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Skip to 0 minutes and 15 secondsHello and welcome back to a step in practice-- we are dealing here with absolute values. In this video, we are going to solve just exercises from 1 to 4, and we leave exercise 5 to the PDF. In exercise 1, we are asked to prove the triangle inequality-- the absolute value of x plus y, less or equal than the sum of the absolute values. Now let us recall that the inequality, absolute value of x less or equal than a is equivalent to the fact that minus a is less than x, and at the same time, x is less than a.
Skip to 0 minutes and 55 secondsThis is the situation-- if a is positive, the elements whose absolute value is less than or equal than a are just those between minus a and a. This is the case a positive. And if a is negative, strictly negative, there are no x satisfying the first inequality, and there are no x satisfying both of these inequalities. So if we put a equal to the sum of the absolute values, it's enough to prove that x plus y lies between a and minus a. Now x is between its absolute value and the opposite of its absolute value. At the same time, y is between its absolute value and the opposite of the absolute value.
Skip to 1 minute and 45 secondsSo if we take the sum, term by term, we get minus the sum of the two absolute values less or equal than x plus y, less or equal than the sum of the absolute values. And thus, this implies that the triangular inequality is true.
Skip to 2 minutes and 7 secondsIn exercise 2, we've got the function, which is the difference of two absolute values-- precisely, the absolute value of x minus 1 and the absolute value of 4x plus 8. Now, x minus 1-- while the absolute value of x minus 1 depends on the sign of x minus 1. And analogously, the absolute value of 4x plus 8 depends on the sign of 4x plus 8. Now x minus 1 is greater or equal than 0, if and only if x is greater than 1. Whereas 4x plus 8 is greater or equal than 0 if and only if 4x is greater or equal than 8. That is, x greater or equal than minus 2.
Skip to 2 minutes and 59 secondsSo we consider three cases, depending whether x is less than minus 2, between minus 2 and 1, or greater than 1. So if x is strictly less than minus 2, then 4x plus 8 is negative. And also x minus 1 is negative. And therefore f of x equals-- well instead of the absolute values of x minus 1, we can write minus x minus 1, minus, minus, 4x plus 8. Which is minus x plus 1, plus 4x plus 8. That is 3x plus 9. If x belongs to the interval between minus 2 and 1, we take, for convenience, minus 2, and we do not take 1.
Skip to 3 minutes and 57 secondsThen well, x is greater than minus 2, so 4x plus 8 is greater or equal than 0. But x minus 1 is negative. And thus, f of x equals minus x minus 1, minus 4x plus 8.
Skip to 4 minutes and 22 secondsAnd this equals minus 5x plus 1, minus 8. So minus 7. Finally, if x is greater or equal than 1, then both terms are positive. 4x plus 8 is greater or equal than 0, x plus 1, x minus 1 is greater or equal than 0. And thus, we get f of x equal to x minus 1, minus 4x plus 8, which is exactly minus 3x, minus 9.
Skip to 5 minutes and 8 secondsIn exercise 3, we are asked to find the formula for the maximum between two real numbers-- a and b.
Skip to 5 minutes and 20 secondsNow notice that if we draw the segment between a and b, well, we can have b be on the right hand side or a on the right hand side. But in any case, there are some quantities that do not depend on the position of a with respect to b, like the midpoint within between a and b, which is always equal to a plus b over 2. And also the distance between a and b, which is always the absolute value of b minus a. We get the maximum between a and b by adding to the midpoint, half of the distance between a and b.
Skip to 6 minutes and 10 secondsTherefore, the maximum between a and b is the midpoint plus the absolute value of b minus a over 2. And this is a formula involving just absolute values and sums and quotients. Now in exercise 4, we've three sets. Let us write them in a more compact way. The set A is the set of numbers from minus 3 to plus infinity, whereas the set B is the set of numbers that are strictly less than 6-- so minus infinity, 6. And C is the set of numbers whose distance to 5 is less than or equal to 2. And so we get the closed interval 5 minus 2, 5 plus 2. That is, the interval 3, 7-- with 3, and 7 on it.
Skip to 7 minutes and 14 secondsNow it is convenient in order to find the intersection of the three sets to draw each set as a subset of the real line. So we draw 3 real lines, and on each of the real lines, we represent the sets A, B, and C. So the values that count here are minus 3, 3, 6, and 7. So we consider minus 3, 3, 6, and 7.
Skip to 7 minutes and 49 secondsAnd in the first real line, we consider the set A, in the second set B, in the third the set C. And we draw one more line here with the intersection of A with B and C. So let us draw the set A is the set from minus 3 to plus infinity. And we take minus 3--
Skip to 8 minutes and 18 secondsthe set B goes from minus infinity to 6, but we do not take 6, so this is excluded.
Skip to 8 minutes and 32 secondsAnd also the set C goes from 3 to 7, so we take a set from 3 to 7. And we take 7. So be careful. We do not take 6, but we take 7, 3 here, and minus 3 here. So the intersection is what is in common between the three sets. So we look in each column. Here, we do not take this column, we do not take this column, but we take this one. For sure we'll take from 3 to 6, but be careful we do not take 6, and we take 3.
Skip to 9 minutes and 20 secondsWe do not take the column between 3, 6, and 7, and we don't take the column between 7 plus infinity. So the intersection of the three sets is the set from 3-- the interval from 3 to 6. We take 3, and we do not take 6. And this ends exercise 4. Exercise 5 is solved just in the PDF. So see you on the next step.
Absolute value in practice
The following exercises are solved in this step.
We invite you to try to solve them
before watching the video.
In any case, you will find below a PDF file with the solutions.
Exercise 1.
Prove the triangle inequality \(|x+y|\le |x|+|y|\) for all \(x,y\in\mathbb R\)
Exercise 2.
Write \(f(x) = |x-1| - |4x + 8|\) as an expression where there are no absolute values.
Exercise 3.
Express the maximum \(\max\lbrace a,b\rbrace\) betweenthe real numbers \(a\) and \(b\) in a formula involving just sums,products and the absolute value of real numbers. (
Hint: what is the distance from \(\max\lbrace a,b\rbrace\) to the midpoint of the segment joining \(a\) to \(b\)?) Exercise 4.
Let \(A=\lbrace x\in\mathbb R:\, x\ge -3\rbrace, B=\lbrace x\in\mathbb R:\, x<6\rbrace,\) \( C=\lbrace x\in \mathbb R:\, |x-5|\le 2\rbrace\). Find the set \(A\cap B\cap C\).
Exercise 5. [Solved only in the PDF file]
Knowing that \(\sqrt 3=1.732050808…\) give an estimate of \(|\sqrt 3-1.73205|\).
© Università degli Studi di Padova |
Paul, PKC and Sukumar, M and Bardi, R and Piazzesi, AM and Valle, G and Toniolo, C and BaIaram, P (1986)
Stereochemically Constrained Peptides. Theoretical and Experimental Studies on the Conformations of Peptides Containing 1-Aminocyclohexanecarboxylic Acid. In: Journal of the American Chemical Society, 108 (20). pp. 6363-6370.
PDF
Stereochemically_constrained_peptides.pdf
Restricted to Registered users only
Download (950kB) | Request a copy
Abstract
Conformational energy calculations on the model system N-acetyl- 1 -aminocyclohexanecarboxylic acid N'methylamide $(Ac-Acc^6-NHMe)$, using an average geometry derived from 13 crystallographic observations, establish that the $Acc^6$ residue is constrained to adopt conformations in the $3_{10}/\alpha$-helical regions of \o \psi space $(\o = \pm 50 \pm 20^0$ , $\psi = \pm 50 \pm 20^0)$. In contrast, the \alpha, \alpha -dialkylated residue with linear hydrocarbon side chains, \alpha, \alpha-di-n-propylglycine favors fully extended backbone structures $(\o \approx \psi \approx = 180^0)$. The crystal structures of two model peptides, Boc-($Acc^6)_3$-OMe (type III \beta -turn at $-Acc^6(1)-Acc^6(2)-)$ and Boc-Pro-$Acc^6$-Ala-OMe (type II \beta -turn at -Pro-$Acc^6$-), establish that $Acc^6$ residues can occupy either position of type III \beta turns and the i + 2 position of type II \beta -turns. The stereochemical rigidity of these peptides is demonstrated in solution by NMR studies, which establish the presence of one intramolecular hydrogen bond in each peptide in $CDCl_3$, and $(CD_3)_2SO$. Nuclear Overhauser effects permit characterization of the \beta -turn conformations in solution and establish their similarity to the solid-state structures. The implications for the use of $Acc^6$ residues in conformational design are considered.
Item Type: Journal Article Additional Information: Copyright of this article belongs to American Chemical Society. Department/Centre: Division of Biological Sciences > Molecular Biophysics Unit Depositing User: sidhartha sahoo Date Deposited: 11 Jul 2008 Last Modified: 19 Sep 2010 04:47 URI: http://eprints.iisc.ac.in/id/eprint/14919 Actions (login required)
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This question already has an answer here:
Show using the Poisson distribution that
$$\lim_{n \to +\infty} e^{-n} \sum_{k=1}^{n}\frac{n^k}{k!} = \frac {1}{2}$$
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This question already has an answer here:
Show using the Poisson distribution that
$$\lim_{n \to +\infty} e^{-n} \sum_{k=1}^{n}\frac{n^k}{k!} = \frac {1}{2}$$
By the definition of Poisson distribution, if in a given interval, the expected number of occurrences of some event is $\lambda$, the probability that there is exactly $k$ such events happening is $$ \frac {\lambda^k e^{-\lambda}}{k!}. $$ Let $\lambda = n$. Then the probability that the Poisson variable $X_n$ with parameter $\lambda$ takes a value between $0$ and $n$ is $$ \mathbb P(X_n \le n) = e^{-n} \sum_{k=0}^n \frac{n^k}{k!}. $$ If $Y_i \sim \mathrm{Poi}(1)$ and the random variables $Y_i$ are independent, then $\sum\limits_{i=1}^n Y_i \sim \mathrm{Poi}(n) \sim X_n$, hence the probability we are looking for is actually $$ \mathbb P\left( \frac{Y_1 + \dots + Y_n - n}{\sqrt n} \le 0 \right) = \mathbb P( Y_1 + \dots + Y_n \le n) = \mathbb P(X_n \le n). $$ By the central limit theorem, the variable $\frac {Y_1 + \dots + Y_n - n}{\sqrt n}$ converges in distribution towards the Gaussian distribution $\mathscr N(0, 1)$. The point is, since the Gaussian has mean $0$ and I want to know when it is less than equal to $0$, the variance doesn't matter, the result is $\frac 12$. Therefore, $$ \lim_{n \to \infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!} = \lim_{n \to \infty} \mathbb P(X_n \le n) = \lim_{n \to \infty} \mathbb P \left( \frac{Y_1 + \dots + Y_n - n}{\sqrt n} \le 0 \right) = \mathbb P(\mathscr N(0, 1) \le 0) = \frac 12. $$
Hope that helps, |
I think I have the right formulas the answer the following question, but there is something conceptual that is throwing me off (see end of question).
A body with mass $m$ is located on a slope with angle $\alpha$ (see picture).
A. What should be the size of friction between the body and the inclined plane if it is known that the body does not move across the inclined plane, when the inclined plane is moving at acceleration $a_0$ rightwards?
B. Create an expression for the minimal size of $\mu_s$ in order for that to happen.
If $a_0$ is "small":
$$ma_x=ma_0\cos\alpha-mg\sin\alpha+F_s=0$$
$$ma_y=N-mg\cos\alpha-ma_0\sin\alpha=0$$
$$F_s=mg\sin\alpha-ma_0\cos\alpha$$
$$N = mg\cos\alpha + ma_0\sin\alpha$$
$$\mu_s = \frac {mg\sin\alpha-ma_0\cos\alpha}{mg\cos\alpha + ma_0\sin\alpha}=\frac {g\sin\alpha-a_0\cos\alpha}{g\cos\alpha + a_0\sin\alpha}$$
If $a_0$ is "big":
$$ma_x=ma_0\cos\alpha-mg\sin\alpha -F_s=0$$
$$ma_y=N-mg\cos\alpha-ma_0\sin\alpha=0$$
$$F_s=ma_0\cos\alpha-mg\sin\alpha$$
$$N = mg\cos\alpha + ma_0\sin\alpha$$
$$\mu_s = \frac {ma_0\cos\alpha-mg\sin\alpha}{mg\cos\alpha + ma_0\sin\alpha}=\frac {a_0\cos\alpha-g\sin\alpha}{g\cos\alpha + a_0\sin\alpha}$$
Here come the questions:
I believe $F_s$ in my calculation inherently refers to the
maximumstatic friction, but I am not sure of this.
I have no idea what is meant by "the minimal size" of $\mu_s$. I thought it depended only on the materials, and therefore always has the same value. How should I interpret the word "minimal" here? |
12005
12025
Deletions are marked like this. Additions are marked like this. Line 63: Line 63: || time min || orbit degrees || rotation rate || sun angle || Illumination || Night Light ||
|| 0 to 60 || 0° to 90° ||$0 ~ \Large\omega$|| 0° || 100% || 0W ||
|| 60 to 100 || 90° to 150° ||$1.5 ~ \Large\omega$|| 0° to 90° || 100% to 0% || 0W ||
|| 100 to 140 || 150° to 210° ||$3 ~ \Large\omega$|| 90° to 270° || Eclipse || 0W ||
|| 140 to 180 || 210° to 270° ||$1.5 ~ \Large\omega$|| 270° to 0° || 0% to 100% || 0W ||
|| 180 to 240 || 270° to 0° ||$0 ~ \Large\omega$|| 0° || 100% || 0W ||
|| time min || orbit degrees || rotation rate || sun angle || Illumination || Night Light ||
|| 0 to 60 || 0° to 90° ||$0 \Large\omega$|| 0° || 100% || 0W ||
|| 60 to 100 || 90° to 150° ||$1.5 \Large\omega$|| 0° to 90° || 100% to 0% || 0W ||
|| 100 to 140 || 150° to 210° ||$3 \Large\omega$|| 90° to 270° || Eclipse || 0W || Start at 3.333 \Large\omega$||
|| 140 to 180 || 210° to 270° ||$1.5 \Large\omega$|| 270° to 0° || 0% to 100% || 0W ||
|| 180 to 240 || 270° to 0° ||$0 \Large\omega$|| 0° || 100% || 0W ||
Night Side Maneuvers
We can minimize night light pollution, and advance perigee against light pressure orbit distortion, by turning the thinsat as we approach eclipse. The overall goal is to perform 1 complete rotation of the thinsat per orbit, with it perpendicular to the sun on the day-side of the earth, but turning it by varying amounts on the night side.
Another advantage of the turn is that if thinsat maneuverability is destroyed by radiation or a collision on the night side, it will come out of night side with a slow tumble that won't be corrected. The passive radar signature of the tumble will help identify the destroyed thinsat to other thinsats in the array, allowing another sacrificial thinsat to perform a "rendezvous and de-orbit". If the destroyed thinsat is in shards, the shards will tumble. The tumbling shards ( or a continuously tumbling thinsat ) will eventually fall out of the normal orbit, no longer get J_2 correction, and the thinsat orbit will "eccentrify", decay, and reenter. This is the fail-safe way the arrays will reenter, if all active control ceases.
Maneuvering thrust and satellite power
Neglecting tides, the synodic angular velocity of the m288 orbit is \Large\omega = 4.3633e-4 rad/sec = 0.025°/s. The angular acceleration of a thinsat is 13.056e-6 rad/sec
2 = 7.481e-4°/s 2 with a sun angle of 0°, and 3.740e-4°/s 2 at a sun angle of 60°. Because of tidal forces, a thinsat entering eclipse will start to turn towards sideways alignment with the center of the earth; it will come out of eclipse at a different velocity and angle than it went in with.
If the thinsat is rotating at \omega and either tangential or perpendicular to the gravity vector, it will not turn while it passes into eclipse. Otherwise, the tidal acceleration is \ddot\theta = (3/2) \omega^2 \sin 2 \delta where \delta is the angle to the tangent of the orbit. If we enter eclipse with the thinsat not turning, and oriented directly to the sun, then \delta = 30° .
Three Strategies and a Worst Case Failure Mode
There are many ways to orient thinsats in the night sky, with tradeoffs between light power harvest, light pollution, and orbit eccentricity. If we reduce power harvest, we will need to launch more thinsats to compensate, which makes more problems if the system fails. I will present three strategies for light harvest and nightlight pollution. The actual strategies chosen will be blend of those.
Tumbling
If things go very wrong, thinsats will be out of control and tumbling. In the long term, the uncontrolled thinsats will probably orient flat to the orbital plane, and reflect very little light into the night sky, but in the short term (less than decades), they will be oriented in all directions. This is equivalent to mapping the reflective area of front and back (2 π R
2 ) onto a sphere ( 4 π R 2 ). Light with intensity I shining onto a sphere of radius R is evenly reflected in all directions uniformly. So if the sphere intercepts π R 2 I units of light, it scatters e I R 2/4 units of light (e is albedo) per steradian in all directions. While we will try to design our thinsats with low albedo ( high light absorption on the front, high emissivity on the back), we can assume they will get sanded down and more reflective because of space debris, and they will get broken into fragments of glass with shiny edges, adding to the albedo. Assume the average albedo is 0.5, and assume the light scattering is spherical for tumbling.
Source for the above animation: g400.c
Three design orientations
All three orientations shown are oriented perpendicularly in the daytime sky.
Max remains perpendicular in the night sky, Min is oriented vertically in the night sky, and Zero is edge on to the terminator in the night sky. All lose orientation control and are tilted by tidal forces in eclipse - the compensatory thrusts are not shown. Min and Zero are accelerated into a slow turn before eclipse, so they come out of the eclipse in the correct orientation. In all cases, there will probably be some disorientation and sun-seeking coming out of eclipse, until each thinsat calibrates to the optimum inertial turn rate during eclipse. So, there may be a small bit of sky glow at the 210° position, directly overhead at 2am and visible in the sky between 10pm and 6am.
Max NLP: Full power night sky coverage, maximum night light pollution
The most power is harvested if the thinsats are always oriented perpendicular to the sun. During the half of their orbit into the night sky, there will be some diffuse reflection to the side, and some of that will land in the earth's night sky. The illumination is maximum along the equator. For the M288 orbit, about 1/6th of the orbit is eclipsed, and 1/2 of the orbit is in daylight with the diffuse (Lambertian) reflection scattering towards the sun and onto the day side of the earth. Only the two "horns" of the orbit, the first between 90° and 150° (6pm to 10pm) and the second between 210° and 270° (2am to 6am) will reflect night into the light sky. The light harvest averages to 83% around the orbit.
This is the worst case for night sky illumination. Though it is tempting to run thinsats in this regime, extracting the maximum power per thinsat, it is also the worst case for eccentricity caused by light pressure, and the thinsats must be heavier to reduce that eccentricity.
Min NLP: Partial night sky coverage, some night light pollution
This maneuver will put some scattered light into the night sky, but not much compared to perpendicular solar illumination all the way into shadow. In the worst case, assume that the surface has an albedo of 0.5 (typical solar cells with an antireflective coating are less than 0.3) and that the reflected light is entirely Lambertian (isotropic) without specular reflections (which will all be away from the earth). At a 60° angle, just before shadow, the light emitted by the front surface will be 1366W/m
2 × 0.5 (albedo) × 0.5 ( cos 60° ), and it will be scattered over 2π steradians, so the illumination per steradian will be 54W/m 2-steradian just before entering eclipse.
Estimate that the light pollution varies from 0W to 54W between 90° and 150° and that the average light pollution is half of 54W, for 1/3 of the orbit. Assuming an even distribution of thinsat arrays in the constellation, that is works out to an average of 9W/m
2-steradian for all thinsats in M288 orbit.
The full moon illuminates the night side of the equatorial earth with 27mW/m
2 near the equator. A square meter of thinsat at 6400km distance produces 9W/6400000 2 or 0.22 picowatts per m 4 times the area of all thinsats. If thinsat light pollution is restricted to 5% of full moon brightness (1.3mW/m 2), then we can have 6000 km 2 of thinsats up there, at an average of 130 W/m 2, or about 780GW of thinsats at m288. That is about a million tons of thinsats.
The orientation of the thinsat over a 240 minute synodic m288 orbit at the equinox is as follows, relative to the sun:
time min
orbit degrees
rotation rate
sun angle
Illumination
Night Light
0 to 60
0° to 90°
0 ~ \Large\omega
0°
100%
0W
60 to 100
90° to 150°
1 ~ \Large\omega
0° to 60°
100% to 50%
0W to 54W
100 to 140
150° to 210°
4 ~ \Large\omega
60° to 300°
Eclipse
0W
140 to 180
210° to 270°
1 ~ \Large\omega
300° to 0°
50% to 100%
54W to 0W
180 to 240
270° to 0°
0 ~ \Large\omega
0°
100%
0W
The angular velocity change at 0° takes 250/7.481 = 33.4 seconds, and during that time the thinsat turns 0.42° with negligible effect on thrust or power. The angular velocity change at 60° takes 750/3.74 = 200.5 seconds, and during that time the thinsat turns 12.5°, perhaps from 53.7° to 66.3°, reducing power and thrust from 59% to 40%, a significant change. The actual thrust change versus time will be more complicated (especially with tidal forces), but however it is done, the acceleration must be accomplished before the thinsat enters eclipse.
The light harvest averages 78% around the orbit.
Zero NLP: Partial night sky coverage, no night light pollution
In this case, in the night half of the sky the edge of the thinsat is always turned towards the terminator. As long as the thinsats stay in control, they will never produce any nighttime light pollution, because the illuminated side of the thinsat is always pointed away from the night side of the earth. The average illumination fraction is around 68%.
