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http://blog.myrank.co.in/instantaneous-velocity-and-speed/

# Instantaneous Velocity and Speed Instantaneous velocity: It is defined as the rate of change of position for a time interval which is very small almost zero. Beside Instantaneous velocity, average velocity is a way of defining how fast a body is moving. But average velocity is only the average of the velocity of body over a given time interval, and not the exact measure of velocity of a body at an instant. The exact measure of velocity of a body at an instant or over a very – very – very short time interval Δt is known as instantaneous velocity. Mathematically Instantaneous velocity of an object is the limiting value of average velocity as Δt tends towards zero. $$\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\frac{\Delta x}{\Delta t}$$ And Instantaneous velocity is also the derivative of the position of the object (x) with respect to time interval (t). Instantaneous velocity =  $$\frac{dx}{dt}$$ In a graph of motion of an object drawn “x” versus “t”, Instantaneous velocity of the object is the slope of tangent line drawn at the position representing that instant. Instantaneous velocity is a vector quantity and thus have its direction associated with it along with its magnitude. We know that the average velocity for a given time interval is total displacement divided by total time. As this time interval approaches zero, the displacement also approaches zero. But the limit of the ratio of displacement and time is non-zero and is called the Instantaneous velocity.If the displacement of the particle varies with respect to time and is given as (6t² + 2t + 4) m, the Instantaneous velocity can be found out at any given time by: s = (6t² + 2t + 4) Velocity (V) =  $$\frac{ds}{dt}$$ $$=\,\frac{d(6{{t}^{2}}+2t+4)}{dt}$$ = 12 t + 2 So, If we have to find out the Instantaneous velocity at t = 5 sec, then we will put the value of t in the obtained expression of velocity. Instantaneous velocity at t = 5 sec = (12 x 5 + 2) = 62 m/sec Let us calculate the average velocity now for 5 seconds. Displacement = (6 x 5² + 2 x 5 + 4) = 164 m Average Velocity =$$\frac{164}{5}$$ = 32.8 m/sec Did the bullet have a speed at the instant this picture was taken? And how fast were you driving at a specific point or at a specific time? To answer these questions, we need to consider a new concept known as INSTANTANEOUS SPEED.Instantaneous Speed: We know that the average speed is for a given time interval is total distance travelled divided by the total time taken. As this time interval approaches zero, the distance travelled also approaches zero. But the limit of the ratio of distance and time is non-zero and is called the Instantaneous speed. To understand it in simple words we can also say that instantaneous speed at any given time is the magnitude of the instantaneous velocity at that time. If distance as a function of time is known to us, we can find out the instantaneous speed at any time. Let’s understand this by means of an example. Distance (s) = 5t³ m Speed (V) =  $$\frac{ds}{dt}$$ $$=\frac{d(5{{t}^{3}})}{dt}$$ = 15 t² Now, we can easily find the Instantaneous speed at any given time by putting the value of t in this Obtained Expression. Real life Applications: 1. The speedometer of a car reveals information about the Instantaneous speed of your car. It shows your speed at a particular instant in time. On the average, your car was moving with a speed of 25 miles per hour. 2. A cheetah who is running with speed of 80 miles per hour then it is his Instantaneous speed because it is shown as in per hour speed. How to find Instantaneous Speed? Example: The displacement of a body is given by the equation D = at². Calculate Instantaneous speed of the body at t = 2sec, If the value of acceleration is 5 m/s²? Solution: We know that, Ins tan taneous velocity (IS) =  $$\underset{T\to t}{\mathop{\lim }}\,\frac{dX}{dT}$$ So, here X = D and hence we have $$IS\,=\,\underset{t\to 2}{\mathop{\lim }}\,\frac{d(a\times {{t}^{2}})}{dt}$$ $$IS=\,\underset{t\to 2}{\mathop{\lim }}\,\,(2at)$$ IS = 2 x 5 x 2 = 20 m/sec Hence, the Instantaneous speed of the body is 20 m/sec.

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https://papers.nips.cc/paper/2018/hash/b4a721cfb62f5d19ec61575114d8a2d1-Abstract.html

#### Authors Jianqiao Zhu, Adam Sanborn, Nick Chater #### Abstract Both resources in the natural environment and concepts in a semantic space are distributed "patchily", with large gaps in between the patches. To describe people's internal and external foraging behavior, various random walk models have been proposed. In particular, internal foraging has been modeled as sampling: in order to gather relevant information for making a decision, people draw samples from a mental representation using random-walk algorithms such as Markov chain Monte Carlo (MCMC). However, two common empirical observations argue against people using simple sampling algorithms such as MCMC for internal foraging. First, the distance between samples is often best described by a Levy flight distribution: the probability of the distance between two successive locations follows a power-law on the distances. Second, humans and other animals produce long-range, slowly decaying autocorrelations characterized as 1/f-like fluctuations, instead of the 1/f^2 fluctuations produced by random walks. We propose that mental sampling is not done by simple MCMC, but is instead adapted to multimodal representations and is implemented by Metropolis-coupled Markov chain Monte Carlo (MC3), one of the first algorithms developed for sampling from multimodal distributions. MC3 involves running multiple Markov chains in parallel but with target distributions of different temperatures, and it swaps the states of the chains whenever a better location is found. Heated chains more readily traverse valleys in the probability landscape to propose moves to far-away peaks, while the colder chains make the local steps that explore the current peak or patch. We show that MC3 generates distances between successive samples that follow a Levy flight distribution and produce 1/f-like autocorrelations, providing a single mechanistic account of these two puzzling empirical phenomena of internal foraging.

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https://asmedigitalcollection.asme.org/turbomachinery/article/145/6/061017/1156453/Experimental-Study-of-Impact-of-In-Service

## Abstract In this paper we experimentally evaluate the impact of in-service deterioration on the aerodynamic performance of heavily film-cooled high-pressure nozzle guide vanes from large civil jet engines. We study 15 mid-life to end-of-life parts removed from operational engines, and compare their performance to those of new parts. Deterioration features included: increased surface roughness; thermal barrier coating spallation; damaged film cooling holes; and trailing edge burn-back. We characterize and present statistics for the surface roughness. Aerodynamic measurements were performed in the high technology readiness level Engine Component AeroThermal (ECAT) facility at the University of Oxford, at engine-representative conditions of exit Mach number, exit Reynolds number, coolant-to-mainstream pressure ratio, and turbulence intensity. We present detailed experimental measurements of the coolant capacity characteristics, downstream loss, and downstream flow structures. The results show that service time has the following effects on high-pressure nozzle guide vanes: increased equivalent sandgrain roughness of (up by 1056% change); reduced coolant flow capacity (maximum change of −6.27% for film cooling holes and −24.7% for the trailing edge slot); increased overall mixed-out kinetic energy loss coefficient by (up to 33% change); leads to greater downstream flow angle variation (change of −6 deg). This is one of the first significant studies of its type in the open literature, and is an important step towards whole-life engine performance assessment. ## Introduction and Related Literature High-pressure (HP) nozzle guide vanes (NGVs) deteriorate with service time due to (see, for example [1]) oxidation, deposition of airborne contaminants, thermal fatigue cracking (e.g., of the thermal barrier coat (TBC) layer), and erosion due to impact-abrasion [2]. These effects lead to both local changes in surface roughness, and global changes in geometry. Thermal barrier coating spallation and cracking is common, as is the partial collapse of cooling holes (smaller effective size), or the blockage or geometry change of holes due to particulate deposition. Eroded trailing edges (TE), or TE burn-back due to overheating are also common. These effects generally reduced the aero-thermal performance of the parts. The purpose of this paper is to perform accurate aerodynamic characterization of engine-run parts. We first review the literature related to the impact of surface roughness and TE damage on aerodynamic performance. ### Impact of Surface Roughness. There have been several cascade experiments to assess the impact of surface roughness on the aerodynamic performance of turbine components. Most of these studies have used simple simulated roughness elements (sandgrains, hemispheres, and truncated cones) [35] or surface finishes arising from manufacturing processes (e.g., machining processes, polishing processes, and coating types) [611], with only one study (to the authors knowledge) attempting to simulate engine-representative roughened surfaces on an airfoil [12]. A comprehensive review of literature (up to 2010) on effects due to roughness was provided by Bons [13]. As expected, the primary effects are earlier laminar-turbulent boundary layer transition, and an increased rate of thickening of the boundary layer (in both states), leading to higher aerodynamic loss. One exception is the case of low Reynolds number (Re) operation (low pressure turbines for instance) where an earlier roughness-induced transition has been shown to prevent a laminar separation on the suction surface (SS) of the airfoil, leading to lower losses than a smooth (separated) airfoil [10,13,14]. This general effect is—of course—well known. We can differentiate between studies in which the surface roughness of the entire airfoil is varied evenly to represent possible manufacturing routes [3,4,610], for example, metal printing techniques, and those in which the roughness on only part of the surface is varied [5,11,12]. These latter studies are more representative of in-service roughening which appears to preferentially affect the pressure surface (PS) and early SS. Kind et al. [5] performed aerodynamic surveys 0.4 axial chords (Cax) downstream of a low-speed linear cascade of turbine blades to study the impact of roughness on profile loss. The inlet ReC (based on tangential chord) was 3.0 × 105. They tested 18 cascade configurations: one smooth cascade configuration in which the airfoil surface (PS and SS) and endwalls are reported to be aerodynamically smooth (roughness height not reported) and 17 other configurations in which spanwise-oriented bands of roughness (application of sand grains) were introduced at different locations around the airfoil. The roughness distribution most relevant to the current study was a roughened band covering approximately 30‒100% of the PS with the rest of the airfoil being smooth. For this distribution, two sandgrain roughness heights (k) were used: k/C = 4.5 × 10−3 and 6.3 × 10−3, where C is the tangential chord. They reported that at the design incidence angle the profile loss increased from the smooth configuration by approximately 30% and 40% for the two roughness heights, respectively. Matsuda et al. [11] performed aerodynamic surveys 0.165 tangential chords downstream of a linear cascade of vanes to study the impact of airfoil and endwall roughness on profile loss and net secondary loss. Net secondary loss was defined as the difference between the total pressure loss measured in the secondary flow region, and the estimated profile loss (based on mid-span measurements). The exit ReC (based on tangential chord) range was 0.30 × 106 < ReC < 1.0 × 106 and inlet turbulence intensity (Tu) was 0.43%. They tested two configurations: a smooth cascade configuration in which the airfoil surface (PS and SS) and endwalls had a normalized maximum peak-to-valley roughness height Rz/C of 5.2 × 10−5 (typical of a milled surface of a new steam turbine vane); and a rough cascade configuration in which the PS and endwalls had Rz/C equal to 84.0 × 10−5 (approximately 16 times larger than the smooth configuration, and typical of a shot blasted surface), and with the SS essentially unchanged from the smooth case with Rz/C equal to 6.0 × 10−5. At ReC = 0.87 × 106, they found that the profile loss and net secondary loss for the rough configuration increased by approximately 36% and 80%, respectively, compared to the smooth configuration. The measured increase in loss was similar in magnitude to the study of Kind et al. [5], despite an almost order-of-magnitude difference in reported surface roughness height. This apparently surprising result may be explained by a combination of two factors: the peak-to-valley roughness height, Rz, and sandgrain roughness height, k, both fail to take account of shape and density of the roughness elements in the profile; the roughness types are different, one surface being artificially roughened (sand grains) and the other being the result of a manufacturing process. Differences in roughness type and quantification make the comparison between studies difficult. Erickson et al. [12] studied the impact of in-service roughening on the aerodynamic performance of turbine vanes in a low-speed linear cascade over a wide range of Tu and Re. Because of the similarity of their study to the current experiment (in terms of roughness type, distribution, modeling method, height, and operating conditions of Re and Tu) we consider the Erickson et al. [12] study an important reference point when it comes to the aerodynamic impact of roughness. At the closest condition to our experiment an increase in midspan total pressure loss coefficient of 12% was measured. We devote a later section of this paper to a more detailed comparison of our work and that of Erickson et al. [12]. The purpose is to allow detailed comparison of very similar studies, in the hope of cross-validation of the results, and in the hope of developing a benchmark loss-enhancement factor for typical engine geometries under engine realistic conditions. Comparison to existing correlations would be desirable, but is problematic, because, as demonstrated in the extensive review of Bons [13], variation in roughness type (simulated, manufacturing-related, and service-related) appears to have a significant effect on the correlation, and therefore a single global correlation remains elusive. It may be the case that in the foreseeable future it is necessary to perform high-fidelity studies of the type presented in this paper, in the hope of direct rules-of-thumb for particular combinations of part-type and deterioration characteristic. ### Impact of Trailing Edge Damage. We now review the literature on the impact of TE damage on the aerodynamic characteristics of turbine components. Sjolander et al. [15] studied the impact of a semi-circular cutout in the TE of a turbine blade (to simulate damage caused by burn-back or erosion) on loss and flow turning angle. Using a low-speed linear cascade, they did aerodynamic surveys downstream of an undamaged blade and two blades with a semi-circular cutout in their TE with diameters equivalent to 15% and 25% of the blade tangential chord. Results showed that the wake downstream of the cutout was significantly modified, with local under-turning of the flow (up to 4.2 deg yaw angle deficit compared to a new blade) in the cutout region. They also observed the formation of counter-rotating streamwise vortices at the edges of the cutout, driven by the sharp gradients (with radius) in yaw angle. Similar flow structures have been observed in computational fluid dynamics (CFD) studies of vanes with TE damage by Di Mare et al. [16] and Meyer et al. [17]. Sjolander et al. reported that the mass-flux-average profile total pressure loss coefficient increased by 56% (compared to undamaged blade) for the deepest cutout. Bouchard et al. [18] did aerodynamic surveys downstream of full rings of uncooled new and in-service deteriorated engine vanes in a transonic annular cascade. The vanes were those of a Rolls-Royce A-250 engine. The TE of the deteriorated vanes had small chordwise cracks, and some TE were significantly thermally distorted. No surface roughness measurements were reported in the paper. Profile loss measurements were taken one half of an axial chord downstream of the vanes at exit Mach number (M) M = 1.2. They reported an increase in profile loss of 51% for the deteriorated vane ring. ### Objective of This Study. Although there is a small body of literature which presents results of studies of simulated deterioration features in isolation, there are very few studies indeed [18] which characterize components with in-service deterioration. Although the overall impact of deterioration could reasonably be expected to be well-modeled by superposition of the individual deterioration effects, for the purpose of building whole-life turbine performance models it is important to characterize real engine parts at various stages through the life cycle. There is very little data available for this task. The purpose of this paper—and the main distinction from previous studies—is to characterize the range of aerodynamic performance that exists for real-engine HP NGVs subject to in-service deterioration. The data provide a reference point for whole-life performance modeling. To this end, for 15 engine-run parts, we present detailed measurements of: surface roughness; coolant capacity; and aerodynamic performance. Measurements were conducted for parts operating under engine-representative conditions of exit M and exit Re. The data are to be used to benchmark whole-life modeling codes. ## Characterization of Tested Components In this study we experimentally characterize a total of 17 real-engine vanes. The vane design was that from a modern large civil jet engine. The vanes were divided in two sets, referred to as set 1 and set 2. Set 1 was composed of one new vane and 12 mid-life vanes and set 2 was composed of one new vane and three end-of-life vanes. The new vane within each set has the same as-new geometry as the deteriorated vanes in the same set and acts as a baseline for that set. The deteriorated vanes within each set are from the same engine, and therefore were operated in the same environment and for the same number of cycles. The two sets operated in a non-sandy, yet not identical, environment. A summary of the operation history of both sets of vanes is given in Table 1. Table 1 Descriptive summary of the operation history of the in-service deteriorated parts in set 1 and set 2 ParameterSet 1Set 2 Number of vanes12 (+1 new)3 (+1 new) Position in life cycleMid-lifeEnd-of-life Operating environmentNon-sandyNon-sandy ParameterSet 1Set 2 Number of vanes12 (+1 new)3 (+1 new) Position in life cycleMid-lifeEnd-of-life Operating environmentNon-sandyNon-sandy Although the vanes in set 1 and set 2 are from the same engine-class, they are from different generations of the same engine and have slightly different geometries. Thus, comparisons between sets should be avoided. All vanes were fully featured, including film cooling holes, TE slots, and TBC. We identify individual vanes using their set number (S1 or S2) and vane number within the set (V1, V2, etc.). Although experimental data were acquired for all vanes in both sets, in this paper we focus on data from four vanes for each set: one new vane and three deteriorated vanes. For the focused study, we chose the vanes with deterioration patterns representative of the full population of vanes. We refer to selected set 1 vanes as: S1-V1 (new); and S1-V2, S1-V3, and S1-V4 (mid-life). The corresponding nomenclature for set 2 is: S2-V1 (new); and S2-V2, S2-V3, and S2-V4 (end-of-life). Although we present detailed measurements for only this sample, we present statistics on the parent set of parts where space allows. We now characterize the deterioration of the tested vanes. First we consider TE burn-back and TBC spallation, then we present surface roughness measurements. ### Trailing Edge Burn-Back and Thermal Barrier Coat Spallation. Figure 1 shows a schematic of the vanes in set 1 (S1) and set 2 (S2). Film cooling holes and the TE slots are omitted for clarity. The vane outline is shown by a solid outline. Solid and dashed lines on the vane surface demarcate regions of TE burn-back and TBC spallation, respectively. Fig. 1 Fig. 1 Close modal We consider first the selection of set 1 parts: one reference vane and three mid-life vanes. Vanes S1-V1 to S1-V4 are illustrated in Fig. 1(a). We see TE burn-back for vanes S1-V2 and S1-V3, with notches at approximately mid-span that extend to a depth equal to approximately 5% of the tangential chord and up to 18% of the span of the vane. There is a small region of TBC spallation at the leading edges (LE) close to the stagnation point for vanes S1-V3 and S1-V4. These regions are narrow and cover less than 10% of the span of the vanes. The TE burn-back and TBC spallation on vanes S1-V2 to S1-V4 are representative of the damage of the other deteriorated vanes in set 1. Now we consider the set 2 parts: one reference vane and three end-of-life vanes. With reference to Fig. 1(b), we see both significant TE burn-back and significant TBC spallation. Looking first at the TE damage, vane S2-V2 was burned back up to 14.2% of the tangential chord and the damage covers 80% of the span of the vane. The TE of vane S2-V3 is burned back up to 9.6% of the tangential chord at midspan and the damage covers 40% of the span of the vane. The TE of vane S2-V4 is slightly cracked and distorted towards the SS, with TE TBC spallation on both the PS and SS extending up to 7% of the tangential chord over 35% of the span of the vane. It is suspected that the TE damage is the result of local overheating rather than erosion (by, for example, sand) because approximately half of the vanes from the same engine had relatively undeteriorated TE. Now consider the LE region damage for the set 2 (end-of-life) vanes. The LE of all vanes in this set were severely damaged with TBC spallation covering up to 65% of the span of the vanes, and extending as far as 30% of the tangential chord on the PS of the vanes (see Fig. 1(b)). The spallation was so severe in some places that the metal was left unprotected. This is thought to be due to a combination of thermal fatigue cracking and oxidation. In the regions of most severe deterioration, primarily in the showerhead region of the parts, there was significant collapse (shrinkage of diameter, and damage to form) of the cooling holes, and cracking between holes. A descriptive summary of the deterioration characteristics of both sets of vanes is given in Table 2. Table 2 Descriptive summary of deterioration characteristics of the set 1 (mid-life) and set 2 (end-of-life) parts VaneLE geometryTE geometry S1-V1NewNew S1-V2IntactSmall notch at midspan S1-V3Minor TBC spallationSmall notch at midspan S1-V4Minor TBC spallationIntact S2-V1NewNew S2-V2Severe TBC spallation; collapsed cooling holes; cracking between holes Burn-back along the entire span S2-V3Severe TBC spallation; collapsed cooling holes; cracking between holes Burn-back at midspan S2-V4Severe TBC spallation; collapsed cooling holes; cracking between holes Minor distortion and cracking; minor TBC spallation VaneLE geometryTE geometry S1-V1NewNew S1-V2IntactSmall notch at midspan S1-V3Minor TBC spallationSmall notch at midspan S1-V4Minor TBC spallationIntact S2-V1NewNew S2-V2Severe TBC spallation; collapsed cooling holes; cracking between holes Burn-back along the entire span S2-V3Severe TBC spallation; collapsed cooling holes; cracking between holes Burn-back at midspan S2-V4Severe TBC spallation; collapsed cooling holes; cracking between holes Minor distortion and cracking; minor TBC spallation ### Surface Roughness Measurements. In this section we present surface roughness measurements for the tested vanes. Profilometry measurements were performed using an Alicona InfiniteFocus profilometer (non-contact optical three-dimensional surface measurement system). Measurements were taken at three span fractions and three streamwise positions on the PS, and at three span fractions and four streamwise positions on the SS. These 21 locations are indicated in Fig. 2. Each profilometry measurement covered a 500 µm × 500 µm area and consisted of 250,000 (500 × 500) discrete points. The measurement resolution in the wall-normal direction (z) direction was 0.1 µm. The macroscopic surface shape was evaluated by cubic surface fitting and was removed from the measurement to only keep the surface profile. Figure 3 shows examples of the measured and processed surface profiles. The positive streamwise direction is along the x axis. The data show good measurement resolution. The new TBC surface profile (Fig. 3(a)) was measured at 50% span at position PS2 (Fig. 2) on vane S1-V1 and is characterized with a high density of shallow and blunt valleys. The deteriorated TBC surface profile (Fig. 3(b)) was measured at 50% span at position PS2 on vane S1-V2 and is characterized by sharper high peaks formed by deposits. The shape of the peaks appears to be independent of direction (streamwise or spanwise). Fig. 2 Fig. 2 Close modal Fig. 3 Fig. 3 Close modal We represent roughness at the measurement locations with an equivalent sandgrain roughness height (ks) using the methodology proposed by Bons [19]. We present ks normalized by the tangential chord C of the vanes. Figure 4 shows ks/C measured on vanes S1-V1 to S1-V4 for set 1 and S2-V1 to S2-V4 for set 2 (entire set). Fig. 4 Fig. 4 Close modal Plots (a) and (b) show data for the PS of the vanes in set 1 and set 2, respectively. Plots (c) and (d) show data for the SS of the vanes in set 1 and set 2, respectively. Position labels indicate: the side of the vane (PS or SS); streamwise position (1–3 on PS and 1–4 on SS, as shown in Fig. 2); and span fraction (20%, 50%, and 80%). Each bar at a given position corresponds to a particular vane measurement. Consider first the vanes in set 1 (mid-life) part data of Figs. 4(a) and 4(c). On the PS and early SS (SS/1/XX locations, where XX = 20, 50, or 80) we see a significant increase in ks/C of the deteriorated vanes compared to the new vane. The mean increase in these 12 locations was 441%, with a maximum increase of 1056%. On the remainder of the SS (locations SS/2–4/XX) the deteriorated and new vanes are more similar, with a mean increase in roughness of 53.3%, and a maximum increase of 192%. This is in accord with the result of Bons et al. [20], who documented the surface roughness on nearly 100 in-service-deteriorated turbine components and noted that the PS and early SS suffer from in-service roughening much more than the late SS. Now consider the set 2 (end-of-life) part data of Figs. 4(b) and 4(d). The early PS (locations PS/1/XX) and early SS (locations SS/1/XX) show a significant increase in roughness compared to the new vane, with a mean increase in roughness of 104% and a maximum increase of 723%. In all other locations on the vane the increase in roughness is smaller, with a mean increase of 49.5%, and a maximum increase of 201%. Taking all vanes within set 1 (mid-life) (entire set of 12 deteriorated vanes and one new vane), the mean values of ks/C across all measurement points were 0.109 × 10−3 for the new vane and 0.399 × 10−3 for the deteriorated vanes (average of 12 vanes). The average increase in roughness (compared to the new vanes) was 266%. This was strongly biased towards the PS and the early SS (locations SS/1/XX). Taking the vanes within set 2 (end-of-life) as a whole (three deteriorated vanes in total and one new vane), the mean values of ks/C across all measurement points were 0.103 × 10−3 for the new vane and 0.185 × 10−3 for the deteriorated vanes (average of three vanes). The average increase in roughness (compared the new vanes) was 79.6%. Again, this was strongly biased towards the PS and the early SS (locations SS/1/XX). Although in this paper we focus on the airfoil surface, for completeness we note that the hub and casing platforms of the deteriorated vanes of both set 1 and set 2 were qualitatively rougher than new platforms, but without major geometrical deviations. ## Overview of the Experimental Facility Aerodynamic measurements were conducted in the Engine Component AeroThermal (ECAT) facility at the University of Oxford [21]. This is a high technology readiness level (TRL) blowdown facility capable of testing an annular cascade of HP NGVs from operating engines at engine-representative conditions of Mach number, Reynolds number, and coolant-to-mainstream pressure ratio. A schematic cross section of the facility working section with the traverse system installed is presented in Fig. 5. Fig. 5 Fig. 5 Close modal ### Operating Conditions. During a test, high-pressure air (initially stored in large tanks) is discharged through the cascade of vanes. Steady operating conditions can be achieved for approximately 60 s by the action of a pressure regulator upstream of the cascade. During this time aerodynamic traverse measurements are taken. The vanes are supplied with coolant from the hub and case (see Fig. 5), feeding the vane film cooling holes and the TE slot. Further details on the operation of the ECAT facility can be found in Refs. [2123]. The operating conditions used in this study are summarized in Table 3. Table 3 ECAT facility operating conditions for aerodynamic measurements of this study ParameterValue Mean vane pressure ratio, p2/p010.58 Mean vane isentropic exit Mach no., M2,is0.92 Exit Reynolds no., ReC1.60 × 106 Vane inlet turbulence intensity, Tu12% Coolant-to-mainstream pressure ratio, p0c/p011.027 Coolant mass ratio, $m˙c/m˙$10% ParameterValue Mean vane pressure ratio, p2/p010.58 Mean vane isentropic exit Mach no., M2,is0.92 Exit Reynolds no., ReC1.60 × 106 Vane inlet turbulence intensity, Tu12% Coolant-to-mainstream pressure ratio, p0c/p011.027 Coolant mass ratio, $m˙c/m˙$10% ### Instrumentation for Aerodynamic Measurements. Mainstream total pressure and total temperature measurements at the cascade inlet (p01 and T01) were performed using probe rakes (approximate axial location shown in Fig. 5) at four circumferential positions around the annulus. Hub and case coolant cavity total pressure and total temperature (p0c and T0c) were also measured at four circumferential positions. The vane exit static pressure (p2,hub and p2,case) was measured on six vanes (from a total of 40) with 28 tappings per vane (84 tappings in total) located on the hub and case platform overhangs approximately 5 mm downstream of the cascade TE plane (see Fig. 5). Area surveys were conducted in an axial plane one-quarter of an axial chord downstream of the HP NGV TE using an automated radial-circumferential hub-mounted traverse gear. The location and approximate geometry of the traverse are shown in Fig. 5. The probe tip is approximately aligned with the mean flow angle, to ensure measurements were in the range of minimum error for the probe system. Aerodynamic measurements were taken using a five-hole probe (tip diameter 2.8 mm) calibrated over a wide Mach number range (0.3 ≤ M ≤ 1.4). The probe was calibrated to allow determination of total and static pressures, Mach number, and flow angles. Data on a full area-traverse plane are built from six discrete blocks (separate runs) which are then stitched together. A full-area traverse map contains approximately 60 traverse passes with a radial step size of approximately 1 mm. Better than 1 mm effective resolution is obtained in the circumferential direction. More details on the instrumentation and measurement processing for the ECAT facility can be found in Refs. [21,23]. ### Coolant Flow Capacity Measurement Process. We define in-situ coolant capacity as the capacity measured in the presence of main flow, i.e., $Γc(p0cp01,p2p01)=m˙c(p0cp01,p2p01)T0cp0c$ (1) where $m˙c$ is the total coolant mass flowrate, and p0c and T0c are the total pressure and total temperature of the coolant flow in the feed plenum. Likewise, ex-situ capacity is that measured in a bench test experiment with an atmospheric back-pressure condition (in the absence of the potential field developed by the vane in the presence of main flow) $Γc(p0cpatm)=m˙c(p0cpatm)T0cp0c$ (2) Ex-situ capacity measurements were performed individually on all 17 vanes (2 new vanes and 15 deteriorated vanes) used in this study. It is necessary to perform ex-situ measurements on individual parts to separate effects on a per-vane basis. It is—of course—impossible in practice to measure the change in in-situ coolant capacity on a vane-by-vane basis because there is a common feed to the entire vane ring. It is possible however, to accurately infer the per-vane in-situ capacity from the per-vane ex-situ capacity measurements. The following process was used: • Ex-situ capacity measurements were performed on individual deteriorated vanes and non-deteriorated vanes. In these experiments the TE slot capacity characteristic and the capacity characteristic for all film cooling rows were separately determined. These experiments define the absolute ex-situ capacity characteristics of the individual vanes. • The ratios between individual film row capacity characteristics were taken from the design intent (validated against earlier experiments). These ratios were assumed to be unchanged with deterioration. The inferred overall in-situ cooling capacity can be shown to be relatively independent of these ratios over a relatively wide range of variation. • A low-order flow-capacity model for the in-situ part was developed in which—for each film row—an average film-row exit pressure at a particular vane operating point (p2/p01) was defined from CFD simulations at that operating point. Absolute capacity (all films and TE slot) come from the ex-situ experiments; ratios between rows come from the design intent. This low-order model allows prediction of individual capacity values for rows operating at particular local pressure ratios p0c/px, where px is the local static pressure defined by the vane pressure ratio and the vane aerodynamics. This allows the in-situ capacity on a per-vane basis to be calculated as a function of vane operating point (p2/p01) and coolant-to-mainstream pressure ratio (p0c/p01). By summation the overall in-situ coolant capacity can be calculated as a function of vane operating point and coolant-to-mainstream pressure ratio. • Separate in-situ experiments of overall coolant capacity were performed and compared to the predicted value of in-situ coolant capacity from the low-order model (based on the ex-situ measurements and CFD boundary conditions). These values agreed to within 1.1% giving confidence in this check-sum validation of the low-order model. • Once the check-sum was performed, the low-order model could be used to predict overall film capacity and TE slot capacity on a per-vane basis. This can be done for any particular combination of p2/p01 and p0c/p01. This enables us to predict per-vane variation in film and TE-slot capacity for the individual deteriorated vanes of this study. This is a simple process, relatively accurate to first-order. In a more sophisticated analysis a fully-coupled internal-external network loss model would be used, with individual deteriorated parts would be characterized on a per-film-row basis in the ex-situ experiments. This would enable extremely accurate predictions for the in-situ environment. Such an approach is discussed in Ref. [24]. The improved process is arduous in practice, however, and for the purpose of studying the overall change in in-situ coolant capacity from the situation for undamaged vane, we believe the simple process to be at least adequate. Bench-test measurements were taken in an experiment in which the NGV was fed from both the hub and case using custom-made (plastic printed) feed plena sealed to the vane with wax sealant (to ensure complete removal). The feed plena were pressure-balanced using variable metered feeds fed from a constant pressure (regulated) supply. Total temperature and total pressure were measured upstream of the NGV coolant inlet using measurement rakes. The flow discharged directly to atmosphere through the film cooling holes and the TE slot. Test data were taken at five coolant-to-mainstream pressure ratios, p0c/patm = 1.1, 1.2, 1.4, 1.6, and 1.8. At the nominal coolant-to-mainstream pressure ratio and nominal vane pressure ratio (Table 3) the coolant-to-overall (mainstream plus coolant) mass flowrate ratio was approximately 10%, of which approximately one-tenth was the TE slot flow. Using the ex-situ capacity measurements we express proportional changes (from the undamaged vane) in ex-situ per-row coolant capacity (e.g., Γc,S1–VXΓc,S1–V1)/Γc,S1–V1, where X = 2, 3, or 4). That is, changes between the individual deteriorated and undeteriorated parts. As these results are approximately independent of pressure ratio, we present average values (from all tested pressure ratios). For unchanged in-situ aerodynamics the proportional change in per-row in-situ capacity is the same as the proportional change in the per-row ex-situ capacity. The same statement is equally true for any group of cooling rows, and independently for the TE slot. In our particular tests we consider two groups: the TE slot; and the entire flow from 11 film cooling rows. These results are presented in a later section. ## Experimental Results The experimental results are now presented. We will consider: coolant capacity measurements; local kinetic energy (KE) loss coefficient distributions; plane-average and mixed-out KE loss coefficients; comparisons to open-literature data of loss enhancement due to increased surface roughness; circumferential profiles of local KE loss coefficient; and radial distributions of flow angle. ### Coolant Flow Capacity Measurements. In this section we consider the measurements of ex-situ coolant capacity which—we have explained—is a reasonable proxy for in-situ coolant capacity in our experiments. The proportional changes in coolant capacity (Γc,S1–VXΓc,S1–V1)/Γc,S1–V1 for set 1 and (Γc,S2–VXΓc,S2–V1)/Γc,S2–V1 for set 2) are presented in Table 4. The measurement process allows separation of the TE slot flow and the film flow. Proportional changes are relative to the reference (new) vane in each set. Table 4 Proportional changes in coolant capacity Set 1 vane(Γc,S1–VXΓc,S1–V1)/Γc,S1–V1 (%) Film holesTE slot S1-V2‒5.37‒13.4 S1-V3‒6.21‒15.0 S1-V4‒3.87‒2.54 Mean S1-V(2–4)‒5.15‒10.3 Mean S1-V(2–12)‒2.20‒4.87 (Γc,S2–VXΓc,S2–V1)/Γc,S2–V1 (%) Set 2 vaneFilm holesTE slot S2-V2‒5.62‒24.7 S2-V3‒5.43‒15.9 S2-V4‒6.27‒9.79 Mean S2-V(2–4)‒5.77‒16.8 Set 1 vane(Γc,S1–VXΓc,S1–V1)/Γc,S1–V1 (%) Film holesTE slot S1-V2‒5.37‒13.4 S1-V3‒6.21‒15.0 S1-V4‒3.87‒2.54 Mean S1-V(2–4)‒5.15‒10.3 Mean S1-V(2–12)‒2.20‒4.87 (Γc,S2–VXΓc,S2–V1)/Γc,S2–V1 (%) Set 2 vaneFilm holesTE slot S2-V2‒5.62‒24.7 S2-V3‒5.43‒15.9 S2-V4‒6.27‒9.79 Mean S2-V(2–4)‒5.77‒16.8 Looking first at the set 1 data, we see a reduction in coolant capacity for all the mid-life deteriorated vanes for both film cooling holes and TE slot. For the film cooling flow the reduction in capacity is between ‒6.21% and ‒3.87% with an average (across three parts) of ‒5.15%, and for the TE slot the reduction is between −15.0% and −2.54% with an average of −10.3% (across three parts). The decrease in film coolant capacity appears to be caused by partial blockage of the holes due to particle deposition. This is marked A in Fig. 6(b). The condition of an undamaged film cooling hole is shown in Fig. 6(a). Close inspection of the film cooling holes of the mid-life vanes shows some level of deposition in all holes. The decrease in TE slots coolant capacity appears to be due to partial collapse of the TE slot, due to overheating and thermal distortion. This is shown in Fig. 6(c), marked B. Most mid-life vanes showed partial collapse in some areas, even on vanes without TE burn-back, but the effect was more advanced on vanes which also showed TE burn-back (e.g., S1-V2 and S1-V3). Taking the average across all vanes in set 1 (total of 12 mid-life vanes; not individually reported in Table 4) we find a mean reduction in film cooling flow capacity of ‒2.20% and a mean reduction in TE flow capacity of ‒4.87%. These average values are presented in Table 4. The difference in the mean reduction in film hole coolant capacity between the three selected vanes of set 1 (‒5.15%) and the parent set of 12 vanes (‒2.20%) is attributed to the random nature of the deposition process (even within the same engine), and the small sample size. Likewise the mean reduction in TE slot coolant capacity between the three selected vanes (‒10.3%) and the parent set of 12 vanes (‒4.87%) is explained by the fact that only three vanes from the parent set have TE burn-back: i.e., in selecting the vanes to be studied in detail, vanes with more significant TE burn-back were preferred, and these had greater associated collapse of the TE slot than average across the parent set. Fig. 6 Fig. 6 Close modal Looking now at the data for set 2 (end-of-life parts) we see a reduction in coolant capacity between ‒6.27% and ‒5.43% for the film cooling holes (mean of ‒5.77% across three parts) and between ‒24.7% and ‒9.79% for the TE slot (mean of ‒16.8% across three parts). In contrast to the set 1 data, the decrease in coolant capacity of the film cooling holes appears to be primarily due to shrinkage of holes in the region of spalled TBC (spallation region shown in Fig. 1(b)), due to local over-heating of the part (similar effect to the TE slot collapse) This is shown in Fig. 6(d), marked C (LE of vane S2-V4). Interestingly, the effect is partially offset by enchained cracking between holes (marked D in Fig. 6(d)), leading to weeping flow through the cracks. As for set 1 parts, the decrease in TE flow for the set 2 parts is due to partial collapse of the TE slot. The radial extent of this was, on average, greater for the set 2 parts than the set 1 parts, on account of more severe TE overheating. This is consistent with greater radial extent of TE burn-back for the set 2 parts. Extreme burn-back (greater than the parts shown in the present study) would potentially lead to a reversing of this effect, due to an increase in slot width in the direction towards the LE of the vane. The impact of a change in film cooling flow capacity and TE flow capacity on aerodynamic loss will be considered in the next section. ### Local Kinetic Energy Loss Coefficient Distributions. Results of the downstream aerodynamic surveys are now considered. We defined a local KE loss coefficient ζ′(r, θ) by $ζ′(r,θ)=1−1−(p2(r,θ)p02(r,θ))χ1−(p2(r,θ)p01)χ$ (3) where χ = (γ − 1)/γ and γ is ratio of specific heats, p01 is the total pressure upstream of the vanes (assumed uniform), and p02(r, θ) and p2(r, θ) are the total and static pressures measured in the plane of interest downstream of the vanes. A discussion of this, and other performance metrics, is given in Ref. [23]. Figure 7 shows the local distributions of ζ′(r, θ) downstream of the mid-life vanes S1-V1 to S1-V4 (top row, frames ad respectively) and the end-of-life vanes S2-V1 to S2-V4 (bottom row, frames eh respectively). Flow turning is clockwise when viewed from downstream, and data are viewed from downstream. The distribution of ζ′(r, θ) is normalized by the maximum value of ζ′(r, θ) measured across all the data presented. Fig. 7 Fig. 7 Close modal Consider first the results for the new vanes, S1-V1 and S2-V1. There is a well-defined wake profile, which has a curve due to the compound lean in the aft region of the vane. The wake has a number of distinct maxima associated with regions of TE coolant ejection (for detailed analysis see [23]), which are intermittent on account of internal webs. The maximum value of ζ′(r, θ) in the wake is near the hub of the vane, where the vane exit Mach number is the highest. Compound lean and sweep in the vane geometry mean the traverse plane is closest to the TE at the hub, and furthest from the vane at mid-span. This is in accord with the result that the wake towards the hub is the least mixed (a second reason for the greatest peak value in this region), and the wake in the mid-span region is most mixed. Now consider the set 1 (mid-life) deteriorated vanes. Vanes S1-V2 and S1-V3 have only minor TE burn-back (see Fig. 1(a)), but the distortion to the downstream wake is significant. In the region of the TE damage, the wake is under-turned with respect to the undamaged vane, leading to turning of the passage flow away from the SS and towards the PS. The regions of under-turned flow (co-incident with the TE burn-back regions of Fig. 1(a)) are marked A and B in Figs. 7(b) and 7(c) respectively. Vane S1-V4 (Fig. 7(d)) did not exhibit TE burn-back and the wake is deeper than that of the reference vane (Fig. 7(a))—due to greater surface roughness—but of relatively similar form. All of the mid-life vanes assessed had greater profile loss than the undamaged vane, which was attributed to the increased surface roughness causing greater boundary layer loss (combination of earlier transition, but also greater growth rate within the turbulent boundary layer). The deteriorated set 2 (end-of-life) vanes have more significant TE burn-back (for extents see Fig. 1(b)), leading to gross disruption of the wake due to severe under-turning, and—in the case of S2-V2 and S2-V3—the formation of counter-rotating vortices (marked C and D in Fig. 7) either side of the more severe burnt-back regions (caused by strong radial gradient of whirl angle). Similar effects have been observed in Refs. [1517]. Between counter-rotating vortices, there are patches of low loss between the vortices, as the lossy flow is rolled-up into the vortex. The small cracks and bends at the TE of vane S2-V4 have caused a significant local broadening of the wake (marked E in Fig. 7). In the profile loss regions undisturbed by TE damage, the wake is thicker and deeper than the reference vane due to increased surface roughness of the part. We now examine the secondary flow regions in more detail. Local distributions of ζ′(r, θ) in the near-endwall regions are shown in Fig. 8. Secondary flow loss cores in the casing region (Figs. 8(a)8(h)) are difficult to identify for two reasons: they are less intense than at the hub due to the lower Mach number; there is downwash on the vane surface, spreading the boundary layer fluid throughout the wake (see e.g., [23]) and causing coalescence of the flow structures. In contrast to the casing data, in the near-hub region (Figs. 8(i)8(p)) the secondary loss cores are well-defined: for the reference vanes S1-V1 (Fig. 8(i)) and S2-V1 (Fig. 8(m)) the SS corner vortex and passage vortex are marked A and B respectively. For the corner vortex and passage vortex, the peak measured values of local normalized KE loss coefficient were approximately 80% and 66%, respectively, for both reference vanes. The radial locations of the peaks were 2% and 10% span fraction respectively. For all deteriorated vanes the secondary loss cores are both more intense and greater in size than for their corresponding reference vane. Fig. 8 Fig. 8 Close modal ### Plane-Average and Mixed-Out Kinetic Energy Loss Coefficient. In this section we consider both plane-average KE loss coefficients, and mixed-out KE loss coefficients for the vanes. The plane-average coefficient may be taken to represent the loss already manifested at the measurement plane. The mixed-out KE loss coefficient can be taken to represent the sum of the loss already manifested, and the unavoidable loss as the result of the mixing out of gradients already present within the flow (and unlikely to contribute usefully to work): i.e., the secondary kinetic energy (SKE) within the flow. Adopting the definitions of [23], we define the plane-average KE loss coefficient for a film-cooled vane by $ζ″″=1−(m˙m+m˙c)[1−(p2¯p02¯)χ]m˙m[1−(p2¯p01)χ]+m˙c[1−(p2¯p0c)χ]$ (4) where $m˙m$ and $m˙c$ are the mainstream and coolant mass flowrates, respectively, and $p02¯$ and $p2¯$ are the mass-flux-average total pressure and area-average static pressure, respectively. The mixed-out KE loss coefficient for a film-cooled vane is defined [23] by $ζ″″′=1−(m˙m+m˙c)[1−(p2¯′p02¯′)χ]m˙m[1−(p2¯′p01)χ]+m˙c[1−(p2¯′p0c)χ]$ (5) where $p02¯′$ and $p2¯′$ are the mixed-out total and static pressures, respectively, and where a prime is used to distinguish these mixed-out variables (result of mixing process) from averages resulting from the in-plane weighting (e.g., area, volume-flux, or mass-flux). The definition (5) is developed in Ref. [23], and the method for calculation of the mixed-out pressures ($p02¯′$ and $p2¯′$) is presented in Ref. [25] and summarized in Ref. [26]. As discussed, the residual SKE is defined as the difference between the mixed-out and plane-average KE loss coefficient $SKE=ζ″″′−ζ″″$ (6) Normalized plane-average $(ζ″″)$ and mixed-out $(ζ″″′)$ KE loss coefficients, and corresponding residual SKE values are summarized in Table 5. The data are normalized by the mixed-out KE loss coefficient of the reference vane within each set ($ζS1−V1″″′$ and $ζS2−V1″″′$). Table 5 Normalized plane-average KE loss coefficient, mixed-out KE loss coefficient, and residual SKE Set 1 vane$ζS1−VX″″ζS1−V1″″′$$ζS1−VX″″′ζS1−V1″″′$$SKES1−VXζS1−V1″″′$ S1-V10.651.000.35 S1-V20.771.120.35 S1-V30.731.070.34 S1-V40.741.080.34 Mean change S1-V(2–4)14.1%8.72%‒2.13% Mean change S1-V(2–12)13.6%7.81%‒2.57% Set 2 vane$ζS2−VX″″ζS2−V1″″′$$ζS2−VX″″ζS2−V1″″′$$SKES2−VXζS2−V1″″′$ S2-V10.651.000.35 S2-V20.861.330.47 S2-V30.721.120.40 S2-V40.771.110.34 Mean change S2-V(2–4)22.7%18.8%14.9% Set 1 vane$ζS1−VX″″ζS1−V1″″′$$ζS1−VX″″′ζS1−V1″″′$$SKES1−VXζS1−V1″″′$ S1-V10.651.000.35 S1-V20.771.120.35 S1-V30.731.070.34 S1-V40.741.080.34 Mean change S1-V(2–4)14.1%8.72%‒2.13% Mean change S1-V(2–12)13.6%7.81%‒2.57% Set 2 vane$ζS2−VX″″ζS2−V1″″′$$ζS2−VX″″ζS2−V1″″′$$SKES2−VXζS2−V1″″′$ S2-V10.651.000.35 S2-V20.861.330.47 S2-V30.721.120.40 S2-V40.771.110.34 Mean change S2-V(2–4)22.7%18.8%14.9% Consider first the data for set 1 (mid-life). For the reference vane, the normalized mixed-out KE loss coefficient is by definition unity. The normalized plane-average KE loss coefficient and SKE are 0.65 and 0.35 (sum to unity). The relatively large SKE (0.35 of the mixed-out loss) is accounted for by significant unmixed wake (traverse plane relatively close to TE). Now consider the set 1 deteriorated vanes. The average changes in $ζ″″$, $ζ″″′$, and SKE for vanes S1-V(2–4) were 14.1%, 8.72%, and ‒2.13%. Corresponding values for the full set of 12 vanes—vanes S1-V(2–12)—are extremely similar: 13.6%, 7.81%, and ‒2.57%, respectively. The vane-to-vane spread in data was very low, suggesting high similarity between parts. The increase in plane-average KE loss coefficient (average of 13.6% across all parts) is thought to be due simply to an increase in roughness, and associated boundary layer thickening. The SKE remains almost constant (average change of only ‒2.57%), which is explained by the fact that the flow structure is essentially unchanged (slight change in wake thickness). The corresponding average change in mixed-out loss is 7.81%. Looking at the data for set 2 (end-of-life), the reference vane, has normalized plane-average KE loss coefficient and SKE values of 0.65 and 0.35. These are identical to the values for set 1 parts, suggesting high consistency between sets. Looking at the set 2 deteriorated vanes (vanes S2-V(2–4)) we find the average increase (across three vanes) in $ζ″″$, $ζ″″′$, and SKE to be 22.7%, 18.8%, and 14.9%, respectively. The loss coefficients changes are significantly greater for set 2 than for set 1 both because of the flow-structure changes caused by severe TE burn-back, and the effects of increased surface roughness. The vane-to-vane spread in data is also higher that for set 1, on account of significant differences in the particular TE damage (see Fig. 1(b)). The increase in plane-average KE loss coefficient (average of 22.7% across all parts; approximately double that for set 1) is associated with both an increase in roughness and an increase in—already manifested—secondary flow loss at the mixing plane. In contrast to the set 1 data, the SKE change is significant (average change of 14.9%), and is caused by significant structural changes in the flow (see discussion above, noting—in particular—streamwise vortices marked C and D in Fig. 7) with the average dominated by the changes for vanes S2-V2 and S2-V3 (these have the most significant TE burn-back; see Fig. 1(b)), 34.3% and 14.3% respectively. The corresponding average change in mixed-out loss is 18.8%, approximately double that for set 1. So far as whole-life turbine performance modeling is concerned, we suggest preliminary (limited data sets) enhancement factors for mixed-out row loss of approximately 8% for mid-life parts, and 19% for end-of-life parts. Taking a typical mixed-out row total pressure loss of 6% [23] for new parts, these would translate to approximately −0.49% and −1.19% on row efficiency, respectively. For stage loss, an additional enhancement factor might be expected for the case of end-of-life parts (but not mid-life parts) on account of the extreme yaw angle variation in the vicinity of the TE burn-back, which would be expected to cause additional losses due to incidence variation in the rotor frame. We now consider the individual contributions of the change in film cooling flow capacity and TE slot flow capacity (due to in-service deterioration) on the overall change in aerodynamic loss. For this purpose we used the Hartsel [27] film cooling flow loss model and a TE slot flow loss model which we develop for our specific vane geometry and operating conditions. We first consider the film cooling loss model. This model is based on that of Hartsel [27], in which the flow is considered to be comprised of a mixing layer and mainstream layer, interacting at constant static pressure in the surface normal direction, but with the possibility of acceleration in the streamwise direction. Relative row capacities were calculated using nominal hole diameters, with a scaling factor (equivalent to a common discharge coefficient) chosen to match the overall coolant capacity of the model to that measured in the experiment. In the model individual rows exhaust to particular static pressures determined using CFD (see Burdett et al. [28] for more details). Changes in film cooling flow capacity due to deterioration were assumed to affect all film cooling holes uniformly. Using the model, we can calculate the predicted change in mixed-out KE loss coefficient (normalized by $ζS1−V1″″′$) as a function of a change in film cooling flow capacity: a so-called exchange rate between the two parameters. This function is shown in Fig. 9. We observe that a reduction of film cooling flow from the design condition leads to a reduction in overall loss. This trend was expected since in the Hartsel [27] model, the mixing loss (between the mainstream and the ejected coolant) is proportional to the coolant-to-mainstream mass flow ratio, which has reduced due to deterioration. If deterioration acts to reduce film capacity and increase roughness and TE damage, the reduced film loss acts to mitigate the increase in aerodynamic loss due to the increase surface roughness and TE burn-back. Fig. 9 Fig. 9 Close modal For the set 1 parts, the mean change in film cooling flow capacity was −2.20% (see Table 4), leading to a predicted change in mixed-out KE loss coefficient $ζ″″′$ of −0.91%. For the set 2 parts the mean change in film cooling flow capacity was −5.77% (see Table 4), leading to a predicted change in mixed-out KE loss coefficient $ζ″″′$ of ‒2.36%. It should be noted that the vanes in set 2 have heavily damaged LE with distorted film cooling holes and cracks (see Fig. 6(d)), and changes in the aerodynamics of the mixing process are beyond the scope of the model. We must regard the results for set 2 parts as therefore only indicative of likely trends rather than quantitatively robust. We now consider the TE slot flow loss model. The model is that presented in Burdett and Povey [29], based on earlier work by Stewart [30] and then Deckers and Denton [31]. The model uses a mass-momentum control volume method to determine the mixed-out loss coefficient for a particular TE geometry and TE slot mass flowrate. We assume that the change in TE slot flow capacity due to deterioration is uniform across the entire span of the slot, and represented by the measured change in TE slot flow capacity. The predicted change in mixed-out KE loss coefficient (normalized by $ζS1−V1″″′$) as a function of the change in TE slot flow capacity is shown in Fig. 9. We observe that a reduction of TE slot flow from the design condition leads to an increase in overall loss. Deckers and Denton [31] showed that at a given coolant ejection pressure ratio, an increase in coolant TE slot flow capacity increases the base region pressure of an airfoil, and thus reduces the mixed-out KE loss coefficient (lower pressure drag) of the vane. The mean changes in TE slot flow capacity for set 1 and 2 parts (see Table 4) were −4.87% and −16.8% respectively, leading to predicted increases in mixed-out KE loss coefficient $(ζ″″′)$ of 2.18% and 7.62%, respectively. For our deteriorated parts both the film capacity and TE slot flow capacity are reduced, but the modeled changes in KE loss coefficient are in opposite directions. Taking the results of these models in combination with the measured overall change in mixed-out KE loss coefficient with deterioration (Table 5), we can estimate the change in mixed-out loss attributable to aerodynamic changes not included in the loss models. That is, effects such as: boundary layer changes caused by surface roughness and spalling; changes in secondary flow behavior caused by TE burn-back; and additional loss introduced at the point of coolant injection not captured in the basic model. The estimated values for additional aerodynamic loss were 6.54% and 13.5% for set 1 and set 2 parts, respectively. The measured overall loss (Table 5) and the predicted (loss models) and inferred (by difference) contributions are represented diagrammatically in Fig. 10. Fig. 10 Fig. 10 Close modal ### Circumferential Profiles of Local Kinetic Energy Loss Coefficient. We now consider the circumferential profiles of local KE loss coefficient, ζ′(θ), downstream of the vanes. Figures 11 and 12 show the circumferential profiles of ζ′(θ) at 10%, 50%, and 90% span downstream of vanes of set 1 (S1-V1 to S1-V4) and set 2 (S2-V1 to S2-V4), respectively. Each profile is radially averaged over ±5% at the relevant span location, and is circumferentially centered at its peak value. The distributions ζ′(θ) are normalized by the maximum value of ζ′(r, θ) measured across all the data presented. Fig. 11 Fig. 11 Close modal Fig. 12 Fig. 12 Close modal First consider the reference vanes: S1-V1 in Fig. 11, and S2-V1 in Fig. 12. As expected, the wake profiles have a quasi-normal distribution. The wake profiles at 10% span are narrower and have a greater peak value of ζ′(θ) than at 50% and 90% span. This is primarily due to spanwise variation of the TE position with respect to the traverse plane (see earlier discussion). The gradient of ζ′(θ) is slightly greater moving away from the SS than the PS. This is thought to be due to differences in the circumferential distributions of the separation loss contribution to overall loss, arising due to greater TE thickness on the PS than on the SS (design is of SS TE overhang style; see, for example, [29]). This effect opposes, and is thought to be more dominant than, the significantly larger boundary layer momentum thickness on the SS at the TE (for further discussion see [23,29]). Now consider wake profiles for the deteriorated vanes of set 1: lines S1-V2 to S1-V4 in Fig. 11. In general, the ζ′(θ) profiles show higher peak loss, but with a similar profile shape as the undeteriorated vane. As discussed, this is due to greater profile loss and greater boundary layer thickening on account of the greater surface roughness. We characterize the circumferential profiles with three metrics: peak height (method of Burdett and Povey [23]); peak width (method of [23]); and integrated loss (IL) defined by $IL=∫−0.50.5ζ′(θ)/ζmax′dθ$ (7) Values for these metrics evaluated at 50% span are presented in Table 6, both for the three deteriorated vanes of Fig. 11, and for the parent set of 12 vanes. Taking an average across all 12 vanes, the mean changes in peak height, peak width, and integrated loss were 23.9%, ‒1.67%, and 29.8% respectively. That is, there is a substantial increase in overall loss, but relatively little thickening of the wake. This is because the wake width at the traverse plane is dominated by the width of the separated TE region (or base region), as opposed to the boundary layer thickness (more details in Ref. [23]), into which both the boundary layer loss and so-called pressure drag of the base region are subsumed. Table 6 Metrics for wake characterization for set 1 (mid-life) and set 2 (end-of-life) vanes at 50% span Vane$(ζ′/ζmax′)peak$Peak widthIL S1-V10.2990.2250.0763 S1-V20.3480.2460.1071 S1-V30.3760.2230.0967 S1-V40.3750.2020.0952 Mean change S1-V(2–4)22.5%‒0.64%30.3% Mean change S1-V(2–12)23.9%‒1.67%29.8% S2-V10.2830.2450.0940 S2-V20.4880.2790.1521 S2-V30.2970.2030.0748 S2-V40.4610.2370.1244 Mean change S2-V(2–4)46.7%‒1.99%24.5% Vane$(ζ′/ζmax′)peak$Peak widthIL S1-V10.2990.2250.0763 S1-V20.3480.2460.1071 S1-V30.3760.2230.0967 S1-V40.3750.2020.0952 Mean change S1-V(2–4)22.5%‒0.64%30.3% Mean change S1-V(2–12)23.9%‒1.67%29.8% S2-V10.2830.2450.0940 S2-V20.4880.2790.1521 S2-V30.2970.2030.0748 S2-V40.4610.2370.1244 Mean change S2-V(2–4)46.7%‒1.99%24.5% Based on this statistically meaningful set of vanes (12 in total) we conclude that mid-life vanes can be characterized by a loss increase of approximately 30%, with no substantial change in the wake width (rotor forcing implications) or profile shape. These numbers could be used in preliminary design for whole-life modeling. Now consider the wake profiles for the deteriorated set 2 (end-of-life) vanes. For this set, the wake profile shapes for the deteriorated vanes differ from the reference vane by more than for the set 1 comparisons, on account of significant vortex activity associated with strong radial whirl angle variation in the regions of the TE burn-back. At the traverse plane, these secondary flows affect all span locations, but with a significant impact at midspan (where the burn-back is most severe). This leads to more local variation in the wake depth and width than for parts without TE burn-back. For all the three tested deteriorated vanes at all span locations, there was an increase in peak height, with the greatest increases in peak height being substantially larger than for the set 1 parts. That is, the peak loss was greater, but there was more randomness in the profile due to significant secondary flow activity. The mean changes (across three vanes) in peak height, peak width, and integrated loss were 46.7%, ‒1.99%, and 24.5% respectively. These are summarized in Table 6. Based on the relatively small sample (three parts) we conclude that end-of-life vanes can be characterized by a loss increases in the range ‒20.4–61.8% (relatively low confidence due to small statistical set). There is significant part-to-part variability in the integrated loss, but also in the radial distribution of loss, the wake shape, and the associated secondary flow, leading to greater variability in the downstream rotor inlet flow. ### Midspan Profile Total Pressure Loss: Comparison With the Study of Erickson et al. In this section we perform a detailed comparison of the measured changes in midspan profile total pressure loss—with increased surface roughness—to the results of study of Erickson et al. [12]. We consider this useful because of the high degree of similarity between the experiments. Erickson et al. tested two configurations: a reference cascade configuration in which the airfoil surface (PS and SS) has an aerodynamically smooth surface (roughness height not reported); and a rough cascade configuration in which the entire PS and the first 10% of the SS had a surface profile designed to simulate in-service roughening (the remaining SS had the same surface finish as the reference case). The reported ks/C for the roughened surface was 0.98 × 10−3, using the same reporting methodology as the present study [19]. For the comparison we consider only two vanes: the reference vane for set 1 (S1-V1); and vane S1-V10, which has no TE burn-back, and minimal changes in film cooling flow capacity and TE slot flow capacity (‒0.31% and ‒1.21% respectively). These vanes were chosen to isolate the effect of roughness without conflating it with effects due to other deterioration features. For part S1-V10, average values of roughness were ks/C = 1.02 × 10−3 for the PS and early SS, and ks/C = 0.29 × 10−3 for the rest of the SS. These values were very similar to those in the Erickson et al. [12] study. Table 7 compares the two studies in terms of: experimental facility; operating conditions; details of tested components including roughness. The primary difference between the studies was unmatched exit Mach number (0.05–0.20 for Erickson et al. versus 0.92 for the present study). Table 7 Comparison of facility details, operating conditions, and tested components between the present study and Erickson et al. [12] ParameterErickson et al. [12]Present study Film coolingNoYes Axial distance of measurement plane downstream of vane TE0.25 Cax0.25 Cax Mean vane isentropic exit Mach no., M2,is0.05–0.200.92 Exit Reynolds no., ReC0.50 × 106–2.00 × 1061.60 × 106 Vane inlet turbulence intensity, Tu0.70–13.5%12% Equivalent sandgrain roughness height calculation methodBons [19]Bons [19] Reference configuration Average equivalent sandgrain roughness height all around airfoil, ks/CNot reported0.11 × 10−3 Rough configuration PS surface distance significantly roughened0–100%0–100% SS surface distance significantly roughened0–10%0–15% Rough surface typeEngine-representativeEngine-representative Average equivalent sandgrain roughness height on PS and early SS, ks/C0.98 × 10−31.02 × 10−3 Average equivalent sandgrain roughness height on the rest of SS, ks/CNot reported0.29 × 10−3 ParameterErickson et al. [12]Present study Film coolingNoYes Axial distance of measurement plane downstream of vane TE0.25 Cax0.25 Cax Mean vane isentropic exit Mach no., M2,is0.05–0.200.92 Exit Reynolds no., ReC0.50 × 106–2.00 × 1061.60 × 106 Vane inlet turbulence intensity, Tu0.70–13.5%12% Equivalent sandgrain roughness height calculation methodBons [19]Bons [19] Reference configuration Average equivalent sandgrain roughness height all around airfoil, ks/CNot reported0.11 × 10−3 Rough configuration PS surface distance significantly roughened0–100%0–100% SS surface distance significantly roughened0–10%0–15% Rough surface typeEngine-representativeEngine-representative Average equivalent sandgrain roughness height on PS and early SS, ks/C0.98 × 10−31.02 × 10−3 Average equivalent sandgrain roughness height on the rest of SS, ks/CNot reported0.29 × 10−3 We compare aerodynamic loss data using the metric of Erickson et al. [12]: a plane-average mid-span profile loss coefficient defined by $Yp=p01−p02¯p01−p2¯$ (8) in which $p02¯$ and $p2¯$ are the plane-average total and static pressures based on data in the 45–55% span region. The method of calculation of the plane-average pressures ($p02¯$ and $p2¯$) is presented in Ref. [26]. For an extended discussion of relative merits of different loss coefficients see [32]. Comparison of plane-average instead of mixed-out loss coefficients is thought to be acceptable in this environment since the measurement plane for both studies is located at the same normalized axial distance from the vane TE (i.e., 0.25 Cax), and therefore the result is less subject to traverse plane specification (see further discussion in Ref. [32]). Figure 13 shows a comparison between the change in Yp (due to increased surface roughness) reported by Erickson et al. and reported in the present study. In the present study an increase in Yp (from reference vane S1-V1) of 12.2% was measured, for Tu = 12%. Linearly interpolating the Erickson et al. data (between Tu = 8.50% and Tu = 13.5%, and between Re = 1.0 × 106 and Re = 2.0 × 106) gives an increase in Yp of 15.2%. Fig. 13 Fig. 13 Close modal We take these results to be in good agreement, with the discrepancy potentially arising due to higher non-dimensional roughness for our reference (ks/C = 0.11 × 10−3) than the vane used in the Erickson et al. study (reported to be aerodynamically smooth, defined by ks/C < 100/ReC, i.e., ks/C < 0.05 × 10−3). That is, we would expect our reference vane loss to be slightly higher than in the Erickson et al. study. We conclude the following: agreement between the results can be taken as evidence of cross-validation; from a whole-life modeling perspective, we propose profile loss enhancement factors in the range 12.2% to 15.2% for mid-life nozzle guide vanes with ks/C ∼ 1.00 × 10−3 on the PS and early SS, where the lower end of the range can be taken for parts with TBC coating, and the upper end for parts which are initially aerodynamically smooth. ### Radial Distributions of Flow Angle. We now consider radial flow-angle distributions downstream of the vanes. Pitch and yaw angles (α2 and β2, respectively) are defined in Fig. 14. Fig. 14 Fig. 14 Close modal Spanwise distributions of circumferentially mass-flux-average flow angle (pitch angle, $α¯2$, and yaw angle, $β¯2$) downstream of vanes in set 1 (S1-V1 to S1-V4) and set 2 (S2-V1 to S2-V4) are presented in Figs. 15 and 16 respectively. Fig. 15 Fig. 15 Close modal Fig. 16 Fig. 16 Close modal First consider the pitch angle results for the set 1 (mid-life) vanes (Fig. 15). Radial distributions of pitch angle, $α¯2$, are shown in Fig. 15(a). For the reference vane (S1-V1) pitch angle is small and positive and decreases slowly with increasing radius. This is thought to be due to the more prominent annulus hade at the hub than the case. That is, an overall annulus contraction, dominated by the hub annulus line hade (see Fig. 2). The pitch angle trend is approximately linear with radius, varying between 1.9 deg < $α¯2$ < 5.3 deg. The mid-life vanes (S1-V2 to S1-V4) have the following characteristic differences from the reference vane: slightly greater mean pitch angle (by 0.2 deg on average); slightly lower spanwise variation (by ‒1.1 deg per unit of span on average); and slightly greater local variability from a linear trend (mean RMS variation of 0.5 deg from a linear trend compared to 0.4 deg for the reference vane). The small increase in both mean pitch angle and radial variation of pitch angle are not thought to arise from deterioration effects, but rather are thought to be due to quasi-random variation in the manufacturing of the vanes. The increased local variation is thought to arise from increased secondary flow activity, associated with surface defects, increased boundary layer thickness (and associated intensification of vortex features), and TE notching. Vane S1-V3 has the largest region of TE burn-back (see Fig. 1(a)), and this is clearly associated with a local deviation in $α¯2$. This is marked A in Fig. 15(a). The estimated value of Δ$α¯2$ (deviation from a locally linear trend for the particular vane) is Δ$α¯2$ = ‒0.7 deg. Now consider radial yaw angle distribution for the mid-life vanes (Fig. 15(b)). Looking at the reference vane, yaw angle increases with radius according to the particular forced vortex design of the vane, varying approximately linearly between 74.1 deg $<β¯2<$ 80.3 deg. The effect of deterioration (S1-V2 to S1-V4) is threefold: to very slightly decrease the mean yaw angle (by 0.5 deg on average); to cause slightly more local variation in hub and case secondary flow regions (2–10% span; 90–98% span); and to cause strong local deviations in regions of TE burn-back. An example of this last effect is marked B in Fig. 15(b), for which the deviation from the locally linear trend for the particular vane was approximately $Δβ¯2$ = ‒1.9 deg. Now consider the corresponding results for the set 2 (end-of-life) vanes. The flow angle distributions downstream of the baseline vane (S2-V2) have similar trends to the baseline vane of set 1 (S1-V1). The aerodynamic vane standard was essentially identical so this was expected. For the deteriorated vanes, in regions of severe TE burn-back (see Fig. 1(b)) there are significant variations in both pitch and yaw angles. The yaw angle variation arises because of under turning in regions of burn-back (see, for example, features E and F in Fig. 16(b)). Pitch angle variation arises because of significant secondary flow associated with the strong shear (and associated secondary flow formation) at the edges of the burn-back regions (see, for example, features C and D in Fig. 16(a)). For the most severe burn-back (vane S2-V2) the local under turning (deviation from local trend) was estimated to be $Δβ¯2$ = ‒6.1 deg. The corresponding variation in local pitch angle was estimated to be $Δα¯2$ = ‒1.8 deg. We now consider the relationship between maximum yaw angle deviation, $(Δβ¯2)max$, measured downstream of a particular burn-back feature, and the corresponding maximum depth of TE burn-back feature, Dmax/C. Three data points from set 1 parts (mid-life) and three data points from set 2 parts (end-of-life) are presented in Fig. 17. The data for both sets of parts were well-correlated with a linear trend of form $(Δβ¯2)max=−47Dmax/C$ (9) Fig. 17 Fig. 17 Close modal The trend was constrained to go through the origin for obvious physical reasons. A slightly better fit of set 1 data than set 2 data, and a slightly higher gradient for set 2 data than set 1 data (when individually fitted, not shown) may be explained by the greater radial extent of set 2 burn-back features, leading to less diminution of the yaw angle deviation effect in the intermediate mixing process (between vane TE and measurement plane). The empirical correlation (9) is proposed for preliminary whole-life performance assessment. The correlation would be improved with further experimental data taking separate account of both burn-back depth and radial extent. ## Conclusions In this paper, 15 real-engine in-service-deteriorated HP NGVs with a broad variety of deterioration features were studied for the purpose of describing and quantifying their aerodynamic performance, as a step towards whole-life engine modeling. The deterioration features included increased surface roughness; thermal barrier coating spallation; damaged film cooling holes; and trailing edge burn-back. The study was split into six areas, from which the following conclusions were drawn: 1. Surface roughness. In-service roughening was found to primarily affect the PS and early SS, but not the late SS. For vanes initially coated with TBC, typical mean equivalent sandgrain roughness height was found to increase on the PS and early SS from ks/C = 0.109 × 10−3 for new vanes, to ks/C = 0.590 × 10−3 and ks/C = 0.222 × 10−3 for mid-life and end-of-life vanes, respectively. These are increases of 441% and 104%—on average—for mid-life and end-of-life vanes in this study. 2. Coolant flow capacity. Coolant flow capacity decreased with service time both due to accumulation of deposits in the film cooling holes, and partial collapse of film cooling holes and TE slots due to overheating. For mid-life vanes, the mean change in coolant capacity was ‒2.20% for the film cooling holes and ‒4.87% for the TE slots. Corresponding values for end-of-life vanes were ‒5.77% and ‒16.8% respectively. 3. Overall aerodynamic loss. In-service deterioration was found to increase aerodynamic loss. The overall mean enhancement for mixed-out row KE loss was found to be approximately 8% for mid-life vanes, and 19% for end-of-life vanes. We propose these enhancement factors as rules of thumb for whole-life engine modeling. Mechanisms for increased mixed-out loss include: boundary layer thickening on the vane surface and endwalls due to increased surface roughness, causing change in transition point, and change in rate of growth of the boundary layer; streamwise vortex generation (and associated higher residual SKE) in the downstream wake caused by TE burn-back leading to high gradients of yaw angle; reduction in TE slot flow leading to increased base loss. In addition to these effects it is believed that the reduced cooling flow reduces loss, partly mitigating these increases (but leading to excessive part temperature). 4. Separated loss components. Using film cooling and TE slot loss models we show that of the 7.81% and 18.8% loss enhancements associated with mid-life and end-of-life deterioration, changes of ‒0.91% and ‒2.36% can be attributed to reduced film cooling flow (reduced loss), and 2.18% and 7.62% can be attributed to reduced TE flow (increased loss). The remaining 6.54% and 13.5%, respectively, are attributable to aerodynamic changes not included in the loss models, for example: boundary layer changes caused by surface roughness and spalling; changes in secondary flow behavior caused by TE burn-back; and additional loss introduced at the point of coolant injection not captured in the basic model. 5. Profile loss for undamaged mid-life parts. Taking the results of the present study, and the study of Erickson et al. [12] together, we conclude that in-service deterioration increase profile loss of mid-life undamaged parts with ks/C ∼ 1.00 × 10−3 on their PS and early SS by between 12.2% and 15.2%, where the lower value corresponds to parts initially TBC coated, and the upper value for parts which were initially aerodynamically smooth. 6. Downstream flow angles: TE burn-back was found to locally reduce downstream pitch and yaw flow angles. A simple empirical correlation between maximum yaw angle deviation and maximum TE burn-back depth was presented, and is proposed for preliminary whole-life performance assessment (stage modeling). This paper addresses a gap in literature, by providing the first detailed analysis of the overall impact of combined in-service deterioration features on the overall aerodynamic performance of HP NGVs. It is hoped that the results from this paper will be a useful step towards whole-life engine performance modeling and assessment. ## Acknowledgment The financial support of Rolls-Royce plc is gratefully acknowledged. Nafiz Chowdhury and Daniel Burdett are thanked for their support with the experiments. David Newman is thanked for his support with the coolant capacity measurements and technical advice. The Laboratory for In-situ Microscopy and Analysis (LIMA) at the University of Oxford is also thanked for support of the surface roughness measurements. ## Conflict of Interest There are no conflicts of interest. ## Data Availability Statement The authors attest that all data for this study are included in the paper. ## Nomenclature ax = axial position, m C = tangential chord, m Cax = axial chord, m D = trailing edge damage depth, m Dmax = maximum trailing edge damage depth, m IL = integrated loss, – k = sandgrain roughness height, m ks = equivalent sandgrain roughness height, m $m˙$ = mass flowrate, kg/s M = Mach number, – p = static pressure, Pa p0 = total pressure, Pa $p¯$ = plane-average static pressure, Pa $p0¯$ = plane-average total pressure, Pa $p¯′$ = mixed-out static pressure, Pa $p0¯′$ = mixed-out total pressure, Pa r = ReC = exit Reynolds number based on tangential chord, – Rz = maximum peak-to-valley roughness height, m SKE = secondary kinetic energy, – T = static temperature, K T0 = total temperature, K Tu = inlet turbulence intensity, – x = streamwise distance, m y = spanwise distance, m Yp = plane-average total pressure loss coefficient, – z = wall-normal distance, m ## Greek Symbols α = pitch angle, ° $α¯$ = circumferentially mass-flux-average pitch angle, ° β = yaw angle, ° $β¯$ = circumferentially mass-flux-average yaw angle, ° γ = ratio of specific heat capacities, – Γ = capacity, kg/s K1/2/Pa ζ = kinetic energy loss coefficient, – ζ′ = local kinetic energy loss coefficient, – ζ″″ = plane-average kinetic energy loss coefficient, – ζ″″ = mixed-out kinetic energy loss coefficient, – θ = normalized (by vane pitch) circumferential position, – χ = (γ − 1)/γ, – ## Subscripts 1 = upstream of HP NGVs 2 = downstream of HP NGVs atm = atmospheric c = coolant is = isentropic ref = reference ## Acronyms CFD = computational fluid dynamics ECAT = Engine Component AeroThermal (facility) EXP = experiment HP = high pressure KE = kinetic energy LE = NGV = nozzle guide vane PS = pressure surface SS = suction surface TBC = thermal barrier coating TE = trailing edge TRL = ## References 1. 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# Double Logarithms Resummation in Exclusive Processes: The Surprising Behavior of DVCS§ T. Altinoluk1, *, B. Pire2, L. Szymanowski3, S. Wallon44 1 Departamento de Fisica de Particulas and IGFAE, Universidade de Santaigo de Compostela, E-15782 Santiago de Compostela, Galicia, Spain 2 CPhT, École Polytechnique, CNRS, F91128 Palaiseau, France 3 National Center for Nuclear Research (NCBJ), Warsaw, Poland 4 LPT, Université Paris-Sud, CNRS, 91405, Orsay, France & UPMC Univ. Paris 06, faculté de physique, 4 place Jussieu, 75252 Paris Cedex 05, France #### Article Metrics 0 ##### Total Statistics: Full-Text HTML Views: 1316 Abstract HTML Views: 546 ##### Unique Statistics: Full-Text HTML Views: 709 Abstract HTML Views: 359 open-access license: This is an open access article licensed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted, non-commercial use, distribution and reproduction in any medium, provided the work is properly cited. * Address correspondence to this author at the Departamento de Fisica de Particulas and IGFAE, Universidade de Santaigo de Compostela, E-15782 Santiago de Compostela, Galicia, Spain; Tel: + 34 881 813 984; Fax: +34 881 814 086; E-mail: tolga.altinoluk@usc.es ## Abstract Double logarithms resummation has been much studied in inclusive as well as exclusive processes. The Sudakov mechanism has often be the crucial tool to exponentiate potentially large contributions to amplitudes or crosssections near phase-space boundaries. We report on a recent work where a very different pattern emerges : the DVCS quark coefficient function Cq (x,ξ) develops near the particular point x =ξ a non-alternate series in ${a}_{s}^{n}\text{}{\mathrm{log}}^{2n}\text{}\left(x–\xi \right)$ which may be resummed in a cosh $\left(k\sqrt{{a}_{s}}\right)\text{}\mathrm{log}\text{}\left(\left(x–\xi \right)\right)$ factor. This result is at odds with the known result for the corresponding coefficient function for the pion transition form factor near the end point Cq (z) although they are much related through a Cq(Z) correspondence. Keywords: Collinear gluon radiation, DVCS, quark coefficient function, resummation, semi-eikonal approximation, soft gluon radiation.. ## 1. INTRODUCTION While perturbative calculations are widely used in quantum field theory, their summation is always a formidable task, unreachable but in the simplest, not to say most simplistic, occurences. >From elementary particle to atomic and to solid state physics, resummation techniques have been developed to go beyond a fixed order perturbation estimate through the sampling and evaluation of an infinite class of diagrams shown to dominate in a limited kinematical region for some given observables. The result of such procedures is often an exponentiated factor as in the famous Sudakov case [1], exp(-Kg2log2z) where g is the coupling constant and z is the ratio of two different characteristic scales, which governs both QED and QCD calculations of exclusive form factors. As explained in detail in [2] we obtain for a specific case of exclusive scattering amplitude, namely the deeply virtual Compton scattering in the generalized cosh(kg log z) orken regime, a very different resummed result of the form cosh(kg log z) where z is a momentum fraction. To our knowledge, this form never previously emerged in field theoretic calculations. The process that we focus on is the most studied case of a class of reactions - exclusive hard hadronic processes - which are under intense experimental investigation. The result presented here provides an important stepping-stone for further developments enabling a consistent extraction of the quantities describing the 3-dimensional structure of the proton. In the collinear factorization framework the scattering amplitude for exclusive processes such as deeply virtual Compton scattering (DVCS) has been shown [3] to factorize in specific kinematical regions, provided a large scale controls the separation of short distance dominated partonic subprocesses and long distance hadronic matrix elements, the generalized parton distributions (GPDs) [4]. The amplitude for the DVCS process ${y}^{*}{\left(q\right)}^{N}\left(p\right)\text{}\to y{\left({q}^{\text{'}}\right)}^{{N}^{\text{'}}}\left({p}^{\text{'}}\right)$ with a large virtuality q2=-Q2, factorizes in terms of perturbatively calculable coefficient functions $C\left(x,\xi ,{\alpha }_{s}\right)$ and GPDs H(x,ξ,t), where the scaling variable in the generalized Bjorken limit is the skewness ξ defined as $\xi =\frac{{Q}^{2}}{\left(p+{p}^{\text{'}}\right).\left(q+{q}^{\text{'}}\right)}.$The calculation of first order perturbative corrections to the partonic amplitude has shown that terms of order $\frac{{\mathrm{log}}^{2}\left(x±\xi \right)}{x±\xi }$ play an important role in the region of small (x±ξ) i.e. in the vicinity of the boundary between the domains where the QCD evolution equations of GPDs take distinct forms (the so-called ERBL and DGLAP domains). We scrutinize these regions and demonstrate that they are dominated by soft fermion and gluon propagation. This explains why they can be exponentiated using quasi-eikonal techniques. ## 2. MAIN STEPS OF OUR ANALYSIS To set up our notations, let us remind the reader of the known results for the NLO corrections to the DVCS amplitude (1), specializing to the quark contribution to its symmetric part. After proper renormalization, it reads ${A}^{µv}\text{}=\text{}{g}_{T}^{µv}\text{}{\int }_{–1}^{1}\mathrm{dx}\text{}\left(\left(\sum _{q}^{{n}_{F}}{T}^{q}\left(x\right){H}^{q}\left(x,\xi ,t\right)\right)\right),$ where the quark coefficient function Tq reads [5] (3)  ${T}^{q}={C}_{0}^{q}+{C}_{1}^{q}+{C}_{\mathrm{coll}}^{q}\mathrm{log}\frac{\left({Q}^{2}\right)}{{}_{F}^{2}},$ ${C}_{0}^{q}={e}_{q}^{2}\left(\left(\frac{1}{x–\xi +i\epsilon }–x\to –x\right)\right),$ (5)  ${C}_{1}^{q}=\frac{{e}_{q}^{2}{}_{s}{C}_{F}}{4\left(x+i\right)}\left({\mathrm{log}}^{2}\left(\frac{x}{2}i\right)93\frac{x}{+x}\mathrm{log}\left(\frac{x}{2}i\right)\right)\left(xx\right).$ The first (resp. second) terms in Eqs. (4) and (5) correspond to the s-channel (resp. u-channel) class of diagrams. One goes from the s-channel to the u-channel by the interchange of the photon attachments. Since these two contributions are obtained from one another by a simple (x↔-x) interchange, we now restrict mostly to the discussion of the former class of diagrams. Let us first point out that in the same spirit as for evolution equations, the extraction of the soft-collinear singularities which dominate the amplitude in the limit x → ± ξ is made easier if one uses the light-like gauge p1. A = 0 with p1 = q' We argue (and verified) that in this gauge the amplitude is dominated by ladder-like diagrams. We expand any momentum in the Sudakov basis p1p2, as $k=\alpha {p}_{1}+\beta {p}_{2}+{k}_{}$ where p2 is the light-cone direction of the two incoming and outgoing partons $\left({p}_{1}^{2}={p}_{2}^{2}=0,2{p}_{1}.\text{}{p}_{2}=s={Q}^{2}2\xi \right)$. In this basis, ${q}_{\stackrel{\text{∗}}{y}}={p}_{1}–2\text{}\xi {p}_{2}.$. We now restrict our study to the limit x → + ξ . The dominant kinematics is given by a strong ordering both in longitudinal and transverse momenta, according to (see Fig. 1) : $x\xi \left(\left(\beta \right)\left(\right)x–\left(\xi \right)\left(\right)x–\xi +\left({\beta }_{1}\right)\left(\right)\left({\beta }_{2}\right)...\right)$ $...\left(x–\xi +{\beta }_{1}+{\beta }_{2}+...+{\beta }_{n–1}\left(\right)\left({\beta }_{n}\right),\right)$ $\left({k}_{1}^{2}\left(\left({k}_{n}^{2}\left(s\text{}{Q}^{2}\right)\right)\right)\right)$ $\left({a}_{1}\left(...\left({a}_{n}\left(1.\right)\right)\right)\right)$ This ordering is related to the fact that the dominant double logarithmic contribution for each loop arises from the region of phase space where both soft and collinear singularities manifest themselves. When x→ξ the left fermionic line is a hard line, from which the gluons are emitted in an eikonal way (which means that these gluons have their all four-components neglected in the vertex w.r.t. the momentum of the emitter), with an ordering in p2 direction and a collinear ordering. For the right fermionic line, eikonal approximation is not valid, since the dominant momentum flow along p2 is from gluon to fermion, nevertheless the collinear approximation can still be applied. When computing the coefficient functions, one faces both UV and IR divergencies. On the one hand, the UV divergencies are taken care of through renormalization, which manifest themselves by a renormalization scale µR dependency. On the other hand, the IR divergencies remain, but factorization proofs at any order for DVCS justify the fact that they can be absorbed inside the generalized parton distributions and result in finite coefficient functions. In our study, we are only interested into finite parts. Thus, using dimensional regularization, in a factorization scheme like $\stackrel{}{\mathrm{MS}}$, any scaleless integral can be safely put to zero although it contains both UV and IR divergencies. Following this line of thought, we can thus safely deal with DVCS on a quark for our resummation purpose. Finally, the issue related to the iε prescription in Eq. (5) is solved by computing the coefficient function in the unphysical region ξ > 1. After analytical continuation to the physical region 0≤ ξ ≤ 1, the final result is then obtained through the shift ξ→ξ-iε. We define Kn as the contribution of a n-loop ladder to the coefficient function. Let us sketch the main steps of the derivation of K1 and then generalize it for Kn. ### 2.1. The Ladder Diagram at Order αs A careful analysis [2] shows that among the one loop diagrams and in the light-like axial gauge p1.A=0, the box diagram is dominant for x→ξ. Starting from the dominant part of the numerator of the Born term which is θ=-2p1, the numerator of the box diagram is $\mathrm{tr}\left({p}_{2}{y}^{µ}\left(k+\left(x–\xi \right)\left({p}_{2}\right)\left(k+\right)\right)\left(x+\xi \right)\left({p}_{2}\right)\left({y}^{v}\right){d}^{µv}.\right)$ In the limit x → ξ, while the left fermionic line is hard with a large p2 momentum, the gluonic line is soft with respect to the left fermionic line. So we perform soft gluon approximation in the numerator by taking $k+\left(x+\xi \right){p}_{2}\to \left(x+\xi \right){p}_{2}.$. The dominant contribution comes from the residue of the gluonic propagator. Thus, the numerator of the on-shell gluon propagator, dµv, is expressed in terms of transverse polarizations, i.e. ${d}^{µv\text{}}–{\sum }_{}{}_{\left(\right)}^{µ}\text{}{}_{\left(\right)}^{v}$ Writing the gluon polarization vectors in the light-like P1.A=0 gauge through their Sudakov decomposition 10  ${}_{\left(\right)}^{µ}={}_{\left(\right)}^{µ}–2\text{}\frac{{}_{\left(\right).\text{}{k}_{}}}{\beta s}{p}_{1}^{µ}$ (10) allows us to define an effective vertex for the gluon and outgoing quark through the polarization sum 11  $\sum _{}{}_{\left(\right).\text{}}{k}_{}{}_{\left(\right)}^{µ}\text{}=\text{}–\text{}{k}_{}^{µ}+2\text{}\frac{{k}_{}^{2}}{\beta s}{p}_{1}^{µ}$ The numerator, (Num)1, is α - independent and reads $\frac{–4\left(x+\xi \right)}{\beta }\mathrm{tr}\left(\left({p}_{2}\left({k}_{}–2\right)\text{}\frac{{k}_{}^{2}}{\beta s}\left({p}_{1}\right)\left(k+\left(x–\xi \right)\left({p}_{2}\right){p}_{1}\right)\right)\right)$ 12  $=–4\text{}\left(x+\xi \right)s\text{}\frac{2{k}_{}^{2}}{\beta }\left(1+\frac{\left(x–\xi \right)}{\beta }\right)$ (12) We now calculate the integral over the gluon momentum k, using dimensional regularization $\int {d}^{d}\text{}k\to \frac{s}{2}\int d\alpha d\beta {d}^{d–2}\underset{}{k},$ , $\left({k}_{}^{2}={\underset{}{k}}^{2}\right)$ The Cauchy integration of the gluonic pole which gives the dominant contribution reads 13  $–2i\frac{s}{2}{\int }_{0}^{\xi –x}\frac{d\beta }{s\beta \text{}}{\int }_{0}^{}{d}^{d–2}\text{}\underset{–}{k}\text{}{\left(\frac{{\left(\mathrm{Num}\right)}_{1}}{{L}_{1}^{2}{R}_{1}^{2}{S}^{2}}\right)}_{\alpha =\frac{{k}^{2}}{{\beta }^{s}}}$ with the denominators ${L}_{1}^{2}=-{\underset{_}{k}}^{2}+\alpha \left(\beta +x+\xi \right)s$ , ${R}_{1}^{2}=-{\underset{_}{k}}^{2}+\alpha \left(\beta +x-\xi \right)s$ , ${S}^{2}=-{\underset{_}{k}}^{2}+\left(\beta +x-\xi \right)s$ and ${k}^{2}=-{\underset{_}{k}}^{2}+\alpha \beta s.$ The relevant region of integration corresponds to small $|\beta +x-\xi |$ . The β and $\underset{_}{k}$ integrations results in our final one-loop expression : 14  ${k}_{1\text{}}=\frac{i}{4}{e}_{q}^{2}\left(–{c}_{F}{\alpha }_{s}\frac{1}{{\left(2\right)}^{2}}\right)\frac{4}{x–\xi }\frac{2i}{2!}{\mathrm{log}}^{2}\left(a\left(x–\xi \right)\right),$ where we kept only the most singular terms in the x→ξ region and have no control of the value of a within our approximation. To fix a, we match our approximated one-loop result with the full one-loop result (5). This amounts to cut the ${\underset{_}{k}}^{2}$ integral at Q2. The ίε term is included according to the same matching. This leads to 15  ${k}_{1\text{}}=\frac{i}{4}{e}_{q}^{2}\left(–{c}_{F}{\alpha }_{s}\frac{1}{{\left(2\right)}^{2}}\right)\frac{4}{x–\xi +i}\frac{2i}{2!}{\mathrm{log}}^{2}\left(\frac{\xi –x}{2\xi }\text{}–i\right),\text{}$ (15) which is the known result. This is a positive test of the validity of our approximation procedure that we now generalize to the n-rung ladder. ### 2.2. The Ladder Diagram at Order ${\alpha }_{s}^{n}$ Let us now turn to the estimation of all ${log}^{2n}\left(x-\xi \right)$ terms in the diagram shown on Fig. (1). Assuming the strong ordering (7, 8) in K and α, the distribution of the poles generates nested integrals in βί as : Fig.(1). The ladder diagrams which contribute in the light-like gauge to the leading ${a}_{s}^{n}{1n}^{2n}\text{}\left(x\right)\left(x\right)$ terms in the perturbative expansion of the DVCS amplitude. The p2 and ᾕ momentum components are indicated. The dashed lines show the dominant momentum flows along the p2 direction. 16  ${\int }_{0}^{\xi –x}d{\beta }_{1}{\int }_{0}^{\xi –x–{\beta }_{1}}{d\beta }_{2}...{\int }_{0}^{\xi –x–{\beta }_{1}–..–{\beta }_{n–1}}d\text{}{\beta }_{n}$ The numerator for the nth order box diagram is obtained as: 17  ${\left(\mathrm{Num}\right)}_{2}=4s{\left(x+\xi \right)}^{n}\frac{2{k}_{1}^{2}}{{\beta }_{1}}\left(1+\left(\frac{2\left(x–\xi \right)}{{\beta }_{1}}\right)\frac{2{k}_{2}^{2}}{{\beta }_{2}}\text{}\left(1+\left(\frac{2\left({\beta }_{1}+x–\xi \right)}{{\beta }_{2}}\right)...\frac{{2k}_{n}^{2}}{{\beta }_{n}}\right)\text{}\left(1+\left(\frac{2\left({\beta }_{n–1}+...+{\beta }_{1}+x–\xi \right)}{{\beta }_{n}}\right)\right)\right)$ and the denominators of propagators are, for ί=1...n, 18  ${R}_{i}^{2}=–{\underset{}{k}}_{i}^{2}{\alpha }_{i}\left({\beta }_{1}\dots {\beta }_{i}x–\xi \right)s\text{},{S}^{2}=–{\underset{}{k}}_{n}^{2}\left({\beta }_{1}\dots {\beta }_{n}x–\xi \right)s\text{}.$ Using dimensional regularization and omitting scaleless integrals, the integral reads: 19  $\begin{array}{c}\text{}{\int }_{0}^{\xi –x}\text{}d{\beta }_{1}\text{}\dots \text{}{\int }_{0}^{\xi –x–\dots –{\beta }_{n–1}}\text{}d{\beta }_{n}\text{}{\int }_{0}^{}\text{}{d}^{d–2}{\underset{}{k}}_{n}\text{}\dots \text{}{\int }_{0}^{{\underset{}{k}}_{2}^{2}}\text{}{d}^{d–2}{\underset{}{k}}_{1}{\left(–1\right)}^{n}\\ \frac{4\text{}s{\left(2i\right)}^{n}}{x–\xi }\frac{1}{{\beta }_{1}x–\xi }\dots \frac{1}{{\beta }_{1}\text{}\text{}\dots \text{}\text{}{\beta }_{n–1}\text{}\text{}x\text{}–\text{}\xi }\\ \frac{1}{{\underset{}{k}}_{1}^{2}}\dots \frac{1}{{\underset{}{k}}_{n}^{2}}\frac{1}{{\underset{}{k}}_{n}^{2}–\left({\beta }_{1}\dots {\beta }_{n}x–\xi \right)s}\text{}\end{array}$ The integrals over ${\underset{_}{k}}_{1}\cdots {\underset{_}{k}}_{n}$ are performed similarly as in the one-loop case, resulting in: 20  ${k}_{n}=\frac{i}{4}{e}_{q}^{2}{\left(–i{C}_{F}{a}_{s}\frac{1}{{\left(2\right)}^{2}}\right)}^{n}\frac{4}{x–\xi +i}\frac{{\left(2i\right)}^{n}}{{\left(2n\right)}^{!}}{\mathrm{log}}^{2n}\left(\frac{\xi –x}{2\xi }–i\xi \right),$ where the matching condition introduced in one-loop case is extended to n-loops. ## 3. THE RESUMMED FORMULA Based on the results Eqs. (15, 20), one can build the resummed formula for the complete amplitude; we get with $D=\sqrt{\frac{{\alpha }_{s}{C}_{F}}{2}}$ 21  $\sum _{n=0}^{}{K}_{n}=\frac{{e}_{q}^{2}}{x–\xi i}cosh\left[Dlog\left(\frac{\xi –x}{2\xi }–i\right)\right]=\frac{1}{2}\frac{{e}_{q}^{2}}{x–\xi i}\left[{\left(\frac{\xi –x}{2\xi }–i\right)}^{D}{\left(\frac{\xi –x}{2\xi }–i\right)}^{–D}\right]\text{}.$ In the absence of a next to leading logarithmic calculation, the minimal and most natural resummed formula which has the same O(αs) expression as the full NLO result, reads, : 22  ${\left({C}_{0}{C}_{1}\right)}^{res}=\frac{{e}_{q}^{2}}{x–\xi i}&\mathrm{lcub};cosh\left[Dlog\left(\frac{\xi –x}{2\xi }–i\right)\right]–\frac{{D}^{2}}{2}\left[93\frac{\xi –x}{x\xi }log\left(\frac{\xi –x}{2\xi }–i\right)\right]&\mathrm{rcub};–\left(x–x\right)\text{}.$ ## 4. THE GLUON COEFFICIENT FUNCTION For several decades the effects of gluons on many high energy processes has been widely studied. Specifically the theory of "Color Glass Condensate" shows that at very high energies the behavior of the scattering amplitudes are dominated by gluons [6]. Recently it was also shown that even at moderate energies, there are significant O(αs) corrections to scattering amplitudes due to gluonic contributions for spacelike and timelike virtual Compton scatterings [7]. With the above mentioned motivations performing a similar resummation procedure for gluon coefficient function of DVCS and TCS would result in a more trustful extraction gluon GPDs. ### SUMMARY AND OUTLOOK We have demonstrated that resummation of soft-collinear gluon radiation effects can be performed in hard exclusive reactions amplitudes. The resulting formula for coefficient function stabilizes the perturbative expansion, which is crucial for a trustful extraction of GPDs from experimental data. A related expression should emerge in various reactions, such as the crossed case of timelike Compton scattering [8] and exclusive meson electroproduction. Giving these results, a question should be raised : what is the physics beyond this result, or in other words, why is the Sudakov resummation [9] familiar to experts of hard exclusive processes not applicable here? An even more precise question may be: how is our analysis compatible with the discussion of soft effects in the pion transition form factor, a quantity which has been much discussed [10] recently thanks to the experimental results of BABAR and BELLE? Let us stress that the coefficient function of this quantity is identical to the ERBL part of the coefficient function of the DVCS amplitude after a rescaling →x/ξ. Our result thus may be applied to the transition form factor. In [11], it has been argued that the ${\alpha }_{s}{log}^{2}\left(1–z\right)$ factor in the one loop expression of the coefficient function had to be understood as the sum of two very distinct terms, one of them exponentiating in a Sudakov form factor. To advocate this fact, the authors allow themselves an excursion outside the colinear factorization framework and use the familiar detour into the coordinate space framework. Our procedure is different and we resum the complete one loop result. In other words, one may ask to the authors of [11]: what happens to the remnant term proportional to ${\alpha }_{s}{log}^{2}\left(1–z\right)$ If indeed the usual resummation procedure of the transition pion form factor must be revised following our new results, one may ask whether the understanding of the meson form factor [12] should also be reconsidered. ## CONFLICT OF INTEREST The authors confirm that this article content has no conflict of interest. ## ACKNOWLEDGEMENTS We thank the organisors and the French CEA (IPhT and DSM) for support. This work is supported by the P2IO consortium, the Polish Grant NCN No. DEC-2011/01/B/ ST2/03915, the French grant grant ANR PARTONS (ANR-12-MONU-0008-01), the Joint Research Activity "Study of Strongly Interacting Matter" (HadronPhysics3, Grant Agree-ment no. 283286) under the 7th Framework Programme of the European Community, the European Research Council grant HotLHC ERC-2001- StG-279579, Ministerio de Ciencia e Innovac on of Spain grants FPA2009-06867-E, Consolider-Ingenio 2010 CPAN CSD2007-00042 and FEDER. ## REFERENCES [1] Sudakov VV. Vertex parts at very high-energies in quantum electrodynamics Sov Phys JETP 1956; 3: 65-71. [2] Altinoluk T, Pire B, Szymanowski L, Wallon S. Resumming soft and collinear contributions in deeply virtual Compton scattering JHEP 2012; 1210: 49-81. [3] (i) Müller D, Robaschik D, Geyer B, Dittes FM, Horesjsi J. Wave functions, evolution equations and evolution kernels from light ray operators of QCD ortsch Phys 1994; 42: 101-41.; (ii) Radyushkin AV. 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On the aysmptotic behavior of the sudakov form-factor Phys Rev D 1979; 20: 2037-62.; (ii) Botts J, Sterman GF. Hard elastic scattering in QCD Leading behavior Nucl Phys B 1989; 325: 62-124.; (iii) Li HN, Sterman GF. The Perturbative pion form-factor with Sudakov suppression Nucl Phys B 1992; 381: 129-40.; (iv) Contopanagos H, Laenen E, Sterman GF. Sudakov factorization and resummation Nucl Phys B 1997; 484: 303-.; (v) Stefanis NG, Schroers W, Kim HC. Analytic coupling and Sudakov effects in exclusive processes Pion and ?*???° form-factors Eur Phys J C 2000; 18: 137-56.; (vi) Korchemsky GP. Sudakov Form-factor in QCD Phys Lett B 1989; 220: 629-39.; (vii) Mueller AH, Xiao BW, Yuan F. Sudakov double logarithms resummation in hard processes in small-x saturation formalism Phys Rev D 2013; 88: 114010-0. [10] Bakulev AP, Mikhailov SV, Pimikov AV, Stefanis NG. Empha-sizing the different trends of the existing data for the ?*???° transition form factor Acta Phys Polon Supp 2013; 6: 137-44. 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https://www.thejournal.club/c/paper/276/

