) (c) 5cm Figure: The piece of clay and rod: (a) before they collide, (b) at the time of the collision, and (c) after they collide. After the collisions the clay-rod system has an angular velocity about the pivot. With respect to the pivot point O , what is the magnitude of the initial angular momen- tum L i of the clay-rod system? Correct answer: 0 . 1445 kg m 2 / s. Explanation: Let : m putty = m = 34 g = 0 . 034 kg , m rod = 2 m = 84 g = 0 . 084 kg , = 5 cm = 0 . 05 m , and v = 1 . 7 m / s . Basic Concepts: Conservation of angular momentum ~ L ~ L = ~r ~p = m~r ~v X ~ ext = d ~ L dt L z = I . Therefore, if the net external torque acting on a system is zero, the total angular momentum of that system is constant. Since the total external torque acting on the clay-rod system is zero, the total angular momentum is a constant of motion. The total initial angular momentum L i is simply the angular momentum of the clay, since the rod is at rest initially L i = k ~r ~p k = m r v = m v 2 (1) = (0 . 034 kg)(1 . 7 m / s)(0 . 05 m) 2 = . 1445 kg m 2 / s . 002 (part 2 of 3) 0 points With respect to the pivot point O , what is the final moment of inertia I f of the clay-rod system? Correct answer: 8 . 75 10- 5 kg m 2 . Explanation: The final moment of inertia I f of the clay- rod system is the moment of inertia of the rod plus the moment of inertia of the clay I f = I rod + I clay = 1 12 2 m 2 + m 2 2 = 5 12 m 2 (2) = 5 12 (0 . 034 kg)(0 . 05 m) 2 = 8 . 75 10- 5 kg m 2 ....
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hw27 - Muraj, Hamza Homework 27 Due: Apr 5 2006, 4:00 am...
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Homework #2 - Part I
Homework #2 - Part I - LAST NAME FIRST NAME Physics 107...
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LAST NAME: __________________________ FIRST NAME: __________________________ Physics 107 – Homework #2 Due: January 30, 2007 Remember: this is the 3 rd edition of PHYSICS, Volume II, by James S. Walker Assigned End of Chapter Problems: 19.52 19.55 20.13 20.15 20.17 20.20 20.31 20.34 20.38 20.39 PLEASE NOTE: Additional multiple-choice (MCAT-style) problems are on the next page. This is part of your homework assignment, so please be sure you do them. Please be aware that there is no partial credit for wrong answers in the case of multiple-choice problems. We recommend you do carefully put any needed work on the separate paper you use for your end-of-chapter problems above, so you can later recall what you did
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when reviewing for exams as well as to help yourself in understanding what went wrong if your answer is incorrect.
Multiple Choice Problems: Please circle the correct answer below. 1. A uniform electric field with a magnitude of 8 x 10 6 N/C is applied to a cube of edge length 0.1 m as
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# Given the following information: Percent of capital structure: Debt 25 % Preferred stock 15 Common...
Given the following information:
Percent of capital structure:
Debt 25 % Preferred stock 15 Common equity 60
Bond coupon rate 9% Bond yield to maturity 7% Dividend, expected common \$ 3.00 Dividend, preferred \$ 10.00 Price, common \$ 50.00 Price, preferred \$ 116.00 Flotation cost, preferred \$ 8.50 Growth rate 6% Corporate tax rate 30%
Calculate the Hamilton Corp.'s weighted cost of each source of capital and the weighted average cost of capital. (Do not round intermediate calculations. Input your answers as a percent rounded to 2 decimal places.)
Debt:
cost of debt=7% --- yield to maturity
after tax cost of debt =cost of debt*(1-tax rate)=7*(1-.3)=4.9%
after tax weighted cost of debt= percentage of capital*after tax cost of debt
=.25*4.9=1.225%
Common Equity:
cost of equity = D1/P(1-F) + g
where D1=dividend expected ; P=Price of share ; F=Floatation cost ; g=growth rate
cost of equity= 3/50 + .06 =12%
weighted cost of equity =.6*12=7.2%
Preferred equity:
cost of preferred equity = D1/P(1-F)
where D1=dividend ; P=Price of share ; F=Floatation cost ;
cost of preferred equity= 10/(116*(1-.085)) =9.42%
weighted cost of preferred share= .15*9.42% =1.413%
WACC= sum of (weighted cost of capitals)
=1.225+7.2+1.413= 9.838%
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# Convert decimal 160 to percent
Convert from percent to decimal. Here is the answer to the question: Convert decimal 160 to percent or how to convert 160% into a decimal equivalent. Use the percent to decimal calculator below to write any percent as a decimal.
### Percent to Decimal Calculator
Enter a percent value:
% Ex.: 0.7, 3/4, 2, 10 ,33, 125, etc.
Decimal result
## How to convert from percent to decimal
Let's see this example:
• Percent means 'per 100'. So, 160% means 160 per 100 or simply 160/100.
• If you divide 160 by 100, you get 1.6 (a decimal number).
• So, to convert from percent to decimal, simply divide by 100 and remove the '%' sign.
• ➥ There is a easy way to convert from percent to decimal:
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# Electric Potential Energy Calculator
Potential energy in common is defined as the capacity of work done by external forces. In electrical terms, the energy stored in the circuits is referred to as electric potential energy or voltage. The stored energy is transferred to other mediums in the forms like heat, light, or motion. The electric potential (voltage) energy at any point in space at a distance of r from a single charge can be calculated using the formula V = kQ/r. Use this electric potential energy calculator to calculate electric potential energy (Voltage) at a point of distance.
## Calculate Voltage at a Point of Distance
q
F/m
Potential energy in common is defined as the capacity of work done by external forces. In electrical terms, the energy stored in the circuits is referred to as electric potential energy or voltage. The stored energy is transferred to other mediums in the forms like heat, light, or motion. The electric potential (voltage) energy at any point in space at a distance of r from a single charge can be calculated using the formula V = kQ/r. Use this electric potential energy calculator to calculate electric potential energy (Voltage) at a point of distance.
#### Formula:
V(r) = kQ / r (i.e) k = 1 / (4 x π x ϵ0) Where, V(r) = Electric Potential Q = Charge k = Scalar Quantity ϵ0 = Permittivity r = Point Distance
### Example:
A system with charge 3q at a point distant of 7. What is its electric potential energy?
#### Solution:
To find PE, we need to calculate scalar quantity, k
k = 1 / (4 x 3.14 x 8.85418e-12)
= 8987559722.746508
V(r) = (8987559722.746508 x 3) / 7
= 3851811309.7485
Similarly, for multiple charges also it can be calculated easily just by adding the electric potential energy at a point from different charges. | 425 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-14 | latest | en | 0.913608 |
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### Inequality of Entire functions
You have$$\cos\left(\frac\pi2\right)=1\quad\text{and}\quad\int_\gamma\cos(z)\,\mathrm dz=0$$if $\gamma\colon\left[-\frac\pi2,\frac\pi2\right]\longrightarrow\Bbb C$ is the path defined by $\gamma(t)=t$....
### Derivative of the complex power function (by definition)
If you are allowed to use the chain rule, one gets (at least in the real case): \begin{align*} f(x) := x^{\alpha} = \exp\left(\alpha\ln(x)\right) \Rightarrow f'(x) = \frac{\alpha\exp\left(\alpha\ln(x)\...
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1 vote
### Evaluate $\int_0^{2 \pi} e^{\sin(e^{i \theta})} \hspace{0.1cm} d \theta$
There is a symmetry around $\theta=\pi$. The real part is even and the imaginary part is odd. So, just remain the integration of the real part tp get $2\pi$ as @Bertrand87 showed in his good answer.
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1 vote
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1 vote
Yes the limit exists and it is 0, we can even set aside the branch cut ambiguity since this one doesn’t depend on that. If you write $z = re^{i\theta}$ and we say $f(z)$ is some branch cut of $\sqrt{... • 2,009 1 vote ### Prove that$\int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta = \frac{\pi}{(p+q-1)2^{p+q-1}B(p,q)}\$
One solution using mostly real methods: \begin{eqnarray*} \int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta &=& \frac{1}{2^{p+q-1}}\Re \int_0^{\pi/2} \left(e^{i\theta} + e^{-i\...
• 2,311
Only top scored, non community-wiki answers of a minimum length are eligible | 529 | 1,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.699807 |
https://reason.town/linear-machine-learning-models/ | 1,701,249,634,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00513.warc.gz | 569,951,519 | 27,417 | # How Linear Machine Learning Models Work
Linear machine learning models are a subset of supervised learning, which is a branch of machine learning where algorithms learn from labeled training data. Linear models are used for regression and classification tasks.
Checkout this video:
## Introduction
Linear models are a very popular and widely used type of machine learning model. In this article, we’ll take a look at how linear models work and some of the key ways they’re used in machine learning.
linear models make predictions by combining a set of weights with a set of input features. The weights determine how much each feature contributes to the final prediction. For example, if we have a linear model that predicts the price of a house based on its size, the weight for size would determine how much the model’s predictions increase when the size of the house increases.
Linear models can be used for regression, which is when we want to predict a continuous value like a price or temperature, or classification, which is when we want to predict which category an item belongs to (e.g. whether an email is spam or not).
There are many different types of linear model, but some of the most popular are logistic regression, linear regression, and support vector machines.
## Linear Regression
Linear regression is a type of machine learning algorithm that is used to predict continuous values. Continuous values are values that can take any value within a certain range, such as price, weight, or height. Linear regression algorithms find the line of best fit for a dataset and use that line to make predictions.
To understand how linear regression works, it is helpful to think about a simple example. Suppose you want to predict the price of a house based on its size. You could use a linear regression algorithm to find the line of best fit for a dataset of house sizes and prices. The line of best fit would be used to make predictions about the price of a house based on its size.
linear regression is a powerful tool that can be used to make predictions about continuous values. However, it is important to remember that linear regression algorithms only find the line of best fit for a dataset. They cannot always accurately predict the value of a data point that lies outside of the range of data points in the dataset.
## Logistic Regression
Logistic regression is a type of linear machine learning model that is used to predict a binary outcome (yes/no, 1/0, True/False). In other words, it can be used to answer the question: ” given some input data, what is the probability that the output will be 1?”
The way logistic regression works is by taking the input data and multiplying it by a set of weights. These weights are then used to calculate a probability (between 0 and 1) that the output will be 1. The final Prediction is then made by thresholding the probability – if it is above a certain threshold (usually 0.5), then the Prediction is 1, otherwise it is 0.
Logistic regression is a powerful tool that can be used for both classification and prediction tasks. It is interprettable, easy to use, and widely available in many software packages.
## Support Vector Machines
Support Vector Machines (SVMs) are a type of linear machine learning model. A linear model makes predictions by combining a set of input variables using a linear function. SVMs are a type of linear model that is particularly well suited for data that is not linearly separable. This means that the data cannot be divided into two groups by drawing a line (or hyperplane in higher dimensions).
SVMs find the best possible line (or hyperplane) to separate the two groups of data points. This line (or hyperplane) is called the decision boundary. The decision boundary is chosen so that it maximizes the margin between the two groups of points. The margin is the distance between the decision boundary and the closest data point from each group. SVMs are said to have good generalization performance because they try to find a decision boundary that is as large as possible.
The decision boundary learned by an SVM can be nonlinear if the data is not linearly separable. To do this, SVMs use something called the kernel trick. The kernel trick transforms the input data so that it becomes linearly separable. This transformation is done automatically by the SVM algorithm and happens behind the scenes.
There are many different types of kernels that can be used with SVMs, but some of the most popular include:
-Linear kernel: Does not transform the input data and therefore can only be used with linearly separable data.
-Polynomial kernel: Can be used with nonlinear data but requires more computational power than other kernels.
-Radial basis function (RBF) kernel: Also known as the Gaussian kernel, this is commonly used in SVMS because it has good generalization performance and is not computationally expensive.
## Naive Bayes
Naive Bayes is a linear machine learning model that is often used for text classification tasks. It is called “naive” because it makes the assumption that all of the features in the data are independent of each other, which is not always true. However, this assumption simplifies the model and makes it easier to train.
To understand how Naive Bayes works, we need to understand two concepts: probability and conditional probability. Probability is a way of quantifying how likely something is to happen. For example, if there is a 20% chance of rain tomorrow, that means that there is a 20% chance that it will rain and an 80% chance that it will not rain.
Conditional probability is a way of quantifying how likely something is to happen given that something else has already happened. For example, if we know that it has rained today, then the conditional probability of it raining tomorrow increases. This is because the conditions for rain (humidity, temperature, etc.) are more likely to be met if it has already rained recently.
Naive Bayes works by using probability and conditional probability to calculate the probability that an input belongs to each class (e.g., spam or not spam). It then predicts the class with the highest probability.
Let’s say we have a dataset with two features: x1 and x2. We want to use Naive Bayes to classify this data into two classes: class 1 and class 2. We’ll start by calculating the prior probabilities for each class, which are just the probabilities of each class without taking any features into account:
P(class 1) = 0.5
P(class 2) = 0.5
Now we need to calculate the conditional probabilities for each feature given each class. For example, P(x1|class 1) is the probability of x1 being true given that the data belongs to class 1:
P(x1|class 1) = 0.8
P(x1|class 2) = 0.6
## Decision Trees
Decision trees are a supervised learning algorithm used for both, classification and regression tasks where we will try to predict a target variable based on several input variables. The idea behind a decision tree is to create a model that predicts the value of a target variable by learning simple decision rules inferred from the data features.
To build a decision tree, we need to follow certain steps:
– Calculate the entropy of the target variable.
– Split the dataset into two parts based on the entropy calculation i.e, one with all the values less than the threshold and other with values greater than or equal to the threshold.
– Repeat step 1 and 2 on both the split datasets.
– Stop when you reach a stage where there is no further improvement in entropy or we have reached a leaf node.
## Ensemble Methods
Ensemble methods are a type of machine learning model that combines the predictions of multiple models. Ensemble methods can be used to improve the accuracy of linear machine learning models, and they are often used in competition to improve the performance of a model on a specific dataset.
## Neural Networks
Neural networks are a type of machine learning algorithm that are used to model complex patterns in data. Neural networks are similar to other machine learning algorithms, but they are composed of a large number of interconnected processing nodes, or neurons, that can learn to recognize patterns of input data.
Neural networks are often used for tasks such as image recognition and classification, natural language processing, and time series prediction.
## Conclusion
In conclusion, linear machine learning models are powerful tools that can be used to make predictions based on data. By understanding how these models work, you can more effectively use them to solve problems and make decisions.
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# Matt Eicholtz
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1 – 50 of 1,232 | 1,084 | 4,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-36 | longest | en | 0.783382 |
http://www.chestnut.com/en/eg/6/mathematics/1/4/examples/ | 1,477,273,045,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719463.40/warc/CC-MAIN-20161020183839-00430-ip-10-171-6-4.ec2.internal.warc.gz | 369,743,906 | 28,653 | # 6.1.4. Ratio between Three Numbers
Ahmed, Mohammed, and Alaa saved 162 LE, 144 LE, and 108 LE respectively. Find the ratio of Ahmed's savings to those of Mohammed to those of Alaa.
• A
• B
• C
• D
Show Solution
Khaled, Ali, and Abdullah saved 216 LE, 120 LE, and 144 LE respectively. Find the ratio of Khaled's savings to those of Ali to those of Abdullah.
• A
• B
• C
• D
Show Solution
A man has three sticks. If the length of the first stick is 2.4 m, that of the second is 2.1 m, and that of the third is 1.5 m, find the ratio between their lengths, respectively.
• A
• B
• C
• D
Show Solution
Determine the ratio between the three numbers , , and given that and .
• A
• B
• C
• D
Show Solution
Determine the ratio between the three numbers , , and given that and .
• A
• B
• C
• D
Show Solution
A girl bought some fruits from the market, where the ratio of the weight of guavas to that of bananas she bought is , and the ratio of the weight of bananas to that of apples is . Find the ratio between the weights of guavas, bananas, and apples.
• A
• B
• C
• D
Show Solution
A girl bought some fruits from the market, where the ratio of the weight of grapes to that of bananas she bought is , and the ratio of the weight of bananas to that of oranges is . Find the ratio between the weights of grapes, bananas, and oranges.
• A
• B
• C
• D
Show Solution
The ratio between the heights of three buildings A, B, and C is . If the height of building A is 60 meters, find the heights of building B and building C.
• A height of Building B m and height of Building C m
• B height of Building B m and height of Building C m
• C height of Building B m and height of Building C m
• D height of Building B m and height of Building C m
Show Solution
The ratio between the heights of three buildings A, B, and C is . If the height of building A is 36 meters, find the heights of building B and building C.
• A height of Building B m and height of Building C m
• B height of Building B m and height of Building C m
• C height of Building B m and height of Building C m
• D height of Building B m and height of Building C m
Show Solution
The ratio between the ages of Ola, Asmaa, and Heba is . If the difference between the ages of Ola and Asmaa is 10 years, determine the Heba's age.
Show Solution
In order to prepare 64 meals, the owner of a restaurant uses 32 kg of meat. How much meat is needed to prepare 8 meals?
• A8 kg
• B16 kg
• C4 kg
• D kg
Show Solution
The ratio between the side lengths of a triangle is . If the perimeter of the triangle equals 105 cm, find the length of its longest side.
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The ratio between the side lengths of a triangle is . If the perimeter of the triangle equals 69 cm, find the length of its longest side.
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The ratio of the height of Khaled to that of Hani is and the ratio of the height of Hani to that of Ahmed is . Find the ratio of the height of Khaled to that of Ahmed.
• A
• B
• C
• D
Show Solution
The ratio of the height of Yasser to that of Hani is and the ratio of the height of Hani to that of Ayman is . Find the ratio of the height of Yasser to that of Ayman.
• A
• B
• C
• D
Show Solution
Put the ratio in its simplest form.
• A
• B
• C
• D
Show Solution
Put the ratio in its simplest form.
• A
• B
• C
• D
Show Solution
A man bought a television, an oven, and a fridge, where the ratio between the prices of the three appliances is respectively. If the television costs LE, determine the prices of each of the oven and the fridge.
• Athe price of the oven LE, the price of the fridge LE
• Bthe price of the oven LE, the price of the fridge LE
• Cthe price of the oven LE, the price of the fridge LE
• Dthe price of the oven LE, the price of the fridge LE
Show Solution
If Hany's share : Samy's share : Tarek's share is , where Hany's share is LE, calculate the share of each of Samy and Tarek.