The orientation of the thinsat over a 240 minute synodic m288 orbit at the equinox is as follows, relative to the sun:
time min
orbit degrees
rotation rate
sun angle
Illumination
Night Light
0 to 60
0° to 90°
0 \Large\omega
0°
100%
0W
60 to 100
90° to 150°
1.5 \Large\omega
0° to 90°
100% to 0%
0W
100 to 140
150° to 210°
3 \Large\omega
90° to 270°
Eclipse
0W
Start at 3.333 \Large\omega$
140 to 180
210° to 270°
1.5 \Large\omega
270° to 0°
0% to 100%
0W
180 to 240
270° to 0°
0 \Large\omega
0°
100%
0W
Pedants and thinsat programmers take note: The actual synodic orbit period is 240 minutes and 6.57 seconds long; that results in 2190.44 rather than 2191.44 sidereal orbits per year, accounting for the annual apparent motion of the sun around the sky.
The light harvest averages 67% around the orbit. Why would a profit maximizing operator settle for 67% when 83% was possible?
Infrared-filtering thinsats reduce launch weight and can use the Zero NLP flip to increase the minimum temperature of a thinsat during eclipses. An IR filtering thinsat in maximum night light pollution mode will have the emissive backside pointed at 2.7K space when it enters eclipse; the thinsat temperature will drop towards 20K if it cannot absorb the 64W/m
2 of 260K black body radiation reaching it from the earth through the 3.5μm front side infrared filter. The thinsat will become very brittle at those temperatures, and the thermal shock could destroy it. If the high thermal emissivity back side is pointed towards the 260K earth, the temperature will drop to 180K - still challenging, but the much higher thermal mobility may heal atomic-scale damage.
Details of the Zero NLP maneuver
In the night sky, assuming balanced thrusters, only tidal forces act on the thinsat, \ddot\theta = -(3/2) \omega^2 \sin( 2 \theta . Integrating numerically from the proper choice of initial rotation rate,
Power versus angle
Night light pollution versus hour
The night light pollution for 1 Terawatt of thinsats at M288. Mirror the graph for midnight to 6am. Some light is also put into the daytime sky (early morning and late afternoon), but it will be difficult to see in the glare of sunlight. The Zero NLP option puts no light in the night sky, so that curve is far below the bottom of this graph.
Source for the above two graphs: nl02.c |
I would like to add another answer to this old question. Considerthe case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$:
$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$
Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in factwe can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the
universal point $S = R[T_0]$, $a_0 = T_0$, we see that the ``universal Taylor expansion'' of $f$ is$$f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T].$$If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as$$1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i$$Looking mod $(1\otimes T-T\otimes 1)^2$ we get:$$1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2}$$
Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by$1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear termsof Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function$f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$.
This also explains nicely what happens in higher degree. We can introduce the algebras$P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the
ring of Taylor expansions of degree $\leq n$ wherethe terms of degree at most $n$ of the Taylor expansion live. There is a natural map$d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion.
This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), becauselocally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and thismap induces an isomorphism on $I/I^2$ and $P^n$ more generally. |
Difference between revisions of "Huge"
m (→Ultrafilter definition)
Line 24: Line 24:
=== Ultrafilter definition ===
=== Ultrafilter definition ===
−
The first-order definition of $n$-huge is somewhat similar to [[measurable|measurability]]. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete [[filter|ultrafilter]], $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $
+
The first-order definition of $n$-huge is somewhat similar to [[measurable|measurability]]. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete [[filter|ultrafilter]], $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\lambda$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n\forall x\subseteq\lambda(\mathrm{ot}(x\cap\lambda_{i+1})=\lambda_i\rightarrow x\in U)$$
$$\forall i<n\forall x\subseteq\lambda(\mathrm{ot}(x\cap\lambda_{i+1})=\lambda_i\rightarrow x\in U)$$
Revision as of 23:34, 29 November 2017
Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $\text{ZFC}$+"there is a $\aleph_2$-saturated $\sigma$-complete ideal on $\omega_1$". [1]
Contents Definitions
Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties $n$-$P_0$ and $n$-$P_1$, $n$-$P_0$ has less consistency strength than $n$-$P_1$, which has less consistency strength than $(n+1)$-$P_0$, and so on. This phenomenon is seen only around the $n$-fold variants as of modern set theoretic concerns. [2]
Although they are very large, there is a first-order definition which is equivalent to $n$-hugeness, so the $\theta$-th $n$-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability.
Elementary embedding definitions $\kappa$ is almost $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subseteq M$). $\kappa$ is $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subseteq M$). $\kappa$ is almost $n$-hugeiff it is almost $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is $n$-hugeiff it is $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is super almost $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost $n$-huge with target $\lambda$ (that is, the target can be made arbitrarily large). $\kappa$ is super $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is $n$-huge with target $\lambda$. $\kappa$ is almost huge, huge, super almost huge, and superhugeiff it is almost $1$-huge, $1$-huge, etc. respectively. Ultrafilter definition
The first-order definition of $n$-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\lambda$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n\forall x\subseteq\lambda(\mathrm{ot}(x\cap\lambda_{i+1})=\lambda_i\rightarrow x\in U)$$
Where $\mathrm{ot}(X)$ is the order-type of the poset $(X,\in)$. [1] This definition is, more intuitively, making $U$ very large, like most ultrafilter characterizations of large cardinals (supercompact, strongly compact, etc.). $\kappa$ is then super $n$-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is $n$-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. [1]
If $j:V\to M$ is such that $M^{j^n(\kappa)}\subseteq M$ (i.e. $j$ witnesses $n$-hugeness) then there is a ultrafilter $U$ as above such that, for all $k\leq n$, $\lambda_k = j^k(\kappa)$, i.e. it is not only $\lambda=\lambda_n$ that is an iterate of $\kappa$ by $j$; all members of the $\lambda_k$ sequence are.
Consistency strength and size
Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the $n$-fold variants) known as the
double helix. This phenomenon is when for one $n$-fold variant, letting a cardinal be called $n$-$P_0$ iff it has the property, and another variant, $n$-$P_1$, $n$-$P_0$ is weaker than $n$-$P_1$, which is weaker than $(n+1)$-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest): measurable = $0$-superstrong = almost $0$-huge = super almost $0$-huge = $0$-huge = super $0$-huge $n$-superstrong $n$-fold supercompact $(n+1)$-fold strong, $n$-fold extendible $(n+1)$-fold Woodin, $n$-fold Vopěnka $(n+1)$-fold Shelah almost $n$-huge super almost $n$-huge $n$-huge super $n$-huge $(n+1)$-superstrong
All huge variants lay at the top of the double helix restricted to some natural number $n$, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of $n$-huge cardinals, for all $n$. [1]
Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every $(n+1)$-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is super $n$-huge". [1]
In terms of size, however, the least $n$-huge cardinal is smaller than the least supercompact cardinal. Assuming both exist, for any $\kappa$ which is supercompact and has an $n$-huge cardinal above it, there are $\kappa$ many $n$-huge cardinals less than $\kappa$. [1]
Every $n$-huge cardinal is $m$-huge for every $m\leq n$. Similarly with almost $n$-hugeness, super $n$-hugeness, and super almost $n$-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1]
References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex |
Are there any signals whose autocorrelation $R(\tau)$ has the following form? Assuming $\tau_c > 0$ and $R_0 > 0$ a constant,
$$R(\tau) = \begin{cases}R_0, \text{ for $|\tau| < \tau_c$} \\ 0, \text{ otherwise}\end{cases}$$
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It is
notpossible for a (WSS) random process $\{X(t)\}$to have an autocorrelation function of the form $$R_X(\tau) = E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & |\tau|\leq T,\\0, & |\tau| > T. \end{cases}$$
Assuming a zero-mean process for simplicity, note that $R_X(0) = E[X^2(t)] = \sigma^2$ where $\sigma^2$ is the common variance of the random variables. Suppose that the highlighted statement above is true and that $T = 2$ and thus $R_X(1) = E[X(t)X(t+1)] = \operatorname{cov}(X(t), X(t+1))$ also has the same value $\sigma^2$. Then, $X(t)$ and $X(t+1)$ are
perfectly correlated random variables and thus are equal in the mean-square sense -- their difference has variance $0$:\begin{align}\operatorname{var}(X(t) - X(t+1)) &= \operatorname{var}(X(t))+\operatorname{var}(X(t+1))- 2\operatorname{cov}(X(t), X(t+1))\\ &= \sigma^2+\sigma^2-2\sigma^2\\&=0.\end{align} and thus $X(t) = X(t+1)$ in the mean-square sense.But then, since $t$ is arbitrary, we have that\begin{align}X(t+1) &= X((t+1) + 1) = X(t+2)\\X(t+2) &= X((t+2) + 1) = X(t+3).\end{align} So, $X(t) = X(t+1) = X(t+2) = X(t+3)$ so that $R_X(3) = E[X(t)X(t+3)]=\sigma^2$ also, contrary to the hypothesis that $R_X(\tau) = 0$ when $|\tau| > 2$.
In summary,$R_X(\tau)$ cannot have a fixed nonzero value in a
finite-length interval including the origin and value $0$ outside this finite-length interval. |
The red-shift of the light of a star in a galaxy or that of a galaxy in a cluster of galaxies is generally interpreted as how fast the star or the galaxy is moving, i.e. it is interpreted in a purely special-relativistic way. However, general relativity predicts that a light produced in a gravitational field gets red-shifted when comes out of the field. I wonder why the excess red-shift (over the Hubble red-shift) of stars and galaxies is only interpreted as how fast the object moves, not that how strong the ambient gravitational field is.
A Galaxy cluster could have $10^{14}$ solar masses within a radius of 5 Mpc.
In this case $GM/Rc^2 \sim 10^{-6}$, equivalent to a velocity shift of less than 1 km/s.
Our own Milky Way has a mass of around $10^{12}$ solar masses within 100 kpc. This gives a gravitational redshift of about 100 m/s.
These are completely negligible compared to cosmological redshifts and the peculiar velocities within clusters or groups of galaxies (the latter are of order 100-1000 km/s).
EDIT: To clarify, as a result of Lubos's comment (see below). Because the Earth is in the Milky Way, light is gravitationally
blueshifted on it's way into the Milky Way and also into the "local group" of galaxies. However that light has emerged from a different galaxy in a different environment that will be redshifted as it makes its way out of that potential. These two shifts will not in general cancel because galaxies and galaxy clusters have a variety of masses, sizes and potential depths. Hence the numbers I give are the correct orders of magnitude for the errors introduced by ignoring gravitational redshift, but any exact correction needs to be calculated on a case-by-case basis.
FURTHER EDIT: The comment above is even more apt in the light of the literature referred to by Pulsar. For example Cappi (1995) model the (more realistic) potentials of rich clusters and show that the redshift is a strong function of where the galaxy is in the cluster, but could be anywhere in the range from less than 1 km/s to 300 km/s at the centres of the most massive clusters. This is a lot larger than my estimate above because densities in clusters vary more steeply than $r^{-2}$. However, this is still
small compared to their velocity dispersions within the same clusters, because more massive clusters also have higher intrinsic velocity dispersions.
The gravitational red shift is only significant for black holes – where the coefficient may grow arbitrarily large in the vicinity of the horizon – and the neutron stars – where the frequency drops to something comparable to 50%.
For all other celestial objects, the red shift is much smaller than one. And only planets and white dwarfs are objects for which the red shift may be easily detectable. It's helpful to calculate the red shift for the Sun. The red shift is given by the gravitational potential $$ \frac{\Delta f}{f} = - \frac{GM}{Rc^2} $$ For the Sun, the relative decrease of the frequency may be calculated e.g. via Wolfram Alpha
and it is just $2\times 10^{-6}$; two parts per million. Note that the Hubble constant is about $10^{-10}$ per year – the inverse of 14 billion years, more precisely, so the distance needed to reduce the frequency relatively by two parts per million is $2\times 10^{-6}$ times 14 billion light years which is just 28,000 light years. That's still inside our galaxy!
For all other galaxies, the cosmological red shift is much greater than the gravitational red shift from Sun-like stars. The Sun is just too large, by radius, too diluted. This smallness of the gravitational potential becomes even more extreme if you consider the "groups" of stars such as galaxies themselves. |
Example 12.18
Consider two–dimensional flat thin plate at an angle of attack of \(4^\circ\) and a Mach number of 3.3. Assume that the specific heat ratio at stage is \(k=1.3\), calculate the drag coefficient and lift coefficient.
Solution 12.18
For \(M=3.3\), the following table can be obtained:
Prandtl — Meyer Input: \(M\) k = 1.4 \(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\) 3.300 62.3113 0.01506 0.37972 0.03965 73.1416
With the angle of attack the region 3 will be at \(\nu \sim 62.31+4\) for which the following table can be obtained (Potto-GDC)
Prandtl — Meyer Input: \(M\) k = 1.4 \(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\) 3.4996 66.3100 0.01090 0.35248 0.03093 74.0528
On the other side, the oblique shock (assuming weak shock) results in
Oblique Shock Input: \(M_x\) k = 1.4 \(M_x\) \({{M_y}_s}\) \({{M_y}_w}\) \(\theta_{s}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 3.300 0.43534 3.1115 88.9313 20.3467 4.0000 0.99676
and the additional information, by clicking on the minimal button, provides
Oblique Shock Input: \(M_x\) k = 1.4 \(M_x\) \({{M_y}_w}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P}_y}{{P}_x}\) \(\dfrac{{T}_y}{{T}_x}\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 3.300 3.1115 20.3467 4.0000 1.1157 1.1066 0.99676
The pressure ratio at point 3 is
\begin{align*} \dfrac{P_3 }{ P_1} = \dfrac{P_3 }{ P_{03} } \, \dfrac{P_{03} }{ P_{01}} \, \dfrac{P_{01} }{ P_{1}} = 0.0109 \times 1 \times \dfrac{1 }{ 0.01506} \sim 0.7238 \end{align*} The pressure ratio at point 4 is \begin{align*} {P_3 \over P_1} = 1.1157 \end{align*} \begin{align*} d_L = {2 \over k P_1 {M_1}^2 }(P_4 - P_3) \cos \alpha = {2 \over k {M_1}^2 } \left( {P_4 \over P_1} - {P_3 \over P_1} \right) \,\cos \alpha \end{align*} \begin{align*} d_L = { 2 \over 1.3 3.3^2 } \left( 1.1157 - 0.7238 \right) \cos 4^\circ \sim .054 \end{align*} \begin{multline*} d_d = \dfrac{2 }{ k\, {M_1}^2 } \left( \dfrac{P_4 }{ P_1} - \dfrac{P_3 }{ P_1} \right) \sin \alpha = \dfrac{ 2 }{ 1.3\, 3.3^2 } \left( 1.1157 - 0.7238 \right) \sin 4^\circ \sim .0039 \end{multline*} This shows that on the expense of a small drag, a large lift can be obtained. Discussion on the optimum design is left for the next versions.
Example 12.19
To understand the flow after a nozzle consider a flow in a nozzle shown in Figure 12.31. The flow is choked and additionally the flow pressure reaches the nozzle exit above the surrounding pressure. Assume that there is an isentropic expansion (Prandtl–Meyer expansion) after the nozzle with slip lines in which there is a theoretical angle of expansion to match the surroundings pressure with the exit. The ratio of exit area to throat area ratio is 1.4. The stagnation pressure is 1000 [kPa]. The surroundings pressure is 100[kPa]. Assume that the specific heat, \(k=1.3\). Estimate the Mach number after the expansion.
Solution 12.19
The Mach number a the nozzle exit can be calculated using Potto-GDC which provides
Isentropic Flow Input: \(\dfrac{A}{A^{\star}}\) k = 1.3 \(M\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\dfrac{A}{A^{\star}}\) \(\dfrac{P}{P_0}\) \(\dfrac{{A\,P}_y}{{A^{\star}\,P_0}_x}\) \(\dfrac{F}{F_0}\) 1.7285 0.69052 0.29102 1.4000 0.20096 0.28134 0.59745
Thus, the exit Mach number is 1.7285 and the pressure at the exit is
\begin{align*} P_{exit} = P_0 \dfrac{P_{\text{exit}}}{P_0} = 1000 \times 0.20096 = 200.96 [kPa] \end{align*} This pressure is higher than the surroundings pressure and an expansion must occur. This pressure ratio is associated with a expansion angle that Potto-GDC provide as
Oblique Shock Input: \(M_x\) k = 1.3 \(M_x\) \({{M_y}_w}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P}_y}{{P}_x}\) \(\dfrac{{T}_y}{{T}_x}\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 3.300 3.1115 20.3467 4.0000 1.1157 1.1066 0.99676
The final pressure ratio ultimately has to be
\begin{align*} \dfrac{ P_{\text{surroundings}}} {P_{0}} = \dfrac{100}{1000} = .1 \end{align*} Hence the information for this pressure ratio can be provided by Potto-GDC as
Oblique Shock Input: \(M_x\) k = 1.3 \(M_x\) \({{M_y}_w}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P}_y}{{P}_x}\) \(\dfrac{{T}_y}{{T}_x}\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 3.300 3.1115 20.3467 4.0000 1.1157 1.1066 0.99676
The change of the angle is
\begin{align*} \Delta \text{angle} = 30.6147 - 20.0641 = 10.5506 \end{align*} Thus the angle, $\beta$ is \begin{align*} \beta = 90 - 10.5506 \sim 79.45 \end{align*} The pressure at this point is as the surroundings. However, the stagnation pressure is the same as originally was enter the nozzle! This stagnation pressure has to go through serious of oblique shocks and Prandtl-Meyer expansion to match the surroundings stagnation pressure. Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
Table of Contents
Coupled coincidence point results for contraction of $C$-class mappings in ordered uniform spaces Article References A. H. Ansari, D. Binbasioglu, D. Turkoglu 3-13
Some weaker sufficient conditions of $L$-index boundedness in direction for functions analytic in the unit ball Article References A. I. Bandura 14-25
Asymptotics of the entire functions with $\upsilon$-density of zeros along the logarithmic spirals Article References M. V. Zabolotskyj, Yu. V. Basiuk 26-32
Representation of a quotient of solutions of a four-term linear recurrence relation in the form of a branched continued fraction Article References I. Bilanyk, D. I. Bodnar, L. Buyak 33-41
Note on bases in algebras of analytic functions on Banach spaces Article References I. V. Chernega, A. V. Zagorodnyuk 42-47
Spectral approximations of strongly degenerate elliptic differential operators Article References M. I. Dmytryshyn, O. V. Lopushansky 48-53
On some of convergence domains of multidimensional S-fractions with independent variables Article References R. I. Dmytryshyn 54-58
Ricci soliton and Ricci almost soliton within the framework of Kenmotsu manifold Article References A. Ghosh 59-69
Interconnection between Wick multiplication and integration on spaces of nonregular generalized functions in the Levy white noise analysis Article References N. A. Kachanovsky, T. Kachanovska 70-88
Algebraic basis of the algebra of block-symmetric polynomials on $\ell_1 \oplus \ell_{\infty}$ Article References V. V. Kravtsiv 89-95
The relationship between algebraic equations and $(n,m)$-forms, their degrees and recurrent fractions Article References I. I. Lishchynsky 96-106
Inverse problem for $2b$-order differential equation with a time-fractional derivative Article References A. O. Lopushansky, H. P. Lopushanska 107-118
Some inequalities for strongly $(p,h)$-harmonic convex functions Article References M. A. Noor, K. I. Noor, S. Iftikhar 119-135
Characterizations of regular and intra-regular ordered $\Gamma$-semihypergroups in terms of bi-$\Gamma$-hyperideals Article References S. Omidi, B. Davvaz, K. Hila 136-151
On a new application of quasi power increasing sequences Article References H. S. Özarslan 152-157
On approximation of homomorphisms of algebras of entire functions on Banach spaces Article References H. M. Pryimak 158-162
On the solutions of a class of nonlinear integral equations in cone $b$-metric spaces over Banach algebras Article References L. T. Quan, T. Van An 163-178
Classification of generalized ternary quadratic quasigroup functional equations of the length three Article References F. M. Sokhatsky, A. V. Tarasevych 179-192
On integral representation of the solutions of a model $\vec{2b}$-parabolic boundary value problem Article References N. I. Turchyna, S. D. Ivasyshen 193-203
The journal is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported. |
Pure three dimensional stream functions exist physically but at present there is no known way to represent then mathematically. One of the ways that was suggested by Yih in 1957 suggested using two stream functions to represent the three dimensional flow. The only exception is a stream function for three dimensional flow exists but only for axisymmetric flow i.e the flow properties remains constant in one of the direction (say z axis).
Advance Material
The three dimensional representation is based on the fact the continuity equation must be satisfied. In this case it will be discussed only for incompressible flow. The \(\nabla \pmb{U} = 0\) and vector identity of \(\nabla \cdot \nabla \pmb{U} = 0 \) where in this case \(\pmb{U}\) is any vector. As opposed to two dimensional case, the stream function is defined as a vector function as
\[ \label{if:eq:3DstreamFun} \pmb{B} = \psi \,\nabla \xi \tag{65} \] The idea behind this definition is to build stream function based on two scalar functions one provide the "direction'' and one provides the the magnitude. In that case, the velocity (to satisfy the continuity equation) \[ \label{if:eq:velocityVector} \pmb{U} = \boldsymbol{\nabla} \boldsymbol{\times} \left( \psi \,\boldsymbol{\nabla} \chi \right) \tag{66} \] where \(\psi\) and \(\chi\) are scalar functions. Note while \(\psi\) is used here is not the same stream functions that were used in previous cases. The velocity can be obtained by expanding equation (66) to obtained \[ \label{if:eq:UstreamFun1} \pmb{U} = \boldsymbol{\nabla}\psi \boldsymbol{\times} \boldsymbol{\nabla}\chi + \psi \,\overbrace{\boldsymbol{\nabla} \boldsymbol{\times}\left( \boldsymbol{\nabla}\chi\right)}^{=0} \tag{67} \] The second term is zero for any operation of scalar function and hence equation (67) becomes \[ \label{if:eq:UstreamFun} \pmb{U} = \boldsymbol{\nabla}\psi \boldsymbol{\times} \boldsymbol{\nabla}\chi \tag{68} \] These derivations demonstrates that the velocity is orthogonal to two gradient vectors. In another words, the velocity is tangent to the surfaces defined by \(\psi = constant\) and \(\chi = constant\). Hence, these functions, \(\psi\) and \(\chi\) are possible stream functions in three dimensions fields. It can be shown that the flow rate is \[ \label{if:eq:stream3DFlowRate} \dot{Q} = \left(\psi_2 - \psi_1\right) \left( \chi - \chi_1 \right) \tag{69} \] The answer to the question whether this method is useful and effective is that in some limited situations it could help. In fact, very few research papers deals this method and currently there is not analytical alternative. Hence, this method will not be expanded here.