#### A variant of the Recoil Growth algorithm to generate multi-polymer systems ##### Florian Simatos The Recoil Growth algorithm, proposed in 1999 by Consta et al., is one of the most efficient algorithm available in the literature to sample from a multi-polymer system. Such problems are closely related to the generation of self-avoiding paths. In this paper, we study a variant of the original Recoil Growth algorithm, where we constrain the generation of a new polymer to take place on a specific class of graphs. This makes it possible to make a fine trade-off between computational cost and success rate. We moreover give a simple proof for a lower bound on the irreducibility of this new algorithm, which applies to the original algorithm as well. arrow_drop_up

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https://homework.cpm.org/category/CC/textbook/CCA2/chapter/Ch4/lesson/4.2.2/problem/4-84

### Home > CCA2 > Chapter Ch4 > Lesson 4.2.2 > Problem4-84 4-84. Solve each of the following inequalities. Express the solutions algebraically and on a number line. 1. $3x-5\le7$ Solve the equation $3x-5\le7$ for $x$. Plot the boundary point on a number line. Test a point on each side of the boundary point. Shade the region that makes the inequality true. $x\le4$ 1. $x^2+6>42$ See part (a). This inequality is quadratic, so look for two boundary points. $x<−6$ or $x>6$

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http://mathoverflow.net/questions/52931/point-on-a-line-nearest-a-point-in-banach-space/52939

# Point on a line nearest a point in Banach space I have a Banach space geometry question (a curiosity-driven spin-off from a research topic). Given a point $x$ on the unit sphere of a Banach space and a vector $y\ne 0$, there is a multiple $t_0y$ of $y$ for which $\|t_0y-x\|$ is minimized (this will be unique if the norm is strictly convex). My question is this: For which Banach spaces $X$ is it guaranteed that $\|t_0y\|\le \|x\|$? My "Euclidean intuition" suggested that this should be the case for all Banach spaces, but a little experimentation showed that this is not the case. You quickly see this is really a question about two dimensions. In fact it seems to fail for every $\ell^p$, $p\ne 2$ (see the attached figure in $p=1.2$). Could it be true that this property characterizes Hilbert space? (I looked at the obvious sources: (MO 11192 and papers mentioned in there and didn't find anything of the sort). - The answer is no in dimension 2 and yes in dimension 3 and higher. The property that the nearest-point projection to a line does not increase the norm is equivalent to the symmetry of orthogonality relation defined as follows: $x$ is orthogonal to $y$ iff $\|x+ty\|\ge\|x\|$ for all $t\in\mathbb R$. It is well-known that symmetry of this orthogonality relation in dimension $\ge 3$ implies that the norm is Euclidean, see e.g. Thompson's "Minkowski geometry", Theorem 3.4.10. This is not the case in dimension 2. There are many counter-examples (I believe they are called Radon planes). Basically you only need to ensure that every unit vector with its unit orthogonal one span a constant parallelogram area, this is easy to satisfy and is equivalent to the symmetry of orthogonality. For a simple explicit example (although non-smooth), consider a norm on the plane whose unit ball is a regular hexagon. - Thanks for the nice answer. Anthony –  Anthony Quas Jan 23 '11 at 21:04

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http://mathhelpforum.com/math-topics/62120-finding-primes.html

# Math Help - finding primes 1. ## finding primes How many primes are in the form 4n^8+1, where n is a positive integer? *show working n explanation, PLS... 2. Hello n=5p (where p is an integer) is a necessary condition but I do not know if it is a sufficient condition If n=5p+1 Then n^8=(5p+1)^8=5P+1 Therefore 4n^8+1=20P+5 is divisible by 5 => not prime If n=5p+2 Then n^8=(5p+2)^8=5P+2^8=5P+256 Therefore 4n^8+1=20P+1025 is divisible by 5 => not prime If n=5p+3 Then n^8=(5p+3)^8=5P+3^8=5P+6561 Therefore 4n^8+1=20P+26245 is divisible by 5 => not prime If n=5p+4 Then n^8=(5p+4)^8=5P+4^8=5P+65536 Therefore 4n^8+1=20P+262145 is divisible by 5 => not prime

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https://www.physicsforums.com/threads/cumulative-distribution-and-density-functions.689221/

# Cumulative distribution and density functions 1. May 1, 2013 ### EngnrMatt 1. The problem statement, all variables and given/known data Let X be a random variable with probability density function: 0.048(5x-x2) IF 0 < x < 5 0 otherwise Find the cumulative distribution function of X a) If x ≤ 0, then F(x) = b) If 0 < x < 5, then F(x) = c) If x ≥ 5, then F(x) = 2. Relevant equations Not quite sure 3. The attempt at a solution The answer to a) is 0. The answer to c) is 1. I am making the reasonable assumption that a) is 0 because there is no probability at that point, and that c) is 1 because after that, all probability has been "used" so to speak. However, integrating the function between 0 and 5 does not work. It seems as if my professor totally skipped over teaching us this particular type of problem. Statistics usually makes a good deal of sense to me, but this is pretty foreign. 2. May 1, 2013 ### Ray Vickson You say "integrating the function between 0 and 5 does not work". What about it does not work? In fact, if we define f(x) = 0 for x < 0 and for x > 5, then the cumulative distribution F(z) is $$F(z) = \int_{-\infty}^z f(x) \, dx \\ = 0 \; \text{ if } z < 0,\\ = \int_0^z (48/1000)(5x - x^2) \, dx \; \text{ if } 0 \leq z \leq 5,\\ = 1 \; \text{ if } z > 5.$$ Do the integration to see what you get. Are you sure your course notes or textbook do not have any similar examples?

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https://cs.nyu.edu/crg/newAbstracts/vinod_abstract_9_17_08.html

Vinod Vaikuntanathan IBM Watson TITLE: Trapdoors for Hard Lattices and New Cryptographic Constructions ABSTRACT: We show how to construct a variety of trapdoor'' cryptographic tools assuming the worst-case hardness of standard lattice problems (such as approximating the shortest nonzero vector to within small factors). The applications include trapdoor functions with \emph{preimage sampling}, simple and efficient hash-and-sign'' digital signature schemes, universally composable oblivious transfer, and identity-based encryption. A core technical component of our constructions is an efficient algorithm that, given a basis of an arbitrary lattice, samples lattice points from a Gaussian-like probability distribution whose standard deviation is essentially the length of the longest vector in the basis. In particular, the crucial security property is that the output distribution of the algorithm is oblivious to the particular geometry of the given basis.

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https://testbook.com/question-answer/the-equation-of-the-line-when-the-portion-of-it-i--5eba8b45f60d5d76ccb8a96e

# The equation of the line, when the portion of it intercepted between the axes is divided by the point (2, 3) in the ratio of 3 : 2 is This question was previously asked in NDA (Held On: 22 April 2018) Maths Previous Year paper View all NDA Papers > 1. Either x + y = 4 or 9x + y = 12 2. Either x + y = 5 or 4x + 9y = 30 3. Either x + y = 4 or x + 9y = 12 4. Either x + y =5 or 9x + 4y = 30 Option 4 : Either x + y =5 or 9x + 4y = 30 Free Electric charges and coulomb's law (Basic) 49600 10 Questions 10 Marks 10 Mins ## Detailed Solution Concept: 1. The intercept form of the line$$\frac{x}{a} + \frac{y}{b} = 1$$ Where a is the x- intercept and b is the y- intercept 1. Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n 2. Let P and Q be the given two points (x1, y1) and (x2, y2) respectively and M(x, y) be the point dividing the line- segment PQ internally in the ratio m: n 1. Internal Section Formula: When the line segment is divided internally in the ration m: n, we use this formula. ⇔ $$\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)$$ 2. External Section Formula: When the point M lies on the external part of the line segment. ⇔ $$\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2} - {\rm{\;}}n{x_1}}}{{m\; - {\rm{\;}}n}},\frac{{m{y_2} - {\rm{\;}}n{y_1}}}{{m\; - {\rm{\;}}n}}} \right)$$ Calculation: Intercept form of the line is $$\frac{x}{a} + \frac{y}{b} = 1$$ At x axis point is A (a, 0) At y axis point is B (0, b) Case 1: Given point P (2, 3) divides A and B in ratio of m : n = 3 : 2 Applying Internal Section Formula, We know that $$\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)$$ $$\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3\left( 0 \right) + \;2\left( a \right)}}{{3 + 2}},\frac{{3\left( b \right) + 2\left( 0 \right)}}{{3 + 2}}} \right)$$ $$\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2a}}{5},\frac{{3b}}{5}} \right)$$ Now, $$\frac{{2a}}{5} = 2\;and\;\frac{{3b}}{5} = 3$$ ⇒ a = 5 and b = 5 ∴ Equation of line is $$\frac{x}{5} + \frac{y}{5} = 1$$ ⇒ x + y = 5 Case 2: Given point P (2, 3) divides A and B in ratio of m : n = 2 : 3 $$\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2\left( 0 \right) + \;3\left( a \right)}}{{3 + 2}},\frac{{2\left( b \right) + 3\left( 0 \right)}}{{3 + 2}}} \right)$$ $$\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3a}}{5},\frac{{2b}}{5}} \right)$$ Now, $$\frac{{3a}}{5} = 2\;and\;\frac{{2b}}{5} = 3$$ ⇒ a = 10/3 and b = 15/2 ∴ Equation of line is $$\frac{x}{\left( {}^{10}\!\!\diagup\!\!{}_{3}\; \right)}+\frac{y}{\left( {}^{15}\!\!\diagup\!\!{}_{2}\; \right)}=1$$ $$\Rightarrow {\rm{\;}}\frac{{3x}}{{10}} + \frac{2}{{15}} = 1$$ ⇒ 9x + 4y = 30 Shortcut Method: • Check through the options: Substitute the point (2, 3) in all options ⇒ Only option 4 satisfied the point (2, 3), so option 4 is correct.

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http://aliquote.org/micro/2019-05-25-20-11-02/

# aliquot ## < a quantity that can be divided into another a whole number of time /> An ode to DEK and TeX: The Lingua Franca of LaTeX (via HN).

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http://blog.cdmansfield.com/tag/partial-derivatives/

February 14, 2012  Tagged with: , , When I took my process thermodynamics course as an undergraduate student, I was told to use what I had initially considered to be a completely redundant notation for partial derivatives for thermodynamic potentials.  The notation involved wrapping the partial derivative in a set of parentheses and noting which variables were “held constant” as a subscript. Ex. $\displaystyle\left(\frac{\partial P}{\partial V}\right)_{T,\vec{N}}$ The derivation leading to this was woefully devoid of the mathematical basis for this apparently redundant notation.  Furthermore, so was the general literature on the subject, most of which consisted of fleeting introductions to their respective application.  As I played with the idea, it became clear why this notation was indeed very necessary.  It was only as I was solving a separate problem on my own that such notation yielded valuable information about the nature of the potentials being described.