• ALE, LE
• BLE, LE
• CLE, LE
• DLE, LE
Show Solution
The ratio between the weights of 3 girls of different ages is . If the weight of the first girl is kg, find the weights of the 2 other girls.
• Akg, kg
• Bkg, kg
• Ckg, kg
• Dkg, kg
Show Solution
Given that the ratio between the weights of three containers is , and the weight of the second container is kg, determine the weights of the first and the third containers respectively.
• Akg, kg
• Bkg, kg
• Ckg, kg
• Dkg, kg
Show Solution
If the ratio of three numbers is , where their sum is 69, find the value of each number.
• A21, 47, 49
• B12, 27, 30
• C14, 44, 53
• D19, 32, 36
Show Solution
The ratio of the money that Hoda has to what Ahmed has to what Samah has is , find how much money each of them has if Hoda's money exceeds what Samah has by LE.
• ALE, LE, LE
• BLE, LE, LE
• CLE, LE, LE
• DLE, LE, LE
Show Solution
The ratio among the production of three factories is . If the production of the third factory exceeds that of the first one by 924 tonnes, find the production of each factory.
• A429 tonnes, 264 tonnes, 660 tonnes
• B353 tonnes, 163 tonnes, 407 tonnes
• C
6 106
tonnes,
2 872
tonnes,
7 030
tonnes
• D
6 006
tonnes,
2 772
tonnes,
6 930
tonnes
Show Solution
Given that the ratio between the productions of three factories of TVs is , and the sum of the productions of the first and second factories is 51 000 sets, find the production of each one.
• A
34 459
,
28 333
,
22 666
• B
31 875
,
19 125
,
15 300
• C
32 875
,
20 125
,
16 300
• D
24 519
,
14 711
,
11 769
Show Solution
The ratio among the measures of the angles of a triangle is . Find the measure of the greatest angle of this triangle.
• A
• B
• C
• D
Show Solution | 1,609 | 5,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-44 | latest | en | 0.932183 |
https://www.cas-ino.co.uk/2023/10/13/how-many-black-numbers-in-roulette/ | 1,701,294,925,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100146.5/warc/CC-MAIN-20231129204528-20231129234528-00042.warc.gz | 801,993,703 | 25,306 | # How Many Black Numbers In Roulette?
Home » How Many Black Numbers In Roulette?
Picture this: you’re at a casino, feeling the adrenaline rush as the roulette wheel spins in anticipation. You’ve probably noticed that the numbers on the wheel alternate between red and black. But have you ever wondered just how many black numbers there are in roulette? Well, wonder no more! In this article, we’ll dive into the exciting world of roulette and answer the burning question: “How Many Black Numbers in Roulette?”
Roulette, with its iconic spinning wheel and suspenseful gameplay, has been captivating gamblers for centuries. And one of the key elements that adds to the thrill is the division of numbers into red and black. It’s like a battle between two contrasting colors, ready to determine your fate with every spin. But exactly how many numbers fall into the black category? Stay with us to find out!
Whether you’re a seasoned roulette enthusiast or a curious novice, understanding the distribution of black numbers is crucial. You might be surprised to learn that there are… Well, let’s leave the suspense for a moment and explore this fascinating topic further. Get ready to unravel the secrets behind the wheel and discover just how many black numbers you’ll encounter in the exhilarating game of roulette! It’s time to take a spin and find out!
## How Many Black Numbers in Roulette? Understanding the Odds
Roulette is a popular casino game that has captivated players for centuries. It is a game of chance and suspense, where players place bets on where they think a ball will land on a spinning wheel. One of the intriguing aspects of roulette is the division of numbers into red and black. In this article, we will explore how many black numbers are in roulette and delve into the odds and probabilities associated with them.
### The Mechanics of a Roulette Wheel
Before we dive into the specifics of black numbers, it’s important to understand the mechanics of a roulette wheel. A standard roulette wheel consists of 36 numbered pockets, alternating between red and black, with an additional pocket or two for the green-colored zero(s). The layout of the wheel may vary depending on the specific type of roulette being played, such as American or European roulette.
In European roulette, which is the most commonly played version, there are 18 black numbers and 18 red numbers. The remaining pockets are reserved for the green zero(s). This means that in European roulette, the odds of the ball landing on a black number are 18 to 37 or approximately 48.6%. In American roulette, the odds are slightly different due to the presence of an additional double zero pocket, resulting in 18 black numbers and 18 red numbers, along with two green zero pockets.
### Understanding the Odds and Probabilities
Now that we know the basics of the roulette wheel, let’s explore the odds and probabilities associated with black numbers. As mentioned earlier, in European roulette, there are 18 black numbers out of a total of 37 pockets. This gives us a probability of approximately 48.6% for the ball to land on a black number. Similarly, in American roulette, with 18 black numbers out of 38 pockets, the probability decreases slightly to approximately 47.4%.
It’s important to note that while the odds of the ball landing on a black number are less than 50% due to the presence of the green zero(s), roulette is still considered a fair game. The house edge, which represents the advantage the casino has over the players, is primarily derived from the presence of the green zero(s). The specific layout of the roulette wheel determines the house edge.
### Strategies for Betting on Black Numbers
Now that we have a better understanding of the odds and probabilities associated with black numbers in roulette, let’s explore some strategies for betting on black. Remember, roulette is a game of chance, and no strategy can guarantee consistent winnings. However, some players prefer to use certain betting systems to enhance their enjoyment and potentially improve their chances of winning.
One popular betting strategy is the Martingale system, where players double their bet after each loss and revert to the initial bet after a win. This strategy relies on the belief that a winning streak is bound to happen eventually, and by doubling the bets, the player can recover their losses. However, it’s important to approach any strategy with caution and set limits to avoid excessive losses.
Another strategy is the Fibonacci system, based on the Fibonacci sequence where each number is the sum of the two preceding ones. In this system, players increase their bet according to the sequence after each loss and revert to the previous two bets after a win. This strategy aims to limit losses while allowing for potential gains during winning streaks.
## Different Types of Bets in Roulette
In addition to understanding the number of black numbers in roulette, it’s important to be familiar with the different types of bets available. Here are some common bets you can place on a roulette table:
### 1. Inside Bets
Inside bets refer to bets placed on specific numbers or a small group of numbers. These bets have higher odds but offer greater payouts. Examples of inside bets include:
– Straight Bet: Betting on a single number.
– Split Bet: Betting on two adjacent numbers.
– Street Bet: Betting on a row of three numbers.
– Corner Bet: Betting on four numbers that form a square on the layout.
### 2. Outside Bets
Outside bets are placed on larger groups of numbers, offering better odds but smaller payouts. Examples of outside bets include:
– Red/Black Bet: Betting on whether the ball will land on a red or black number.
– Even/Odd Bet: Betting on whether the ball will land on an even or odd number.
– High/Low Bet: Betting on whether the ball will land on a high (19-36) or low (1-18) number.
– Dozens Bet: Betting on whether the ball will land on one of the three dozens (1-12, 13-24, 25-36).
### 3. Combination Bets
Combination bets involve placing multiple bets at once to cover various numbers on the layout. These bets provide a combination of higher and lower odds and payouts. Examples of combination bets include:
– Basket Bet: Betting on 0, 1, 2, and 3 (available in American roulette).
– Six Line Bet: Betting on two adjacent rows of numbers.
– Column Bet: Betting on one of the three columns of numbers.
In conclusion, understanding the number of black numbers in roulette is essential in strategizing and placing bets. In European roulette, there are 18 black numbers, while in American roulette, this number remains the same. By familiarizing yourself with the odds and probabilities, exploring different betting strategies, and being aware of the various types of bets, you can enhance your roulette experience and make informed choices at the roulette table.
## Key Takeaways: How Many Black Numbers in Roulette?
• In the game of roulette, there are 18 black numbers out of a total of 38 numbers.
• Black numbers on a roulette wheel include 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, and 35.
• The remaining 20 numbers are either red or green on the roulette wheel.
• The black numbers and red numbers alternate on the roulette wheel.
• Understanding the distribution of black and red numbers can help players plan their betting strategies.
Roulette is a popular casino game that involves spinning a wheel with numbered compartments. Black and red numbers are alternately placed around the wheel, along with a green zero or double zero. If you’re wondering about the number of black numbers in roulette, here are some frequently asked questions that will help clarify things.
### 1. Are there more black or red numbers in roulette?
In European roulette, there are 18 red numbers and 18 black numbers, along with one green zero. So, there is an equal number of black and red numbers. However, in American roulette, there is an additional double zero, making a total of 38 numbered compartments. In this case, there are still 18 red numbers, but now there are 18 black numbers as well, along with two green zeroes. So, in American roulette, there are more black numbers than red numbers.
### 2. How many black numbers are there on a roulette wheel?
On a standard European roulette wheel, there are 18 black numbers. These numbers are distributed across the wheel and include: 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, and 35. The remaining numbers are red, except for the green zero. In American roulette, with the additional double zero, there are also 18 black numbers, but they are different from the ones in European roulette.
### 3. How are black numbers determined in roulette?
The colors of the numbers on a roulette wheel are determined to provide an equal distribution between black and red, with the addition of the green zero(s) to give the house an edge. The precise placement of black and red numbers on the wheel is part of the game’s design, aiming to ensure fairness and randomness. Roulette wheels are carefully manufactured to meet specific standards, and the position of each number is meticulously calibrated to ensure a balanced outcome in terms of color distribution.
### 4. Can you bet on black or red numbers in roulette?
Absolutely! Betting on black or red numbers in roulette is one of the most popular and straightforward bets you can make. If you place your chips on the “black” section of the betting layout, you are betting that the next winning number will be black. Likewise, if you place your chips on the “red” section, you are betting on the next winning number being red. These bets offer a 1:1 payout, meaning you will win an equal amount to your bet if you predict the correct color.
### 5. Are black numbers more likely to appear in roulette?
No, black numbers are not more likely to appear in roulette, nor are red numbers. The outcome of each spin is entirely random and independent of previous spins. The odds of a black or red number appearing on the next spin are always the same, assuming a fair and properly functioning roulette wheel. While there are strategies and betting systems that may attempt to predict or exploit patterns, ultimately, roulette is a game of chance where each number has an equal probability of being the result of the next spin.
## Summary
So, here’s a quick recap: In roulette, there are 18 black numbers out of 38 total numbers on the wheel. This means that the chances of landing on a black number are almost 50%. It’s important to remember that each spin is independent, so past outcomes don’t affect future ones. Ultimately, roulette is a game of luck, and whether you bet on black or any other color, it’s all about having fun and enjoying the excitement! | 2,292 | 10,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | longest | en | 0.902114 |
https://fr.slideserve.com/search/junior-physics-ppt-presentation | 1,638,957,845,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00194.warc.gz | 331,811,453 | 6,651 | # 'Junior physics' diaporamas de présentation
## Fig. 2
Propagating Ultrasound in Oscillating Quantum Solid 4 He Harry Kojima, Rutgers University New Brunswick, DMR 1005325.
(136 views)
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## Sound Junior Science: Physics
Sound Junior Science: Physics. Revised Science Syllabus Extract. Learning Outcomes for 3B5 and 3B 6. On completion of this section, students should be able to. OP40 show that sound is a form of energy, and understand that sound is produced by vibrations .
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## Tyler Junior College Physics 1405 Elementary Physics
Tyler Junior College Physics 1405 Elementary Physics. Section Cero SCIENTIFIC LITERACY. Important Concepts. Measurements & Units Your Need for Scientific Literacy Science & Pseudoscience Scientific Reasoning A Brief History of Western Science. PHYSICS. Physics . . . .
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## Junior
Junior. Program Enrollment Form Grades 7 & 8. What is Junior School Counts ! ?. For many middle school students, making the connection between what they learn in the classroom and what employers expect on the job can seem like an impossible task.
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Fa bulously British. Junior.
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## Junior til Junior Konceptet
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Junior. Fabulously British.
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## Junior Health
Junior Health. Sexually transmitted diseases. Do now. Daily recording calendar Think about it: How many sexually active young people will get an STD by the age of 25? How many people catch an STD everyday in America?. How do stds spread?.
By enye (120 views) | 745 | 3,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.719267 |
https://www.tutorialspoint.com/finding-even-and-odd-numbers-in-a-set-of-elements-dynamically-using-c-language | 1,638,548,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00596.warc.gz | 1,125,803,989 | 11,207 | # Finding even and odd numbers in a set of elements dynamically using C language
CServer Side ProgrammingProgramming
## Problem
To compute sum of even numbers and odd numbers in a set of elements using dynamic memory allocation functions.
## Solution
In this program, we are trying to find even and odd numbers in a set of numbers.
The logic used to find even numbers in a set elements is given below −
for(i=0;i<n;i++){
if(*(p+i)%2==0) {//checking whether no is even or not
even=even+*(p+i); //calculating sum of even all even numbers in a list
}
}
The logic used to find odd numbers in a set elements is given below −
for(i=0;i<n;i++){
if(*(p+i)%2==0) {//checking number is even or odd
even=even+*(p+i);
}
Else {//if number s odd enter into block
odd=odd+*(p+i); //calculating sum of all odd numbers in a list
}
}
## Example
Live Demo
#include<stdio.h>
#include<stdlib.h>
void main(){
//Declaring variables, pointers//
int i,n;
int *p;
int even=0,odd=0;
//Declaring base address p using malloc//
p=(int *)malloc(n*sizeof(int));
printf("Enter the number of elements : ");
scanf("%d",&n);
/*Printing O/p -
We have to use if statement because we have to check if memory
has been successfully allocated/reserved or not*/
if (p==NULL){
printf("Memory not available");
exit(0);
}
//Storing elements into location using for loop//
printf("The elements are : \n");
for(i=0;i<n;i++){
scanf("%d",p+i);
}
for(i=0;i<n;i++){
if(*(p+i)%2==0){
even=even+*(p+i);
}
else{
odd=odd+*(p+i);
}
}
printf("The sum of even numbers is : %d\n",even);
printf("The sum of odd numbers is : %d\n",odd);
}
## Output
Enter the number of elements : 5
The elements are :
34
23
12
11
45
The sum of even numbers is : 46
The sum of odd numbers is : 79
Published on 08-Mar-2021 13:42:37 | 498 | 1,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-49 | longest | en | 0.626169 |
https://www.cakecentral.com/forum/t/653173/how-much-does-a-cup-of-butter-weigh | 1,506,220,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689823.92/warc/CC-MAIN-20170924010628-20170924030628-00546.warc.gz | 759,314,132 | 9,109 | ## How Much Does A Cup Of Butter Weigh?
By sugarandslice Updated 26 Sep 2009 , 3:37pm by brincess_b
sugarandslice Posted 26 Sep 2009 , 4:00am
post #1 of 7
I want to try the NFSC recipe here on CC but I don't really work in cup measures, I weigh everything. If someone could tell me how much a cup of butter weighs, preferably in metric! i'd really appreciate it.
TIA
Emma
6 replies
jenng1482 Posted 26 Sep 2009 , 4:10am
post #2 of 7
I just double checked with butter i have in the freezer: my 1 pound (453 g)block is 2 cups
CookieD-oh Posted 26 Sep 2009 , 4:10am
post #3 of 7
OK, I can't believe I just got up and weighed some butter, just because I was curious!
Anyway, I got 8 ounce = 226.796185 gram
And here's a conversion link for you: http://www.onlineconversion.com/weight_volume_cooking.htm
Elise87 Posted 26 Sep 2009 , 4:14am
post #4 of 7
thanks i wondered that too!
sugarandslice Posted 26 Sep 2009 , 4:30am
post #5 of 7
Thanks CookieD-oh and jenng1482. That really helps.
Now I'm going to make a batch of NFSC and see how they compare to my regular butter cookie recipe.
Thanks again for your super-quick responses.
Emma
brendabaker Posted 26 Sep 2009 , 3:18pm
post #6 of 7
1/2 Cup is 4 oz
1 cup 8 oz
3/4 Cup 1 1/2 stick
1 stick is 1/2 cup =4 oz
so forth and so on yadda yadda yadda (smiles)
hope its helps someone !! brendabaker
brincess_b Posted 26 Sep 2009 , 3:37pm
post #7 of 7
i just stick conversions into google, it usually works.
xx
Quote by @%username% on %date%
%body% | 496 | 1,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-39 | latest | en | 0.929455 |
https://www.hackmath.net/en/math-problem/29061?tag_id=61 | 1,603,412,890,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00526.warc.gz | 754,854,548 | 12,560 | # Part-time workers
Three part-time workers received CZK 1,235 for their work. The first received 20% less than the second and the third received 45 CZK more than the second. Determine how many crowns (CZK) each of them received.
Correct result:
a = 340
b = 425
c = 470
#### Solution:
a+b+c=1235
a = b-0.20b
c = 45 + b
a+b+c=1235
a = b-0.20•b
c = 45 + b
a+b+c = 1235
a-0.8b = 0
b-c = -45
a = 340
b = 425
c = 470
Our linear equations calculator calculates it.
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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CZK 2,400 cost clothes, a sweater by 150% than a T-shirt, trousers 2x more than a sweater, a jacket as a sweater and trousers together. What did the individual clothes cost? | 971 | 3,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-45 | latest | en | 0.945182 |
https://www.equationsworksheets.net/one-step-equations-subtraction-worksheet/ | 1,708,965,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00795.warc.gz | 749,701,220 | 15,523 | # One Step Equations Subtraction Worksheet
One Step Equations Subtraction Worksheet – The goal of Expressions and Equations Worksheets is to assist your child in learning more efficiently and effectively. These worksheets are interactive and challenges based on sequence of operations. Through these worksheets, kids are able to grasp simple and advanced concepts in a brief amount of duration. Download these free documents in PDF format. They will aid your child in learning and practice math concepts. These resources are beneficial for students who are in the 5th-8th grades.
## Free Download One Step Equations Subtraction Worksheet
A few of these worksheets are designed for students in the 5th-8th grade. These two-step word problems comprise fractions as well as decimals. Each worksheet contains ten problems. You can access them through any print or online resource. These worksheets can be used to exercise rearranging equations. In addition to practicing the art of rearranging equations, they aid your student in understanding the properties of equality and reverse operations.
These worksheets are designed for fifth and eight grade students. These worksheets are perfect for students who are struggling to calculate percentages. There are three different types of questions. You have the choice to either work on single-step problems which contain whole numbers or decimal numbers, or use words-based methods to solve fractions and decimals. Each page will contain 10 equations. These worksheets for Equations can be used by students from 5th-8th grades.