End Advance Material
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
This article is cited in scientific papers (total in 2 2 papers) Germs of mappings $\omega$-determined with respect to a given group G. R. Belitskii Abstract: Let $ J(n,p)$ be the space of germs of $C^\infty$-mappings $F\colon(R^n,0)\to(R^p,0)$ and $\mathfrak G$ a group operating on $J(n,p)$. The germ $F\in J(n,p)$ is called finitely determined with respect to $\mathfrak G$ if there exists an integer $k$ such that the orbit of the germ $F$ under the action of $\mathfrak G$ is uniquely determined by the $k$-jet of the germ $F$. The germ $F$ is called $\omega$-determined with respect to the group $\mathfrak G$ if each germ $G\in J(n,p)$ that has the same formal series as $F$ at the origin lies in the orbit of $F$ under the action of $\mathfrak G$.
In this work, sufficient conditions are stated for $\omega$-determinedness. Examples are given of $\omega$-determined germs which are not finitely determined.
Bibliography: 5 titles. Full text: PDF file (1719 kB) References: PDF file HTML file English version: Mathematics of the USSR-Sbornik, 1974, 23:3, 425–440 Bibliographic databases: UDC: 519.46 MSC: 58A20, 58C25 Received: 20.11.1973 Citation: G. R. Belitskii, “Germs of mappings $\omega$-determined with respect to a given group”, Mat. Sb. (N.S.), 94(136):3(7) (1974), 452–467; Math. USSR-Sb., 23:3 (1974), 425–440 Citation in format AMSBIB
\Bibitem{Bel74}
\by G.~R.~Belitskii \paper Germs of mappings $\omega$-determined with respect to a~given group \jour Mat. Sb. (N.S.) \yr 1974 \vol 94(136) \issue 3(7) \pages 452--467 \mathnet{http://mi.mathnet.ru/msb3692} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=358853} \zmath{https://zbmath.org/?q=an:0315.58002} \transl \jour Math. USSR-Sb. \yr 1974 \vol 23 \issue 3 \pages 425--440 \crossref{https://doi.org/10.1070/SM1974v023n03ABEH002182} Linking options: http://mi.mathnet.ru/eng/msb3692 http://mi.mathnet.ru/eng/msb/v136/i3/p452 Citing articles on Google Scholar: Russian citations, English citations Related articles on Google Scholar: Russian articles, English articles This publication is cited in the following articles: G. R. Belitskii, “Normal forms for formal series and germs of $C^\infty$-mappings with respect to the action of a group”, Math. USSR-Izv., 10:4 (1976), 809–821 G. R. Belitskii, “Equivalence and normal forms of germs of smooth mappings”, Russian Math. Surveys, 33:1 (1978), 107–177
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If the spectrogram was computed as the magnitude of short time fourrier transforms from overlapping windows, then the spectrogram contains implicitly some phase information.The following iterations do the job :$$x_{n+1} = \text{istft}(S\cdot\exp(i\cdot\text{angle}(\text{stft}(x_n))))$$$S$ is the spectrogram, $\text{stft}$ is the forward-short time ...
No. Why do you think it would? First of all, the human brain works very different then any human constructed computer (to date); so the assumption that it runs mathematical "algorithms" is somewhat dicey..Here is roughly how it works:Sound wiggles the air drumThat vibration is transferred by the ossicles to the cochlea. The ossicles act mainly as a ...
Conjugate symmetric means$$f(-x) = f^{\ast}(x)$$i.e. the sign of the imaginary part is opposite when $x<0$The FFT of a real signal is conjugate symmetric. One half of the spectrum is the positive frequencies, and the other half is the negative. The negative coefficients are conjugate of the positive.So if you do filtering, your envelope must do ...
The concept is based on the convolution theorem, which states that for two signals $x(t)$ and $y(t)$, the product of their Fourier transforms $X(f)$ and $Y(f)$ is equal to the Fourier transform of the convolution of the two signals. That is:$$\mathcal{F}\{x(t) * y(t)\} = \mathcal{F}\{x(t)\}\mathcal{F}\{y(t)\}$$You can read more on the derivation of ...
The (linear or aperiodic) convolution of two vectors $\mathbf x = (x[0], x[1], \ldots, x[N-1])$ and $\mathbf y = (y[0], y[1], \ldots, y[N-1])$ is a vector $$\mathbf z = {\mathbf x}\star {\mathbf y} = (z[0], z[1], \ldots, z[2N-1]).$$On the other hand, their cross-correlation is a vector$$\mathbf w = {\mathbf x}\otimes {\mathbf y} = (w[-(N-1)], w[-(N-2)], \...
You can say the null subcarriers have their own bandwidth if you define an alphabet including "zeros" and use null carriers to transmit these "zeros". As the comment of MBaz,If we start with the discrete-time OFDM symbol s[n], and increase thesampling rate (defined by the time interval between successive samplesof s[n]), then the OFDM symbol ...
Regarding example 1:first of all, either the fft or the ifft needs to be normalised by the number of sampling points as you have done (actually you can normalise each by a factor of the square root of the number of points, it is just a meter of definition). However, in your case the ifft is half the length of the fft you performed. Hence your signal is ...
The FFT of the FFT bascially gets you the original signal again, it's just scaled and time flipped, i.e.$$FFT\left \{ FFT \left \{ x(n) \right \} \right \} = N \cdot x(-n)$$That's a simple consequence of the fact that the forward and inverse Fourier Transforms are almost the same.I have no idea what all the acronyms in the second question mean.
you don't have to set $X\left(\frac{N}{2} \right)=0$ if you don't want to. it will correspond to this component:$$ X(k)\frac{1}{N}e^{j 2 \pi \frac{nk}{N}}\bigg|_{k=\frac{N}{2}} = X\left(\frac{N}{2} \right)\frac{1}{N}(-1)^n $$but when you sample some $x[n]$, FFT it and find that $X\left(\frac{N}{2} \right) \ne 0$, you do not know the phase of that ...
The scaling convention used by Matlab is common in DSP. You could also use the unitary DFT where both the DFT and the IDFT are scaled by a factor of $1/\sqrt{N}$. You could also use the factor $1/N$ for the DFT and the factor $1$ for the IDFT. As long as you're consistent it doesn't really matter (apart from numerical considerations, especially when using ...
Whether or not to scale the forward FFT by 1/N depends on which result you want for further analysis: energy (preserving Parseval's identity), or amplitude (measuring height or volts, etc.).If you want to measure or analyze energy, then don't scale by 1/N, and a longer sinusoid of the same amplitude will produce a larger FFT result, proportional to the ...
What you're looking for is called a pruned DFT. In principle, it is possible to calculate a subset of outputs from a DFT using fewer mathematical operations. In practice, however, existing highly-optimized FFT implementations like FFTW are designed for full-output transforms. You'll find in many cases, unless you're only concerned with a very small ...
The basic concept of OFDM is to divide a high-bitrate datastream into $N$ low-bitrate datastreams and to multiplex these low-bitrate datastreams in frequency. That is, every datastream is assigned to a distinct frequency band (so-called subcarrier) that does not interfere with the other frequency bands. Orthogonality allows the frequency bands to be packed ...
Re-construction using just the magnitude (and an assumed phase of zero) will only work for exactly symmetric images. The phase is needed to make the top look different from the bottom and left look different from the right side of the image, etc.That's because all the cosine waveform DFT basis vectors are exactly symmetric around the center of an FFT or ...
If you'd like to think analog, an OFDM signal can be written as a sum of weighted complex sinusoidals:$$x(t)=\sum_{k=0}^{N-1}X_k \exp{\left( \mathrm j 2\pi\frac{kf_\mathrm{s}}{N}t \tag{1}\right)},$$where $N$ is the number of subcarriers, $f_\mathrm{s}$ is the sampling frequency and $X_k$ denotes the subcarrier values. For a digital implementation, (1) is ...
Say you want to transmit complex symbols $s_0,s_2,\ldots,s_{N-1}$ using frequency-division multiplexing. In the "bank of modulators" approach, you would assign a carrier frequency $f_k$ to each symbol and create the signal$$x(t) = \sum_{k=0}^{N-1} s_k \exp(j2\pi f_kt).$$Contrast this with the definition of the inverse discrete Fourier transform:$$x[n] =...
particularly since this is a question about convention, i will not reinforce the ridiculous convention of MATLAB and will answer only with the right and proper convention or conventions. i.e. MATLAB's indexing for the DFT is not right and proper, but i am pretty much agnostic about the which of the three common scaling conventions.also, i am not ...
Your code uses the phase for the reconstruction. Have a look at the output of fft2(x); they are complex numbers, i.e. the contain phase and magnitude. Have a look at this code:%%[x,map] = imread('http://www.cs.cmu.edu/~chuck/lennapg/lena_std.tif'); %import image%%i = rgb2gray(x); % to greyscaleI = fft2(i); % 2D FFTI = fftshift(I); % centremag = ...
You must keep in mind that for a real-valued signal, second half of your spectrum is a complex conjugate of all values below the Nyquist frequency. In your case:X(2) = -32X(N) = 32as you can see the coefficients are not the complex conjugate of each other. Because of that you are ending up with round-off errors, since two frequency components are ...
An FFT possesses linear features. Let us use a linear analogy. If one tells you that the sum of two numbers is $a+b=3$, what can you say about these numbers? Not much. More generally, $n$ linear equations on $m$ variables is underdetermined when $n<m$: you do not have enough information to retrieve the variables without more conditions. This can be ...
The Windowing method for filter design uses samples of the impulse response for the filter coefficients, truncated to the filter length (and then possibly, but not required, tapered by a higher performing window).The Frequency Sampling method for filter design uses the IDFT (Inverse Discrete Fourier Transform) of the desired frequency response for the ...
Suppose, I have this image in my hand and nothing else. Clearly, this is a Magnitude-plot of some unknown image.Sounds like you're losing the phase information in the signal; you only have the pixel values, correct?Is it possible to apply an Inverse Fast Fourier Transform (I-FFT) operation to recover the original Grayscale image from this image?The ...
$N=1000$ and your signal frequency is $10000$. So there should be a peak at bin $k=10$. And another one at $N-k=1000-10=990$. However since MATLAB starts indexing at $1$ instead of zero, you see peaks at $11$ and $991$.
In your link I cannot find a reference to the time-reversal. However, here are some additions:The time reversal $y(t)$ of a continuous signal $x(t)$ is given by $y(t)=x(-t)$. Similarly, for a discrete signal we have $y[n]=x[-n]$. However, here the delicate issue occurs that $n=0...N-1$, where $N$ is the signal length. Now, noting that in the finite ...
First note that when you use a logarithmic function you shall avoid negative arguments if it's output should be real valued. Then consider the following relation:$$ y[n] = \ln( 1 + x[n] )$$where the input $x[n]$ is conditioned such that $ 0 < 1 + x[n] \leq2$ is maintained. Then a Maclaurin series expansion of this expression will be$$ y[n] = \ln(1 + ...
The problem appears to be that your windowing function doesn't preserve complex conjugate symmetry. So ifft(y1) has a significant imaginary part. By discarding this through the real() operation you induce a significant error which results in the discrepancy. To verify tryz = ifft(y1); plot(imag(z));
This is tricky. The phase information impacts the time domain form very much. For example white noise and a delta impulse have exactly the same amplitude spectrum, the only difference is in the phase.Without any further assumptions or extra knowledge this can't be done. For example if you know that the signal is stationary, you could just try a random ...
I had a bit of a hard time to understand the answer of @edouard, which is doing the right thing. Compare to https://dsp.stackexchange.com/a/3410/9031 , which I used to implement my reconstruction.Note that $i$ is the imaginary number, and $x_n$ is the reconstructed signal at the $n^{\text{th}}$ iteration. Start with $x_0$ being a random vector of length ... |
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences). |
The terms used in photometry and radiometry have specific meanings that may not match the meanings that the words have in other context. (The same is true of the words "heat" and "work", which mean different things outside of a physics context.) Furthermore, not everyone uses the same words. The astronomy community may have their own standard usage that differs from the optics community. (I hope not, but I'm not familiar with what astronomers do.)
The first thing to straighten out is that your question concerns
radiometry, not photometry. Photometric quantities are weighted by the spectral response of the eye. It does not appear that your question involves what a human sees with his eyes.
Intensity in radiometry is radiant power per unit solid angle. The word
radiant here specifies that we are talking about a radiometric quantity, not a photometric quantity.
You can get the radiant intensity of your star by dividing $L$ by $4\pi$, the total solid angle subtended by a sphere. (Assuming the star radiates isotropically.) $$I=\frac{L}{4\pi}$$
The quantity $F(\lambda)$ might be better called
spectral radiant power or spectral power (the word "radiant" being redundant, but makes it clear that we are not talking about photometry.) Calling $L$ "flux" is ok, but the word "flux" is one of those words that different people give different meanings. To avoid that problem, I usually call $L$ radiant power, although if I'm writing in a context where radiant flux has already been established as the definition, I'll use "radiant flux".
What you call
apparent luminosity I would call irradiance, the total power incident on a unit area, which in this case, is a unit area on earth. ("Apparent luminosity might be a term astronomers use. I don't know.) Watts per square meter. It is obtained by multiplying the intensity by the solid angle at the star subtended by one square meter on earth: $$ \ell = I\Omega = I \frac{1}{d^2} = \frac{L}{4\pi d^2}$$
It can all be very confusing. I think if you use the standard terms as found, for example, on Wikipedia, everyone, including astronomers, will understand you and not get confused by conflicting terms. |
A person will go to the town center if they can determine that they must be a member of some organization. If they can determine that there is at least one letter they did not receive, then they must be a member of some organization.
If they are not a member of any organization, then they received the membership of every organization. This allows them to do a simple proof by contradiction. They start off by assuming that they are not a member of any organization (and therefore have perfect knowledge of the situation), and then wait until something unexpected happens.
The simplest case is that a person receives no letters. Assuming that they are not a member of any organization implies there are no organizations, which they know to be false. Therefore, they must be a member of at least one organization.
Consider how this plays out with organizations consisting only of single members: A, B, and C
A gets letters about B and C, B gets letters about A and C, and C gets letters about A and B.
On day 1, as everyone received letters, nobody goes to the town center.
On day 2, A assumes that there are only two organizations — B and C. B received a letter about C, and C received a letter about B. Thus, neither of them should have gone to the town center. As neither of them did, A has no reason to believe that there are any other organizations. B and C follow similar reasonings.
On day 3, A assumes that there are only two organizations — B and C. On day 2, B would have reasoned as follows: "Suppose there is only one organization, with C as its only member. Then C should have gone to the town center on day 1. However, C did not, so I must be a member of some group. So I will go to the town center today." However, B did not go to the town center on day 2, so there must be an organization that A does not know about that A is a member of. So A will go to the town center. By symmetric reasoning, B and C will also go to the town center.
We know that it is not the case that everyone will always be able to determine that they are a member of some organization. Consider the case of organizations A, AB, ABC, and ABCD.
On day 1, A, having received no letters, goes to the town center.
Every day after that, A goes to the town center while B, C, and D all stay home. B thinks there's only one organization A, and only A going to the town center confirms that. C knows there's also an organization AB, but expects B to stay at home because B doesn't know about that one. D knows there's also an organization ABC, but knows that C and B are unaware of its existence. There will never be an event that disrupts anyone's expectations, so this is a stable situation.
There are two questions we still need to answer — will there always be at least one proper meeting, and will everyone who goes eventually be in a proper meeting?
First, let's establish a base condition - there will always be at least one person who goes to the town center. I'll use a group of A, B, C, and D to help explain my proof.
If anyone does not receive a letter, then they will go the first day. This is the only way in which someone will go on the first day.
Now assume that everyone received at least one letter (and so nobody went on the first day). Then on the second day A, assuming that she is not a part of any group and therefore knows the membership of every group, will consider every other person. If any of B, C, and D is a member of all the groups, then that person should not have received any letters and gone on the first day. But nobody went the first day, so A must not have perfect information, and therefore must be a member of some group.
If nobody goes on the second day, then A will go one level deeper in her reasoning. She first considers B — he knows about all the groups for which he is not a member. If C or D is a member of all of those groups, then B would have expected them to go on the first day. When he observed that they did not, he should have gone on the second day. Because he did not go on the second day, A knows that there must be some group that she does not know about and therefore is a member of. She follows the same reasoning for C and D.
If nobody goes on the third day, A will reach the deepest level of reasoning necessary for four people. She will again consider B's reasoning. Assuming that she knows about all the groups that exist, B knows only about groups that consist of C and/or D. So he, on the third day, would have considered C's reasoning for the second day. He would have assumed that he knew about all the groups that exist, and since he only knows about groups the have only C and D in them, C must have only known about a group of which D is the only member. So C, on the second day, would have thought that only D was a member of a group, and having not witnessed D go on the first day, should have gone on the second day. Thus, B, having not witnessed C go on the second day, should have gone on the third day. Thus A, if she does not see B go on the third day, will go on the fourth day.
In more formal terms, on day $k+1$ person $p_0\in P$ will determine if $\exists p_{i1}\in (P-\{p_0\}) \exists p_{i2}\in (P-\{p_0,p_{i1}\})...\exists p_{ik}\in (P-\{p_0,...,p_{i(k-1)}\})\forall g\in G: \{p_{i1},...,p_{ik}\}\cap g \ne \emptyset$
where $G$ is the set of all groups that $p_0$ knows about. If this condition is met, then there are no groups that do not contain at least one of the people considered. This means that on day $k+1$, $p_0$ expected each person in the chain $p_{ij}$ to expect the next person in the chain to have left on the previous day (with the last person expected to leave on the first day). Because the expectation was not met, $p_0$ will go to the town center on day $k+1$.
Everyone who goes will eventually be a part of a proper meeting:
Let $p_0$ be a person who will go to the town center. If $p_0$ is the sole member of an organization, then a proper meeting can be held immediately. Otherwise, $\exists g_0,p_1\mid p_0,p_1\in g_0$. Because $p_0$ and $p_1$ are both members of that organization, neither received a letter about it. Suppose there are no organizations of which $p_0$ is a member but $p_1$ is not. Then, when $p_0$ goes to the town center, $p_1$ will know there is at least one organization that they did not receive a letter for, and must therefore be a member of.
Otherwise $\exists g_1\mid p_0\in g_1\land p_1\notin g_1$. By our above reasoning, we know that anyone who only belongs to organizations $p_0$ does will go to the town center one day after $p_0$ at the latest.
For each person $p_i\in g_1$ who is a member of $g_i\mid p_0\notin g_i$, $p_1$ can follow the same chain of logic we used to prove that someone will eventually go. I'm having trouble writing out the formal logic for it, but here's the basic idea — $p_1$ will expect $p_i$ to expect etc. In the end it boils down to two cases - either an entire organization goes, or $p_1$ will go, and because $p_1$ was an arbitrary member of $g_0$ that means everyone in $g_0$ will go. |
Seeing that in the Chomsky Hierarchy Type 3 languages can be recognised by a DFA (which has no stacks), Type 2 by a DFA with one stack (i.e. a push-down automaton) and Type 0 by a DFA with two stacks (i.e. with one queue, i.e. with a tape, i.e. by a Turing Machine), how do Type 1 languages fit in...
Considering this pseudo-code of an bubblesort:FOR i := 0 TO arraylength(list) STEP 1switched := falseFOR j := 0 TO arraylength(list)-(i+1) STEP 1IF list[j] > list[j + 1] THENswitch(list,j,j+1)switched := trueENDIFNEXTIF switch...
Let's consider a memory segment (whose size can grow or shrink, like a file, when needed) on which you can perform two basic memory allocation operations involving fixed size blocks:allocation of one blockfreeing a previously allocated block which is not used anymore.Also, as a requiremen...
Rice's theorem tell us that the only semantic properties of Turing Machines (i.e. the properties of the function computed by the machine) that we can decide are the two trivial properties (i.e. always true and always false).But there are other properties of Turing Machines that are not decidabl...
People often say that LR(k) parsers are more powerful than LL(k) parsers. These statements are vague most of the time; in particular, should we compare the classes for a fixed $k$ or the union over all $k$? So how is the situation really? In particular, I am interested in how LL(*) fits in.As f...
Since the current FAQs say this site is for students as well as professionals, what will the policy on homework be?What are the guidelines that a homework question should follow if it is to be asked? I know on math.se they loosely require that the student make an attempt to solve the question a...
I really like the new beta theme, I guess it is much more attractive to newcomers than the sketchy one (which I also liked). Thanks a lot!However I'm slightly embarrassed because I can't read what I type, both in the title and in the body of a post. I never encountered the problem on other Stac...
This discussion started in my other question "Will Homework Questions Be Allowed?".Should we allow the tag? It seems that some of our sister sites (Programmers, stackoverflow) have not allowed the tag as it isn't constructive to their sites. But other sites (Physics, Mathematics) do allow the s...