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https://educatingphysics.com/gcse/electricity/experimenting-with-resistance/

Objectives: • To be able to define resistance • To be able to select and use the equation for resistance  $R = \frac{V}{I}$ • To state and use Ohm’s law • To describe the I–V characteristics of a resistor at constant temperature and a bulb where the temperature varies (the bulb will largely will covered in the next lesson) Resistors serve a vital role in electronic circuits, and that is to limit the amount of current flowing through the circuits. The larger the resistance the stronger resistance there is to the flow of electrons. During this lesson you will have worked on trying to investigate the difference between a number of different unknown resistors. In order to do this we set up the diagram as shown below; For each resistor placed in the circuit, the voltage supplied was varied and a number of results were recorded. The following is a document showing these results which can be used for your some practice if you did not record a sufficient amount of data.

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https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-14-partial-derivatives-14-1-functions-of-several-variables-14-1-exercises-page-942/61

## Calculus 8th Edition This function is presented on the graph $C$ and its level curves are on the graph II. This is because this function is periodical in the product of $xy$. This product grows most rapidly along the lines $x=y$ and $x=-y$ and we see that this function oscillates with greatest frequency along these lines. This is also reflected on the graph II because there the level curves get denser along the mentioned lines.

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https://phys.libretexts.org/Bookshelves/Astronomy_and_Cosmology_TextMaps/Map%3A_Stellar_Atmospheres_(Tatum)/7%3A_Atomic_Spectroscopy/7.13%3A_LS-coupling

$$\require{cancel}$$ # 7.13: LS-coupling Each of the several electrons in an atom has an orbital angular momentum $$\textbf{l}$$ and a spin angular momentum $$\textbf{s}$$, and there are numerous conceivable ways in which the various angular momenta can be coupled together to result in the total electronic angular momentum of the atom. (The total angular momentum of the atom may also include a small contribution from the nucleus. This contribution is usually quite tiny, but measurable. We'll ignore it for the time being; in any case, many nuclides (including most of those that have even numbers of protons and neutrons) have zero nuclear spin. One of the simplest coupling schemes is called LS-coupling (or sometimes Russell-Saunders coupling). In this scheme (which may be regarded as one extreme of a host of conceivable coupling schemes), all the orbital angular momenta $$\textbf{l}$$ of the several electrons are strongly coupled together to form the total electronic orbital angular momentum of the atom, which is denoted by $$\textbf{L}$$. This can be represented symbolically by $\sum{\textbf{I}}= \textbf{L} . \label{7.13.1} \tag{7.13.1}$ The summation indicated is a vector summation. The magnitude of $$\textbf{L}$$ is $$\sqrt{L(L+1)} \hbar$$, and $$L$$ can have nonnegative integral values, 0, 1, 2, 3, etc. Similarly, all the spin angular momenta $$\textbf{s}$$ of the several electrons are strongly coupled together to form the total electronic spin angular momentum of the atom, which is denoted by $$\textbf{S}$$. This can be represented symbolically by $\sum{\textbf{s}} = \textbf{S} \label{7.13.2} \tag{7.13.2}$ The magnitude of $$\textbf{S}$$ is $$\sqrt{S(S+1)}\hbar$$. If there is an even number of electrons in the atom, $$S$$ can have nonnegative integral values. If there is an odd number of electrons in the atoms, the value of S is a positive odd integral number times $$1/2$$, such as $$1/2, \ 3/2, \ 5/2$$,… etc. The total electronic orbital angular momentum of the atom, $$\textbf{L}$$, then couples weakly to the total electronic spin angular momentum of the atom, $$\textbf{S}$$, to form the total (orbital plus spin) electronic angular momentum of the atom, denoted by $$\textbf{J}$$. This is denoted symbolically by $\textbf{L} + \textbf{S} = \textbf{J} \label{7.13.3} \tag{7.13.3}$ The magnitude of $$\textbf{J}$$ is $$\sqrt{J(J+1)}\hbar$$. If there is an even number of electrons, $$J$$ can take any of the $$2 \ \text{min}\left\{ L , S\right\} + 1$$ nonnegative integral values from $$|L − S|$$ to $$L + S$$. If there is an odd number of electrons, $$J$$ can have any of the $$2 \ \text{min} \left\{ L , S \right\} + 1$$ odd-half-integral values from $$|L − S |$$ to $$L + S$$. The $$z$$-component of $$\textbf{J}$$ is $$M \hbar$$. If $$J$$ is integral (i.e. if there is an even number of electrons), $$M$$ can have any of the $$2J+1$$ integral values from $$−J$$ to $$+J$$. If $$J$$ is odd-half-integral, $$M$$ can have any of the $$2J+1$$ odd-half-integral values from $$−J$$ to $$+J$$. In many of the lighter elements near the beginning of the periodic table, the coupling of the angular momenta is close to that of ideal $$LS$$-coupling. There are appreciable departures from this simple scheme higher up in the periodic table. We shall discuss other coupling schemes a little later.

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https://www.general-relativity.net/2018/07/commentary-22-xx3-is-c2-function.html

## Friday, 6 July 2018 ### Commentary 2.2 ф(x)=|x^3| is a C^2 function In section 2.2 on "What is a manifold?" under equation (2.8) he says ф(x)=|x3| is a C2 function because it is infinitely differentiable everywhere except at x=0 where it is differentiable twice but not three times. I had to think about that one and had a quick look at this video on the Khan academy to refresh my memory. I then got to use my own graph paper to test the assertion for myself. What fun! Now I have put it in Excel. Here is the result. ф(x)=|x3| is in blue. It's gradient is obviously negative for x<0 and positive for x>0. So For x<0, ф'=-3x2 and for x>=0 ф'=3x2. The gradient of ф' is always positive. So ф''=|6x|. We can now see the problem at x=0 in green. It is hardly necessary to plot ф''' which is -6 for x<0, 6 for x>0, but undefined at x=0. Resources

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http://mathhelpforum.com/calculus/64234-average-cost-problem-simple-but-having-some-trouble.html

# Math Help - average cost problem - simple, but having some trouble 1. ## average cost problem - simple, but having some trouble Hello everyone! I am having some trouble working out this question: If a marginal cost function nis c'(q)=6q^2+880 dollars per unit, then the cost to increase production from 6 units to 9 units is ... I did this question in 2 ways. One of which I may have found a fault in my steps. 1) Take q=9 and q=6, sub into the marginal cost function and subtract q=9 from q=6. Personally, I don't see how that can give a reasonable answer, but it was worth a shot. 2) Integrate (although this type of question was around before I learned it) and sub in q=6 and q=9 for c(q) and subtract. So I was wondering, if that is how you should go about doing the questions or what I should do. And on the lines of that, another question was bugging me. If for all real x we assume f(g(x)) = x and f'(x) = 1+ [f(x)]^2, then g'(0) equals .. I came up with the answer of: 1 Thanks! 2. Dear finalfantasy, If for all real x we assume f(g(x)) = x and f'(x) = 1+ [f(x)]^2, then g'(0) equals .. I came up with the answer of: 1 What is the derivative of the function f(g(x)) at the point a? The chain rule says g'(a) * f'(g(a)). So $g'(0) = 1 / (1 + g^2(0))$ If we knew g(0)=0 you would be right. 3. increase in cost is $C(9) - C(6) = \int_6^9 C'(q) \, dq$ 4. Originally Posted by skeeter increase in cost is $C(9) - C(6) = \int_6^9 C'(q) \, dq$ That's what I did, but how do you know what the C is, or in other words, fixed cost? I got \$3666 by using FTC BUT that's assuming FC were 0. 5. $C'(q) = 6q^2 + 880$ antiderivative ... $C(q) = 2q^3 + 880q + k$ , where $k$ = fixed cost $C(9) - C(6) = [2(9)^3 + 880(9) + k] - [2(6)^3 + 880(6) + k]$ what happens to the fixed cost, $k$ ? doesn't matter what the fixed cost is, does it?

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https://www.hepdata.net/record/88960

Event-shape engineering for the D-meson elliptic flow in mid-central Pb-Pb collisions at $\sqrt{s_{\rm NN}} =5.02$ TeV No Journal Information The collaboration Abstract (data abstract) CERN-LHC. The production yield of prompt D mesons and their elliptic flow coefficient v_{2} were measured with the Event-Shape Engineering (ESE) technique applied to mid-central (10-30% and 30-50% centrality classes) Pb-Pb collisions at the centre-of-mass energy per nucleon pair sqrt{s_{NN}}=5.02 TeV, with the ALICE detector at the LHC. The ESE technique allows the classification of events, belonging to the same centrality, according to the azimuthal anisotropy of soft particle production in the collision. The reported measurements give the opportunity to investigate the dynamics of charm quarks in the Quark-Gluon Plasma and provide information on their participation in the collective expansion of the medium. D mesons were reconstructed via their hadronic decays at mid-rapidity, |eta|<0.8, in the transverse momentum interval 1<p_{T}<24 GeV/c. The v_{2} coefficient is found to be sensitive to the event-shape selection confirming a correlation between the D-meson azimuthal anisotropy and the collective expansion of the bulk matter, while the per-event D-meson yields do not show any significant modification within the current uncertainties.

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https://verification.asmedigitalcollection.asme.org/fluidsengineering/article-abstract/113/2/290/410785/Unsteady-Gaseous-Diffusion-Associated-With-a-Fully?redirectedFrom=fulltext

This paper gives an analysis of convective gaseous diffusion into a full cavity behind an oscillating flat-plate hydrofoil in a turbulent flow. The unsteady diffusion theory accounts for fluctuations of cavity gas pressure and length which are assumed to be harmonic oscillations but are not necessarily in phase with the hydrofoil motion. A diffusive lag function is found which, for a given reduced frequency, determines the instantaneous diffusion rate as a product of the lag function and the quasisteady mass diffusion. The present results can be used to study the rate of gas entrainment from the cavity into the wake behind the oscillating cavity. This content is only available via PDF.

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https://stats.stackexchange.com/questions/386025/why-is-correlation-formula-the-way-it-is-or-say-how-it-formed?noredirect=1

Why is correlation formula the way it is? Or Say how it formed? [duplicate] I am really being confused by why the correlation formula is called the correlation of two variables $$X$$ and $$Y$$. Also how is it derived? The part where we divide covariance by product of standard derivation of $$X$$ and standard derivation of $$Y$$ is the most confusing for me. Please explain the reasons or provide some good source for such things. marked as duplicate by Stefan, user158565, kjetil b halvorsen, Michael Chernick, Juho KokkalaJan 8 at 7:07 • One way of thinking about the correlation and it's specific representation is that it is a unitless quantity. So by dividing the product moment estimator $E(XY)$ by the SD of X and Y, you get something that doesn't depend on the scale of either variable (in a sense). – AdamO Jan 7 at 17:22 • Correlation equals covariance of standardized variables. – Michael M Jan 7 at 17:23 • Have a look here maybe that helps: stats.stackexchange.com/questions/256344/… Or here: stats.stackexchange.com/questions/70969/… – Stefan Jan 7 at 17:24 Maybe going back to the notion of covariance would help. Say we have two random variables $$X$$ and $$Y$$, with a certain number $$n$$ of independent realizations $$x_1,x_2,\dots x_n$$ and $$y_1,y_2,\dots y_n$$. We know that the formula for the sample covariance is $$\sigma_{xy} =\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$$ where $$\bar{x}$$ and $$\bar{y}$$ are respectively sample means for $$X$$ and $$Y$$. Now, thanks to the Cauchy-Schwarz inequality, we have that the sample covariance is bounded by the product of the standard deviations of the two random variables, which I will denote with $$\sigma_x$$ and $$\sigma_y$$. We have then that $$-\sigma_x\sigma_y \leq \sigma_{xy}\leq \sigma_x\sigma_y$$ Now divide all terms in the inequality by $$\sigma_x\sigma_y$$ and you have the formula for correlation $$-\frac{\sigma_x\sigma_y}{\sigma_x\sigma_y} \leq \frac{\sigma_{xy}}{\sigma_x\sigma_y}\leq \frac{\sigma_x\sigma_y}{\sigma_x\sigma_y}$$ $$-1 \leq \frac{\sigma_{xy}}{\sigma_x\sigma_y} \leq 1$$ with the correct bounds, $$-1$$ and $$1$$. If you grasp the notion of covariance, then you'll surely see that it is simply a standardized version of the latter. • standardized version means? – Vicrobot Jan 7 at 18:02 • It means that it gives you the same information as the covariance, but on a scale that varies from $-1$ to $1$. As others have put it, it is indeed unitless, while covariance is expressed in the units in which the variables were measured. – Easymode44 Jan 7 at 18:06 • So how does that division ensures that resultant amount (i.e. corr.) will still increase in magnitude as there will be a more linear relationship? – Vicrobot Jan 7 at 19:46 • This behavior is already an inherent quality of covariance. CS ensures the inequality, division only gives a measure of linear relationship that is not dependent on units of measure (which can be a problem, especially when dealing with big magnitudes) – Easymode44 Jan 7 at 20:01 • One last thing; what makes it pretty sure that when the relationship tends to linearity; the cov. tends to be equal to magnitude(σ X σ Y) ?? – Vicrobot Jan 7 at 20:35

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http://mathhelpforum.com/advanced-applied-math/280809-does-limit-approximate-zero-set-converge-zero-set.html

# Thread: Does limit of "approximate zero set" converge to the zero set? 1. ## Does limit of "approximate zero set" converge to the zero set? Let $\displaystyle f:\mathbb{R}^m\rightarrow\mathbb{R}^m$. Define the zero set by $\displaystyle \mathcal{Z}\triangleq\{x\in\mathbb{R}^m | f(x)=\mathbf{0}\}$ and an $\displaystyle \epsilon$-approximation of this set by $\displaystyle \mathcal{Z}_\epsilon\triangleq\{x\in\mathbb{R}^m|~ ||f(x)||\leq\epsilon\}$ for some $\displaystyle \epsilon>0$. Clearly $\displaystyle \mathcal{Z}\subseteq \mathcal{Z}_\epsilon$. Can one assume any condition on the function $\displaystyle f$ so that $\displaystyle \lim_{\epsilon\rightarrow 0}~\max_{x\in \mathcal{Z}_\epsilon}~\text{dist}(x, \mathcal{Z})=0,$ holds? I know in general this doesn't hold by this example (function of a scalar variable): \displaystyle f(x)=\left\{\begin{align} 0,\quad{x\leq 0}; \\ 1/x,\quad x>0. \end{align} \right. I really appreciate any help or hint. Thank you. 2. ## Re: Does limit of "approximate zero set" converge to the zero set? If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion. 3. ## Re: Does limit of "approximate zero set" converge to the zero set? Actually $\displaystyle f$ here is the gradient of a non-convex function $\displaystyle g$, i.e. $\displaystyle f=\nabla g$ which is not monotone, and the zero set is the set of critical points. However, I assume $\displaystyle g$ is $\displaystyle \mathcal{C}^\infty$. If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion. Actually $\displaystyle f$ here is the gradient of a non-convex function $\displaystyle g$, i.e. $\displaystyle f=\nabla g$ which is not monotone, and the zero set is the set of critical points. However, I assume $\displaystyle g$ is $\displaystyle \mathcal{C}^\infty$.

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http://science.sciencemag.org/content/early/2012/08/29/science.1228380

Report # Kepler-47: A Transiting Circumbinary Multiplanet System See allHide authors and affiliations Science  28 Aug 2012: 1228380 DOI: 10.1126/science.1228380 ## Abstract We report the detection of Kepler-47, a system consisting of two planets orbiting around an eclipsing pair of stars. The inner and outer planets have radii 3.0 and 4.6 times that of Earth, respectively. The binary star consists of a Sun-like star and a companion roughly one-third its size, orbiting each other every 7.45 days. With an orbital period of 49.5 days, 18 transits of the inner planet have been observed, allowing a detailed characterization of its orbit and those of the stars. The outer planet's orbital period is 303.2 days, and although the planet is not Earth-like, it resides within the classical “habitable zone,” where liquid water could exist on an Earth-like planet. With its two known planets, Kepler-47 establishes that close binary stars can host complete planetary systems. View Full Text

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https://planetmath.org/restricteddirectproduct

# restricted direct product Let $\{G_{v}\}_{v\in V}$ be a collection of locally compact topological groups. For all but finitely many $v\in V$, let $H_{v}\subset G_{v}$ be a compact open subgroup of $G_{v}$. The of the collection $\{G_{v}\}$ with respect to the collection $\{H_{v}\}$ is the subgroup $G:=\left\{\left.(g_{v})_{v\in V}\in\prod_{v\in V}G_{v}\ \right|\ g_{v}\in H_{v% }\text{ for all but finitely many v\in V}\right\}$ of the direct product $\prod_{v\in V}G_{v}$. We define a topology on $G$ as follows. For every finite subset $S\subset V$ that contains all the elements $v$ for which $H_{v}$ is undefined, form the topological group $G_{S}:=\prod_{v\in S}G_{v}\times\prod_{v\notin S}H_{v}$ consisting of the direct product of the $G_{v}$’s, for $v\in S$, and the $H_{v}$’s, for $v\notin S$. The topological group $G_{S}$ is a subset of $G$ for each such $S$, and we take for a topology on $G$ the weakest topology such that the $G_{S}$ are open subsets of $G$, with the subspace topology on each $G_{S}$ equal to the topology that $G_{S}$ already has in its own right. Title restricted direct product RestrictedDirectProduct 2013-03-22 12:35:38 2013-03-22 12:35:38 djao (24) djao (24) 5 djao (24) Definition msc 11R56 msc 22D05

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https://betterexplained.com/articles/intuitive-trigonometry/

Trig mnemonics like SOH-CAH-TOA focus on computations, not concepts: TOA explains the tangent about as well as x2 + y2 = r2 describes a circle. Sure, if you’re a math robot, an equation is enough. The rest of us, with organic brains half-dedicated to vision processing, seem to enjoy imagery. And “TOA” evokes the stunning beauty of an abstract ratio. I think you deserve better, and here’s what made trig click for me. • Visualize a dome, a wall, and a ceiling • Trig functions are percentages to the three shapes ## Motivation: Trig Is Anatomy Imagine Bob The Alien visits Earth to study our species. Without new words, humans are hard to describe: “There’s a sphere at the top, which gets scratched occasionally” or “Two elongated cylinders appear to provide locomotion”. After creating specific terms for anatomy, Bob might jot down typical body proportions: • The armspan (fingertip to fingertip) is approximately the height • A head is 5 eye-widths wide Well, when Bob finds a jacket, he can pick it up, stretch out the arms, and estimate the owner’s height. And head size. And eye width. One fact is linked to a variety of conclusions. Even better, human biology explains human thinking. Tables have legs, organizations have heads, crime bosses have muscle. Our biology offers ready-made analogies that appear in man-made creations. Now the plot twist: you are Bob the alien, studying creatures in math-land! Generic words like “triangle” aren’t overly useful. But labeling sine, cosine, and hypotenuse helps us notice deeper connections. And scholars might study haversine, exsecant and gamsin, like biologists who find a link between your fibia and clavicle. And because triangles show up in circles… …and circles appear in cycles, our triangle terminology helps describe repeating patterns! Trig is the anatomy book for “math-made” objects. If we can find a metaphorical triangle, we’ll get an armada of conclusions for free. ## Sine/Cosine: The Dome Instead of staring at triangles by themselves, like a caveman frozen in ice, imagine them in a scenario, hunting that mammoth. Pretend you’re in the middle of your dome, about to hang up a movie screen. You point to some angle “x”, and that’s where the screen will hang. The angle you point at determines: • sine(x) = sin(x) = height of the screen, hanging like a sign • cosine(x) = cos(x) = distance to the screen along the ground [“cos” ~ how “close”] • the hypotenuse, the distance to the top of the screen, is always the same Want the biggest screen possible? Point straight up. It’s at the center, on top of your head, but it’s big dagnabbit. Want the screen the furthest away? Sure. Point straight across, 0 degrees. The screen has “0 height” at this position, and it’s far away, like you asked. The height and distance move in opposite directions: bring the screen closer, and it gets taller. ## Tip: Trig Values Are Percentages Nobody ever told me in my years of schooling: sine and cosine are percentages. They vary from +100% to 0 to -100%, or max positive to nothing to max negative. Let’s say I paid \$14 in tax. You have no idea if that’s expensive. But if I say I paid 95% in tax, you know I’m getting ripped off. An absolute height isn’t helpful, but if your sine value is .95, I know you’re almost at the top of your dome. Pretty soon you’ll hit the max, then start coming down again. How do we compute the percentage? Simple: divide the current value by the maximum possible (the radius of the dome, aka the hypotenuse). That’s why we’re told “Sine = Opposite / Hypotenuse”. It’s to get a percentage! A better wording is “Sine is your height, as a percentage of the hypotenuse”. (Sine becomes negative if your angle points “underground”. Cosine becomes negative when your angle points backwards.) Let’s simplify the calculation by assuming we’re on the unit circle (radius 1). Now we can skip the division by 1 and just say sine = height. Every circle is really the unit circle, scaled up or down to a different size. So work out the connections on the unit circle and apply the results to your particular scenario. Try it out: plug in an angle and see what percent of the height and width it reaches: The growth pattern of sine isn’t an even line. The first 45 degrees cover 70% of the height, and the final 10 degrees (from 80 to 90) only cover 2%. This should make sense: at 0 degrees, you’re moving nearly vertical, but as you get to the top of the dome, your height changes level off. ## Tangent/Secant: The Wall But can we make the best of a bad situation? Sure. What if we hang our movie screen on the wall? You point at an angle (x) and figure out: • tangent(x) = tan(x) = height of screen on the wall • distance to screen: 1 (the screen is always the same distance along the ground, right?) • secant(x) = sec(x) = the “ladder distance” to the screen We have some fancy new vocab terms. Imagine seeing the Vitruvian “TAN GENTleman” projected on the wall. You climb the ladder, making sure you can “SEE, CAN’T you?”. (Yeah, he’s naked… won’t forget the analogy now, will you?) Let’s notice a few things about tangent, the height of the screen. • It starts at 0, and goes infinitely high. You can keep pointing higher and higher on the wall, to get an infinitely large screen! (That’ll cost ya.) • Tangent is just a bigger version of sine! It’s never smaller, and while sine “tops off” as the dome curves in, tangent keeps growing. • Secant starts at 1 (ladder on the floor to the wall) and grows from there • Secant is always longer than tangent. The leaning ladder used to put up the screen must be longer than the screen itself, right? (At enormous sizes, when the ladder is nearly vertical, they’re close. But secant is always a smidge longer.) Remember, the values are percentages. If you’re pointing at a 50-degree angle, tan(50) = 1.19. Your screen is 19% larger than the distance to the wall (the radius of the dome). (Plug in x=0 and check your intuition that tan(0) = 0, and sec(0) = 1.) ## Cotangent/Cosecant: The Ceiling Amazingly enough, your neighbor now decides to build a ceiling on top of your dome, far into the horizon. (What’s with this guy? Oh, the naked-man-on-my-wall incident…) Well, time to build a ramp to the ceiling, and have a little chit chat. You pick an angle to build and work out: • cotangent(x) = cot(x) = how far the ceiling extends before we connect • cosecant(x) = csc(x) = how long we walk on the ramp • the vertical distance traversed is always 1 Tangent/secant describe the wall, and COtangent and COsecant describe the ceiling. Our intuitive facts are similar: • If you pick an angle of 0, your ramp is flat (infinite) and never reachers the ceiling. Bummer. • The shortest “ramp” is when you point 90-degrees straight up. The cotangent is 0 (we didn’t move along the ceiling) and the cosecant is 1 (the “ramp length” is at the minimum). ## Visualize The Connections A short time ago I had zero “intuitive conclusions” about the cosecant. But with the dome/wall/ceiling metaphor, here’s what we see: Whoa, it’s the same triangle, just scaled to reach the wall and ceiling. We have vertical parts (sine, tangent), horizontal parts (cosine, cotangent), and “hypotenuses” (secant, cosecant). (Note: the labels show where each item “goes up to”. Cosecant is the full distance from you to the ceiling.) Now the magic. The triangles have similar facts: From the Pythagorean Theorem (a2 + b2 = c2) we see how the sides of each triangle are linked. And from similarity, ratios like “height to width” must be the same for these triangles. (Intuition: step away from a big triangle. Now it looks smaller in your field of view, but the internal ratios couldn’t have changed.) This is how we find out “sine/cosine = tangent/1”. I’d always tried to memorize these facts, when they just jump out at us when visualized. SOH-CAH-TOA is a nice shortcut, but get a real understanding first! ## Gotcha: Remember Other Angles Psst… don’t over-focus on a single diagram, thinking tangent is always smaller than 1. If we increase the angle, we reach the ceiling before the wall: The Pythagorean/similarity connections are always true, but the relative sizes can vary. (But, you might notice that sine and cosine are always smallest, or tied, since they’re trapped inside the dome. Nice!) ## Summary: What Should We Remember? For most of us, I’d say this is enough: • Trig explains the anatomy of “math-made” objects, such as circles and repeating cycles • The dome/wall/ceiling analogy shows the connections between the trig functions • Trig functions return percentages, that we apply to our specific scenario You don’t need to memorize 12 + cot2 = csc2, except for silly tests that mistake trivia for understanding. In that case, take a minute to draw the dome/wall/ceiling diagram, fill in the labels (a tan gentleman you can see, can’t you?), and create a cheatsheet for yourself. In a follow-up, we’ll learn about graphing, complements, and using Euler’s Formula to find even more connections. ## Appendix: The Original Definition Of Tangent You may see tangent defined as the length of the tangent line from the circle to the x-axis (geometry buffs can work this out). As expected, at the top of the circle (x=90) the tangent line can never reach the x-axis and is infinitely long. I like this intuition because it helps us remember the name “tangent”, and here’s a nice interactive trig guide to explore: Still, it’s critical to put the tangent vertical and recognize it’s just sine projected on the back wall (along with the other triangle connections). ## Appendix: Inverse Functions Trig functions take an angle and return a percentage. sin(30) = .5 means a 30-degree angle is 50% of the max height. The inverse trig functions let us work backwards, and are written sin-1 or arcsin (“arcsine”), and often written asin in various programming languages. If our height is 25% of the dome, what’s our angle? Now what about something exotic, like inverse secant? Often times it’s not available as a calculator function (even the one I built, sigh). Looking at our trig cheatsheet, we find an easy ratio where we can compare secant to 1. For example, secant to 1 (hypotenuse to horizontal) is the same as 1 to cosine: $\displaystyle{\frac{sec}{1} = \frac{1}{cos}}$ Suppose our secant is 3.5, i.e. 350% of the radius of the unit circle. What’s the angle to the wall? \begin{align*} \frac{\sec}{1} &= \frac{1}{\cos} = 3.5 \\ \cos &= \frac{1}{3.5} \\ \arccos(\frac{1}{3.5}) &= 73.4 \end{align*} ## Appendix: A Few Examples Example: Find the sine of angle x. Ack, what a boring question. Instead of “find the sine” think, “What’s the height as a percentage of the max (the hypotenuse)?”. First, notice the triangle is “backwards”. That’s ok. It still has a height, in green. What’s the max height? By the Pythagorean theorem, we know \begin{align*} 3^2 + 4^2 &= \text{hypotenuse}^2 \\ 25 &= \text{hypotenuse}^2 \\ 5 &= \text{hypotenuse} \end{align*} Ok! The sine is the height as a percentage of the max, which is 3/5 or .60. Follow-up: Find the angle. Of course. We have a few ways. Now that we know sine = .60, we can just do: $\displaystyle{\arcsin(.60) = 36.9}$ Here’s another approach. Instead of using sine, notice the triangle is “up against the wall”, so tangent is an option. The height is 3, the distance to the wall is 4, so the tangent height is 3/4 or 75%. We can use arctangent to turn the percentage back into an angle: $\displaystyle{\tan = \frac{3}{4} = .75 }$ $\displaystyle{\arctan(.75) = 36.9}$ Example: Can you make it to shore? You’re on a boat with enough fuel to sail 2 miles. You’re currently .25 miles from shore. What’s the largest angle you could use and still reach land? Also, the only reference available is Hubert’s Compendium of Arccosines, 3rd Ed. (Truly, a hellish voyage.) Ok. Here, we can visualize the beach as the “wall” and the “ladder distance” to the wall is the secant. First, we need to normalize everything in terms of percentages. We have 2 / .25 = 8 “hypotenuse units” worth of fuel. So, the largest secant we could allow is 8 times the distance to the wall. We’d like to ask “What angle has a secant of 8?”. But we can’t, since we only have a book of arccosines. We use our cheatsheet diagram to relate secant to cosine: Ah, I see that “sec/1 = 1/cos”, so \begin{align*} \sec &= \frac{1}{\cos} = 8 \\ \cos &= \frac{1}{8} \\ \arccos(\frac{1}{8}) &= 82.8 \end{align*} A secant of 8 implies a cosine of 1/8. The angle with a cosine of 1/8 is arccos(1/8) = 82.8 degrees, the largest we can afford. Not too bad, right? Before the dome/wall/ceiling analogy, I’d be drowning in a mess of computations. Visualizing the scenario makes it simple, even fun, to see which trig buddy can help us out. In your problem, think: am I interested in the dome (sin/cos), the wall (tan/sec), or the ceiling (cot/csc)? Happy math. Update: The owner of Grey Matters put together interactive diagrams for the analogies (drag the slider on the left to change the angle): Thanks!

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http://physics.stackexchange.com/questions/45158/about-the-gauge-invariance-of-chern-simons-theory-in-local-coordinates

# About the gauge invariance of Chern-Simons' theory (in local coordinates) I am aware of the differential form language proof of the fact that for arbitrary gauge transformations the Chern-Simons' term shifts by a WZW term (on the boundary). But I am getting confused if in local coordinates I try to prove that under infinitesimal gauge transformations the Chern-Simons' term is invariant (may be upto total derivatives) So I took my infinitesimal gauge transformation as $\delta A^a_\mu = \partial _\mu \epsilon ^a + f^{abc}A_\mu^b \epsilon ^c$ where my structure constants $f^{abc}$ are totally antisymmetric and cyclic. Then I look at the two terms that I have in the CS form, $\epsilon^{\mu \nu \lambda}A^a_\mu \partial_\nu A^a_\lambda$ and $\epsilon^{\mu \nu \lambda}f^{abc}A^a_\mu A^b_\nu A^c_\lambda$ and I vary them infinitesimally to get, $\delta (\epsilon^{\mu \nu \lambda}A^a_\mu \partial_\nu A^a_\lambda ) = 2\epsilon^{\mu \nu \lambda}( \partial_\mu \epsilon ^a \partial _\nu A^a_\lambda + f^{abc}A_\mu ^b \partial _\nu A_\lambda ^a \epsilon ^c)$ $\delta (\epsilon^{\mu \nu \lambda}f^{abc}A^a_\mu A^b_\nu A^c_\lambda) = 3\epsilon^{\mu \nu \lambda}f^{abc} ( \partial _\mu \epsilon^a A_\nu ^b A_\lambda ^c + f^{adf} A_\mu ^d A_\nu ^b A_\lambda ^c \epsilon ^f)$ • Now I can't see how any linear combination of the above two terms can make their combined variation 0 (or a total derivative). I would like to know as to what am I missing here. - Use the Jacobi identity $f^{abc}f^{adf}+f^{adb}f^{acf}+f^{acd}f^{abf}=0$ to the last term.

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http://mathoverflow.net/questions/87600/calculate-all-the-equivalent-martingale-measures?sort=oldest

# calculate all the equivalent martingale measures Under the assumption of no arbitrage without vanish risk, in an incomplete market $(\Omega,{\cal F}, P)$, the set of equivalent martingale measure is NOT empty, i.e. $\mathcal{P} = \{Q: Q \sim P\}\neq \emptyset$ My question is: in the following simplified market with one stock which is driving by two independent Brownian Motions and one bond, i.e. $$dS_t = S_t(\mu dt + \sigma_1 dW_1(t) + \sigma_2 dW_2(t))$$ $$dB_t = rB_tdt, \mbox{ } B_0 = 1$$ How to calculate all the equivalent martingale ${\cal P}.$ We suppose that $\mu,\sigma_1,\sigma_2, r$ are constants. One approach in my mind is using another stock to complete the market, i.e. we suppose there is another stock $\tilde{S}$ with parameters, $\tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}$ such that $$d\tilde{S_t} = \tilde{S_t}(\tilde{\mu} dt + \tilde{\sigma_1} dW_1(t) + \tilde{\sigma_2} dW_2(t)).$$ Then, following the classic method, we could get the equivalent martingale measures described by parameters, $\mu,\sigma_1,\sigma_2, r, \tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}.$ But, how could I know the equivalent martingale measure obtained by above approach are the set of all the equivalent martingale measures in this financial market? Any suggestion, reference books, or papers are welcome. Thanks. - The question might be suitable for the sister stackexchange site devoted to Quantitative Finance: quant.stackexchange.com –  Andrey Rekalo Feb 5 '12 at 21:21 One possible approach is to use the fact that the density process $\left. Z_t =\frac{d\mathcal{Q}}{d\mathcal{P}} \right\vert_{\mathcal{F}_t}$ for every equivalent local martingale measure $\mathcal{Q}$ is a true martingale in the Brownian filtration and you can characterize it via the martingale representation theorem. As a guiding example you may look e.g. on Appendix A of R. Frey's paper "Derivative Asset Analysis in Models with Level Dependent and Stochastic Volatility", CWI Quaterly 10, no 1 (special issue on the Mathematics of Finance) p 1-34 which he links on his webpage: http://www.math.uni-leipzig.de/~frey/vol_survey.ps

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https://quantgirl.blog/fan-charts/

# Fan Charts Starting on 7 November 2019, the Bank of England‘s Inflation Report became the Monetary Policy Report. This quarterly publication communicates economic analysis and inflation projections that the Monetary Policy Committee uses to make its interest rate decisions. Since 1996 the Bank of England (BoE) inflation forecast has been published as a probability distribution and presented in what is now known as ‘the fan chart’. The first fan charts that were published by the BoE can be found in its Inflation Report-February 1996. A digital copy of the printed report (which used to be sold for £3!) can be found here (see page 48 for the charts). In this post, we will review the ideas behind the fan charts, as well as details on how they are constructed and interpreted. We will use the fanchart library to plot them with Python! ## Motivation The BoE introduced the fan charts aiming to communicate a more accurate representation of their forecast for medium term inflation. In particular, the charts have two key objectives: • To convey the uncertainty in their forecasts. This is, to focus attention on the the forecast distribution, rather than only on small changes to the central projection. • To promote discussion of the risks to the economic outlook, and thus contribute to a wider debate about economic policy. Fan charts help to make it clear that monetary policy is about making decisions instead of knowing the exact rate of inflation in two years time. ## Choice of the Two-Piece Normal Historical observations showed that inflation outcomes were not symmetrically dispersed around a central value, with the values closer to the centre being more likely than those further away. This led to the choice of the  two-piece normal distribution, which can capture asymmetry through a skewness parameter, for the forecast model. As we know, the probability density function (pdf) of the two-piece normal (see my previous post for more details) is defined as $$s(x) := s\left(x; \mu,\sigma_1,\sigma_2\right) = \begin{cases} \dfrac{2}{\sigma_1+\sigma_2}f\left(\dfrac{x-\mu}{\sigma_1}\right), \qquad \mbox{if } x < \mu, \\ \dfrac{2}{\sigma_1+\sigma_2}f\left(\dfrac{x-\mu}{\sigma_2}\right), \qquad \mbox{if } x \geq \mu, \\ \end{cases}$$ where $f: \mathbb{R} \mapsto \mathbb{R}_{+}$ is the pdf of the standard normal distribution. However, the Bank of England uses the following re-parametrisation:$$\sigma_1 = \dfrac{\sigma}{\sqrt{1-\gamma}} ; \qquad \sigma_2 = \dfrac{\sigma}{\sqrt{1+\gamma}},$$ where $\sigma >0$ and $\gamma \in (-1, 1)$. Thus, the pdf of the two-piece normal can be written as$$s(x) := s\left(x; \mu,\sigma,\gamma\right) = \begin{cases} \dfrac{A}{\sqrt{2\pi}\sigma} \exp \left\{ -\dfrac{1-\gamma}{2\sigma^2} \left[(x-\mu)^2\right] \right\}, \qquad \mbox{if } x < \mu, \\ \dfrac{A}{\sqrt{2\pi}\sigma} \exp \left\{ -\dfrac{1+\gamma}{2\sigma^2} \left[(x-\mu)^2\right] \right\}, \qquad \mbox{if } x \geq \mu, \\ \end{cases}$$ where $A = \dfrac{2}{\frac{1}{\sqrt{1-\gamma}}+\frac{1}{\sqrt{1+\gamma}}}$. So, in order to derive the forecast distribution for each quarter ahed, three parameters need to be estimated: • $\mu$ : a measure of the central tendency for inflation • $\sigma$ : a view on the degree of uncertainty • $\gamma$ : a view on the balance of the risks, to get a measure of the skew. Once these parameters are set, the two-piece normal is completely defined. BoE publishes these parameters as part of its quarterly report as: • Mode • Uncertainty • Skewness You can download all the data related to the fan charts by using the “Download the chart slides and data (ZIP)” option in the Monetary Policy Report website. ## Python Implementation We will use the fanchart library to reproduce the fan charts. This library contains a copy of the BoE data and parameters (already in the format required by the functions) as of November 2019. from fanchart.plot import * ### Quarter Fan Chart First, we plot a variant of the fan charts which uses only the projection associated to a single quarter. For this purpose, we will use the function fan_single which provides functionality for two kinds of charts, namely the probability density function (pdf) and the cumulative distribution function (cdf). In order to use the fan_single function, we need to provide: • The parameters $\mu, \sigma, \gamma$ (loc, sigma, and gamma) for the two-piece distribution corresponding to the quarterly projection. • A set of probabilities which define the bands of the chart • The kind of plot, namely ‘pdf’ or ‘cdf ‘ For illustration purposes, we use the parameters corresponding to the projection for 2020-10-01. You can find parameters corresponding to other dates here. Remember that the parameters are called mode, uncertainty, and skewness by the BoE. prob = [0.05, 0.20, 0.35, 0.65,0.80, 0.95] fan_single(loc=1.51, sigma=1.34, gamma=0.0, p=prob, kind='pdf'); fan_single(loc=1.51, sigma=1.34, gamma=0.0, p=prob, kind='cdf'); prob = np.arange(0.10, 1, 0.05) fan_single(loc=1.51, sigma=1.34, gamma=0.0, p=prob, kind='pdf'); fan_single(loc=1.51, sigma=1.34, gamma=0.0, p=prob, kind='cdf'); ### Projection Fan Chart Now, we will plot the fan chart that illustrates the entire projection ahead. For this we will use the function fan which requires: • A data frame containing the parameters $\mu, \sigma, \gamma$ (loc, sigma, and gamma) for every quarter in the projection. You can load the parameters corresponding to the Nov-2019 Report from the fanchart library. • A data frame with the observed historical inflation. This serves to produce the solid line before the fan part. You can load the historical data corresponding to the Nov-2019 Report from the fanchart library. • A set of probabilities which define the bands of the chart. history = load_boe_history() probs = [0.05, 0.20, 0.35, 0.65,0.80, 0.95] fan(data=parameters, p=probs, historic=history[history.Date >='2015'].iloc[::3,]); probs = np.arange(0.10, 1, 0.05) fan(data=parameters, p=probs, historic=history[history.Date >= '2015'].iloc[::3,]); Note that at any particular point in the forecast period (which has a grey background) the shading of the bands gets lighter as the probability of inflation lying further away from the central projection decreases. Besides, as predictions become increasingly uncertain over time, these forecast ranges spread out, creating the distinctive”fan” or wedge shapes which originated the “fan chart” term. ### Surface Fan Chart As we mentioned, uncertainty increases over time so we can think of looking at the whole density function over time. This is done in the following three-dimensional chart, which includes the probability density on the z-axis.

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https://www.physicsforums.com/threads/a-power-series-question.84456/

# A power series question 1. Aug 7, 2005 ### complexhuman Hi Assuming tan(x) is given by a power series with coefficiants (An). How can it be shown that An = 0 whenever n is even. Thanks 2. Aug 7, 2005 ### LeonhardEuler I assume you mean a power series centered about x=0. tan(x) is an odd function, i.e. tan(-x)=-tan(x). If it had nonzero coefficients to its even powered terms this could not be the case. 3. Aug 7, 2005 ### complexhuman thanks a lot :)

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https://dsp.stackexchange.com/questions/1930/how-to-calculate-the-gain-in-a-bivariate-fft-in-r

How to calculate the gain in a bivariate fft in R? In Statistica gain is defined as follows: Gain. The gain value is computed by dividing the cross-amplitude value by the spectrum density estimates for one of the two series in the analysis. Consequently, two gain values are computed, which can be interpreted as the standard least squares regression coefficients for the respective frequencies. However, spec.pgram (the engine behind spectrum) in R does not return the cross-amplitude value (as far as I can tell). How can I calculate gain for these signals? Example: makewave <- function(freq,phase,amp,Nsamples=length*samplerate,samplerate,time=Nsamples/samplerate,as.time.series=TRUE) { time <- Nsamples/samplerate phase <- phase*(2*pi)/180 wavetimes <- seq(0+phase,time*freq*pi*2+phase,length.out=Nsamples) #plot(1:samples/samplerate,amp*sin(wavetimes),type="l",xlab="Time") if (as.time.series) {res <- ts(amp*sin(wavetimes),deltat=1/samplerate)} else {res <- amp*sin(wavetimes)} return(res) } signal1 <- makewave(15,30,1,180,60) signal2 <- makewave(15.01,60,1,180,60) signal.union <- ts.union(signal1,signal2) sp <- spectrum(signal.union,span=c(3),taper=0) plot(sp) plot(sp,plot.type="phase") plot(sp,plot.type="coh") • I'm sorry if this question is too simple for this site. If so, I'll delete it. – russellpierce Apr 3 '12 at 22:04 • It's not too simple, it's just not very clear- at least to me. What do you mean by "bivariate fft"? It looks like you are calculating the fft of two signals added together. Is that what "bivariate fft" means? What does "cross-amplitude value" mean? – Jim Clay Apr 4 '12 at 1:10 • @JimClay I believe that bivariate FFT is simply a two-dimensional FFT. – Phonon Apr 4 '12 at 1:44 • "The cross amplitude values are computed as the square root of the sum of the squared cross-density and quad-density values. The cross-amplitude can be interpreted as a measure of covariance between the respective frequency components in the two series." (documentation.statsoft.com/STATISTICAHelp.aspx?path=TimeSeries/…) – Jim Clay Apr 4 '12 at 13:33 • I am/was shaky on the correct terminology. I have two signals, and I want to compare them to see how much the 2nd matches the first (coh), how much the 2nd lags behind the 1st (phase), and the extent to which the 2nd is under/overshooting the peak amplitudes of the 1st. – russellpierce Apr 4 '12 at 15:55 1 Answer I am/was shaky on the correct terminology. I have two signals, and I want to compare them to see how much the 2nd matches the first (coh), how much the 2nd lags behind the 1st (phase) Normalized cross-correlations are good at that. I don't know what the relevant command would be in Statistica, but in Matlab you would use the command xcorr. and the extent to which the 2nd is under/overshooting the peak amplitudes of the 1st. To answer that in a useful way it would help a great deal to see the signals that you are talking about. Are you just interested in the peaks, or the signal in general? If you are talking about the signal in general then you could time-align the signals using the information from your cross-correlation, and then divide one of the signals by the other. If you are just interested in the peaks you could do the same and then just extract the results at the peak locations. • I am interested in the power in each FFT bin in relation to the power in the other bin. Statistica reports some kind of ratio here as "gain", but I am not sure how it is calculated. I'll provide sample data later today. – russellpierce Apr 12 '12 at 16:32

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https://web2.0calc.com/questions/knowing-the-diameter-of-6-feet-what-would-be-the-circumference

+0 # Knowing the diameter of 6 feet, what would be the circumference? 0 148 1 Knowing the diameter of 6 feet, what would be the circumference? Guest Jun 5, 2017 #1 +1 $$\pi * d$$ $$\pi *6$$ $$6\pi$$ 6π feet! Of course, this is assuming that you're talking about a circle. Guest Jun 6, 2017 edited by Guest  Jun 6, 2017 Sort: #1 +1 $$\pi * d$$ $$\pi *6$$ $$6\pi$$ 6π feet! Of course, this is assuming that you're talking about a circle. Guest Jun 6, 2017 edited by Guest  Jun 6, 2017 ### 5 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details

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https://socratic.org/questions/how-do-you-simplify-16-1-2-2

Algebra Topics # How do you simplify (-16)^(-1/2) ? Mar 28, 2016 $\pm \frac{1}{4 i}$ #### Explanation: Consider the example ${x}^{- \frac{1}{2}}$. This is the same as $\frac{1}{\sqrt{x}}$ Now use this approach for your question. $\text{ } {\left(- 16\right)}^{\frac{1}{2}} = \frac{1}{\sqrt{- 16}}$ Write $- 16 \text{ as } - 1 \times 16$ giving $\text{ } \frac{1}{\sqrt{16} \times \sqrt{- 1}}$ The square root of negative 1 gives rise to the complex number context. So we have: $\text{ } \frac{1}{\pm 4} \times \frac{1}{i}$ $\text{ } \pm \frac{1}{4 i}$ '~~~~~~~~~~~~~~~~~~~~~~~~~ If the question had been $- {\left(16\right)}^{- \frac{1}{2}}$ Then the solution would have been: $- \frac{1}{4}$ ##### Impact of this question 2155 views around the world

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http://math.stackexchange.com/questions/83926/probability-why-my-solution-doesnt-work-out-p-of-drawing-a-pair

# Probability, why my solution doesn't work out? (P of drawing a pair) The task is simple, the probability of drawing a pair of cards. You draw two cards from a stack, what is the chance that you get two kings or two fours. My idea was the following. There are 13 different valued cards. The probability of getting lets say a pair of two is following. $\frac{1}{52} \cdot \frac{3}{51}$ The chance of getting first card 2 is $\frac{1}{52}$, the chance of getting one of the three other cards that would make this a pair is $\frac{3}{51}$. That sounds reasonable to me. And now to account for all 13 different types you just multiply this with 13. Or add it up 13 times. $13(\frac{1}{52} \cdot \frac{3}{51})$ This is wrong and I don't understand why. The probability of getting a random pair should be the sum of getting every type of pair. A correct solution would be the following. $\frac{52 \cdot 3}{52 \cdot 51}$ I try to stick with the 'count the number of beneficial outcomes in every step and multiply method' but I got stuck on why my way of thinking didn't work out here. - The probability that the first card is a 2, is $4/52$, not $1/52$. Your reasoning works out, with this correction. –  David Mitra Nov 20 '11 at 13:14 Silly me. Lalalala. Thanks. –  Algific Nov 20 '11 at 13:16 It doesn't matter what first card drawn is. Probability of matching is $3/51$. –  André Nicolas Nov 20 '11 at 15:56 Shouldn't it be $\frac{2}{13}\cdot\frac{3}{51}$ where the $\frac{2}{13}$ is the probability that the first card is a four or a king? –  Henning Makholm Nov 20 '11 at 16:00 @David, why not post that as an answer, so it may be accepted? –  Mr.Wizard Nov 22 '11 at 9:54

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https://www.intmath.com/plane-analytic-geometry/8-curves-polar-coordinates.php

# 8. Curves in Polar Coordinates Don't miss the Polar graphs interactive applet. We'll plot the graphs in this section using a computer. You'll also learn how to sketch some of them on paper because it helps you understand how graphs in polar coordinates work. Don't worry about all the difficult-looking algebra in the second part of the answers - it's just there to demonstrate that polar coordinates are much simpler than rectangular coordinates for these graphs. We convert them using what we learned in the last section, Polar Coordinates. Curves in polar coordinates work very similarly to vectors. See: Vector concepts ### Need Graph Paper? (Polar graph paper included.) Sketch each of the following functions using polar coordinates, and then convert each to an equation in rectangular coordinates. Example 1: r = 2 + 3 sin θ (This polar graph is called a limacon from the Latin word for "snail".) Here's another example of a limacon: Example 2: r = 3 cos 2θ Example 3: r = sin θ − 1 (This one is called a cardioid because it is heart-shaped. It is a special case of the limacon.) Continues below Example 4: r = 2.5 Example 5: r = 0.2 θ This is an interesting curve, called an Archimedean Spiral. As θ increases, so does r. Later, we'll learn how to find the Length of an Archimedean Spiral. Example 6: r = sin (2θ) − 1.7 This is the face I drew at the top of this page. We're not even going to try to find the equivalent in rectangular coordinates! You can play with this graph in the following interactive applet. ## Interactive Graph You an explore the above graphs using this interactive graph. Use the slider below the graph to trace out the curves. See what happens as you go beyond the normal domain for these graphs (i.e. when theta is less than 0 or greater than theta = 2pi). Change the function using the select box at the top of the graph. Choose function: ### Application Check out Polar Coordinates and Cardioid Microphones for an application of polar coordinates. top ### Online Algebra Solver This algebra solver can solve a wide range of math problems. ### Math Lessons on DVD Easy to understand math lessons on DVD. See samples before you commit.

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https://www.physicsforums.com/threads/an-intro-to-real-analysis-question-eazy.431949/

# An intro to real analysis question. eazy? 1. Sep 24, 2010 ### eibon 1. The problem statement, all variables and given/known data Let f : A -> B be a bijection. Show that if a function g is such that f(g(x)) = x for all x ϵ B and g(f(x)) = x for all x ϵ A, then g = f^-1. Use only the definition of a function and the definition of the inverse of a function. 2. Relevant equations 3. The attempt at a solution well since f is a bijection then there exist an f^-1 that remaps f back to A and since g does that then g is the unique inverse of f, or something like that please help im not very good that this stuff 2. Sep 25, 2010 ### losiu99 Yes, you only need to prove uniqueness of an inverse function. If f(g(x))=x, then g(x) must be the unique argument a for which f(a) = x. And so, this condition completely defines g for all the arguments. g is then obviously equal to f-1. 3. Sep 25, 2010 ### eibon thanks losiu for your response. but how do you prove that it is the unique inverse?

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http://www.physicsforums.com/printthread.php?t=199487

Physics Forums (http://www.physicsforums.com/index.php) -   Calculus (http://www.physicsforums.com/forumdisplay.php?f=109) -   -   What are derivatives and integrals? (http://www.physicsforums.com/showthread.php?t=199487) The_Z_Factor Nov20-07 10:30 AM What are derivatives and integrals? What are they? In my book Im studying limits and it has mentioned a few times before and in the current chapter Derivatives and Integrals, but hasnt explained them. Could anybody explain what these two things are, exactly? SiddharthM Nov20-07 10:36 AM if your asking for a formal definition then goto www.wikipedia.com and search for derivative and separately integration. Geometric calculus interpretation: If f(x) is a line and can be represented by mx +b then the slope is m, but this is a line and the slope is the same throughout the real numbers. Now consider y= x^2, what is the slope? it changes at each point, the slope at any point is the derivative of the function evaluated at that point. the integral of a real function gives you the area under the curve of the function. The_Z_Factor Nov20-07 10:41 AM Quote: Quote by SiddharthM (Post 1511842) If f(x) is a line and can be represented by mx +b then the slope is m, but this is a line and the slope is the same throughout the real numbers. Now consider y= x^2, what is the slope? it changes at each point, the slope at any point is the derivative of the function evaluated at that point. So does this mean that there can be as many derivatives as there are points? Gib Z Nov20-07 08:31 PM Quote: Quote by The_Z_Factor (Post 1511846) So does this mean that there can be as many derivatives as there are points? Different functions can be differentiated a different number of times. Mute Nov21-07 12:24 AM Quote: Quote by The_Z_Factor (Post 1511846) So does this mean that there can be as many derivatives as there are points? It means that in general the derivative of a function of x is itself a function of x. i.e., the slope of a function is different at each point on that function. For example, the derivative of x^2 is 2x. This means that on the curve y = x^2, at the point x = 4, the slope of the curve at x = 4 (or, perhaps more precisely, the slope of the line tangent to the curve at x = 4) is 2*4 = 8. Similarly, the slope at the point x = -5 is -10. The_Z_Factor Nov21-07 01:52 AM Ah, thanks for clearing that up for me everybody, that explains it. I think as I'm beginning to learn more about simple calculus I'm beginning to like it more. Haha, it just might turn into a hobby once I learn enough. All times are GMT -5. The time now is 12:51 AM.