These worksheets can be used to test fraction calculations as well as other concepts in algebra. Some of the worksheets let users to select from three types of challenges. It is possible to select the one that is word-based, numerical or a mix of both. The type of the problem is importantas each will present a different problem kind. There are ten issues in each page, and they’re excellent for students in the 5th through 8th grade.
These worksheets will help students comprehend the connections between numbers and variables. They provide students with the chance to practice solving polynomial expressions, solving equations, and understanding how to apply them in daily life. If you’re in search of an educational tool that will help you learn about expressions and equations and equations, start by exploring these worksheets. They will teach you about the different kinds of mathematical issues and the different types of symbols used to describe them.
These worksheets are great for children in the first grade. These worksheets can teach students how to solve equations as well as graph. These worksheets are ideal for practice with polynomial variables. They can help you understand how to factor them and simplify these variables. There are a variety of worksheets that can be used to teach kids about equations. The best way to learn about equations is to do the work yourself.
You will find a lot of worksheets that teach quadratic equations. There are several levels of equations worksheets for each level. These worksheets are designed to assist you in solving problems in the fourth level. Once you’ve finished a level, you’ll be able to go on to solving different kinds of equations. Once you have completed that, you are able to work to solve the same problems. You could, for instance solve the same problem in a more extended form. | 641 | 3,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.941251 |
https://undergroundmathematics.org/trigonometry-triangles-to-functions/r8386/suggestion | 1,534,866,907,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218357.92/warc/CC-MAIN-20180821151743-20180821171743-00672.warc.gz | 797,095,469 | 5,286 | Review question
# How many solutions does $\cos (\sin x)=\frac{1}{2}$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8386
## Suggestion
In the range $0 \le x < 2\pi$ the equation $\cos(\sin x) = \frac{1}{2}$ has
1. no solutions;
2. one solution;
3. two solutions;
4. three solutions.
First consider $\cos y = \frac{1}{2};$ what are the solutions of this equation?
From this, can you find the values of $\sin x$ that satisfy $\cos(\sin x) = \frac{1}{2}?$
Do all or any of these values of $\sin x$ give a solution for $x$? | 181 | 626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-34 | latest | en | 0.826861 |
https://justbeginningsflowers.com/how-many-ounces-in-a-gallon-garden-10/ | 1,702,065,332,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00801.warc.gz | 369,228,687 | 21,181 | Garden
Experienced gardeners share their insights in answering this question :
128 US ounces to a US gallon
## How to Identify and Fix Common Gardening Problems ?
We provide a variety of viewpoints on how to identify and fix common gardening problems. Our sources include academic articles, blog posts, and personal essays from experienced gardeners :
Plain soil from a garden can weigh 12 pounds per 1 gallon. Add water, which weighs 8.3 pounds per gallon at room temperature, and a large container can become an immovable object. Soilless potting mixtures, depending on their composition, can weigh a few ounces to 1 pound per gallon.
When calculating liter to gallon conversions, remember than there are 128 oz in a gallon, which conveniently is pretty close to recommended daily fluid intake for men and women: Men: about 1 gallon of water per day.
8.5 inch pot (22 cm) = 2 gallon (7.5L) = 0.3 cu. ft.
Generally speaking, according to Fabric Pot-Pots, a 1-gallon pot holds . 13 to . 15 cubic feet of dirt or dry soil, or about 4 quarts of soil.
About 0.7 cubic feet of soil mix will be sufficient to fill a five-gallon pot. It`s the rule of thumb you should go with.
Is 1 gallon the same as 64 oz? No, one gallon equals 128 fluid ounces; 64 US fl. oz make ½ gallon.
A gallon equals 4 quarts, 8 pints, or 128 fluid ounces.
If you`re doing yard work, you might be wondering how much dirt you can haul in a 5 gallon bucket. A gallon of dirt weighs approximately 11 pounds. That means that 5 gallons of dirt weighs about 55 pounds.
1 Gallon Plants
Expect the plant size to be between 6 inches to 2 feet tall and 6 to 18 inches wide. 1 gallon shrubs establish and grow quickly once planted in the ground and watered properly.
How do I Calculate Volume? It is easiest to calculate volume in cubic metres or litres. To do this simply measure the length, width and depth of the area you need to fill with soil in metres and multiply the three figures together to get the volume in cubic metres.
To estimate soil volume for any area, all you need is a tape measure. “The basic formula is simple: Length x Width x Height = Volume,” says Michael Dean, co-founder of Pool Research. Then divide the number of cubic feet by 27. So one cubic yard = 27 cubic feet = 1,728 cubic inches.
Compost and soil generally comes in 60 Litre bags, so if you want to know how many bags it will take to fill your raised beds, just divide the litre figures below by 60.
Yes, 20 gallons is 80 quarts. 20 gallons is also 2.67 cubic feet (which potting mixes are sometimes measured in).
20-gallon grow bag: 2.67 to 2.75 cubic feet of soil.
In U.S. measurements, a gallon is 3.785 liters, 128 fluid ounces, 4 quarts, 8 pints, or 16 cups. A gallon of water weighs 8.34 pounds. What is this?
How many gallons is 50 ounces? 50 ounces equals approximately 0.390625 gallon.
How many 12 ounces are in a gallon? You will need 10.6667 twelve-ounce measurements to make one gallon or 128 fluid ounces.
Answer and Explanation: There are 128 ounces in one gallon, which means that to find how many gallons 60 ounces are we divide 60 by 128. This means that 60 ounces is or 0.46875 gallons.
In the US Customary System, there are 128 fluid ounces in 1 US gallon. This conversion can also be expressed as 128 fl oz = 1 (US) gal. While in the Imperial System, there are 160 (UK) fluid ounces in 1 Imperial (UK) gallon. The Imperial conversion can be expressed as 160 fl oz = 1 (UK) gal.
One US gallon is equal to 3.785411784 liter.
There are 128 oz in a gallon. = 5.33 – 24 oz bottles.
The total soil volume is the combined volume of solids and pores which may contain air (Vair) or water (Vwater), or both (figure 1). The average values of air, water and solid in soil are easily measured and are a useful indication of a soils physical condition.
For a square or rectangular garden bed, use this formula: length in meters x width in meters x (depth in centimeters / 100) = volume in cubic meters. Multiply the volume value by 1000 to get liters. Do you fill the whole planter with soil?
So, how much does a yard of soil weigh? The weight of a cubic yard of topsoil is typically between 1,500 and 3000 lbs. However, the weight can vary depending on the type of topsoil and the moisture, debris, and rocks it may contain.
How many ounces per gallon is 50:1 – Ryobi Three Trimmer Combo
ANSWER : 3.2 ounces to a UK gallon or 2.56 ounces to a US gallon.
Read Full Q/A … : Garden
The directions got wet to my round up..its not the pump its the bottle that needs to be concentrated….how much to a gallon do i need to use?
ANSWER : Add 6 fluid ounces per 1 gallon of warm water. Spray the ivy or brush you want to kill until it is thoroughly wet.
Read Full Q/A … : Garden
What is the gas oil ratio for ryobia ZR08570 leaf blower
ANSWER : Add 5.5 ounces (one bottle) of 2 cycle oil to 2 gallons of gas.
Read Full Q/A … : Garden
What type sae 30 oil do I put in my bolens weedwackewr
ANSWER : If its two cycle then you have to use two cycle oil. You can buy the small container which you use in a 1 gallon gas can or you can buy a larger can and mix your gas yourself. The larger can works out cheaper and I have the small can which I keep refilling from the bigger can. If you buy your gas by the liter you put one small can in 4.5 liters of gas which is a gallon.
Read Full Q/A … : Garden
Oil mixture for 1985f series 2 cycle lawn-boy mower
ANSWER : Contact your dealer ! but 32:1 sounds correct. 8 ounces of Lawnboy oil to two gallons of gasoline ….is 32:1, 8 ounces of oil to one gallon of fuel is 16:1 for the real old lawnboys,
When Toro bought out OMC,and started using Tecumseh mowers,the ratio changed in the late 80’s or early nineties ! IIRC :o)
Read Full Q/A … : Garden
What is the gas and oil ratio? – Husqvarna Forest & Garden 125l 18" Gas Trimmer
ANSWER : 1:50 (2%)1 gallon of gas to 2.5 fluid ounces of 2-stroke oil.
Read Full Q/A … : Garden
How many fluid ounces in a cup – Roundup "" Weed & Grass Killer
ANSWER : A cup is eight (8) ounces.
Read Full Q/A … : Garden
Gas oil mixture – Ryobi 26CC, 22 In. Hedge Trimmer
ANSWER : Use a good quality 2 cycle oil at 50:1 thats about 2.6 ounces of oil per gallon if you are using Red Armour or most other high quality oils. I have ran everything that calls for 40:1 at the 50:1 ratio with NO failures using Opti 2 premix…. all modern hand held equipment is usually 50:1
Read Full Q/A … : Garden | 1,693 | 6,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-50 | latest | en | 0.902164 |
http://mathhelpforum.com/differential-equations/228741-can-t-find-my-mistake-dif-equation.html | 1,480,899,345,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00451-ip-10-31-129-80.ec2.internal.warc.gz | 167,079,304 | 12,686 | # Thread: Can't find my mistake in dif equation
1. ## Can't find my mistake in dif equation
I am trying to solve this equation for 2 days and I can't do it.... I solved it but wolfram alpha gives different answers. Please help me to find my mistake
- this is my solution. $y = c_1x^3-\dfrac{1}{2x}+c_2$ - Wolfram alpha answer
2. ## Re: Can't find my mistake in dif equation
in wolfram alpha input u have given rhs as -2/x^3 while in ur soln u have computed for -2/x^2
3. ## Re: Can't find my mistake in dif equation
$xy''-y'=-\dfrac{2}{x^2}$
The integrating factor is $\dfrac{1}{x^3}$, so multiply both sides by that:
$\dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}$
$\left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}$
Integrating both sides:
$\dfrac{y'}{x^2} = -\int \dfrac{2}{x^5}dx = \dfrac{1}{2x^4}+C_1$
Multiply both sides by $x^2$:
$y' = \dfrac{1}{2x^2}+C_1x^2$
Integrate both sides:
$y = -\dfrac{1}{2x}+\dfrac{C_1x^3}{3}+C_2$
Fold the $\dfrac{C_1}{3}$ into a single constant:
$y = c_1x^3-\dfrac{1}{2x}+c_2$
When you entered the problem into WolframAlpha, you entered $xy''-2y' = -2/x^3$. Notice the RHS of that differential equation has $x^3$ in the denominator. Your original problem has $x^2$ in the denominator.
4. ## Re: Can't find my mistake in dif equation
You are correct that the general solution to the associated homogeneous equation is $y= C_1x^3+ C_2$ but you are doing the "variation of parameters" incorrectly.
Yes, we look for a solution to the entire equation of the form $y= C_1(x)x^3+ C_2(x)$, allowing the "parameters", $C_1$ and $C_2$ to "vary".
Then we have $y'= C_1'x^3+ 3C_1x^2+ C_2'$. We simplify the equation (there are many different possible functions for $C_1$ and $C_2$) by requiring that $C_1'x^3+ C_2'= 0$ leaving $y'= 3C_1x^2$. Differentiating again, $y''= 3C_1'x^2+ 6C_1x$ and now the differential equation becomes $xy''- 2y'= 3C_1'x^3+ 6C_1x^2- 6C_1x^2= 3C_1'x^3= -\frac{2}{x^2}$.
There is your error- you have, although you don't say how you arrived at it, $x^2$ on the left rather than $x^3$.
5. ## Re: Can't find my mistake in dif equation
Can you explaint this part? And why you wrote y' instead of 2y' and got right answer?
$\dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}$
$\left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}$
6. ## Re: Can't find my mistake in dif equation
Sorry I can't understand integrating factor method, can you say where is error in my solution? ;/
7. ## Re: Can't find my mistake in dif equation
Originally Posted by darjiaus7
Can you explaint this part? And why you wrote y' instead of 2y' and got right answer?
$\dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}$
$\left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}$
HallsofIvy explained the error with your solution. But, you are correct that I wrote the problem incorrectly. Check post #4 above for HallsofIvy's explanation of where you went wrong.
8. ## Re: Can't find my mistake in dif equation
Ok forget about my solution Now can you explain how $\dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y'$ became $\left(\dfrac{y'}{x^2}\right)^\prime$
9. ## Re: Can't find my mistake in dif equation
this is easy use product rule of differentiation d(fg)=fd(g)+gd(f)
here f=1/(x^2) and g=y' | 1,200 | 3,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 33, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-50 | longest | en | 0.868383 |
http://simple.wikipedia.org/wiki/Principle_of_relativity | 1,432,354,822,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927185.70/warc/CC-MAIN-20150521113207-00090-ip-10-180-206-219.ec2.internal.warc.gz | 226,646,598 | 9,801 | Principle of relativity
In the Physics of Aristotle, heavy objects fall faster than objects that are not heavy. The natural science of Aristotle was most popular in Western thought for 2,000 years.
Galileo proved that all objects fall with the same acceleration. Therefore, the longer an object moves with constant acceleration the faster its final velocity is. Also, if different objects each having a different mass are dropped from rest (initial velocity is zero) at the same height in a vacuum, they will all hit the ground at the same velocity regardless of their mass. The experimental discoveries of Galileo and the Laws of Motion developed mathematically by Newton gave birth to modern science.
Galileo's principle of relativity states "It is impossible by mechanical means to say whether we are moving or staying at rest". If two trains are moving at the same speed in the same direction, then a passenger in either car will not be able to notice that either train is moving. However, if the passenger takes a fixed frame of reference, a fixed point, like the earth, he will then be able to notice the motion of either train. Another thing, if one stands on the earth one will not be able to see that it is moving.
This principle is just taken from observation. For example, if we are travelling by airplane at a constant speed, we can walk through the inside of the airplane without noticing anything special.
From a practical point of view, this means that Newton's laws of motion are valid in all inertial systems,[1] which means those at rest or those moving with constant speed relative to one considered at rest. This is the law of inertia: a body at rest continues at rest and a body in motion continues in motion in a straight line unless influenced by an external force. A Galilean coordinate system is one where the law of inertia is valid. The laws of mechanics of Galileo and Newton are valid in a Galilean coordinate system. If K is a Galilean coordinate system, then every other system K' is a Galileian coordinate system if it lies at rest or moves according to the law of inertia relative to K. Relative to K', the mechanical laws of Galileo and Newton are as valid as they are relative to K.
``` If, relative to K, K' is a coordinate
system moving according to the law of
inertia and is devoid of rotation, then
the laws of nature obey the same general
principles in K' as they do in K. This
statement is known as the Principle of
Relativity.
```
In other words, if a mass m is at rest or is moving with constant acceleration (the constant acceleration could be equal to zero in which case the velocity would remain constant) in a straight line relative to a Galilean coordinate system K, then it will also be at rest or moving with constant acceleration in a straight line relative to a second coordinate system K' provided the law of inertia is valid in system K' (in other words, provided it is a Galilean coordinate system).
Therefore, if we want to observe an effect in a moving system at constant speed, we can apply the Newton laws directly. If the moving system speeds up (or we speed up relative to it, like looking at the stars from the earth) then we will have to introduce imaginary forces to compensate this effect.
These fictitious forces are called centrifugal force and coriolis force.[2]
Newton's Laws of Motion are mechanically accurate for speeds that are slow compared with the velocity of light. For speeds that approach the speed of light, it is necessary to apply the discoveries of Einstein's Special Theory of Relativity.
In order to describe what happens mechanically in the universe, physicists use mass, length and time. In the physics of Galileo and Newton, these quantities remain the same throughout the universe.
With Einstein's Special Theory of Relativity, these quantities can change. | 800 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2015-22 | latest | en | 0.937986 |
https://cstheory.stackexchange.com/questions/12106/generating-a-point-in-a-rational-polytope-p-subseteq-rk-given-a-point-in-p | 1,719,344,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00414.warc.gz | 171,043,456 | 41,444 | # Generating a point in a rational polytope $P \subseteq R^k$ given a point in $P^\epsilon$
Consider a rational polytope $P$ that is defined by means of a separation oracle. That is, $P$ can be described implicitly as $P = \{x \in R^k: Ax \leq b, A \in Z^{m \times k}, b \in Z^m \}$, but since $m$ is very large, we use an oracle, that given a point $x \in R^k$, either says $x \in P$ or returns a half-space such that $x \notin S$.
My goal is to find a point in $P$ or determine that $P$ is empty. I'm aiming for a polynomial running time in the representation size of $U$ and $k$, where $U$ is the largest absolute value in $A$. That is, the algorithm should make only polynomial many calls to the separation oracle.
In general, $P$ might be contained in a hyper-plane of lower dimension and thus it is problematic to use the ellipsoid method. So, as in Khachiyan's trick, I alter $P$ (and the separation oracle) to use $P^\epsilon$, where $\epsilon$ is something like $1/U$. Intuitively, the half-spaces that define $P^\epsilon$ are the same as the ones that define $P$ only that they are translated by $\epsilon$. The polytope $P^\epsilon$ has the following properties: $P^\epsilon$ is empty iff $P$ is empty, and if $P$ is not empty, $P^\epsilon$ is full-dimensional.
My question is as follows: Assume the algorithm finds a point $p \in P^\epsilon$. Is it possible to generate a point in $P$ using $p$?
From any choice of a polytope $P$ in ${\mathbb R}^k$, $\epsilon$, and a point $q$ in ${\mathbb R}^k$ it is possible to find a polytope $\hat P$ in ${\mathbb R}^{k+1}$, together with an embedding of ${\mathbb R}^k$ into ${\mathbb R}^{k+1}$, such that $\hat P$ is within $\epsilon$ Hausdorff distance of (the embedded image of) $P$ and such that (the embedded image of) $q$ belongs to $\hat P^\epsilon$. To do this, simply make the facets of $\hat P$ be nearly parallel to the embedded image of ${\mathbb R}^k$, so that translating them by $\epsilon$ in ${\mathbb R}^{k+1}$ causes their intersection with ${\mathbb R}^k$ to move away from $\hat P$ by a much greater distance.
Because $q$ was arbitrary, the knowledge of $q$ is of no use in finding a point in or near $P$; everything you could do with it you could do without it. But, because $\hat P$ and $P$ are so close, finding a point near $P$ is equivalent to finding a point near $\hat P$. Therefore, the knowledge of $q$ (a point in $\hat P^\epsilon$) is of no use in finding a point near $\hat P$.
• Thanks! I added an assumption that the polytope is rational, so (hopefully) now, there is hope of finding a point in P.