There have been many questions on CST that were either closed, or just not answered because they weren't considered research level. May those questions (as long as they are of good quality) be reposted or moved here?I have a particular example question in mind: http://cstheory.stackexchange.com...
Ok, so in most introductory Algorithm classes, either BigO or BigTheta notation are introduced, and a student would typically learn to use one of these to find the time complexity.However, there are other notations, such as BigOmega and SmallOmega. Are there any specific scenarios where one not...
Many textbooks cover intersection types in the lambda-calculus. The typing rules for intersection can be defined as follows (on top of the simply typed lambda-calculus with subtyping):$$\dfrac{\Gamma \vdash M : T_1 \quad \Gamma \vdash M : T_2}{\Gamma \vdash M : T_1 \wedge T_2}...
I expect to see pseudo code and maybe even HPL code on regular basis. I think syntax highlighting would be a great thing to have.On Stackoverflow, code is highlighted nicely; the schema used is inferred from the respective question's tags. This won't work for us, I think, because we probably wo...
Sudoku generation is hard enough. It is much harder when you have to make an application that makes a completely random Sudoku.The goal is to make a completely random Sudoku in Objective-C (C is welcome). This Sudoku generator must be easily modified, and must support the standard 9x9 Sudoku, a...
I have observed that there are two different types of states in branch prediction.In superscalar execution, where the branch prediction is very important, and it is mainly in execution delay rather than fetch delay.In the instruction pipeline, where the fetching is more problem since the inst...
Is there any evidence suggesting that time spent on writing up, or thinking about the requirements will have any effect on the development time? Study done by Standish (1995) suggests that incomplete requirements partially (13.1%) contributed to the failure of the projects. Are there any studies ...
NPI is the class of NP problems without any polynomial time algorithms and not known to be NP-hard. I'm interested in problems such that a candidate problem in NPI is reducible to it but it is not known to be NP-hard and there is no known reduction from it to the NPI problem. Are there any known ...
I am reading Mining Significant Graph Patterns by Leap Search (Yan et al., 2008), and I am unclear on how their technique translates to the unlabeled setting, since $p$ and $q$ (the frequency functions for positive and negative examples, respectively) are omnipresent.On page 436 however, the au...
This is my first time to be involved in a site beta, and I would like to gauge the community's opinion on this subject.On StackOverflow (and possibly Math.SE), questions on introductory formal language and automata theory pop up... questions along the lines of "How do I show language L is/isn't...
This is my first time to be involved in a site beta, and I would like to gauge the community's opinion on this subject.Certainly, there are many kinds of questions which we could expect to (eventually) be asked on CS.SE; lots of candidates were proposed during the lead-up to the Beta, and a few...
This is somewhat related to this discussion, but different enough to deserve its own thread, I think.What would be the site policy regarding questions that are generally considered "easy", but may be asked during the first semester of studying computer science. Example:"How do I get the symme...
EPAL, the language of even palindromes, is defined as the language generated by the following unambiguous context-free grammar:$S \rightarrow a a$$S \rightarrow b b$$S \rightarrow a S a$$S \rightarrow b S b$EPAL is the 'bane' of many parsing algorithms: I have yet to enc...
Assume a computer has a precise clock which is not initialized. That is, the time on the computer's clock is the real time plus some constant offset. The computer has a network connection and we want to use that connection to determine the constant offset $B$.The simple method is that the compu...
Consider an inductive type which has some recursive occurrences in a nested, but strictly positive location. For example, trees with finite branching with nodes using a generic list data structure to store the children.Inductive LTree : Set := Node : list LTree -> LTree.The naive way of d...
I was editing a question and I was about to tag it bubblesort, but it occurred to me that tag might be too specific. I almost tagged it sorting but its only connection to sorting is that the algorithm happens to be a type of sort, it's not about sorting per se.So should we tag questions on a pa...
To what extent are questions about proof assistants on-topic?I see four main classes of questions:Modeling a problem in a formal setting; going from the object of study to the definitions and theorems.Proving theorems in a way that can be automated in the chosen formal setting.Writing a co...
Should topics in applied CS be on topic? These are not really considered part of TCS, examples include:Computer architecture (Operating system, Compiler design, Programming language design)Software engineeringArtificial intelligenceComputer graphicsComputer securitySource: http://en.wik...
I asked one of my current homework questions as a test to see what the site as a whole is looking for in a homework question. It's not a difficult question, but I imagine this is what some of our homework questions will look like.
I have an assignment for my data structures class. I need to create an algorithm to see if a binary tree is a binary search tree as well as count how many complete branches are there (a parent node with both left and right children nodes) with an assumed global counting variable.So far I have...
It's a known fact that every LTL formula can be expressed by a Buchi $\omega$-automata. But, apparently, Buchi automata is a more powerful, expressive model. I've heard somewhere that Buchi automata are equivalent to linear-time $\mu$-calculus (that is, $\mu$-calculus with usual fixpoints and onl...
Let us call a context-free language deterministic if and only if it can be accepted by a deterministic push-down automaton, and nondeterministic otherwise.Let us call a context-free language inherently ambiguous if and only if all context-free grammars which generate the language are ambiguous,...
One can imagine using a variety of data structures for storing information for use by state machines. For instance, push-down automata store information in a stack, and Turing machines use a tape. State machines using queues, and ones using two multiple stacks or tapes, have been shown to be equi...
Though in the future it would probably a good idea to more thoroughly explain your thinking behind your algorithm and where exactly you're stuck. Because as you can probably tell from the answers, people seem to be unsure on where exactly you need directions in this case.
What will the policy on providing code be?In my question it was commented that it might not be on topic as it seemes like I was asking for working code. I wrote my algorithm in pseudo-code because my problem didnt ask for working C++ or whatever language.Should we only allow pseudo-code here?... |
FAQ - Frequently Asked Questions When was the cup anemometer invented?
It is not certain who actually devised the instrument. But is was first described in 1846 by the Irish astronomer T. R. Robinson. The first cup anemometers had 4 cups. Patterson (1926) found that if the rotor had only 3 cups the torque is more uniform through the entire revolution.
Why is the calibration linear?
The short answer is that it is an empirical fact. In all wind tunnel calibrations of modern cup anemometers it is not possible to detect any deviation from linearity. Brazier (1914) found that the ratio of the cup diameter to that of the circle described by the centers of the cups is an important parameter and that the best linearity is obtained when this ratio is about 0.5.
How many pulses per one full rotor rotation is required to obtain a reliable signal?
In principle one pulse suffices. It is not possible to obtain a better velocity measurement corresponding to just a fraction of a revolution. The instrument averages over one full revolution, which will then be the resolution. For each revolution a column of air with a fixed length \(\mathit\ell\) passes through the rotor. This length is called the calibration length. Summing all these columns of length \(\mathit\ell\) over a particular period, often 10 minutes, we obtain the mean-wind speed in this period. For practical purposes we have chosen 2 pulses for each revolution, but it is really overkill.
How fast is the response of a cup anemometer?
The reaction time is determined by the so-called distance constant \(\ell_\circ\).The response time is obtained by \(\ell_\circ\) and the mean-wind speed \(U\) as \(\tau_\circ=\ell_\circ/U\). Let us consider a situation where we have a constant wind speed \(V\) and then have it changed instantaneously to \(V+\Delta{v}\). Then, after the time \(\tau_\circ\), the anemometer will show 63 percent (=1-\(\mbox{e}^{-1}\)) of the final value \(V+\Delta{v}\). We can also express this by stating that a column of air of length \(\ell_\circ\) must pass through the anemometer to get to the same result. The distance constant \(\ell_\circ\) is an instrument constant. However, the time constant is not an instrument constant, since it is inversely proportional to the mean-wind speed \(U\). The instrument has a slower response to changes in the wind speed the larger the distance constant.
What is overspeeding anyway? Is there a simple explanation?
Overspeeding is a positive bias on the measured mean wind speed \(U\). There are two basic reasons, the atmospheric turbulence and the asymmetric construction of the rotor. This asymmetry exists simply due to the fact that the wind forces are larger with the wind blowing into the cups than with the wind blowing on their back sides. This is essential for the working of the cup anemometer because without this difference the anemometer rotor would never start moving. The reason that the turbulence plays an important role is that the anemometer is calibrated in a wind tunnel with no, or very little, turbulence. So when exposed to turbulence in the atmosphere the cup anemometer will react to the wind fluctuations from a whole spectrum of eddy sizes. An anemometer with a small distance constant \(\ell_\circ\) can easily follow the larger eddies, but when eddies are of sizes equal to or smaller than the distance constant the anemometer cannot follow the fluctuations. However, these will cause the rotor to spend more time on the plus side than on the minus side of the mean wind speed \(U\) than it would have been measured in the wind tunnel with no turbulence. A special, but quite common, case is the signal from an anemometer in a windy surface layer. The overspeeding relative to the mean-wind speed is \(C(\sigma^2/U^2)(\ell_\circ/z)^{2/3}\). Here \(\sigma^2\) is the variance of the wind speed or the square of the standard deviation, \(z\) the height and \(C\) a dimensionless constant of the order of one.
It is possible to use the cup anemometer to measure the variance and higher-order moments. But these moments will show no bias.
What is the relation between IEC 61400-12-1, ISO 17025 and MEASNET?
Here is how it works:
IEC 61400-12-1 Power performance measurements of electricity producing wind turbines This IEC standard specifies the procedures for measuring the power curve. A key element of power performance testing is the measurement of wind speed for which the standard prescribes the use of cup anemometers. Hence, it also specifies the procedure for cup anemometer calibration. http://www.iec.ch ISO/IEC 17025 General requirements for the competence of testing and calibration laboratories This is the main ISO standard used by testing and calibration laboratories. In most major countries, ISO/IEC 17025 is the standard for which most labs must hold accreditation in order to be deemed technically competent. In many cases, suppliers and regulatory authorities will not accept test or calibration results from a lab that is not accredited. ISO 17025 does not relate to the anemometer but to the requirements for the competence of the laboratories calibrating it and utilizing it in a power performance test, respectively. http://www.iso.org Measnet MEASNET is a co-operation of companies which are engaged in the field of wind energy and want to ensure high quality measurements, uniform interpretation of standards and recommendations as well as interchangeability of results. http://www.measnet.com Who is doing what? WindSensor manufactures anemometers (BTW, the best on the market). WindSensor anemometers are classified according to IEC 61400-12-1:2005 Annex I.
WindSensor anemometers are calibrated in the SOH wind tunnel. The SOH wind tunnel is accredited according to ISO 17025 to calibrate anemometers according to the IEC 61400-12-1:2005 Annex F (BTW, with the lowest uncertainty on the market).
A number of testing institutes are accredited according to ISO 17025 to carry out power performance testing according to IEC 61400-12-1 using calibrated anemometers.
What is the service life of WindSensor anemometers?
There are basically two parameters which can influence the calibration of a cup anemometer: the geometry and the bearing friction. Assuming that the geometry, and hence the aerodynamics, is constant over the lifetime of an anemometer the critical parameter is bearing friction.
Bearing friction may among others be affected by lightning, mechanical impact, wear and corrosion. By design, WindSensor anemometers are tolerant of lightning due to the non-conducting property of the cup rotor, which tends to direct the lightning current along the surface to the body of the anemometer rather than through the bearings. Furthermore, the bearings utilize a special steel type which is several times more resistant to wear and corrosion than steel types otherwise used in the industry. Combined with the most durable lubrication in the industry WindSensor anemometers have proven to operate in excess of 10 years under ideal conditions.
However, as cup anemometers are operated worldwide under different conditions in all types of environment it is not possible to specify rigid service intervals.
We specify a service interval of 1 to 2 years depending on the operating conditions. For example, it is common practice to replace cup anemometers on offshore towers at 1-year intervals due to the extremely high costs of accessing the site. We recommend that our customers should define a maximum service interval depending on the conditions on the specific site.
However, given the risk that instruments might be damaged by handling already during the installation or at any time after installation for example by static discharges and falling ice etc., the most important precaution in operating a meteorological tower is to continuously monitor the performance by correlating measurements from two identical anemometers installed at the same level as well as correlating measurements from identical anemometers installed at different levels.
It cannot be justified deploying different models of anemometers on a meteorological tower as different types of anemometers have different responses to air temperature, air density, inflow angle and turbulence resulting in different measurements, even if installed at the same level. Do you offer refurbishment of cup anemometers?
Yes, we offer refurbishment together with subsequent recalibration.
However, in the case of an operating meteorological tower it will typically be more cost effective to replace the used anemometers with new sensors in a single mission and discard the old ones, rather than splitting the job into two missions with traveling, climbing, packing and shipping twice adding to the cost of refurbishment and recalibration.
Furthermore, by replacing with new anemometers, data collection will be interrupted only for a few minutes rather than the several weeks it will take to ship, recondition, run-in, recalibrate, return and reinstall. |
Development
given dimensions, number of water and land nodes and land percentage, water percentage inferred from land percentage place initial nodes, make sure they are well mixed land nodes or water nodes can be centralized, depending on the focus of the map, or they may be placed at random determine the closest distance that nodes may be to one another, regardless of their type determine the closest distance that nodes of the same types may be to one another, say twice type insensitive distance randomly place the nodes grow nodes close to the seeds, do not worry about balance yet grow nodes in general. nodes should be sufficiently close to other nodes. randomly select the node type, unless the land percentage becomes top high or too low; force the type appropriately until rebalanced
(1)
\begin{align} d = \sqrt{\frac{x y} {8 \pi n}} \end{align}
page revision: 33, last edited: 02 Nov 2007 18:05 |
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences). |
I think there are two legitimate sources of complaint. For the first, I will give you the anti-poem that I wrote in complaint against both economists and poets. A poem, of course, packs meaning and emotion into pregnant words and phrases. An anti-poem removes all feeling and sterilizes the words so that they are clear. The fact that most English speaking humans cannot read this assures economists of continued employment. You cannot say that economists are not bright.
Live Long and Prosper-An Anti-Poem
May you be denoted as $k\in{I},I\in\mathbb{N}$, such that $I=1\dots{i}\dots{k}\dots{Z}$
where $Z$ denotes the most recently born human.
$\exists$ a fuzzy set $Y=\{y^i:\text{Human Mortality Expectations}\mapsto{y^i},\forall{i\in{I}}\},$
may $y^k\in\Omega,\Omega\in{Y}$ and $\Omega$ is denoted as "long"
and may $U(c)$, where c is the matrix of goods and services across your lifetime
$U$ is a function of $c$, where preferences are well-defined and $U$ is qualitative satisfaction,
be maximized $\forall{t}$, $t$ denoting time, subject to
$w^k=f'_t(L_t),$ where $f$ is your production function across time
and $L$ is the time vector of your amount of work,
and further subject to $w^i_tL^i_t+s^i_{t-1}=P_t^{'}c_t^i+s^i_t,\forall{i}$
where $P$ is the vector of prices and $s$ is a measure of personal savings across time.
May $\dot{f}\gg{0}.$
Let $W$ be the set $W=\{w^i_t:\forall{i,t}\text{ ranked ordinally}\}$
Let $Q$ be the fuzzy subset of $W$ such that $Q$ is denoted "high".
Let $w_t^k\in{Q},\forall{t}$
The second is mentioned above, which is the misuse of math and statistical methods. I would both agree and disagree with the critics on this. I believe that most economists are not aware of how fragile some statistical methods can be. To provide an example, I did a seminar for the students in the math club as to how your probability axioms can completely determine the interpretation of an experiment.
I proved using real data that newborn babies will float out of their cribs unless nurses swaddle them. Indeed, using two different axiomatizations of probability, I had babies clearly floating away and obviously sleeping soundly and securely in their cribs. It wasn't the data that determined the result; it was axioms in use.
Now any statistician would clearly point out that I was abusing the method, except that I was abusing the method in a manner that is normal in the sciences. I didn't actually break any rules, I just followed a set of rules to their logical conclusion in a way that people do not consider because babies don't float. You can get significance under one set of rules and no effect at all under another. Economics is especially sensitive to this type of problem.
I do belive that there is an error of thought in the Austrian school and maybe the Marxist about the use of statistics in economics that I believe is based on a statistical illusion. I am hoping to publish a paper on a serious math problem in econometrics that nobody has seemed to notice before and I think it is related to the illusion.
This image is the sampling distribution of Edgeworth's Maximum Likelihood estimator under Fisher's interpretation (blue) versus the sampling distribution of the Bayesian maximum a posteriori estimator (red) with a flat prior. It comes from a simulation of 1000 trials each with 10,000 observations, so they should converge. The true value is approximately .99986. Since the MLE is also the OLS estimator in the case, it is also Pearson and Neyman's MVUE.
Note how relatively inaccurate the Frequency based estimator is compared to the Bayesian. Indeed, the relative efficiency of $\hat{\beta}$ under the two methods is 20:1. Although Leonard Jimmie Savage was certainly alive when the Austrian school left statistical methods behind, the computational ability to use them didn't exist. The first element of the illusion is inaccuracy.
The second part can better be seen with a kernel density estimate of the same graph.
In the region of the true value, there are almost no examples of the maximum likelihood estimator being observed, while the Bayesian maximum a posteriori estimator closely covers .999863. In fact, the average of the Bayesian estimators is .99987 whereas the frequency based solution is .9990. Remember this is with 10,000,000 data points overall.
Frequency based estimators are averaged over the sample space. The missing implication is that it is unbiased, on average, over the entire space, but possibly biased for any specific value of $\theta$. You also see this with the binomial distribution. The effect is even greater on the intercept.
The red is the histogram of Frequentist estimates of the itercept, whose true value is zero, while the Bayesian is the spike in blue. The impact of these effects are worsened with small sample sizes because the large samples pull the estimator to the true value.
I think the Austrians were seeing results that were inaccurate and didn't always make logical sense. When you add data mining into the mix, I think they were rejecting the practice.
The reason I believe the Austrians are incorrect is that their most serious objections are solved by Leonard Jimmie Savage's personalistic statistics. Savages
Foundations of Statistics fully covers their objections, but I think the split had effectively already happened and so the two have never really met up.
Bayesian methods are generative methods while Frequency methods are sampling based methods. While there are circumstances where it may be inefficient or less powerful, if a second moment exists in the data, then the t-test is always a valid test for hypotheses regarding the location of the population mean. You do not need to know how the data was created in the first place. You need not care. You only need to know that the central limit theorem holds.
Conversely, Bayesian methods depend entirely on how the data came into existence in the first place. For example, imagine you were watching English style auctions for a particular type of furniture. The high bids would follow a Gumbel distribution. The Bayesian solution for inference regarding the center of location would not use a t-test, but rather the joint posterior density of each of those observations with the Gumbel distribution as the likelihood function.
The Bayesian idea of a parameter is broader than the Frequentist and can accomodate completely subjective constructions. As an example, Ben Roethlisberger of the Pittsburgh Steelers could be considered a parameter. He would also have parameters associated with him such as pass completion rates, but he could have a unique configuration and he would be a parameter in a sense similar to Frequentist model comparison methods. He might be thought of as a model.
The complexity rejection isn't valid under Savage's methodology and indeed cannot be. If there were no regularities in human behavior, it would be impossible to cross a street or take a test. Food would never be delivered. It may be the case, however, that "orthodox" statistical methods can give pathological results that have pushed some groups of economists away. |
Think of the logarithm as giving you back the
angle between the positive real axis and the number you have plugged in (times a factor $i$). You were not surprised to see
$$\log(-1)=\pi i.$$
However, this statement just means that the angle between the positive axis, and the negative axis (represented by $-1$) is $180^°$ or (in radians) $\pi$. And when you look at the complex plain, you would see that $i$ is "above" the zero, hence there is a $90^°$ angle between the positive axis and the complex axis (represented by $i$), which expressed by radians is $\pi/2$. Hence
$$\log(i)=\frac\pi 2 i.$$
There is some more math and some more subtleties hidden inside. But this should suffice as a first explanation.
Maybe also this might be interesting. You probably know that logarithms "convert" multipication into addition in the sense $\log(ab)=\log(a)+\log(b)$. And you you know that $i\cdot i=-1$. So it would be quite natural to assume that
$$\log(i)+\log(i)=\log(-1).$$
And thats exactly what you found. |
Fix positive integers $m, n, d$. In what follows, the height of an algebraic number will mean the absolute multiplicative height.
Let $V \subset \bar{\mathbb{Q}}^n$ be an affine algebraic variety such that $$ V = \{ \mathbf{x} \in \bar{\mathbb{Q}}^n \mid f_1(\mathbf{x}) = \dotsb = f_m(\mathbf{x}) = 0 \} $$ for some polynomials $f_1, \dotsc, f_m$ each of degree at most $d$, with coefficients in $\bar{\mathbb{Q}}$ of height at most $H$. We do not require that the ideal $(f_1, \dotsc, f_m)$ should be radical.
Let $\mathop{\mathrm{Sing}} V$ denote the set of singular points of $V$.
Do there exist constants $m', d', a, b$ depending only on $m, n, d$ such that: $$ \mathop{\mathrm{Sing}} V = \{ \mathbf{x} \in \bar{\mathbb{Q}}^n \mid g_1(\mathbf{x}) = \dotsb = g_{m'}(\mathbf{x}) = 0 \} $$ for some polynomials $g_1, \dotsc, g_{m'}$ each of degree at most $d'$, with coefficients in $\bar{\mathbb{Q}}$ of height at most $aH^b$?
Note that I don't care how $m', d', a, b$ depend on $m, n, d$.
If the ideal $I = (f_1, \dotsc, f_m)$ is radical, then the answer is easily yes because then $\mathop{\mathrm{Sing}} V$ is defined by a condition on the Jacobians of $f_1, \dotsc, f_m$. In the general case, I suspect that one might be able to use Gröbner bases to show that the radical of $I$ is generated by a set of polynomials of constant-bounded degrees and polynomially-bounded heights. But I don't know much about Gröbner bases and I haven't found anything which discusses height bounds for Gröbner bases.