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http://docplayer.net/355519-Flow-by-mean-curvature-of-convex-surfaces-into-spheres.html

# FLOW BY MEAN CURVATURE OF CONVEX SURFACES INTO SPHERES Size: px Start display at page: Download "FLOW BY MEAN CURVATURE OF CONVEX SURFACES INTO SPHERES" ## Transcription 1 J. DIFFERENTIAL GEOMETRY 20 (1984) FLOW BY MEAN CURVATURE OF CONVEX SURFACES INTO SPHERES GERHARD HUISKEN 1. Introduction The motion of surfaces by their mean curvature has been studied by Brakke [1] from the viewpoint of geometric measure theory. Other authors investigated the corresponding nonparametric problem [2], [5], [9]. A reason for this interest is that evolutionary surfaces of prescribed mean curvature model the behavior of grain boundaries in annealing pure metal. In this paper we take a more classical point of view: Consider a compact, uniformly convex w-dimensional surface M = M o without boundary, which is smoothly imbedded in R π+1. Let M o be represented locally by a diffeomorphism F o : R" D U -> F 0 (U) c M o c R w+1. Then we want to find a family of maps F(-,t) satisfying the evolution equation γ t F(x,t) = Δ t F(x 9 t) 9 F(,0) = F 0, ίeί/, where Δ, is the Laplace-Beltrami operator on the manifold Λf,, given by F(,0 Wehave Δ,F(x, 0 = -H(x 9 t) v(x 9 t) 9 where H(, t) is the mean curvature and v(, t) is the outer unit normal on M r With this choice of sign the mean curvature of our convex surfaces is always positive and the surfaces are moving in the direction of their inner unit normal. Equation (1) is parabolic and the theory of quasilinear parabolic differential equations guarantees the existence of F(, /) for some short time interval. Received April 28,1984. 2 238 GERHARD HUISKEN We want to show here that the shape of M t approaches the shape of a sphere very rapidly. In particular, no singularities will occur before the surfaces M t shrink down to a single point after a finite time. To describe this more precisely, we carry out a normalization: For any time t, where the solution F(- 9 t) of (1) exists, let \p(t) be a positive factor such that the manifold M t given by has total area equal to M 0, the area of M o : M t dμ = Af o for all t. After choosing the new time variable t(t) = / 0 ' Ψ 2 (τ) dτ it is easy to see that F satisfies j ^ r (2) ^7 ' «r ^' '' where h= f H 1 dμ/f dμ J M is the mean value of the squared mean curvature on M t (see 9 below). 1.1 Theorem. Let n ^ 2 and assume that M o is uniformly convex, i.e., the eigenvalues of its second fundamental form are strictly positive everywhere. Then the evolution equation (1) has a smooth solution on a finite time interval 0 < / < Γ, and the M 9 t s converge to a single point ) as t -> T. The normalized equation (2) has a solution M~ t for all time 0 < t < oo. The surfaces M~ t are homothetic expansions of the M 9 t s, and if we choose as the origin ofr n + 1, then the surfaces M~ t converge to a sphere of area \M 0 \ in the C 00 -topology ast^> oo. Remarks, (i) The convergence of M- t in any C^-norm is exponential. (ii) The corresponding one-dimensional problem has been solved recently by Gage and Hamilton (see [4]). The approach to Theorem 1.1 is inspired by Hamiltons paper [6]. He evolved the metric of a compact three-dimensional manifold with positive Ricci curvature in direction of the Ricci curvature and obtained a metric of constant curvature in the limit. The evolution equations for the curvature quantities in our problem turn out to be similar to the equations in [6] and we can use many of the methods developed there. In 3 we establish evolution equations for the induced metric, the second fundamental form and other important quantities. In the next step a lower ' J M 3 FLOW OF CONVEX SURFACES 239 bound independent of time for the eigenvalues of the second fundamental form is proved. Using this, the Sobolev inequality and an iteration method we can show in 5 that the eigenvalues of the second fundamental form approach each other. Once this is established we obtain a bound for the gradient of the mean curvature and then long time existence for a solution of (2). The exponential convergence of the metric then follows from evolution equations for higher derivatives of the curvature and interpolation inequalities. The author wishes to thank Leon Simon for his interest in this work and the Centre for Mathematical Analysis in Canberra for its hospitality. 2 Notation and preliminary results In the following vectors on M will be denoted by X = {X*}, covectors by Y= {Yi} and mixed tensors by T = {Ttf}. The induced metric and the second fundamental form on M will be denoted by g = {g /7 } and A = {Λ /y } We always sum over repeated indices from 1 to n and we use brackets for the inner product on M: (τ/ k, sj k ) = g is g jr g ku τ ks s ru, \τ\ 2 = (τ k, τ; k ). In particular we use the following notation for traces of the second fundamental form on M: - MI 4. By (* > *) w e denote the ordinary inner product in R n+1 Λί M is given locally by some F as in the introduction, the metric and the second fundamental form on M can be computed as follows: x e R w, where v(x) is the outer unit normal to M at F(x). The induced connection on M is given by 1 fel so that the covariant derivative on M of a vector X is J dxj Jk 4 240 GERHARD HUISKEN The Riemann curvature tensor, the Ricci tensor and scalar curvature are given by Gauss' equation &ijkl = "ik"jl ~ "il"jk> K ik = Hh ik - h u g% k9 R = H 2 - \A\ 2. With this notation we obtain, for the interchange of two covariant derivatives, V,V,X Λ - V,V,** = Rl Jk X" = (h,jh, k - h lk h u )g-'x k, v,vjΐ k - v,v,n = R ijkl g""y m = {h lk hj, - h u h jk )g""y m. The Laplacian ΔΓ of a tensor Ton Mis given by AT' = σ mn T7 V7 T i whereas the covariant derivative of T will be denoted by vt = { V{T Jk }. Now we want to state some consequences of these relations, which are crucial in the forthcoming sections. We start with two well-known identities. im 2.1 Lemma, (i) ΔΛ,.. = v t VjH + Hh ilg h mj - Λ 2 /* 0, (ii) ±Δ Λ 2 = (h^vfjh) + \VA\ 2 + Z. Proof. The first identity follows from the Codazzi equations Vih kl = V k h u = V/Λ Λ and the formula for the interchange of derivatives quoted above, whereas (ii) is an immediate consequence of (i). The obvious inequality \VH\ 2 < w V^4 2 can be improved by the Codazzi equations. 2.2 Lemma.(i) \VA\ 2 > 3/(n + 2) V# 2. (ii) VΛ 2 - \vh\ 2 /n > 2{n - l) vλ 2 /3n. Proof. Similar as in [6, Lemma 11.6] we decompose the tensor va: where = E ijk Then we can easily compute that \E\ 2 = 3 vh\ 2 /(n + 2) and ( E ijk> F ijk) = ( E ijk^i h jk - I>Λ) =» i.e., ^ and F are orthogonal components of va. Then which proves the lemma. 5 FLOW OF CONVEX SURFACES 241 If M tj is a symmetric tensor, we say that M ly is nonnegative, M tj > 0, if all eigenvalues of M tj are nonnegative. In view of our main assumption that all eigenvalues of the second fundamental form of M o are strictly positive, there is some ε > 0 such that the inequality (3) h tj > εh gij holds everywhere on M Q. It will be shown in 4 that this lower bound is preserved with the same ε for all M t as long as the solution of (1) exists. The relation (3) leads to the following inequalities, which will be needed in Lemma. IfH> 0, and (3) is valid with some ε > 0, then (i) Z > nε 2 H 2 (\A\ 2 - H 2 /n). (ϋ) IVA/ H - V z # h kl \ 2 > \ε 2 H 2 \vh\ 2. Proof, (i) This is a pointwise estimate, and we may assume that g zy = δ /y and K 2 0 In this setting we have Z = HC- \A\ 4 n \ I n i-l and the conclusion follows since (ϋ) We have μ 2_I i f 2 = 1 IA Σ ( κ κ )\ n n r*. v 7/ ' ί<7 H - \{V,H h kl + V k H h u ) - Kv,iί h kl - V k H h u )\ 2 H - \(v,h h kl + V k H A, 7 ) 2 + \\v,h h kl - V k H hf >\\v i H-h kl - V k H-hf, 6 242 GERHARD HUISKEN since V t h kl is symmetric in (/, k) by the Codazzi equations. Now we have only to consider points where the gradient of the mean curvature does not vanish. Around such a point we introduce an orthonormal frame e l9,e n such that e λ = vh/\vh\. Then in these coordinates. Therefore 10, i > 2, T Σ {v,h'h kl - V k H-h u f h 22 -V 2 H h u f since any eigenvalue, and thus any trace element of h tj is greater than εh. 3. Evolution of metric and curvature In this and the following sections we investigate equation (1) which is easier to handle than the normalized equation (2). The results will be converted to the normalized equation in Theorem. The evolution equation (1) has a solution M t for a short time with any smooth compact initial surface M = M o at t = 0. This follows from the fact that (1) is strictly parabolic (see for example [3, III.4]). From now on we will assume that (1) has a solution on the interval 0 < t < T. Equation (1) implies evolution equations for g and A, which will be derived now. 3.2 Lemma. The metric ofm t satisfies the evolution equation ( 4 ) γ t Su = ~ 2Hh u Proof. The vectors df/dx t are tangential to M, and thus 7 FLOW OF CONVEX SURFACES 243 From this we obtain 9F 3 df\ JdF 3 = -2Hh tj. 3.3 Lemma. The unit normal to M, satisfies dp/dt Proof. This is a straightforward computation: 3ί" ~ \ 3ί"' 3x, Jdxj 8 \ V ' dt Now we can prove 3.4 Theorem. The second fundamental form satisfies the evolution equation lh,j = ΔΛ, 7-2Hh u g lm h mj + \A\ 2 h tj. Proof. We use the Gauss-Weingarten relations 3 2 F rjt 3F, 3, /m 3F = 1 h v v = h,q x3x, l 'Jdx h k 'J V ' dx/ h to conclude J> 8 dx _3_ 3 3/ ' 7 3ί - v,v/r - Hh,,g'"h mj. Then the theorem is a consequence of Lemma 2.1. 8 244 GERHARD HUISKEN 3.5 Corollary. We have the evolution equations: (i) ^H = AH+ \A\ 2 H, (ϋ) j^\a\ 2 = A\A\ 2-2 VΛ 2 + 2\A\\ Proof. We get, from Lemma 3.2, and the first identity follows from Theorem 3.4. To prove the second equation, we calculate + 2g ik gj'h kl {Ah u - 2Hh im g m»h nj + \A\ 2 h,j) = 2g ik gj'h k Ah ij + 2\A\\ = g kl V k V,{g"g m»h pm h qι,) = 2gPig m "h pm Ah qn + 2\VA\\ The last identity follows from (ϋ) and γ f H 2 = 2if(Δi/ + μ 2^) = Δ^2-2\VH\ Corollary, (i) If dμ t = μ,(3c) dx is the measure on M n then μ = Jdet g tj and dμ t /dt = -H 2 μ r In particular the total area \M t \ of M t is decreasing. (ii) // the mean curvature of M o is strictly positive everywhere, then it will be strictly positive on M t as long as the solution exists. Proof. The first part of the corollary follows from Lemma 3.2, whereas the second part is a consequence of the evolution equation for H and the maximum principle. 4. Preserving convexity We want to show now that our main assumption, that is inequality (3), remains true as long as the solution of equation (1) exists. For this purpose we need the following maximum principle for tensors on manifold, which was 9 FLOW OF CONVEX SURFACES 245 proved in [6, Theorem 9.1]: Let M^bea vector field and let g ij9 M {j and N tj be symmetric tensors on a compact manifold M which may all depend on time /. Assume that N tj = p(m ip gtj) is a polynomial in M i} formed by contracting products of M tj with itself using the metric. Furthermore, let this polynomial satisfy a null-eigenvector condition, i.e. for any null-eigenvector X of M tj we have N ij X i X J > 0. Then we have 4.1 Theorem (Hamilton). Suppose that on 0 < / < T the evolution equation holds, where N έj = p(m ij9 g /y ) satisfies the null-eigenvector condition above. If M tj > 0 at t = 0, then it remains soono < t < T. An immediate consequence of Theorems 3.4 and 4.1 is 4.2 Corollary. J/λ, 7 > 0 at t = 0, then it remains so for 0 < / < Γ. Proo/. Set M tj = A l7, «Λ ^ 0 and N tj = -2Hh u g""h mj + μ4 2 *, y. We also have the following stronger result. 4.3 Theorem. If εhg^ < Λ /y < βhgij, and H > 0 at the beginning for some constants 0 < ε < \/n < β < 1, ίaew r/zw remains soono < / < Γ. Proof. To prove the first inequality, we want to apply Theorem 4.1 with M,j = - - ε gij, u k = jjg kl V,H, N ij =2εHh ij -2h im g'"'h IJ. With this choice the evolution equation in Theorem 4.1 is satisfied since ^ i m g lj ' H 8 It remains to check that N^ is nonnegative on the null-eigenvectors of M tj. Assume that, for some vector X = {X 1 }, u hijx j = εhx r Then we derive NyX'XJ = lεhhtjx'xj - 2h im g mi h lj X i X J = 2ε 2 H 2 \X\ 2-2ε 2 H 2 \X\ 2 = 0. That the second inequality remains true follows in the same way after reversing signs. 10 246 GERHARD HUISKEN 5. The eigenvalues of A In this section we want to show that the eigenvalues of the second fundamental form approach each other, at least at those points where the mean curvature tends to infinity (for the unnormalized equation (1)). Following the idea of Hamilton in [6], we look at the quantity which measures how far the eigenvalues K t oi A diverge from each other. We show that \A\ 2 H 2 /n becomes small compared to H Theorem. There are constants δ > 0 and C o < oo depending only on M o, such that for all times 0 < t < T. Our goal is to bound the function/, = {\A\ 2 - H 2 /n)/h 2 ~ σ small σ. We first need an evolution equation for/ σ. 5.2 Lemma. Let a = 2 σ. Then, for any σ, for sufficiently (2 - a)(a - +(2 - a)\aff a. Proof. We have, in view of the evolution equations for \A\ 2 and H, 1/ =AίMl!_I ί /2 dt J dt\h a n = HA\A\2 - a\a\ 2 ΔH _ (2 - «) fj1 -. H a+1 11 FLOW OF CONVEX SURFACES 247 Furthermore I 2 - «MI 2 v,ff (2-α), (2-<*),-, n (5) 1 (2-α)(l-α), 2 F Viί and now the conclusion of the lemma follows from reorganizing terms and the identity Iv,* w # - v,^ AJ 2 = H 2 \VA\ 2 + μ 2 VH\ 2 - (v,μ 2, V, Unfortunately the absolute term (2 α) ^4 2 / σ in this evolution equation is positive and we cannot achieve our goal by the ordinary maximum principle. But from Theorem 4.3 and Lemma 2.3(ϋ) we get 5.3 Corollary. For any σ the inequality (6) l t f σ < Δ/ σ + ^Zi) ( V H> VίΛ ) _ ε 2^ vtf 2 + o\a\ 2 f σ holds ono < t < T. The additional negative term in (6) will be exploited by the divergence theorem: 5.4 Lemma. Let p > 2. Then for any η > 0 and any 0 < σ < \ we have the estimate nε 2 ffph 2 dμ < (2ηp + 5)/ -^ff-^vh] 2 dμ Proof. Let us denote by Λ^ the trace-free second fundamental form 12 248 GERHARD HUISKEN In view of Lemma 2.1(ϋ), the identity (5) may then be rewritten as H ~ W A *Ί 2 " f Now we multiply the inequality by//" 1 and integrate. Integration by parts yields 0 > (P - l)ff. p - 2 \vf a \ 2 dμ + f jp ί ZfΓ 1 dμ -2(a-l)f jjfγ'ivj^v where we used the Codazzi equation. Now, taking the relations (7) ab < f α 2 + J-6 2, α < 2, 2 2i) into account, we derive, for any η > 0, / jp fγ'zdμ < (2η/> + 5)/ ^//-^ The conclusion then follows from Lemma 2.3(i) and Theorem 4.3. 13 FLOW OF CONVEX SURFACES 249 Now we can show that high ZAnorms of f σ are bounded, provided σ is sufficiently small. 5.5 Lemma. There is a constant C λ < oo depending only on M o, such that, for all (8) p > looε" 2, σ < \έp~ X/ \ the inequality holds ono < t < T. Proof. We choose (ί J M t and it is then sufficient to show : ( Af + 1) sup (sup/ ) x o σ [01/2] J σe [0,1/2] V Λ/ o ( To accomplish this, we multiply inequality (6) by/?//" 1 and obtain f / S! dμ + p(p ~ 1)/ //" < 2(β - where the last term on the left-hand side occurs due to the time dependence of dμ as stated in Corollary 3.6(i). In view of (7) we can estimate 2(a-l)pf j and since/? - 1 > looε" 2-1 > 4ε" 2, ^4 2 < H 2, we conclude + hp{p ~ \)jfγ 2 \Vf dμ + Wpj jpfγλvh?dμ H 2 fidμ. 14 250 GERHARD HUISKEN The assumption (8) on σ and Lemma 5.4 yield + \p(p - l)jfγ 2 \vf.\ 2 dμ + Wpj jpfγ^h? dμ 5)/ ±. fr i\ V H\ 2 for any η > 0. Then (9) follows if we choose η = εp~ 1/2 / Corollary. If we assume then we have \ 1/p on 0 < t < T. Proof. This follows from Lemma 5.5 since with n i _ Ί n. i n n ε σ' = σ + < ε 3 /? 7 + mp 1/2 < -ε 3 p 1/2. /? 16 m 16 8 We are now ready to bound f σ by an iteration similar to the methods used in [2], [5]. We will need the following Sobolev inequality from [7]. 5.7 Lemma. For all Lipschitz functions υ on Mwe have If \v\ n/n ~ 1 dμ) <c(n)l[ \w\dμ+ f H\υ\dμ). \ J M I \ J M J M ) Proof of Theorem 5.1. Multiply inequality (6) by pfj-'^, where f ak = max(/ σ k,0) for all k > k 0 = sup M / σ, and denote by A(k) the set where f σ > k. Then we derive as in the proof of Lemma 5.5 for/? > 100ε~ 2 l Z k \ ) j σpf H 2 fp- k 15 FLOW OF CONVEX SURFACES 251 On A(k) we have and thus we obtain with υ = 2 3 TΓ-ί v 2 dμ+( \Vυ\ 2 dμ^op( H 2 fξ dμ. 0 1 J A(k) J A{k) J Λ{k) Let us agree to denote by c n any constant which only depends on n. Then Lemma 5.7 and the Holder inequality lead to where [ V 2 «dμ\ < c j \VΌ\ dμ + c n [j H" dμ) [I v 2^ dμ), \ J M ) J M \/suppi; I \ J M ] [n/{n- 2), n>2, \ < oo, «= 2. Since supp v <z A(k),we have in view of Corollary 5.6 / \ 2/n I \ 2/n \[ H n dμ\ ^k- 2p/n \( H n f p dμ\ < k~ 2p/n C 2p/ \ \ j s\xpvυ I \ J A(k) I provided p > 2 ε~, σ < ~TΣ ε P~ Thus, under this assumption we conclude for k > k λ = A: X (A: O, C\, π, ε) that sup f υ 2 dμ + c n ( f υ 2q dμ dt [0, T) J A(k) J 0 \ J A(k) I <σp Γ f H 2 f p dμdt. Now we use interpolation inequalities for L ^-spaces \l/<7o If ) V<7o / \α /^/ \ ( A(k) < \( υ 2q dμ\ \f v 2 dμ\ \ J A(k) I \ J A(k) I (), with a = l/ί 0 such that 1 < q 0 < q. Then we have 0 f 0 f \ 1/q ί T f f v 2 * dμdt\ <c n σpf [ H 2 fξdμdt 0 J A(k) A(k) I J 0 J A(k) ί f ~ 1/r (ί T f H 2r V J A(k) Ό J A(k) 16 252 GERHARD HUISKEN where r > 1 is to be chosen and IM(^)II = / / dμdt. Again using the Holder inequality we obtain J 0 Γ ί f^dμdt < c n ap\\a{k)\γ /q ~ 1/r [ Γ ί H*'f m»dμ dt\"'. J A(k) \ J 0 J A(k) I If we now choose r so large that 2 - l/q 0 - \/r = γ > 1, then r only depends on n and we may take (10) p > rε , σ < ε^v 1 / 2 such that by Corollary 5.6 for all h > k > k v By a well-known result (see e.g. [8, Lemma 4.1]) we conclude for some/? and σ satisfying (10). Since dμ < \M t \ < Af o by Corollary 3.6(i), it remains only to show that Γis finite. 5.8 Lemma. T < oo. Proof. The mean curvature H satisfies the evolution equation γ t H = Δ// + H\A\ 2 > AH + \H\ Then let φ be the solution of the ordinary differential equation ~«Γ = ~Φ 3 9 φ(0) = ^min(o) > 0. If we consider φ as a function on M X [0, Γ), we get such that by the maximum principle H ^ φ on 0 < / < T. On the other hand φ is explicitly given by φ(0- 17 FLOW OF CONVEX SURFACES 253 And since φ -> oo as / -> («/2)i/^(0), the result follows. Moreover, in the case that M o is a sphere, φ describes exactly the evolution of the mean curvature and so the bound T < (w/2)ϋq^(0) is sharp. This completes the proof of Theorem A bound on I V// In order to compare the mean curvature at different points of the surface M n we bound the gradient of the mean curvature as follows. 6.1 Theorem. For any η > 0 there is a constant C(η, M o, n) such that Proof. First of all we need an evolution equation for the gradient of the mean curvature. 6.2 Lemma. We have the evolution equation v 2 i/ 2 + 2\A\ 2 \VH\ 2 + 2( v,h h mj, VjHvh im ) 6.3 Corollary. ^ vtf 2 < Δ V# 2-2\V 2 H\ 2 + 4M 2 V# 2 + 2H( V t H, V Proof of Lemma 6.2. Using the evolution equations for H and g we obtain = 2H(h u, V,# VjH) + ig'jv The result then follows from the relations Δ V# 2 = 2g*'Δ( V,^) V,H Δ( v k H) = vλδtf) + g'jvmmicj ~ h km g mn h nj ). 6.4 Lemma. We have the inequality 18 254 GERHARD HUISKEN Proof. We compute 9 / \VH\ 2 \ HA\VH\ 2 -\VH\ 2 AH dt\ H and the result follows from Schwarz' inequality. We need two more evolution equations. 6.5 Lemma. We have (i) ^Jϊ 3 = Δi/ 3-6H\VH\ 2 + 3\A\ 2 H\ w/yλ α constant C 3 depending onn, C o and 8, i.e., only on M Q. Proof. The first identity is an easy consequence of the evolution equation for H. To prove the inequality (ϋ), we derive from Corollary 3.5(iii) Now, using Theorem 5.1 and (7) we estimate 2 and the conclusion follows from Lemma 2.2(ii). C(n, Co, 8)\VA\\ 19 We are now going to bound the function FLOW OF CONVEX SURFACES 255 /= l^l + N(\ A \ 2 _ ±H 2 \H + NC 3 \A\ 2 - ηh 3 for some large N depending only on n and 0 < η < 1. From Lemmas 6.4 and 6.5 we obtain dt 6ηH\vH\ N\A\ 2 H(\A\ 2 - ±H 2 \ - 3τ,\A\ 2 H 3. Since (ϊ/n)h 2 < \A\ 2 < H 2, V# 2 < n V^ 2 and η < 1 we may choose JV depending only on n so large that By Theorem 5.1 we have f t < Δ/+ 2NC,H* + 3NH*(\A\ 2 - \H 2 ] - \^H\ 2NC 3 H 4 + 3NH 3 l\a\ 2 - ^H 2 \ < 2NC 3 H 4 + 3NC 0 H 5 δ and hence df/dt < Δ/ 4- C(η, M o ). This imphes that max/(o < max/(0) + C(η, M 0 )t, and since we already have a bound for Γ, / is bounded by some (possibly different) constant C(η, M o ). Therefore IVH\ 2 < τ?if 4 + C(η 9 M 0 )H < 2η/ί 4 + C(η, M o ) which proves Theorem 6.1 since η is arbitrary. 7. Higher derivatives of A As in [6] we write S * T for any linear combination of tensors formed by contraction on S and Γby g. The mth iterated covariant derivative of a tensor T will be denoted by V"T. With this notation we observe that the time derivative of the Christoffel symbols Γ^ is equal to 9 _.. 1 u ( ί 9 \ / 9 \ / 9 97 Γ^ - 2 g { V A **») + **(**") ~ V \T = -g il { Vj(Hh kl ) + V k {Hh β ) - 20 256 GERHARD HUISKEN in view of the evolution equation for g = {g /7 }. Then we may proceed exactly as in [6, 13] to conclude 7.1 Theorem. For any m we have an equation j 2 \ 2-2\v m+ι A\ 2 i +j r + k = m Now we need the following interpolation inequality which is proven in [6, 12]. 7.2 Lemma. // T is any tensor and if 1 < i < m 1, then with a constant C(n, m) which is independent of the metric g and the connection Γ we have the estimate /I This leads to 73 Theorem. d f, dt JM, m \v ι T\ dμ < C - max T\ M We have the estimate 4\ dμ + 2 I V m+λ A\ dμ < dμ, where C only depends on n and the number of derivatives m. Proof. By integrating the identity in Theorem 7.1 and using the generalised Holder inequality we derive M t fj M t i/2m, χ j/2m { k/2m. v 1/2 with i + j + k = m. The interpolation inequality above gives //2m.. i/2m 1/ { ) and if we do the same withy and k, the theorem follows. 1 } 21 FLOW OF CONVEX SURFACES The maximal time interval We already stated that equation (1) has a (unique) smooth solution on a short time interval if the uniformly convex, closed and compact initial surface M o is smooth enough. Moreover, we have 8.1 Theorem. The solution of equation (1) exists on a maximal time interval 0 < / < T < oo and max M \A\ 2 becomes unbounded as t approaches T. Proof. Let 0 < t < T be the maximal time interval where the solution exists. We showed in Lemma 5.8 that T < oo. Here we want to show that if max M \A\ 2 ^ C for t -> T, the surfaces M t converge to a smooth limit surface M τ. We could then use the local existence result to continue the solution to later times in contradiction to the maximality of T. In the following we suppose (11) max \A\ 2 < C on 0 < t < T, M t and assume that as in the introduction M t is given locally by F(x 9 t) defined for x e U c R" and 0 < t < T. Then from the evolution equation (1) we obtain for 0 < σ < p < T. Since H is bounded, F(- 9 t) tends to a unique continuous limit F(,T) as/ -> T. In order to conclude that F( 9t) represents a surface M T9 we use [6, Lemma 14.2]. 8.2 Lemma. Let g tj be a time dependent metric on a compact manifold M for 0 < t < T < oo. Suppose / max dt < C < oo. Then the metrics gij(t)for all different times are equivalent, and they converge as t -> T uniformly to a positive definite metric tensor gij(t) which is continuous and also equivalent. Here we used the notation dt 8ij In our case all the surfaces M t are diffeomorphic and we can apply Lemma 8.2 in view of Lemma 3.2, assumption (11) and the fact that T < oo. It remains only to show that M τ is smooth. To accomplish this it is enough to prove that 22 258 GERHARD HUISKEN all derivatives of the second fundamental form are bounded, since the evolution equations (1) and (4) then imply bounds on all derivatives of F. 8.3 Lemma. If (II) holds ono < t < T and T < oo, then \ V m A\ < C m for all m. The constant C m depends on n, M o and C. Proof. Theorem 7.3 immediately implies since the inequality dg/dt < eg on a finite time interval gives a bound on g in terms of its initial data. Then Lemma 7.2 yields J M t for all m and p < oo. The conclusion of the lemma now follows if we apply a version of the Sobolev inequality in Lemma 5.7 to the functions g m = V m A\ 2. Thus the surfaces M t converge to M τ in the C -topology as t -> T. By Theorem 3.1 this contradicts the maximality of T and proves Theorem 8.1. We now want to compare the maximum value of the mean curvature # max to the minimum value H πήn as t tends to T. Since \A\ 2 < H 2, we obtain from Theorem 8.1 that i/ max is unbounded as / approaches T. 8.4 Theorem. We have H m3λ /H πύn -» 1 as t -» T. Proof. We will follow Hamiltons idea to use Myer's theorem. 8.5 Theorem (Myers). If R tj^ (n V)Kg tj along a geodesic of length at least πk~ 1/2 on M, then the geodesic has conjugate points. To apply the theorem we need 8.6 Lemma. Ifh^ > εhgjj holds on M with some 0 < ε < l/n, then R ij >{n-\)ε 2 H 2 g ij. Proof of Lemma 8.6. This is immediate from the identity R tj = Hh u - h im g mn h nj. Now we obtain from Theorem 6.1 that for every η > 0 we can find a constant c(η) with \VH\ < \tfh 2 + C(η) on 0 < / < T. Since i/ max becomes unbounded as t -> Γ, there is some θ < T with C(η) < Wϋmax at ί = fl. Then (12) Ivi/NηΉLc at time t = θ. Now let x be a point on M^, where H assumes its maximum. Along any geodesic starting at x of length at most tf ι ll^ we have H ^ (1 η)h max. In view of Lemma 8.6 and Theorem 8.5 those geodesies then reach any point of M θ if η is small and thus (13) H mϊn >(l-η)h max onm θ. 23 Since H^^ is nondecreasing we have FLOW OF CONVEX SURFACES 259 #max(0 > 2 H {θ) On θ < t < T, maol and hence the inequalities (12) and (13) are true on all of θ < / < T which proves Theorem 8.4. We need the following consequences of Theorem Theorem. We have /H ^ ^ τ) dτ = oo. Proof. Look at the ordinary differential equation ff = #maχg, g(0) = H maχ. We get a solution since H^ is continuous in t. Furthermore we have and therefore T^H = ΔH + \A\ H < Δ/f + H max H, g^(jϊ " g) < Δ(JΪ - g) + ^ax(^ " g) So we obtain i/ < g for 0 < t < T by the maximum principle, and g -» oo as t -> Γ. But now we have jγ H^(τ) dτ = log{g(ί)/g(0)} - oo as / -> Γ, which proves Theorem Corollary. //, as /n /Λe introduction, h is the average of the squared mean curvature then = f H 2 dμ/f dμ J 9 M / J M / h(τ) dτ = oo. Proof. This follows from Theorems 8.4 and 8.7 since H^ ^ h < H^. 8.9 Corollary. We have \A\ 2 /H 2 - \/n -* 0 as t -> T. Proof. This is a consequence of Theorem 5.1 since H min -> oo by Theorem 8.4. Obviously M t stays in the region of R w+1 which is enclosed by M t for t λ > t 2 since the surfaces are shrinking. By Theorem 8.4 the diameter of M t tends to zero as t -> T. This implies the first part of Theorem 1.1. 24 260 GERHARD HUISKEN 9. The normalized equation As we have seen in the last sections, the solution of the unnormalized equation (1) γ t F=ΔF= -Hv shrinks down to a single point ) after a finite time. Let us assume from now on that ) is the origin of R π+1. Note that ) lays in the region enclosed by M t for all times 0 < t < T. We are going to normalize equation (1) by keeping some geometrical quantity fixed, for example the total area of the surfaces M r We could as well have taken the enclosed volume which leads to a slightly different normalized equation. As in the introduction multiply the solution F of (1) at each time 0 < t < T with a positive constant ψ(/) such that the total area of the surface M t given by is equal to the total area of M o : (14) f dμ= \M 0 \ on 0 < t < T. J M t Then we introduce a new time variable by such that dt/dt = ψ 2. We have / and so on. If we differentiate (14) for time ί, we obtain ψ dt n jdμ Now we can derive the normalized evolution equation for F on a different maximal time interval 0 < t < T: dp dp,_ 2,_ 7 ~ ^ \b = ψ όt ut = -HP + -hp n 25 FLOW OF CONVEX SURFACES 261 as stated in (2). We can also compute the new evolution equations for other geometric quantities. 9.1 Lemma. Suppose the expressions P and Q, formed from g and A, satisfy op/ot = ΔP + Q, and P has "degree" a, that is, P = ψ P. Then Q has degree (a 2) and dt n Proof. We calculate with the help of (15) = ψ- 2 {^AP + ψ«δp = -hp + ΔP + Q. n The results in Theorem 4.3, Theorem 8.4 and Corollary 8.9 convert unchanged to the normalized equation, since at each time the whole configuration is only dilated by a constant factor. 9.2 Lemma. We have (i) (ϋ) hy > eϊlgij, h (iii) ^ ^ \ as t - f. Now we prove 9.3 Lemma. There are constants C 4 and C 5 such that for 0 < t < f 0 < C 4 < H^ < H max < C 5 < oo. Proof. theorem The surface M encloses a volume V which is given by the divergence Since the origin is in the region enclosed by M- t for all times as well, we have that Fv is everywhere positive on M- t. By the isoperimetric inequality we have = c n \M 0 \ 26 262 GERHARD HUISKEN On the other hand we get from the first variation formula \M 0 \ = M? = \/ H{Fv) dμ < H max Pj, which proves the first inequality in view of Lemma 9.2(ii). To obtain the upper bound we observe that in view of h tj > εh rrύn g ij the enclosed volume V can be estimated by the volume of a ball of radius (ε//^)" 1 : The first variation formula yields V< c (FH r (w+1) V, > -^rjh^f {Fv) Hdμ > ^y^llmol, which proves the upper bound again in view of Lemma 9.2(ii). 9.4 Corollary, f = oo. Proof. We have dt/dt = ψ 2 and H 2 = ψ" 2 /ί 2 such that I h(τ) dτ = I h(τ) dτ = oo by Corollary 8.8. But by Lemma 9.3 we have h < H^ < C 52 and therefore f = oo. 10. Convergence to the sphere We want to show that the surfaces M- t converge to a sphere in the C topology as / -» oo. Let us agree in this section to denote by 8 > 0 and C < oo various constants depending on known quantities. We start with 10.1 Lemma. There are constants 8 > 0 and C < oo such that 7ι2 1 ~ o,_o f \A\ 2,-δt - - Proof. Let/be the function/= \A\ 2 /H 2 - \/n which has degree 0. Then we conclude as in the proof of Lemma 5.5 that, for some large/? and a small 8 depending on ε, -δffp\a\ 2 dμ + / (A - H 2 )f" dμ, since d/dt dμ = (h H 2 )άμ. In view of Lemma 9.2(ii) and Lemma 9.3 we have for all times t larger than some t 0 d 27 FLOW OF CONVEX SURFACES 263 with a different δ. Thus where C now depends on t 0 as well. The conclusion of the lemma then follows from the Holder inequality \M}\ = \M 0 \ and Lemma 9.3. Now let us denote by h the mean value of the mean curvature on M: 10.2 Lemma. We have h= I Hdμ/ί dμ f (H- hfdμ = j H 2 -h 2 dμ^ Ce~ sl. Proof. In view of the Poincare inequality it is enough to show that / \vh\ 2 dμ decreases exponentially. Note that the constant in the Poincare inequality can be chosen independently of t since we got control on the curvature in Lemma 9.2 and Lemma 9.3. Look at the function where N is a large constant depending only on n. The degree of g is -3, and from the results in 6 we obtain for all times larger than some t v Here we used that the term becomes small compared to H\vΛ\ 2 as t -> oo since \h Q kl\ = (\A\ 2 H 2 /n) ι/2 tends to zero. Now using Lemma 10.1 and C 4 < H < C 5 we conclude for t > ϊ l9 j,j gdμ < -δf gdfi 4- Ce- δl + / (h - H 2 )gdμ. Since (h H 2 ) -+ 0 as t -* ooby Lemma 9.2(ii), we have for all t larger than some 1 2 and therefore ### A PRIORI ESTIMATES FOR SEMISTABLE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS. 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Photoacoustic and Thermoacoustic Tomography with a variable sound speed FIELDS-MITACS Conference on the Mathematics of Medical Imaging Photoacoustic and Thermoacoustic Tomography with a variable sound speed Gunther Uhlmann UC Irvine & University of Washington Toronto, Canada, ### 1 Scalars, Vectors and Tensors DEPARTMENT OF PHYSICS INDIAN INSTITUTE OF TECHNOLOGY, MADRAS PH350 Classical Physics Handout 1 8.8.2009 1 Scalars, Vectors and Tensors In physics, we are interested in obtaining laws (in the form of mathematical ### Notes V General Equilibrium: Positive Theory. 1 Walrasian Equilibrium and Excess Demand Notes V General Equilibrium: Positive Theory In this lecture we go on considering a general equilibrium model of a private ownership economy. In contrast to the Notes IV, we focus on positive issues such ### Metric Spaces. Chapter 7. 7.1. Metrics Chapter 7 Metric Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y X. The purpose of this chapter is to introduce metric spaces and give some ### Vectors, Gradient, Divergence and Curl. Vectors, Gradient, Divergence and Curl. 1 Introduction A vector is determined by its length and direction. They are usually denoted with letters with arrows on the top a or in bold letter a. We will use ### 2.3 Convex Constrained Optimization Problems 42 CHAPTER 2. FUNDAMENTAL CONCEPTS IN CONVEX OPTIMIZATION Theorem 15 Let f : R n R and h : R R. Consider g(x) = h(f(x)) for all x R n. The function g is convex if either of the following two conditions ### Theory of Sobolev Multipliers Vladimir G. Maz'ya Tatyana O. Shaposhnikova Theory of Sobolev Multipliers With Applications to Differential and Integral Operators ^ Springer Introduction Part I Description and Properties of Multipliers ### BANACH AND HILBERT SPACE REVIEW BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but ### The Heat Equation. Lectures INF2320 p. 1/88 The Heat Equation Lectures INF232 p. 1/88 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3) ### LINEAR ALGEBRA W W L CHEN LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, ### A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS A QUIK GUIDE TO THE FOMULAS OF MULTIVAIABLE ALULUS ontents 1. Analytic Geometry 2 1.1. Definition of a Vector 2 1.2. Scalar Product 2 1.3. Properties of the Scalar Product 2 1.4. Length and Unit Vectors ### MICROLOCAL ANALYSIS OF THE BOCHNER-MARTINELLI INTEGRAL PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Pages 000 000 S 0002-9939(XX)0000-0 MICROLOCAL ANALYSIS OF THE BOCHNER-MARTINELLI INTEGRAL NIKOLAI TARKHANOV AND NIKOLAI VASILEVSKI ### ISOMETRIES OF R n KEITH CONRAD ISOMETRIES OF R n KEITH CONRAD 1. Introduction An isometry of R n is a function h: R n R n that preserves the distance between vectors: h(v) h(w) = v w for all v and w in R n, where (x 1,..., x n ) = x ### 4. Expanding dynamical systems 4.1. Metric definition. 4. Expanding dynamical systems Definition 4.1. Let X be a compact metric space. A map f : X X is said to be expanding if there exist ɛ > 0 and L > 1 such that d(f(x), f(y)) Ld(x, ### Invariant Metrics with Nonnegative Curvature on Compact Lie Groups Canad. Math. Bull. Vol. 50 (1), 2007 pp. 24 34 Invariant Metrics with Nonnegative Curvature on Compact Lie Groups Nathan Brown, Rachel Finck, Matthew Spencer, Kristopher Tapp and Zhongtao Wu Abstract. ### 3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space. ### RESULTANT AND DISCRIMINANT OF POLYNOMIALS RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results ### x if x 0, x if x < 0. Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the ### Inner Product Spaces and Orthogonality Inner Product Spaces and Orthogonality week 3-4 Fall 2006 Dot product of R n The inner product or dot product of R n is a function, defined by u, v a b + a 2 b 2 + + a n b n for u a, a 2,, a n T, v b, ### t := maxγ ν subject to ν {0,1,2,...} and f(x c +γ ν d) f(x c )+cγ ν f (x c ;d). 1. Line Search Methods Let f : R n R be given and suppose that x c is our current best estimate of a solution to P min x R nf(x). A standard method for improving the estimate x c is to choose a direction ### MATH 425, PRACTICE FINAL EXAM SOLUTIONS. MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator ### Mathematical Physics, Lecture 9 Mathematical Physics, Lecture 9 Hoshang Heydari Fysikum April 25, 2012 Hoshang Heydari (Fysikum) Mathematical Physics, Lecture 9 April 25, 2012 1 / 42 Table of contents 1 Differentiable manifolds 2 Differential ### A matrix over a field F is a rectangular array of elements from F. The symbol Chapter MATRICES Matrix arithmetic A matrix over a field F is a rectangular array of elements from F The symbol M m n (F) denotes the collection of all m n matrices over F Matrices will usually be denoted ### APPLICATIONS OF TENSOR ANALYSIS APPLICATIONS OF TENSOR ANALYSIS (formerly titled: Applications of the Absolute Differential Calculus) by A J McCONNELL Dover Publications, Inc, Neiv York CONTENTS PART I ALGEBRAIC PRELIMINARIES/ CHAPTER ### Differentiating under an integral sign CALIFORNIA INSTITUTE OF TECHNOLOGY Ma 2b KC Border Introduction to Probability and Statistics February 213 Differentiating under an integral sign In the derivation of Maximum Likelihood Estimators, or ### Payment streams and variable interest rates Chapter 4 Payment streams and variable interest rates In this chapter we consider two extensions of the theory Firstly, we look at payment streams A payment stream is a payment that occurs continuously, ### Similarity and Diagonalization. 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Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas, ### The Matrix Elements of a 3 3 Orthogonal Matrix Revisited Physics 116A Winter 2011 The Matrix Elements of a 3 3 Orthogonal Matrix Revisited 1. Introduction In a class handout entitled, Three-Dimensional Proper and Improper Rotation Matrices, I provided a derivation ### Math 497C Sep 9, Curves and Surfaces Fall 2004, PSU Math 497C Sep 9, 2004 1 Curves and Surfaces Fall 2004, PSU Lecture Notes 2 15 sometries of the Euclidean Space Let M 1 and M 2 be a pair of metric space and d 1 and d 2 be their respective metrics We say ### Duality of linear conic problems Duality of linear conic problems Alexander Shapiro and Arkadi Nemirovski Abstract It is well known that the optimal values of a linear programming problem and its dual are equal to each other if at least ### CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. 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Dr. Doreen De Leon Math 152, Fall 2015 Determinants Dr. Doreen De Leon Math 52, Fall 205 Determinant of a Matrix Elementary Matrices We will first discuss matrices that can be used to produce an elementary row operation on a given matrix A. ### OPTIMAL CONTROL OF A COMMERCIAL LOAN REPAYMENT PLAN. E.V. Grigorieva. E.N. Khailov DISCRETE AND CONTINUOUS Website: http://aimsciences.org DYNAMICAL SYSTEMS Supplement Volume 2005 pp. 345 354 OPTIMAL CONTROL OF A COMMERCIAL LOAN REPAYMENT PLAN E.V. Grigorieva Department of Mathematics ### SOME PROPERTIES OF FIBER PRODUCT PRESERVING BUNDLE FUNCTORS SOME PROPERTIES OF FIBER PRODUCT PRESERVING BUNDLE FUNCTORS Ivan Kolář Abstract. Let F be a fiber product preserving bundle functor on the category FM m of the proper base order r. We deduce that the r-th ### Lecture 13 Linear quadratic Lyapunov theory EE363 Winter 28-9 Lecture 13 Linear quadratic Lyapunov theory the Lyapunov equation Lyapunov stability conditions the Lyapunov operator and integral evaluating quadratic integrals analysis of ARE discrete-time ### 5 =5. Since 5 > 0 Since 4 7 < 0 Since 0 0 a p p e n d i x e ABSOLUTE VALUE ABSOLUTE VALUE E.1 definition. The absolute value or magnitude of a real number a is denoted by a and is defined by { a if a 0 a = a if a ### MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. Norm The notion of norm generalizes the notion of length of a vector in R n. Definition. Let V be a vector space. A function α ### Rotation Rate of a Trajectory of an Algebraic Vector Field Around an Algebraic Curve QUALITATIVE THEORY OF DYAMICAL SYSTEMS 2, 61 66 (2001) ARTICLE O. 11 Rotation Rate of a Trajectory of an Algebraic Vector Field Around an Algebraic Curve Alexei Grigoriev Department of Mathematics, The ### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization ### Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: Chapter 7 Div, grad, and curl 7.1 The operator and the gradient: Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: ( ϕ ϕ =, ϕ,..., ϕ. (7.1 x 1 ### Section 4.4 Inner Product Spaces Section 4.4 Inner Product Spaces In our discussion of vector spaces the specific nature of F as a field, other than the fact that it is a field, has played virtually no role. In this section we no longer ### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca ### DETERMINANTS. b 2. x 2 DETERMINANTS 1 Systems of two equations in two unknowns A system of two equations in two unknowns has the form a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 This can be written more concisely in ### Coefficient of Potential and Capacitance Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay We know that inside a conductor there is no electric field and that ### 1 Fixed Point Iteration and Contraction Mapping Theorem 1 Fixed Point Iteration and Contraction Mapping Theorem Notation: For two sets A,B we write A B iff x A = x B. So A A is true. Some people use the notation instead. 1.1 Introduction Consider a function ### Chapter 7. Lyapunov Exponents. 7.1 Maps Chapter 7 Lyapunov Exponents Lyapunov exponents tell us the rate of divergence of nearby trajectories a key component of chaotic dynamics. For one dimensional maps the exponent is simply the average ### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22 FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22 RAVI VAKIL CONTENTS 1. Discrete valuation rings: Dimension 1 Noetherian regular local rings 1 Last day, we discussed the Zariski tangent space, and saw that it ### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper. FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is ### equations Karl Lundengård December 3, 2012 MAA704: Matrix functions and matrix equations Matrix functions Matrix equations Matrix equations, cont d and and Karl Lundengård December 3, 2012 Solving General, Contents of todays lecture and (Kroenecker product) Solving General, Some useful from calculus and : f (x) = x n, x C, n Z + : f (x) = n x, x R, ### IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible ### 8.1 Examples, definitions, and basic properties 8 De Rham cohomology Last updated: May 21, 211. 8.1 Examples, definitions, and basic properties A k-form ω Ω k (M) is closed if dω =. It is exact if there is a (k 1)-form σ Ω k 1 (M) such that dσ = ω. ### Stochastic Inventory Control Chapter 3 Stochastic Inventory Control 1 In this chapter, we consider in much greater details certain dynamic inventory control problems of the type already encountered in section 1.3. In addition to the ### 9 More on differentiation Tel Aviv University, 2013 Measure and category 75 9 More on differentiation 9a Finite Taylor expansion............... 75 9b Continuous and nowhere differentiable..... 78 9c Differentiable and nowhere monotone...... ### Chapter 7. Continuity Chapter 7 Continuity There are many processes and eects that depends on certain set of variables in such a way that a small change in these variables acts as small change in the process. Changes of this ### Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin e-mail: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider two-point ### MATHEMATICS OF FINANCE AND INVESTMENT MATHEMATICS OF FINANCE AND INVESTMENT G. I. FALIN Department of Probability Theory Faculty of Mechanics & Mathematics Moscow State Lomonosov University Moscow 119992 g.falin@mail.ru 2 G.I.Falin. Mathematics ### 1. Introduction. PROPER HOLOMORPHIC MAPPINGS BETWEEN RIGID POLYNOMIAL DOMAINS IN C n+1 Publ. Mat. 45 (2001), 69 77 PROPER HOLOMORPHIC MAPPINGS BETWEEN RIGID POLYNOMIAL DOMAINS IN C n+1 Bernard Coupet and Nabil Ourimi Abstract We describe the branch locus of proper holomorphic mappings between ### A Primer on Index Notation A Primer on John Crimaldi August 28, 2006 1. Index versus Index notation (a.k.a. Cartesian notation) is a powerful tool for manipulating multidimensional equations. However, there are times when the more ### Quasi-static evolution and congested transport Quasi-static evolution and congested transport Inwon Kim Joint with Damon Alexander, Katy Craig and Yao Yao UCLA, UW Madison Hard congestion in crowd motion The following crowd motion model is proposed ### The Dirichlet Unit Theorem Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if ### 18.4. Errors and Percentage Change. Introduction. Prerequisites. Learning Outcomes Errors and Percentage Change 18.4 Introduction When one variable is related to several others by a functional relationship it is possible to estimate the percentage change in that variable caused by given ### Further Study on Strong Lagrangian Duality Property for Invex Programs via Penalty Functions 1 Further Study on Strong Lagrangian Duality Property for Invex Programs via Penalty Functions 1 J. Zhang Institute of Applied Mathematics, Chongqing University of Posts and Telecommunications, Chongqing ### ON FIBER DIAMETERS OF CONTINUOUS MAPS ON FIBER DIAMETERS OF CONTINUOUS MAPS PETER S. LANDWEBER, EMANUEL A. LAZAR, AND NEEL PATEL Abstract. We present a surprisingly short proof that for any continuous map f : R n R m, if n > m, then there ### Extrinsic geometric flows On joint work with Vladimir Rovenski from Haifa Paweł Walczak Uniwersytet Łódzki CRM, Bellaterra, July 16, 2010 Setting Throughout this talk: (M, F, g 0 ) is a (compact, complete, any) foliated, Riemannian ### Extremal equilibria for reaction diffusion equations in bounded domains and applications. Extremal equilibria for reaction diffusion equations in bounded domains and applications. Aníbal Rodríguez-Bernal Alejandro Vidal-López Departamento de Matemática Aplicada Universidad Complutense de Madrid, ### Load balancing of temporary tasks in the l p norm Load balancing of temporary tasks in the l p norm Yossi Azar a,1, Amir Epstein a,2, Leah Epstein b,3 a School of Computer Science, Tel Aviv University, Tel Aviv, Israel. b School of Computer Science, The ### A RIGOROUS AND COMPLETED STATEMENT ON HELMHOLTZ THEOREM Progress In Electromagnetics Research, PIER 69, 287 304, 2007 A RIGOROU AND COMPLETED TATEMENT ON HELMHOLTZ THEOREM Y. F. Gui and W. B. Dou tate Key Lab of Millimeter Waves outheast University Nanjing, ### Brief Review of Tensors Appendix A Brief Review of Tensors A1 Introductory Remarks In the study of particle mechanics and the mechanics of solid rigid bodies vector notation provides a convenient means for describing many physical ### Lecture 3: Linear Programming Relaxations and Rounding Lecture 3: Linear Programming Relaxations and Rounding 1 Approximation Algorithms and Linear Relaxations For the time being, suppose we have a minimization problem. Many times, the problem at hand can ### Rolle s Theorem. q( x) = 1 Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question ### Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x) SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f ### The Henstock-Kurzweil-Stieltjes type integral for real functions on a fractal subset of the real line The Henstock-Kurzweil-Stieltjes type integral for real functions on a fractal subset of the real line D. Bongiorno, G. Corrao Dipartimento di Ingegneria lettrica, lettronica e delle Telecomunicazioni, ### Sign changes of Hecke eigenvalues of Siegel cusp forms of degree 2 Sign changes of Hecke eigenvalues of Siegel cusp forms of degree 2 Ameya Pitale, Ralf Schmidt 2 Abstract Let µ(n), n > 0, be the sequence of Hecke eigenvalues of a cuspidal Siegel eigenform F of degree ### Examination paper for Solutions to Matematikk 4M and 4N Department of Mathematical Sciences Examination paper for Solutions to Matematikk 4M and 4N Academic contact during examination: Trygve K. Karper Phone: 99 63 9 5 Examination date:. mai 04 Examination ### Some Notes on Taylor Polynomials and Taylor Series Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited ### Prime Numbers and Irreducible Polynomials Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry. ### LINE INTEGRALS OF VECTOR FUNCTIONS: GREEN S THEOREM. Contents. 2. Green s Theorem 3 LINE INTEGRALS OF VETOR FUNTIONS: GREEN S THEOREM ontents 1. A differential criterion for conservative vector fields 1 2. Green s Theorem 3 1. A differential criterion for conservative vector fields We

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http://cms.math.ca/cmb/msc/26D15?fromjnl=cmb&jnl=CMB

location:  Publications → journals Search results Search: MSC category 26D15 ( Inequalities for sums, series and integrals ) Expand all        Collapse all Results 1 - 5 of 5 1. CMB 2011 (vol 55 pp. 355) Nhan, Nguyen Du Vi; Duc, Dinh Thanh Convolution Inequalities in $l_p$ Weighted Spaces Various weighted $l_p$-norm inequalities in convolutions are derived by a simple and general principle whose $l_2$ version was obtained by using the theory of reproducing kernels. Applications to the Riemann zeta function and a difference equation are also considered. Keywords:inequalities for sums, convolutionCategories:26D15, 44A35 2. CMB 2011 (vol 54 pp. 630) Fiorenza, Alberto; Gupta, Babita; Jain, Pankaj Mixed Norm Type Hardy Inequalities Higher dimensional mixed norm type inequalities involving certain integral operators are characterized in terms of the corresponding lower dimensional inequalities. Keywords:Hardy inequality, reverse Hardy inequality, mixed norm, Hardy-Steklov operatorCategories:26D10, 26D15 3. CMB 2010 (vol 53 pp. 327) Luor, Dah-Chin Multidimensional Exponential Inequalities with Weights We establish sufficient conditions on the weight functions $u$ and $v$ for the validity of the multidimensional weighted inequality $$\Bigl(\int_E \Phi(T_k f(x))^q u(x)\,dx\Bigr)^{1/q} \le C \Bigl (\int_E \Phi(f(x))^p v(x)\,dx\Bigr )^{1/p},$$ where 0<$p$, $q$<$\infty$, $\Phi$ is a logarithmically convex function, and $T_k$ is an integral operator over star-shaped regions. The condition is also necessary for the exponential integral inequality. Moreover, the estimation of $C$ is given and we apply the obtained results to generalize some multidimensional Levin--Cochran-Lee type inequalities. Keywords:multidimensional inequalities, geometric mean operators, exponential inequalities, star-shaped regionsCategories:26D15, 26D10 4. CMB 2005 (vol 48 pp. 333) Alzer, Horst Monotonicity Properties of the Hurwitz Zeta Function Let $$\zeta(s,x)=\sum_{n=0}^{\infty}\frac{1}{(n+x)^s} \quad{(s>1,\, x>0)}$$ be the Hurwitz zeta function and let $$Q(x)=Q(x;\alpha,\beta;a,b)=\frac{(\zeta(\alpha,x))^a}{(\zeta(\beta,x))^b},$$ where $\alpha, \beta>1$ and $a,b>0$ are real numbers. We prove: (i) The function $Q$ is decreasing on $(0,\infty)$ if{}f $\alpha a-\beta b\geq \max(a-b,0)$. (ii) $Q$ is increasing on $(0,\infty)$ if{}f $\alpha a-\beta b\leq \min(a-b,0)$. An application of part (i) reveals that for all $x>0$ the function $s\mapsto [(s-1)\zeta(s,x)]^{1/(s-1)}$ is decreasing on $(1,\infty)$. This settles a conjecture of Bastien and Rogalski. Categories:11M35, 26D15 5. CMB 1999 (vol 42 pp. 478) Pruss, Alexander R. A Remark On the Moser-Aubin Inequality For Axially Symmetric Functions On the Sphere Let $\scr S_r$ be the collection of all axially symmetric functions $f$ in the Sobolev space $H^1(\Sph^2)$ such that $\int_{\Sph^2} x_ie^{2f(\mathbf{x})} \, d\omega(\mathbf{x})$ vanishes for $i=1,2,3$. We prove that $$\inf_{f\in \scr S_r} \frac12 \int_{\Sph^2} |\nabla f|^2 \, d\omega + 2\int_{\Sph^2} f \, d\omega- \log \int_{\Sph^2} e^{2f} \, d\omega > -\oo,$$ and that this infimum is attained. This complements recent work of Feldman, Froese, Ghoussoub and Gui on a conjecture of Chang and Yang concerning the Moser-Aubin inequality. Keywords:Moser inequality, borderline Sobolev inequalities, axially symmetric functionsCategories:26D15, 58G30

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https://www.physicsforums.com/threads/electromagnetic-waves-radie-antenna-help.289586/

# Electromagnetic waves radie antenna HELP 1. Feb 3, 2009 ### MEAHH 1. The problem statement, all variables and given/known data A radio-frequency EM plane wave propagates in the +z-direction. A student finds that her portable radio obtains the best reception of the wave when the antenna is parallel to the x--y plane making an angle of 60 degrees with respect to the y-axis . (a) Consider an instant when the fields are non-zero at the location of the antenna. Draw and lable the direction of the electric field and the direction og the magnetic field.Explain (b)How would your answers to part a be different if the wave were propagating in the -z-direction.Explain 3. The attempt at a solution a) i drew the E field parallel to the antenna and the b field perpendiculat to it so that the oscillation will be at a max...is this right? b) i am unsure of what changes 2. Feb 4, 2009 ### LowlyPion I don't think anything changes if you reverse the direction between +z and -z. The e and the h waves are still perpendicularly polarized with respect to the orientation of the antenna is my understanding. Similar Discussions: Electromagnetic waves radie antenna HELP

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http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-5-section-5-1-exponents-and-scientific-notation-exercise-set-page-262/22

## Intermediate Algebra (6th Edition) We are asked to evaluate the expression $-5x^{0}$. In general, $a^{0}=1$ when a does not equal 0. Therefore, $-5x^{0}=-5(1)=-5$. Since the term -5 is not in parentheses with the base of the exponent, we evaluate it after evaluating the exponent.