– Guy
Commented Jul 26, 2012 at 11:39
• That restriction doesn't help; it's easy to make $\hat{P}$ rational in David's construction. Commented Jul 27, 2012 at 3:05
• I am guessing the questioner is really asking how to find a point $p \in P$ when $P$ is not full dimensional. Commented Jul 28, 2012 at 0:56
If your goal is to find a point in $P$ or determine that $P$ is empty, why don't you do the following.
Let $H$ be a set of half-spaces, initially empty.
Let $x$ be a point, initially equal to $0^k$.
1. Give $x$ to the oracle.
2. If the oracle said $x \in P$, you've done.
3. Otherwise, let $S$ be the violated half-space returned by the oracle. Let $y$ be the orthogonal projection of $x$ on $S$.
• If there exists at least one $T \in H$ such that $y \not \in T$, then you've done: $P$ is empty.
• Otherwise set $H := H \cup \{S\}$, and set $x := y$.
4. Go back to 1.
• Thanks for the response. I think it will work, but, unless I'm missing something, the running time will be porportional to $m$, the number of half-spaces that define $P$. I'm hoping for a running time that is polynomial in $k$ and the representation of $U$, which is the absolute largest value in $A$. That is, only polynomial many calls to the oracle.
– Guy
Commented Jul 26, 2012 at 17:30
• You are welcome! Are you sure that such requirement about the running time is evident by reading the question? Commented Jul 26, 2012 at 18:19
• You're right. I'll add it.
– Guy
Commented Jul 26, 2012 at 18:26 | 1,207 | 4,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.875584 |
https://www.khanacademy.org/math/get-ready-for-ap-statistics/xc9bacb4afa74e6e9:get-ready-for-exploring-one-variable-quantitative-data/xc9bacb4afa74e6e9:box-plots/v/another-example-constructing-box-plot | 1,718,541,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00189.warc.gz | 736,422,196 | 77,237 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Get ready for AP® Statistics>Unit 2
Lesson 2: Box and whisker plots
# Worked example: Creating a box plot (even number of data points)
Box-and-whiskers plots help visualize data ranges and medians. First, arrange your numbers from least to greatest. The smallest and largest numbers form the 'whiskers'. The median of the entire data set splits the 'box' in the middle. The medians of the top and bottom halves of the numbers form the 'box' boundaries.
## Want to join the conversation?
• can i get some help what does IQR mean
• The interquartile range. Where you subtract Quartile 3 and Quartile 1.
• Shouldn't the upper quartile be 7.5, the average of 7 and 8? Considering the upper quartile consists of six numbers?
• it has 7:5 5 6 7 8 8 10
the first five stays since it is not the median
• The question says to exclude the median when calculating the quartiles. In this video they have included median.
Can you explain what does exclude means here ?
what I think is, Q3 will be 7.5 and Q1 will be 2.
• When the question states to "exclude the median when computing the quartiles," it means that when you're finding the first quartile (Q1) and the third quartile (Q3), you should not include the median value in the calculations. In other words, the median is not considered when determining the quartiles. Your interpretation is correct: Q1 will be the median of the lower half of the data, and Q3 will be the median of the upper half of the data. So, if the median is 4.5, then Q1 will be the median of the numbers less than 4.5, and Q3 will be the median of the numbers greater than 4.5.
To summarize:
Q1 is the median of the lower half of the data (excluding the median).
Q3 is the median of the upper half of the data (excluding the median).
Based on your calculation, if the median is 4.5, then Q1 could indeed be 2 and Q3 could be 7.5. However, these values might vary depending on the specific distribution of the data points. If you'd like, I can guide you through the process of finding Q1 and Q3 using the given data set.
• I'm confused. In the last worked example when we had an odd number of data, we were taught to eliminate the mean when calculating the upper and lower means. Does that rule not apply with even numbers of data? That was not clear.
• We want the median to divide the data set into two equal halves.
However, with an odd number of data points the two halves can't be equal in size which is why we remove the median before we calculate the upper and lower quartiles.
With an even number of data points we don't have this problem and don't have to remove the median.
• sleep ideas or can you give me some
• Bro dis hard
• Good luck
P.S Just trying to help
• im abt to fall asleep listening to this ngl
• On the previous video, we are told to exclude the median when computing the quartiles, so Sal does. On this video Sal includes them when it says to exclude them. I'm confused!
• How does outliers affect mean
• Outliers tend to skew the mean to the left or right of the center (according to where the outlier is).
For example take this data set {2,6,6,8,9,11} the mean is (2+6+6+8+9+11)/6 = 42/6 = "7"
If we replaced the 11 in our data set with an outlier "41" the new data set becomes {2,6,6,8,9,41} and the new mean becomes (2+6+6+8+9+41)/6 = 72/6 = "12"
Notice that our new mean "12" is outside our original data set, so what that outlier "41" did is it skewed the mean to the right.
On a side note: This effect of outliers does not happen in median (or it does not change that much); notice that the median is the arithmetic mean of 6 and 8, i.e, the median is 7 in both data sets.
(1 vote)
• I dont know how to create a box plot from a histogram. Help please! | 1,014 | 3,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-26 | latest | en | 0.900723 |
https://whatisconvert.com/85-pounds-in-metric-tons | 1,653,805,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00434.warc.gz | 682,013,114 | 7,168 | # What is 85 Pounds in Metric Tons?
## Convert 85 Pounds to Metric Tons
To calculate 85 Pounds to the corresponding value in Metric Tons, multiply the quantity in Pounds by 0.00045359237 (conversion factor). In this case we should multiply 85 Pounds by 0.00045359237 to get the equivalent result in Metric Tons:
85 Pounds x 0.00045359237 = 0.03855535145 Metric Tons
85 Pounds is equivalent to 0.03855535145 Metric Tons.
## How to convert from Pounds to Metric Tons
The conversion factor from Pounds to Metric Tons is 0.00045359237. To find out how many Pounds in Metric Tons, multiply by the conversion factor or use the Mass converter above. Eighty-five Pounds is equivalent to zero point zero three eight six Metric Tons.
## Definition of Pound
The pound or pound-mass (abbreviations: lb, lbm, lbm, ℔) is a unit of mass with several definitions. Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces.
## Definition of Metric Ton
The tonne (SI unit symbol: t), commonly referred to as the metric ton in the United States, is a non-SI metric unit of mass equal to 1,000 kilograms; or one megagram (Mg); it is equivalent to approximately 2,204.6 pounds, 1.10 short tons (US) or 0.984 long tons (imperial). Although not part of the SI per se, the tonne is "accepted for use with" SI units and prefixes by the International Committee for Weights and Measures.
## Using the Pounds to Metric Tons converter you can get answers to questions like the following:
• How many Metric Tons are in 85 Pounds?
• 85 Pounds is equal to how many Metric Tons?
• How to convert 85 Pounds to Metric Tons?
• How many is 85 Pounds in Metric Tons?
• What is 85 Pounds in Metric Tons?
• How much is 85 Pounds in Metric Tons?
• How many tonne are in 85 lb?
• 85 lb is equal to how many tonne?
• How to convert 85 lb to tonne?
• How many is 85 lb in tonne?
• What is 85 lb in tonne?
• How much is 85 lb in tonne? | 553 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.896862 |
http://blog.csdn.net/princeyuaner/article/details/7403757 | 1,503,428,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112539.18/warc/CC-MAIN-20170822181825-20170822201825-00398.warc.gz | 57,040,741 | 15,679 | # K - Goldbach's Conjecture解题报告
275人阅读 评论(0)
K - Goldbach's Conjecture
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Every even number greater than 4 can be
written as the sum of two odd prime numbers.
For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.
Input
The input will contain one or more test cases.
Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.
Output
For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."
Sample Input
8
20
42
0
Sample Output
8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
#include<iostream>
#include<time.h>
short prime[1000000];
using namespace std;
int main()
{
int i,j,s=0,n,num;
memset(prime,0,sizeof(prime));
for(i=2;i<1000000;i++)
for(j=2;i*j<1000000;j++)
{
if(prime[i*j]==0)
prime[i*j]=1;
}
while(scanf("%d",&n)&&n!=0)
{
for(i=3;i*2<=n;i+=2)
if(prime[i]==0&&prime[n-i]==0)
{
printf("%d = %d + %d\n",n,i,n-i);
goto end;
}
printf("Goldbach's conjecture is wrong.\n");
end:;
}
return 0;
}
1
0
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评论排行 | 669 | 1,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-34 | longest | en | 0.711789 |
https://www.scribd.com/document/149783013/16th-Note-Grid-Double-Accent-Forward | 1,540,072,820,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513441.66/warc/CC-MAIN-20181020205254-20181020230754-00047.warc.gz | 1,075,484,957 | 25,628 | # 16th Note Grid - Double Accent
Snare
RL r l RL r l RL RL LR LR LR LR RL RL RL RL R
SnareScience.com - Universal
LR LR LR L
R L R L LR LR R L R LR LR L RL R L LR LR R L R LR LR L
RL LR RLR L RL LR RLR L RL LR RLR L RL LR RLR L R
Purpose:
As with most grid exercises, this exercise presents some tricky accent pattern / pulse relationships. The pattern that usually presents the most difficulty is when the accent is on the 'e' and the 'and'. The key to this excercise is MARKING TIME! Once you have this exercise mastered, you should know exactly how your feet (i.e. the pulse) line up with the accents. Some things to think about: Do your sticks move exactly the same way for every tap? How about for every accent? Does every tap stroke sound the same? How about from left to right hand? Make sure you play every tap with the same exact volume. The same goes for all of the accents. Some common mistakes: a) Playing the first tap on a hand after an accent at too low of a height. You need to control the rebound of the stick so that it comes up to the correct tap height. b) Pounding every accent. Don't squeeze the stick into the head. The accented stroke should be identical to an eight on a hand stroke. After the note has been played, control the rebound to the correct hand with RELAXED (not clenched) fingers. c) Pounding the very last note of the exercise. d) Playing the right hand accents louder and higher than the left... don't be the guy with the weak left!
1) Stick motion: 2) Sound Quality:
3) Rhythm and Timing:
How perfect is your sixteenth note rhythm? Some common areas where mistakes occur: a) First two notes of the exercise: The first note defines the exercise starting point and the second note defines the tempo. You must internalize the tempo before you start so that these two notes are 'perfect'. b) Underlying sixteenth note pulse: Don't let the accents affect the rhythm!
4) Variations to this exercise: a) Play one hand on drum and one hand on rim to check for stroke and quality of sound consistency on each individual hand. b) Play quarter notes on one hand to keep the pulse, while playing the accents only with the other hand. --- independence!!! | 1,365 | 2,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-43 | latest | en | 0.741704 |
https://math.stackexchange.com/questions/2600951/polynomial-approximation-to-formal-power-series-matrix | 1,713,632,813,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00043.warc.gz | 344,504,568 | 37,543 | # Polynomial approximation to formal power series matrix
I noticed that starting with a $2 {\times} 2$ matrix $M$ with a handful of polynomial entries in two variables such that $\det M$ is invertible in $\mathbb C [x] [[y]]$ I can add infinitely many terms of higher orders in $y$ to the entries of $M$ to make $\det M$ a constant.
Further, I noticed that with a bit of tinkering it's actually possible to add finitely many terms to $M$ and still achieve that $\det M$ is constant.
I am now wondering if this is true in general:
Let $M$ be a $2 {\times} 2$ matrix with entries in the ring $\mathbb C [x][[y]]$ and constant determinant. Given $k > 0$, does there exist a matrix $N$ with entries in $\mathbb C [x, y]$ such that $M \,\mathrm{mod} \, (y^{k+1}) = N \,\mathrm{mod} \, (y^{k+1})$ and $\det M = \det N$?
(If it helps, I would be content with assuming that one of the entries is a multiple of $y$ — e.g. if the $(1,2)$ entry is a multiple of $y$, then the determinant being constant implies that the $(1,1)$ and $(2,2)$ entries have a constant term and are both of the form $1 + O (y)$.)
I give a simple example to illustrate:
Example. Let $$M = \begin{pmatrix} 1 - y & 0 \\ 0 & 1 + y + y^2 + \dotsb \end{pmatrix}$$ Then $\det M = 1$ (as the $(1,1)$ entry is invertible in $\mathbb C [[y]]$ with inverse $(2,2)$).
Now let $$N = \begin{pmatrix} 1-y & -y^{k+1} \\ y^{k+1} & 1 + y + \dotsb + y^{2k+1} \end{pmatrix}$$ Then $\det M = \det N = 1$ and $$M \, \mathrm{mod} (y^{k+1}) = N \, \mathrm{mod} (y^{k+1}) = \begin{pmatrix} 1 - y & 0 \\ 0 & 1 + y + \dotsb + y^k \end{pmatrix}. \qquad \diamond$$
This example works equally well if you replace $y$ by a polynomial $p = p (x,y)$ which is a multiple of $y$.
I would also be interested to hear about (textbook/paper) examples computing with matrices of formal power series to see what kind of techniques can be used.
Edit. I have discovered a truly tedious proof of this, which this edit section is too narrow to contain.
It is a "proof by brute force" (solving linear equations for each monomial that may appear in the expression of $\det M \mod (y^{n+1})$), and I think writing up the details should best be left as an Exercise To The Reader.
There ought to be a more elegant proof and although I haven't disclosed the full proof, at least I now believe the result to be true, which might be a reason to look for such a proof.
• Is $u=y$? If so, what have you tried? Jan 11, 2018 at 15:19
• @Mohan Yes, $u$ was a typo for $y$. For a range of "simple" matrices (2-3 summands per entry up to $y^3$ or $y^4$) I was able to do it by hand. I tried to translate the general case into a different way of thinking about it — e.g. do there exist polynomials $a,b,c,d$ in $x,y$ such that for given fixed polynomials $A,B,C,D,P$ we have that $P = ad-bc + Ad - Bc + aD - bC$ — but the problem doesn't appear any simpler in this formulation... Jan 11, 2018 at 17:43
• "it's actually possible to add finitely many terms to $M$ and still achieve that $\det M$ is constant". Let $M=\begin{pmatrix} 1-y & 0 \\ 0 & 1 \end{pmatrix}$. Then $\det M=1-y=\left(\sum_{n=0}^\infty y^n\right)^{-1}$ is invertible in $\mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$. Let $P=\|p_{ij}\|$ be an arbitrary $2\times 2$ matrix over $\mathbb C[x,y]$. Then $\det (M+y^2P)=1-y+y^2p(x,y)$, where $p\in\Bbb C[x,y]$, so it is not invertible in $C[x,y]$. Jan 20, 2018 at 1:58
• Of course, $M+\begin{pmatrix} y & 0 \\ 0 & 0 \end{pmatrix}=I$, but this carry no deep idea, because if $M=M_0+yM_1$, where $M_0\in\Bbb C[x]$ and $M_1\in \mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$ are arbitrary matrices such that $\det M$ is invertible in $\mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$ then $\det M_0$ is a non-zero constant. Jan 20, 2018 at 1:58
Notation: For a ring $$A$$ and ideal $$\mathfrak{a}$$, let $$\operatorname{SL}_{r}(A,\mathfrak{a})$$ denote the kernel of the group homomorphism $$\operatorname{SL}_{r}(A) \to \operatorname{SL}_{r}(A/\mathfrak{a})$$.
Lemma: Let $$r \ge 2$$, let $$A$$ be a ring, let $$\mathfrak{a}$$ be an ideal of $$A$$. For any $$\ell \ge 1$$, the map \begin{align} \varphi_{\ell} : \operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell}) \end{align} is surjective.
Proof: We proceed by induction on $$\ell$$; for $$\ell = 1$$, we have $$\operatorname{SL}_{r}(A/\mathfrak{a},\mathfrak{a}) = \{\mathsf{id}_{r}\}$$. Suppose $$\ell \ge 2$$ and let $$\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell})$$; let $$\mathsf{W}_{\ell-1} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell-1},\mathfrak{a}/\mathfrak{a}^{\ell-1})$$ be the image of $$\mathsf{W}_{\ell}$$; by induction on $$\ell$$, there exists $$\mathbf{M}_{\ell-1} \in \operatorname{SL}_{r}(A,\mathfrak{a})$$ such that $$\varphi_{\ell-1}(\mathbf{M}_{\ell-1}) = \mathsf{W}_{\ell-1}$$; after replacing $$\mathsf{W}_{\ell}$$ by $$(\varphi_{\ell}(\mathbf{M}_{\ell-1}))^{-1} \cdot \mathsf{W}_{\ell}$$, we may assume that $$\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell} , \mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$$. Write $$\mathsf{W}_{\ell} = \mathsf{id}_{r} + \mathsf{F}$$ with $$\mathsf{F} = (\mathsf{F}_{i,j}) \in \operatorname{Mat}_{r \times r}(\mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$$. Suppose $$\mathsf{F}_{1,1} \ne 0$$. Choose a lift $$\mathbf{f}_{1,1} \in \mathfrak{a}^{\ell-1}$$ of $$\mathsf{F}_{1,1}$$; we may replace $$\mathsf{W}_{\ell}$$ by $$\varphi_{\ell}(\mathbf{E}_{1}) \cdot \mathsf{W}_{\ell}$$ where $$\mathbf{E}_{1} \in \operatorname{SL}_{r}(A)$$ is the matrix which differs from the identity $$\mathrm{id}_{r}$$ by \begin{align} \begin{bmatrix} 1 - \mathbf{f}_{11} & \mathbf{f}_{11} \\ -\mathbf{f}_{11} & 1+\mathbf{f}_{11} \end{bmatrix} \end{align} in the upperleft $$2 \times 2$$ submatrix. This ensures that $$\mathsf{F}_{1,1} = 0$$. If $$\mathsf{F}_{1,1} = 0$$, then we may add multiples of the first row (i.e. multiply on the left by elementary matrices) to ensure that the first column is equal to the 1st standard basis vector. In this way we may assume that (for $$i = 1,\dotsc,r-1$$) the $$i$$th column of $$\mathsf{W}_{\ell}$$ is equal to the $$i$$th standard basis vector; for $$\mathsf{F}_{r,r}$$, the condition $$\det \mathsf{W}_{\ell} = 1$$ implies that necessarily $$\mathsf{F}_{r,r} = 0$$ as well; then as before we can make the $$r$$th column the $$r$$th standard basis vector using elementary row operations.
Remark: IIRC I got the idea for the above from Cohn's example $$\begin{bmatrix} 1+xy & x^{2} \\ -y^{2} & 1-xy \end{bmatrix}$$ which is a matrix that cannot be expressed as a product of elementary matrices with coefficients in $$\mathbb{Q}[x,y]$$.