I would appreciate comments from experts on whether this will work, or has been done before. Other suggestions are also welcome.
This came up when trying to prove a result about the existence of real points of bounded height on an affine variety defined over $\bar{\mathbb{Q}} \cap \mathbb{R}$. My proof does not work if all the real points of the variety are contained in the singular locus, so I would like to use the above to say that in this case we can replace our initial variety by its singular locus while retaining control of heights of the defining equations. |
Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it with your professor) but I think it helped me.
So, I think, maybe I can understand this as an analogy with linear momentum, because linear momentum is conserved too. I remember the problem of a car of mass m, traveling toward the right at velocity v, and on the same road an identical car traveling to the left at velocity –v. They collide head-on and smash and stick together. Velocities after the crash – zero. Momentum after the crash – zero and of course, momentum is conserved. The total momentum of the system was zero before and after.
Let say your container filled with water is a long annulus with thick steel walls. The flow is initially a circular flow around the axis (i.e. 2D flow.) What is initial total angular momentum of the system? Eventually the fluid stops moving, so the final total angular momentum of the system must be zero. How do we show that the initial angular momentum is zero too?
At this point you need to recognize that the vorticity vector in the moving fluid is everywhere parallel to the axis of the container. And you need to use Stokes’ theorem to write an integral equation with a line integral on the LHS (the circulation) and a surface integral on the RHS (vorticity integrated over the container cross section. )
\begin{align*}\oint_{C} v \cdot dl = \int_{S} w \cdot dS\\end{align*}
Take your integration path (the closed path C) entirely inside the steel wall of your container. The velocity inside the container wall is always zero, and so the circulation along the path is always zero, and so the total vorticity across the cross-sectional area (the area S) of the container and fluid is always zero too.
You can calculate for yourself that the boundary layer vorticity is equal and opposite to the bulk viscosity using a similar approach. Imagine spinning the container about its axis at a constant angular velocity. Eventually the entire viscous-fluid and container system will be rotating like a rigid body around the axis. Every point has the same angular velocity, and there is now a vortex located at the center. Compute the circulation for any closed path that includes the vortex inside it – this will be the strength of the vortex, and the magnitude and sign of vorticity in the bulk fluid. You can show yourself that the strength of this vortex is the total vorticity. Compute the circulation around any path that does not include the vortex, this will always come out to be zero. Pick a path near the fluid-container boundary, s’, so half of it is in fluid and half is inside the container wall, as long as the container and fluid are still rotating together the circulation around this path will be zero too. Now stop the container’s rotation. The fluid continues to move. Compute the circulation around the path s’ again, it is no longer zero and is the vorticity at the boundary layer. It’s sign is opposite that of the vortex at the center. Every point along the fluid-boundary can be associated with path like s’ and a small amount of boundary layer vorticity. Integrate around the entire boundary and the sum will be equal in magnitude and opposite in sign to the strength of the vortex at the center. Eventually, boundary layer vorticity will diffuse towards the center and annihilate the center vortex.
@Isopycnal_Oscillation is correct to point out that in 3D, and particularly near turbulent conditions, vorticity is not conserved. The second term on the RHS of your ‘transport equation’ says that the stretching and tilting of vortex tubes can change vorticity too. However, I expect that in the classes where your professor is fond of saying that “vorticity cannot be destroyed” turbulent flow is seldom if ever encountered.
Finally, assuming that the LHS of your ‘transport equation’ equals zero does not necessarily require that the fluid be inviscid or that the problem be 2D – you are assuming that the terms on the RHS happen to cancel exactly and the vorticity is fortuitously ‘steady-state.’ So yes, that is a very strong assumption to accept. |
Epimorphism Preserves Identity Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity element $e_S$.
Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$. Proof
Then:
$\forall x \in S: x \circ e_S = x = e_S \circ x$ Thus:
\(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi {x \circ e_S}\) $e_S$ is an identity for $\circ$ \(\displaystyle \) \(=\) \(\displaystyle \map \phi x * \map \phi {e_S}\) Morphism property of $\circ$ under $\phi$
and:
\(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi {e_S \circ x}\) $e_S$ is an identity for $\circ$ \(\displaystyle \) \(=\) \(\displaystyle \map \phi {e_S} * \map \phi x\) Morphism property of $\circ$ under $\phi$
The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.
$\blacksquare$
Also see Epimorphism Preserves Associativity Epimorphism Preserves Commutativity Epimorphism Preserves Inverses |
Before discussing the homogeneous models, it is worthwhile to appreciate the complexity of the flow. For the construction of fluid basic equations, it was assumed that the flow is continuous. Now, this assumption has to be broken, and the flow is continuous only in many chunks (small segments). Furthermore, these segments are not defined but results of the conditions imposed on the flow. In fact, the different flow regimes are examples of typical configuration of segments of continuous flow. Initially, it was assumed that the different flow regimes can be neglected at least for the pressure loss (not correct for the heat transfer). The single phase was studied earlier in this book and there is a considerable amount of information about it. Thus, the simplest is to used it for approximation. The average velocity (see also equation (13)) is
\[ U_m = \dfrac{Q_L + Q_G} {A} = U_{sL} + U_{sG} = U_m \label{phase:eq:Uavarge} \tag{22} \] It can be noted that the continuity equation is satisfied as
Averaged Mass Rate
\[
\label{phase:eq:m} \dot{m} = \rho_m \,U_m \, A \tag{23} \]
Example 13.1
Under what conditions equation (23) is correct?
Solution 13.1
Under Construction
The governing momentum equation can be approximated as
\[ \dot{m}\, \dfrac{dU_m}{dx} = - A\, \dfrac{dP}{dx} - S\, \tau_w - A\,\rho_m\,g\,\sin\theta \label{phase:eq:momentum} \tag{24} \] or modifying equation (24) as
Averaged Momentum
\[
\label{phase:eq:Pmomentum} - \dfrac{dP}{dx} = - \dfrac{S}{A} \, \tau_w - \dfrac{\dot{m}}{A} \, \dfrac{dU_m}{dx} + \rho_m\,g\,\sin\theta \tag{25} \]
The energy equation can be approximated as
Averaged Energy
\[
\label{phase:eq:energy} \dfrac{dq}{dx} - \dfrac{dw}{dx} = \dot{m}\, \dfrac{d}{dx} \left( h_m + \dfrac{{U_m}^2}{2} + g\,x\,\sin\theta \right) \tag{26} \]
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
From wikipedia:
For some approximation algorithms it is possible to prove certain properties about the approximation of the optimum result. For example, a $ρ$-approximation algorithm $A$ is defined to be an algorithm for which it has been proven that the value/cost, $f(x)$, of the approximate solution $A(x)$ to an instance $x$ will not be more (or less, depending on the situation) than a factor $ρ$ times the value, OPT, of an optimum solution.
$$\begin{cases}\mathrm{OPT} \leq f(x) \leq \rho \mathrm{OPT},\qquad\mbox{if } \rho > 1; \\ \rho \mathrm{OPT} \leq f(x) \leq \mathrm{OPT},\qquad\mbox{if } \rho < 1.\end{cases}$$
For a maximization problem, a $\rho$-approximation algorithm means that, for an instance $x$, $\rho \mathrm{OPT} \leq f(x) \leq \mathrm{OPT}$, for any $\rho<1$. Is this mean that
in the worst case the algorithm produces a solution that is at least $\rho \mathrm{OPT}$? Or is this in the average case? Because there are some instances where the algorithm find the optimal solution.
I mean, if the algorithm is executed, say $10000$ times (with randomly generated instances) and produces $\mathrm{OPT}$, say $9995$ times and $5$ times it produces a value that is greater than $\rho\mathrm{OPT}$. What can we say about the algorithm then? |
Homework 7 collaboration area
Question: What exactly is 6.2, #9 asking when it says to use another method to find the laplace transform for Prob 1? (AM, 07-Oct)
Answer: I think they just want you to show that it can be computed in two ways. In problem 1, you probably used the identity
L[f'] = s F(s) - f(0).
To compute the same Laplace transform a second way, you could integrate directly from the definition of the Laplace transform, or maybe you could use
L[f"] = s^2 F(s) - s f(0) - f'(0)
to get the same answer as problem 1.
For 6.2, #9 (prob 1 part) I followed the book's hint and used shifting. See Theorem 2 pp. 224. Let f(t)=t, and use $ e^{kt} $ for shifting. --Rekblad 06:57, 11 October 2010 (UTC)
p. 232, #9, part II: What does the problem mean? It says to express cos^2(.5t) in terms of cosine and by using problem 3. What does this mean?
Answer: I think it means to use
$ \cos^2\theta=\frac{1}{2}(1+\cos(2\theta)) $
to write
$ \cos^2(t/2)=\frac{1}{2}(1+\cos t)) $.
Then use that to compute the Laplace Transform. Another way to compute the same thing, is to use
$ \cos^2\theta+\sin^2\theta =1 $
to get
$ \cos^2(t/2)=1-\sin^2(t/2) $,
and now problem 3 can be used to compute the Laplace Transform.
Sec6.2 P232 #31: I've factored out the s in the denominator so it looks like
$ \frac{1}{s}\ \frac{5}{s^2-5} $
But I'm not sure how to proceed from there.
Answer: You will need to use the integration formula on p. 239:
$ \mathcal{L}[\ \int_0^t f(\tau)d\tau \ ]=\frac{1}{s}F(s), $
using F(s) = 5/(s^2 - 5). Find f(t) and integrate as shown to find the inverse transform of the given function.
Question: How do you find the inverse Laplace of (5/(s^2 - 5))?
Answer: There are two ways to do that. One way is to use the table on page 224 to get
$ \sqrt{5}\sinh(\sqrt{5}\, t) $.
The other way is to use partial fractions to get
$ \frac{5}{s^2-5}=\frac{A}{s-\sqrt{5}} + \frac{B}{s+\sqrt{5}} $
and take it from there.
Sec6.3 P240 #8: I have it written out as
f(t)=[u(t-0)-u(t-pi)]*(1-e^(-t)).
I'm stuck on how to work out (1-e^(-t)). In the previous problem, #5, it was easy to make t^2 into
[(t-1)+1]^2 or [(t-2)+2]^2
and essentially not change the function. However, that's not the case with (1-e^(-t)) and I don't know what to do with it.
Answer: Do the same thing:
$ 1-e^{-t}=1-e^{-[(t-\pi)+\pi]}=1-e^{-\pi}e^{-(t-\pi)} $
Sec6.3 #5: Yes, I agree it is easy to make t^2 into [(t-1)+1]^2 or [(t-2)+2]^2. Doing this, the book's solution is 2/s^3 + 2/s^2 + 1/s but I get: ... - 1/s by expanding upon the previous square. Any thoughts as to why the sign difference? Thanks.
Question: Pg. 232 - #9: The back of the book states that (cos a)^2 = 1/2 + 1/2 cos 2a.....where does this come from?
Answer: (cosx)^2=1-(sinx)^2, (sinx)^2=1/2-1/2cos(2x) => (cosx)^2=1-[1/2-1/2cos(2x)]=1/2+1/2cos(2x)
Question: pg247 #7: Does anyone think that there might be a typo in the back of the book? in the "u(t-4)" term, I don't see how they have 1/3sin(3t-12), I think it should be 2/3. The 1/3e^-t*sin3t makes sense because you have a (1-2/3)e^-t*sin33t, but there are no other sine terms to combine with in the "u(t-4" area. This problem was heavy on the book keeping and I might have dropped something, but I can't find it.
Follow Up Question to Above: The (1/3)sin3t term inside u(t-4) multiplier must have an exp(-t+4) multiplied to it, right? I dont see how this has vanished in the Book answer.--Sdhar 23:04, 12 October 2010 (UTC)
Follow Up Answer: You are right...the (1/3)sin3t term does have an exp(-t+4) component. Look at the book answer again. You will see that the exp(-t+4) term was factored out of both the cos and sine terms for the u(t-4) part.
Answer: I actually made the same mistake my first go around. The problem I encountered was after solving for the partial fractions of the "u(t-4)" term, I had:
$ \frac{-1}{s}\ + \frac{s+2}{(s+1)^2+9}\ $
This should expand to:
$ \frac{-1}{s}\ + \frac{s+1}{(s+1)^2+9}\ + \frac{1}{(s+1)^2+9}\ $
This will give you the answer the book has. Our error came from violating the s-shifting rules and incorrectly solving for the inverse Laplace using the following:
$ \frac{-1}{s}\ + \frac{s}{(s+1)^2+9}\ + \frac{2}{(s+1)^2+9}\ $
As you can see, the inverse laplace cannot be taken of the second term.... Hope this helps you.
It does, I can't believe we both made the same mistake. Thank you.
P.240 #27: Given the piecewise function r(t), I converted this into a function of u(t-pi) and input that as the RHS of the system. I go through and solve for the answer that matches the back of the book, but I notice that it is only valid between 0 and pi and that there is a different answer when t > pi. I thought that by creating the function 8sin(t)*(1-u(t-pi)) we took care of the fact that the function is zero after pi... I am assuming that we obtain the second solution by setting the RHS to zero, but can anyone explain why we have to do this after building the piecewise function? Thanks!
P.240 #27: I don't really understand the question above, but maybe someone can answer me this: At the end of the evaluation of the H(s) functions through partial fractions, shouldn't
$ \mathcal{L}^{-1}[ \frac{-1}{s^2+3^2}\ ] = -\frac{1}{3}sin(3t) $?
Yet in the solution the 1/3 has disappeared. I've spent hours trying to track down where I went wrong and I can't find where I am missing a 3 to cancel that term. Anybody have an idea what I'm missing here?
Answer: You should have a 4/3 sin(3t) component as part of your y(t). This accounts for the "missing" -1/3 sin(3t) component. Think of it as y(t) = sin(t) -1/3 sin(3t) + 4/3 sin(3t) which becomes [sin(t) + 3/3 sin(3t)] or [sin(t) + sin(3t)] for 0 < t < pi. For t > pi, y(t) = sin(t) - 1/3 sin(3t) - sin(t) - (-1/3) sin(3t) + 4/3 sin(3t) which simplifies to 4/3 sin(3t). Hope this helps.
Page 241, Problem 45: Can any of you EE's explain how to get this started? Do I need to go learn the integro-differential equation from sec 2.9?
Answer: check out example 2 on page 132. The switch in our problem is like the battery and switch in the example. Reply: Thanks. Yeah, I just needed to follow that walk-through.
p. 240:
QUESTION: I'm confused by the directions. It says that what is given is the Laplace transform of f(t) (that is, L[f]). But it looks to me what is given is L[(f)*u(t-a)]), not just L[f]. Can someone please clarify what is going on here? Furthermore, the back of the book gives answers for f(t) for various intervals, but for #21, it doesn't. Please help!! Thanks.
From Bell: Here is a little Flash video hint about 247:16.
And here is a Flash video hint about 240: 45.
You can find PDFs of the aftermath of the videos at Bell's Jing things
p. 240, #45: Since 1 kV = 1000 V, shouldn't there be a factor of 1000 somewhere in the answer? Thank you. Re: I agree. Since the answer in the back does not have any units associated with it, I left the factor of 1000 in my i(t) and put [A] at the end. Figured that would cover my bases ---Rayala 20:42, 12 October 2010 (UTC)
Re: I was thinking the same. I believe they probably factored out the 1000 and their current is in terms of mA instead of A. -Artesha
From Bell: Yes, I agree with Artesha. Don't sweat about that factor of 1000. |
In the discussion above it was assumed that the liquid is pure. In this short section a discussion about the bulk modulus averaged is presented. When more than one liquid are exposed to pressure the value of these two (or more liquids) can have to be added in special way. The definition of the bulk modulus is given by equation (34) or (35) and can be written (where the partial derivative can looks as delta \(\Delta\) as \[\partial{V} = \frac{V\partial{P}}{B_T} \cong \frac{V \Delta P}{B_T}\tag{45}\] The total change is compromised by the change of individual liquids or phases if two materials are present. Even in some cases of emulsion (a suspension of small globules of one liquid in a second liquid with which the first will not mix) the total change is the summation of the individuals change. In case the total change isn't, in special mixture, another approach with taking into account the energy-volume is needed. Thus, the total change is \[\partial{V} = \partial{V_{1}} + \partial{V_{2}} + \cdot \cdot \cdot + \partial{V_{i}} \cong \Delta{V_{1}} + \Delta{V_{2}} + \cdot \cdot \cdot + \Delta{V_{i}}\tag{46}\]
Substituting equation (45) into equation (46) results in \[ \label{intro:eq:DeltaVexpli} \partial V = \dfrac{V_1\,\partial P }{{B_T}_1} + \dfrac{V_2\,\partial P }{{B_T}_2} + \cdots + \dfrac{V_i\,\partial P }{{B_T}_i} \cong \\ \dfrac{V_1\,\Delta P }{{B_T}_1} + \dfrac{V_2\,\Delta P }{{B_T}_2} + \cdots + \dfrac{V_i\,\Delta P }{{B_T}_i} \,\, \] Under the main assumption in this model the total volume is comprised of the individual volume hence, \[ \label{intro:eq:v-vi} V = x_1 \, V + x_1 \,V + \cdots + x_i \, V \] Where \(x_1\), \(x_2\) and \(x_i\) are the fraction volume such as \(x_i= V_i/V\). Hence, using this identity and the fact that the pressure is change for all the phase uniformly the previous equation can be written as \[ \label{intro:eq:v-vi1} \partial V = V \,\partial P \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i} \right) \cong \\ V\, \Delta P \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots +\dfrac{x_i}{{B_T}_i} \right) \] Rearranging it yields \[ \label{intro:eq:newBTstart} v\, \dfrac{\partial P}{\partial v} \cong v\, \dfrac{\Delta P}{\Delta v} = \dfrac{1}{ \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i} \right) } \] This equation suggested an averaged new bulk modulus \[ \label{intro:eq:BTmixDef} {B_T}_{mix} = \dfrac{1}{ \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i} \right) } \] In that case the equation for mixture can be written as \[ \label{intro:eq:BTmix} v\, \dfrac{\partial P}{\partial v} = {B_T}_{mix} \]
End Caution: advance matherial
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
The Annals of Probability Ann. Probab. Volume 41, Number 2 (2013), 848-870. Nonconcentration of return times Abstract
We show that the distribution of the first return time $\tau$ to the origin, $v$, of a simple random walk on an infinite recurrent graph is heavy tailed and nonconcentrated. More precisely, if $d_{v}$ is the degree of $v$, then for any $t\geq1$ we have
\[\mathbf{P} _{v}(\tau\ge t)\ge\frac{c}{d_{v}\sqrt{t}}\]
and
\[\mathbf{P} _{v}(\tau=t\mid\tau\geq t)\leq\frac{C\log(d_{v}t)}{t}\]
for some universal constants $c>0$ and $C<\infty$. The first bound is attained for all $t$ when the underlying graph is $\mathbb{Z}$, and as for the second bound, we construct an example of a recurrent graph $G$ for which it is attained for infinitely many $t$’s.
Furthermore, we show that in the
comb product of that graph $G$ with $\mathbb{Z}$, two independent random walks collide infinitely many times almost surely. This answers negatively a question of Krishnapur and Peres [ Electron. Commun. Probab. 9 (2004) 72–81] who asked whether every comb product of two infinite recurrent graphs has the finite collision property. Article information Source Ann. Probab., Volume 41, Number 2 (2013), 848-870. Dates First available in Project Euclid: 8 March 2013 Permanent link to this document https://projecteuclid.org/euclid.aop/1362750944 Digital Object Identifier doi:10.1214/12-AOP785 Mathematical Reviews number (MathSciNet) MR3077528 Zentralblatt MATH identifier 1268.05183 Subjects Primary: 05C81: Random walks on graphs Citation
Gurel-Gurevich, Ori; Nachmias, Asaf. Nonconcentration of return times. Ann. Probab. 41 (2013), no. 2, 848--870. doi:10.1214/12-AOP785. https://projecteuclid.org/euclid.aop/1362750944 |
The Astrophysical Multimessenger Observatory Network (AMON) has been working to link the world's high-energy and multimessenger observatories together into a single network in order to evoke discovery of multimessenger sources, exploit these sources for purposes of astrophysics, fundamental physics, and cosmology, and explore project datasets for evidence of multimessenger source populations.
AMON has been working to commission multiple multimessenger alert streams, including gravitational wave + gamma-ray (GW+$\gamma$) and high energy neutrino + gamma-ray ($\nu$+$\gamma$) coincidence alerts.
One such $\nu$+$\gamma$ alert stream, now in an advanced stage of development, will search in near real-time for statistically-rare coincidences between $\sim$TeV gamma-rays observed by the High-Water Altitude Cherenkov Observatory (HAWC) and $\geqslant$TeV neutrinos detected by the IceCube Neutrino Observatory.
We describe the statistical design, calibration, and validation of these HAWC and IceCube $\nu$+$\gamma$ alerts, which will be commissioned soon and made available to AMON follow-up partners under terms of the AMON MoU.
With a median delay to alert distribution of six hours and angular uncertainties of $\leqslant$1\arcdeg, the alerts should be well-suited for deep electromagnetic follow-up observations. |
Electronic Journal of Statistics Electron. J. Statist. Volume 10, Number 2 (2016), 1927-1972. Sharp minimax tests for large covariance matrices and adaptation Abstract
We consider the detection problem of correlations in a $p$-dimensional Gaussian vector, when we observe $n$ independent, identically distributed random vectors, for $n$ and $p$ large. We assume that the covariance matrix varies in some ellipsoid with parameter $\alpha >1/2$ and total energy bounded by $L>0$.