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https://dms.umontreal.ca/~mathapp/Archives/Year1516/abs1516/JeffCalder.html

## Numerical Schemes for the Hamilton-Jacobi Equation Continuum Limit of Non-dominated Sorting ### Jeff CalderDepartment of Mathematics, University of California, Berkeley Non-dominated sorting arranges a set of points in n-dimensional Euclidean space into layers by repeatedly peeling away the coordinatewise minimal elements. It was recently shown that non-dominated sorting of random points has a Hamilton-Jacobi equation continuum limit. The obvious numerical scheme for this PDE has a slow convergence rate of $O(h^{1/n})$. In this talk, we introduce two new numerical schemes that have formal rates of $O(h)$ and we prove the usual $O(\sqrt{h})$ theoretical rates.

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https://www.itlive.pk/mod/page/view.php?id=122

## 20 - Document Map This feature of Word 2003 will teach you how you can navigate within your document using document maps.

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http://pruffle.mit.edu/~ccarter/3.21/Lecture_14/

Last Time The Position of a Particle Executing a Random Walk The Probability of Finding a Particle at a Position after a Random Walk Treating the Concentration as Time-Dependent Probability Distribution Relating the Self-Diffusivity to a Random Walk A Puzzle: Why for a random walk is ? 3.21 Spring 2001: Lecture 14 The Successful Jump Frequency as an Activated Process The treatment of diffusion as a statistical process permitted a physical correspondence between the macroscopic diffusivity and microscopic parameters for, average jump distance , jump correlation , and the average frequency at which a jumper makes a finite jump . In this lecture, the statistical evaluation of microscopic process will be applied to the successful jump frequency . A physical correspondence for that is related to microscopic processes of attempt or natural atomic vibration frequency and the difference in energy between the potential energy of site and the maximum value of the minimum potential energy (the saddle energy) as the atom moves from one equilibrium site to the next. The result that will be obtain, that the frequency of successful hops, (14-1) is related to the natural frequency multiplied by a Boltzmann factor has remarkable general application. Distribution of Energy among Particles A fundamental result from statistical mechanics is that for an ensemble of atoms at a fixed temperature , that the energies of the atoms has a characteristic probability distribution: Below, the rate of successful jumps for simple models of activated processes will be derived. Each derivation will depend on the distribution of energies given above. It will be supposed that a single atom will assume all possible values of energy with probabilities given by the Boltzmann distribution over time (the ergodic assumption). In other words, the distribution is considered to apply to the atoms at a time scale that is rapid compared to the natural frequency of the atoms--no correlation is made for the loss (or gain) of energy as an atom hops from one equilibrium site to the next. Activation Processes in Square Wells Consider an ensemble of particles with distributed energies moving about on the following energy landscape: The characteristic time it takes a particle to cross the activated state is (14-2) where and is the mass of the particle with characteristic thermal energy . The total rate, , that particles cross the barrier is (14-3) where and are the partition functions for the activated and minimum states. The rate that single particle crosses, , is: (14-4) (14-5) Therefore, (14-6) The term that multiplies the Arrhenius factor (the 1/T exponential) is the characteristic time it takes a particle to make an attempt at the activated state. Activation Processes in Harmonic Wells Consider the following modification of the above simple case, the minima are treated as harmonic wells: The minima can be approximated by (14-7) The analysis is similar to the case of the square wells, but for the ratio of the partition functions: (14-8) Approximating, (14-9) and carrying out the integration, (14-10) where is the characteristic oscillating frequency at the minima of a particle with mass sitting in a well of curvature . Many-Body Theory of Activated Processes at Constant Pressure In a real system, an atom or a vacancy does not make a successful hop without affecting (or getting effected by) its neighbors--all of the particles are vibrating and saddle point energy is an oscillating target produced by the random vibrations of all the atoms. The energy-surface that an atom, interstitial, or vacancy travels upon is a complicated and changing surface. If there are spherical particles, then there are -degrees of freedom to this surface, but it will be assumed that the momentum variables can be averaged out so that only a -dimensional potential surface remains: The minima, or equilibrium values of momenta and positions, can be approximated by harmonic wells: (14-11) This is approximated by a one-dimensional problem by assuming that all states during a hop lie on the surface in the figure. Let the first coordinate be in the direction of the crossing (parallel to ), then the average (rms) momentum in that direction is related to the an average rate of attempts. The result that was derived for the harmonic potential can be re-used in this case: (14-12) where is recogonized to be the width in Figure. 14-4. However, in this case, the particle may have a different volume in the activated state compared to the equilibrium state:1 (14-13) For the case where the volume may vary, but pressure is constant, the canonical constant pressure partition function must be used: (14-14) Therefore picks up an additional factor: (14-15) It remains to evaluate the partition functions by summing over all energies: . The partition function is evaluated by passing to the classical limit by dividing up the quantum phase space into cells of side-length equal to Planck's constant, : Because of the uncertainty principle: (14-16) Each elementary volume, , in phase space must be considered to have degeneracy: (14-17) Therefore, (14-18) In the classical limit, (14-19) Using the Harmonic approximations for the minima and carrying out the integration and letting all the masses have the same value: (14-20) Carrying out the same process for the activated state (which has one less degree of freedom) and adding the momentum near the activated state to the integral: (14-21) The products over the vibrational modes can be related to the entropies of the states, i.e., (14-22) Putting this all back into the expression for the rate of jumps, (14-23)

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https://physics.stackexchange.com/questions/708932/what-is-the-role-of-hermitian-hamiltonians-in-relativistic-qft/708933

What is the role of Hermitian Hamiltonians in relativistic QFT? In single-particle quantum mechanics, the probability of finding the particle in all space is conserved due to the hermiticity of the Hamiltonians (and remains equal to unity for all times, if normalized). But in relativistic quantum field theory, particle numbers are not conserved. For example, in QED, an initial state consisting of an electron-positron pair can annihilate into two photons in the final state. There is no trace of the initial electron and the positron in the final state. Similarly, in $$\beta$$-decay, an initial neutron is converted into a proton, an electron and an anti-electron neutrino in the final state. There is no neutron after the decay takes place though the Fermi theory Hamiltonian is Hermitian. So it seems that hermitian Hamiltonian in QFT is not responsible for the conservation of the probability. I seem to have a conceptual glitch here which I would like to be clarified. What is the role of hermitian Hamiltonians in relativistic QFT in the time development of states? There must be some constraining feature of hermitian QFT Hamiltonian. Sorry if the question sounds dumb. Some more thoughts on this for clarification In single-particle quantum mechanics, the norm of a quantum state is the probability of finding the particle in all of space. Is there a similar probability interpretation of the norm of a state in QFT? Since QFT hamiltonians are hermitian, the norm of the state remains preserved under time development but the state itself can change. This confuses me. To take the example given above the norm of the initial state neutron does not change under time development but the state itself can into other states. How do we interpret this in QFT?

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https://www.whsmith.co.uk/products/advanced-calculus/9780130652652

By: Gerald B. Folland (author)Hardback 2 - 4 weeks availability £68.99 Description For undergraduate courses in Advanced Calculus and Real Analysis. This text presents a unified view of calculus in which theory and practice reinforce each other. It covers the theory and applications of derivatives (mostly partial), integrals, (mostly multiple or improper), and infinite series (mostly of functions rather than of numbers), at a deeper level than is found in the standard advanced calculus books. Contents 1. Setting the Stage. Euclidean Spaces and Vectors. Subsets of Euclidean Space. Limits and Continuity. Sequences. Completeness. Compactness. Connectedness. Uniform Continuity. 2. Differential Calculus. Differentiability in One Variable. Differentiability in Several Variables. The Chain Rule. The Mean Value Theorem. Functional Relations and Implicit Functions: A First Look. Higher-Order Partial Derivatives. Taylor's Theorem. Critical Points. Extreme Value Problems. Vector-Valued Functions and Their Derivatives. 3. The Implicit Function Theorem and Its Applications. The Implicit Function Theorem. Curves in the Plane. Surfaces and Curves in Space. Transformations and Coordinate Systems. Functional Dependence. 4. Integral Calculus. Integration on the Line. Integration in Higher Dimensions. Multiple Integrals and Iterated Integrals. Change of Variables for Multiple Integrals. Functions Defined by Integrals. Improper Integrals. Improper Multiple Integrals. Lebesgue Measure and the Lebesgue Integral. 5. Line and Surface Integrals; Vector Analysis. Arc Length and Line Integrals. Green's Theorem. Surface Area and Surface Integrals. Vector Derivatives. The Divergence Theorem. Some Applications to Physics. Stokes's Theorem. Integrating Vector Derivatives. Higher Dimensions and Differential Forms. 6. Infinite Series. Definitions and Examples. Series with Nonnegative Terms. Absolute and Conditional Convergence. More Convergence Tests. Double Series; Products of Series. 7. Functions Defined by Series and Integrals. Sequences and Series of Functions. Integrals and Derivatives of Sequences and Series. Power Series. The Complex Exponential and Trig Functions. Functions Defined by Improper Integrals. The Gamma Function. Stirling's Formula. 8. Fourier Series. Periodic Functions and Fourier Series. Convergence of Fourier Series. Derivatives, Integrals, and Uniform Convergence. Fourier Series on Intervals. Applications to Differential Equations. The Infinite-Dimensional Geometry of Fourier Series. The Isoperimetric Inequality. APPENDICES. A. Summary of Linear Algebra. Vectors. Linear Maps and Matrices. Row Operations and Echelon Forms. Determinants. Linear Independence. Subspaces; Dimension; Rank. Invertibility. Eigenvectors and Eigenvalues. B. Some Technical Proofs. The Heine-Borel Theorem. The Implicit Function Theorem. Approximation by Riemann Sums. Double Integrals and Iterated Integrals. Change of Variables for Multiple Integrals. Improper Multiple Integrals. Green's Theorem and the Divergence Theorem. Answers to Selected Exercises. Bibliography. Index. Product Details • publication date: 21/12/2001 • ISBN13: 9780130652652 • Format: Hardback • Number Of Pages: 480 • ID: 9780130652652 • weight: 640 • ISBN10: 0130652652 Delivery Information • Saver Delivery: Yes • 1st Class Delivery: Yes • Courier Delivery: Yes • Store Delivery: Yes Calculus and AnalysisView More Prices are for internet purchases only. Prices and availability in WHSmith Stores may vary significantly Close

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http://www.contrib.andrew.cmu.edu/~ryanod/?p=1517

# Chapter 10: Advanced hypercontractivity In this chapter we complete the proof of the Hypercontractivity Theorem for uniform $\pm 1$ bits. We then generalize the $(p,2)$ and $(2,q)$ statements to the setting of arbitrary product probability spaces, proving the following: The General Hypercontractivity Theorem Let $(\Omega_1, \pi_1), \dots, (\Omega_n, \pi_n)$ be finite probability spaces, in each of which every outcome has probability at least $\lambda$. Let $f \in L^2(\Omega_1 \times \cdots \times \Omega_n, \pi_1 \otimes \cdots \otimes \pi_n)$. Then for any $q > 2$ and $0 \leq \rho \leq \frac{1}{\sqrt{q-1}} \cdot \lambda^{1/2-1/q}$, $\|\mathrm{T}_\rho f\|_q \leq \|f\|_2 \quad\text{and}\quad \|\mathrm{T}_\rho f\|_2 \leq \|f\|_{q'}.$ (In fact, the upper bound on $\rho$ can be slightly relaxed to the value stated in Theorem 17 of this chapter.) We can thereby extend all the consequences of the basic Hypercontractivity Theorem for $f : \{-1,1\}^n \to {\mathbb R}$ to functions $f \in L^2(\Omega^n, \pi^{\otimes n})$, except with quantitatively worse parameters depending on “$\lambda$”. We also introduce the technique of randomization/symmetrization and show how it can sometimes eliminate this dependence on $\lambda$. For example, it’s used to prove Bourgain’s Sharp Threshold Theorem, a characterization of boolean-valued $f \in L^2(\Omega^n, \pi^{\otimes n})$ with low total influence which has no dependence at all on $\pi$. ### 2 comments to Chapter 10: Advanced hypercontractivity • Is $q’$ defined here? • It’s the Holder conjugate of $q$ (i.e., the number satisfying $1/q + 1/q’ = 1$). Its definition is made in a few other places in the book. Thanks.

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https://proofwiki.org/wiki/Vector_Space_has_Basis_between_Linearly_Independent_Set_and_Spanning_Set

# Vector Space has Basis between Linearly Independent Set and Spanning Set ## Theorem Let $V$ be a vector space over a field $F$. Let $L$ be a linearly independent subset of $V$. Let $S$ be a set that spans $V$. Suppose that $L \subseteq S$. Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$. ### Corollary Let $K$ be a division ring. Let $V$ be a vector space over $K$. Then $V$ has a basis. ## Outline of Proof We use Zorn's Lemma to construct a maximal linearly independent subset. ## Proof Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion. Note that $L \in \mathscr I$, so $\mathscr I \ne \varnothing$. Let $\mathscr C$ be a nest in $\mathscr I$. Let $C = \bigcup \mathscr C$. Aiming for a contradiction, suppose that $C$ is linearly dependent. Then there exist $v_1, v_2, \ldots, v_n \in C$ and $r_1, r_2, \ldots, r_n \in F$ such that $r_1 \ne 0$: $\displaystyle \sum_{k \mathop = 1}^n r_k v_k = 0$ Then there are $C_1, C_2, \ldots, C_n \in \mathscr C$ such that $v_k \in C_k$ for each $k \in \set {1, 2, \ldots, n}$. Since $\mathscr C$ is a nest, $C_1 \cup C_2 \cup \cdots \cup C_n$ must equal $C_k$ for some $k \in \set {1, 2, \ldots, n}$. But then $C_k \in \mathscr C$ and $C_k$ is linearly dependent, which is a contradiction. Thus $C$ is linearly independent. By Zorn's Lemma, $\mathscr I$ has a maximal element $M$ (one that is not contained in any other element). Since $M \in \mathscr I$, $M$ is linearly independent. All that remains is to show that $M$ spans $V$. Suppose, to the contrary, that there exists a $v \in V \setminus \map {\operatorname {span} } M$. Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s \notin \map {\operatorname {span} } M$. Then $M \cup \set s$ is linearly independent. Thus $M \cup \set s \supsetneq M$, contradicting the maximality of $M$. Thus $M$ is a linearly independent subset of $V$ that spans $V$. Therefore, by definition, $M$ is a basis for $V$. $\blacksquare$ ## Axiom of Choice This theorem depends on the Axiom of Choice, by way of Zorn's Lemma. Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial. Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted. However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

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https://cmurandomized.wordpress.com/2011/02/01/lecture-7-the-local-lemma/

# CMU Randomized Algorithms Randomized Algorithms, Carnegie Mellon: Spring 2011 ## Lecture #7: The Local Lemma 1. The Local Lemma: Various Forms The symmetric version of the local lemma we saw in class: Theorem 1 (Symmetric Form I) Given a collection of events ${B_1, B_2, \ldots, B_m}$ such that ${\Pr[B_i] \leq p}$, and each ${B_i}$ is independent of all but ${d}$ of these events, if ${epd < 1}$ then ${\Pr[ \cap \overline{B_i} ] > 0}$. Very often, the sample space is defined by some experiment which involves sampling from a bunch of independent random variables, and the bad events each depend on some subset of these random variables. In such cases we can state the local lemma thus: Theorem 2 (Symmetric Form II) Suppose ${X_1, X_2, \ldots, X_n}$ are independent random variables, and events ${B_1, B_2, \ldots, B_m}$ such that each ${B_i}$ that depends only on some subset ${\{X_j : j \in S_i\}}$ of these variables. Moreover, suppose ${\Pr[ B_i ] \leq p}$ and each ${S_i}$ intersects at most ${d}$ of the ${S_j}$‘s. If ${epd < 1}$ then ${\Pr[ \cap \overline{B_i} ] > 0}$. Note that both the examples from class (the ${k}$-SAT and Leighton Maggs and Rao results) fall into this setting. Finally, here’s the asymmetric form of the local lemma: Theorem 3 (Asymmetric Form) Given events ${B_1, B_2, \ldots, B_m}$ with each ${B_i}$ independent of all but the set ${\Gamma_i}$ of these events, suppose there exist ${x_i \in (0,1)}$ such that $\displaystyle \Pr[ B_i ] \leq x_i \prod_{j \in \Gamma_i \setminus \{i\}} (1 - x_j).$ Then ${\Pr[ \cap \overline{B_i} ] \geq \prod_i (1 - x_i) > 0}$. Occasionally one needs to use the asymmetric form of the local lemma: one example is Uri Feige’s result showing a constant integrality gap for the Santa Claus problem, and the resulting approximation algorithm due to Heupler, Saha and Srinivasan. 1.1. Proofs of the Local Lemma The original proof of the local lemma was based on a inductive argument. This was a non-constructive proof, and the work of Beck gave the first techniques to make some of the existential proofs algorithmic. In 2009, Moser, and then Moser and Tardos gave new, intuitive, and more algorithmic proofs of the lemma for the case where there is an underlying set of independent random variables, and the bad events are defined over subsets of these variables. (E.g., the version of the Local Lemma given in Theorem~2, and its asymmetric counterpart). Check out notes on the proofs of the Local Lemma by Joel Spencer and Uri Feige. The paper of Heupler, Saha and Srinivasan gives algorithmic versions for some cases where the number of events is exponentially large. 1.2. Lower Bounds The local lemma implies that if ${d < 2^k/e}$ then the formula is satisfiable. This is complemented by the existence of unsatisfiable E${k}$-SAT formulas with degree ${d = 2^k(\frac1e + O(\frac{1}{\sqrt{k}}))}$: this is proved in a paper of Gebauer, Szabo and Tardos (SODA 2011). This shows that the factor of ${e}$ in the local lemma cannot be reduced, even for the special case of E${k}$-SAT. The fact that the factor ${e}$ was tight for the symmetric form of the local lemma was known earlier, due to a result of Shearer (1985). 2. Local Lemma: The E${k}$-SAT Version Let me be clearer, and tease apart the existence question from the algorithmic one. (I’ve just sketched the main ideas in the “proofs”, will try to fill in details later; let me know if you see any bugs.) Theorem 4 If ${\varphi}$ is a E${k}$-SAT formula with ${m}$ clauses, ${n}$ variables, and where the degree of each clause is at most ${d \le 2^{k-3}}$, then ${\varphi}$ is satisfiable. Proof: Assume there is no satisfying assignment. Then the algorithm we saw in class will run forever, no matter what random bits it reads. Let us fix ${M = m \log m + 1}$. So for every string ${R}$ of ${n+Mk}$ bits the algorithm reads from the random source, it will run for ${M}$ iterarations. But now one can encode the string ${R}$ thus: use ${m \log m}$ bits to encode the clauses at the roots of the recursion trees, ${M(\log d + 2)}$ to encode the clauses lying within these recursion trees, and ${n}$ bits for the final settings of the variables. As we argued, this is a lossless encoding, we can recover the ${n+Mk}$ bits from this encoding. How long is this encoding? It is ${M(\log d + 2) + n + m \log m}$, which is strictly less than ${n+Mk}$ for ${M = m \log m + 1}$ and ${d \leq 2^{k-3}}$. So this would give us a way to encode every string of length ${n+Mk}$ into strings of shorter lengths. But since for every length ${\ell}$, there are ${2^\ell}$ strings of length ${\ell}$ and ${1 + 2 + \ldots + 2^{\ell - 1} = 2^{\ell} - 1}$ strings of length strictly less than ${\ell}$, this is impossible. So this contradicts our assumption that there is no satisfying assignment.$\Box$ Now we can alter the proof to show that the expected running time of the algorithm is small: Theorem 5 If ${\varphi}$ is a E${k}$-SAT formula with ${m}$ clauses, ${n}$ variables, and where the degree of each clause is at most ${d \le 2^{k-3}}$, then the algorithm FindSat finds a satisfying assignment in ${O(m \log m)}$ time. Proof: Assume that we run for at least ${M + t}$ steps with probability at least ${1/2^s}$. (Again, think of ${M = m \log m}$.) Then for at least ${1/2^s}$ of the ${2^{n+(M+t)k}}$ strings, we compress each of these strings into strings of length ${(M+t)(\log d + 2) + n + m \log m}$. But if we have any set of ${2^{n+(M+t)k} \cdot 2^{-s}}$ strings, we must use at least ${n + (M+t)k-s}$ bits to represent at least one of them. So $\displaystyle n + (M+t)k - s \leq n + (M+t)(\log d + 2) + M.$ If ${d \leq 2^{k-3}}$, we have ${k - \log d - 2 \geq 1}$, and $\displaystyle (M+t)(k - \log d - 2) - s \leq M$ or $\displaystyle M+t-s \leq M \implies s \geq t.$ So we get that the probability of taking more than ${M+t}$ steps is at most ${1/2^t}$, which implies an expected running time of ${M + O(1)}$. $\Box$

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http://math.stackexchange.com/questions/137011/futures-pricing-and-futures-price-process-under-the-real-world-measure/194828

# Futures pricing and futures price process under the real world measure This is something that keeps bothering me about the Benchmark approach of Platen, which (very) shortly is as follows: Compare the development of an economic value with a growth optimal portfolio. Taking the expectation in the real world measure $\mathbb{P}$, conditioned on today's information, this should yield a fair and in some sense arbitrage free price, because you compare to the best possible (in expectation). Now, I want to know the price of a future on $X_T$ (hence $F(T,T)=X_T$). In discrete time, one just has to take the discounted future payments. For a time interval partition $\sigma$, this yields: $$\Sigma_t^{(\sigma)} = \sum_{i=1}^N\frac{1}{S^{\pi^*}_{t_i}}( F(t_i,T) - F(t_{i-1},T) )\\ = {\sum_{i=1}^N\frac{1}{S^{\pi^*}_{t_{i-1}}}( F(t_i,T) - F(t_{i-1},T) )} +{\sum_{i=1}^N\left(\frac{1}{S^{\pi^*}_{t_i}}-\frac{1}{S^{\pi^*}_{t_{i-1}}}\right)( F(t_i,T) - F(t_{i-1},T) )}$$ Going to continuous time, this converges to $\int_t^{T} \left(\frac{1}{S^{\pi^*}_{s-}}\right) dF(s,T)+\int_t^{T}d\left[\left(\frac{1}{S^{\pi^*}}\right),F\right]_s$. Since the net position is zero and using the product rule for semimartingales, we get for the futures price at time t: $$\mathbb{E}\left[ \frac{F(T,T)}{S^{\pi^*}_{T}}-\int_t^{T}F(s-,T)d\left( \frac{1}{S^{\pi^*}_s} \right) \big|\mathcal{F}_t\right] = \mathbb{E}\left[ \frac{F(t,T)}{S^{\pi^*}_{t}} \big|\mathcal{F}_t\right] = \frac{F(t,T)}{S^{\pi^*}_{t}}$$ To me it seems intractable. Is there any way to come the futures price process even close? I read some extensions to the usual equivalent martingale measure, cont., FV interest process one takes usually and which yields $F(t,T)=\mathbb{E}^{\mathbb{Q}}[F(T,T)|\mathcal{F}_t]$. But none of these can be applied here. Any ideas? - @ user13655 : I think this question should be displaced to the quant.stackexchange.com forum. Best regards. – TheBridge May 10 '12 at 15:38 Well, at least there is a solution for deterministic interest rate. As an example take a BS model: $$dS_t = S_t (rdt+\theta^2dt+\theta dW_t)$$ implies $$d\left(\frac{1}{S_t}\right) = \frac{1}{S_t} (-rdt-\theta dW_t)$$ Define $\bar{F}=\bar{F}(t,T):=\frac{F(t,T)}{S_t}$, which implies $\bar{F}(T,T)=\frac{F(T,T)}{S_T}$ and $H_t = \mathbb{E}[\frac{h(L_t)}{S_T}|\mathcal{F}_t]$. $H_t$ is a martingale. The solution of $Z_t = 1+\int_0^t Z_{s}dXs$ is $\mathcal{E}(-X)_t$ which is just the reciprocal of the savings account, i.e. $\mathcal{E}(-X)_t=\frac{1}{B_t}$. Note that $\mathcal{E}(-X)_t dX_t = -\mathcal{E}(X)_t r dt = -\frac{1}{B_t}r dt = d\frac{1}{B_t} = d\left( \mathcal{E}(-X)_t \right)$. We get: $$0=H_t + \mathbb{E}\left[ -\bar{F}(t,T) - \int_t^{T}\bar{F}(s-,T)dX_s \big|\mathcal{F}_t\right],$$ which looks very similar to a OU-SDE (except it doesn't run from $0$ to $t$). After looking carefully at the solution of the usual OU-SDE (see Revuz and Yor p. 378 Prop. 2.3), I tried $$\bar{F}(t,T)=\mathbb{E}\left[ \mathcal{E}(-X)_t\left( \frac{H_t}{\mathcal{E}(-X)_T}+\int_t^T\mathcal{E}(-X)^{-1}_s(dH_s-d\langle H,X\rangle_s) \right)|\mathcal{F}_t\right]$$ Since $H_t$ is a martingale and $X_t$ deterministic, we can forget about the integral and obtain: $$\bar{F}(t,T)=\frac{B_t}{B_T}\mathbb{E}\left[ \frac{h(L_T)}{S_T} |\mathcal{F}_t \right]\Rightarrow F(t,T)=S_t \frac{B_t}{B_T}\mathbb{E}\left[ \frac{h(L_T)}{S_T} |\mathcal{F}_t \right]$$ Plugging in shows that this is indeed a solution. I am not sure of uniqueness though. No solution for random $X_t$ yet, let alone if $\langle X,H\rangle\neq0$. -

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https://cypenv.info/and-relationship/expected-return-and-standard-deviation-relationship-to-variance.php

# Expected return and standard deviation relationship to variance ### Measures of Risk - Variance and Standard Deviation Learning Objectives. Calculate the expected return of an investment portfolio . Learning Objectives. Explain the importance of a stock's variance and standard deviation Variance in Relation to Expected Return. In the discussion of expected. Measures of Risk - Variance and Standard Deviation. Risk reflects the chance that the actual return on an investment may be very different than the expected. Standard deviation is a measure of how much an investment's returns can vary For math-oriented readers, standard deviation is the square root of the variance. return, then square those differences (that is, multiply each difference by itself): of risk that an investment will not meet the expected return in a given period. Сквозь строй - лучший антивирусный фильтр из всех, что я придумал. Через эту сеть ни один комар не пролетит. • Measures of Risk - Variance and Standard Deviation Выдержав долгую паузу, Мидж шумно вздохнула.

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https://www.physicsforums.com/threads/pressure-and-force-analysis.534667/

# Pressure and force analysis 1. Sep 28, 2011 ### Misr I don't understand what makes us analyise a certain force in to two perpendicular components http://library.thinkquest.org/C001429/statics/pressure.htm [URL]http://library.thinkquest.org/C001429/images/cos.jpg[/URL] why do we calculate pressure in this image from the relation P = F cos q/A why not P=F/A?? Is this beacuse pressure is always perpendicular?and if so why is pressure perpendicular,when the force makes an angle with the normal? Last edited by a moderator: Apr 26, 2017 2. Sep 28, 2011 ### Studiot If you apply a force at some angle to a surface as shown in your picture, that force has two components. One component is perpendicular to the area. This is called the normal stress or direct stress or pressure. This is Fcos($\theta$) /A The other component is parallel to the area and is called the shear stress. It is not a pressure, since it does not 'press' on the area, but drags along the surface. This is Fsin($\theta$/A F is the resultant of these two forces the normal force and the shear force. go well 3. Sep 29, 2011 ### Misr What is a shear stress? I don't understand why is F the resultant of those two forces 4. Sep 29, 2011 ### Studiot Good morning, misr. When you resolve a force in some direction, as you have done in your diagram, you get two components (in 2 dimensions). I assume you understand this since you showed Fcos($\theta$) The component you showed is called the normal force. This acts perpendicular to the area. The other one is called the shear force. This acts parallel to the area. Do you understand this so far? 5. Sep 29, 2011 ### Misr Yeah,I understand this but is this normal force due to gravity or something else? because the object is pressing upon the surface,so the surface reacts by this normal force?correct? 6. Sep 29, 2011 ### Studiot I really have no idea what the source of F is, you drew the diagram after all. A few basic ideas. In mechanics we need to distinguish between Force and stress. Stress is Force divided by area. I think we both know that Force has a specific line of action and a point of application. We distinguish between body forces such as gravity, which act throughout a body and surface forces which are externally applied by some agent (eg the tension in a string). We consider the body force of gravity to conform to my rule above by saying that this force is acts at the centre of gravity and is called the weight. This force cannot directly exert a pressure it is an internal force. If a block is resting on a (horizontal) table its weight exerts a downward pressure on the table. This pressure equals the weight divided by the area of the block sitting on the table. I have sketched this in fig1. As noted above this is also called the stress imposed on the table by the block. Since the weight acts perpendicular to the table that is all there is to it. Now suppose I replaced the block with a light plate and a spring compressed so as to exert a force on the plate equal to W, the weight of the block. Although the source of the force on the table is quite different the effect is the same ie the pressure is the same, so long as the spring force is exerted vertically. I have sketched this in Fig2. However we tilt the spring to apply its force at an angle, as shown in Fig3. What do you think the pressure on the table would be if the angle was 90 degrees as shown in fig4? Obviously there is no vertical force now being applied to the table via the plate. There is, however a still force being applied. This is called the shear force (in this case it is also the friction force). If we return to Fig3 and apply our spring force at some intermediate angle we have two components, one vertical and one horizontal. This is a more general situation. does this help? #### Attached Files: • ###### pre1.jpg File size: 12.3 KB Views: 128 7. Sep 30, 2011 ### Misr I don't understand this.Is the plate fixed to the table and you are pulling the spring? why is the pressure the same for both fig(1) and (2) why do we define the pressure as the vertical force per unit area?why not the net force per unit area? 8. Sep 30, 2011 ### Studiot Now why do you suppose I said the spring was compressed? I do hope you are reading this fully. No the plate just sits on the table. I said it is a light plate so that means it has no weight of its own. The spring pushed down against the plate. Well I did say the force the spring pushed down with was set to be the same as the weight of the block. Now try reading my post again and see if it makes more sense. 9. Oct 5, 2011 ### Misr Certainly,It makes more sense now I guess I understand what are you trying to say,the pressure becomes less when it is not perpendicular Now I want to ask another question: Why do we define pressure as "AVERAGE" force acting normally on unit area at this point as in the page provided above? 10. Oct 5, 2011 ### Studiot Yes that's right. If two equal forces are acting (on equal surfaces) one perpendicular and one at an angle - the perpendicular force exerts more pressure than the angled one. Most people just accept what they are told, but you are obviously a thinking person so here is some extra detail. The terms 'normal stress' and 'pressure' refer to the same physical phenomenon. 'Normal stress' is usually used in conection with solids and 'pressure' in connection with fluids. We sometimes talk about pressure in connection with solids when we are considering contact stresses between two solids for instance 'foundation pressure' or 'bearing pressure'. In the first sketch I have a 1kg weight sitting on a block of concrete, which is much bigger than the weight. On the surface (section AA) where the weight is sitting the weight is concentrated only over the area of the contact surface, not over the whole area of AA. As we go deeper into the concrete the 1kg spreads out over a wider and wider part of the concrete until we can say that the weight exterts an average pressure of 1kg divided by the area of the concrete block at section CC. At intermediate section BB the pressure exerted by the weight is intermediate between that at AA and CC. So what would happen if the block of concrete extended much further? Well in sketch 2 I have shown the foundation pressure under a building of weight W. You can see a series of 'bowls of soil' that get larger and larger in area as we get further from the building. So W is distributed over an increasing area and the pressure gradually diminishes over these increasing areas. Back to fluids, for although the pressure is the same in all directions at a point in a fluid, it can still vary from point to point. So in sketch 3 I have shown the steadily increasing pressure of the water on the back of a dam. This increases linearly from nothing at the surface to a maximum at the base. As a result I have shown a triangle of forces. I do not know if you have yet covered centre of gravity? The 'average' pressure is the pressure at the centre of gravity (properly called the centre of pressure) of the triangle. The force on the dam equals this average pressure times the wetted area of the back of the dam. go well #### Attached Files: • ###### pre2.jpg File size: 8.4 KB Views: 109 Last edited: Oct 5, 2011 11. Oct 6, 2011 ### Misr I can't imagine this :( In fig (1)As we go deeper ,the pressure on a certain surace is not affected according to the relation P=h*rho*g where h is the height of the block 12. Oct 6, 2011 ### Studiot A concrete block is not a fluid. However the concrete at each section also experiences the pressure due to the weight of the concrete above it, just like a fluid. But I am only considering the effect of the 1kg weight - or if you like the extra effect of the 1kg weight. 13. Oct 6, 2011 ### Misr This one is okay,but if so how could we calculate the total pressure on the back of the dam,if each point on the back of the dam has a different pressure? 14. Oct 6, 2011 ### Misr You are right..sorry for this stupidness but I still can't imagine 15. Oct 6, 2011 ### Studiot There is no such animal as 'total pressure' - There is only average pressure over a whole area or specific pressure at a point. Pressure does not 'add up' over the area. You are thinking of Force. If I exert 10 pascal over 1 square metre and 10 pascal over another half a square metre there is no total pressure, just a pressure of 10 pascal acting. The force, however is 10 newton in the first case and 15 in the second Have a good look at the response by Halls of Ivy to your question about the fluid in the tank, it is similar to my dam example. It is fundamental and very important to distinguish between force and pressure. 16. Oct 6, 2011 ### Studiot It's coming up to 1 am where I am and, as you can see from my passport photo, I desperately need my beauty sleep so we will have to continue this another day, but please come back and confirm you have conquered the difference (and link as they are intimately connected) between force and pressure. go well 17. Oct 9, 2011 ### Misr Well,the main problem here is that i can't imagine how pressure is distributed over a larger area on going deeper.. Isn't the contact area the same? What are you talking about?What is "the centre of gravity"? Yeah 18. Oct 9, 2011 ### Studiot Is this a language problem or a physics problem? We really need to know the extent of your knowledge to be able to help as a centre of gravity is a pretty basic concept in physics. I also said more, as did Redbelly in another thread, that it is fundamental you understand the difference between force and pressure. This is fine if we need to start there (or even further back), but progress cannot be made without it. 19. Oct 11, 2011 ### Misr It's a physics problem and I think I know the difference between force and pressure,pressure is the force divided by area,anything else? 20. Oct 11, 2011 ### Studiot The 'Centre of Gravity' of any body is the point through which all the weight of that may be be considered to act as a single concentrated force. For instance consider a canon ball. We take the weight to be one force - say W, acting vertically downwards at the geometric centre of the ball. Things may get much more complicated however - take for instance an L shaped metal bracket. Now the weight of a body is just the sum of all the weights of the individual small elements added up W = sum of welements Are you familiar with sigma notation? W = Ʃw W and w are, of course, forces. We can add forces from a different cause that are distributed over a body, area or volume in the same way to obtain a 'centre of force' for that particular force. The 'Centre of Pressure' is just such a calculation.

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https://www.physicsforums.com/threads/how-to-find-the-plane-at-which-two-hyperplanes-intersect.601258/

# How to find the plane at which two hyperplanes intersect. 1. Apr 28, 2012 ### jenny_shoars I know that to find the line at which two planes intersect, you can take the cross product of their normal vectors. This gives you a vector parallel to the line. Then you can just find a point which lies on both planes and that position plus the vector is your line. How do you do the equivalent for the plane at which two hyperplanes intersect? I would initially think you could do something like take the determinant of the two hyperplanes in the same way that the cross product is from the determinant of the the two planes. However, the two hyperplanes don't give a square matrix the same way the two regular planes do. Also, how would you go about finding a point which lies on both hyperplanes in order to get the fully determined plane? Thank you for your time! 2. Apr 29, 2012 ### homeomorphic One way to look at it is that it's just a linear algebra problem. The first plane is a set that satisfies some linear equations, so is the second plane and the intersection is the set the satisfies both sets of equations. Cross products don't really exist in higher dimensions. An appropriate analog would be the wedge product, which could also be used to find the intersection. By the way, hyperplane usually means one dimension less than the ambient space, not just higher dimensional planes. 3. Apr 29, 2012 ### jenny_shoars Of course. You're right on both accounts. Thank you much! Similar Discussions: How to find the plane at which two hyperplanes intersect.

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https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=44116

## Zero Order Reaction $\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$ Emma Scholes 1L Posts: 62 Joined: Fri Sep 28, 2018 12:18 am ### Zero Order Reaction What is an example of a zero order reaction? Cole Elsner 2J Posts: 88 Joined: Fri Sep 28, 2018 12:25 am ### Re: Zero Order Reaction An example I like to think of is having to reactants, A and B, and a few different experiments changing the concentrations of each, and analyzing the change in the rate. When A is increased, let's say that the reaction shows no change. You do the same for B, still no change. I think of zero order as any increase or decrease of the reactant with zero order, there is ZERO change to rate. mcredi Posts: 63 Joined: Fri Sep 28, 2018 12:16 am ### Re: Zero Order Reaction For example 10mg of a drug maybe eliminated per hour, this rate of elimination is constant and is independent of the total drug concentration in the plasma caseygilles 1E Posts: 73 Joined: Fri Sep 28, 2018 12:18 am ### Re: Zero Order Reaction a zero order reaction is one whose rate is independent of the concentration. An example is when nitrous oxide decomposes to nitrogen and oxygen. In the presence of a catalyst, platinum, we see that changing concentration of N2O has no effect on the rate of decomposition so we know the rate only depends on the rate constant, k. More precisely, as long as there is sufficient n2o to react with platinum the rate is not determined by concentration of n2o. Brian Chang 2H Posts: 65 Joined: Fri Sep 28, 2018 12:17 am ### Re: Zero Order Reaction The Haber process is an example of a 0th order rxn. Kyither Min 2K Posts: 60 Joined: Wed Oct 03, 2018 12:15 am ### Re: Zero Order Reaction I think this was mentioned in class but zero order reactions occur when the concenctration is super high that there is barely a difference in concentration respect to time.

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https://asknigeria.com.ng/topic/4151/teaching-children-how-to-engage-in-useful-discussions

Discussion involves the exchange between people any given idea or topic. It can be formal or informal and can be in large groups or small groups. It is an effective teaching technique which promote the sharing of information and it involves the student. One of the key ways to promote communication is through discussion. How to use Discussion • You must be prepared to encourage and begin your discussion when appropriate . it is a way to take advantage of a teachable moment. • Moment – interesting discussion which are unplanned may arise. Formal discussions requires some planning and possible informal discussion can be planned but the spontaneous discussion can be equally effective. The teacher will have to decide whether discussion take place in small groups or involve the whole class. • The teacher should have a reason for utilizing discussion in the classroom. Checklist for Utilizing Meaningful Discussion • How will the discussion enhance the teaching experience of all students. • What will be emphasized in terms of the discussion. The content of the discussion or both. • How will students be evaluated. • How will the teacher get all students to participate in the discussion. • The teacher should make clear any rules the student will be required to show prior to the discussion. • There must be a recorder for each group but if the whole class is to serve as a group, the teacher can act as an impartial recorder and record all the activities on the board. • The teacher should encourage the participation of all students. Things teachers must consider before creating a discussion. • Establish a stress free environment • Establish or arrange their sitting positions in a way that they can have a eye contact • Never encourage the idea that teachers is all knowing. You should as much as possible learn. • Teachers must be supportive and refrain from dominating and discussion. • Teacher should make research before the class. • Decide which form of discussion is appropriate for the topic being discussed either formal, informal, large, small group discussion. • The discussion must have purpose and focus. • Consider the group dynamic, lay down the rules e.g. be considerate in the class, do not interrupt, do not insult, ignore the teacher or student, you encourage rather than discourage them. • Make sure you put it in mind that one or two students may want to dominate the discussion.

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https://www.physicsforums.com/threads/rl-circuit-fun-am-i-drawing-this-circuit-correct-at-different-times.116177/

# RL circuit fun! am I drawing this circuit correct at different times? 1. Apr 1, 2006 ### mr_coffee Hello eveeryone! exam time monday! Just doing some last checks to make sure i'm understanding this correctly. Here is the directions and circuit: THe circuit shown below has beeen in the form shown for a very long time. THe switchopens at t =0. FInd iR at t equal to (a)0-, (b) 0+, (c) infinity, and (d) 1.5 I'm showing u my work for (a) and (b). FOr par (a) WHat i'm confused on is, U see that wire that says at time t =0, its going to switch open. At time t < 0, does that mean there is just a wire there? Does that mean all the current is going to go through that wire and say screw all the other compoents like the 60, and 40 ohm resistors? THe answer in the back of the book is iR = 0; But is it 0 because all the current will go into that wire with no resistance? I remember the professor said, if the inductor is shorted(in this case it would be because the circuit has been sitting there for a long time) then anything in parellel with that shorted inductor is also shorted out? For part (b) iR = 10mA Is this iR 10mA because when that wire is opened that means the conductor isn't shorted but acts as a huge resistor. Not wanting any current to go through it, so it makes all the current go through the 60 ohm resistor? For partt (d) 1.5ms I need to find x = L/Req, to find Req, ur suppose to "look" into the inductor and take out all power supplies, in this case I think you would be left with 60 and 40 in parellel, so i got Req = 4mA, L = .1H I know i'm going to use the equation: iL(t) = io*e^t/x, where x = L/R, L being the indcutor, R being the equivlent resistance, but what io do i use? io is the initial current i though, usually i use the io i found at time i(0-) but in this case its 0! The book is getting an answer of 5.34mA. Thanks! Last edited: Apr 1, 2006 2. Apr 1, 2006 ### nrqed Yes. The switch is closed which means it acts as an ideal wire. well, yes...but please be careful with the language. I might not be the only one who is easily offended Yes, anything parallel with a wire with no resistance is shorted Yes The current will be of the general form C_1 + C_2 e^(-t/tau). You must impose that at t=0 this reproduces the result at t=0+, so 10 mA. At t=infinity, find the current by replacing the inductor by a wire. That gives you a second condition which will fix C_1 and C_2. Then you can find the current at any time. 3. Apr 1, 2006 ### mr_coffee Thanks for the responce, sorry about the language somtimes when I type i don't realize what i'm actually typing I had no idea there was another form like that. Makes sense though! I used 10ma as C1, and C2, i used the value i got by evaulating the circuit at t(infinity) and i got the following value of iR = (10mA)(40)/(60+40) = 4mA. Which is what the book has. But when i put it into the forumla: 10E-3 + 4E-3*e^(-240*1.5E-3) = .012791A or 12.79mA but the book has: 5.34mA. Any idea where i misunderstood? 4. Apr 1, 2006 ### nrqed No problem, I am probably too sensitive Watch out. If you set t=0 in the equation, you get C_1 + C_2 = 10 mA. If you set t= infinity, you get C_1 = 4 mA (the exponential is zero). So C_2 = 6 mA. Don`t jump to the conclusion that C_1 is the current at t=0 and C_2 is the current at t=infinity! Pat 5. Apr 1, 2006 ### nrqed Also (I had not checked that part of your calculation) but if you take out all power sources as you said, the inductor will seetwo resistors in series so your equivalent resistance should be 1000 ohms. Soyou get I(1.5 ms) = 4 mA + 6 mA e^(-100/.1 * 1.5E-3) = 5.34 mA Patrick 6. Apr 1, 2006 ### mr_coffee ahh thanks again! it worked out nicely and great explanation! But i'm having problems visualing what happens when u "look" through a inductor and simpify from there. For example, in this case would it look like this?

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http://schlitt.info/opensource/blog/0736_highlight_source_code_lines_latex.html

Highlight source code lines in LaTeX - Blog - Open Source - schlitt.info # schlitt.info - php, photography and private stuff ## Highlight source code lines in LaTeX I love LaTeX for any kind of text writing (actually typesetting), simply because it creates so nice looking and consistent layouts. And, of course, because I can write it in my favorite text editor. We use LaTeX especially for presentation slides at Qafoo, since the beamer package provides such a convenient environment. Combined with listings package, presenting source code snippets with nice syntax highlighting has never been easier. However, there was one problem we did not solve, yet, until some days ago: Highlighting certain source code lines of a listing on different slides. So, let me give an example on when you want to highlight certain lines of a listing on different slides of your beamer presentation. The following listing shows how you can convert an XHTML document into a PDF, using the Apache Zeta Components Document component: ```<?php require 'autoload.php'; // Convert some web page to PDF \$xhtml = new ezcDocumentXhtml(); \$xhtml->setFilters( array( new ezcDocumentXhtmlElementFilter(), new ezcDocumentXhtmlXpathFilter( '//div[@class="content"]' ), ) ); \$xhtml->loadFile( 'consulting.html' ); // Load the docbook document and create a PDF from it \$pdf = new ezcDocumentPdf(); \$pdf->options->errorReporting = E_PARSE | E_ERROR | E_WARNING; // Load a custom style sheet \$pdf->loadStyles( 'custom.css' ); // Add a customized header \$pdf->registerPdfPart( new ezcDocumentPdfHeaderPdfPart( new ezcDocumentPdfFooterOptions( array( 'showPageNumber' => false, 'height' => '10mm', ) ) ) ); \$pdf->createFromDocbook( \$xhtml->getAsDocbook() ); file_put_contents( __FILE__ . '.pdf', \$pdf );``` The actual content of the listing is not important here, what really matters is its length and complexity. Of course the code is not highly complex in itself, but it is, if you are watching a presentation and suddenly a slide appears which shows the code. Using the LaTeX listings package, you already get a nicely highlighted visualization out of the box, including line numbers and possible other goodies: LaTeX beamer highlighting You can click on the image to enlarge it, so you can better see how nicely the lisiting is typeset with custom highlighting colors and line numbers. So, when presenting such a listing, it is likely to overwhelm people. Their focus will be on reading the full listing and understanding it and it is hard to draw their attention to the specific parts you are talking about at a given moment. You can try by pointing at the specific lines using a laser pointer or your finger or even just by naming the specific line numbers. However, have a clear visual indication on your slides is much more effective. Our idea was therefore, for a longer time now, to visually highlight certain lines by changing their background color. This is not an easy task in LaTeX. One way to solve this issue is to put additional LaTeX commands into the listing, using the `lstlisting` escape character. This works out, but basically makes your listing code unmaintainable, even unreadable. In addition, you cannot use the `lstinputlisting` command any more, which allows you to include lisitings directly from a source file, which is what you usually want to be doing instead of pasting the listing into the LaTeX file itself. I have to admit that it took more than two years until I finally found a really nice solution to this problem. To highlight certain lines of code, we now use the following command: ```\qalisting[fontsize=\tiny]{code/02_create_pdf_styled.php}{ \only<2>{ \qahigh{5,...,10} } \only<3>{ \qahigh{13,14} } }``` This sources the listing `code/02_create_pdf_styled.php` and displays it in font size `\tiny`. But instead of just generating a single beamer slide, it actually generates three: On the first slide, just the pure listing is shown. On the second one, the source code lines 5 to 10 are highlighted, and on the third one, lines 13 and 14. Simple, isn't it? You can see the results below (again click the images to see a larger variant). Highlighted lines 5 to 10 Highlighted lines 13 and 14 So how does it work internally? OK, I don't really want to talk about this, since it is really hackish. In short: I use a TiKZ image where the listing is embedded as a node and then create additional nodes on the background layer of this image, using the line height of the listing font size. I put up the source code of the highlighting commands to Github, so you can use it in your presentations, if you want. Beware, the commands are not really configurable and you will need to adjust the code manually to suite your presentation style. Furthermore, it does only work with inclusion of external source code files and is stuck to PHP code for now (easily adjustable to other languages). Maybe its still useful for you. If you know some LaTeX, I would love if you contribute additional options, like settings for the listing package or configurable styling. If you are a LaTeX guru and know how to fix some of the bigger issues, I would pretty much appreciate if you take some time, fork the code on Github and send me a pull request, or if you just send me a patch! Thanks in advance! :) If you liked this blog post or learned something, please consider using flattr to contribute back: . • #### Christoph Thank you very much for sharing this piece of code! Some days ago I wondered how you did that line highlighting in your slides, really good idea. Definitively useful when explaining source code :) • #### christian Hi, I have a problem with the use of the qalisting. The error was: and I don't find that file on internet. any suggestion? thanks! • #### louboutin Therefore, it is a small perturbation people don't understand the truth said anti-corruption commissioner abuse. • #### girlfriend activation system just to let you know I like visiting your blog, because the information you provide here contains really beneficial information that will satisfy readers and can clarify things

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https://hilbertthm90.wordpress.com/2009/03/05/lying-over-and-going-up/

# Lying Over and Going Up If you haven’t heard the terms in the title of this post, then you are probably bracing yourself for this to be some weird post on innuendos or something. Let’s first do some motivation (something I’m not often good at…remember that Jacobson radical series of posts? What is that even used for? Maybe at a later date we’ll return to such questions). We can do ring extensions just as we do field extensions, but they tend to be messier for obvious reasons. So we want some sort of property that will force an extension to be with respect to prime ideals. Two such properties are “lying over” and “going up.” Let $R^*/R$ be a ring extension. Then we say it satisfies “lying over” if for every prime ideal $\mathfrak{p}\subset R$ in the base, there is a prime ideal $\mathfrak{p}^*\subset R^*$ in the extension such that $\mathfrak{p}^*\cap R=\mathfrak{p}$. We say that $R^*/R$ satisfies “going up” if in the base ring $\mathfrak{p}\subset\mathfrak{q}$ are prime ideals, and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then there is a prime ideal $\mathfrak{q}^*\supset \mathfrak{p}^*$ which lies over $\mathfrak{q}$. (Remember that Spec is a contravariant functor). Note that if we are lucky a whole bunch of posts of mine will finally be tied together and this was completely unplanned (spec, primality, localization, even *gasp* the Jacobson radical). First, let’s lay down a Lemma we will need: Let $R^*$ be an integral extension of R. Then i) If $\mathfrak{p}$ a prime ideal of R and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then $R^*/\frak{p}^*$ is integral over $R/\mathfrak{p}$. ii) If $S\subset R$, then $S^{-1}R^*$ is integral over $S^{-1}R$. Proof: By the second iso theorem $R/\frak{p}=R/(\frak{p}^*\cap R)\cong (R+\frak{p}^*)/\frak{p}^*\subset R^*/\frak{p}^*$, so we can consider $R/\frak{p}$ as a subring of $R^*/\frak{p}^*$. Take any element $a+\frak{p}^*\in R^*/\frak{p}^*$. By integrality there is an equation $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$ with the $r_i\in R$. Now just take everything $\mod \frak{p}^*$ to get that $a+\frak{p}^*$ integral over $R/\frak{p}$. This yields part (i). For part (ii), let $a^*\in S^{-1}R^*$, then $a^*=a/b$, where $a\in R^*$ and $b\in\overline{S}$. By integrality again we have that $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$, so we multiply through by $1/b^n$ in the ring of quotients to get $(a/b)^n+(r_{n-1}/b)(a/b)^{n-1}+\cdots +r_0/b^n=0$. Thus $a/b$ is integral over $S^{-1}R$. I’ll do two quick results from here that will hopefully put us in a place to tackle the two big results of Cohen and Seidenberg next time. First: If $R^*/R$ is an integral ring extension, then $R^*$ is a field if and only if $R$ is a field. If you want to prove this, there are no new techniques from what was done above, but you won’t explicitly use the above result, so I won’t go through it. Second: If $R^*/R$ is an integral ring extension, ten if $\frak{p}$ is a prime ideal in R and $\frak{p}^*$ is a prime ideal lying over $\frak{p}$, then $\frak{p}$ is maximal if and only if $\frak{p}^*$ is maximal. Proof: By part (i) of above, $R^*/\frak{p}^*$ is integral over $R/\frak{p}$ and so as a corollary to “First” we have one is a field if and only if the other is. This is precisely the statement that $\frak{p}$ is maximal iff $\frak{p}^*$ is maximal.