Corollary: Let $$r \ge 2$$, let $$A$$ be a ring, let $$\mathfrak{a}$$ be an ideal of $$A$$, let $$A^{\wedge}$$ be the $$\mathfrak{a}$$-adic completion of $$A$$. Then the group homomorphism $$\operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A^{\wedge},\mathfrak{a}A^{\wedge})$$ is dense for the $$\mathfrak{a}$$-adic topology. | 2,581 | 7,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 56, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.85951 |
https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Odd_Squares | 1,695,679,476,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00535.warc.gz | 519,384,213 | 10,631 | # Integer as Sum of Three Odd Squares
## Theorem
Let $r$ be a positive integer.
Then:
$r \equiv 3 \pmod 8$
$r$ is the sum of $3$ odd squares.
## Proof
### Sufficient Condition
From Integer as Sum of Three Squares, every positive integer not of the form $4^n \paren {8 m + 7}$ can be expressed as the sum of three squares.
Hence every positive integer $r$ such that $r \equiv 3 \pmod 8$ can likewise be expressed as the sum of three squares.
From Square Modulo 8, the squares modulo $8$ are $0, 1$ and $4$.
Thus for $r$ to be the sum of three squares, each of those squares needs to be congruent modulo $8$ to $1$.
Thus each square is odd, and $r$ can be expressed in the form $8 n + 3$ as the sum of $3$ odd squares.
$\Box$
### Necessary Condition
Suppose $r$ is the sum of $3$ odd squares.
From Odd Square Modulo 8, each of these odd squares is congruent to $1 \pmod 8$.
Therefore:
$r \equiv 1 + 1 + 1 \pmod 8 \equiv 3 \pmod 8$
$\blacksquare$ | 298 | 961 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-40 | latest | en | 0.841873 |
http://mathhelpforum.com/calculus/94688-integral-sq-root-u-bottom-question-print.html | 1,495,915,207,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609054.55/warc/CC-MAIN-20170527191102-20170527211102-00260.warc.gz | 281,030,778 | 4,171 | # Integral with Sq Root of u on bottom question
• Jul 8th 2009, 07:46 PM
Integral with Sq Root of u on bottom question
Going through my book I've found this problem which I can't seem to find the answer to.
Directions: Evaluate the definite integral of the algebraic function.
--------------------------
Problem: Inegral symbol (the bottom of the symbol there is a 1 and at the top a 4)
u-2/ u^(1/2) .
----------------------
Can anyone help me in answering this? This probably seems like a simple and dumb question, but it seems that my algebra may be a bit weak.
Thanks
• Jul 8th 2009, 07:56 PM
TKHunny
You may wish to note this:
$
\frac{u-2}{\sqrt{u}}\;=\;\sqrt{u} - \frac{2}{\sqrt{u}}\;=\; u^{1/2} - 2u^{-1/2}
$
Looking easier, yet?
• Jul 8th 2009, 08:01 PM
AlephZero
$\int \frac{u-2}{u^{1/2}} du = \int u^{1/2}du - 2\int u^{-1/2}du.$
Should be enough for you to find the answer.
• Jul 8th 2009, 08:25 PM
I can see what both of you did, and thank you. My last question would be what exactly do I do to bring the u^(1/2) to the numerator?
AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still. (Headbang)
Thanks to both of you again :)
• Jul 8th 2009, 08:40 PM
AlephZero
Quote:
My last question would be what exactly do I do to bring the u^(1/2) to the numerator?
AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still.
I'm not exactly clear on what you mean by "bringing it to the numerator," but $\frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2}$ by the basic rules of exponents. So $\frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.$
The integral portion follows from the basic rule
$\int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.$
Make sense?
• Jul 8th 2009, 08:47 PM
Quote:
Originally Posted by AlephZero
I'm not exactly clear on what you mean by "bringing it to the numerator," but $\frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2}$ by the basic rules of exponents. So $\frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.$
The integral portion follows from the basic rule
$\int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.$
Make sense?
It all clicked after reading that! Thank you so much :D
• Jul 10th 2009, 06:36 PM
TKHunny
Technical note: Is $\frac{u}{\sqrt{u}} = \sqrt{u}$?
Really? Truly? Absolutely in every way?
It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?
• Jul 11th 2009, 10:38 AM
HallsofIvy
Quote:
Originally Posted by TKHunny
Technical note: Is $\frac{u}{\sqrt{u}} = \sqrt{u}$?
Really? Truly? Absolutely in every way?
It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?
Well, obviously not if the the denominator were 0. In the same sense that $\frac{x^2- 4}{x- 2}= x+ 2$ as long as $x\ne 2$.
If the lower limit of the integral were 0, that is, if it were $\int_0^4 \frac{u}{\sqrt{u}}du$, the integrand is not defined at u= 0 and so the integral is an improper integral. Your point is a good one but we can do the following:
Replace $\int_0^4 \frac{u}{\sqrt{u}}du$ with $\lim_{\epsilon\rightarrow 0}^4 u^{1/2} du$. We can do that because u is not 0 there. That gives $\left[\frac{2}{3}u^{3/2}\right]_\epsilon^4= \frac{2}{3}\left(8- \epsilon^{3/2}\right)$. And that goes to $\frac{16}{3}$ as $\epsilon$ goes to 0, the same as if we had must integrated $\int_0^4 u^{1/2} du$
• Jul 11th 2009, 05:28 PM
TKHunny
I was just wondering if anyone ever had introduced Shadoowned to Domain considerations. Ploughing through without validity is bad business. One should at least think about it, I think. | 1,257 | 3,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-22 | longest | en | 0.935159 |
https://www.investopedia.com/terms/e/exponential-growth.asp | 1,686,217,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00793.warc.gz | 884,488,170 | 55,367 | # Exponential Growth: Definition, Examples, Formula To Calculate
## What Is Exponential Growth?
Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function.
For example, suppose a population of mice rises exponentially by a factor of two every year starting with 2 in the first year, then 4 in the second year, 8 in the third year, 16 in the fourth year, and so on. The population is growing by a factor of 2 each year in this case. If mice instead give birth to four pups, you would have 4, then 16, then 64, then 256.
Exponential growth (which is multiplicative) can be contrasted with linear growth (which is additive) and with geometric growth (which is raised to a power).
### Key Takeaways:
• Exponential growth is a pattern of data that shows sharper increases over time.
• In finance, compounding creates exponential returns.
• Savings accounts with a compounding interest rate can show exponential growth.
## Understanding Exponential Growth
In finance, compound returns cause exponential growth. The power of compounding is one of the most powerful forces in finance. This concept allows investors to create large sums with little initial capital. Savings accounts that carry a compound interest rate are common examples of exponential growth.
### Applications of Exponential Growth
Assume you deposit $1,000 in an account that earns a guaranteed 10% rate of interest. If the account carries a simple interest rate, you will earn$100 per year. The amount of interest paid will not change as long as no additional deposits are made.
If the account carries a compound interest rate, however, you will earn interest on the cumulative account total. Each year, the lender will apply the interest rate to the sum of the initial deposit, along with any interest previously paid. In the first year, the interest earned is still 10% or $100. In the second year, however, the 10% rate is applied to the new total of$1,100, yielding $110. With each subsequent year, the amount of interest paid grows, creating rapidly accelerating, or exponential, growth. After 30 years, with no other deposits required, your account would be worth$17,449.40.
### The Formula for Exponential Growth
On a chart, this curve starts slowly, remains nearly flat for a time before increasing swiftly to appear almost vertical. It follows the formula:
$V=S\times(1+R)^T$
The current value, V, of an initial starting point subject to exponential growth, can be determined by multiplying the starting value, S, by the sum of one plus the rate of interest, R, raised to the power of T, or the number of periods that have elapsed.
## Special Considerations
While exponential growth is often used in financial modeling, the reality is often more complicated. The application of exponential growth works well in the example of a savings account because the rate of interest is guaranteed and does not change over time. In most investments, this is not the case. For instance, stock market returns do not smoothly follow long-term averages each year.
Other methods of predicting long-term returns—such as the Monte Carlo simulation, which uses probability distributions to determine the likelihood of different potential outcomes—have seen increasing popularity. Exponential growth models are more useful to predict investment returns when the rate of growth is steady.
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The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. | 755 | 3,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.930445 |
https://funloop.org/post/2017-05-13-tower-of-hanoi.html | 1,701,622,654,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00217.warc.gz | 320,307,388 | 6,635 | # The Math Behind the Tower of Hanoi Problem
In the very first chapter of the book Concrete Mathematics 2ed there is a discussion about the Tower of Hanoi. This post is a distillation of that discussion.
# The Problem
There are 3 rods, with 8 discs (with holes) resting on one rod; the discs are sorted in size like a pyramid, with the smallest disc on top. We want to move all discs to another rod, but with the following rules: (1) a move consists of moving a single disc onto a rod; (2) you may never place a bigger disc on top of a smaller one. A question arises — how many steps are required to move the entire tower of disks onto another rod?
# Finding the Recurrence
First consider the simplest case, without any discs. Because there are no discs to move, we cannot make any moves, and so the number of steps required is 0. We can write this as
$S_0 = 0$
with $S$ meaning the number of steps and the subscript representing the number of discs in the tower.
Now let’s consider how the problem scales. With 1 disc, the answer is a single step since the one disc is itself the entire tower. With 2 discs, the answer is three steps — one step to move the top (small) disc to another rod, one step to move the big disc to the destination rod, and lastly one step to move the small disc on top of the big disc. With 3 discs, the answer is seven steps — the insight here is that we treat the top two discs exactly the same as the previous problem; so we need 3 moves to move the top two to another rod, then one move to move the biggest disc to the destination rod, then again 3 moves to move the 2-disc sub-tower to the destination rod.
The example with 3 discs is quite telling. We can use the insights gained there to set an upper bound to the number of steps required for the general case of $n$ discs; if we take more steps than this upper bound, we would know that we made mistakes. For a tower of size $n$, we require $S_{n - 1}$ steps to move all discs except the biggest one, then move the biggest disc, then move the sub-tower on top of that disc with (again) $S_{n - 1}$ steps. So the upper bound is
$$$\label{eq:recurrence} S_n = \begin{cases} 0 & \text{if } n = 0 \\ 2 * (S_{n - 1}) + 1 & \text{if } n > 0. \end{cases}$$$
If that’s the upper bound, then is there a separate formula for the lower bound (optimal solution)? Nope! It’s because there must come a time in solving the puzzle where we move the biggest disc to the destination rod. To get to the biggest disc, we must have moved all discs on top of it to another rod (the sub-tower); and, after having moved the biggest disc, we must move this sub-tower back on top of that rod (back onto the biggest disc). Because of these constraints stemming the definition of the puzzle itself, we know that for $n$ > 0 we must take at least $2 * (S_{n - 1}) + 1$ steps.
The upper and lower bounds agree in their formulation, and this formulation (Equation $\ref{eq:recurrence}$) is our recurrence. In mathematics, a recurrence relation is basically a recursively-defined equation, where a base case in the recurrence defines the starting point. In Equation $\ref{eq:recurrence}$, the base case is $n = 0$; for $n > 0$, we define the number of steps required in a recursive manner.
In our discussion of finding the upper and lower bounds, there were two key concepts — the need to move the biggest disc, and the need to move the sub-tower twice (before and after moving the biggest disc). Our recurrence clearly agrees with these two concepts. The “$+ 1$” in the non-base case is the step of moving the biggest disc, whereas the $2 * (S_{n - 1})$ is the number of steps required to move the sub-tower twice.
# Simplifying the Recurrence
Recurrences are great, but they are painful to compute. For example, it’s not immediately clear what $S_{11}$ or $S_{54}$ evaluates to. It would be really nice if we could avoid defining $S_n$ recursively.
And this is where math meets science. In the scientific method, we have to come up with a hypothesis and then test that hypothesis with one or more experiments. We can do the same thing here by trying to guess the solution to the recurrence.
For one thing, we know that $S_n$ grows as $n$ grows (it will never be the case that $S_n$ somehow plateaus or decreases down the road). The more discs there are, the more work we have to do, right? So let’s look at small cases to see how the numbers grow, and see if there is a pattern to the growth rate of $S_n$.
$n$ $S_n$
0 0
1 1
2 3
3 7
4 15
5 31
6 63
7 127
8 255
We don’t have to actually simulate the puzzle to derive these values; using the recurrence Equation $\ref{eq:recurrence}$ we start off from the first row (the base case) and then calculate our way down, reusing $S_n$ from the previous row as $S_{n - 1}$. 1
Anyway, the values of $S_n$ sure look familiar — especially if we use base 2.
$n$ binary($S_n$)
0 $0_2$
1 $1_2$
2 $11_2$
3 $111_2$
4 $1111_2$
5 $11111_2$
6 $111111_2$
7 $1111111_2$
8 $11111111_2$
It looks like our recurrence simplifies to just
$$$\label{eq:solution} S_n = 2^n - 1 \quad \text{for } n \geq 0,$$$
except it is no longer a recurrence as there is no need to define a base case. We’ll call it a solution to the recurrence.
# Proving the Solution
Although the empirical evidence looks very good, we have not formally proved that the solution (Equation $\ref{eq:solution}$) holds for all $n$. It’s one thing to say that something is true for all observed cases (scientific experiment), and quite another to say that something is true for all cases (mathematical proof).
Can we prove it? Yes! Fortunately for us, Equation $\ref{eq:recurrence}$ lends itself to proof by induction. Induction requires you to first prove some number $k_0$ as a starting point (the base case) using some proposition $P$. Then you prove that $P$ holds for $k + 1$ (the next number); i.e., show that going from $k$ to $k + 1$ does not change $P$. This is the inductive step. In this way, we prove the “totality” of $P$ as it applies to all numbers in the range $[k_0, k_{m}]$ and we are done. 2
Here we want to prove that Equation $\ref{eq:solution}$ holds for all $n$ (all natural numbers). 3 For this proof let’s rewrite Equation $\ref{eq:solution}$ to use $k$ instead of $n$:
$$$\label{eq:proposition} S_k = 2^k - 1 \quad \text{for } k \geq 0.$$$
Equation $\ref{eq:proposition}$ is our proposition $P$. The base case is easy enough to prove: $S_0 = 0$ because there are no disks to move. For the inductive step, we use the non-base part of our recurrence from Equation $\ref{eq:recurrence}$ to get
\begin{align} S_k &= 2 * (S_{k - 1}) + 1 \label{eq:induct1} \end{align}
and rewrite it in terms of $k + 1$:
\begin{align} S_{k + 1} &= 2 * (S_{k}) + 1. \label{eq:induct2} \end{align}
Now the critical part: we replace $S_k$ with Equation $\ref{eq:proposition}$ (our proposition), because we assume that our proposition is true for all steps up to $k$ (but not $k + 1$, which is what we’re trying to prove):
\begin{align} S_{k + 1} &= 2 * (2^k - 1) + 1. \end{align}
In case you forgot algebra, $2 * 2^k = 2^1 * 2^k = 2^{k + 1}$ and we can use this to simplify our equation.
\begin{align} S_{k + 1} &= 2 * (2^k - 1) + 1\\ &= [2 * (2^k - 1)] + 1\\ &= [(2 * 2^k - 2)] + 1\\ &= (2^{k + 1} - 2) + 1\\ &= 2^{k + 1} - 1 \label{eq:induct3}. \end{align}
And now we can see that Equation $\ref{eq:induct3}$ (our “evolved” proposition $P$, if you will) is the same as our solution (Equation $\ref{eq:solution}$), even though we increased $k$ to $k + 1$! This is because simple substitution allows us to replace “$k + 1$” with “$n$”. We have completed our proof by induction. 4
# Alternate Recurrence and Solution
The book goes on to offer an alternate recurrence to Equation $\ref{eq:recurrence}$, by adding 1 to both sides:
\begin{align} (S_n) + 1 &= \begin{cases} 0 + 1 & \text{if } n = 0 \\ 2 * (S_{n - 1}) + 1 + 1 & \text{if } n > 0 \\ \end{cases}\\ &= \begin{cases} 1 & \text{if } n = 0 \\ 2 * (S_{n - 1}) + 2 & \text{if } n > 0. \label{eq:recurrence2} \end{cases} \end{align}
This recurrence is the same as the original, except that it adds 1 to the answer. Now we let $W_n = (S_n) + 1$ and $W_{n - 1} = (S_{n - 1}) + 1$ and rewrite everything in terms of $W$:
\begin{align} W_n &= \begin{cases} 1 & \text{if } n = 0 \\ 2 * (W_{n - 1}) & \text{if } n > 0. \label{eq:recurrence3} \end{cases} \end{align}
Notice how the “$+ 2$” in Equation $\ref{eq:recurrence2}$ goes away, because the coefficient $2$ in Equation $\ref{eq:recurrence3}$ will multiply with the “$+ 1$” from $W_{n - 1}$ to get it back. Using this alternate recurrence, it’s easy to see that the solution is just $W_n = 2^n$, because $W$ can only grow by multiplying $2$ to itself! Hence
\begin{align} W_n = (S_n) + 1 = 2^n \end{align}
and subtracting 1 from all sides gives us
\begin{align} (W_n) - 1 =S_n = 2^n - 1. \end{align}
The lesson here is that if it is difficult to find the solution to a recurrence, we can use basic algebra rules to transform the recurrence to something more amenable. In this case, all it took was adding 1 to the original recurrence.
# Conclusion
I thoroughly enjoyed figuring this stuff out because possibly for the first time in my life I used my programming experience (recurrence/recursion, memoization) to help myself understand mathematics — not the other way around. The other way around was never enjoyable — calculating what i was in some $n$th iteration of a for-loop never really excited me.