We propose a test procedure based on a U-statistic of order 2 which is weighted in an optimal way. The weights are the solution of an optimization problem, they are constant on each diagonal and non-null only for the $T$ first diagonals, where $T=o(p)$. We show that this test statistic is asymptotically Gaussian distributed under the null hypothesis and also under the alternative hypothesis for matrices close to the detection boundary. We prove upper bounds for the total error probability of our test procedure, for $\alpha>1/2$ and under the assumption $T=o(p)$ which implies that $n=o(p^{2\alpha})$. We illustrate via a numerical study the behavior of our test procedure.
Moreover, we prove lower bounds for the maximal type II error and the total error probabilities. Thus we obtain the asymptotic and the sharp asymptotically minimax separation rate $\widetilde{\varphi}=(C(\alpha,L)n^{2}p)^{-\alpha/(4\alpha +1)}$, for $\alpha>3/2$ and for $\alpha >1$ together with the additional assumption $p=o(n^{4\alpha -1})$, respectively. We deduce rate asymptotic minimax results for testing the inverse of the covariance matrix.
We construct an adaptive test procedure with respect to the parameter $\alpha$ and show that it attains the rate $\widetilde{\psi}=(n^{2}p/\ln\ln(n\sqrt{p}))^{-\alpha/(4\alpha +1)}$.
Article information Source Electron. J. Statist., Volume 10, Number 2 (2016), 1927-1972. Dates Received: November 2015 First available in Project Euclid: 18 July 2016 Permanent link to this document https://projecteuclid.org/euclid.ejs/1468849967 Digital Object Identifier doi:10.1214/16-EJS1143 Mathematical Reviews number (MathSciNet) MR3522665 Zentralblatt MATH identifier 1346.62086 Citation
Butucea, Cristina; Zgheib, Rania. Sharp minimax tests for large covariance matrices and adaptation. Electron. J. Statist. 10 (2016), no. 2, 1927--1972. doi:10.1214/16-EJS1143. https://projecteuclid.org/euclid.ejs/1468849967 |
Yes, there are differences in accuracy since with machine numbers the usual properties of arithmetics don't hold.
Machine numbers are defined as $$ F(\beta,t,m,M)= \{ 0 \} \cup \{ x \in \mathbb{R} : x = sign(x)\beta^{p} \sum_{i=1}^{t}d_i\beta^{-i},\ 0 \leq d_i \lt \beta\ ,\ d_1\ne 0\ , -m \le p \le M \} $$ and represent the subset of $\mathbb{R}$ that your machine is able to represent.
All other numbers must be approximated with a number in this subset (usually by truncating the numbers or rounding them).
Let's assume we are in $F(10,2,m,M)$ meaning we are working in base 10 with two digits.
Let $x=0.11*10^1$, $y=0.31*10^{1}$ and $z=0.25*10^{1}$.
The associative property of multiplication doesn't hold:
$(x*y)*z = 0.34*10^1 * z = 0.85*10^1$
and
$x*(y*z) = x * (0.78 * 10^1) = 0.86*10^1$
other properties that don't hold are: |
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Find the value of cos(2cos−1x+sin−1x)\cos(2\cos^{-1}x + \sin^{-1}x)cos(2cos−1x+sin−1x) at x=15x=\frac {1}{5}x=51, where 0≤cos−1x≤π0≤\cos^{-1}x≤\pi0≤cos−1x≤π and −π2≤sin−1x≤π2\frac {-\pi}{2} ≤ \sin^{-1}x ≤ \frac {\pi}{2}2−π≤sin−1x≤2π.
If your answer is in the form of −abc-\frac {a\sqrt{b}}{c}−cab, where bbb is a square free integer, enter the value of a+b+ca+b+ca+b+c.
This problem is part of the set Trigonometry.
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In the solution to Making more easy the itemized of item with tabulation system, I am counting up the number of spaces in order to determine the type of leading character to insert. However, if the I replace the leading spaces with a tab character, the solution does not work.
If I could detect the tab character in the
literate, then I could have
\ProcessSpace increment the counter
NumOfContigousSpaces appropriately, but I don't know how to test for it?
I thought adding
tabsize=4, keepspaces=true would do the job but this is not quite enough. So I attempted to use
lstag@tabulator from How to automatically skip leading white spaces in listings, but was not able to get that to work.
The code below has a 4 leading spaces before the
W in the first line and a tab as the leading character before the
W in the second line. This produces no bullet for the line with a tab:
The correct output can be seen by using 4 spaces before the
W in both lines:
Note: It appears the posting a code snippet here replaces a tab with 4 spaces. So to use the MWE below you will need to replace the four leading spaces before the
Wxxxwith a
tabcharacter.
Code:
\documentclass{article}\usepackage{pgf}\usepackage{xstring}\usepackage{listings}\newcounter{NumOfContigousSpaces}%\setcounter{NumOfContigousSpaces}{0}%\newcommand{\Width}{1}%\newcommand*{\AddApproriateBulletIfFirstChar}[1]{% \pgfmathtruncatemacro{\BulletType}{\arabic{NumOfContigousSpaces}/4}% \IfEqCase{\BulletType}{% {0}{\gdef\Width{1}} {1}{\gdef\Width{3}$\bullet$ } {2}{\gdef\Width{3}$\circ$ } {3}{\gdef\Width{3}$\times$ } {4}{\gdef\Width{3}$\star$ } {5}{\gdef\Width{3}$-$ } }[\gdef\Width{3}$\bullet$ ]% #1% \setcounter{NumOfContigousSpaces}{0}%}%\newcommand*{\ProcessSpace}{% \addtocounter{NumOfContigousSpaces}{1}% \space%}%\newcommand*{\ProcessTab}{% \addtocounter{NumOfContigousSpaces}{4}% \space\space\space\space%}%\makeatletter\lstdefinestyle{MyItemize}{% basicstyle=\ttfamily, columns=flexible, tabsize=4, keepspaces=true, literate=% {\ }{{{\ProcessSpace}}}1% Count contigous spaces {lstag@tabulator}{{{\ProcessTab}}}4% ??? how detect a tab? % %--- much code removed here (See https://tex.stackexchange.com/questions/57939/making-more-easy-the-itemized-of-item-with-tabulation-system for full code) {W}{{{\AddApproriateBulletIfFirstChar{W}}}}\Width {x}{{{\AddApproriateBulletIfFirstChar{x}}}}\Width}%\makeatother\begin{document}\begin{lstlisting}[style=MyItemize] Wxxx xxx Wxxx xx xx\end{lstlisting}\end{document} |
Obviously, from the given inequality we gain
$$f( 2^{2^k}) < ck f(2).$$
Then, let $x \in \mathbb N$. There exists a unique $k = k(x)$ such that $x \in [2^{2^k}, 2^{2^{k+1}})$. Then
$$f(x) \le f(2^{2^k}) + f(x - 2^{2^k}).$$
From this we see by induction that $f(x) \le f(2^{2^{k+1}}) < c(k+1) f(2)$. But
$$k(x) = \lfloor \log_2(\log_2(x)) \rfloor,$$
proving the first claim. The given sum will be bounded by
$$\log(n) \sum_{k=0}^{\log(\log(n))} \frac{c(k+1)}{2^k} f(2)$$
by the above estimates. One can consider the polynomial$$p_n(x) := 4 \log(n) \sum_{k=0}^{\log(\log(n))} cx^k f(2)$$
and observe that the sum comes from a derivative of that polynomial, inserting $x = 1/2$. Then one should be able to apply calculus to get some further bounds. |
So, I am at the moment working on gauge-fixing a path integral. The procedure involves adding a delta function $\delta g$ to the path integral (together with the faddeev-popov determinant, but that is not important for this discussion). Once the delta function has been added, we make the argument that the gauge condition is arbitrary, and we are free to add any function $\omega$ to it as long as it does not affect the Faddeev-Popov deteminant, which it will not as it does not depend on the gauge variable.
Now comes the part where I find my understanding to be lacking. In the literature, we say that we average over all the arbitrary functions around $\omega = 0$, invoking a gauge weight. This argument seems a bit arbitrary for me; We should be free to use any weighting function so far as I am aware, since $\omega$ is arbitrary.
It looks like this: $$Z = \int DA_\mu\, \delta(g-\omega) e^{iS}\,,$$ $$Z = N(\xi)\int DA_\mu D\omega\, e^{-i\int dx\, \omega^2/2\xi} \delta(g-\omega) e^{iS}\,,$$ $$Z = N(\xi)\int DA_\mu\, e^{-i\int dx\, g^2/2\xi} e^{iS}\,.$$
I find no specific argument for why the exponential that gets added has the form that it has. It sure is convenient since it gives us the gauge condition in quadratic form, so that it integrates nicely with the Lagrangian we needed to gauge-fix in the first place. However, my current understanding is that this is an arbitrary choice; We might have just used a cubic weighing function, and ended up with a gauge-fixing Lagrangian of a completely different form.
Please tell me if my understanding is correct, and if it is not, please tell me the argument for the use of Gaussian weights.
Thanks. |
Here's one 5 piece solution:To make this into a square:To make it into an equilateral triangle:Here's a different 5 piece solution:Here's an answer withYou will see thatHowever, which is still a problem, it's possible to just solve it the original way without any flips. To solve this, you can simply take a symmetrical sliver out of the blue piece ...
@Daniel_Mathias gave a very helpful link which has all the 12x5 solutions in a text file. So some simple code allows us to see that of the 1010 12x5 solutions, there are 264 with 1 straight cut. But, sadly, none with 2 or more cuts. A few examples of the former are:12FFPPP IIIIINNLFFPP ZZWNNNTLFXUU VZWWTTTLXXXU VZZWWYTLLXUU VVVYYYY81...
Here's simple 2-D pattern that seems to tile quite efficiently:The area of the each tile (blue square) is $21\times21 = 441$ tiles, and it contains $4\times14=56$ generators tiles, for a ratio of $\frac{56}{441} \approx 12.7\%$The trick here is thatThe final pattern looks like this:POST-TICK EDIT: managed to find an even better pattern with $\mathbf{...
I decided to place the wind generators together pinwheel-fashion:The repeated section looks like this:To calculate the efficiency:There are $70+70+16+9 = 165$ empty squares, $4\cdot14 = 56$ filled squares, for an efficiency of $\frac{56}{221} = 25.339\%$.Sadly this is not quite as efficient as Bass's solution, but ever so close.In my first attempt I ...
As mentioned in the question, here is an example of an answer:This 20x7 tile setup can be validly tile-replicated, as the required open space of the left generator which is "out of bounds" correctly loops to the right to coincide with the open space of the right one:Since we've established validity, the ratio is: 28 generator tiles/140 total tiles=20% |
An equilateral triangle of side length 2 is inscribed in a circle. A rectangle, with side lengths x and y, is inscribed in another circle of the same size as the first one. If the area of the rectangle is the same as that of the triangle, what are the sizes of x and y?
.
1: Cos(30) = 1/r therefore r = 2/sqrt3 and r^2 = 4/3 2:Area of equil Triang = (1/2)*(2)*sqrt((2^2)-(1^2))=sqrt3 therefore Area of rectangle = xy = sqrt3 3 Draw lines to the opposite vertices of the rectangle then add the area of the 4 triangles you just made inside the rectangle = total area:
2*area Tri A= 2*[(1/2)(x)sqrt(4/3 - (x^2)/4)
2*area Tri B= 2*[(1/2)(y)(x/2)]
4, Solve simultaneous equation to get x and y:
xy = sqrt3
(x)sqrt(4/3 - (x^2)/4) + (1/2)xy = sqrt3
--> If doing manually you substitute the first equation into the second then solve for x, followed by y, but since I'm lazy I used Wolfram alpha to do this for me and got
x≈2.16662 and y≈0.799424
I thought this was a really cool question. Thanks for providing it Alan.
Isn't any one of you high school students going to show us what you are made of and answer it.
You could at least give it a try :/
1: Cos(30) = 1/r therefore r = 2/sqrt3 and r^2 = 4/3 2:Area of equil Triang = (1/2)*(2)*sqrt((2^2)-(1^2))=sqrt3 therefore Area of rectangle = xy = sqrt3 3 Draw lines to the opposite vertices of the rectangle then add the area of the 4 triangles you just made inside the rectangle = total area:
2*area Tri A= 2*[(1/2)(x)sqrt(4/3 - (x^2)/4)
2*area Tri B= 2*[(1/2)(y)(x/2)]
4, Solve simultaneous equation to get x and y:
xy = sqrt3
(x)sqrt(4/3 - (x^2)/4) + (1/2)xy = sqrt3
--> If doing manually you substitute the first equation into the second then solve for x, followed by y, but since I'm lazy I used Wolfram alpha to do this for me and got
x≈2.16662 and y≈0.799424
Correct numerical answer Bro, though it would be nice to see your full expressions for x and y! The reason for this is that there are several ways to solve this problem, each of which can result in expressions that look different, but are, of course, equivalent.
.
I did like Brodudedoodebrodude in principle (even if I´m a little late.)
x:
$${\frac{{\sqrt{{\mathtt{3}}}}}{{\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}}}$$
y:
$${\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}$$
This has probably been done to death but too bad.
The area of the triangle is
$$\\3\times \frac{1}{2}\times r^2*sin \frac{2\pi}{3}\\\\
=\frac{3r^2}{2}sin\frac{\pi}{3}\\\\ =\frac{3r^2}{2}\times\frac{\sqrt{3}}{2}\\\\ =\frac{3\sqrt{3}r^2}{4}\\\\$$
$$\\Area\\
=4\times\frac{1}{2}\times r^2sin\theta\\\\ =2r^2sin\theta\\\\ so\\\\ 2r^2sin\theta=\frac{3\sqrt3*r^2}{4}\\\\ sin\theta=\frac{3\sqrt3}{8}\\\\ If\;opp=3\sqrt3\;\;and\;\;hyp=8\;\;then\;\;adj=\sqrt{64-27}=\sqrt{37}\\\\ so \;\;cos\theta=\frac{\sqrt{37}}{8}\\\\ -----\\ x^2=2r^2-2r^2cos\theta\\ x^2=2r^2-2r^2\times\frac{\sqrt{37}}{8}\\ x^2=\frac{8r^2}{4}-\frac{r^2\sqrt{37}}{4}\\ x^2=\frac{r^2}{4}\left(8-\sqrt{37}\right)\\\\ x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies$$
$$\\xy=\frac{3\sqrt3r^2}{4}\\\\
y=\frac{3\sqrt3r^2}{4}\div x\\\\ y=\frac{3\sqrt3r^2}{4}\times \frac{1}{x} \\\\ y=\frac{3\sqrt3r^2}{4}\times \frac{2}{r\sqrt{8-\sqrt{37}}} \\\\ y=\frac{3\sqrt3r}{2}\times \frac{1}{\sqrt{8-\sqrt{37}}} \\\\ y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\$$
Is this answer the same as the others?
This is fine as far as it goes Melody, and if you calculate r you can take it all the way!
.
Yes I forgot about that Alan. Maybe this is better
$$\\4=2r^2-2r^2cos120\\
4=2r^2-2r^2*-0.5\\ 4=2r^2+r^2\\ 4=3r^2\\ \frac{4}{3}=r^2\\ r=\frac{2}{\sqrt3}\\\\ y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{3*\frac{2}{\sqrt3}\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{6}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{3}{\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\\\ x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\ x=\frac{\frac{2}{\sqrt3}*\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\ x=\frac{\sqrt{3\left(8-\sqrt{37}\right)}}{3}\;\;unit\;thingies$$
The area of the equilateral triangle = (1/2) 4 sin120 = 2(√3)/ 2 = √3
The radius of both circles is given by √[ 1^2 + (1/√3)^2 ] = √[1 + 1/3] = 2/ √3
So....the diagonal of the rectangle = 2 ( 2/ √3 ) = 4 / √3
And, using the Pythagorean Thorem, "y" in the rectangular figure = √ [( 4/ √3) ^2 - x^2 ] = √[16/3 - x^2]
So......since the area of the triangle and the rectangle is equal, we have...
y * x = √3
√[16/3 - x^2] * x = √3 square both sides
(16/3)x^2 - x^4 = 3 multiply through by 3
16x^2 - 3x^4 = 9 rearrange
3x^4 - 16x^2 + 9 = 0 let m= x^2 .... so we have
3m^2 - 16m + 9 = 0
Using the quadratic formula [and an assist from the on-site solver], we have
$${\mathtt{3}}\left[{{m}}^{{\mathtt{2}}}\right]{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{m}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,-\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\
{\mathtt{m}} = {\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{0.639\: \!079\: \!156\: \!567\: \!260\: \!1}}\\ {\mathtt{m}} = {\mathtt{4.694\: \!254\: \!176\: \!766\: \!073\: \!2}}\\ \end{array} \right\}$$
So x = about .7994 or x = about 2.1666
If x = .799, y = √[16/3 - .7994^2] = .2166 ...but this is impossible....since y is smaller than x
If x = 2.1666, y = √[16/3 - 2.1666^2] = about . 7994
So....x =about .2166 and y = about .7994
Nice answer Chris
Yours looks a bit easier than mine.
I suppose I should include my approx answers so that the answersc an be compared. |
November 21st, 2017, 06:03 PM
# 1
Newbie
Joined: Nov 2017
From: uk
Posts: 3
Thanks: 0
I don't understand the use of these trig functions for a force vector
Hi
I am trying to learn mechanics with static particles.
I have a vector that points up and left with an angle theta with respect to the vertical axis.
The answer sheet says that the direction is therefore:
V = −|V| sinθi + |V| cosθj
But I don't see how it's a negative sin trig for i component and positive cos trig for j component.
I can understand it more like:
V = |V|cos(θ+90)i + |V|sin(θ+90)j
I was hoping some one could explain how they got their answer ?
Last edited by skipjack; November 22nd, 2017 at 01:36 PM.
November 21st, 2017, 06:32 PM
# 3
Newbie
Joined: Nov 2017
From: uk
Posts: 3
Thanks: 0
Thank you for the reply.
I've not seen it done like that before. We have only been taught doing it like this so far:
Are there other ways to visualise it without thinking about changing the i and j axis signs?
It's still not 100% obvious to me why the vertical j in your picture is using "cos" and the horizontal negative i is "sin" ? I would naturally had used "-cos" for i and "sin" for j in that image.
November 21st, 2017, 06:55 PM
# 4
Math Team
Joined: Jul 2011
From: Texas
Posts: 3,033
Thanks: 1621
Look at the position of $\theta$ in the right triangle ...
... the horizontal component is opposite to $\theta$ and the vertical component adjacent.
Remember SOHCAHTOA?
Last edited by skipjack; November 22nd, 2017 at 01:34 PM.
November 27th, 2017, 06:15 AM
# 6
Math Team
Joined: Jan 2015
From: Alabama
Posts: 3,264
Thanks: 902
Since in the xi+ yj form, the angle is typically measured from the positive x-axis, you can also write that as $\displaystyle cos(\theta+ \pi/2)i+ sin(\theta+ \pi/2)j$. But $\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$ and $\displaystyle cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)cos(\phi)$
So $\displaystyle sin(\theta+ \pi/2)= sin(\theta)cos(\pi/2)+ cos(\theta)sin(\pi/2)= sin(\theta)(0)+ cos(\theta)(1)= cos(\theta)$ and $\displaystyle cos(\theta+ \pi/2)= cos(\theta)(cos(\pi/2)- sin(\theta)sin(\pi/2)= cos(\theta)(0)- sin(\theta)(1)= -sin(\theta)$.
$\displaystyle cos(\theta+ \pi/2)i + sin(\theta+ \pi/2)j= -sin(\theta)i + cos(\theta)j$.
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Strongly compact cardinal
The strongly compact cardinals have their origins in the generalization of the compactness theorem of first order logic to infinitary languages, for anuncountable cardinal $\kappa$ is
strongly compact if the infinitary logic $L_{\kappa,\kappa}$ exhibits the $\kappa$-compactness property. It turns out that this model-theoretic concept admits fruitful embedding characterizations, which as with so many large cardinal notions, has become the focus of study. Strong compactness rarefies into a hierarchy, and a cardinal $\kappa$ is strongly compact if and only if it is $\theta$-strongly compact for every ordinal $\theta\geq\kappa$.
The strongly compact embedding characterizations are closely related to that of supercompact cardinals, which are characterized by elementary embeddings with a high degree of closure: $\kappa$ is $\theta$-supercompact if and only if there is an embedding $j:V\to M$ with critical point $\kappa$ such that $\theta<j(\kappa)$ and every subset of $M$ of size $\theta$ is an element of $M$. By weakening this closure requirement to insist only that $M$ contains a small cover for any subset of size $\theta$, or even just a small cover of the set $j''\theta$ itself, we arrive at the $\theta$-strongly compact cardinals. It follows that every $\theta$-supercompact cardinal is $\theta$-strongly compact and so every supercompact cardinal is strongly compact. Furthermore, since every ultrapower embedding $j:V\to M$ with critical point $\kappa$ has $M^\kappa\subset M$, for $\theta$-strong compactness we may restrict our attention to the case when $\kappa\leq\theta$.
Contents 1 Diverse characterizations 1.1 Strong compactness characterization 1.2 Strong compactness embedding characterization 1.3 Cover property characterization 1.4 Fine measure characterization 1.5 Filter extension characterization 1.6 Discontinuous ultrapower characterization 1.7 Discontinuous embedding characterization 1.8 Ketonen characterization 1.9 Regular ultrafilter characterization 2 Strongly compact cardinals and forcing 3 Relation to other large cardinal notions 4 References Diverse characterizations
There are diverse equivalent characterizations of the strongly compact cardinals.