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http://math.stackexchange.com/users/23793/user23793?tab=activity&sort=all

user23793 Reputation 361 Next privilege 500 Rep. Access review queues 1d comment Proving harmonic function is zero I think you meant the boundary of omega. Hint: use a connectivity argument. 1d comment Exchanging Series with Integrals: when is it possible? Note that by linearity, it's always possible to do this for finite sums. For infinite sums, it becomes a matter of exchanging a limit and an integral which you can use Fubini's convergence theorem, monotone convergence theorem, Vitali's convergence theorem, etc. 1d comment Convergence of $\sum_{-\infty}^{\infty}e^{-\pi tn^2}$ Thank you very much. I had an inkling that it was this simple. 1d accepted Convergence of $\sum_{-\infty}^{\infty}e^{-\pi tn^2}$ 1d asked Convergence of $\sum_{-\infty}^{\infty}e^{-\pi tn^2}$ Sep 28 awarded Popular Question Sep 7 awarded Notable Question Jul 6 comment Prove that $V$ is the direct sum of $W_1, W_2 ,\dots , W_k$ if and only if $\dim(V) = \sum_{i=1}^k \dim W_i$ I think you mean the sum of the dimension of the span of $B_i$ is the dimension of $W_i$. Jul 4 comment Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. My question would be if this proof is correct. I saw this problem in an old analysis book and so I thought I would try it out. Sorry, I should have been more clear. Jul 4 revised Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. added 148 characters in body Jul 4 comment Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. Sorry about that, I have corrected the statement in my edit. Jul 4 revised Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. added 98 characters in body Jul 4 asked Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. Jun 19 accepted If, $\lim x_n$ exists and finite then there is a function $f$ that is continuous Jun 19 asked If, $\lim x_n$ exists and finite then there is a function $f$ that is continuous Jun 6 accepted Difficult limits every grad should be able to do Jun 5 asked Difficult limits every grad should be able to do May 27 accepted $\lim_{n\to \infty} n^{1/n^2}$ May 27 comment $\lim_{n\to \infty} n^{1/n^2}$ Ahh, this is much better. Thanks! May 27 asked $\lim_{n\to \infty} n^{1/n^2}$

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http://assert.pub/papers/1907.06549

##### Subgroups of simple primitive permutation groups defined by unordered relations The problem of describing the invariance groups of unordered relations, called briefly \emph{relation groups}, goes back to classical work by H. Wielandt. In general, the problem turned out to be hard, and so far it has been settled only for a few special classes of permutation groups. The problem have been solved, in particular, for the class of primitive permutation groups, using the classification of finite simple groups and other deep results of permutation group theory. In this paper we show that, if $G$ is a finite simple primitive permutation group other then the alternating group $A_n$, then each subgroup of $G$, with four exceptions, is a relation group. ###### NurtureToken New! Token crowdsale for this paper ends in ###### Authors Are you an author of this paper? Check the Twitter handle we have for you is correct. ###### Subcategories #1. Which part of the paper did you read? #2. The paper contains new data or analyses that is openly accessible? #3. The conclusion is supported by the data and analyses? #4. The conclusion is of scientific interest? #5. The result is likely to lead to future research? User: Repo: Stargazers: 0 Forks: 0 Open Issues: 0 Network: 0 Subscribers: 0 Language: None Views: 0 Likes: 0 Dislikes: 0 Favorites: 0 0 ###### Other Sample Sizes (N=): Inserted: Words Total: Words Unique: Source: Abstract: [5, 6, 7, 8, 9] 07/15/19 06:02PM 2,800 902 ###### Tweets mathGRbot: Mariusz Grechand, Andrzej Kisielewicz : Subgroups of simple primitive permutation groups defined by unordered relations https://t.co/lXkiH2hYxp https://t.co/KemoCpqt4j

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https://www.physicsforums.com/threads/particle-in-a-potential-well-gre-93.9954/

# Particle in a potential well Gre. 93 1. Nov 30, 2003 ### yxgao What concepts are involved here? 93. A particle of mass m moves in the potential shown here. The period of the motion when the particle has energy E is The potential is V = 1/2kx^2 for x <0 and V= mgx for x > 0. A. Sqrt[k/m] B. 2*pi*Sqrt[m/k] c . 2*Sqrt[2E/(mg^2)] D. pi*Sqrt[m/k] + 2*Sqrt[2*E/(m*g^2)] E. 2*pi*Sqrt[m/k] + 4*Sqrt[2*E/(mg^2)] 2. Nov 30, 2003 ### arcnets Harmonic oscillator and ballistic motion. 3. Nov 30, 2003 ### yxgao What is ballistic motion? How do you arrive at the answer? 4. Dec 1, 2003 ### arcnets Ballistic motion is when a body moves in a field of constant gravity. You can look up the formulae in any basic mechanics book (or basic mechanics website). Sorry, I'm really too lazy to type it all down here for you. Similar Discussions: Particle in a potential well Gre. 93

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https://www.physicsforums.com/threads/how-to-solve-a-problem-involving-modulus.675807/

# How to solve a problem involving modulus 1. Mar 2, 2013 ### hackish Even though this deals with programming an encryption algorithm I feel this is more math based so I'm asking it here. x=((y mod 380951)*3182) mod 380951 How can I solve for y? 1) the math involved here is limited to integers, so for example division by 3182 must result in an integer. 2) the number 380951 is a prime number. 3) the result must be smaller than 2^24 4) 3182-1 is also prime but I don't think this helps the solution. 5) x is an integer in the range of 0..2^16 I've read all the wiki pages on fermat's little theorem and the extended euclidean algorithm but the concepts they describe are beyond my math abilities. I understand that there are multiple correct answers to this: EX: y=291905 ;x=83172 or y=1434758;x=83172 The first one will do and would be preferred since it has the greatest chance of fitting within the 24 bit result. At present I've been exhaustively calculating it by trying every possible multiple of the prime number + the result / 3182 and it works so I know a solution is possible. Relating to the example above if you multiply 380951*2834+83172 gives 928841710 then divide by 3182 gives 291905. Can anyone give steps to calculate the answer I need? 2. Mar 2, 2013 ### chiro Hey hackish and welcome to the forums. You may want to try writing y as y = pq + r where p = 380951 and r is between 0 and 380950 inclusive. Then do the same sort of thing for the outer modulus and bring the results together. 3. Mar 3, 2013 ### hackish Ok, someone else solved it... thanks. Turns out you can apply (3182^(380951-2)) mod 380951 then multiply that by x, then mod 380951 and you get y. Wow. A little math wizardry and encrypting a file now takes 250ms instead of 23 minutes. Last edited: Mar 3, 2013 Similar Discussions: How to solve a problem involving modulus

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http://hal.in2p3.fr/in2p3-01301197

# Nuclear excitations as coupled one and two random-phase-approximation modes Abstract : We present an extension of the random-phase approximation (RPA) where the RPA phonons are used as building blocks to construct the excited states. In our model, that we call double RPA (DRPA), we include up to two RPA phonons. This is an approximate and simplified way, with respect to the full second random-phase approximation (SRPA), to extend the RPA by including two-particle–two-hole configurations. Some limitations of the standard SRPA model, related to the violation of the stability condition, are not encountered in the DRPA. We also verify in this work that the energy-weighted sum rules are satisfied. The DRPA is applied to low-energy modes and giant resonances in the nucleus O16. We show that the model (i) produces a global downwards shift of the energies with respect to the RPA spectra and (ii) provides a shift that is, however, strongly reduced compared to that generated by the standard SRPA. This model represents an alternative way of correcting for the SRPA anomalous energy shift, compared to a recently developed extension of the SRPA, where a subtraction procedure is applied. The DRPA provides results in good agreement with the experimental energies, with the exception of those low-lying states that have a dominant two-particle–two-hole nature. For describing such states, higher-order calculations are needed. http://hal.in2p3.fr/in2p3-01301197 Contributor : Sophie Heurteau <> Submitted on : Monday, April 11, 2016 - 5:35:38 PM Last modification on : Thursday, January 11, 2018 - 6:12:41 AM ### Citation D. Gambacurta, F. Catara, M. Grasso, M. Sambataro, M. V. Andrés, et al.. Nuclear excitations as coupled one and two random-phase-approximation modes. Physical Review C, American Physical Society, 2016, 93 (2), pp.024309. ⟨10.1103/PhysRevC.93.024309⟩. ⟨in2p3-01301197⟩ Record views

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https://leanprover-community.github.io/mathlib_docs/algebra/category/Mon/limits.html

# mathlibdocumentation algebra.category.Mon.limits # The category of (commutative) (additive) monoids has all limits # Further, these limits are preserved by the forgetful functor --- that is, the underlying types are just the limits in the category of types. @[instance] @[instance] def Mon.monoid_obj {J : Type u} (F : J Mon) (j : J) : Equations add_submonoid (Π (j : J), (F.obj j)) The flat sections of a functor into AddMon form an additive submonoid of all sections. def Mon.sections_submonoid {J : Type u} (F : J Mon) : submonoid (Π (j : J), (F.obj j)) The flat sections of a functor into Mon form a submonoid of all sections. Equations @[instance] @[instance] def Mon.limit_monoid {J : Type u} (F : J Mon) : Equations limit.π (F ⋙ forget AddMon) j as an add_monoid_hom. def Mon.limit_π_monoid_hom {J : Type u} (F : J Mon) (j : J) : limit.π (F ⋙ forget Mon) j as a monoid_hom. Equations def Mon.has_limits.limit_cone {J : Type u} (F : J Mon) : Construction of a limit cone in Mon. (Internal use only; use the limits API.) Equations (Internal use only; use the limits API.) def Mon.has_limits.limit_cone_is_limit {J : Type u} (F : J Mon) : Witness that the limit cone in Mon is a limit cone. (Internal use only; use the limits API.) Equations (Internal use only; use the limits API.) @[instance] The category of monoids has all limits. @[instance] @[instance] The forgetful functor from monoids to types preserves all limits. (That is, the underlying types could have been computed instead as limits in the category of types.) Equations @[instance] def CommMon.comm_monoid_obj {J : Type u} (F : J CommMon) (j : J) : Equations @[instance] @[instance] @[instance] def CommMon.limit_comm_monoid {J : Type u} (F : J CommMon) : Equations @[instance] We show that the forgetful functor CommMon ⥤ Mon creates limits. All we need to do is notice that the limit point has a comm_monoid instance available, and then reuse the existing limit. Equations @[instance] def CommMon.limit_cone {J : Type u} (F : J CommMon) : A choice of limit cone for a functor into CommMon. (Generally, you'll just want to use limit F.) Equations A choice of limit cone for a functor into CommMon. (Generally, you'll just want to use limit F.) The chosen cone is a limit cone. (Generally, you'll just want to use limit.cone F.) def CommMon.limit_cone_is_limit {J : Type u} (F : J CommMon) : The chosen cone is a limit cone. (Generally, you'll just want to use limit.cone F.) Equations @[instance] The category of commutative monoids has all limits. @[instance] @[instance] The forgetful functor from commutative monoids to monoids preserves all limits. (That is, the underlying monoid could have been computed instead as limits in the category of monoids.) Equations @[instance] The forgetful functor from commutative monoids to types preserves all limits. (That is, the underlying types could have been computed instead as limits in the category of types.) Equations

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http://physics.stackexchange.com/users/4686/warren-p?tab=reputation

Warren P Reputation Top tag Next privilege 250 Rep. 5 Impact ~885 people reached • 0 posts edited +5 04:12 upvote How can I explain the scientific basis of the constant speed of light to a $c$-decay proponent? +5 03:05 upvote How can I explain the scientific basis of the constant speed of light to a $c$-decay proponent? +5 02:40 upvote How can I explain the scientific basis of the constant speed of light to a $c$-decay proponent?

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https://www.physicsforums.com/threads/explanation-of-superfluidity-in-he-4.586414/

# Explanation of superfluidity in He-4 1. Mar 12, 2012 ### sam_bell Hi. I was just reading the explanation of superfluidity in He-4 (from the beginning of QFT Methods in Statistical Physics by Abrikosov et. al.). There is something I don't understand. At finite temperatures there is a "gas of excitations", which they take to be moving at an average velocity v relative to the stationary liquid. They then derive that the quasi-momentum of this gas (per unit volume) is P = (const.) v. They claim this constant represents a mass and therefore there is mass transfer and that this part of the liquid is "normal". The rest of the mass is taken to be in the ground state superfluid. OK, my question: If we are talking about *quasi-*momentum, how can we be sure that there is really mass transfer? After all, a single quasi-particle has quasi-momentum, but this doesn't correspond to mass transfer as a drift of He-4 atoms. I suppose this is related to diffraction experiments, where is deflection of photon has a conversation law written in terms of quasi-momentum. 2. Mar 13, 2012 ### DrDu I would say that the quasi-momentum and the true momentum coincide in this case. A liquid does not break translation invariance whence momentum is well defined. 3. Mar 13, 2012 ### sam_bell There is still a consistency here that I can't follow. For simplicity, I imagine the case of a linear chain of oscillators. In this case the total momentum operator is P = sum(i = 1..N, P(i)) where P(i) is the momentum operator of the ith body in the chain. Expading P(i) in terms of normal modes gives = sum(n = 1..N, sum(all k, f(k) (a(k) exp(inb) - a(k)* exp(-inb)) where f(k) ~ 1/sqrt(energy) and b is the periodicity of the lattice. This doesn't look like the crystal momentum operator P = sum(all k, k a*(k) a(k)). Nevertheless, since b --> 0, if we excite a phonon of crystal momentum hk, then the external environment loses "real" momentum hk. Alternately, this means the linear chain gains a "real" momentum hk. But calculating <k|P|k> = 0 because none of the P(i) conserves phonon number. In going from |0> to |k> the real momentum of the chain didn't change? 4. Mar 14, 2012 ### DrDu Stop! thats not correct. There should be a k in the exponents. The sum over n then gives a delta function in k and only the k=0 components are left. Either in a or in a^* the k should read -k. Of course in true harmonic oscillator eigenstates the expectation of momentum always vanishes. However in the limit k=0 (and omega=0!), coherent states with unsharp number of quanta become alternative true eigenstates and have non-vanishing momentum. You may replace a(0) by its expectation value on these states. (You are not forced to do so. A state with fixed number of quanta would correspond to a macroscopic superposition of states with opposite momenta.) Due to the f(k) factor it diverges (an infinite long moving chain will have an infinite momentum) and you should divide by sqrt(L) to calculate the finite momentum per length. Note the strong analogy to your previous thread. Here, we have an example how a finite momentum density breaks symmetry (Galilean symmetry). I am not totally sure how the crystal momentum enters. I think we need to take coupling to the lattice into account to describe state with like momentum but unlike velocity.

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https://mathoverflow.net/questions/205821/around-vop%C4%9Bnka-accessible-category-with-small-full-discrete-subcategories-of-ar

# Around Vopěnka: Accessible category with small full discrete subcategories of arbitrary size? I believe the model-theoretic version of the question is: is there a theory in finitary first-order logic which has, for each cardinal $\lambda$, a set $C_\lambda$ of $\lambda$-many models, such that if $M,N \in C_\lambda$ then there is no elementary embedding from $M$ to $N$ or vice versa (in ZFC)? One statement of the large cardinal axiom Vopěnka's Principle is that no accessible category has a full subcategory which is both large and discrete. Adámek and Rosický point out (Remark 6.2(2)) that for any cardinal $\lambda$, it's trivial to come up (in ZFC) with an accessible category with a full discrete subcategory with $\lambda$-many objects. They use the example of the theory $\mathbf{Rel}_\lambda$, with $\lambda$-many unary relation symbols, and the set of objects $A_i$, each carried by the one-point set, where $A_i$ has just the $i$th relation turned "on". Here the accessible category in question is allowed to vary with the cardinal $\lambda$. But, in ZFC, is there one single accessible category $\mathcal{K}$ which has a full, discrete subcategory $\mathcal{K}_\lambda \subset \mathcal{K}$ of cardinality $\lambda$, for each cardinal $\lambda$? Of course, the union $\cup_\lambda \mathcal{K}_\lambda$ is large, so (assuming that Vopěnka's principle is consistent over ZFC), if such a category exists, then one won't be able to show that $\cup_\lambda \mathcal{K}_\lambda$ is discrete. But it could be that all of its morphisms go from objects of one $\mathcal{K}_\lambda$ to another $\mathcal{K}_{\lambda'}$, and the $\mathcal{K}_\lambda$'s themselves might all be discrete. Bonus question: in your example, are there clearly morphisms between objects in different $\mathcal{K}_\lambda$'s, or is your example a candidate to become a counterexample to Vopenka in some models (in which connection, this question may be relevant)? • I've accepted Joel's answer for sheer elegance. Thanks to Jiří, too -- it's important to know that examples also flow naturally from the existing theory of accessible categories. – Tim Campion May 6 '15 at 17:22 • Another point is that $\mathsf{Gph}$ apparently also embeds fully into familiar categories like Fields (and hence Rings), Groups, and Partial Orders, so these categories also have this property. Actually, the linked answer (of Joel's, ironically) discusses this with elementary embedding as the morphisms; I'm not sure whether the same goes for homomorphisms as morphisms. – Tim Campion May 6 '15 at 20:09 The answer is yes. One can do this with pointed directed graphs. Specifically, for any infinite cardinal $\lambda$, let $C_\lambda$ consist of all structures of the form $\langle V_{\lambda+2},{\in},\beta\rangle$, where $\beta<\lambda$ and $V_{\lambda+2}$ consists of the sets of von Neumann rank at most $\lambda+1$. So this is a pointed directed graph. Since there are $\lambda$ many choices for the constant $\beta$, we have $\lambda$ many models here. But there can be no elementary embedding between any two such structures, since any such embedding would give rise to a nontrivial elementary embedding $j:V_{\lambda+2}\to V_{\lambda+2}$, which is impossible by the Kunen inconsistency. • Wow! I was not expecting an answer from set theory, despite the set-theoretical nature of the question. Now let me advertise my ignorance and ask: in ZFC are there any elementary embeddings $V_{\lambda+2} \to V_{\lambda'+2}$ for $\lambda < \lambda'$? Or is this an example that might become a counterexample to Vopenka in some models? – Tim Campion May 6 '15 at 5:32 • @TimCampion Thanks! The existence of an elementary embedding $j:V_{\lambda+2}\to V_{\lambda'+2}$ is exactly connected with the extendible cardinals, which are fairly high in the large cardinal hierarchy, and so you cannot prove that in ZFC alone. Indeed, this is how one can show that Vopenka's principle has large cardinal strength, by considering the class of all structures $\langle V_\theta,\in\rangle$, and then getting elementary embeddings $j:V_\theta\to V_\lambda$. – Joel David Hamkins May 6 '15 at 10:12 • With a similar idea as in my post, you can get rid of the points, by considering $\langle V_{\lambda+\beta},\in\rangle$ for $2\leq \beta<\lambda$. There can be no elementary embedding $j:V_{\lambda+\alpha}\to V_{\lambda+\beta}$ for such distinct $\alpha,\beta<\lambda$. – Joel David Hamkins May 6 '15 at 10:42 • One can also get rid of the need for a special point simply by adding a self-edge on that point; it will be the only one. – Joel David Hamkins May 7 '15 at 1:51 Another, less elegant, but not so set-theoretical positive answer using graphs: Any accessible category has an accessible full embedding to graphs. • I have a question about this construction. Thinking model-theoretically, the idea here, I assume, is to code a given first-order structure $M$ with a graph $G_M$ in such a way that an elementary embedding $j:G_M\to G_N$ amounts to an elementary embedding $j^*:M\to N$. This is a common construction, and one can easily do this in the case where the signature of the structure $M$ is small, such as when it is countable.... – Joel David Hamkins May 6 '15 at 12:34 • ...But when the language of $M$ is enormous, then in order to code all the various relations on $M$ and ensure that maps between the $G_M$ actually respect those relations, it seems to me that one needs at bottom to produce large families of graphs that are rigid in the sense of having no elementary self-embeddings or elementary embeddings between them. In this case, I worry that the construction is circular, since it seems that we have come around to the very same question of the post again.... – Joel David Hamkins May 6 '15 at 12:34 • Or can one undertake the coding-into-graphs construction with first-order models in an arbitrary language without already having examples as in the original question? – Joel David Hamkins May 6 '15 at 12:34 • I see, it is as I suspected, since that fact already answers the question, if you look at pointed graphs like that, on a set of size $\lambda$. Is the proof of that fact non-set-theoretic? I can prove it using set-theoretic ideas... – Joel David Hamkins May 6 '15 at 13:57 • A certain graph is built on $\lambda$ by viewing it as the ordinal $\lambda+2$. Some care is needed to ensure that cardinals with cofinality $\omega$ and certain sequences approaching them are fixed by any endomorphism $f$, and then by considering the sup of the iterates of the critical point of $f$ (which has cofinality $\omega$) a contradiction is obtained. The appearance of $\lambda+2$ and an iteration argument (which, I gather from wikipedia appears in the proof of Kunen inconsistency) suggest that maybe this is a similar idea to that used in Kunen inconsistency? – Tim Campion May 6 '15 at 15:18

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http://math.stackexchange.com/questions/92223/uniqueness-of-solution-to-1st-order-pdes

# Uniqueness of solution to 1st order pdes I am given a 1st order partial differential equation $y{\partial \psi\over\partial x}+x{\partial \psi\over\partial y}=0$ subjected to boundary condition $\psi(x,0)=\exp(-x^2)$. I have found that a solution is $\psi(x,y)=\exp(y^2-x^2)$. But I am asked when the solution is unique. Could someone please explain how to answer this? Thanks. - It would be useful if you could tell us how you came up with your solution. Perhaps by the method of characteristics? – Jeff Dec 18 '11 at 17:21 Also, please include the domain of your PDE. Is it all of $\mathbb{R}^2$? The answer to your question will depend on the domain. – Jeff Dec 18 '11 at 18:13 Consider the parametric curves $x = A e^t + B e^{-t}$, $y = A e^t - B e^{-t}$, which satisfy $x' = y$, $y' = x$. Along such a curve any solution $\psi$ must be constant, according to the chain rule: $$\frac{d}{dt} \psi(x(t),y(t)) = \psi_x \frac{dx}{dt} + \psi_y \frac{dy}{dt} = 0$$ Now the curve intersects $y=0$ if and only if $A$ and $B$ are either both positive (i.e. $x > |y|$), both negative ($x < -|y|$), or both $0$ ($x=y=0$). So a boundary condition on $y=0$ produces uniqueness only in the regions $|x| \ge |y|$. In the region $|y| > |x|$ the solution is not unique. For example, you could add $f(y^2 - x^2)$ to $\psi(x,y)$ where $f$ is differentiable with $f(s) = 0$ for $s \le 0$. - how do we know that if the characteristic curve intersects the initial curve then there is a unique solution?. i.e., how do we know that there is a unique solution in this case iff the curve intersects $y = 0$? – user27182 Mar 26 '13 at 18:26 Since the solution must be constant on the curve, the value on $y=0$ determines the solution on the curve if the curve intersects $y=0$. Note by the way that (except for the trivial case $A=B=0$) the curve will intersect $y=0$ at only one point $t = -\ln(A/B)/2$. – Robert Israel Mar 29 '13 at 23:03 Uniqueness can be addressed in the following way. Let us suppose that exist another solution $\phi(x,y)$ such that $\phi(x,0)=e^{-x^2}$ then, being your equation linear then also $\phi'(x,y)=a\psi(x,y)+b\phi(x,y)$, with two arbitrary coefficients $a$ and $b$ is a solution. The boundary condition will give $a+b=1$. So, unless you do not give another condition on $y$ your solution cannot be unique. Of course, you have also the other condition in the given problem. From the fact that $\psi(x,0)=e^{-x^2}$ and from the other fundamental result that your equation has the general solution (characteristic method cited in the comments) $\psi(x,y)=\psi(x^2-y^2)$ for you is enough to set $\psi(0,0)=1$ and your solution is unique. Finally, I would like to point out the simple way the solution OP proposed can be found. One have to search for a solution in the form $$\psi(x,y)=\phi(y)e^{-x^2}$$ with $\phi(0)=1$ and the solution is immediately obtained, consistent with the characteristic method. - You are assuming the existence of another solution $\phi \neq \psi$. – Jeff Dec 18 '11 at 17:06 If I want to prove uniqueness I have to guess that another solution does exist and then, to prove that this is not independent from the other. This is standard matter and I do not see the reason to downvote unless my argument is wrong. – Jon Dec 18 '11 at 17:32 Maybe I am mistaken, but your argument seems to be: assume another solution exists, try to prove it is linearly dependent with the first solution, and then if you can't succeed, it must imply non-uniqueness? This is wrong. – Jeff Dec 18 '11 at 18:09 I think you have some difficulties with foundations en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem. Here the idea is taken from the fixed point uniqueness. I just repeat: Before to downvote, think! – Jon Dec 18 '11 at 18:14 Your solution is still wrong, and has nothing to do with the Picard Lindelof Theorem. Try to think about why your solution would imply non-uniqueness (and remember this is not an ODE). – Jeff Dec 18 '11 at 18:19

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http://mathhelpforum.com/algebra/132698-solving-equation-print.html

Solving an Equation • Mar 8th 2010, 09:36 AM StephenPoco Solving an Equation Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. • Mar 8th 2010, 10:00 AM e^(i*pi) Quote: Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Solve for what If $fg(x)=0$ use the quadratic formula • Mar 8th 2010, 10:01 AM masters Quote: Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Hi StephenPoco, I'm not sure what fg(x) is, but I would set the expression = 0 and use the quadratic formula to solve. $3x^2-6x+17=0$ $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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http://nickjames.co.nz/equestrian-brands-zpqv/54-kg-to-lbs-7feb41

Step 1: Convert from kilograms to pounds. kg or lbs The SI base unit for mass is the kilogram. Here is one of the Mass conversion : 54 lbs in kg Convert 518.54 kg to pounds. To use this calculator, simply type the value in any box at left or at right. 54.28 kg to lbs. 14.54 KG to Lbs – Unit Definition. What is 518.54 kg in pounds. Thus, for 54 kilograms in pound we get 119.04962158 lbs. Mass is defined as the tendency of objects at rest to remain so unless acted upon by a force. Therefore, another way would be: - 54 kilograms is equal to how many pounds. It accepts fractional values. Step 1: Convert from kilograms to pounds. The kilogram or kilogramme (symbol: kg) is the base unit of mass in the International System of Units (SI). In this article you will find everything about kilogram to pound conversion - both theoretical and practical. The 54.9 kg in lbs formula is [lb] = 54.9 * 2.2046226218. 54 kilograms equal 119.04962158 pounds (54kg = 119.04962158lbs). It accepts fractional values. Note that rounding errors may occur, so always check the results. How to convert 10.54 lbs to kg? 54 Kilograms to Pounds, 54 Kilograms in Pounds, 54 Kilogram to lb, 54 Kilogram in lb, 54 kg to lbs, 54 kg in lbs, 54 kg to Pound, 54 kg in Pound, 54 Kilograms to lb, 54 Kilograms in lb, 54 Kilograms to lbs, 54 Kilograms in lbs, 54 Kilograms to Pound, 54 Kilograms in Pound, 54 Kilogram to lbs, 54 Kilogram in lbs, 54 Kilogram to Pound, 54 Kilogram in Pound, ‎54 Χιλιόγραμμο σε λίμπρα, ‎54 কিলোগ্রাম মধ্যে পাউন্ড, ‎54 किलोग्राम से पाउण्ड, ‎54 กิโลกรัมปอนด์, ‎54 કિલોગ્રામ પાઉન્ડ. 14.54 pounds it is equal 6.5952330598 kilograms, so 14.54 lb is equal 6.5952330598 kgs. 518.54kg to pounds. If messing around with numbers and multiplying and dividing are not your thing, our 10.54 kg to lbs conversion chart can do it for you. But if you’re just looking for a rounded off figure, you can also use the 5.54 kg to lbs conversion chart above. kg to pounds kg to lb + oz. Kilograms: Pounds (lb) = Detailed result here. To convert 54.5 kg to lbs, multiply 54.5 by 2.205. One pound (lb), the international avoirdupois pound, is legally defined as exactly 0.45359237 kilograms. Kilograms to Pounds Converter. A gram is equal to 1/1000 of a kilogram, and its SI symbol is K, and kilo may also be used. How do I convert lbs to kg? Mass conversion provides conversion between measure of mass. To convert 54.9 kg to lbs multiply the mass in kilograms by 2.2046226218. It accepts fractional values. To use this calculator, simply type the value in any box at left or at right. - 54 kilograms is equal to how many stones and pounds? Converting 54.6 kg to lb is easy. 53 kg to lbs to convert 53 kilograms to pounds and find out how many pounds is 53 kg. The Kg to Pounds Conversion Formula to convert 60.54 kg to lbs To know how many pounds in a kilogram, you can use the following formula to convert kg to lbs : X(lb) = Y(kg) / 0.45359237 - 53 kg is equal to 116.84 pounds. Whether you opt for a 54.11 kilograms to pounds conversion chart or a 54.11 kg to lbs converter, there is no questioning the need for them. It is important to use the converter weight lbs to kg twice or even more to get the exact result. 54 pounds equal 24.49398798 kilograms (54lbs = 24.49398798kg). 1.54 kilograms or 1540 grams equals 3.40 pounds. How many pounds in 0.54 Kilograms? Q: How many Kilograms in 1 Pounds? It is also needed/We also want to point out that whole this article is devoted to a specific amount of kilograms - this is one kilogram. If you want more accurate results down to the decimals, you should try our 5.54 kg to lbs converter. How to convert 54.4 Pounds (lbs) to kilograms (kg). M (lb) = 2.204622621849 × M (kg) = 2.204622621849 × 3 = 6.613867865546 lbs. If M (kg) represents mass in kilograms and M (lb) represents mass in pounds, then the formula for converting kg to lbs is: M (lb) = 2.204622621849 × M (kg) Example. 54 kg are equal to 54 x 2.20462262 = 119.049622 pounds. The answer is 0.453592. Easily enter your kilogram weight and instantly get the result in pounds. How to convert 54 kilograms to pounds To convert 54 kg to pounds you have to multiply 54 x 2.20462, since 1 kg is 2.20462 lbs . kg to pounds kg to lb + oz. Q: How many Pounds in 1.54 Kilograms? While the emphasis here is on 14.54 kg, you can use other kilograms. Learn how to convert from lb to kg and what is the conversion factor as well as the conversion formula. 119.07. The answer is 119.050 . Thus, for 54.1 kilograms in pound we get 119.270083842 lbs. The kg to lbs conversion calculator is based on formulas which is not errorless. Kg Stones st and pounds; 40kg: 6.3st: 6st 4.2lb: 40.5kg: 6.38st: 6st 5.3lb: 41kg: 6.46st: 6st … If messing around with numbers and multiplying and dividing are not your thing, our 54.8 kg to lbs conversion chart can do it for you. 1 kilogram is equal to 2.204622621849 pounds or lbs and … One pound (symbol: lb), the international avoirdupois pound, is legally defined as exactly 0.45359237 kilograms. ›› Quick conversion chart of kg to lbs. 0.54 kg to lbsto convert 0.54 kilograms to pounds and find out how many pounds is 0.54 kg. The kilogram (kg) is the SI unit of mass. Step 2: Convert the decimal part of pounds to ounces An answer like "3.388 pounds" might not mean much to you because you may want to express the decimal part, which is in pounds, in ounces which is a smaller unit. 54 Pounds to Kilograms Conversion breakdown and explanation 54 lbs to kg conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). 1 kg to lbs = 2.20462 lbs. 1 kilogram = 2.2 x pounds, so, 54.5 x 1 kilogram = 54.5 x 2.2 pounds (rounded), or. Approximation An approximate numerical result would be: fifty-four kilograms is about one hundred and nineteen point zero four pounds , or alternatively, a pound is about zero point zero one times fifty-four kilograms . Convert 54 Kilograms to Pounds To calculate 54 Kilograms to the corresponding value in Pounds, multiply the quantity in Kilograms by 2.2046226218488 (conversion factor). To convert 53 kg to lbs, multiply 53 by 2.205. But if you’re just looking for a rounded off figure, you can also use the 10.54 kg to lbs conversion chart above. Kilograms: Pounds (lb) = Detailed result here. We do not take the responsibility for errors caused by lbs into kg converter. Kilograms to Pounds Converter. 54 kg are equal to 54 x 2.20462262 = 119.049622 pounds. - 54 kg is equal to 119.05 pounds. kg to pounds kg to lb + oz. But if you’re just looking for a rounded off figure, you can also use the 54.8 kg to lbs conversion chart above. There are 16 lb 9 15/16 oz (ounces) in 7.54 kg. From people to cars to everyday items, kg is the standard. 54.6 kilograms equal 120.372395153 pounds (54.6kg = 120.372395153lbs). Use this page to learn how to convert between kilograms and pounds. 5.54 kg to lbs. 54.8 kilograms equal 120.813319677 pounds (54.8kg = 120.813319677lbs). The 54 kg in lbs formula is [lb] = 54 * 2.2046226218. Simply use our calculator above, or apply the formula to change the length 54.8 kg to lbs. Use this page to learn how to convert between kilograms and pounds. How many pounds in 5.54 Kilograms? Convert 0.54 kg to pounds. The factor 2.20462262 is the result from the division 1 / 0.45359237 (pound definition). … If you want more accurate results down to the decimals, you should try our 54.8 kg to lbs converter. Kilograms [kg] The kilogram , or kilogramme, is … Definition of kilogram. 54.00. It can also be expressed as: 54 kilograms is equal to 1 0.0083998587037037 pounds. Kilogram abbreviation: “kg”, Pound abbreviation: “lb. A kilogram (also kilograms and abbreviated as kg), is a unit of mass. One kilogram is a unit of masss (not weight) and equals approximately 2.2 pounds. 54.5 Kilograms to Pounds shows you how many pounds are equal to 54.5 kilograms as well as in other units such as grams, metric tons, milligrams, micrograms, stones and ounces. 54.5 kg to lbs to convert 54.5 kilograms to pounds and find out how many pounds is 54.5 kg. 10 kg to lbs = 22.04623 lbs. You can also press the arrow so you can select other weight units that you could convert. One pound equals 16 ounces exactly. How to convert 0.54 to pounds? A kilogram (also kilograms and abbreviated as kg), is a unit of mass. Thus, for 54 kilograms in pound we get 119.04962158 lbs. So, if you want to calculate how many pounds are 54 … or lbs”. Converting 54.8 kg to lb is easy. Convert 54 kg to pounds. To use this calculator, simply type the value in any box at left or at right. 5.54 kilograms or 5540 grams equals 12.21 pounds. In many parts of the world, kilogram is the unit used to measure weight and mass. 1 kilogram is equal to 2.2046226218488 lbs. To convert 54.1 kg to lbs multiply the mass in kilograms by 2.2046226218. Converting 54 lb to kg is easy. One pound (symbol: lb), the international avoirdupois pound, is legally defined as exactly 0.45359237 kilograms. Convert kg to lbs; 54 Kilograms to Pounds; Convert 54 Kilograms to Pounds. Step 2: Convert the decimal part of pounds to ounces. Formula for converting kilogram to ounces . An infographic chart is further down the page (60kg to 130kg).. Kg to stone, pounds ›› Quick conversion chart of kg to lbs. How many is 518.54 kilograms in pounds. Using our kilograms to stones and pounds converter you can get answers to questions like: - How many stones and pounds are in 54 kg? How to convert 54.4 Pounds (lbs) to kilograms (kg) Kilograms [kg] The kilogram, or kilogramme, is the base unit of weight in the Metric system. Do you want to know how much is 2.54 kg equal to lbs and how to convert 2.54 kg to lbs? Formula for converting kilogram to ounces 1 kilogram is equal to 2.204622621849 pounds or lbs and 1 pound is equal to 16 ounces or oz. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. 119.04962 Pounds (lb) Kilograms : The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Kilograms to Pounds Conversions. Kilograms to Pounds Converter. More information from the unit converter. This method is easy, quick and reliable. lb. Here you go. One kilogram equals 2.20462262 pounds, to convert 54 kg to pounds we have to multiply the amount of kg by 2.20462262 to obtain amount in pounds. What is 54.5 kg in pounds? - 54 kg is equal to 119.05 pounds. If messing around with numbers and multiplying and dividing are not your thing, our 5.54 kg to lbs conversion chart can do it for you. This prototype is a platinum-iridium international prototype kept at the International Bureau of Weights and Measures. Use these charts to quickly look up common weight conversions for kilograms to stone and pounds. Using our kilograms to stones and pounds converter you can get answers to questions like: - How many stones and pounds are in 54.54 kg? The final formula to convert 54 Kg to Lb is: [Kg] = 54 / 0.453592 = 119.05 Kilogram is the SI unit of mass. 54.5 kilograms = 119.9 pounds. 54.01. Here you go. 54.02. How many lbs and oz in 7.54 kilograms? 3.54 kg to lbs. How to convert 14.54 lbs to kg? Kilograms to Pounds Converter. To calculate a kilogram value to the corresponding value in pounds, just multiply the quantity in kilogram by 2.20462262 (the conversion factor). 0.54 kilograms are equal to 0.24494 pound. 10.54 pounds it is equal 4.7808635798 kilograms, so 10.54 lb is equal 4.7808635798 kgs. Kilograms: Pounds (lb) = Detailed result here. Type in your own numbers in the form to convert the units! lbm. If messing around with numbers and multiplying and dividing are not your thing, our 65.54 kg to lbs conversion chart can do it for you. A single kilogram is equal to 2.20 lbs. How to convert 54 kg to lb To calculate a value in kg to the corresponding value in lb, just multiply the quantity in kg by 2.2046226218488 (the conversion factor). 24.04 kg: 54 lb: 24.49 kg: 55 lb: 24.95 kg: 56 lb: 25.40 kg: 57 lb: 25.85 kg: 58 lb: 26.31 kg: 59 lb: 26.76 kg: Onces en Grammes; Grammes en Onces; Onces en Livres; Livres en Onces; Table de conversion métrique Application pour iPhone & Android Poids Température Longueur Superficie Volume Vitesse Temps Monnaie. 0.54 kg to lbs What is 0.54 kg in pounds? It is also needed/We also want to emphasize that all this article is devoted to a specific amount of kilograms - that is one kilogram. Until 20 May 2019, it remains defined by a platinum alloy cylinder, the International Prototype Kilogram (informally Le Grand K or IPK), manufactured in 1889, and carefully stored in Saint-Cloud, a suburb of Paris. 2.54 kg to lbs. Converting 54 kg to lb is easy. 2.54 kilograms = 5.588 pounds. How Heavy Is 54.4 Pounds in Kilograms? 1 lbs = 0.453592 kg. Thus, for 54.9 kilograms in pound we get 121.03378194 lbs. Remember that our calculator from lbs to kg or kg to lbs located on the site can sometimes show the wrong results. 518.54 kg to lbs. One kg is approximately equal to 2.20462262 pounds. One kilogram equals 2.20462262 pounds, to convert 54.7 kg to pounds we have to multiply the amount of kg by 2.20462262 to obtain amount in pounds. 54.7 kg to lbs conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). 54 pounds equal 24.49398798 kilograms (54lbs = 24.49398798kg). 54 kg to lbs to convert 54 kilograms to pounds and find out how many pounds is 54 kg. The answer is 0.453592. 20 kg to lbs = 44.09245 lbs. Type in your own numbers in the form to convert the units! - 54.4 kilograms is equal to how many stones and pounds? 54.5kg to lbs. Whether you opt for a 54.4 kilograms to pounds conversion chart or a 54.4 kg to lbs converter, there is no questioning the need for them. Simply use our calculator above, or apply the formula to change the length 54.6 kg to lbs. The 54.1 kg in lbs formula is [lb] = 54.1 * 2.2046226218. 3870 Liters to Kilograms 181200 Liters to Pounds 1.8 Milligram to Grams 64 Pounds to Kilograms 385 Milliliters to Pounds 188.48 Pounds to Liters 147.15 Pounds to Liters 14.26 Pounds to Liters 76.32 Pounds to Liters 146.46 Pounds to Liters 256 Ounces to Pounds How many pounds in 3.54 Kilograms? Defined as being equal to the mass of the International Prototype Kilogram (IPK), that is almost exactly equal to the mass of one liter of water. To use this calculator, simply type the value in any box at left or at right. - 54.5 kg is equal to 120.15 pounds. Others Weight and Mass converter. Converting 54 lb to kg is easy. To convert 54 kg to lbs multiply the mass in kilograms by 2.2046226218. If you need to be super precise, you can use one kilogram = 2.2046226218488 pounds. 54 Kilograms (kg) =. 3.44 kilograms equals 7.58 pounds: 3.54 kilograms equals 7.80 pounds: 3.64 kilograms equals 8.02 pounds: 3.74 kilograms equals 8.25 pounds: 3.84 kilograms equals 8.47 pounds: 3.94 kilograms equals 8.69 pounds: 4.04 kilograms equals 8.91 pounds: 4.14 kilograms equals 9.13 pounds: 4.24 kilograms equals 9.35 pounds: 4.34 kilograms equals 9.57 pounds Kilograms: Pounds (lb) = Detailed result here. The kilogram (kg) is the SI unit of mass. Step 2: Convert the decimal part of pounds to ounces. Easily convert Kilograms to pounds, with formula, conversion chart, auto conversion to common weights, more 119.05. Use our free metric conversion tool to convert 54 lb (pounds) to kg (kilograms) KGtoLBS.com Convert kilograms into pounds quickly. Definition of kilogram. The answer is 3.395119 The kilogram (kg) is the SI unit of mass. If messing around with numbers and multiplying and dividing are not your thing, our 54.11 kg to lbs conversion chart can do it for you. Kilograms [kg] The kilogram , or kilogramme, is … How Many Pounds in a Kilogram? kg to pounds kg to lb + oz. If M (kg) represents mass in kilograms, M (lb) represents nass in pounds and M (oz) represents mass in ounces, then the formula for converting kg … So finally 54 kg = 119.04962157983 lbs. Supose you want to convert 54 kg into lb. Conclusion: 54 ⁢ kg ≈ 119.0496204 ⁢ lb Conversion in the opposite direction The inverse of the conversion factor is that 1 pound is equal to 0.0083998587037037 times 54 … 106.7 kg to stones and lbs Disclaimer While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. 25 kg to lbs = 55.11557 lbs Here is the formula: Value in lb = value in kg × 2.2046226218488. Kilograms to Pounds Converter. What is a Kilogram? If we want to calculate how many Pounds are 54 Kilograms we have to multiply 54 by 100000000 and divide the product by 45359237. Simply use our calculator above, or apply the formula to change the length 54 kg to lbs. In this case we should multiply 54 Kilograms by 2.2046226218488 to get the equivalent result in Pounds: Kilograms: Pounds (lb) = Detailed result here. Using our kilograms to stones and pounds converter you can get answers to questions like: - How many stones and pounds are in 54.4 kg? The 54 kg in lbs formula is [lb] = 54 * 2.2046226218. kg to pounds kg to lb + oz. It is the approximate weight of a cube of water 10 centimeters on a side. Kilograms to Stone and Pounds Chart. (some results rounded) kg. - 54.54 kilograms is … But if you’re just looking for a rounded off figure, you can also use the 65.54 kg to lbs conversion chart above. More information from the unit converter. To convert 54 kg to lbs, multiply 54 by 2.205. Defined as being equal to the mass of the International Prototype Kilogram (IPK), that is almost exactly equal to the mass of one liter of water. What is 53 kg in pounds? 5 kg to lbs = 11.02311 lbs. What is 54 kg in pounds? In many parts of the world, kilogram is the unit used to measure weight and mass. But if you’re just looking for a rounded off figure, you can also use the 54.11 kg to lbs conversion chart above. 54 kg to lbs to convert 54 kilograms to pounds and find out how many pounds is 54 kg. It is equal to the mass of the international prototype of the kilogram. Once this is very close to 2.2 pounds, you will almost always … How many pounds in 1.54 Kilograms? To use this calculator, simply type the value in … What is 54 kg in pounds? 0.54 kg to lbs. 15 kg to lbs = 33.06934 lbs. To convert 54 kg to lbs, multiply 54 by 2.205. Use our calculator below to transform any kg or grams value in lbs and ounces. 54 Kilograms (kg) = 119.04962 Pounds (lb) Kilograms : The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Q: How many Kilograms in 1 Pounds? 54.9 Kilogram Conversion Table Do you need to know how much is 54.28 kg equal to lbs and how to convert 54.28 kg to lbs? Kg to Lbs converter. 1 kilogram is equal to 1000 grams, and 1 gold bar is equivalent to 1 kg. 3.54 kilograms or 3540 grams equals 7.80 pounds. To convert 3 kilograms to pounds: M (kg) = 3 . So for 54 we have: (54 × 100000000) ÷ 45359237 = 5400000000 ÷ 45359237 = 119.04962157983 Pounds. It accepts fractional values. 54 Kilograms (kg) = 119.050 Pounds (lbs) 1 kg = 2.204623 lbs. - 0.54 kg is equal to 1.19 pounds. Simply use our calculator above, or apply the formula to change the length 54 lbs to kg. How many pounds in 518.54 kilograms. One kilogram equals 2.20462262 pounds, to convert 54 kg to pounds we have to multiply the amount of kg by 2.20462262 to obtain amount in pounds. One pound (symbol: lb), the international avoirdupois pound, is legally defined as exactly 0.45359237 kilograms. 0.54 kilogram or 540 grams equals 1.19 pounds. Q: How many Pounds in 54 Kilograms? 1.54 kilograms = 3.388 pounds. To convert 54 kg to lbs multiply the mass in kilograms by 2.2046226218. 1.54 kg to lbs. Convert 54.7 kg to pounds. Simply use our calculator above, or apply the formula to change the length 54 lbs to kg. 0.54 kilogram or 540 grams equals 1.19 pounds. From people to cars to everyday items, kg is the standard. 1 kilogram = 2.2 x pounds, so, 2.54 x 1 kilogram = 2.54 x 2.2 pounds (rounded), or. It is part of the Standard International (SI) System of Units. 54 KG (Kilograms) = 119.04962158 LBS (Pounds) Two Decimal Point Results 54 KG (Kilograms) is equal to 119.05 LBS (Pounds) $$36 kg*{2.2046 lbs \over 1 kg} = 79.123 lbs$$ The Kilogram. 54.7 kg are equal to 54.7 x 2.20462262 = 120.592857 pounds. Convert: (Please enter a number) From: … Kilogramme ( symbol: lb ), or the formula to change the length 54 to! Everything about kilogram to pound conversion - both theoretical and practical try our 5.54 kg to lbs multiply mass! So for 54 kilograms to pounds and find out how many pounds 54! 10.54 lb is equal to how many pounds lb ] = 54 2.2046226218. 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http://mathhelpforum.com/advanced-algebra/194500-separable-extensions.html

1. ## Separable Extensions Prove that if L/G and G/F are separable algebraic extensions (not necessarily finite), then L/F is also separable. I could do it for the finite case, but I'm not sure what to do here? 2. ## Re: Separable Extensions i don't think this is true, without more information on F. for one can devise algebraic, but not separable, extensions of F and then G, in which case L is NOT separable over F. perhaps F is a field of characteristic 0? 3. ## Re: Separable Extensions Ohh I'm sorry, that should have said, "if L/G and G/F are *separable* algebraic extensions". And we already proved in class that if L/G and G/F are both algebraic extensions (not necessarily finite) then L/F is an algebraic extension. So I just need to show L/F is separable too.

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http://mathoverflow.net/questions/63402/martins-philosophical-issues-about-the-hierarchy-of-sets

# Martin's “Philosophical Issues about the Hierarchy of Sets” Some months ago (October 2010), in the context of the Workshop on Set Theory and the Philosophy of Mathematics, Professor Donald A. Martin gave a talk entitled "Philosophical issues about the hierarchy of sets". Abstract: I will discuss some philosophical questions about the cumulative hierarchy of sets, its levels, and their theories. Some examples: (1) It is sometimes asserted one cannot quantify over everything. A related assertion is that each of our statements about the universe of sets can from a different perspective be seen as a statement about some Va. Thus the class-set distinction is really a relative one. Does this make sense? Is it right? (2) Is the first order theory of V determinate? Does every sentence have a truth value? Are there levels of the hierarchy whose first order theories are indeterminate? If so, what is the lowest such level? What about L and the constructibility hierarchy? (3) There are many examples of proofs of a statement about one level of the hierachy that use principles about a higher level. Under what conditions and in what sense do these count as establishing the lower level statement? I will discuss these questions mainly from a viewpoint that takes mathematics to be about basic mathematical concepts, e.g., those of natural number, real number, and set. I am highly interested in learning how these questions might be answered (as you may problably know from previous questions of mine here in MO), so I would be grateful if anyone could give any information in this respect, especially for those questions of 1 and 3 (I am afraid it is almost impossible to do justice to 2 in a few lines). - Have you written to Professor Donald A. Martin? –  j.c. Apr 29 '11 at 13:20 @jc Yes, I have, but with no success by now. –  Marc Alcobé García Apr 29 '11 at 20:25 ## 1 Answer Of course there are no universally agreed-upon answers to these philosophical questions, and if you are interested in Martin's views specifically, then I suggest that you read his articles. Meanwhile, allow me simply to explain a few of the issues arising in the specific questions you mention. • "One cannot quantify over everything." This is a reference to the predicative/impredicative debate in the philosophy of set theory. One of the objections to the replacement and collection axioms is that they are used to describe sets by means of properties of a totality of which they themselves are a member. That is, you define a subset of $B=\{a\in A\mid \varphi(a)\}$, but $\phi(a)$ may be a very complicated property that quantifies over the entire universe, referring to objects and properties of objects, including $B$ itself. But also, it can be a reference to the cumulative view of set theory as building up more and more sets in a process that is never completed, and in this case, it may not be sensible to form sets by means of properties holding in the entire universe, as though it were completed. • "Each of our statements about the universe of sets can from a different perspective be seen as a statement about some $V_\alpha$." The Levy reflection theorem shows that for any assertion $\sigma(x)$, there is an ordinal $\alpha$ such that $\sigma(x)$ is true if and only if it is true in $V_\alpha$, for any $x\in V_\alpha$. That is, $\sigma$ is absolute between $V_\alpha$ and $V$. Going a bit beyond this, consider the theory denoted "$V_\delta\prec V$", which asserts, in the language with a constant for $\delta$, that $\forall x\in V_\delta\, [\varphi(x)\iff \varphi^{V_\delta}(x)]$. This is the scheme asserting that $V_\delta$ is an elementary substructure of the universe. Although some set theorists are surprised to hear it, this scheme is equiconsistent with ZFC, and any model $M$ of set theory can be elementarily embedded into a model of this theory. (This is done by a simple compactness argument; one writes down the theory $V_\delta\prec V$ plus the elementary diagram of $M$, and observes that the reflection theory shows that it is finitely consistent.) Finally, note that in a model of $V_\delta\prec V$, every sentence can be viewed as an assertion about $V_\delta$, rather than about $V$, since they have exactly the same theory. • "Is the first order theory of $V$ determinate?". This question is asking whether there is a fact of the matter in regard to our set-theoretic questions. For example, does it make sense to say that there is ultimately an answer to the question of whether the Continuum Hypothesis is really true? Or whether large cardinals exist? This question is connected in my mind with issues about whether there is a unique structure that we are investigating when we do set theory---the universe $V$ of all sets---or is there instead a multiverse of possibilities? In other words, is there a final truth of the matter in set theory, or is set theory instead something more like geometry, having a plethora of diverse Euclidean and non-Euclidean worlds? In the slides for my talk at the same conference, I explore the multiverse view in detail. • "Are there levels of the hierarchy whose first order theories are indeterminate? What is the lowest such level?" Some set theorists may view questions about the Continuum Hypothesis to be a source of indeterminateness, in the sense that there is no fact of the matter about CH. But CH is a statement expressible in $H_{\omega_2}$, or alternatively in $V_{\omega+2}$. Martin is asking whether we might expect indeterminateness at lower levels. In his talk at the workshop you mention, I recall him saying that he found it unacceptable to think that there would be indeterminateness arising at the level of $V_\omega$, and that arithmetic truth was absolute in some very strong sense. • "There are many examples of proofs of a statement about one level of the hierachy that use principles about a higher level." This is referring to the fact that mathematicians routinely use higher level objects in order to make conclusions about lower level objects. For example, one might use infinite objects (such as automorphisms of field extensions) in order to make conclusions about finite objects, or very large function spaces or ultrafilters in order to make conclusions about a lower level object. Part of Martin's point was the philosophical concern that if there is indeterminism about features of the higher level objects, then they might seem unsuited for this purpose. - I remember also an argument that he made or considered (and I've heard him make this argument in other forums) that any two instantiations $V$ and $\bar V$ of the full set concept must agree; the idea is that one inductively shows that they agree at every level of the hiearchy, essentially since if they agree up to $V_\alpha$ and each is claiming to have all of the subsets of $V_\alpha$, then they agree up to $V_{\alpha+1}$. –  Joel David Hamkins Apr 30 '11 at 16:28 Thank you vey much, Joel. Do you know where could I read more about $V_\delta\prec V$ and its properties? Also, I have googled for Martin's articles and the most recent that I have found is "Multiple Universes of Sets and Indeterminate Truth Values" (2001). –  Marc Alcobé García Apr 30 '11 at 18:07 I use the axiom $V_\delta\prec V$ in my article on the Maximality Principle (J. D. Hamkins, "A simple maximality principle," Journal of Symbolic Logic, vol. 68, pp. 527--550, June 2003), where I give a brief account of it. It is necessary in the forcing construction that is used to obtain the Maximality Principle. Also, I believe that Solomon Feferman has used this axiom in some of his work, in order to provide an alternative weaker foundation for the use of universes in category theory. One can have a whole proper class club of such $\delta$, still just with consistency strength ZFC. –  Joel David Hamkins Apr 30 '11 at 18:52 Joel, as usual a very informative and interesting answer. As for your first comment, what do you mean by "full set concept"? –  Asaf Karagila Apr 30 '11 at 19:46 I suppose that something counts as an instantiation of the full set concept if its powersets contain every conceivable subset (if this really means something), but also if it contains every conceivable ordinal (again, if this really means something). –  Marc Alcobé García Apr 30 '11 at 20:08

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http://www.zora.uzh.ch/48761/

# Recursive contracts, lotteries and weakly concave pareto sets - Zurich Open Repository and Archive Cole, Harold; Kübler, Felix (2012). Recursive contracts, lotteries and weakly concave pareto sets. Review of Economic Dynamics, 15 (4)(10-038):479-500. ## Abstract Marcet and Marimon (1994, revised 1998) developed a recursive saddle point method which can be used to solve dynamic contracting problems that include participation, enforcement and incentive constraints. Their method uses a recursive multiplier to capture implicit prior promises to the agent(s) that were made in order to satisfy earlier instances of these constraints. As a result, their method relies on the invertibility of the derivative of the Pareto frontier and cannot be applied to problems for which this frontier is not strictly concave. In this paper we show how one can extend their method to a weakly concave Pareto frontier by expanding the state space to include the realizations of an end of period lottery over the extreme points of a flat region of the Pareto frontier. With this expansion the basic insight of Marcet and Marimon goes through - one can make the problem recursive in the Lagrangian multiplier which yields significant computational advantages over the conventional approach of using utility as the state variable. The case of a weakly concave Pareto frontier arises naturally in applications where the principal's choice set is not convex but where randomization is possible. ## Abstract Marcet and Marimon (1994, revised 1998) developed a recursive saddle point method which can be used to solve dynamic contracting problems that include participation, enforcement and incentive constraints. Their method uses a recursive multiplier to capture implicit prior promises to the agent(s) that were made in order to satisfy earlier instances of these constraints. As a result, their method relies on the invertibility of the derivative of the Pareto frontier and cannot be applied to problems for which this frontier is not strictly concave. In this paper we show how one can extend their method to a weakly concave Pareto frontier by expanding the state space to include the realizations of an end of period lottery over the extreme points of a flat region of the Pareto frontier. With this expansion the basic insight of Marcet and Marimon goes through - one can make the problem recursive in the Lagrangian multiplier which yields significant computational advantages over the conventional approach of using utility as the state variable. The case of a weakly concave Pareto frontier arises naturally in applications where the principal's choice set is not convex but where randomization is possible. ## Citations 3 citations in Web of Science® 4 citations in Scopus® ## Altmetrics Detailed statistics Item Type: Journal Article, refereed, original work 03 Faculty of Economics > Department of Banking and Finance 330 Economics C61, C63 English 15 October 2012 21 Jul 2011 10:40 05 Apr 2016 14:57 Elsevier 30 1094-2025 https://doi.org/10.1016/j.red.2012.05.001

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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-7-algebraic-fractions-7-4-addition-and-subtraction-of-algebraic-fractions-and-simplifying-complex-fractions-problem-set-7-4-page-300/61

## Elementary Algebra $\frac{m}{40}$ Assuming a constant velocity, she is completing 1/40 of the course every minute. Thus, multiplying by m minutes, she completes $m/40$ of the course in m minutes. We know that in fourty minutes, she will have completed the whole course, so it makes sense that when we plug in 40 for m, we get one.