I hope this explanation helps you better understand the first few pages of Concrete Mathematics; I had to read that part three times over to really “get it” (never having learned what induction is). And henceforth, I will never look at a string of consecutive 1’s in binary the same way again. 😃
1. In computer science, this process of avoiding the recalculation of previously known values is called memoization and is useful in generating the first N values of a recursive algorithm in $O(N)$ (linear) time.↩︎
2. Note that if $k_0 = 0$, then $[k_0, k_{m}]$ is the set of all natural numbers (zero plus the positive integers).↩︎
3. There is no need to prove the recurrence (Equation $\ref{eq:recurrence}$) as we have already proved it in the process of deriving it.↩︎
4. In Concrete Mathematics 2 ed. p. 3 (where the book uses $T_n$ instead of $S_n$), the proof is simply a one-liner: $T_n = 2(T_{n - 1}) + 1 = 2(2^{n - 1} - 1) + 1 = 2^n - 1.$ But I find it a bit too terse for my tastes.↩︎ | 3,066 | 10,492 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 8, "equation": 3, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-50 | latest | en | 0.926768 |
https://www.cfd-online.com/Forums/openfoam/75540-compressible-flow-solver-of1-6-a.html | 1,718,958,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00051.warc.gz | 611,870,873 | 18,045 | # compressible flow solver in OF1.6
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April 28, 2010, 09:33 compressible flow solver in OF1.6 #1 Senior Member KGN Join Date: Oct 2009 Location: Chennai, India Posts: 121 Rep Power: 16 Hi all, Why PISO & SIMPLE algorithms are used in OF for compressible flow problems. These algorithms are for pressure linked equation i.e. incompressible flows and for low Mach no flows. Even density based solver "rhoSonicFoam" uses PISO algorithm why? In compressible flows computation, we solve momentum equation to get velocity, energy equation to get temperature, density from continuity equation and pressure from perfect gas equation. correct me if i am wrong. If this is the case, then why PISO & SIMPLE come into picture. Thanks mecbe
April 29, 2010, 11:04 #2 Member Ngoc-Minh Truong Join Date: Feb 2010 Location: Toulouse, France Posts: 42 Rep Power: 16 You'll find some answers here: http://www.cfd-online.com/Forums/ope...mple-piso.html Enjoy
April 29, 2010, 12:52 #3 Senior Member KGN Join Date: Oct 2009 Location: Chennai, India Posts: 121 Rep Power: 16 Hi, I read the thread and it was mentioned that steady state problem has to be solved by Pseudo Transient for stability. Can you give some details on it, because i am in the process of converting the "rhoSonicFoam" inviscid density based transient compressible flow solver(this doesn't uses PISO algorithm). thanks mecbe
April 30, 2010, 01:07 #4 Senior Member Alberto Passalacqua Join Date: Mar 2009 Location: Ames, Iowa, United States Posts: 1,912 Rep Power: 36 Please, define the problem you want to solve, or we cannot give any specific answer. What is the Mach number, for example? In addition, what do you mean with "converting" rhoSonicFoam? It is already a transient solver, and it doesn't use the PISO algorithm because it is density based. Best, __________________ Alberto Passalacqua GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM - An open-source implementation of quadrature-based moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using.
April 30, 2010, 01:17 #5 Senior Member KGN Join Date: Oct 2009 Location: Chennai, India Posts: 121 Rep Power: 16 Hi Alberto, I want to solve transonic & supersonic inviscid steady flow over missile. Mach No -0.9,1.2,1.8 & 2.5. I saw the "rhoSonicFoam" code, it doesnt uses PISO algorithm, but in tutorial case file, it was there. sorry for that. I want to convert it to steady state solver, bcz transient solver takes lot of time to converge. One more question, can you explain about the slip boundary condition (other than given in the user guide). Also, i want to implement some more inviscid flux discretization schemes like AUSM, HLLC.. Can you give some advice on how to carry on. Thanks mecbe
April 30, 2010, 01:40
#6
Senior Member
Alberto Passalacqua
Join Date: Mar 2009
Location: Ames, Iowa, United States
Posts: 1,912
Rep Power: 36
Quote:
Originally Posted by mecbe2002 Hi Alberto, I want to solve transonic & supersonic inviscid steady flow over missile. Mach No -0.9,1.2,1.8 & 2.5. I saw the "rhoSonicFoam" code, it doesnt uses PISO algorithm, but in tutorial case file, it was there. sorry for that. I want to convert it to steady state solver, bcz transient solver takes lot of time to converge.
You might want to take a look also to rhoCentralFoam and rhopSonicFoam.
Quote:
One more question, can you explain about the slip boundary condition (other than given in the user guide).
What is not clear?
Best,
__________________
Alberto Passalacqua
GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541)
OpenQBMM - An open-source implementation of quadrature-based moment methods.
To obtain more accurate answers, please specify the version of OpenFOAM you are using.
April 30, 2010, 02:37 #7 Senior Member KGN Join Date: Oct 2009 Location: Chennai, India Posts: 121 Rep Power: 16 I will look into those two schemes. What about implementing new inviscid flux computation schemes like AUSM, HLLC. thanks mecbe
April 30, 2010, 12:06
#8
Senior Member
Alberto Passalacqua
Join Date: Mar 2009
Location: Ames, Iowa, United States
Posts: 1,912
Rep Power: 36
Quote:
Originally Posted by mecbe2002 I will look into those two schemes. What about implementing new inviscid flux computation schemes like AUSM, HLLC. thanks mecbe
There's quite some work to do to be able to implement those. You might want to take a look at aeroFoam:
http://www.cfd-online.com/Forums/ope...-aerofoam.html
Best,
__________________
Alberto Passalacqua
GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541)
OpenQBMM - An open-source implementation of quadrature-based moment methods.
To obtain more accurate answers, please specify the version of OpenFOAM you are using.
April 30, 2010, 12:50 #9 Senior Member KGN Join Date: Oct 2009 Location: Chennai, India Posts: 121 Rep Power: 16 Hi Alberto, Thanks for the thread. I downloaded the solver. This will help a lot for my work. thanks mecbe
December 25, 2010, 09:17 #10 New Member senthilkumar.R Join Date: Sep 2010 Location: Ranchi, India Posts: 20 Rep Power: 15 Hi mecbe, Please reply with AeroFOAM (its written for OF V 1.4.1) compatibility problem in OF V 1.6 + .
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Contact Us - CFD Online - Privacy Statement - Top | 1,750 | 6,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.893206 |
https://socratic.org/questions/how-do-you-find-the-domain-of-f-x-1-x-3-1 | 1,585,663,603,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00109.warc.gz | 684,913,246 | 6,016 | # How do you find the domain of f (x) = -1 / ( x + 3)?
The domain of a function is a set of all x, for which the function exists or $f$ be defined
In our case, the only value for $f \left(x\right)$ does'nt exist is $- 3$
En efect: if $x = - 3$ then $f \left(- 3\right) = - \frac{1}{- 3 + 3} = - \frac{1}{0}$. There is no number which multiplied by 0 gives -1
For this reason the domain is $\mathrm{do} m \left(f \left(x\right)\right) = \left\{x \in \mathbb{R} / x \ne - 3\right\}$ | 180 | 481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-16 | latest | en | 0.663309 |
http://physics.stackexchange.com/questions/32345/determining-orbital-velocity | 1,469,640,321,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826916.34/warc/CC-MAIN-20160723071026-00074-ip-10-185-27-174.ec2.internal.warc.gz | 198,824,585 | 17,293 | Determining Orbital Velocity
Is there any way to determine the orbital velocity of a point around another stationary point, if I don't know the mass of either of the points but know the force that gravity exerts and the distance between them. I'd like to know at what velocity I should push the non-stationary point perpendicular to the stationary point, so that it will maintain a circular orbit.
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1 Answer
It sounds and looks like a school problem. If the orbit is circular, write the expression of the acceleration (normal component only, assume v=cte). a=V^2/R Then use Newton's 2nd law: f=m*a and ....work out the answer
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The masses are unknown in the OP's problem. There is probably not enough information. – Emilio Pisanty Jul 19 '12 at 10:59
I ended up just giving my points a mass. And solving the problem using Kepler's Laws. – Sina Ghaffari Jul 19 '12 at 14:56 | 214 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-30 | latest | en | 0.902033 |
https://forums.fast.ai/t/why-do-you-need-to-divide-by-255-when-using-stack-function/93728 | 1,653,826,594,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00326.warc.gz | 302,383,428 | 3,861 | # Why do you need to divide by 255 when using stack function?
I’m trying to understand the concept behind dividing by 255 when you call the stack function when placing all images into a rank-3 tensor. This is related to the 04_mnist_basics code.
stacked_sevens = torch.stack(seven_tensors).float()/255
stacked_threes = torch.stack(three_tensors).float()/255
There was no explanation as to why the number to divide by is 255 and I can’t figure out where the math for this comes into play.
found the answer to my own question.
As the pixel values range from 0 to 256, apart from 0 the range is 255. So dividing all the values by 255 will convert it to range from 0 to 1.
1 Like | 170 | 681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.901074 |
http://mathoverflow.net/questions/19240?sort=votes | 1,369,454,160,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705407338/warc/CC-MAIN-20130516115647-00082-ip-10-60-113-184.ec2.internal.warc.gz | 166,232,468 | 14,863 | ## Algebraic Proof of 4-Colour Theorem?
4-Colour Theorem. Every planar graph is 4-colourable.
This theorem of course has a well-known history. It was first proven by Appel and Haken in 1976, but their proof was met with skepticism because it heavily relied on the use of computers. The situation was partially remedied 20 years later, when Robertson, Sanders, Seymour, and Thomas published a new proof of the theorem. This new proof still relied on computer analysis, but to such a lower extent that their proof was actually verifiable. Finally, in 2005, Gonthier and Werner used the Coq proof assistant to formalize a proof, so I suppose only the most die hard skeptics remain.
My question stems from reading this paper by Robin Thomas. In it, he describes several interesting reformulations of the 4-colour theorem. Here is one:
Note that the cross-product on vectors in $\mathbb{R}^3$ is not an associative operation. We therefore define a bracketing of a cross-product $v_1 \times \dots v_n$ to be a set of brackets which makes the product well-defined.
Theorem. Let $i, j, k$ be the standard unit vectors in $\mathbb{R}^3$. For any two different bracketings of the product $v_1 \times \dots \times v_n$, there is an assignment of $i,j,k$ to $v_1, \dots, v_n$ such that the two products are equal and non-zero.
The surprising fact is that this innocent looking theorem implies the 4-colour theorem.
Question. Is anyone working on an algebraic proof of the 4-colour theorem (say by trying to prove the above theorem)? If so, what techniques are involved? What partial progress has been made? Or do most people consider the effort/reward ratio of such an endeavor to be too high?
I think it would be interesting to have an algebraic proof, even a very long one, particularly if the algebraic proof does not use computers. Given its connection to many other areas (Temperley-Lieb Algebras), the problem seems to be amenable to other forms of attack.
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On the other hand, if it turns out that a proof of the four-color theorem via this fact also reduced to checking some very large number of cases, that might be evidence that there is some amount of irreducible complexity in the proof of 4CT. – Michael Lugo Mar 24 2010 at 22:03
Noah Snyder has some ideas about what must be checked --- I can't remember if they're here on MO or over at Secret Blogging Seminar. See also R. Penrose, "Applications of Negative Dimensional Tensors", Combinatorial mathematics and its applications, 1971. – Theo Johnson-Freyd Mar 25 2010 at 3:18
I attempted a purely algebraic proof of the result for a research paper in a MATHEMATICA course a few years ago,Tony-nothing came of it. I still think it's an avenue worth chasing down. – Andrew L Mar 25 2010 at 5:50
It's not accurate to say that any of those ideas are mine. As to the original question, it's certainly something that lots of quantum topolgists think about on and off (Kauffman's written multiple papers on it, Bar Natan wrote a paper mentioned below, Vaughan Jones says "the worst thing you can say about planar algebras is that they haven't yielded a proof of the 4-color theorem yet" etc.). If anyone actually solves it we'll all hear about it. It's certainly something in the back of everyone's mind. – Noah Snyder Mar 25 2010 at 5:53
My view is that at the moment this is an isolated problem and that to make progress in needs to be understood in the right context as one of a class of similar problems. – Bruce Westbury Mar 25 2010 at 6:11
There is a classical approach by Birkhoff and Lewis, which remained dormant for decades. It was recently revived by Cautis and Jackson (start here and proceed here), using the Temperley-Lieb algebra.
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Does "Lie Algebras and the Four Color Theorem" by Dror Bar-Natan qualify ?
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At first glance I would say no, although I will have to read the paper more carefully. It seems to me that the bulk of the paper is spent proving that the 4-colour theorem is equivalent to a 'reasonable' statement about Lie algebras. This equivalence is certainly important, but to me the relevant question is why aren't people trying to prove this 'reasonable' statement about Lie algebras. What is the point of translating a theorem into the language of another field if one doesn't plan to use the methods of that other field to prove the original theorem? – Tony Huynh Mar 25 2010 at 5:17
To be provacative: would the Taniyama-Shimura conjecture have been proven if Fermat's Last Theorem had already been proven with the aid of computers? – Tony Huynh Mar 25 2010 at 5:23
@Tony: I'm not sure myself if it qualifies (that's why I asked). However, if you want an algebraic proof, you have to start by translating it into some algebraic setting. – David Lehavi Mar 25 2010 at 5:51
I believe this paper by Bar-Natan is mentioned in the paper of R.Thomas linked in the original post, with a comment that there is an equivalence there, not an independent proof... – Vladimir Dotsenko Mar 25 2010 at 8:59
I must admit I'm a bit baffled about what the question is here, and about why so many people have voted it up. What are you looking for in an answer? I don't think it's appropriate to post speculation on the internet about which mathematicians are privately working on which big problems. As to public work, you seem to have a weirdly restrictive view of what "working on" and "partial progress" mean that don't fit with my understanding of how mathematics works. Several papers have been written on the subject of possible algebraic proofs of the 4-color theorem (look at google scholar or Mathscinet for papers which cite the Saleur-Kauffman paper mentioned in the paper you're reading), but if the Bar-Natan paper doesn't count for you then you're likely to be disappointed by all of them.
The long and short of it is that everyone in quantum topology would love to prove the 4-color theorem and occasionally thinks about it. There's lots of tantalizing clues that an algebraic argument has promise, but if anyone knew how to prove it they'd have done so. As far as I know there isn't anyone who is holed up in their attic thinking about only the 4-color theorem, instead there's a lot of people who every time they find a new tool think "hrm, I wonder if this tool would work on the 4-color theorem?"
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Hi Noah. I think a reasonable answer was given by Igor, and I'll probably accept his answer once I've looked at the Cautis and Jackson paper more carefully. In another sense, I think the question is also a general probe into the current state of affairs in areas related to the 4-colour theorem. So, your second paragraph was quite helpful. Indeed, if you could elaborate or provide a link on what you feel remains to be checked that would be very helpful. – Tony Huynh Mar 25 2010 at 19:04 Thanks for not taking my criticism personally. I think "what remains to be checked" is not quite the right way of thinking about it. What you have is a tensor category given by generators and relations, and you want to understand why all closed diagrams evaluate to something positive. There are lots of other tensor categories given by generators and relations, and people find new techniques for dealing with them all the time. The hope is that someday the right new technique will crack the 4-color theorem. – Noah Snyder Mar 25 2010 at 23:39
An article in Scientific American, Jan 2003 offered a supposed counterexample, that sparked my interest in the problem. That and the complexity of the Appel and Haken proof motivated me to do my own study. It has minimal math, but is consistent, and approaches the problem from an entirely different direction. If interested it's at insight.awardspace.info.
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This Scientific American article is about a paper by Hud Hudson in the May 2003 Math Monthly. It constructs a "map" somewhat like the well-known Lakes of Wada (en.wikipedia.org/wiki/Lakes_of_Wada) which cannot be four-colored. So here the author is generalizing the notion of map beyond the usual one considered in the four color theorem. Although he points out, interestingly, that his example contradicts various statements which have been claimed to be equivalent to the four color theorem. – Dan Ramras Apr 3 2010 at 0:01
Algebraic Proof of 4-Colour Theorem? see please http://www.math.accent.kiev.ua/article/00/4ct-2-.htm
- | 1,975 | 8,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2013-20 | latest | en | 0.949492 |
https://kodlogs.com/33984/the-truth-value-of-an-array-with-more-than-one-element-is-ambiguous-use-a-any-or-a-all | 1,624,237,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00286.warc.gz | 313,350,278 | 9,527 | # the truth value of an array with more than one element is ambiguous. use a.any() or a.all()
2.4k views
## Problem :
I got following rather cryptic error message:
the truth value of an array with more than one element is ambiguous. use a.any() or a.all()
## Solution :
I had the same problem with indexing with multiple-conditions while finding the data in a certain date range. The (a-b).any() OR (a-b).all() was not working for me at all.
I found the solution which works perfectly for my functionality.
Simply used the numpy.logical_and(a,b) and it worked for me. Following is the way to rewrite the code :
`selected = r[numpy.logical_and(r["dt"] >= startdate, r["dt"] <= enddate)]`
38.6k points
The error message explains itself very well.
``````ValueError: The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all().``````
# Solutions:
## Array comparison return boolean:
Make sure your array comparison returns a boolean array. Methods any() and all() reduce values over the array either the value is logical_or or logical_and. Probably you also don’t need to check for equality.
## What should bool return?
The main question is what should the bool(np.array([False, False, True])) have to return. There are many arguments.
• True, because bool(np.array(x)) should have to return same as bool(list(x)), and non empty lists.
• True, Because at least one element has to be true.
• False, because all the elements are not false value.
Since, the truth value of an array having more than one element is ambiguous so you have to use .any() or .all().
### Example:
``````>>> v = np.array([1, 2, 3]) == np.array([1, 2, 4])
>>> v
Array([True, True, False], dtype=bool)
>>> v.any()
True
>>> v.all()
False``````
## Float values:
You have to use np.allclose, if you are making comparisons of array for float data type.
### Example:
``````>>> np.allclose(np.array([1, 2, 3_1e-8]), np.array([1, 2, 3]))
True.``````
I hope you can understand and this answer will help to solve your problem.
3.9k points | 529 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | latest | en | 0.801877 |
https://en.sorumatik.co/t/design-an-experiment-that-shows-how-the-activation-energy-can-be-altered-with-temperature-using-a-glow-stick-describe-the-items-needed-how-the-experiment-would-be-done-and-the-measurments-that-we-need/3901 | 1,726,237,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00029.warc.gz | 207,960,275 | 6,404 | # Design an experiment that shows how the activation energy can be altered with temperature using a glow stick. describe the items needed, how the experiment would be done and the measurments that we need
design an experiment that shows how the activation energy can be altered with temperature using a glow stick. describe the items needed, how the experiment would be done and the measurments that we need to be recorded to achieve the desired results. if you have a glow stick avaliable, do the experiment and send your ta a picture.