Strong compactness characterization
An uncountable cardinal $\kappa$ is
strongly compact if every $\kappa$-satisfiable theory in the infinitary logic $L_{\kappa,\kappa}$ is satisfiable. The signature of an $L_{\kappa,\kappa}$ language consists, just as in the first order context, of a set of finitary function, relation and constant symbols. The $L_{\kappa,\kappa}$ formulas, however, are built up in an infinitary process, by closing under infinitary conjunctions $\wedge_{\alpha<\delta}\varphi_\alpha$ and disjunctions $\vee_{\alpha<\delta}\varphi_\alpha$ of any size $\delta<\kappa$, as well as infinitary quantification $\exists\vec x$ and $\forall\vec x$ over blocks of variables $\vec x=\langle x_\alpha\mid\alpha<\delta\rangle$ of size less than $\kappa$. A theory in such a language is satisfiable if it has a model under the natural semantics. A theory is $\kappa$-satisfiable if every subtheory consisting of fewer than $\kappa$ many sentences of it is satisfiable. First order logic is precisely $L_{\omega,\omega}$, and the classical compactness theorem asserts that every $\omega$-satisfiable $L_{\omega,\omega}$ theory is satisfiable. Similarly, an uncountable cardinal $\kappa$ is defined to be strongly compact if every $\kappa$-satisfiable $L_{\kappa,\kappa}$ theory is satisfiable (and we call this the $\kappa$-compactness property}). The cardinal $\kappa$ is weakly compact, in contrast, if every $\kappa$-satisfiable $L_{\kappa,\kappa}$ theory, in a language having at most $\kappa$ many constant, function and relation symbols, is satisfiable. Strong compactness embedding characterization
A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an elementary embedding $j:V\to M$ of the set-theoretic universe $V$ into a transitive class $M$ with critical point $\kappa$, such that $j''\theta\subset s\in M$ for some set $s\in M$ with $|s|^M\lt j(\kappa)$. [1]
Cover property characterization
A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an ultrapower embedding $j:V\to M$, with critical point $\kappa$, that exhibits the
$\theta$-strong compactness cover property, meaning that for every $t\subset M$ of size $\theta$ there is $s\in M$ with $t\subset s$ and $|s|^M<j(\kappa)$. Fine measure characterization
An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a fine measure on $\mathcal{P}_\kappa(\theta)$. The notation $\mathcal{P}_\kappa(\theta)$ means $\{\sigma\subset\theta\mid |\sigma|<\kappa\}$. [1]
Filter extension characterization
An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if every $\kappa$-complete filter of size at most $\theta$ on a set extends to a $\kappa$-complete ultrafilter on that set. [1]
Discontinuous ultrapower characterization
A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an ultrapower embedding $j:V\to M$ with critical point $\kappa$, such that $\sup j''\lambda<j(\lambda)$ for every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$. In other words, the embedding is discontinuous at all such $\lambda$.
Discontinuous embedding characterization
A cardinal $\kappa$ is $\theta$-strongly compact if and only if for every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$, there is an embedding $j:V\to M$ with critical point $\kappa$ and $\sup j''\lambda<j(\lambda)$.
Ketonen characterization
An uncountable regular cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a $\kappa$-complete uniform ultrafilter on every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$. An ultrafilter $\mu$ on a cardinal $\lambda$ is
uniform if all final segments $[\beta,\lambda)= \{\alpha<\lambda\mid \beta\leq\alpha\}$ are in $\mu$. When $\lambda$ is regular, this is equivalent to requiring that all elements of $\mu$ have the same cardinality. Regular ultrafilter characterization
An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a $(\kappa,\theta)$-regular ultrafilter on some set. An ultrafilter $\mu$ is
$(\kappa,\theta)$-regular if it is $\kappa$-complete and there is a family $\{X_\alpha\mid\alpha<\theta\}\subset \mu$ such that $\bigcap_{\alpha\in I}X_\alpha=\emptyset$ for any $I$ with $|I|=\kappa$. Strongly compact cardinals and forcing
If there is proper class-many strongly compact cardinals, then there is a generic model of $\text{ZF}$ + "all uncountable cardinals are singular". If each strongly compact cardinal is a limit of measurable cardinals, and if the limit of any sequence of strongly compact cardinals is singular, then there is a forcing extension V[G] that is a symmetric model of $\text{ZF}$ + "all uncountable cardinals are singular" + "every uncountable cardinal is both almost Ramsey and a Rowbottom cardinal carrying a Rowbottom filter". This also directly follows from the existence of a proper class of supercompact cardinals, as every supercomact cardinal is simultaneously strongly compact and a limit of measurable cardinals.
Relation to other large cardinal notions
Strongly compact cardinals are measurable. The least strongly compact cardinal can be equal to the least measurable cardinal, or to the least supercompact cardinal, by results of Magidor. [2] (It cannot be equal to both at once because the least measurable cardinal cannot be supercompact.)
Even though strongly compact cardinals imply the consistency of the negation of the singular cardinal hypothesis, SCH, for any singular strong limit cardinal $\kappa$ above the least strongly compact cardinal, $2^\kappa=\kappa^+$ (also known as "SCH holds above strong compactness"). [2]
If there is a strongly compact cardinal $\kappa$ then for all $\lambda\geq\kappa$ and $A\subseteq\lambda$, $\lambda^+$ is ineffable in $L[A]$.
It is not currently known whether the existence of a strongly compact cardinal is equiconsistent with the existence of a supercompact cardinal. The ultrapower axiom gives a positive answer to this, but itself isn't known to be consistent with the existence of a supercompact in the first place.
Every strongly compact cardinal is tall, although the existence of a strongly compact cardinal is equiconsistent with "the least measurable cardinal is the least strongly compact cardinal, and therefore the least tall cardinal" meaning strongly compact cardinals aren't necessarily limits of tall cardinals.
References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Jech, Thomas J. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex Set Theory. |
Logic can be applied to draw conclusions from a set of premises. A premise is just a proposition that is known to be true or that has been accepted to be true for the sake of argument, and a conclusion is a proposition that can be deduced logically from the premises. The idea is that if you believe that the premises are true, then logic forces you to accept that the conclusion is true. An “argument” is a claim that a certain conclusion follows from a given set of premises. Here is an argument laid out in a traditional format:
If today is Tuesday, then this is Belgium
Today is Tuesday ____________________________________ ∴ This is Belgium
The premises of the argument are shown above the line, and the conclusion below. The symbol ∴ is read “therefore.” The claim is that the conclusion, “This is Belgium,” can be deduced logically from the two premises, “If today is Tuesday, then this is Belgium” and “Today is Tuesday.” In fact, this claim is true. Logic forces you to accept this argument. Why is that? Let p stand for the proposition “Today is Tuesday,” and let q stand for the proposition “This is Belgium.” Then the above argument has the form
\(p \to q\)
\(p\) ________ \(∴q\)
Now, for any propositions \(p\) and \(q\)—not just the ones in this particular argument—if \(p → q\) is true and p is true, then q must also be true. This is easy to check in a truth table:
The only case where both \(p → q\) and p are true is on the last line of the table, and in this case, q is also true. If you believe \(p → q\) and \(p\), you have no logical choice but to believe \(q\). This applies no matter what p and q represent. For example, if you believe “If Jill is breathing, then Jill pays taxes,” and you believe that “Jill is breathing,” logic forces you to believe that “Jill pays taxes.” Note that we can’t say for sure that the conclusion is true, only that if the premises are true, then the conclusion must be true.
This fact can be rephrased by saying that \((p → q)∧p → q\) is a tautology. More generally, for any compound propositions \(P\) and \(Q\), saying “\(P → Q\) is a tautology” is the same as saying that “in all cases where \(P\) is true, \(Q\) is also true”.11 We will use the notation \(P ⇒ Q\) to mean that \(P → Q\) is a tautology. Think of \(P\) as being the premise of an argument or the conjunction of several premises. To say \(P ⇒ Q\) is to say that \(Q\) follows logically from \(P\). We will use the same notation in both propositional logic and predicate logic. (Note that the relation of ⇒ to → is the same as the relation of ≡ to ↔.)
Definition 1.9
Let \(P\) and \(Q\) be any formulas in either propositional logic or predicate logic. The notation \(P ⇒ Q\) is used to mean that \(P → Q\) is a tautology. That is, in all cases where \(P\) is true, \(Q\) is also true. We then say that \(Q\) can be
logically deduced from \(P\) or that \(P\) l ogically implies \(Q\).
An argument in which the conclusion follows logically from the premises is said to be a
valid argument. To test whether an argument is valid, you have to replace the particular propositions or predicates that it contains with variables, and then test whether the conjunction of the premises logically implies the conclusion. We have seen that any argument of the form
\(p→q\)
\(p\) _____ \(∴q\)
is valid, since \((p → q) ∧ p → q\) is a tautology. This rule of deduction is called
modus ponens. It plays a central role in logic. Another, closely related rule is modus tollens, which applies to arguments of the form
\(p→q\)
\(¬q\) _______ \(∴ ¬p\)
To verify that this is a valid argument, just check that \((p \to q) ⇒ ¬p\), that is, that \((p \to q) \land \neg q) \to \neg p\) is a tautology. As an example, the following argument has the form of modus tollens and is therefore a valid argument:
If Keanu Reeves is a good actor, then I’m the king of France
I am not the king of France ___________________________________________________ ∴ Keanu Reeves in not a good actor
You should note carefully that the validity of this argument has nothing to do with whether or not Keanu Reeves can act well. The argument forces you to accept the conclusion only if you accept the premises. You can logically believe that the conclusion is false, as long as you believe that at least one of the premises is false.
Another named rule of deduction is the
Law of Syllogism, which has the form
\(p→q\)
\(q→r\) ________ \(∴ p→r\)
For example:
If you study hard, you do well in school
If you do well in school, you get a good job ____________________________________ ∴ If you study hard, you get a good job
There are many other rules. Here are a few that might prove useful. Some of them might look trivial, but don’t underestimate the power of a simple rule when it is combined with other rules
\(p \lor q\)
\(p\)
\(p \land q\)
\(p\)
Logical deduction is related to logical equivalence. We defined \(P\) and \(Q\) to be logically equivalent if \(P ↔ Q\) is a tautology. Since \(P ↔ Q\) is equivalent to \((P → Q) ∧ (Q → P)\), we see that \(P ≡ Q\) if and only if both \(Q ⇒ P and P ⇒ Q\). Thus, we can show that two statements are logically equivalent if we can show that each of them can be logically deduced from the other. Also, we get a lot of rules about logical deduction for free—two rules of deduction for each logical equivalence we know. For example, since \(¬(p∧q) ≡ (¬p∨¬q)\), we get that \(¬(p∧q) ⇒ (¬p∨¬q)\). For example, if we know “It is not both sunny and warm,” then we can logically deduce “Either it’s not sunny or it’s not warm.” (And vice versa.)
In general, arguments are more complicated that those we’ve considered so far. Here, for example, is an argument that has five premises:
\((p ∧ r) → s\)
\(q→p\) \(t→r\) \(q\) \(t\) ____________ \(∴s\)
Is this argument valid? Of course, you could use a truth table to check whether the conjunction of the premises logically implies the conclusion. But with five propositional variables, the table would have 32 lines, and the size of the table grows quickly when more propositional variables are used. So, in general, truth tables are not practical.
Fortunately, there is another way to proceed, based on the fact that it is possible to chain several logical deductions together. That is, if \(P ⇒ Q\) and \(Q ⇒ R\), it follows that \(P ⇒ R\). This means we can demonstrate the validity of an argument by deducing the conclusion from the premises in a sequence of steps. These steps can be presented in the form of a proof:
Definition 1.10
A
formal proof that an argument is valid consists of a sequence of propositions such that the last proposition in the sequence is the conclusion of the argument, and every proposition in the sequence is either a premise of the argument or follows by logical deduction from propositions that precede it in the list.
The existence of such a proof shows that the conclusion follows logically from the premises, and therefore that the argument is valid. Here is a formal proof that the argument given above is valid. The propositions in the proof are numbered, and each proposition has a justification.
1. \(q→p\)
premise 2. \(q\) premise 3. \(p\) from 1 and 2 ( modus ponens)
4. \(t→r\)
premise
5. \(t\)
premise
6. \(r\)
from 4 and 5 ( modus ponens)
7. \(p∧r\)
from 3 and 6
8. \((p∧r)→s\)
premise
9. \(s\)
from 7 and 8 ( modus ponens)
Once a formal proof has been constructed, it is convincing. Unfortunately, it’s not necessarily easy to come up with the proof. Usually, the best method is a combination of working forward (“Here’s what I know, what can I deduce from that?”) and working backwards (“Here’s what I need to prove, what other things would imply that?”). For this proof, I might have thought: I want to prove \(s\). I know that \(p ∧ r\) implies \(s\), so if I can prove \(p∧r\), I’m OK. But to prove \(p∧r\), it’ll be enough to prove \(p\) and \(r\) separately. . . .
Of course, not every argument is valid, so the question also arises, how can we show that an argument is invalid? Let’s assume that the argument has been put into general form, with all the specific propositions replaced by propositional variables. The argument is valid if in all cases where all the premises are true, the conclusion is also true. The argument is invalid if there is even one case where all the premises are true and the conclusion is false. We can prove that an argument is invalid by finding an assignment of truth values to the propositional variables which makes all the premises true but makes the conclusion false. For example, consider an argument of the form:
\(p→q\)
\(q → (p ∧ r)\) \(r\) _______________ \(∴p\)
In the case where \(p\) is false, \(q\) is false, and \(r\) is true, the three premises of this argument are all true, but the conclusion is false. This shows that the argument is invalid.
To apply all this to arguments stated in English, we have to introduce propositional variables to represent all the propositions in the argument. For example, consider:
John will be at the party if Mary is there and Bill is not there. Mary will be at the party if it’s on Friday or Saturday. If Bill is at the party, Tom will be there. Tom won’t be at the party if it’s on Friday. The party is on Friday. Therefore, John will be at the party.
Let \(j\) stand for “John will be at the party,” \(m\) for “Mary will be there,” \(b\) for “Bill will be there,” \(t\) for “Tom will be there,” \(f\) for “The party is on Friday,” and s for “The party is on Saturday.” Then this argument has the form
\((m ∧ ¬b) → j\)
\((f ∨ s) → m\) \(b→t\) \(f → ¬t\) \(f\) _________________ \(∴ j\)
This is a valid argument, as the following proof shows:
1.\(f→¬t\) premise
2.\(f\) premise 3. \(¬t\) from 1 and 2 (modus ponens) 4.\(b→t\) premise 5. \(¬b\) from 4 and 3 (modus tollens) 6.\(f∨s\) from 2 7. \((f∨s)→m\) premise 8.\(m\) from 6 and 7 (modus ponens) 9.\(m∧¬b\) from 8 and 5 10. \((m∧¬b)→j\) premise 11.\( j\) from 10 and 9 (modus ponens)
So far in this section, we have been working mostly with propositional logic. But the definitions of valid argument and logical deduction apply to predicate logic as well. One of the most basic rules of deduction in predicate logic says that \((∀xP (x)) =⇒ P (a)\) for any entity a in the domain of discourse of the predicate \(P\) . That is, if a predicate is true of all entities, then it is true of any given particular entity. This rule can be combined with rules of deduction for propositional logic to give the following valid arguments
\(∀x(P (x) → Q(x))\)
\(∀x(P (x) → Q(x))\)
These valid arguments go by the names of
modus ponens and modus tollens for predicate logic. Note that from the premise \(∀x(P(x) → Q(x))\) we can deduce \(P(a) → Q(a)\). From this and from the premise that \(P(a)\), we can deduce \(Q(a)\) by modus ponens. So the first argument above is valid. The second argument is similar, using modus tollens.
The most famous logical deduction of them all is an application of
modus ponens for predicate logic:
All humans are mortal
Socrates is human ____________________ ∴ Socrates is mortal
This has the form of
modus ponens with \(P(x)\) standing for “\(x\) is human,” \(Q(x)\) standing for “\(x\) is mortal,” and \(a\) standing for the noted entity, Socrates.
There is a lot more to say about logical deduction and proof in predicate logic, and we’ll spend the rest of this chapter on the subject.
Exercises
Verify the validity of
modus tollensand the Law of Syllogism.
Each of the following is a valid rule of deduction. For each one, give an example of a valid argument in English that uses that rule.
\(p \lor q\)
\(\neg p\) ___________ \(∴q\)
\(p\)
\(q\) _____ \(∴p∧q\)
\(p \land q\)
____________ \(∴p\)
\(p\)
_______ \(∴p∨q\)
There are two notorious invalid arguments that look deceptively like
modus ponensand modus tollens:
\(p→q\)
\(q\) ______ \(∴ p\)
\(p \to q\)
\(\neg p\) __________ \(∴ ¬q\)
Show that each of these arguments is invalid. Give an English example that uses each of these arguments.
Decide whether each of the following arguments is valid. If it is valid, give a formal proof. If it is invalid, show that it is invalid by finding an appropriate assignment of truth values to propositional variables.
For each of the following English arguments, express the argument in terms of propositional logic and determine whether the argument is valid or invalid.
a) If it is Sunday, it rains or snows. Today, it is Sunday and it’s not raining. Therefore, it must be snowing. b) If there are anchovies on the pizza, Jack won’t eat it. If Jack doesn’t eat pizza, he gets angry. Jack is angry. Therefore, there were anchovies on the pizza. c) At 8:00, Jane studies in the library or works at home. It’s 8:00 and Jane is not studying in the library. So she must be working at home. |
Example 12.2
In Figure 12.9 exhibits wedge in a supersonic flow with unknown Mach number. Examination of the Figure reveals that it is in angle of attack. 1) Calculate the Mach number assuming that the lower and the upper Mach angles are identical and equal to \(\sim 30^\circ\) each (no angle of attack). 2) Calculate the Mach number and angle of attack assuming that the pressure after the shock for the two oblique shocks is equal. 3) What kind are the shocks exhibits in the image? (strong, weak, unsteady) 4) (Open question) Is there possibility to estimate the air stagnation temperature from the information provided in the image. You can assume that specific heats, \(k\) is a monotonic increasing function of the temperature.
Solution 12.2
Part (1) The Mach angle and deflection angle can be obtained from the Figure 12.9. With this data and either using equation (59) or potto-GDC results in
Oblique Shock Input: \(\theta_w\) and \(\delta\) k = 1.4 \(M_1\) \(M_x\) \({{M_y}_s}\) \({{M_y}_w}\) \(\theta_{s}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 2.6810 2.3218 0 2.24 0 30 10 0.97172
The actual Mach number after the shock is then
\begin{align*} M_2 = \dfrac{{M_2}_n}{\sin\left(\theta-\delta\right)} = \dfrac{0.76617}{\sin(30-10)} = 0.839 \end{align*} The flow after the shock is subsonic flow. Part (2) For the lower part shock angle of \(\sim 28^\circ\) the results are
Oblique Shock Input: \(\theta_w\) and \(\delta\) k = 1.4 \(M_1\) \(M_x\) \({{M_y}_s}\) \({{M_y}_w}\) \(\theta_{s}\) \(\theta_{w}\) \(\delta\) \(\dfrac{{P_0}_y}{{P_0}_x}\) 2.9168 2.5754 0 2.437 0 28 10 0.96549
From the last table, it is clear that Mach number is between the two values of 2.9168 and 2.6810 and the pressure ratio is between 0.96549 and 0.97172. One of procedure to calculate the attack angle is such that pressure has to match by "guessing'' the Mach number between the extreme values.
Part (3) The shock must be weak shock because the shock angle is less than \(60^\circ\). Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
I have earlier posted the same question here on
math stackexchange but without any answer. As the question concerns tensors, I guess that I have come to the right place i.e. to physicists.
Say I have the following equation of motion in the
Cartesian coordinate system for a typical mass spring damper system:
$$M \; \ddot{x} + C \; \dot{x} + K \; x = 0$$
where the dot $^\dot{}$ represents differentiation with respect to time.
Now I would like to convert this equation to
Polar coordinates. So I introduce
$$x=r \; \cos{\theta}$$ to obtain
$$\dot{x}=\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}$$
and $$\ddot{x}=\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}$$
I can insert $x, \; \dot{x} \; \text{and} \; \ddot{x}$ in my original equation in the
Cartesian coordinate system to yield
$$M \; (\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}) + C \; (\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}) + K \; (r \; \cos{\theta}) = 0$$
Note: I am just showing the equation and derivatives in the x-direction. But the full system has both $x$ and $y$ components.
I wonder if the above way of thinking is right. I am
very new to tensors and I after reading about covariant derivatives, I am now thinking that one should include consider the basis vectors of the
Polarcoordinate system (a
non-Cartesiancoordinate system) also since unlike the basis vectors of the
Cartesian coordinate system which
do not change direction in the 2D space,
Polar coordinate basis vectors
change direction depending on the angle $\theta$.
I am thinking about
covariant derivatives as the conversion process includes differentiation with respect to the bases. For example if $x=r \; \cos{\theta}$, then
$$\dot{x}=\frac{dx}{dt}=\frac{\partial{x}}{\partial{r}} \cdot \frac{dr}{dt} + \frac{\partial{x}}{\partial{\theta}} \cdot \frac{d\theta}{dt}$$
So we have terms like $\frac{\partial{x}}{\partial{r}}$ and $\frac{\partial{x}}{\partial{\theta}}$ that concern basis vectors both in the
Cartesian and in the
Polar coordinate systems.