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https://dsp.stackexchange.com/questions/49322/instantaneous-frequency-estimation-by-hilbert-transform-theoretical-justificat?noredirect=1

# Instantaneous Frequency Estimation by Hilbert Transform - Theoretical Justification and Proof I would like to better understand why the instantaneous frequency estimation by Hilbert transformation works (and especially why it doesn't work / lead to precise results in many cases). The motivation is to estimate signal $x(t)$ by decomposing it into an amplitude envelope $m(t)$ and phase of cosine $\omega_c (t)$ (or carrier waveform): $$x(t) = m(t) \cos\left(\omega_c (t)\right)$$ Now, assume that $x(t)$ indeed is a result from such a process. Questions: 1) There are two "parameters" to be estimated for any $t$, as such some constraints are needed. What are the constraints regarding $m(t)$ and $\omega_c (t)$ that are selected when applying the Hilbert transform decomposition? 2) Is there a proof available somewhere that given the constraints, the estimation indeed finds the correct amplitude envelope and carrier (for continuous and also discrete case)? • I believe you’d have to assume that $m(t)$ is very slowly varying with respect to the “center” frequency of $\omega_c(t)$ — which is a little mis-named. As you’ve written it, it’s a phase, not a frequency (which is usually what $\omega$ is used for). – Peter K. May 19 '18 at 18:50 See my earlier comments here: Meaning of Hilbert transform Common fractal noise isn't an analytic signal (infinitely differentiable). And a Hilbert transform re-creates the imaginary component of an analytic signal if you have the real component of the complex analytic signal (which one rarely has from real-world data). However, sufficiently band-pass filtered data might be similar to a finite length segment of something that looks like an infinite length analytic signal (e.g. is from a source whose behavior can be modeled or estimated by a 2nd order linear differential equation). • You are giving the definition for an analytic function, which is not identical to that of analytic signal. There is a non-obvious relation between the two, but they are not identical. For example $t\mapsto\exp(-t^2) \sin(t)$ is an analytic function, but it's not an analytic signal. May 19 '18 at 20:01 • I think the OP asked about the actual process of estimating the instantaneous frequency using Hilbert Transform. I think you should point into that in your answer. – Royi Jun 18 '18 at 20:40

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https://www.physicsforums.com/threads/instantaneous-displacement-of-a-sound-wave.794364/

# Instantaneous Displacement of a Sound Wave Tags: 1. Jan 26, 2015 ### Mtscorpion12 1. The problem statement, all variables and given/known data A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x,t) = 2.00cos(15.7x - 858t) where s is in micrometers, x is in meters, and t is in seconds. Find a) the amplitude, b) the wavelength, and c) the speed of this wave. D) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x=0.050 m and t=3.00 ms. E) Determine the maximum speed of the element's oscillatory motion. 2. Relevant equations 3. The attempt at a solution I have already figured out A, B, and C but cannot figure out D or E. I believe D is just to find the derivative and plug in the given X and T, but I do not know how to find the derivative of a three variable function. Also, I believe E just requires setting the derivative equal to 0 and finding when it is a maximum. 2. Jan 26, 2015 ### Nathanael No need; it asks for the instantaneous displacement. The given function s(x,t) gives you the displacement at every position and every time. That would give you the maximum displacement, (not the maximum speed,) but you don't even need to differentiate to find that; it's simply the amplitude. 3. Jan 26, 2015 ### Mtscorpion12 Clearly I overthought this problem way too much. I got it now. Thank you very much. Draft saved Draft deleted Similar Discussions: Instantaneous Displacement of a Sound Wave

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https://mathoverflow.net/questions/99111/orbifold-vs-alexandrove-space-vs-limit-of-manifolds

# Orbifold vs Alexandrove space vs Limit of manifolds I have a question about some defitions : Orbifold, Alexandrov space, limit of manifolds in Gromov-Hausdorff distance sense. Consider following example. Let $r> 0$ $L_c = \{ (x cos \theta, x sin \theta, cx) | 0 \leq x$ and $0 \leq \theta < 2\pi \}$ $S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$ $T$ : $(z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$ If $R$ is sufficiently large, then we have a two dimensional sphere $U_c$ enclosed by $L_c$, $S$, and $T$. First notice the following. $lim_{r \rightarrow 0} U_c$ is an orbifold for some $c$ Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ? If this question is wide, we can restrict to the case of nonnegatively curved orbifolds. : Is a nonnegatively curved $n$-orbifold a limit of positively curved $n$-manifolds ? Question 2: In the following paper, a space with curvature $\geq k$ is defined. M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51. Is a $n$-dimensional space with curvature $\geq k$, which is smooth except finite points, is a limit of $n$-manifolds of positive sectional curvature $\geq k$ ? I believe that this question is trivial and it is true. I do not think that all orbifolds or spaces with curvature $\geq k$ are limits of manifolds. However I can not deny it. Since ${\bf R}^3={\bf R}^4 /S^1 = lim_{k \rightarrow \infty} {\bf R}^4/{\bf Z}_k$, orbifolds are different from spaces with curvature $\geq k$. But they are obtained from the sequences of manifolds. MOTIVATION : Hsiang-Kleiner classified positively curved manifolds with $S^1$-action. I want to extend this result to positively curved orbifolds with $S^1$-action. If orbifold is a limit of manifolds then the problem is simple. Accordingly I want to know the questions. Thank you for your attention. • Every compact length space is a Gromov-Hausdorff limit of two-dimensional Riemannian manifolds. You need to be more precise in your question. Do the approximating manifolds have to have the same lower curvature bound? Same dimension? – Sergei Ivanov Jun 8 '12 at 12:38 • @ IVanov : Thank you. Your suggestion is helpful for me. – Hee Kwon Lee Jun 8 '12 at 13:26 • One technical point: Orbifolds are different animals than topological spaces. An orbifold is a space plus some extra structure. That said, I think that your question is fairly self explanatory (at least in this regard). – Spice the Bird Jun 8 '12 at 19:53 Q1. Note that one oriented orthonormal frame bundle $FO$ over a smooth orbifold $O$ is a smooth manifold. This frame bundle admits a one-parameter family of metrics which collapse to the original manifold and its curvature can be made bounded from below. Proof. Equip $FO$ with $SO(n)$-invariant metric, consider product space $[\varepsilon\cdot SO(n)]\times FO$ and factorize along diagonal action. (This often called Cheeger's trick, but I think it was known before Cheeger.) Q2. It is an open question. It is expected that cones over some positively curved manifolds can not be approximated. Vitali Kapovitch has examples of such $n$ dimesional cones which can not be approximated by $m$-dimensional manifolds with $m< n+8$. By Perelman's Stability Theorem, if a (compact) limit of $n$-dimensional Alexandrov spaces of curvature $\ge k$ has the same dimension, then the convergent spaces are eventually homeomorphic to the limit space. So in this context a limit of manifolds is always a topological manifold. And there exist non-manifold examples of Alexandrov spaces, e.g., $\mathbb R^3/\{id,-id\}$ which is the cone over $\mathbb {RP}^2$. Or, if you want a compact example, take $S^3/\mathbb Z_2$ where $\mathbb Z_2$ acts on $S^3$ with two fixed points (think of $S^3$ lying in $\mathbb R^4$ and consider the reflection in some 1-dimensional axis). This is a compact orbifold of curvature $\ge 1$ and it is not a limit of manifolds of the same dimension with a uniform lower curvature bound.

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https://documentation.aimms.com/language-reference/optimization-modeling-components/automatic-benders-decomposition/benders-decomposition-textbook-algorithm.html

# Benders’ Decomposition - Textbook Algorithm Master problem The basic Benders’ decomposition algorithm as explained in several textbooks (e.g., [NW88], [Mar99]) works as follows. After introducing an artificial variable $$\eta = d^Ty$$, the master problem relaxation becomes: \begin{split}\begin{align} & \text{minimize} & & c^Tx + \eta \\ & \text{subject to} & & A x \leq b & & \\ &&& \eta \geq \overline{\eta} & & \\ &&& x \in \mathbb{Z}^n_+ & & \\ \end{align}\end{split} Here $$\overline{\eta}$$ is a lower bound on the variable $$\eta$$ that AIMMS will automatically derive. For example, if the vector $$d$$ is nonnegative then we know that 0 is a lower bound on $$d^Ty$$ since we assumed that the variable $$y$$ is nonnegative, and therefore we can take $$\overline{\eta} = 0$$. We assume that the master problem is bounded. Subproblem After solving the master problem we obtain an optimal solution, denoted by $$(x^*,\eta^*)$$ with $$x^*$$ integer. This solution is fixed in the subproblem which we denote by $$PS(x^*)$$: \begin{split}\begin{align} & \text{minimize} & & d^Ty \\ & \text{subject to} & & Q y \leq r - Tx^* & & \\ &&& y \in \mathbb{R}^m_+ & & \\ \end{align}\end{split} Note that this subproblem is a linear programming problem in which the continuous variable $$y$$ is the only variable. Dual subproblem Textbooks that explain Benders’ decomposition often use the dual of this subproblem because duality theory plays an important role, and the Benders’ optimality and feasibility cuts can be expressed using the variables of the dual problem. The dual of the subproblem $$PS(x^*)$$ is given by: \begin{split}\begin{align} & \text{maximize} & & r - \pi^T(Tx^*) \\ & \text{subject to} & & \pi^TQ \geq d^T & & \\ &&& \pi \geq 0 & & \\ \end{align}\end{split} We denote this problem by $$DS(x^*)$$. Optimality cut If this subproblem is feasible, let $$z^*$$ denote the optimal objective value and $$\overline{\pi}$$ an optimal solution of $$DS(x^*)$$. If $$z^* \leq \eta^*$$ then the current solution $$(x^*,\eta^*)$$ is a feasible and optimal solution of our original problem, and the Benders’ decomposition algorithm only needs to solve $$PS(x^*)$$ to obtain optimal values for variable $$y$$. If $$z^* > \eta^*$$ then the Benders’ optimality cut $$\eta \geq \overline{\pi}^T (r - Tx)$$ is added to the master problem and the algorithm continues by solving the master problem again. Feasibility cut If the dual subproblem is unbounded, implying that the primal subproblem is infeasible, then an unbounded extreme ray $$\overline{\pi}$$ is selected and the Benders’ feasibility cut $$\overline{\pi}^T (r - Tx) \leq 0$$ is added to the master problem. Modern solvers like CPLEX and Gurobi can provide an unbounded extreme ray in case a LP problem is unbounded. After adding the feasibility cut the Benders’ decomposition algorithm continues by solving the master problem.

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https://www.physicsforums.com/threads/rotational-kinematics-acceleration.138380/

# Rotational kinematics - acceleration 1. Oct 14, 2006 ### physgirl so there's a wheel with radius R and mass M. there's also a hub attached to the wheel's center with radius r and mass m. there's also a mass X suspended from a massless string that's wound around the hub. if the axle has negligible radius and mass and both wheel and the hub are solid with uniform density, how would you find the acceleration of the suspended mass after its released? i thought what it was askng for was the tangential acceleration, which i found to be equal to (F*r^2)/I.... (since tan acc is r*alpha where alpha is r*F/I because torque = I*alpha and also r*F)... so i tried doing: [r^2*X*g]/0.5[M*R+m*r] but that doesn't work and i'm not sure what im doing wrong... can someone point me in the right direction? thanks! 2. Oct 14, 2006 ### Staff: Mentor Two problems: (1) The force F pulling on the hub does not equal the weight of the hanging mass. It does equal the tension in the string. (2) The rotational inertial of a disk is 0.5MR^2. Set up equations (Newton's 2nd law) for both wheel/hub and hanging mass and solve them together to get the acceleration. 3. Oct 14, 2006 ### physgirl 1- isn't the tension of the string the same thing as mg? also, what do you mean by "set up equations... to get the acceleration"? the 2nd law equation for the wheel/hub would be (m+M)alpha=F and for the hanging mass it would be F=Xg?...? i'm lost :( 4. Oct 15, 2006 ### Staff: Mentor No. Think about it: if the tension in the string was equal to mg, then the net force on the mass would be zero. It would just sit there. (This is what would happen if you hung the mass from a string that was fixed to the roof, say.) But since that string moves, the tension will be less than mg. Here are the equations you need: (a) Torque = I alpha ==> Fr = I_total*alpha = I_total*a/r (b) Xg - F = Xa Note: We know that the acceleration is down, so I take down to be positive. 5. Oct 15, 2006 ### physgirl I see, this is probably a dumb question but why did you go from Fr=I_total*alpha => F = I_total*a/r (that is, alpha to a?) and if you didn't mean to switch over... I know I'm supposed to be solving for acceleration, so how do i know what alpha is? lastly... I don't really understand the concept of "I"... the mass and radius of what object is supposed to be involved...? is it everything that's involved in the whole system? or just the wheel that's actually doing the turning? or just the hub that the string is directly attached to? or both? 6. Oct 15, 2006 ### Staff: Mentor Why? Because we are trying to find "a", not alpha. Since the wheel/hub is connected to the hanging mass via the string, alpha can be related to "a" via: a = alpha*r (where r is the radius of the hub that the string wraps around). They are directly related. (See above.) The whole thing turns as one piece, so you must use the "I" for the entire wheel/hub object--which is just the sum of I_wheel and I_hub. Similar Discussions: Rotational kinematics - acceleration

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https://www.zora.uzh.ch/id/eprint/92213/

# Search for the standard model Higgs boson produced in association with a W or a Z boson and decaying to bottom quarks CMS Collaboration; Chatrchyan, S; Khachatryan, V; Sirunyan, A M; et al; Chiochia, V; Kilminster, B; Robmann, P (2014). Search for the standard model Higgs boson produced in association with a W or a Z boson and decaying to bottom quarks. Physical Review D (Particles, Fields, Gravitation and Cosmology), 89(1):012003. ## Abstract A search for the standard model Higgs boson (H) decaying to bb¯ when produced in association with a weak vector boson (V) is reported for the following channels: W(μν)H, W(eν)H, W(τν)H, Z(μμ)H, Z(ee)H, and Z(νν)H. The search is performed in data samples corresponding to integrated luminosities of up to 5.1 inverse femtobarns at s√=7  TeV and up to 18.9  fb−1 at s√=8  TeV, recorded by the CMS experiment at the LHC. An excess of events is observed above the expected background with a local significance of 2.1 standard deviations for a Higgs boson mass of 125 GeV, consistent with the expectation from the production of the standard model Higgs boson. The signal strength corresponding to this excess, relative to that of the standard model Higgs boson, is 1.0±0.5. ## Abstract A search for the standard model Higgs boson (H) decaying to bb¯ when produced in association with a weak vector boson (V) is reported for the following channels: W(μν)H, W(eν)H, W(τν)H, Z(μμ)H, Z(ee)H, and Z(νν)H. The search is performed in data samples corresponding to integrated luminosities of up to 5.1 inverse femtobarns at s√=7  TeV and up to 18.9  fb−1 at s√=8  TeV, recorded by the CMS experiment at the LHC. An excess of events is observed above the expected background with a local significance of 2.1 standard deviations for a Higgs boson mass of 125 GeV, consistent with the expectation from the production of the standard model Higgs boson. The signal strength corresponding to this excess, relative to that of the standard model Higgs boson, is 1.0±0.5. ## Statistics ### Citations Dimensions.ai Metrics 197 citations in Web of Science® 175 citations in Scopus®

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https://www.math.princeton.edu/events/quantum-oracle-classification-case-group-structure-2016-12-12t210012

# Quantum Oracle Classification: The Case of Group Structure - Mark Zhandry, Princeton University Fine Hall 214 The Quantum Oracle Classification (QOC) problem is to classify a function, given only quantum black box access, into one of several classes without necessarily determining the entire function. Generally, QOC captures a very wide range of problems in quantum query complexity. However, relatively little is known about many of these problems. In this work, we analyze the a subclass of the QOC problems where there is a group structure. That is, suppose the range of the unknown function A is a commutative group G, which induces a commutative group law over the entire function space. Then we consider the case where A is drawn uniformly at random from some subgroup A of the function space. Moreover, there is a homomorpism f on A, and the goal is to determine f(A). This class of problems is very general, and covers several interesting cases, such as oracle evaluation; polynomial interpolation, evaluation, and extrapolation; and parity. These problems are important in the study of message authentication codes in the quantum setting, and may have other applications. We exactly characterize the quantum query complexity of every instance of QOC with group structure in terms of a particular counting problem. That is, we provide an algorithm for this general class of problems whose success probability is determined by the solution to the counting problem, and prove its exact optimality. Unfortunately, solving this counting problem in general is a non-trivial task, and we resort to analyzing special cases. Our bounds unify some existing results, such as the existing oracle evaluation and parity bounds. In the case of polynomial interpolation and evaluation, our bounds give new results for secret sharing and information theoretic message authentication codes in the quantum setting.

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https://www.physicsforums.com/threads/how-to-define-the-b-c-s-for-the-em-field-of-a-perfectly-contucting-surface.391735/

How to define the b.c.s for the EM field of a perfectly contucting surface? 1. Apr 2, 2010 sith If you have a 3 dimensional perfectly conducting body the conditions at the boundary for the EM field is as follows: $$\boldsymbol{E}_{\parallel} = 0, B_{\perp} = 0, E_{\perp} = \frac{\sigma}{\epsilon_0}, \boldsymbol{B}_{\parallel} = \mu_0 \boldsymbol{j} \times \boldsymbol{\hat{n}}$$ where $$\sigma$$ and $$\boldsymbol{j}$$ are the surface charge and current density at the boundary respectively. $$\parallel$$ / $$\perp$$ denotes the component parallel/perpendicular to the surface of the body, with normal vector $$\boldsymbol{\hat{n}}$$. The derivations of these boundary conditions comes from assuming that the EM field vanishes inside the conductor and using Maxwell's equations. But when one assume that the body is a planar conducting surface, there is no longer a finite volume in which the EM field vanish. If you for instance take a conducting box, and then study the limit where the height goes to zero, then the top and bottom surface charge and current densities can no longer be separated, and they are unified in one overall surface carge and current density. Can one still assume that these conditions are true, or should they be modified in some way? Could it be possible that these conditions are instead only valid for the differences between the fields on each side of the surface? Last edited: Apr 2, 2010 2. Apr 2, 2010 clem If it is a perfect conductor, then the BC hold for thickness-->0. If it has a very high conductivity, then the "surface current" and the E and B fields will vanish exponentially. Then as the thickness approaches the skin depth, the BC would get modified.

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http://physics.stackexchange.com/questions/117407/relating-energy-to-wavelength-in-curved-space

# Relating Energy to Wavelength in curved space Consider a curved space, e.g. Schwarzschild: \begin{align*} ds^2 = -\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2 \end{align*} Now, the energy of a photon is $E = \hbar \omega$, and $|\mathbf{k}|= \frac{2\pi}{\lambda}$, but am I correct in assuming that $\omega \neq |\mathbf{k}|$? Because if $k^\mu = (\omega,\mathbf{k})$ then $k_\mu k^\mu = 0$ implies that: \begin{align*} g_{tt} \omega^2 + g_{rr}(k^1)^2+g_{\theta\theta}(k^2)^2+g_{\phi\phi}(k^3)^2 =0 \end{align*} So basically, is it correct that the relationship between $\omega$ and $|\mathbf{k}|$ will vary in curved space? (And so relationships like $E = \frac{h}{\lambda}$ no longer hold?) - The main thing to note is that the definition of the wavenumber you cite above is dependent on the underlying function satisfying the standard wave equation, because any function that satisfies $$\eta^{ab}\partial_{a}\partial_{b}\phi(x) = 0$$ Will have its Fourier transform, $\Phi(k) = \int d^{4}x e^{ik^{a}x_{a}}\phi(x)$ satisfy $$k^{a}k_{a}\Phi(k) = 0$$ But this is no longer true, because for the case of curved spacetime, the wave equation is $$g^{ab}\partial_{a}\partial_{b}\phi(x) - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi(x) = 0$$ And this will require some modification to the Fourier transform to work. More physically, you have effects like gravitational lensing that cause light to interact with the gravitational field, so you don't get simple straight-line propogation of monochromatic modes. Note, however, that it is always possible to locally transform to a coordinate system where $\Gamma_{ab}{}^{c} =0$ and $g_{ab} = \eta_{ab}$, and there, you will be able to have a well defined wavenumber and frequency which satisfies $\omega^{2} = k^{i}k_{i}$. This just won't work outside of your local neighborhood. - $\Gamma_{ab}{}^c$ can in general only be made 0 at a point, not in a neighborhood. The latter is possible iff space is flat in that neighborhood. –  Robin Ekman Jun 8 at 16:20 @RobinEkman: you're right, of course. In my mind, I always imagine this as taking some tolerance for the size of the christoffel symbols, and adjusting the size of the neighborhood so that the christoffel symbols are below this tolerance, and then imagine spacetime as "approximately/essentially" flat in this region -- so that my picture of the equivalence principle is a little more physical than just talking about a single point, which is obviously not realizable physically. –  Jerry Schirmer Jun 8 at 16:31 but I shouldn't use my personal language like that when communicating with the public. –  Jerry Schirmer Jun 8 at 16:43 I would think what you are saying is correct, the dispersion relation for light in vacuum $\omega = kc$ is derived by considering plane wave solutions to Maxwell's equations $\partial_{\mu} F^{\mu \nu}=0$. For curved spacetime, you would have to replace the usual derivative by a covariant derivative. -

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https://paramanands.blogspot.com/?widgetType=BlogArchive&widgetId=BlogArchive1&action=toggle&dir=open&toggle=YEARLY-1293820200000&toggleopen=MONTHLY-1467311400000

# The General Binomial Theorem: Part 2 In the previous post we established the general binomial theorem using Taylor's theorem which uses derivatives in a crucial manner. In this post we present another approach to the general binomial theorem by studying more about the properties of the binomial series itself. Needless to say, this approach requires some basic understanding about infinite series and we will assume that the reader is familiar with ideas of convergence/divergence of an infinite series and some of the tests for convergence of a series. # The General Binomial Theorem: Part 1 ### Introduction One of most basic algebraic formulas which a student encounters in high school curriculum is the following $$(a + b)^{2} = a^{2} + 2ab + b^{2}$$ and its variant for $(a - b)^{2}$. And after many exercises and problems later one encounters another formula of similar nature namely $$(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$$ and one wonders if there are similar formulas for higher powers of $(a + b)$. # Theories of Circular Functions: Part 3 Continuing our journey from last two posts we present some more approaches to the development of the theory of circular functions. One approach is based on the use of infinite series and requires basic knowledge of theory of infinite series. This approach is particularly well suited for treating circular functions as functions of a complex variable, but we will limit ourselves to the case of real variables only.

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https://science.purdue.edu/xenon1t/?tag=background

XENON1T presented at Rencontres de Moriond Electroweak Last week I had the opportunity to present the XENON1T experiment at the Recontres de Moriond electroweak conference in La Thuile Italy in the beautiful Aosta Valley. This meeting is one of the most important meetings for LHC physics, but has slowly expanded to encapsulate a variety of topics, including the hunt for dark matter. The conference program and slides are available on indico. The XENON1T presentation focused on our dark matter search results from last spring as well as the upcoming result using about a factor of 10 more exposure, which is under intense preparation for release. The whole presentation is available from the indico page but here is one slide from it: Here we discuss how we were able to increase the amount of liquid xenon we use for our dark matter search from ~1000kg to ~1300kg. The top left plot shows an example larger search volume (red) compared to the smaller volume used for the first result. But it’s not so simple as just adding volume. While our inner detector is completely free of WIMP-like background, the outer radii contain background components that can mimic WIMPs. This is illustrated in the bottom right plot where the background-free inner volume (right) is contrasted with the full search volume containing the outer radial sections (left). The full volume has a contribution from PTFE (Teflon) surface background (green contour and points) that is absent as soon as we consider only the inner volume. Our statistical interpretation has been updated so it is smart enough to take this into account. We parameterize our entire search region in both radial and spatial dimensions with expected signal and background distributions described at each location. This allows us to fully exploit the sensitivity of our innermost background-free volumes while also gaining a modest improvement from the outermost ones. The energy spectrum and resolution of XENON1T The search for new physics with a large underground xenon detector is like listening to your favorite song in a quiet room with high end headphones for the first time. Even if you have listened to the song a thousand times, you will be surprised by all the small nuances that have been there all along and that you did not hear before. This is either because it was too loud around you or because your headphones were not good enough. The quiet room in this analogy is the xenon detector that has been made from materials selected for their ultra-low radioactivity and that is shielded by a water tank, a mountain and ultimately the xenon in the detector itself. The high end headphones on the other hand are the extremely sensitive photomultipliers, data acquisition system and tailor-made software to read out the signals produced by particles interacting inside the detector. As you may have read before on this blog (we love to point this out…) XENON1T is the lowest background dark matter detector in the world. But the fact that the detector is so quiet does not mean that it does not measure anything. As a very sensitive instrument it is able to detect even the faintest signals from radioactive decays in the detector materials or the xenon itself. Over the course of one year these decays amount to a sizeable amount of data. The picture below shows what this looks like. A preliminary energy spectrum from electronic recoil background data for the second science run of the XENON1T experiment. The x-axis denotes the energies of particles measured with the XENON1T detector. These are mostly electrons, x-rays and higher energy -rays. The y-axis shows how many of these particles have been counted over the whole measurement time of the last science run of the experiment. In order to have a better comparability with similar experiments, the event count has been divided by the live time of the experiment, its mass and the step size on the energy axis (the binning) in which we count. One can see that even in the highest peaks we measure less than one event per kilogram detector material and day of measurement time in a 100 keV energy window. A quiet room, indeed. And the features in the spectrum are all those nuances that one could not see before. So what are they? One can divide the spectrum into several regions. Only the small portion of data in the very left of the plot next to the first grey-shaded region is relevant to the standard dark matter search. The heavy and non-relativistic WIMP is expected to only deposit very little energy, so it resides here. The following grey region is blinded, which means it has deliberately been made inacessible to XENON analysers. The reason for this is that it might contain traces of a rare nuclear decay of Xe-124, the two neutrino double electron capture, that has not been observed until now, and we do not want to bias ourselves in looking for it. The large region from about 100-2300 keV contains multiple peaks. Each of these peaks belongs to a monoenergetic -line of a radioactive isotope contained in the detector materials or the extremely pure xenon itself. One can easily see that the peaks are sitting on an irregular continuous pedestal. This is created by -rays depositing only part of their total energy due to Compton scattering inside or outside the detector, decays of radioisotopes inside the detector, and the two neutrino double -decay of Xe-136. The latter produces a continuous energy spectrum over the whole energy range that ends at 2458 keV. The decay is rare, but becomes relevant due to the large amount of Xe-136 in the detector and the relative smallness of other background contributions. Xe-136 is also responsible for the second gray-shaded region at high energies which might contain an experimental signature of its neutrinoless double -decay. This hypothetical decay mode would produce a monoenergetic line centered at the end of the aforementioned spectrum at 2458 keV. The observation of this decay would be a gateway to new physics and complements the physics program of XENON1T. As their signatures have to be distinguished from other background components the energy resolution of the detector becomes crucial. Preliminary energy resolution of the XENON1T experiment as a function of the measured particle energy. To grasp the concept of energy resolution one can imagine the following situation in the energy spectrum. If you have two peaks next to one another, one your sought-after signal and one a pesky background, how far do they have to be apart in order to be seen as individual peaks? This of course relies on how wide they are. Thus, the energy resolution in XENON1T is characterized by the width of peaks in the energy spectrum relative to their measured energy. By fitting Gaussian functions to all the peaks in the spectrum at the top one obtains the ratio of peak width to peak center. This is what the above plot shows for several liquid xenon dark matter experiments. One can see that with an increase in particle energy the resolution improves. It is also evident that XENON1T leads the pack over a wide energy range. This is underlines that XENON1T is the astroparticle physics equivalent of high-end headphones. With these the XENON collaboration is in the position to pursue several exciting physics channels apart from weakly interacting massive particles. So stay tuned for the analyses to come. Intrinsic backgrounds from Rn and Kr in the XENON100 experiment XENON1T is currently the largest liquid xenon detector in the search for dark matter. To fully exploit the capabilities of the ton-scale target mass, a thorough understanding of radioactive background sources is required. In this paper we use the full data of the main science runs of the XENON100 experiment that were taken over a period of about 4 years to asses the target-intrinsic background sources radon (Rn-222), thoron (Rn-220) and krypton (Kr-85). We derive distributions of the individual radionuclides inside the detector (see Figure below) and quantify their abundances during the main three science runs. We find good agreement with external measurements of radon emanation and krypton concentrations, and report an observed reduction in concentrations of radon daughters that we attribute to the plating-out of charged ions on the negatively biased cathode. The preprint of the full study is available on arXiv:1708.03617. Figure: Spatial distributions of the various radon populations identified in XENON100. XENON1T, the most sensitive detector on Earth searching for WIMP dark matter, releases its first result [Press Release May 2017 – for immediate release. Preprint is on the arxiv] The best result on dark matter so far! … and we just got started!”. This is how scientists behind XENON1T, now the most sensitive dark matter experiment world-wide, hosted in the INFN Laboratori Nazionali del Gran Sasso, Italy, commented on their first result from a short 30-day run presented today to the scientific community. XENON1T installation in the underground hall of Laboratori Nazionali del Gran Sasso. The three story building on the right houses various auxiliary systems. The cryostat containing the LXeTPC is located inside the large water tank on th left, next to the building. (Photo by Roberto Corrieri and Patrick De Perio) Dark matter is one of the basic constituents of the Universe, five times more abundant than ordinary matter. Several astronomical measurements have corroborated the existence of dark matter, leading to a world-wide effort to observe directly dark matter particle interactions with ordinary matter in extremely sensitive detectors, which would confirm its existence and shed light on its properties. However, these interactions are so feeble that they have escaped direct detection up to this point, forcing scientists to build detectors that are more and more sensitive. The XENON Collaboration, that with the XENON100 detector led the field for years in the past, is now back on the frontline with the XENON1T experiment. The result from a first short 30-day run shows that this detector has a new record low radioactivity level, many orders of magnitude below surrounding materials on Earth. With a total mass of about 3200kg, XENON1T is at the same time the largest detector of this type ever built. The combination of significantly increased size with much lower background implies an excellent dark matter discovery potential in the years to come. Scientists assembling the XENON1T time projection chamber. (Photo by Enrico Sacchetti) The XENON Collaboration consists of 135 researchers from the US, Germany, Italy, Switzerland, Portugal, France, the Netherlands, Israel, Sweden and the United Arab Emirates. The latest detector of the XENON family has been in science operation at the LNGS underground laboratory since autumn 2016. The only things you see when visiting the underground experimental site now are a gigantic cylindrical metal tank, filled with ultra-pure water to shield the detector at his center, and a three-story-tall, transparent building crowded with equipment to keep the detector running, with physicists from all over the world. The XENON1T central detector, a so-called Liquid Xenon Time Projection Chamber (LXeTPC), is not visible. It sits within a cryostat in the middle of the water tank, fully submersed, in order to shield it as much as possible from natural radioactivity in the cavern. The cryostat allows keeping the xenon at a temperature of -95°C without freezing the surrounding water. The mountain above the laboratory further shields the detector, preventing it to be perturbed by cosmic rays. But shielding from the outer world is not enough since all materials on Earth contain tiny traces of natural radioactivity. Thus extreme care was taken to find, select and process the materials making up the detector to achieve the lowest possible radioactive content. Laura Baudis, professor at the University of Zürich and professor Manfred Lindner from the Max-Planck-Institute for Nuclear Physics in Heidelberg emphasize that this allowed XENON1T to achieve record “silence”, which is necessary to listen with a larger detector much better for the very weak voice of dark matter. The spin-independent WIMP-nucleon cross section limits as a function of WIMP mass at 90% confidence level (black) for this run of XENON1T. In green and yellow are the 1- and 2σ sensitivity bands. Results from LUX (red), PandaX-II (brown), and XENON100 (gray) are shown for reference. A particle interaction in liquid xenon leads to tiny flashes of light. This is what the XENON scientists are recording and studying to infer the position and the energy of the interacting particle and whether it might be dark matter or not. The spatial information allows to select interactions occurring in the central 1 ton core of the detector. The surrounding xenon further shields the core xenon target from all materials which already have tiny surviving radioactive contaminants. Despite the shortness of the 30-day science run the sensitivity of XENON1T has already overcome that of any other experiment in the field, probing un-explored dark matter territory.  “WIMPs did not show up in this first search with XENON1T, but we also did not expect them so soon!” says Elena Aprile, Professor at Columbia University and spokesperson of the project. “The best news is that the experiment continues to accumulate excellent data which will allow us to test quite soon the WIMP hypothesis in a region of mass and cross-section with normal atoms as never before. A new phase in the race to detect dark matter with ultra-low background massive detectors on Earth has just began with XENON1T. We are proud to be at the forefront of the race with this amazing detector, the first of its kind.” As always, feel free to contact the XENON collaboration at contact@xenon1t.org. Material radioassay and selection for XENON1T To attain the high sensitivity needed to detect a dark matter particle with a xenon time-projection chamber, all other sources of particle interactions need to be eliminated or minimized. These interactions are classified as background events. Radiogenic backgrounds, in particular, come from radioactive isotopes within the detector materials that decay and lead to alpha, beta, or gamma emissions. Neutrons from spontaneous fission of heavy isotopes or from secondary reactions within the detector materials also contribute to the radiogenic background and can mimic a dark matter signal. To minimize the radiogenic background, the goal of the XENON1T radioassay program is to measure the radioactivity of all materials that are needed to build the detector and to select only the most radiopure materials for the final construction. To do this, we use mass spectrometry techniques and high-purity germanium spectrometers that are capable of measuring radioactivity at the level of 10-6 decays per second in a kilogram of material (Bq/kg). As comparison, a typical banana has an activity of ~102 Bq/kg! Because natural radioactivity is present in the soil, the water, and in the air, it is also present in the XENON detector materials. The Figure shows a measurement obtained with a germanium spectrometer of the gamma rays emitted from a sample of photomultiplier tubes. The background (purple spectrum) is subtracted from the sample (pink spectrum) in order to quantify the expected activity from a XENON1T component or material sample. A high-purity germanium spectrometer measurement of gamma rays emitted from a sample of XENON1T photosensors. Some prominent isotopes from different sources are labeled: primordial uranium and thorium decay chains (green), potassium (red), man-made (orange) and cosmogenic (orange) isotopes. The most common radioactive isotopes present in the Earth are primordial uranium and thorium, each of which decays into a series of other radioactive isotopes (marked in green in the Figure). Potassium (red) is also a common, primordial isotope that is found in soil, and subsequently in food and in your body. Other isotopes that are found in detector materials come from interactions with cosmic rays (yellow) or from man-made activities (blue), i.e. industrial or medical use, nuclear power plant emissions, nuclear accidents, and military testing. The measured activities of each material selected for detector construction are used in simulations of XENON1T to determine the expected background. This allows for a prediction of the attainable sensitivity of the detector to dark matter interactions. The radioassay measurement results from over 100 material samples are presented in our new paper “Material radioassay and selection for the XENON1T dark matter experiment”. XeSAT2017: Online krypton and radon removal for the XENON1T experiment This talk by Michael Murra (slides) was presented at the XeSAT2017 conference in Khon Kaen, Thailand, from 3. – 7. April 2017. The  main background for the XENON1T experiment are the intrinsic contaminants krypton and radon in the xenon gas. Instead of purifying the xenon once before starting the science run we were able to operate our distillation column in a closed loop with the XENON1T detector system running during its commissioning phase. This resulted into reducing the krypton concentration quickly below 1 ppt (parts per trillion, 1 ppt = 10^(-12) mol/mol) without emptying and refilling of the detector. In addition, the column was operated in the same closed loop in inverse mode in order to reduce Rn-222 by about 20% during the first science run. This so-called online removal for both noble gases along with the working principle of the distillation system are presented within this talk. Water Tank Filling We started to fill the water tank: In this view from the top, the cryostat with the actual detector is visible on the left. Photomultiplier tubes of the water Cherenkov muon veto are seen at the bottom and side of the water tank, to the right of the image. The water acts as a passive shielding against external radioactivity. In addition, using the photomultipliers that can be seen towards the right of the picture, the water acts as an active muon detector. Muons may induce events in the xenon detector that may mimic dark matter signals. We therefore turn a blind eye (“veto”) for a short time whenever a muon travels through the water tank. Lowering the radioactivity of the XENON1T photosensors E. Aprile et al (XENON Collaboration), Lowering the radioactivity of the XENON1T photosensors, arXiv:1503.07698, Eur. Phys. J. C75 (2015) 11, 546. The XENON1T experiment employs 242 photomultiplier tubes (PMTs) in the time projection chamber, arranged into two circular arrays. Because the overall background goal of the detector is incredibly low, with less than 1 expected event in a tonne of liquid xenon and one full year of data, the PMTs must be made out of ultra-pure materials. These materials were selected for their content in traces of 238-U, 232-Th, 40-K, 60-Co, 137-Cs and other long-lived radionuclides. The XENON collaboration joined efforts with Hamamatsu to produce a photosensor that meets the strict requirements of our experiment. The sensor is a 3-inch diameter tube that operates stably at -100 C and at a pressure of 2 atmospheres. It has a high quantum efficiency, with a mean around 35%, for the xenon scintillation light at 178 nm and 90% photon collection efficiency. The sensor, shown schematically in the left picture, features a VUV-transparent quartz window, with a low-temperature bi-alkali photocathode deposited on it. A 12-dynode electron multiplication system ensures a signal amplification of ~3 millions, which is a crucial feature to detect the tiny signals induced by the rare collisions of dark matter particles with xenon nuclei. Before the tubes were ready to be manufactured, the construction materials were inspected with gamma-ray spectroscopy and glow-discharge mass spectroscopy (GDMS). For the former, we employed the world’s most sensitive high-purity germanium detectors, GeMPI and Gator, operated deep underground at the Gran Sasso Laboratory. GDMS can detect trace impurities in solid samples and the results were compatible with those from germanium screening. We measured many samples to select the final materials for the PMT production. As an example, specific 226-Ra activities around or below 0.3 mBq/PMT were seen in most of the inspected materials. Such an activity corresponds to 3 x 10-4 226-Ra decays per second and tube, or about 26 decays per day. The relative contribution of the selected materials to the trace contaminations in U, Th, K, Co and Cs of the final product, seen in the left picture, also tells us how to improve further sensor versions for the XENONnT upgrade. Most of the nuclides in the 238-U and 232-Th chains, especially dangerous for their emission of alpha particles, that can the produce fast neutrons in (alpha,n) reactions, are located in the ceramic stem of the tube. In consequence, finding a new material to replace the ceramic might drastically improve the background expectations. Once the final production started, and the tubes were delivered in several batches to our collaboration, they were measured in the Gator detector. Its inner chamber can accommodate 15 PMTs at a time, as seen in the left picture. Each batch was screened for about 15 days, and theobserved activities were mostly consistent from batch to batch. For all measured PMTs, we obtain contaminations in uranium and thorium below 1 mBq/PMT. While 60-Co was at the level of 0.8 mBq/PMT, 40-K dominates the gamma activity with about 13 mBq/PMT. The information from screening was considered in the final arrangement of the PMTs in the XENON1T arrays. PMTs with somewhat higher activities are placed in the outer rings, where they are more distant from the central, fiducial xenon region of the detector. The average activities per PMT of all trace isotopes served as input contaminations to a full Monte Carlo simulation of the expected backgrounds in XENON1T. The results show that the PMTs will provide about 1% and 6% of the total electronic and recoil background of the experiment, respectively. We can therefore safely conclude that the overall radioactivity of the sensors is sufficiently low, and they will certainly not limit the dark matter sensitivity of the XENON1T experiment. Measuring Kr Contamination with an Atom Trap Prof. Elena Aprile and Graduate Student Luke Goetzke work on the ATTA system at Columbia University The Krypton Problem One of the many advantages of using xenon as a dark matter target is that xenon has no naturally occurring long-lived radioactive isotopes. However, when xenon is distilled from air, about 1 krypton atom per billion xenon atoms is also gathered. A very small fraction of these krypton atoms, only one in one hundred billion, are the radioactive isotope 85-Kr. The decay of 85-Kr releases an electron which can then scatter in the xenon detector. These electronic recoil events can potentially obscure even rarer signals from interactions with dark matter. Thus, for dark matter detectors using liquid xenon, the krypton needs to be removed. This is done by passing the xenon through a cryogenic distillation column specifically designed for removing krypton. After going through the krypton column, the xenon is very clean. For XENON100, there are only ~10 krypton atoms per trillion xenon atoms. Finding one of those krypton atoms is like picking out one single star from the entire Milky Way galaxy. XENON1T has 10 times even less krypton in the xenon. Measuring the Krypton Contamination Measuring such a tiny amount of krypton is not trivial. One way is to look for the decay signature of 85-Kr using the XENON detector itself. However, due to its relatively long half life (~11 years), it takes many months to get an accurate estimate with this method. So, how do we measure the tiny amount of krypton relatively quickly and accurately? An atom trapping device has has been developed by the group at Columbia University to do exactly that (see E. Aprile, T. Yoon, A. Loose, L. W. Goetzke, and T. Zelevinsky, “An atom trap trace analysis system for measuring krypton contamination in xenon dark matter detectors”, Rev. Sci. Instrum., 84, 093105 (2013), arXiv:1305.6510). The method, called Atom Trap Trace Analysis (ATTA), was originally developed at Argonne National Lab for the purpose of radioactive dating. It has been adapted to measure samples of xenon gas taken directly from the XENON detectors. All ATTA devices have the same operating principle: traditional laser cooling and trapping techniques are employed to selectively cool and trap the element of interest present in the sample. The trapped atoms emit light which is detected by a photo detector, in our case an avalanche photodiode. The trapped atoms can thus be counted. The Columbia ATTA device is designed to be sensitive to single trapped atoms, since for clean samples the average number of krypton atoms in the trap at any given time is close to zero. The rate at which the atoms are loaded into the trap is the number we are after. The device is calibrated carefully in order to find the trapping efficiency, i.e. the fraction of krypton atoms that get trapped and counted successfully. Multiplying the measured loading rate for a given sample by the known trapping efficiency gives the total number of krypton atoms flowing through the system. Finally, measuring how many xenon atoms flow through the system at the same time allows the krypton fraction to be calculated. The entire measurement can be completed in one working day. The Columbia ATTA device allows the xenon used in XENON1T to be assayed for krypton contamination quickly and accurately, thus ensuring that krypton levels are safe before beginning a dark matter run, and during the run itself. And it looks pretty cool, too! The neutron background of the XENON100 dark matter search experiment In order to search for dark matter, it is imperative that background signals in particular from neutrons are well under control. We describe the successful techniques and leading results from our efforts in a dedicated publications: E. Aprile et al. (XENON100), The neutron background of the XENON100 dark matter search experiment, arXiv:1306.2303. The paper is also published in Journal of Physics G 40 (2013), 115201.

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https://www.investopedia.com/ask/answers/061715/how-do-i-calculate-standard-error-using-matlab.asp

It's possible to calculate the standard error in MATLAB by running a one-line command. MATLAB is a programming platform from MathWorks that's designed for and used by scientists and engineers. ## What Is Standard Error? In statistics, the standard error is the standard deviation of the sampling statistical measure, and it's most commonly used for the sample mean. The standard error measures how accurately the sample represents the actual population from which the sample was drawn. Since there could be different samples drawn from the population, there exists a distribution of sampled means. The standard error measures the standard deviation of all sample means drawn from the population. The formula for calculating the standard error of the mean is the sample standard deviation divided by the square root of the sample size. ## The Command for Standard Error in MATLAB To calculate the standard error of the mean in a sample, the user needs to run a one-line command in MATLAB:  \begin{aligned} &\text{stderror} = \text{ std( data ) / sqrt( length( data ))}\\ &\textbf{where:}\\ &\text{data} = \text{An array with sample values}\\ &\text{std} = \text{The MATLAB function that computes standard} \\ &\text{deviation of the sample}\\ &\text{sqrt} = \text{The MATLAB function that computes the square} \\ &\text{root of a non-negative number}\\ &\text{length} = \text{The MATLAB function that computes the total} \\ &\text{number of observations in the sample}\\ \end{aligned} ## Example of Calculating Standard Error in MATLAB Consider a sample of annual household incomes drawn from the general population of the United States. The sample contains five observations and consists of values $10,000,$100,000, $50,000,$45,000 and $35,000. First, the user needs to create an array called "data" containing these observations in MATLAB. Next, the user can calculate the standard error of the mean with the command "stderror = std( data ) / sqrt( length )". The result of this command says that the mean of this sample, which is$48,000, has a standard error of \$13,161.

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https://destevez.net/2018/11/measuring-the-stability-of-a-tcxo/

# Measuring the stability of a TCXO Lately, I have been testing the GPSDO that I will use to discipline my Es’hail 2 groundstation. One of the tests I have done is to measure the frequency of the TCXO that I use in my Hermes-Lite 2.0beta2 over a few days. Here I show the details of the measurement process and how to process the data in Python. The GPSDO in question is a DF9NP 10MHz GPSDO that I use to drive a 27 MHz PLL, also made by Dieter DF9NP. The 27MHz signal is used as an external reference for a Ku-band LNBF. Since the 27MHz signal is quite strong, it leaks considerably into my Hermes-Lite HF transceiver. By measuring the frequency of this signal, as received in the Hermes-Lite, I can derive the frequency of the TCXO used in the Hermes-Lite. The TCXO in question is an Abracon ASTXR-12-38.400MHZ-514054-T, which is rated for a stability of 500ppb over a temperature range of -40 to 85ºC. To measure the frequency of the 27MHz signal, I have used WSJT-X 2.0.0-rc3 in frequency measurement mode. This produces frequency measurements with a measurement interval of a few seconds (for some reason the measurement interval is irregular, usually ranging from 2 to 5 seconds). The measurements are stored into a text file called fmt.all. I have made a Jupyter notebook to read this file and do some plots. The notebook can be found in fmt.ipynb, together with the fmt.all file I have used. I am using xarray, a library I have discovered recently and which provides a good way to handle most of the data I’m typically using. The plot below shows the frequency offset in ppb of the 38.4MHz TCXO. The measurements have been done over the course of almost eight days, from 2018-11-06 to 2018-11-14. The graph shows some daily variations, mostly due to temperature variations in the house. An overall decreasing trend is also observed. The rate of change is about 1.8 ppb/day. As I have already mentioned in the past, crystals usually go down in frequency when they age, due to mass transfer. Back in August 2017, when I did that experiment, I measured the frequency of my 38.4MHz TCXO to be 110ppb low in frequency. Approximately 450 days have passed since then, which accounts for an average rate of 1.14 ppb/day. This agrees to some extent with the value of 1.8 ppb/day I’ve eyeballed on the graph above. The daily variations are best studied in the graph below, which plots the frequency for each day with a different colour. The local time in Spain this time of the year is UTC + 1. The frequency variations are mostly due to the temperature changes as the house heating comes on and off during the day. Between 22:00 UTC and 6:00 UTC the frequency is quite stable, because the heating is off and people are sleeping. Around 6:30 the heating system comes on (later on days 10 and 11, which were the weekend) and the temperature and the TCXO frequency raises during the morning. Throughout the day one can see some repetitive patterns that most likely correspond to the usual human activity in the house and the cycles used by the heating system control to keep the house temperature as programmed. I find it quite interesting to see all this in the frequency of a simple crystal. Before you ask, I have no clue what happened on day 10 at 18:00 UTC. The next graph is the same as above, but with the daily average removed. This counteracts most of the decreasing trend and shows much better the daily variations, which are quite consistent in the morning, but not so consistent in the afternoon. Finally, I have computed the Allan deviation, which is a one-liner function using the xarray resample function. The deviation is as expected for a TCXO. It is around $$10^{-10}$$ for $$\tau$$ between 10 and 100 seconds, owing to the good short-term stability of the TCXO. Then it starts increasing for higher $$\tau$$s, due to temperature variations and crystal ageing. This site uses Akismet to reduce spam. Learn how your comment data is processed.

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http://mathhelpforum.com/math-topics/39893-summing-series-2-questions.html

# Math Help - Summing Series -2 questions. 1. ## Summing Series -2 questions. Hey, I've got 2 unsolved problems that need help! 1. Find the sum of the multiples of 7 which are less than 10000. My answer is 714264285 but the book says 7142142? 2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n. A step by step guide would be really appreciated for this one. 2. Hello, Originally Posted by Nyoxis Hey, I've got 2 unsolved problems that need help! 1. Find the sum of the multiples of 7 which are less than 10000. My answer is 714264285 but the book says 7142142? There are 1428 multiples of 7 between 1 and 10000, because $\frac{10000}{7} \approx 1428.571428571429$. Therefore, you can write the sum this way : $S=7 \cdot 1+7 \cdot 2+7 \cdot 3+\dots+7 \cdot 1427+7 \cdot 1428$ $S=7 \left(1+2+\dots+1427+1428\right)$ What's in brackets is the sum of the 1428 first integers. We know that $1+\dots+n=\frac{n(n+1)}{2}$ So here, $1+2+\dots+1428=\frac{1428 \cdot 1429}{2}=\boxed{1020306}$ Hence $S=7 \cdot 1020306=\boxed{7142142}$ How did you find 714264285 ? 3. Originally Posted by Nyoxis 2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n. A step by step guide would be really appreciated for this one. Write it this way : $S=\sum_{k=0}^{n-1} (k+1)(n-k)$ $S=\sum_{k=0}^{n-1} \left[nk+n-k-k^2\right]=\sum_{k=0}^{n-1} (n-1)k+\sum_{k=0}^{n-1} n-\sum_{k=0}^{n-1} k^2$ $S=(n-1) \sum_{k=0}^{n-1} k+n \sum_{k=0}^{n-1} 1-\sum_{k=0}^{n-1} k^2$ We know that $\sum_{k=0}^n k=\sum_{k=1}^n k=\frac{n(n+1)}{2}$. Be careful, here it's n-1, not n. We also know that $\sum_{k=0}^n k^2=\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$. Be careful here too There may be a simplier way, but I just can't see it..