Designing an Experiment to Show the Effect of Temperature on Activation Energy using a Glow Stick
To design an experiment that demonstrates how the activation energy can be altered with temperature using a glow stick, you will need the following items:
1. Glow stick
2. Thermometer
3. Beaker or cup
4. Hot water source (e.g., kettle)
5. Cold water source (e.g., ice)
6. Stopwatch or timer
7. Safety goggles and gloves (to ensure safe handling)
Procedure:
1. Start by putting on your safety goggles and gloves to ensure your safety throughout the experiment.
2. Set up a work area where you can easily observe the glow stick.
3. Take the glow stick and carefully activate it by bending and shaking it, following the manufacturer’s instructions.
4. Measure the initial temperature of the glow stick and record it as T1.
5. Place the glow stick in a beaker or cup, ensuring that it is stable and upright.
6. Prepare two water baths: one with hot water and another with cold water. The temperature of the hot water bath should be around 50-60 degrees Celsius, while the cold water bath should be around 10-15 degrees Celsius. These temperatures should be adjusted based on the manufacturer’s instructions for the glow stick.
7. Immerse the beaker or cup containing the glow stick into the hot water bath. Start the stopwatch or timer simultaneously.
8. Observe the changes in the intensity or duration of the glow stick’s brightness over a specific interval of time (e.g., every 30 seconds or 1 minute). Record your observations and the corresponding time intervals.
9. After a sufficient amount of time has passed (e.g., 5-10 minutes), remove the beaker or cup from the hot water bath and immediately immerse it in the cold water bath.
10. Continue observing the glow stick’s brightness and record your observations and time intervals as before.
11. Repeat the experiment multiple times at different temperature intervals to obtain reliable and consistent results.
Measurements:
To achieve the desired results, you will need to record the following measurements:
1. Initial temperature of the glow stick (T1)
2. Temperature of the hot water bath
3. Temperature of the cold water bath
4. Time intervals at which observations are made (e.g., every 30 seconds or 1 minute)
5. Observations of the glow stick’s brightness or any changes in intensity or duration
By comparing the observations and measurements made at different temperature intervals, you will be able to analyze how the activation energy of the glow stick is affected by temperature changes. Additionally, you can measure the time taken for the glow stick to reach maximum brightness and compare it across different temperatures.
Unfortunately, as an AI assistant, I am unable to perform physical experiments or provide pictures. However, by following the steps outlined above, you should be able to conduct the experiment successfully and obtain the necessary data to understand the relationship between temperature and activation energy using a glow stick. | 720 | 3,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.878525 |
https://www.enotes.com/homework-help/polynomial-whose-coefficients-real-numbers-find-290386 | 1,701,342,485,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00521.warc.gz | 836,567,489 | 16,604 | # A polynomial of degree 6 has real coefficients.What are the remaining roots if the roots given are: -8 + 11*i, -7 + 17*i, 16 - i*sqrt 2
The polynomial of degree 6 has 6 roots. The polynomial that is being considered has roots -8 + 11*i, -7 + 17*i, 16 - i*sqrt 2.
All complex roots of a polynomial are found in conjugate pairs to ensure that the coefficients are real. If this were not the case, the coefficients would not be real numbers.
For -8 + 11*i, the polynomial has a root -8 - 11*i
-7 + 17*i gives another root complex root -7 - 17*i and as 16 - i*sqrt 2 is a root there is another root equal to 16 + i*sqrt 2
The other roots of the polynomial are : -8 - 11*i, -7 - 17*i, 16 + i*sqrt 2 | 217 | 700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-50 | latest | en | 0.916747 |
https://electronics.stackexchange.com/questions/430908/how-can-i-determine-the-output-in-this-circuit | 1,561,631,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001089.83/warc/CC-MAIN-20190627095649-20190627121649-00239.warc.gz | 437,687,336 | 32,414 | # How can I determine the output in this circuit? [closed]
I have such circuit:
simulate this circuit – Schematic created using CircuitLab
Vin is either equal to Vcc or to 0.
How do I go about determining Vout?
## closed as too broad by Bimpelrekkie, laptop2d, PeterJ, RoyC, FinbarrApr 8 at 0:10
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• You can start by ignoring R1, does that help? However, note that "open" is not the same as 0V. – Spehro Pefhany Apr 5 at 12:43
• How can I determine the output in this circuit? By learning about circuit analysis. Hundreds of books have been written on the subject. I see no reason why anyone would need to explain that here as well. If this is asked of you then surely you must have had some education on the subject. If not then get studying. – Bimpelrekkie Apr 5 at 12:47
The first thing you need to know is that the capacitor charges fully to the Vcc at steady state i.e the capacitor maintains 3.3V with respect to ground.And once the capacitor is fully charged, it acts like a open switch which stops the flow of current. Since the Vout is directly connected to the capacitor, the output voltage is equal to the voltage across capacitor that is 3.3V.
Now, when the Vin is equal to 3.3V, nothing happens to the circuit.
Please ignore the values of the components I have in my circuit. Also ignore the names in your circuit and look at mine now. I tried editing the circuit but the 'edit' button has disappeared in the preview section. Sorry for the inconvenience.
In this case. node 1 is at 3.3V. This means that there is no any potential difference across the resistors R1 and R2. So no any current flows through any of the resistors. Also the capacitor doesn't discharge due to the fact that there are no any nodes at lower potential with respect to the capacitor's potential.
When the Vin goes to zero, something happens. Since the node 1 is at zero volt, the capacitor finds 'somewhere' to dump its potential. So the capacitor stats to discharge through the resistor 1(in my figure). Due to the discharge of the capacitor, the voltage across it starts to decrease as well. Once the capacitor is fully discharged, the output voltage is zero. For the instantaneous voltage across the capacitor, I suggest you reading this topic. Once you get the instantaneous voltage across the capacitor, you get the instantaneous Vout
• Thank you for the answer. It is what I was looking for. The main issue for me is that I was used to seeing closed loop circuits rather than such schematics with open ends and therefore couldn't analyse it properly. Thanks! – Aivaras Kazakevičius Apr 5 at 13:44
I'm guessing you need to know Vout when the circuit has reached it's stationary state.
When that happens the capacitor charges to a voltage so that no current flow through it. So you can "remove" it from your circuit.
Then you need to evaluate both cases, when Vin = 0V and when Vin = 3.3V. | 751 | 3,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-26 | longest | en | 0.947794 |
https://www.fuelcellstore.com/blog-section/model-validation-using-residuals | 1,596,570,298,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735882.86/warc/CC-MAIN-20200804191142-20200804221142-00208.warc.gz | 614,053,579 | 18,201 | # Model Validation Using Residuals
Model validation is the most important step in the model building process; however, it is often neglected. Even when the model is validated, it is often not done adequately. It often consists of taking a few experimental data points and plotting these points on the same graph as the model. There are two different types of models: engineering or scientific models and statistical models. Engineering and scientific models are often built using equations from the literature and from derived equations. In this model type, there is often a combination of equations that are used to calculate the natural phenomena that is occurring -- and there is often little to no available data to help build the model.
Figure 1. Comparison between fuel cell engineering model and experiments at 298 K and 1 bar.
The other model type is a statistical or empirical model. These models are generated from known datasets. Often mathematical model validation may only consist of quoting the R2 statistic from the fitted line or curve. Unfortunately, a high R2 value does not mean that the data actually fits the model well. If the model does not fit the data well, this negates the purpose of building the model in the first place.
There are many statistical tools that can be used for model validation. One of the most useful tools is graphical residual analysis. There are many types of plots of residuals that allow the model accuracy to be evaluated. There are also several methods that are important to confirm the adequacy of graphical techniques. To help interpret a borderline residual plot, a lack-of-fit test for assessing the correctness of the functional part of the model can be used. The number of plots that can be used for model validation is limited when the number of parameters being estimated is relatively close to the size of the data set. This occurs when there are designed experiments. Residual plots are often difficult to interpret because the number of unknown parameters.
Residuals
The residuals from a fitted model are the differences of the responses at each combination of variables, and the predicted response using the regression function. The definition of the residual for the ith observation in the data set can be written as:
with denoting the ith response in the data set and represents the list of explanatory variables, each set at the corresponding values found in the ith observation in the data set. The distance from the line at 0 is how bad the prediction are for that value. Since the Residual = Observed – Predicted, then the positive values for the residual (on the y-axis) mean the prediction was too low, and negative values mean the prediction was too high; 0 means the guess was exactly correct.
Figure 2. Graphs of predicted, actual values, and standardized residuals. (http://docs.statwing.com/interpreting-residual-plots-to-improve-your-regression/#the_top)
If a model is adequate, the residuals should have no obvious patterns or systematic structure. The primary method of determining whether the residuals have any particular pattern is by studying the scatterplots. Scatterplots of the residuals are used to check the assumption of constant standard deviation of random errors. The residual pattern is good is they are:
(1) symmetrically distributed, tending to cluster towards the middle of the plot
(2) clustered around the lower single digits of the y-axis (e.g., 0.5 or 1.5, not 30 or 150)
(3) there are no clear patterns
Figure 3. Example of good residual plot. (http://docs.statwing.com/interpreting-residual-plots-to-improve-your-regression/)
If the plots are not evenly distributed vertically, they have an outlier, or they have a shape to them. If you can detect a clear pattern or trend in your residuals, then your model has room for improvement. Most of the time a decent model is better than none at all. So take your model, try to improve it, and then decide whether the accuracy is good enough to be useful for your purposes.
Figure 4. Examples of bad residual plots. (http://docs.statwing.com/interpreting-residual-plots-to-improve-your-regression/)
Drifts in the measurement process
Drifts in the measurement process can be checked by creating a “run order” or “run sequence” plot of the residuals. These are scatterplots where each residual is plotted versus an index that indicates the order (in time) in which the data were collected. This is useful when the data have been collected in a randomized run order, or an order that is not increasing or decreasing in any of the predictor variables. If the data are increasing or decreasing with the predictor variables, then the drift in process may not be separated from the functional relationship between the predictors and the response -- this is why randomization is encouraged when planning out the design of experiments.
Figure 5. Example Run Order Plot.
Independent random errors
A lag plot of residuals helps to assess whether the random errors are independent from one to the next. If the errors are independent, the estimate of the error in the standard deviation will be biased, which leads to improper inferences about the process. The lag plot works by plotting each residual value versus the value of the successive residual. Due to the way that the residuals are paired, there will be one less point than most other types of residual plots.
There will be no pattern or structure in the lag plot if the errors are independent. The points will appear randomly scattered across the plot, and if there is a significant dependence between errors, there will be some sort of deterministic pattern that is evident.
Figure 6. Example Lag Plot.
Potential Model Problems
Figure 7. Potential Model Issues Exposed by Residuals.
When we fit a model to a particular data set, one or more problems may occur. Most common among these are the following:
1. Non-linearity of the response-predictor relationships.
2. Correlation of error terms.
3. Non-constant variance of error terms.
4. Outliers.
5. High-leverage points.
6. Collinearity.
An assumption of many models is that the error terms have constant variance. However, it is often the case were variances are not constant and can increase with the value of the response. One common scenario is when the residual plots have a funnel shape. The funnel shape indicates that the residuals increase with the fitted values.
Figure 8. Non-constant Variance of Error Terms.
Missing Model Terms
Residual plots are the most valuable tool for assessing whether variables are missing in the functional part of the model. However, if the results are nebulous, it may be helpful to use statistical tests for the hypothesis of the model. One may wonder if it may be more useful to jump directly to the statistical tests (since they are more quantitative), however, residual plots provide the best overall feedback of the model fit. These quantitative tests are termed “lack-of-fit” tests, and there are many of them in any statistics textbook.
The most commonly used strategy is to compare the amount of variation in the residuals with an estimate of the random variation in the model using an additional data set. If the random variation is similar, then it can be assumed that no terms are missing from the model. If the random variation from the model is larger than the random variation from the independent data set, then terms may be missing or unspecified in the functional part of the model.
Comparing the variation between experimental and model data sets is very useful, however, there are many instances where a replicate measurement are not available. If this is the case, then the lack-of-fit statistics can be calculated by partitioning the residual standard deviation into two independent estimators of the random variation in the process.
One estimator depends upon the model and the means of the replicated sets of data (σm), and the other estimator is a standard deviation of the variation observed in each set of replicated measurements (σr). The squares of these two estimators are often called “mean square for lack-of-fit”. The model estimator can be calculated by [10]:
where p is the number of unknown parameters in the model, n is the sample size of the data set used to fit the model, nu is the number of combinations of predictor variable levels, is the number of replicated observations at the ith combination of predictor variable levels.
If the model is a good fit, the value of the function would be a good estimate of the mean value of response for every combination of predictor variable values. If the function provides good estimates of the mean response at the ith combination, then σm should be close in value to σr and should also be a good estimate of σ. If the model is missing any important terms, or any of the terms are correctly specified, then the function will provide a poor estimate of the mean response for some combination of predictors, and σm will probably be greater than σr.
The model dependent estimator can be calculated using [10]:
Since σr depends only on the data and not on the functional part of the model, this indicates that σr will be a good estimator of σ, regardless of whether the model is a complete description of the process. Typically, if σm > σr, then one or more parts of the model may be missing or improperly specified. Due to random error in the model, sometimes σm will be greater than σr even when the model is accurate. To insure that the model hypothesis is not rejected by accident, it is necessary to understand how much greater can σr possible be. This will insure that the hypothesis is only rejected when σm is greater than σr. A ratio that can be used when the model fits the data is [10]:
The probability of rejecting the hypothesis is controlled by the probability distribution that describes the behavior of the statistic, L. One method of defining the cut-off value is using the value of L when it is greater than the upper-tail cutoff value from the F distribution. This allows a quantitative method of determining when σm is greater than σr.
The probability specified by the cutoff value from the F distribution is called the “significance level” of the test. The most commonly used significance value is α = .05, which means that the hypothesis of an adequate model will only be rejected in 5% of tests for which the model really is adequate. The cut-off values can be calculated using the F distribution described in most statistics textbooks.
Unnecessary Terms in the Model
Sometimes models fit the data very well, but there are additional unnecessary terms. These models are said to “over fit” the data. Since the parameters for any unnecessary terms in the model usually have values near zero, it may seem harmless to leave them in the model. However, if there are many extra terms in the model, there could be occurrences where the error from the model may be larger than necessary and may affect conclusions drawn from the data.
Over-fitting often occurs when developing purely empirical models for experimental data, with little understanding of the total and random variation in the data. This happens when regression methods fit the data set instead of using functions to describe the structure in the data. There are models that sometimes are made to fit very complex patterns, which actually may be finishing structure in the noise if the model is analyzed carefully.
To determine if a model has too many terms, statistical tests can also be used. The tests for overfitting of the data are one area in which statistical tests are more effective than residual plots. In this case, individual tests for each parameter in the model are used rather than a single test. The test statistics for testing whether or not each parameter is zero are typically based on T distribution. Each parameter estimate in the model is measured in terms of how many standard deviations it is from its hypothesized value of zero. If the parameter’s estimated value is close enough to the hypothesized values that any additional deviation can be attributed to random error, then, the hypothesis that the parameter’s true value is not zero is accepted. However, if the parameter’s estimated value is so far away from the hypothesized value that the deviation cannot be plausibly explained by random error, the hypothesis that the true value of the parameter is zero is rejected.
The test statistic for each of these tests is simply the estimated parameter value divided by its estimated standard deviation:
Equation provides a measure of the distance between the estimated and hypothesized values of the parameter in standard deviations. Since the random errors are normally distributed, and the value of the parameter is zero, the test statistic has a Student’s t distribution with n - p degrees of freedom. Therefore, the cut-off values from the t distribution can used to determine the amount of variable that is due to random error. These tests should each be used with cutoff values with a significance level of α/2 since these tests are generally used to simultaneously test whether or not a parameter value is greater than or less than zero. This will insure that the hypothesis of each parameter equals zero will be rejected by chance with probability α.
Conclusion
Fuel cell validation is the most important step in the model-building process. However, little attention is usually given to this important step. A fast method for analyzing the validity of a model is look at plots of residuals versus the experimental factors, run plots and lag plots. These plots give a good feel for how accurately a model fits the experimental data, and how dependable it is. If residual scatterplots are used with one or more common statistical tests to discern fit, there will be substantial evidence that a model is a good fit to the experimental data.
Posted by Dr. Colleen Spiegel
Dr. Colleen Spiegel is a mathematical modeling and technical writing consultant (President of SEMSCIO) and Professor holding a Ph.D. and an MSc degree in Engineering. She has seventeen years of experience in engineering, statistics, data science, research & technical writing work for many companies as a consultant, employee, and independent business owner. She is the author of ‘Designing and Building Fuel Cells’ (McGraw-Hill, 2007) and ‘PEM Fuel Cell Modeling and Simulation Using MATLAB’ (Elsevier Science, 2008). She previously owned Clean Fuel Cell Energy, LLC, which was a fuel cell organization that served scientists, engineers, and professors world-wide.
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There has been a lot of emphasis on the development of long-lasting, efficient and portable, power sources for further technology improvement in commercial electronics devices, medical diagnostic equipment, mobile communication and military applications. These systems all require... | 3,249 | 16,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-34 | latest | en | 0.951364 |
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Disclaimer: Mometrix Media LLC, the publisher of this product, is not affiliated with or endorsed by any official testing organization. All organizational and test names are trademarks of their respective owners. | 497 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-30 | longest | en | 0.807077 |
http://www.bkgm.com/rgb/rgb.cgi?view+1382 | 1,550,427,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482347.44/warc/CC-MAIN-20190217172628-20190217194628-00544.warc.gz | 320,599,574 | 3,801 | Cube Handling
Robertie's rule
From: Chuck Bower Address: bower@astro.indiana.edu Date: 7 September 2006 Subject: Re: Online Match -- Thu, 07 Sep 2006 Forum: GammOnLine
```X on roll. Cube action?
24 23 22 21 20 19 18 17 16 15 14 13
143 +---+---+---+---+---+---+---+---+---+---+---+---+---+
| O O O | | O O O X |
| O O O | | X |
| | | X |
| | | |
| | | | +---+
| | | | | 1 |
| X | | O | +---+
| X | | X O |
| X | | X X O |
| X | | X X O |
| O X | | X X O |
133 +---+---+---+---+---+---+---+---+---+---+---+---+---+
1 2 3 4 5 6 7 8 9 10 11 12
I used Robertie Rule (5-1 for us; 4-3 for Kit) and felt the resulting
position was near borderline.
> What's Robertie's Rule?
Give yourself an above average roll (but not your best). Then give your
opponent a below average roll (but not his/her worst). If the resulting
position is still a Take, then don't double, otherwise double.