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https://stats.libretexts.org/Bookshelves/Probability_Theory/Applied_Probability_(Pfeiffer)/12%3A_Variance_Covariance_and_Linear_Regression/12.02%3A_Covariance_and_the_Correlation_Coefficient

# 12.2: Covariance and the Correlation Coefficient $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ The mean value $$\mu_X = E[X]$$ and the variance $$\sigma_X^2 = E[(X - \mu_X)^2]$$ give important information about the distribution for real random variable $$X$$. Can the expectation of an appropriate function of $$(X, Y)$$ give useful information about the joint distribution? A clue to one possibility is given in the expression $$\text{Var}[X \pm Y] = \text{Var} [X] + \text{Var} [Y] \pm 2(E[XY] - E[X]E[Y])$$ The expression $$E[XY] - E[X]E[Y]$$ vanishes if the pair is independent (and in some other cases). We note also that for $$\mu_X = E[X]$$ and $$\mu_Y = E[Y]$$ $$E[(X - \mu_X) (Y - \mu_Y)] = E[XY] - \mu_X \mu_Y$$ To see this, expand the expression $$(X - \mu_X)(Y - \mu_Y)$$ and use linearity to get $$E[(X - \mu_X) (Y - \mu_Y)] = E[XY - \mu_Y X - \mu_X Y + \mu_X \mu_Y] = E[XY] - \mu_Y E[X] - \mu_X E[Y] + \mu_X \mu_Y$$ which reduces directly to the desired expression. Now for given $$\omega$$, $$X(\omega) - \mu_X$$ is the variation of $$X$$ from its mean and $$Y(\omega) - \mu_Y$$ is the variation of $$Y$$ from its mean. For this reason, the following terminology is used. Definition: Covariance The quantity $$\text{Cov} [X, Y] = E[(X - \mu_X)(Y - \mu_Y)]$$ is called the covariance of $$X$$ and $$Y$$. If we let $$X' = X - \mu_X$$ and $$Y' = Y - \mu_Y$$ be the ventered random variables, then $$\text{Cov} [X, Y] = E[X'Y']$$ Note that the variance of $$X$$ is the covariance of $$X$$ with itself. If we standardize, with $$X^* = (X - \mu_X)/\sigma_X$$ and $$Y^* = (Y - \mu_Y)/\sigma_Y$$, we have Definition: Correlation Coefficient The correlation coefficient $$\rho = \rho [X, Y]$$ is the quantity $$\rho [X,Y] = E[X^* Y^*] = \dfrac{E[(X - \mu_X)(Y - \mu_Y)]}{\sigma_X \sigma_Y}$$ Thus $$\rho = \text{Cov}[X, Y] / \sigma_X \sigma_Y$$. We examine these concepts for information on the joint distribution. By Schwarz' inequality (E15), we have $$\rho^2 = E^2 [X^* Y^*] \le E[(X^*)^2] E[(Y^*)^2] = 1$$ with equality iff $$Y^* = cX^*$$ Now equality holds iff $$1 = c^2 E^2[(X^*)^2] = c^2$$ which implies $$c = \pm 1$$ and $$\rho = \pm 1$$ We conclude $$-1 \le \rho \le 1$$, with $$\rho = \pm 1$$ iff $$Y^* = \pm X^*$$ Relationship between $$\rho$$ and the joint distribution • We consider first the distribution for the standardized pair $$(X^*, Y^*)$$ • Since $$P(X^* \le r, Y^* \le s) = P(\dfrac{X - \mu_X}{\sigma_X} \le r, \dfrac{Y - \mu_Y}{\sigma_Y} \le s)$$ $$= P(X \le t = \sigma_X r + \mu_X, Y \le u = \sigma_Y s + \mu_Y)$$ we obtain the results for the distribution for $$(X, Y)$$ by the mapping $$t = \sigma_X r + \mu_X$$ $$u = \sigma_Y s + \mu_Y$$ Joint distribution for the standardized variables $$(X^*, Y^*)$$, $$(r, s) = (X^*, Y^*)(\omega)$$ $$\rho = 1$$ iff $$X^* = Y^*$$ iff all probability mass is on the line $$s = r$$. $$\rho = -1$$ iff $$X^* = -Y^*$$ iff all probability mass is on the line $$s = -r$$. If $$-1 < \rho < 1$$, then at least some of the mass must fail to be on these lines. Figure 12.2.1. Distance from point $$(r,s)$$ to the line $$s = r$$. The $$\rho = \pm 1$$ lines for the $$(X, Y)$$ distribution are: $$\dfrac{u - \mu_Y}{\sigma_Y} = \pm \dfrac{t - \mu_X}{\sigma_X}$$ or $$u = \pm \dfrac{\sigma_Y}{\sigma_X}(t - \mu_X) + \mu_Y$$ Consider $$Z = Y^* - X^*$$. Then $$E[\dfrac{1}{2} Z^2] = \dfrac{1}{2} E[(Y^* - X^*)^2]$$. Reference to Figure 12.2.1 shows this is the average of the square of the distances of the points $$(r, s) = (X^*, Y^*) (\omega)$$ from the line $$s = r$$ (i.e. the variance about the line $$s = r$$). Similarly for $$W = Y^* + X^*$$. $$E[W^2/2]$$ is the variance about $$s = -r$$. Now $$\dfrac{1}{2} E[(Y^* \pm X^*)^2] = \dfrac{1}{2}\{E[(Y^*)^2] + E[(X^*)^2] \pm 2E[X^* Y^*]\} = 1 \pm \rho$$ Thus $$1 - \rho$$ is the variance about $$s = r$$ (the $$\rho = 1$$ line) $$1 + \rho$$ is the variance about $$s = -r$$ (the $$\rho = -1$$ line) Now since $$E[(Y^* - X^*)^2] = E[(Y^* + X^*)^2]$$ iff $$\rho = E[X^* Y^*] = 0$$ the condition $$\rho = 0$$ is the condition for equality of the two variances. Transformation to the $$(X, Y)$$ plane $$t = \sigma_X r + \mu_X$$ $$u = \sigma_Y s + \mu_Y$$ $$r = \dfrac{t - \mu_X}{\sigma_X}$$ $$s = \dfrac{u - \mu_Y}{\sigma_Y}$$ The $$\rho = 1$$ line is: $$\dfrac{u - \mu_Y}{\sigma_Y} = \dfrac{t - \mu_X}{\sigma_X}$$ or $$u = \dfrac{\sigma_Y}{\sigma_X} (t - \mu_X) + \mu_Y$$ The $$\rho = -1$$ line is: $$\dfrac{u - \mu_Y}{\sigma_Y} = \dfrac{t - \mu_X}{\sigma_X}$$ or $$u = -\dfrac{\sigma_Y}{\sigma_X} (t - \mu_X) + \mu_Y$$ $$1 - \rho$$ is proportional to the variance abut the $$\rho = 1$$ line and $$1 + \rho$$ is proportional to the variance about the $$\rho = -1$$ line. $$\rho = 0$$ iff the variances about both are the same. Example $$\PageIndex{1}$$ Uncorrelated but not independent Suppose the joint density for $$\{X, Y\}$$ is constant on the unit circle about the origin. By the rectangle test, the pair cannot be independent. By symmetry, the $$\rho = 1$$ line is $$u = t$$ and the $$\rho = -1$$ line is $$u = -t$$. By symmetry, also, the variance about each of these lines is the same. Thus $$\rho = 0$$, which is true iff $$\text{Cov}[X, Y] = 0$$. This fact can be verified by calculation, if desired. Example $$\PageIndex{2}$$ Uniform marginal distributions Figure 12.2.2. Uniform marginals but different correlation coefficients. Consider the three distributions in Figure 12.2.2. In case (a), the distribution is uniform over the square centered at the origin with vertices at (1,1), (-1,1), (-1,-1), (1,-1). In case (b), the distribution is uniform over two squares, in the first and third quadrants with vertices (0,0), (1,0), (1,1), (0,1) and (0,0), (-1,0), (-1,-1), (0,-1). In case (c) the two squares are in the second and fourth quadrants. The marginals are uniform on (-1,1) in each case, so that in each case $$E[X] = E[Y] = 0$$ and $$\text{Var} [X] = \text{Var} [Y] = 1/3$$ This means the $$\rho = 1$$ line is $$u = t$$ and the $$\rho = -1$$ line is $$u = -t$$. a. By symmetry, $$E[XY] = 0$$ (in fact the pair is independent) and $$\rho = 0$$. b. For every pair of possible values, the two signs must be the same, so $$E[XY] > 0$$ which implies $$\rho > 0$$. The actual value may be calculated to give $$\rho = 3/4$$. Since $$1 - \rho < 1 + \rho$$, the variance about the $$\rho = 1$$ line is less than that about the $$\rho = -1$$ line. This is evident from the figure. c. $$E[XY] < 0$$ and $$\rho < 0$$. Since $$1 + \rho < 1 - \rho$$, the variance about the $$\rho = -1$$ line is less than that about the $$\rho = 1$$ line. Again, examination of the figure confirms this. Example $$\PageIndex{3}$$ A pair of simple random variables With the aid of m-functions and MATLAB we can easily caluclate the covariance and the correlation coefficient. We use the joint distribution for Example 9 in "Variance." In that example calculations show $$E[XY] - E[X]E[Y] = -0.1633 = \text{Cov} [X,Y]$$, $$\sigma_X = 1.8170$$ and $$\sigma_Y = 1.9122$$ so that $$\rho = -0.04699$$. Example $$\PageIndex{4}$$ An absolutely continuous pair The pair $$\{X, Y\}$$ has joint density function $$f_{XY} (t, u) = \dfrac{6}{5} (t + 2u)$$ on the triangular region bounded by $$t = 0$$, $$u = t$$, and $$u = 1$$. By the usual integration techniques, we have $$f_X(t) = \dfrac{6}{5} (1 + t - 2t^2)$$, $$0 \le t \le 1$$ and $$f_Y (u) = 3u^2$$, $$0 \le u \le 1$$ From this we obtain $$E[X] = 2/5$$, $$\text{Var} [X] = 3/50$$, $$E[Y] = 3/4$$, and $$\text{Var} [Y] = 3/80$$. To complete the picture we need $$E[XY] = \dfrac{6}{5} \int_0^1 \int_t^1 (t^2 u + 2tu^2)\ dudt = 8/25$$ Then $$\text{Cov} [X,Y] = E[XY] - E[X]E[Y] = 2/100$$ and $$\rho = \dfrac{\text{Cov}[X,Y]}{\sigma_X \sigma_Y} = \dfrac{4}{30} \sqrt{10} \approx 0.4216$$ APPROXIMATION tuappr Enter matrix [a b] of X-range endpoints [0 1] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density (6/5)*(t + 2*u).*(u>=t) Use array operations on X, Y, PX, PY, t, u, and P EX = total(t.*P) EX = 0.4012 % Theoretical = 0.4 EY = total(u.*P) EY = 0.7496 % Theoretical = 0.75 VX = total(t.^2.*P) - EX^2 VX = 0.0603 % Theoretical = 0.06 VY = total(u.^2.*P) - EY^2 VY = 0.0376 % Theoretical = 0.0375 CV = total(t.*u.*P) - EX*EY CV = 0.0201 % Theoretical = 0.02 rho = CV/sqrt(VX*VY) rho = 0.4212 % Theoretical = 0.4216 Coefficient of linear correlation The parameter $$\rho$$ is usually called the correlation coefficient. A more descriptive name would be coefficient of linear correlation. The following example shows that all probability mass may be on a curve, so that $$Y = g(X)$$ (i.e., the value of Y is completely determined by the value of $$X$$), yet $$\rho = 0$$. Example $$\PageIndex{5}$$ $$Y = g(X)$$ but $$\rho = 0$$ Suppose $$X$$ ~ uniform (-1, 1), so that $$f_X (t) = 1/2$$, $$-1 < t < 1$$ and $$E[X] = 0$$. Let $$Y = g(X) = \cos X$$. Then $$\text{Cov} [X, Y] = E[XY] = \dfrac{1}{2} \int_{-1}^{1} t \cos t\ dt = 0$$ Thus $$\rho = 0$$. Note that $$g$$ could be any even function defined on (-1,1). In this case the integrand $$tg(t)$$ is odd, so that the value of the integral is zero. Variance and covariance for linear combinations We generalize the property (V4) on linear combinations. Consider the linear combinations $$X = \sum_{i = 1}^{n} a_i X_i$$ and $$Y = \sum_{j = 1}^{m} b_j Y_j$$ We wish to determine $$\text{Cov} [X, Y]$$ and $$\text{Var}[X]$$. It is convenient to work with the centered random variables $$X' = X - \mu_X$$ and $$Y' = Y - \mu_Y$$. Since by linearity of expectation, $$\mu_X = \sum_{i = 1}^{n} a_i \mu_{X_i}$$ and $$\mu_Y = \sum_{j = 1}^{m} b_j \mu_{Y_j}$$ we have $$X' = \sum_{i = 1}^{n} a_i X_i - \sum_{i = 1}^{n} a_i \mu_{X_i} = \sum_{i = 1}^{n} a_i (X_i - \mu_{X_i}) = \sum_{i = 1}^{n} a_i X_i'$$ and similarly for $$Y'$$. By definition $$\text{Cov} (X, Y) = E[X'Y'] = E[\sum_{i, j} a_i b_j X_i' Y_j'] = \sum_{i,j} a_i b_j E[X_i' E_j'] = \sum_{i,j} a_i b_j \text{Cov} (X_i, Y_j)$$ In particular $$\text{Var} (X) = \text{Cov} (X, X) = \sum_{i, j} a_i a_j \text{Cov} (X_i, X_j) = \sum_{i = 1}^{n} a_i^2 \text{Cov} (X_i, X_i) + \sum_{i \ne j} a_ia_j \text{Cov} (X_i, X_j)$$ Using the fact that $$a_ia_j \text{Cov} (X_i, X_j) = a_j a_i \text{Cov} (X_j, X_i)$$, we have $$\text{Var}[X] = \sum_{i = 1}^{n} a_i^2 \text{Var} [X_i] + 2\sum_{i <j} a_i a_j \text{Cov} (X_i, X_j)$$ Note that $$a_i^2$$ does not depend upon the sign of $$a_i$$. If the $$X_i$$ form an independent class, or are otherwise uncorrelated, the expression for variance reduces to $$\text{Var}[X] = \sum_{i = 1}^{n} a_i^2 \text{Var} [X_i]$$ This page titled 12.2: Covariance and the Correlation Coefficient is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Paul Pfeiffer via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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http://mathhelpforum.com/trigonometry/209614-cos-3pi-5-a.html

1. ## Cos 3pi/5 I don't understand the step where 2Sin^2(3pi/10) = 2((sqrt5+1)/4) 2. ## Re: Cos 3pi/5 Hey Eraser147. Try setting up a right angled triangle or using sums or differences formula to get the value. 3. ## Re: Cos 3pi/5 Consider the right-angled triangle with acute angles $\frac{3\pi}{10}$ and $\frac{2\pi}{10}$ and note that $\sin{\left(\frac{3\pi}{10}\right)}=\cos{ \left( \frac{2\pi}{10}\right) }.$ To ease typing, let $\frac{\pi}{10}=A,$ then $\sin 3A=\cos 2A.$ Use the standard identities, $\sin 3A =3\sin A-4\sin^{3} A .............(1) \text{ and } \cos 2A= 1-2\sin^{2} A ............(2)$ and we have the cubic $4\sin^{3} A -2\sin^{2}A-3\sin A +1 = 0.$ $(\sin A -1)$ is a factor, remove this and we are left with the quadratic $4\sin^{2} A +2\sin A -1=0,$ and solving this, (for the angle in the $0,\pi/2$ range), $\sin A = \frac{-1+\sqrt{5}}{4}.$ Substituting this into (1) and simplifying leads to $\sin 3A = \sin{ \ \left( \frac{3\pi}{10} \right)} = \frac{1+\sqrt{5}}{4}.$

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https://math.stackexchange.com/questions/1899206/show-there-exists-a-sequence-of-polynomials-which-converge-uniformly-to-a-contin

# Show there exists a sequence of polynomials which converge uniformly to a continuous $f$. Let $f : [0,1] \to \mathbb{R}$ be a continuous function that vanishes at $x = 1.$ Show that there exists a sequence of polynomials, each vanishing at $x = 1$, which converges to $f$ uniformly on $[0,1].$ It feels like a Stone-Weierstrass question, but after looking over the Stone-Weierstrass proof several times I am not sure if it truly applies. My second thought is Arzela-Ascoli - namely that the family of functions you consider are the collection of polynomials, call this collection $P$, that vanish at $x = 1.$ My problem then becomes that this family is not uniformly bound over the interval, and I do not know if we can assert that a sequence of these polynomials converges to a specific $f,$ I think Arzela-Ascoli only proves that a uniformly convergent sequence in $P$ exists. Anyone have any insight? Thanks in advance. Another trick may be the following one: take a sequence of polynomials $q_1(x),q_2(x),\ldots$ that uniformly approximate $g(x)=\frac{f(x)}{1-x}$ over $(0,1)$ then consider $p_n(x)=(1-x)q_n(x)$. It is very easy to show that the sequence $p_1(x),p_2(x),\ldots$ meets the given constraints. • Hello Jack. Is it true that there is a sequence of polynomials that uniformly approximates $g$? What if $f$ is slower than $x \mapsto 1-x$, so that $g(x) \to \infty$ as $x\to 1$? An example would be: $f(x) = \sqrt{1-x}$. – user00000 Aug 24 '16 at 11:54 • @user00000: well, that is an issue, but we may circumvent it by first approximating (for instance, by convolution) the original function with a differentiable one that vanishes at $x=1$, hence it is not really restrictive to assume that the $g$ function above is continuous over $(0,1)$. Thanks for pointing that out. Out of curiosity, are you a somewhat special user? That string of zeroes in your user name striked me :D – Jack D'Aurizio Aug 24 '16 at 11:56 • However, how can we approximate the original function with a differentiable one which vanishes at $x =1$ without resorting to polynomials? – user00000 Aug 24 '16 at 12:10 • @user00000: by considering the convolution (en.wikipedia.org/wiki/Convolution) with a smooth kernel. The trick is that the convolution between a $C^0$ and a $C^1$ function is a $C^1$ function, and if the kernel is concentrated enough, that convolution also gives a uniform approximation. – Jack D'Aurizio Aug 24 '16 at 12:12 • Ah, excuse me, I just saw your updated comment now. Convolution would have never hit my mind. And no, I am not by any means a special user! :) – user00000 Aug 24 '16 at 12:16 Take the Stone Weierstrass polynomials $\{p_n\}$, and modify them. Let $p_n(1) = a_n$ and $q_n(x) = p_n(x) - a_n$. Since $a_n = p_n(1) \rightarrow f(1) = 0$, $\{q_n\}$ are convergent to $f$ • Are $p_n$'s the Bernstein polynomials? – Merkh Aug 21 '16 at 18:23 • @Merkh anything converging uniformly to $f$ on $[0,1]$ will do.. Bernstein or not – user290300 Aug 21 '16 at 18:24 As already noted, if $p_n$ is any sequence of polynomials converging uniformly to $f$ then $q_n=p_n-p_n(1)$ works. Another way to look at this is the version of S-W for $C_0(X)$, where $X$ is locally compact Hausdorff; you have $f\in C_0([0,1))$, and that version of S-W says that the polynomials vanishing at $1$ are dense in $C_0([0,1))$.

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https://www.physicsforums.com/threads/oscillary-motion.52381/

# Homework Help: Oscillary Motion 1. Nov 11, 2004 The Problem is: A particle executes simple harmonic motion with an amplitude of 2.25 cm. At what position does its speed equal half its maximum speed? I've been looking at this for a while, and I cant see how to solve it without more information. If I knew the energy and mass I could solve it, if I knew the frequency or period I could solve it, if I knew the spring constant and mass I could solve it, but I dont have any of this, what's the trick? 2. Nov 11, 2004 ### Tide HINT: For what value of z does cos z equal half its maximum value? 3. Nov 11, 2004 ### ehild The position of the particle x=Asin(wt), A= 2.25 cm. The velocity of the particle v= Awcos(wt). The maximum velocity is Aw. You look for the position when v=0.5 Aw. 0.5 Aw = Aw cos(wt) ----> cos(wt)=0.5. You only ned to find sin(wt) to get x. ehild 4. Nov 12, 2004 ### CartoonKid You can apply the general velocity formula for SHM here. since maximum velocity is v=rw, then take the half of it and equal to general formula. Work it out and you should be able to find the positions.

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http://mathhelpforum.com/algebra/81405-using-logarithm-find-earthquake-strength-print.html

# Using logarithm to find earthquake strength • Mar 30th 2009, 01:34 AM vidalex Using logarithm to find earthquake strength Word Problem that I'm stuck on The common logarithm is used to measure the intensity of an earthquake on the Richter scale. The Richter scale rating of an earthquake of intensity I is given by log(I) - log(Io), where Io is the intensity of a small "benchmark" earthquake. Write the Richter scale rating as a single logarithm. What is the Richter scale rating of an earthquake for which I = 1000*Io? • Mar 30th 2009, 02:23 AM mathsquest log (I/I0). For the second part, log1000 • Mar 30th 2009, 04:02 AM mr fantastic Quote: Originally Posted by vidalex Word Problem that I'm stuck on The common logarithm is used to measure the intensity of an earthquake on the Richter scale. The Richter scale rating of an earthquake of intensity I is given by log(I) - log(Io), where Io is the intensity of a small "benchmark" earthquake. Write the Richter scale rating as a single logarithm. What is the Richter scale rating of an earthquake for which I = 1000*Io? You should know the rule $\log A - \log B = \log \frac{A}{B}$.

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https://jspinvandy.home.blog/tag/mass-of-light/

# How Much Does Light Weigh? Light is made of photons. And photons are massless particles, which means they have no invariant/resting mass. Therefore, light has no mass and no weight. End of story, right? ### The Force of Light Did you know that light exerts pressure on objects? This force can even increase an objects weight, albeit to a small degree. For instance, Vsauce explains how much a landmass might weigh covered in sunlight (for the city of Chicago, sunlight only adds 300 lbs). Although insignificant on a small scale, scientists must account for this force – called radiation pressure or solar radiation pressure – in planing space missions. Solar radiation pressure plays a role in the formation of galaxies, stars and clusters, and solar/planetary systems. Additionally, solar radiation shapes the tails of comets. ### A Common Misconception Light carries energy and momentum. We know that energy, momentum and mass are related. Can we assume that light also has mass? $E = mc^{2}$ This misconception stems from Einstein’s mass-energy equivalence formula. Einstein proposed that an object with energy has an equivalent amount of mass. Einstein’s formula only applies to objects with invariant/resting mass. Since photons have no resting mass, we have to use a different formula: $E = pc$ where p is the momentum of the particle. Therefore, we can observe momentum and energy for massless particles. tl;dr Light does not have weight or mass. Light can push an object or increase its weight, to a minimal degree. ### We Weigh Less in the Dark, technically Do we actually weigh more in sunlight? Functionally, no. But, we can still estimate an upper bound. First, solar radiation pressure is applied to objects in the direction of sunlight. For this problem, let us pretend that we are shaped like rectangular solar panels. Second, we learned from Vsauce that light exerts a force of pressure on the surface of Earth of about 1e-9 lbs per square inch. Considering the average human has a surface area of 1.9 meters squared, we can calculate the following: $\frac{1.0*10^{-9} lbs}{in^{2}} * \frac{1550 in^{2}}{m^{2}}\ * 1.9m^{2}$ $= 3.0 * 10^{-6}lbs$ So the next time you weigh yourself, turn off the lights. You might not notice a difference, but you just shaved a couple millionths of a pound.

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http://mathhelpforum.com/advanced-algebra/137703-orthogonal-matrices.html

1. ## Orthogonal matrices i have found two eigenvalues for a matrix i've been working on, both are lambda = 2... when i substitute the eigenvalue back into the matrix and row reduce i get y= 0 and x = z so i said my first eigenvector, V1 is a 3*1 column matrix with entries 1, 0, 1. My problem is finding the 2nd eigenvector... I need to find a vector V2 such that V1.V2 = 0 and V2 (hat) = 1. I think i need another vector V3 as well which is orthogonal to V1 and V2 and is itself a unit vector.. Does this mean i should have a 3rd eigenvalue? Im using these vectors as columns of matrix O... 2. Delete this thread please Mods, i reposted the question using LaTex as i think my wording here was a bit messy Sorry if any hassle caused Tekken

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https://www.physicsforums.com/threads/non-constant-hazard-rates-calculating-the-modal-failure-rate-of-hard-drive.561927/

# Non Constant Hazard Rates - Calculating the modal failure rate of hard drive 1. Dec 21, 2011 ### orangeIV Hi, I'm currently trying to work through a problem about calculating the most likely time for a hard disk to fail: Hard disks fail with a probability per unit time: $\alpha (t) = \alpha _0 t$ where $\alpha_0 = 0.5$ years. I know that the answer is $t_{modal} = \frac{1}{\sqrt{\alpha_0}}$, but am having problems deriving this. Here is what I've done so far: The probability distribution can be calculated as follows: $f(x) = \alpha (t) e^{-\int \alpha (t) dt} = \alpha (t) e^{-\frac{1}{2} \alpha_0 t^2}$ The most likely time for the disk to fail will be when $\frac{df}{dt} = 0$. So when $0 = -{\alpha_0}^2 t^2 e^{-\frac{1}{2} \alpha_0 t^2}$ This is where I get stuck. Is this the correct approach? Any ideas about how how I might proceed :) Thanks 2. Dec 21, 2011 ### Stephen Tashi Did you use the product rule when you computed this derivative? Similar Discussions: Non Constant Hazard Rates - Calculating the modal failure rate of hard drive

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http://mathhelpforum.com/calculus/139162-help-mean-value-theorem.html

# Thread: Help with the mean value theorem 1. ## Help with the mean value theorem I went in to my professor for "extra help" on this problem and left more confused than before I went in. Can someone on here try to explain this to me! Question: For what values of a, m, and b does the function satisfy the hypotheses of the Mean Value Theorem on the inverval [0, 2] Piecewise function: f(x)= 3, x=0 f(x)=-x^2 + 3x + a, 0<x<1 f(x)=mx + b 1</= x </= 2 Thank you so much for any help! 2. Originally Posted by KarlosK I went in to my professor for "extra help" on this problem and left more confused than before I went in. Can someone on here try to explain this to me! Question: For what values of a, m, and b does the function satisfy the hypotheses of the Mean Value Theorem on the inverval [0, 2] Piecewise function: f(x)= 3, x=0 f(x)=-x^2 + 3x + a, 0<x<1 f(x)=mx + b 1</= x </= 2 Thank you so much for any help! In order for the mean value theorem to be applicable, f(x) needs to be continuous on the entire closed interval [0;2], and differentiable in its interior ]0;2[. Continuity at x=0 requires a=3, and you can determine the requisite values of b and m from the tangent to the graph of $y=-x^2+3x+3$ at $x_0=1$. Because, for f(x) to be differentiable at $x_0=1$ the graph $y=mx+b$ must be that tangent itself...

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http://www.emathzone.com/tutorials/math-results-and-formulas/formulas-for-area-of-a-triangle.html

# Formulas for Area of a Triangle 1. $A = \frac{1}{2}b \cdot h$, where $b$ is the base and $h$ is the altitude of the triangle. 2. Area of an equilateral triangle , where $a$ is the length of each side of the triangle. 3. Area of a triangle when two adjacent sides and the included angle is given by 4. Area of a triangle when length of all sides are given where 5. Area of a triangle with vertices $A\left( {{x_1},{y_1}} \right),\,B\left( {{x_2},{y_2}} \right),\,C\left( {{x_3},{y_3}} \right)$ is given by the formulas

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https://link.springer.com/article/10.1007/s42379-018-00019-w?error=cookies_not_supported&code=bda16864-0c63-420a-bfcb-38e98cca4654

# Changes in family structure in China: the impact of residence patterns and demographic factors ## Abstract Based on census data from China, this paper uses SOCSIM microscopic simulation method to decompose the key factors of family transition into demographic and residence pattern factors. The former are further broken down into demographic inertia, fertility, mortality and marriage rate. The results indicate that the current demographic changes are relatively small and, thus, the small fluctuations caused by demographic changes contribute much less than residence pattern to the family transition. Among the demographic factors, demographic inertia and fertility have a greater effect on family transition and the impacts caused by fertility and marriage rate are consistent with the direction of the overall influence of demographic factors: increasing the proportion of single-person households, one-generation households, and two-generation households, and decreasing the proportion of three-generation or more households. In contrast, the effect of mortality rate was opposite to that of fertility rate, which increased the proportion of populations living in single-person, one-generation, and two-generation households, but the decreased the proportion of the population living in three-generation or more households. ## Introduction Family size and family structure reflect the status of a family from both quantitative and qualitative perspectives. As social and economic development move forward in China, the structure of Chinese families has undergone continuous changes. During the decades following the founding of New China, the impacts of the cooperative movement and socio-economic development were behind significant changes to the types and the intergenerational structure of Chinese households. In the year 2000, China’s family structures could be divided into several categories, including relatively stable families (such as three-generation households), emerging families (such as single-person households, grandparent headed household) and diminishing families (such as impaired nuclear families) (Wang 2006). In 2010, the cumulative proportion of one-generation and two-generation households exceeded 80%, with the proportion of one-generation households going up 11% from 2000 and that of two-generation households going down 10% from 2000 (Hu and Peng 2014). According to Western theories of family modernization, family transition is influenced by both socio-economic factors and demographic factors. Socio-economic factors determine people’s residence patterns refers to the agent variable of economic and social factors and take into consideration issues such as attitudes and dwelling conditions; these factors lead directly to family transition. Demographic factors have major impacts on family size and structure (Jiang and O’Neill 2007). In particular, demographic factors such as fertility and mortality have a significant impact on family structure and kinship (Hammel 2005). In view of this, this paper first divides the factors influencing the transition in family structure into two types (demographic factors and residence patterns), and then further breaks down the demographic factors into demographic inertia, fertility rate, mortality rate and marriage rate in an effort to explore their specific impacts. Specifically, the paper applies the SOCSIM microscopic simulation methodFootnote 1 to data from the first to the sixth national census in China to analyze family structures in the past decades, and then studies how and to what extent different factors have contributed to the transition of family structures. ## Impacts from demographic factors and residence patterns We use $$D_{effect}$$ to imply the effect of demographic factors (D refers to the total population) and $$P_{effect}$$ to imply the effect of residence patterns (P refers to the proportion of population groups with different residence patterns to the total population). By multiplying the D vector by the P vector which falls within the same time range, we obtain the size of the population group adopting its residence pattern in the time range under consideration, namely: $${\text{h}}_{t} = f\left( {D_{t} ,P_{t} } \right) = D_{t} *P_{t}$$ When the factors are decomposed, the effect of demographic factors is shown as the variation in h after controlling the residence pattern ($$P_{effect}$$), and the effect of residence pattern is the variation in h after controlling the demographic factors ($$D_{effect}$$). The formulas are as follows: $$D_{effect} = \frac{{\left( {D_{1} *P_{1} - D_{2} *P_{1} } \right) + \left( {D_{1} *P_{2} - D_{2} *P_{2} } \right)}}{2}$$ (1) $$P_{effect} = \frac{{\left( {D_{1} *P_{1} - D_{1} *P_{2} } \right) + \left( {D_{2} *P_{1} - D_{2} *P_{2} } \right)}}{2}$$ (2) where $${\text{h}}_{1}$$, $$D_{1}$$ and $$P_{1}$$ refer, respectively, to the size of the population living in a certain type of household, the total population and the proportion of the population living in this type of household at the first time-point; $${\text{h}}_{2}$$, $$D_{2}$$ and $$P_{2}$$ refer, respectively, to the size of the population living in a certain type of household, the total population and the proportion of the population living in this type of household at the second time-point. Taking single-person households as an example, we analyzed the change in the proportion of the population living in single-person households from 2000 to 2010, and found that the proportion doubled from 2.40% in the Fifth Census in 2000 to 4.95% in the Sixth Census in 2010. Based on the previous analysis, first this paper assumes that the proportion of the population living in single-person households in 2010 is the same as that in 2000, namely 2.40%, and the population size changes are based on the actual situations (Fig. 1). We then subtract the size of the population living in single-person households in 2000 ($$D_{2000} *P_{2000}$$) from the size of that population in 2010 ($$D_{2010} *P_{2000}$$) to analyze the effect of demographic transition on the basis of the proportion in 2000 ($$D_{2010} *P_{2000} - D_{2000} *P_{2000}$$). Next, the paper assumes that the proportion of population living in single-person households in 2000 is the same as that in 2010, namely 4.95%, and the population size changes are based on actual situations. We then subtract the size of population living in single-person households in 2000 ($$D_{2000} *P_{2000}$$) from the size of that population in 2010 ($$D_{2010} *P_{2000}$$) to analyze the effect of demographic transition on the basis of the proportion in 2010 ($$D_{2010} *P_{2000} - D_{2000} *P_{2000}$$). Subsequently, the two differences are summed and averaged to obtain the proportion of the variation in the size of population living in single-person households to the total variation caused by the demographic factors. Finally, the paper multiplies this proportion by the variation (− 1.70%) in the proportion of the population living in single-person households from 2000 to 2010 to locate the part of the variation from 2000 to 2010 that is caused by the demographic factors. In the same way, we can calculate the extent of the effect of the residence pattern variable on the proportion of the population living in single-person households when the demographic factors are controlled. Using the same method, principle and process, the paper decomposes the demographic factors and residence pattern factor based on the variations in the proportions of the population living in single-person, one-generation, two-generation and three-generation or above households from 2000 to 2010 (see Table 1). In general, during the period from 2000 to 2010, we find that for changes in the proportions of all types of households the effect of the change in residence pattern is generally greater than the effect of the changes in demographic factors. From 2000 to 2010, the proportion of the population living in single-person households increased by 2.55%, of which 2.48% was caused by change in residence patterns; demographic factors only made a minor contribution to the change in proportion. It can be concluded that residence pattern has the greatest impact on people’s choice to live alone. Among demographic factors, the gender difference in the mortality rate of the elderly population and the change in marriage rate will have certain impacts on the proportion of the population living in single-person households, but such changes are mainly determined by residence pattern. Unmarried or widowed people may choose to live with their parents or children. Therefore, the change in residence pattern plays a very important role in the changing proportion of single-person households. The proportion of the population living in one-generation households increased by 5.81% from 2000 to 2010, and there was little difference between the effect of residence pattern and that of demographic factors on this increase. Among the demographic factors, the postponement of childbearing has led to an increase in the proportion of the population living in one-generation households. The effect of residence pattern was slightly greater than that of demographic factors. Whether or not to choose to live with the previous or the next generation can make a major difference on the proportion of the population living in one-generation households. As part of the process of family nuclearization, the proportion of two-generation households decreased, with the proportion of the population living in two-generation households going down by 6.32% from 2000 to 2010. Residence pattern was the only contributor to this decline; in fact, demographic factors had an opposite effect, increasing the proportion of the population living in two-generation households. The postponement of childbearing among married couples living with their parents and the late marriages of adult children increases the proportion of the population living in two-generation households. On the other hand, the proportion of three-generation households as a household type has decreased slightly, and yet the increase in total size of the population has led to an increased proportion of the population living in three-generation households. Residence pattern had a significant impact on this change while demographic factors had a relatively minor effect. It can be seen that residence pattern has contributed a great deal to family transition. A booming economy and rapid social change during the years 2000 to 2010 led to a major shift in attitudes toward residence patterns and an improvement in housing conditions. The result of this was a major shift in preferred residence patterns, and this has had a significant impact on family transition. On the other hand, demographic transition was relatively stable during the period from 2000 to 2010, and contributed less to family transition as a result of this stability. ### Further decomposition of the effect of demographic factors In this section, the paper decomposes the demographic factor into fertility rate, mortality rate, marriage rate and other aspects of demographic inertia. Different from the decomposition of demographic factors and residence patterns, the analysis of the impact of fertility rate, mortality rate and marriage rate needs to maintain these rates at levels of last time period and project the population status (i.e., D vector) under these rates. On this basis further analysis is carried out by taking account of the proportions of respective residence patterns (i.e., P vector). The population vector D is a function of fertility rate (F), mortality rate (M) and marriage rate (N), namely: $$D = g(F,M,N)$$ Taking the fertility influencing factor ($${\text{F}}_{effect}$$) as the example, under two scenarios in which the fertility rate is kept unchanged at the previous level ($$F_{fix}$$) and changes as per the actual situations ($$F_{chg}$$), SOCSIM projects the difference in population size ($$D_{fix - chg}$$) and multiplies it by residence pattern (P) to obtain the variations in the sizes of the populations living in different households. With regard to the vector of the residence pattern, this paper uses the mean value before and after the period ($$P_{avg}$$). In the decomposition of its effect on fertility rate, we must also pay attention to the effects of different scenarios for mortality rate and marriage rate variations. This paper uses SOCSIM to perform eight simulations to calculate the variations in demographic factors as caused by the fertility rates in the four scenarios. Specifically: 1. 1. The population difference caused by the change in fertility rate when the mortality rate and marriage rate are kept unchanged at the previous level: $${\text{g}}\left( {F_{fix} ,M_{fix} ,N_{fix} } \right) - {\text{g}}\left( {F_{chg} ,M_{fix} ,N_{fix} } \right)$$ 1. 2. The population difference caused by the change in fertility rate when the mortality rate and the marriage rate change pursuant to the actual situations: $${\text{g}}\left( {F_{fix} ,M_{chg} ,N_{chg} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{chg} } \right)$$ 1. 3. The population difference caused by the change in fertility rate when the mortality rate is kept unchanged at the previous level and the marriage rate changes pursuant to the actual situations: $${\text{g}}\left( {F_{fix} ,M_{fix} ,N_{chg} } \right) - {\text{g}}\left( {F_{chg} ,M_{fix} ,N_{chg} } \right)$$ 1. 4. The population difference caused by the change in fertility rate when the mortality rate changes pursuant to the actual situations and the marriage rate is kept unchanged at the previous level: $${\text{g}}\left( {F_{fix} ,M_{chg} ,N_{fix} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{fix} } \right)$$ On the basis of the results obtained under different scenarios, the population differences are multiplied by the mean value of residence pattern, divided by the corresponding value, and finally summed up. In this way, we can obtain the variation (i.e., $${\text{F}}_{effect}$$) in the size of population living in a certain type of household during such a period when the fertility rate changes. The specific formula is as follows: $$F_{effect} = \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{fix} ,N_{fix} } \right) - {\text{g}}\left( {F_{chg} ,M_{fix} ,N_{fix} } \right)} \right]*P_{avg} }}{3} + \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{chg} ,N_{chg} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{chg} } \right)} \right]*P_{avg} }}{3} + \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{fix} ,N_{chg} } \right) - {\text{g}}\left( {F_{chg} ,M_{fix} ,N_{chg} } \right)} \right]*P_{avg} }}{6} + \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{chg} ,N_{fix} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{fix} } \right)} \right]*P_{avg} }}{6}$$ (3) Similar to the fertility influencing factor ($${\text{F}}_{effect}$$), the mortality influencing factor ($${\text{M}}_{effect}$$) is calculated using the following formula: \begin{aligned} M_{effect} &= \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{fix} ,N_{fix} } \right) - {\text{g}}\left( {F_{fix} ,M_{chg} ,N_{fix} } \right)} \right]*P_{avg} }}{3}\\ &\quad + \frac{{\left[ {{\text{g}}\left( {F_{chg} ,M_{fix} ,N_{chg} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{chg} } \right)} \right]*P_{avg} }}{3}\\ &\quad + \frac{{\left[ {{\text{g}}\left( {F_{fix} ,M_{fix} ,N_{chg} } \right) - {\text{g}}\left( {F_{fix} ,M_{chg} ,N_{chg} } \right)} \right]*P_{avg} }}{6}\\ &\quad + \frac{{\left[ {{\text{g}}\left( {F_{chg} ,M_{fix} ,N_{fix} } \right) - {\text{g}}\left( {F_{chg} ,M_{chg} ,N_{fix} } \right)} \right]*P_{avg} }}{6} \end{aligned} (4) Similarly, the marriage rate influencing factor ($${\text{N}}_{effect}$$) is calculated using the following formula: \begin{aligned} {N_{effect}} &= \frac{{\left[ {{\rm{g}}\left( {{F_{fix}},{M_{fix}},{N_{fix}}} \right) - {\rm{g}}\left( {{F_{fix}},{M_{fix}},{N_{chg}}} \right)} \right]*{P_{avg}}}}{3}\\ &\quad + \frac{{\left[ {{\rm{g}}\left( {{F_{chg}},{M_{chg}},{N_{fix}}} \right) - {\rm{g}}\left( {{F_{chg}},{M_{chg}},{N_{chg}}} \right)} \right]*{P_{avg}}}}{3}\\ &\quad + \frac{{\left[ {{\rm{g}}\left( {{F_{fix}},{M_{chg}},{N_{fix}}} \right) - {\rm{g}}\left( {{F_{fix}},{M_{chg}},{N_{chg}}} \right)} \right]*{P_{avg}}}}{6}\\ &\quad + \frac{{\left[ {{\rm{g}}\left( {{F_{chg}},{M_{fix}},{N_{fix}}} \right) - {\rm{g}}\left( {{F_{chg}},{M_{fix}},{N_{chg}}} \right)} \right]*{P_{avg}}}}{6} \end{aligned} (5) Taking single-person households as the example, the previous analysis pointed out that the proportion of the population living in single-person households to total population increased by 2.55 percentage points from 2000 to 2010. The demographic factors contributed 0.07 percentage points from 2000 to the increase in the proportion of the population living in single-person households. We further decompose the demographic factor effect into the effects of fertility rate, mortality rate, marriage rate and demographic inertia. Based on the effect of fertility rate, the following formula is obtained by referencing the above formulas: \begin{aligned} F_{effect} & = \frac{{\left[ {{\text{g}}\left( {F_{2000} ,M_{2000} ,N_{2000} } \right) - {\text{g}}\left( {F_{2010} ,M_{2000} ,N_{2000} } \right)} \right]*P_{avg} }}{3} \\ & \quad + \frac{{\left[ {{\text{g}}\left( {F_{2000} ,M_{2010} ,N_{2010} } \right) - {\text{g}}\left( {F_{2010} ,M_{2010} ,N_{2010} } \right)} \right]*P_{avg} }}{3} \\ & \quad + \frac{{\left[ {{\text{g}}\left( {F_{2000} ,M_{2000} ,N_{2010} } \right) - {\text{g}}\left( {F_{2010} ,M_{2000} ,N_{2010} } \right)} \right]*P_{avg} }}{6} \\ & \quad + \frac{{\left[ {{\text{g}}\left( {F_{2000} ,M_{2010} ,N_{2000} } \right) - {\text{g}}\left( {F_{2010} ,M_{2010} ,N_{2000} } \right)} \right]*P_{avg} }}{6} \\ \end{aligned} The proportion of population living in the single-person households (P) is the mean value of 2000 and 2010, namely $$P_{avg} = 3.67$$. The $${\text{F}}_{effect}$$ obtained is divided by the $${\text{D}}_{effect}$$ to get the proportion of the effect of fertility rate to the overall effect of demographic factors. Then, we multiply this proportion by the change brought about by the demographic factors (i.e., 0.07) to obtain the effect of fertility rate. Using the same method, principle and process, changes in the proportion of population living in single-person, one-generation, two-generation and three-generation or above households from 2000 to 2010 are decomposed into demographic inertia, fertility rate, mortality rate and marriage rate (see Table 2). The effect of demographic factors on family transition during the period from 2000 to 2010 has been analyzed in the previous section. The findings showed the proportions of population living in single-person, one-generation, two-generation and three-generation or above households changed by 0.07%. 2.75%, 1.29% and − 1.58% respectively during the years from 2000 to 2010. Next, we will analyze the specific changes in detail. The change in fertility rate from 2000 to 2010 increased the proportions of the population living in single-person, one-generation and two-generation households by 0.21, 0.04 and 3.96%, respectively, but decreased the proportion of the population living in three-generation or above households by 7.91%. It can be seen that fertility rate influences family transition in the same way that demographic factors affect family transition. During the period from 2000 to 2010, the change in fertility rate increased the proportions of the populations living in single-person, one-generation and two-generation households, but reduced the proportion of the population living in three-generation or above households. With the decline in the fertility rate and childbearing postponement, a relatively large number of young married couples form single-couple nuclear families and this increases the proportion of one-generation households. There is a drop in the children number of one family along with the decline in fertility rate. As increasingly few children leave their families, the time at which families enter the empty-nest stage is moved forward, and the duration of the empty-nest stage is extended, thereby increasing the proportion of one-generation households. However, comparatively speaking, the change in fertility rate has had a minor effect on changes in the proportions of single-person and one-generation households, but significant impact on two-generation and three-generation households. The difficulties obtaining suitable housing and the convenience of daily life have made some young couples decide to live with their parents. In addition to being a factor causing an increase of the proportion of the population living in two-generation households (young couples living with their parents), the decline in fertility rate has also reduced the proportion of the population living in three-generation or above households. Next, the effect of mortality rate is decomposed. The change in mortality rate in the period from 2000 and 2010 reduced the proportions of population living in single-person, one-generation and two-generation households by 0.04%, 0.01% and 0.81%, respectively, but increased the proportion of population living in three-generation or above households by 1.61%. In fact the effect of mortality rate on demographic transition was opposite to that of fertility rate, which increased the proportion of populations living in single-person, one-generation, and two-generation households, but the decreased the proportion of the population living in three-generation or more households. Theories of family transition point out that fertility rate and mortality rate have a combined effect on family transition, i.e., the two jointly affect family transition. In the process of China demographic transition, the mortality rate has dropped from an initially high level to its presently low level. After the infant mortality rate drops to a certain level, change in the mortality rate of the elderly becomes the principal factor affecting mortality change. Decline in the mortality rate of the elderly, especially the decline in the mortality rate of elderly men, leads to a decrease in the size of the widowed elderly population, thereby reducing the proportion of the population living in single-person households. The decline in the elderly mortality rate also extends the time the elderly may live with their children, and this also reduces the proportion of the population living in single-person households. The change in mortality rate has a relatively minor effect on single-person and one-generation households, but a significant impact on two-generation and three-generation or above households. The decline in the elderly mortality rate has extended the time that the elderly can live with their children and grandchildren, thereby reducing the proportion of the population living in two-generation households and increasing the proportion of the population living in three-generation or above households. According to theories on family transition, demographic factors affect not only fertility rate and mortality rate, but the marriage rate as well. In its decomposition of demographic factors, this paper also analyzes the marriage rate separately. The change in marriage rate in the period from 2000 to 2010 increased the proportion of the population living in single-person, one-generation and two-generation households by 0.05%, 0.01% and 0.99%, respectively, but decreased the proportion of the population living in three-generation or above households by 1.98%. The marriage rate affected the proportions of populations living in all four household types in the same way the fertility rate and demographic transition factors did. Late marriage and overall decline in the marriage rate have increased the proportion of the population living in single-person households and the probability that unmarried children continue to live with their parents, thereby increasing the proportion of the population living in two-generation households. Late marriage has also resulted in childbearing postponement, which in turn reduces the proportion of the population living in three-generation households. We understand that the inertia of the population itself has a great impact on the demographic transition. Therefore, when decomposing the effects of demographic factors on family transition, after excluding the effects of fertility rate, mortality rate and marriage rate, the remaining effect is that of demographic inertia. Demographic inertia has reduced the proportions of the populations living in single-person and two-generation households by 0.15% and 2.86%, respectively, and increased the proportions of the populations living in one-generation and three-generation or above households by 2.71% and 6.70%, respectively. It can be seen that during the period from 2000 to 2010, demographic inertia has a relatively large effect on the proportions of populations living in different types of households, but in some cases increasing and in others decreasing the proportions of populations living in particular types of households. ## Discussion and conclusion This paper divides the causes of family transition into two aspects: residence pattern (the most direct cause) and demographic factors. From the decomposition results, we found that residence pattern has a greater effect on demographic transition than demographic factors. This is evidence that the residence choices of Chinese families are of crucial importance in determining family structure in Chinese society. As a matter of fact, during the period of booming development, Chinese society has gone through in recent years, people’s attitudes toward certain things have changed rapidly. With regard to residence patterns, more and more young couples are choosing not to live with their parents after they become adults. This has a great impact on family transition. On the other hand, China’s current demographic transition has shifted from the traditional pattern featuring “high fertility, high mortality and low population growth” to a new pattern characterized by “low fertility, low mortality and low population growth”, and has entered a relatively stable stage. The result of this shift is that, compared to residence pattern, demographic factors are having a relatively weaker impact on family transition. According to modern Western theories of family transition, fertility rate, mortality rate and marriage rate all have an impact on family transition. In this paper, these three factors are decomposed from demographic factors and examined individually. The results demonstrate the interaction between fertility rate and mortality rate in affecting family transition, and shows that the “combined” effects of the two are actually “conflicting” effects. Marriage and childbearing are interconnected, and in China the effect of marriage on family transition is consistent with the effect of childbearing. It should be emphasized that in China, fertility level is largely determined by fertility policy, and in the future further, in-depth study must be carried out to analyze the effects of the “selective two-child policy” and “universal two-child policy” on the changes in family structure. How can the coverage of current family services be expanded in the future so they better contribute to family development? This shall become a target of the future work of government authorities (such as the newly established Family Development Department of the National Health and Family Planning Commission). Today, we cannot ignore the impacts of family transition on the development of important consumer markets. Plans to develop housing and urban planning in general must take family transition into consideration. Planners much also be aware of the impacts of family transition on demands for education, and medical and health care. Moreover, most Chinese people have traditional attitudes towards support of the elderly, believing such support should come primarily from the family. The transformation in family structure is certain to have an impact on the development elderly support patterns and the care service industry for elderly people. Therefore, continuous research on the influencing factors (and corresponding mechanisms) of the transformation in family structure is of vital significance for China to promote the development of important consumer markets in the future (such as real estate planning and development of the elderly care service industry) and for the government to formulate relevant family policies in order to meet the society’s reasonable expectations for stable development. ## Notes 1. 1. The SOCSIM microscopic simulation method developed by Eugene A. Hammel and Kenneth W. Wachte in the 1970 s has been in use for nearly 40 years, during which time continuous support has been provided by the National Institute on Aging, the National Science Foundation and the National Institute of Child Health and Human Development. There are reviews of the microscopic simulation of population in the book Family Demography (edited by Bongaarts, Burch and Wachter) and in an article written by Devos and Palloni. The article has been included in the Population Index. ## References 1. Hammel, E. A. (2005). Demographic dynamics and kinship in anthropological populations. Proceedings of the National academy of Sciences of the United States of America, 102(6), 2248–2253. 2. Hu, Z., & Peng, X. (2014). Household changes in contemporary China: An analysis based on census data. Sociological Studies, 3, 145–166. 3. Jiang, L., & O’Neill, B. C. (2007). Impacts of demographic trends on US household size and structure. Population and Development Review, 33(3), 567–591. 4. Wang, Y. (2006). The changing family structure in contemporary China: An analysis. Social Sciences in China, 1, 96–108. ## Author information Authors ### Corresponding author Correspondence to Shenghui Yang. ## Rights and permissions Reprints and Permissions Yang, S., Chen, W. Changes in family structure in China: the impact of residence patterns and demographic factors. China popul. dev. stud. 2, 401–411 (2019). https://doi.org/10.1007/s42379-018-00019-w

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http://link.springer.com/article/10.1007/s10652-014-9340-9

Environmental Fluid Mechanics , Volume 14, Issue 6, pp 1335–1355 On the periodicity of atmospheric von Kármán vortex streets Original Article DOI: 10.1007/s10652-014-9340-9 Nunalee, C.G. & Basu, S. Environ Fluid Mech (2014) 14: 1335. doi:10.1007/s10652-014-9340-9 Abstract For over 100 years, laboratory-scale von Kármán vortex streets (VKVSs) have been one of the most studied phenomena within the field of fluid dynamics. During this period, countless publications have highlighted a number of interesting underpinnings of VKVSs; nevertheless, a universal equation for the vortex shedding frequency ($$N$$) has yet to be identified. In this study, we have investigated $$N$$ for mesoscale atmospheric VKVSs and some of its dependencies through the use of realistic numerical simulations. We find that vortex shedding frequency associated with mountainous islands, generally demonstrates an inverse relationship to cross-stream obstacle length ($$L$$) at the thermal inversion height of the atmospheric boundary layer. As a secondary motive, we attempt to quantify the relationship between $$N$$ and $$L$$ for atmospheric VKVSs in the context of the popular Strouhal number ($$Sr$$)–Reynolds number ($$Re$$) similarity theory developed through laboratory experimentation. By employing numerical simulation to document the $$Sr{-}Re$$ relationship of mesoscale atmospheric VKVSs (i.e., in the extremely high $$Re$$ regime) we present insight into an extended regime of the similarity theory which has been neglected in the past. In essence, we observe mesoscale VKVSs demonstrating a consistent $$Sr$$ range of 0.15–0.22 while varying $$L$$ (i.e, effectively varying $$Re$$). Keywords Island wakes Marine boundary layer Stably stratified flows  Strouhal number Von Kármán vortex street

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https://planetmath.org/ergodicityofamapintermsofitsinducedoperator

# ergodicity of a map in terms of its induced operator Let $(X,\mathfrak{B},\mu)$ be a probability space and $T:X\longrightarrow X$ a measure-preserving transformation. The following statements are equivalent: 1. 1. - $T$ is ergodic. 2. 2. - If $f$ is a measurable function and $f\circ T=f$ a.e. (http://planetmath.org/AlmostSurely), then $f$ is constant a.e. 3. 3. - If $f$ is a measurable function and $f\circ T\geq f$ a.e., then $f$ is constant a.e. 4. 4. - If $f\in L^{2}(X)$ and $f\circ T=f$ a.e., then $f$ is constant a.e.. 5. 5. - If $f\in L^{p}(X)$, with $p\geq 1$, and $f\circ T=f$ a.e., then $f$ is constant a.e. $\,$ Let $U_{T}$ denote the operator induced by $T$ (http://planetmath.org/OperatorInducedByAMeasurePreservingMap), i.e. the operator defined by $U_{T}f:=f\circ T$. The statements above are statements about $U_{T}$. The above theorem can be rewritten as follows: $\,$ Theorem - Let $(X,\mathfrak{B},\mu)$ be a probability space and $T:X\longrightarrow X$ a measure-preserving transformation. The following statements are equivalent: 1. 1. - $T$ is ergodic. 2. 2. - The only fixed points of $U_{T}$ are the functions that are constant a.e. 3. 3. - If $f$ a measurable function and $U_{T}f\geq f$ a.e., then $f$ is constant a.e. 4. 4. - The eigenspace of $U_{T}$ (seen as an operator in $L^{p}(X)$, with $p\geq 1$) associated with the eigenvalue $1$, is one-dimensional and consists of functions that are constant a.e. Title ergodicity of a map in terms of its induced operator ErgodicityOfAMapInTermsOfItsInducedOperator 2013-03-22 17:59:22 2013-03-22 17:59:22 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 47A35 msc 37A30 msc 37A25 msc 28D05

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