```
Cube Handling
Against a weaker opponent (Kit Woolsey, July 1994)
Closed board cube decisions (Dan Pelton+, Jan 2009)
Cube concepts (Peter Bell, Aug 1995)
Early game blitzes (kruidenbuiltje, Jan 2011)
Early-late ratio (Tom Keith, Sept 2003)
Endgame close out: Michael's 432 rule (Michael Bo Hansen+, Feb 1998)
Endgame close out: Spleischft formula (Simon Larsen, Sept 1999)
Endgame closeout: win percentages (David Rubin+, Oct 2010)
Evaluating the position (Daniel Murphy, Feb 2001)
Evaluating the position (Daniel Murphy, Mar 2000)
How does rake affect cube actions? (Paul Epstein+, Sept 2005)
How to use the doubling cube (Michael J. Zehr, Nov 1993)
Liveliness of the cube (Kit Woolsey, Apr 1997)
PRAT--Position, Race, and Threats (Alan Webb, Feb 2001)
Playing your opponent (Morris Pearl+, Jan 2002)
References (Chuck Bower, Nov 1997)
Robertie's rule (Chuck Bower, Sept 2006)
Rough guidelines (Michael J. Zehr, Dec 1993)
The take/pass decision (Otis+, Aug 2007)
Too good to double (Michael J. Zehr, May 1997)
Too good to double--Janowski's formula (Chuck Bower, Jan 1997)
Value of an ace-point game (Raccoon+, June 2006)
Value of an ace-point game (Øystein Johansen, Aug 2000)
Volatility (Chuck Bower, Oct 1998)
Volatility (Kit Woolsey, Sept 1996)
When to accept a double (Daniel Murphy+, Feb 2001)
When to beaver (Walter Trice, Aug 1999)
When to double (Kit Woolsey, Nov 1994)
With the Jacoby rule (KL Gerber+, Nov 2002)
With the Jacoby rule (Gary Wong, Dec 1997)
Woolsey's law (PersianLord+, Mar 2008)
Woolsey's law (Kit Woolsey, Sept 1996)
Words of wisdom (Chris C., Dec 2003) | 938 | 2,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | latest | en | 0.715564 |
https://newpathworksheets.com/math/grade-4/division-multiplication/south-dakota-common-core-standards | 1,632,278,540,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00170.warc.gz | 448,154,401 | 8,373 | ## ◂Math Worksheets and Study Guides Fourth Grade. Division/Multiplication
### The resources above correspond to the standards listed below:
#### South Dakota Content Standards
4.NBT. Number and Operation in Base Ten
4.NBT.B. Use place value understanding and properties of operations to perform multi-digit arithmetic.
4.NBT.B.5. Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
4.NBT.B.6. Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. | 190 | 953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-39 | latest | en | 0.826711 |
https://community.smartsheet.com/discussion/81768/changing-currency-on-a-column-where-the-if-formula-has-been-used | 1,717,060,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00264.warc.gz | 148,152,948 | 107,499 | # Changing currency on a column where the "if" formula has been used.
Options
Hello.
I am making a sheet where I use an "if" formula:
=IF([Responsible]@row = "Anders Larsen"; [Total Amount]@row; "") + IF([Responsible no.2]@row = "Anders Larsen"; [Total Amount]@row; "")
This Gives me number, however I am unable to change that number into a currency. why is that, and/or how can I fix it?
sidenote:
I am aware that I could use the "sumifs" formula. however, it is very important that there aren't any false values (by false I mean values = 0), and "sumif" automatically writes in 0 when the criteria's aren't met. That is why the "if" ends with "", so it remains blank if it isn't above zero.
thank you in advance.
• ✭✭✭✭✭✭
edited 08/05/21 Answer ✓
Options
Glad we got it working! (Me & Bassam)
Here's the formula for everyone.
```=IF(VALUE(IF([Project PTA]@row = "Anders Larsen", [Total Amount]@row, "") +
IF([Project PTA Secondary]@row = "Anders Larsen", [Total Amount]@row, "")) = 0, "",
VALUE(IF([Project PTA]@row = "Anders Larsen", [Total Amount]@row, "") +
IF([Project PTA Secondary]@row = "Anders Larsen", [Total Amount]@row, "")))
```
Please support the Community by marking the post(s) that helped or answered your question or solved your problem with the accepted answer/helpful. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
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• ✭✭✭✭✭✭
edited 08/03/21
Options
I hope you're well and safe!
It's because it's interpreted as text after the formula.
Try adding the VALUE function before the Total Amount parts in the formula.
Did that work/help?
I hope that helps!
Be safe and have a fantastic week!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
Did my post(s) help or answer your question or solve your problem? Please support the Community by marking it Insightful/Vote Up or/and as the accepted answer. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭✭✭
Options
Hope you are fine, please try the following formula:
```=IF([Responsible]@row = "Anders Larsen"; Value([Total Amount]@row); "")+
IF([Responsible no.2]@row = "Anders Larsen"; Value([Total Amount]@row); "")
```
bassam.khalil2009@gmail.com
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• Options
Hello and thank you for the suggestion.
However it doesn't appear to change anything, I am still not able to change the currency.
• ✭✭✭✭✭✭
Options
if you like to fix the formula directly on your sheet please share me as an admin on a copy of your sheets ( Source & Destination ) and i will write the exact formula for you then you can copy it to your original sheet.
My Email for sharing : Bassam.k@mobilproject.it
bassam.khalil2009@gmail.com
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• ✭✭✭✭✭✭
Options
I'd be happy to take a quick look.
Can you maybe share the sheet(s)/copies of the sheet(s)? (Delete/replace any confidential/sensitive information before sharing) That would make it easier to help. (share too, andree@workbold.com)
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• Options
Hello Bassam, I have now invited you to a copy of the sheet. I have hidden the irrelevant columns. It is the column called "Rest ALA" the formula is in.
• ✭✭✭✭✭✭
Options
Please stay with me in the sheet conversation
bassam.khalil2009@gmail.com
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• ✭✭✭✭✭✭
Options
Please check it, i fix it
bassam.khalil2009@gmail.com
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• ✭✭✭✭✭✭
edited 08/05/21 Answer ✓
Options
Glad we got it working! (Me & Bassam)
Here's the formula for everyone.
```=IF(VALUE(IF([Project PTA]@row = "Anders Larsen", [Total Amount]@row, "") +
IF([Project PTA Secondary]@row = "Anders Larsen", [Total Amount]@row, "")) = 0, "",
VALUE(IF([Project PTA]@row = "Anders Larsen", [Total Amount]@row, "") +
IF([Project PTA Secondary]@row = "Anders Larsen", [Total Amount]@row, "")))
```
Please support the Community by marking the post(s) that helped or answered your question or solved your problem with the accepted answer/helpful. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭✭✭
Options
Good work done by you as always.
bassam.khalil2009@gmail.com
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• ✭✭✭✭✭✭
Options
Thanks, and by you as well!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
## Help Article Resources
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# Python Math Game Error
P: 16 I made a math game and wanted it to print out which questions you got right at the end. How do i do that? This is my code. # Welcome To The Math Game print "Welcome To The Math Game." print "------------------------" #Rules print print "Just answer the problems as well as you can and do not use a calculator." print "At the end you will get your score." print "You will be asked 4 problems one multiplication one adding one subtracting and one dividing." print "Lets Start." print # Start It answer0 = 0 if answer0 == 0: answersright = 0 else: blah = 0 # Problem1 print "34*32" answer1 = input() if answer1 == 1088: answersright=answersright+1 else: answersright=answersright #problem 2 print "234/13" answer2 = input() if answer2 == 18: answersright = answersright+1 else: answersright=answersright #problem 3 print "412+2345" answer3 = input() if answer3 == 2768: answersright = answersright+1 else: answersright=answersright #problem 4 print "234-2548" answer4 = input() if answer4 == -2314: answersright=answersright+1 else: answersright=answersright #End Of Problems print "Congratulations you just finished the math game" print "You will now receive your score" print "You Got..." from time import sleep sleep(3) if answersright == 0: print "0 right" else: bob = 1 if answersright == 1: print"1 Right" else: blah = 1 if answersright == 2: print "2 right" else: blah = 2 if answersright == 3: print "3 right" else: blah = 3 if answersright == 4: print "4 right" problem0 = 0 else: blah2 = 1 Nov 30 '07 #1
Share this Question
2 Replies
P: 56 Hello there! I think the first thing you should do is to get your program to work with a 'for' loop , so you wont have to write the same things amny times... Before that , you must prepare a 'list' of 'lists' with the question,the correct answer and a 'flag' reference , which your program will change according to wether the answer was wright or wrong. Then at the end , you can sum all the questions which have a "correct answer" flag , and print them. Here is a guideline: Expand|Select|Wrap|Line Numbers question_list=[["10*10",100,0],["20*20",400,0]] for question in question_list: answer = raw_input "how much is "+question[0]+" ?" if answer==question[1]: question[2]=1 This code modifies the third item of each sublist to "1" if the answer was correct , so all you have to do is check which sublists have "1" as third item. Hope i helped a little.... Elias Dec 2 '07 #2
P: 16 Thanks for you help. I got it fixed with the following code: # Welcome To The Math Game print "Welcome To The Math Game." print "------------------------" #Rules print print "Just answer the problems as well as you can and do not use a calculator." print "At the end you will get your score." print "You will be asked 4 problems one multiplication one adding one subtracting and one dividing." print "Lets Start." print # Start It answer0 = 0 if answer0 == 0: answersright = 0 else: blah = 0 # Problem1 print "34*32" answer1 = input() if answer1 == 1088: answersright=answersright+1 q1=1 else: answersright=answersright q1=0 #problem 2 print "234/13" answer2 = input() if answer2 == 18: answersright = answersright+1 q2=1 else: answersright=answersright q2=0 #problem 3 print "412+2345" answer3 = input() if answer3 == 2768: answersright = answersright+1 q3=1 else: answersright=answersright q3=0 #problem 4 print "234-2548" answer4 = input() if answer4 == -2314: answersright=answersright+1 q4=1 else: answersright=answersright q4=0 #End Of Problems print "Congratulations you just finished the math game" print "You will now receive your score" print "You Got..." from time import sleep sleep(3) if answersright == 0: print "0 right" else: bob = 1 if answersright == 1: print"1 Right" else: blah = 1 if answersright == 2: print "2 right" else: blah = 2 if answersright == 3: print "3 right" else: blah = 3 if answersright == 4: print "4 right" problem0 = 0 else: blah2 = 1 if q1>0: print "You got question 1 right" else: print "You got question 1 wrong" if q2>0: print "You got question 2 right" else: print "You got question 2 wrong" if q3>0: print "You got question 3 right" else: print "You got question 3 wrong" if q4>0: print "You got question 4 right" else: print "You got question 4 wrong" Dec 2 '07 #3
### Post your reply
Sign in to post your reply or Sign up for a free account. | 1,322 | 4,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-05 | latest | en | 0.785839 |
https://www.mrexcel.com/board/threads/finding-the-position-of-the-nth-match-in-an-array-with-multiple-criteria.1108395/ | 1,586,063,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00052.warc.gz | 1,063,720,829 | 16,685 | # Finding the position of the Nth match in an array with multiple criteria
#### el saxo
##### New Member
Hello all I hope someone can help me please, I've tried searching lots of sites for an answer but so far I haven't been successful...
In my worksheet, each row contains one of certain text values in columns CL to CV - either 'Transferred', 'Abandoned', 'Consult', 'Call Ended' or 'NULL', as illustrated below:
CL CM CN CO CP CQ CR CS CT CU CV 2 Transferred Consult Consult Abandoned Transferred Consult Call Ended NULL NULL NULL NULL
<tbody>
</tbody>
Assuming column CL always contains 'Transferred', I want to find out the position (in this array) of the next column to contain EITHER 'Transferred' or 'Call Ended'. I.e. in the example above I would expect to return a 5 because column CP contains 'Transferred' and is the fifth column in the array CL:CV.
I can already find the position of the 'Nth' match based on a single criteria - for example if I just wanted to know the position of the next column after CL to contain 'Transferred' I would use this array formula:
{=SMALL(IF(CL2:CV2="Transferred",COLUMN(CL2:CV2)-COLUMN(CL2)+1),2)}
So, I tried introducing an 'OR' argument as shown below, but this doesn't work because it always returns a 2:
{=SMALL(IF(OR(CL2:CV2="Transferred",CL2:CV2="Call Ended"),COLUMN(CL2:CV2)-COLUMN(CL2)+1),2)}
With a bit of googling I think I've learned that 'OR' just doesn't work with SMALL IF, but this is where I'm stumped.
Can anyone help me solve this please?
Thanks
### Excel Facts
How to total the visible cells?
From the first blank cell below a filtered data set, press Alt+=. Instead of SUM, you will get SUBTOTAL(9,)
#### jasonb75
##### Well-known Member
When you use OR it only gives a single result, not an array.
See if this one does what you need. Note that if you are in a country where you have to use ; instead of , then you might need to change the ; that is already in the formula to \
=AGGREGATE(15,6,COLUMN(CL2:CV2)/(CL2:CV2={"Transferred";"Call Ended"}),2)-COLUMN(CL2)+1
#### el saxo
##### New Member
When you use OR it only gives a single result, not an array.
See if this one does what you need. Note that if you are in a country where you have to use ; instead of , then you might need to change the ; that is already in the formula to \
=AGGREGATE(15,6,COLUMN(CL2:CV2)/(CL2:CV2={"Transferred";"Call Ended"}),2)-COLUMN(CL2)+1
Hi Jason, unfortunately that doesn't seem to work once it's copied down all rows - for example on a row containing 'Call Ended' in column CM (i.e. the 2nd column), it returns a 3.
However, it gave me an idea...
I swapped the 'OR' argument out of my original attempt and instead used semicolon-separated values in curly brackets (like you used in your AGGREGATE formula) for the conditional test of the IF statement, like so:
{=SMALL(IF(CL2:CV2={"Transferred";"Call Ended"},COLUMN(CL2:CV2)-COLUMN(CL2)+1),2)}
and it does exactly what I wanted because it returns an array of results instead of a single result.
So thanks very much for your reply because, without seeing it in your solution, I probably wouldn't have thought to try that!
#### jasonb75
##### Well-known Member
I don't see why one would work and not the other, aggregate(15,6 is exactly the same as {small(if(
I just tried 'Call Ended' in CM and it returned 2 as expected, the only exception should be is if CL didn't contain Transferred.
Not critical though, as long as you have something that does what you need it to, that's the important thing.
#### el saxo
##### New Member
I don't see why one would work and not the other, aggregate(15,6 is exactly the same as {small(if(
I just tried 'Call Ended' in CM and it returned 2 as expected, the only exception should be is if CL didn't contain Transferred.
Not critical though, as long as you have something that does what you need it to, that's the important thing.
Apologies, I just tried your formula again and compared the results, and you're quite right! I think I know what I did - I must have offset the formula by one row by mistake, i.e. pasted it in row 3. I blame a lack of caffeine... :wink:
Looks like both formulas are equally valid, and I can wrap them in an IF statement to handle any instances in my data where CM doesn't contain 'Transferred' (which I'm not interested in anyway). | 1,107 | 4,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.883005 |
https://git-ie.linaro.org/people/linus.walleij/linux-stericsson.git/commit/arch/alpha/kernel?id=1e9d90dbed120ec98517428ffff4dacd9797e39d | 1,642,691,018,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00303.warc.gz | 344,972,150 | 3,627 | aboutsummaryrefslogtreecommitdiff log msg author committer range
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author committer Nicolin Chen 2020-09-01 15:16:45 -0700 Christoph Hellwig 2020-09-03 18:12:15 +0200 1e9d90dbed120ec98517428ffff4dacd9797e39d (patch) d743be6b54427bb62e36e6d1ae2f6fc243cd70b7 /arch/alpha/kernel 2281f797f5524abb8fff66bf8540b4f4687332a2 (diff) linux-stericsson-1e9d90dbed120ec98517428ffff4dacd9797e39d.tar.gz
dma-mapping: introduce dma_get_seg_boundary_nr_pages()
We found that callers of dma_get_seg_boundary mostly do an ALIGN with page mask and then do a page shift to get number of pages: ALIGN(boundary + 1, 1 << shift) >> shift However, the boundary might be as large as ULONG_MAX, which means that a device has no specific boundary limit. So either "+ 1" or passing it to ALIGN() would potentially overflow. According to kernel defines: #define ALIGN_MASK(x, mask) (((x) + (mask)) & ~(mask)) #define ALIGN(x, a) ALIGN_MASK(x, (typeof(x))(a) - 1) We can simplify the logic here into a helper function doing: ALIGN(boundary + 1, 1 << shift) >> shift = ALIGN_MASK(b + 1, (1 << s) - 1) >> s = {[b + 1 + (1 << s) - 1] & ~[(1 << s) - 1]} >> s = [b + 1 + (1 << s) - 1] >> s = [b + (1 << s)] >> s = (b >> s) + 1 This patch introduces and applies dma_get_seg_boundary_nr_pages() as an overflow-free helper for the dma_get_seg_boundary() callers to get numbers of pages. It also takes care of the NULL dev case for non-DMA API callers. Suggested-by: Christoph Hellwig <hch@lst.de> Signed-off-by: Nicolin Chen <nicoleotsuka@gmail.com> Acked-by: Niklas Schnelle <schnelle@linux.ibm.com> Acked-by: Michael Ellerman <mpe@ellerman.id.au> (powerpc) Signed-off-by: Christoph Hellwig <hch@lst.de>
Diffstat (limited to 'arch/alpha/kernel')
-rw-r--r--arch/alpha/kernel/pci_iommu.c7
1 files changed, 1 insertions, 6 deletions
diff --git a/arch/alpha/kernel/pci_iommu.c b/arch/alpha/kernel/pci_iommu.cindex 81037907268d..6f7de4f4e191 100644--- a/arch/alpha/kernel/pci_iommu.c+++ b/arch/alpha/kernel/pci_iommu.c@@ -141,12 +141,7 @@ iommu_arena_find_pages(struct device *dev, struct pci_iommu_arena *arena, unsigned long boundary_size; base = arena->dma_base >> PAGE_SHIFT;- if (dev) {- boundary_size = dma_get_seg_boundary(dev) + 1;- boundary_size >>= PAGE_SHIFT;- } else {- boundary_size = 1UL << (32 - PAGE_SHIFT);- }+ boundary_size = dma_get_seg_boundary_nr_pages(dev, PAGE_SHIFT); /* Search forward for the first mask-aligned sequence of N free ptes */ ptes = arena->ptes; | 818 | 2,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.598721 |