Hope that someone can shed some light on this. |
19 0
I am currently studying the Massive Thirring Model (MTM) with the Lagrangian
$$ \mathcal{L} = \imath {\bar{\Psi}} (\gamma^\mu {\partial}_\mu - m_0 )\Psi - \frac{1}{2}g: \left( \bar{\Psi} \gamma_\mu \Psi \right)\left( \bar{\Psi} \gamma^\mu \Psi \right): . $$ and Hamiltonian $$ \int \mathrm{d}x \imath \Psi^\dagger \sigma_z \partial_x \Psi + m_0 \Psi^\dagger \Psi + 2g \Psi^\dagger_1 \Psi^\dagger_2 \Psi_2\Psi_1\\ $$ Due to the infinite set of conservation laws, particle production is said to be absent from this theory. However why isn't it sufficient to show that particle production is absent if the number operator $$N=\int \mathrm{d}x \Psi^\dagger \Psi$$ commutes the the Hamiltonian? Also, by particle production being absent, is that just a statement that all Feynman diagrams with self energy insertions evaluate to 0 but all other Feynman diagrams are possible?
$$
\mathcal{L} = \imath {\bar{\Psi}} (\gamma^\mu {\partial}_\mu - m_0 )\Psi - \frac{1}{2}g: \left( \bar{\Psi} \gamma_\mu \Psi \right)\left( \bar{\Psi} \gamma^\mu \Psi \right): .
$$
and Hamiltonian
$$
\int \mathrm{d}x \imath \Psi^\dagger \sigma_z \partial_x \Psi + m_0 \Psi^\dagger \Psi + 2g \Psi^\dagger_1 \Psi^\dagger_2 \Psi_2\Psi_1\\
$$
Due to the infinite set of conservation laws, particle production is said to be absent from this theory. However why isn't it sufficient to show that particle production is absent if the number operator $$N=\int \mathrm{d}x \Psi^\dagger \Psi$$ commutes the the Hamiltonian? Also, by particle production being absent, is that just a statement that all Feynman diagrams with self energy insertions evaluate to 0 but all other Feynman diagrams are possible? |
How to choose a point uniformly from a convex polytope $P \subset [0,1]^n$ defined by some inequalities, $Ax < b$? (Here $A$ is an $m \times n$ matrix, $x \in \mathbb{R}^n$, and $b \in \mathbb{R}^m$.) I imagine that you could start with a uniformly chosen point in the cube and do some process to get a point with $Ax < b$.
See the answers to this question — uniform sampling from polytopes forms the basis for the known algorithms for calculating their volumes. The methods from the papers mentioned in those answers mostly take the form of a random walk inside the polytope; they differ in the details of the walk and in the analysis of its mixing time.
Hit and run sampling will perform much better than rejection sampling for higher dimensions.
It converges to a uniform distribution in polynomial time. The basic idea is to start at any point $x_0$ inside the polytope, then follow the following procedure iteratively:
pick a direction $\alpha$ uniformly at random. find the minimum and maximum value of $\theta$ such that $x_n+\theta\cdot\alpha$ is contained in the polytope pick a $\theta^*$ uniformly at random from $[\theta_{min},\theta_{max}]$ your new point is $x_{n+1}=x_n+\theta^*\cdot\alpha$
Rejection sampling will definitely work. Take a hypercube that you know contains the polytope, sample from the hypercube, and accept only those samples that belong in the polytope. However if the relative volume of the polytope is small you'll end up rejecting most samples, and the method might get painfully slow. Depending on your needs you might want to find, e.g., the smallest enclosing ball first, so that you can draw your uniform samples from that instead of the hypercube.
See Boyd & Vanderberghe's book on Convex Optimisation (it's online) for finding smallest enclosing sets.
If you use MATLAB cprnd on their File Exchange solves the problem.
Rejection sampling definitely works if you are able to find a superset $Q$ of the polytope $P$ from which you can sample. If you sample a point from that superset, the probability that it gets accepted is equal to the ratio $\frac{\text{Vol}(P)}{\text{Vol}(Q)}$, so $Q$ should be as small as possible. For instance, it is sample to sample from $Q$ if it is a box or a ball.
In the case where the polytope is specified as a list of inequalities, finding the smallest enclosing ball can be quite hard.
Contrary to what Simon Barthelmé mentions, Boyd and Vandenberghe do not deal with this problem. Actually they deal with the case where the vertices of the polytope are available. Going from the list of inequalities to the set of vertices is also hard (I am actually looking for a MATLAB implementation of that).
One possible approach is to find a small box enclosing the polytope. The box is defined by a set of coordinates $(b_i^{\text{min}},b_i^{\text{max}}), i=1\dots n$, and each coordinate can be found by : $$ b_i^{\text{min}} = \arg \min_x x_i \quad \text{subject to } A x \leq b $$ $$ b_i^{\text{max}}= \arg \max_x x_i \quad \text{subject to } A x \leq b $$ Those are linear programs for which you can use your favourite solver.
A case with a fast simple method: to sample the "right simplex" $\ \sum{x_i} \le 1,\ x_i \ge 0$:
sample in $\sum{x_i} = 1$ by taking i.i.d. exponentials scaled to sum 1 scale by random-uniform$^\frac{1}{dim}$.
(I have no idea how to generalize this.)
In Python with NumPy, this is
def random_simplex_sum1( N, dim ): """ N uniform-random points >= 0, sum x_i == 1 """ X = np.random.exponential( size=(N,dim) ) X /= X.sum(axis=1)[:,np.newaxis] return Xdef random_simplex_le1( N, dim ): """ N uniform-random points >= 0, sum x_i <= 1 """ return random_simplex_sum1( N, dim ) \ * (np.random.uniform( size=N ) ** (1/dim)) [:,np.newaxis] |
This question is experimentally accessible, despite the feebleness of the weak interaction, because the strong and electromagnetic interactions are symmetric under parity transformations and the weak interaction is not.
The contribution to the binding energy is small enough that it's not a good way to think of things. Better is to continue the process of trying to describe nuclear energy eigenstates as linear combinations of different spin-orbit states. For instance, the deuteron ground state has isospin zero and spin, parity $J^P=1^+$, and so must be a linear combination of the even-$L$ spin triplets $\left|{}^3S_1{}^{T=0}\right>$ and $\left|{}^3D_1{}^{T=0}\right>$; the d-wave component famously contributes about 4% of the wavefunction and was the first evidence for the tensor nature of the nuclear force. But because the weak interaction contributes to the nuclear interaction, the ground state isn't an
exact eigenstate of the parity operator (or, for that matter, of isospin) and there's a little bit of p-wave mixed in:$$\left|\text{deuteron}\right>=\sqrt{0.96}\left|{}^3S_1{}^{T=0}\right>+\sqrt{0.04}\left|{}^3D_1{}^{T=0}\right>+\epsilon_0\left|{}^3P_1{}^{T=0}\right>+\epsilon_1\left|{}^1P_1{}^{T=1}\right>$$
In the formation of deuterium by neutron capture on hydrogen, you get interference between parity-allowed capture to the $S$- and $D$-wave states and parity-forbidden capture to the $P$-wave states. These interferences manifest as asymmetries or spontaneous polarizations in the photons emitted during capture which are more or less linear in the amount of $P$-wave mixing; typical asymmetries are a few parts per billion.
In heavier nuclei (e.g. helium & beyond) you lose the luxury of a ground-state wavefunction which can be described in a paragraph, or even at all. However, a perturbation-theory way of describing the influence of the weak interaction is to say that a particular physical eigenstate with, say, positive parity $\left|\psi^+_\text{physical}\right>$ will be
mostly given by a strong-force eigenstate with definite parity, but contain contributions from nearby opposite-parity states due to the weak interaction:$$\left| \psi^+_\text{physical} \right>=\left| \psi^+ \right>+\sum_i\left| \psi^-_i \right>\frac{\left< \psi^-_i \middle|H_\text{weak}\middle| \psi^+ \right>}{E_i - E_+}$$
In heavy nuclei with a dense forest of excited states, you sometimes find same-spin, opposite-parity states which have very different lifetimes and very similar energies; these states are prime candidates to exhibit parity mixing due to the weak interaction.There's a famous excitation in lanthanum which decays by emitting photons with a 10% parity-forbidden directional asymmetry.
Microscopically, your other answers are correct that the nucleus is too large and the overlap between nucleons too small for appreciable exchange of $W$ and $Z$ bosons. But you can of course say the same thing about nucleons and exchange of gluons. The effective theory of the weak interaction between nucleons models the nuclear force as an exchange of strong mesons (the $\pi,\rho,\omega$) where each nucleon-nucleon-meson vertex with a given set of quantum numbers has a particular parity-nonconserving amplitude. (There was some effort a few years ago to move into the twenty-first century and come up with an "effective field theory" which described the nucleon-nucleon weak interaction without mesons; a big pile of work seems to have produced a one-to-one relationship between the coupling constants in the modern effective field theory and the coupling constants in the old meson theory.)
This has been a pretty long-winded preparation for my answer to your question: the contribution of the weak interaction to the
energy of any particular nuclear state is pretty small, for the same reason that the Coulomb-force contribution to the energies of light nuclei can generally be neglected. What's more interesting is to try an use the short-range nature of the weak interaction to peek at high-energy physics hiding inside of stable nuclei. |
A few days ago I started studying classical mechanics and today I'm stuck with the integral formula for work. I didn't take the formula for granted and i tried to understand it, however, the presence of both dt and ds didn't allow me to grasp it deeply
$W=\int m\frac{d\mathbf{v}}{dt}\cdot d\mathbf{s}$
So I tried to elaborate this formula, and I get:
$dW=mvdv$
This means that the infinitesimal work is simply the dv of a body with momentum mv, and is very intuitive. Why in books the work formula is always presented as $W=\int \mathbf {F}\cdot d\mathbf{s}$ ? Wouldn't be better to explain first of all the momentum and then the fact that an infinitesimal change of energy is $dW=mvdv$ ? |
The value of the quantity P, where P=$\int\limits_0^1xe^x\;dx$ is equal to
Divergence of the three-dimensional radial vector field r→ is
The period of the signal xt=8sin0.8πt+π4 is
The system represented by the input-output relationship $\style{font-size:14px}{y\left(t\right)=\int\limits_{-\infty}^{5t}x\left(\tau\right)d\tau,t>0}$ is
The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+, the current through the 1μF capacitor is
The second harmonic component of the periodic waveform given in the figure has an amplitude of
As shown in the figure, a 1Ω resistance is connected across a source that has a load line v+ i = 100. The current through the resistance is
A wattmeter is connected as shown in the figure. The wattmeter reads
An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω. In order to change the range to 0 - 25 A, we need to add a resistance of
As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately
For the system 2s+1 the approximate time taken for a step response to reach 98% of its final value is
If the electrical circuit of figure (b) is an equivalent of the coupled tank system of figure (a), then
A single-phase transformer has a turns ratio of 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1A, and the secondary current is 1A. If core losses and leakage reactances are neglected, the primary current is
Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltages VAB and VCD are as indicated in the figure, and I > 0, then
A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by rs, rr, xs , xr and Xm , respectively. The magnitude of the starting current of the motor is given by
Consider a step voltage wave of magnitude 1pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is
Consider two buses connected by an impedance of (0+j5) Ω. The bus 1 voltage is 100∠30°V, and bus 2 voltage is 100∠ 0o100∠0°V.The real and reactive power supplied by bus 1, respectively, are
A three-phase, 33kV oil circuit breaker is rated 1200A, 2000MVA, 3s. The symmetrical breaking current is
Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coil is
The zero-sequence circuit of the three phase transformer shown in the figure is
This is not the official website of GATE.
It is our sincere effort to help you. |
Under the auspices of the Computational Complexity Foundation (CCF)
We study the complexity of arithmetic in finite fields of characteristic two, $\F_{2^n}$.
We concentrate on the following two problems:
Iterated Multiplication: Given $\alpha_1, \alpha_2,..., \alpha_t \in \F_{2^n}$, compute $\alpha_1 \cdot \alpha_2 \cdots \alpha_t \in \F_{2^n}$.
Exponentiation: Given $\alpha \in \F_{2^n}$ and a $t$-bit integer $k$, compute $\alpha^k \in \F_{2^n}$.... more >>> |
Under the auspices of the Computational Complexity Foundation (CCF)
We study the problem of computing the $p\rightarrow q$ norm of a matrix $A \in R^{m \times n}$, defined as \[ \|A\|_{p\rightarrow q} ~:=~ \max_{x \,\in\, R^n \setminus \{0\}} \frac{\|Ax\|_q}{\|x\|_p} \] This problem generalizes the spectral norm of a matrix ($p=q=2$) and the Grothendieck problem ($p=\infty$, $q=1$), and has been widely studied in various regimes. When $p \geq q$, the problem exhibits a dichotomy: constant factor approximation algorithms are known if $2 \in [q,p]$, and the problem is hard to approximate within almost polynomial factors when $2 \notin [q,p]$.
The regime when $p$ is less than $q$, known as hypercontractive norms, is particularly significant for various applications but much less well understood. The case $p = 2$ and $q > 2$ was studied by [Barak et al, STOC'12] who gave sub-exponential algorithms for a promise version of the problem (which captures small-set expansion) and also proved hardness of approximation results based on the Exponential Time Hypothesis. However, no NP-hardness of approximation is known for these problems for any $p < q$.
We study the hardness of approximating matrix norms in both the above cases and prove the following results:
- We show that for any $1< p < q < \infty$ with $2 \notin [p,q]$, $\|A\|_{p\rightarrow q}$ is hard to approximate within $2^{O(\log^{1-\epsilon}\!n)}$ assuming $NP \not\subseteq BPTIME(2^{\log^{O(1)}\!n})$. This suggests that, similar to the case of $p \geq q$, the hypercontractive setting may be qualitatively different when $2$ does not lie between $p$ and $q$.
- For all $p \geq q$ with $2 \in [q,p]$, we show $\|A\|_{p\rightarrow q}$ is hard to approximate within any factor smaller than $1/(\gamma_{p^*} \cdot \gamma_q)$, where $\gamma_r$ denotes the $r^{th}$ norm of the standard Gaussian, and $p^*$ is the dual norm of $p$. |
Here is a “bare-hands” approach.
Let $X$ be your group generated by $x^2,y^2$ and $xy$, $A$ be the free group on three generators $a_1,a_2,a_3$. Thereis a unique group homomorphism $f: A \to X$, sending $a_1$ to $x^2$,$a_2$ to $y^2$, $a_3$ to $xy$. That homomorphism is obviouslysurjective. So all you need to show is that $f$ is injective, i.e.$f(a)\neq e$ whenever $a\neq e$.
In both $X$ and $A$, every element has a unique reduced writing (i.e.expression not containing terms of the form $tt^{-1}$ or $t^{-1}t$). This allowsus to identify each element with a unique word, called its “normal form”, andsaying that an element ends with a certain word.
Let $a\in A,a\neq e$. Then $a$ must end with something, there aresix cases, and I claim that
(1) If $a$ ends with $a_1$ (in $A$), then $f(a)$ ends with one of $x^2,yx,y^{-1}x,y^2$ (in $X$).
(2) If $a$ ends with $a_2$ (in $A$), then $f(a)$ ends with $y^2$ (in $X$).
(3) If $a$ ends with $a_3$ (in $A$), then $f(a)$ ends with one of $xy,x^{-1}y$ (in $X$).
(4) If $a$ ends with $a_1^{-1}$ (in $A$), then $f(a)$ ends with $(x^{-1})^2$ (in $X$).
(5) If $a$ ends with $a_2^{-1}$ (in $A$), then $f(a)$ ends with one of $xy^{-1},x^{-1}y^{-1},(y^{-1})^2$ (in $X$).
(6) If $a$ ends with $a_3^{-1}$ (in $A$), then $f(a)$ ends with one of $yx^{-1},y^{-1}x^{-1}$ (in $X$).
Once this property is stated, its verification by induction on the length of $a$ and by case disjunction is purely mechanical. I can supply further details if you need them.
So we have by this disjunction in six cases, that $f(a)\neq e$ : $f$ is injective, which concludes the proof. |
What is the difference between logistic and logit regression? I understand that they are similar (or even the same thing) but could someone explain the difference(s) between these two? Is one about odds?
The logit is a link function / a transformation of a parameter. It is the logarithm of the odds. If we call the parameter $\pi$, it is defined as follows:
$$ {\rm logit}(\pi) = \log\bigg(\frac{\pi}{1-\pi}\bigg) $$ The logistic function is the inverse of the logit. If we have a value, $x$, the logistic is: $$ {\rm logistic}(x) = \frac{e^x}{1+e^x} $$ Thus (using matrix notation where $\boldsymbol X$ is an $N\times p$ matrix and $\boldsymbol\beta$ is a $p\times 1$ vector), logit regression is: $$ \log\bigg(\frac{\pi}{1-\pi}\bigg) = \boldsymbol{X\beta} $$ and logistic regression is: $$ \pi = \frac{e^\boldsymbol{X\beta}}{1+e^\boldsymbol{X\beta}} $$ For more information about these topics, it may help you to read my answer here: Difference between logit and probit models.
The odds of an event is the probability of the event divided by the probability of the event not occurring. Exponentiating the logit will give the odds. Likewise, you can get the odds by taking the output of the logistic and dividing it by 1 minus the logistic. That is:
$$ {\rm odds} = \exp({\rm logit}(\pi)) = \frac{{\rm logistic}(x)}{1-{\rm logistic}(x)} $$ For more on probabilities and odds, and how logistic regression is related to them, it may help you to read my answer here: Interpretation of simple predictions to odds ratios in logistic regression. |
Weakly measurable cardinals The weakly measurable cardinals were introduced by Jason Schanker in [1], [2]. As their name suggests, they provide a weakening of the large cardinal concept of measurability. If the GCH holds at $\kappa$, then the property of the weak measurability of $\kappa$ is equivalent to that of the full measurability of $\kappa$, but when $\kappa^+\lt 2^\kappa$, these concepts can separate. Nevertheless, the existence of a weakly measurable cardinal is equiconsistent with the existence of a measurable cardinal, since if $\kappa$ is weakly measurable, then it is measurable in an inner model. Contents Formal Definition
A cardinal $\kappa$ is
weakly measurable if and only if for every family $\mathcal{A}\subset P(\kappa)$ of size at most $\kappa^+$, there is a nonprincipal $\kappa$-complete filter on $\kappa$ measuring every set in $\mathcal{A}$. (i.e., For every subset $A \in \mathcal{A}$, either $A$ or $\kappa \setminus A$ is in the filter.) Embedding characterizations of weak measurability
If $(\kappa^+)^{{<}\kappa} = \kappa^+$, then weak measurability can also be equivalently characterized in several different ways in terms of elementary embeddings.
Weak embedding characterization For every $A \subseteq \kappa^+$, there exists a transitive $M \vDash \text{ZFC}^-$ with $A, \kappa \in M$, a transitive $N$ and an elementary embedding $j: M \longrightarrow N$ with critical point $\kappa$. Embedding characterization For every transitive set $M$ of size $\kappa^+$ with $\kappa \in M$, there exists a transitive $N$ and an elementary embedding $j: M \longrightarrow N$ with critical point $\kappa$. Normal embedding characterization For every transitive $M \vDash \text{ZFC}^-$ of size $\kappa^+$ closed under ${<}\kappa$ sequences with $\kappa \in M$, there exists a transitive $N$ of size $\kappa^+$ closed under ${<}\kappa$ sequences and a cofinal elementary embedding $j: M \longrightarrow N$ with critical point $\kappa$ such that $N = \{j(f)(\kappa)| f \in M; f: \kappa \longrightarrow M\}$. Normal ZFC embedding characterization For every $A \subseteq H_{\kappa^+}$ of size $\kappa^+$, there exists a transitive $M \vDash \text{ZFC}$ of size $\kappa^+$ closed under ${<}\kappa$ sequences with $A \subseteq M$ and $\kappa \in M$, a transitive $N$ of size $\kappa^+$ closed under ${<}\kappa$ sequences, and a cofinal elementary embedding $j: M \longrightarrow N$ with critical point $\kappa$ such that $N = \{j(f)(\kappa)| f \in M; f: \kappa \longrightarrow M\}$. Weakly measurable cardinals and inner models
Weakly measurable cardinals are incompatible with the axiom $V = L$ since such cardinals are fully measurable if the GCH holds, and the constructible universe cannot contain nonprincipal countably complete ultrafilters. By the same reasoning, the Dodd-Jensen core model $K^{DJ}$ will not have any cardinals that it thinks are weakly measurable. If $\kappa$ is weakly measurable, then we can always find a countably complete normal $K^{DJ}$-ultrafilter $U$ whereby $\kappa$ will be measurable in $L[U]$ ([3], Lemma 3.36). Under certain anti-large cardinal hypotheses, a weakly measurable cardinal will be measurable in the suitable core model. For example, if $\kappa$ is weakly measurable and there is no inner model with a measurable cardinal $\lambda$ having Mitchell order $\lambda^{++}$, then $\kappa$ will be measurable in Mitchell's core model $K^m$ ([4], Theorem 35.17).
Weakly measurable cardinals and forcing
Weakly measurable cardinals $\kappa$ are invariant under forcing of size less than $\kappa$ and forcing that adds no new subsets of $\kappa^+$. Many other preservation results for these large cardinals are unknown. For example, it is an open question as to whether we can always force to an extension where a weakly measurable cardinal $\kappa$ from the ground model remains weakly measurable and becomes indestructible by the further forcing to add a Cohen subset of $\kappa$. However, if $\kappa$ is measurable in the ground model, we inherit all of the indestructibility results we can get for its weak measurability from its full measurability and more. In particular, we will be able to force to an extension where $\kappa$ is measurable, the GCH holds, and the
weak measurability of $\kappa$ is preserved by the further forcing to add any number of Cohen subsets of $\kappa$. Starting with a measurable cardinal $\kappa$, this result allows us to force to an extension where we preserve the weak measurability of $\kappa$ and yet make the GCH fail first at $\kappa$. Since the GCH cannot fail first at a measurable cardinal, this will also be a forcing extension where $\kappa$ is no longer measurable. Place in the large cardinal hierarchy
In terms of consistency strength, weakly measurable cardinals occupy the same place as measurable cardinals in the large cardinal hierarchy. In terms of size, the possibilities for these large cardinals are still being investigated. Because measurable cardinals must be weakly measurable, and weakly measurable cardinals must be weakly compact, we are provided with strict upper and lower bounds on their sizes with respect to these large cardinal notions. In the presence of the GCH, weakly measurable cardinals and measurable cardinals coincide so their sizes are the same in this case. At the opposite extreme, it was left as an open question in [1] and [2] as to whether the least weakly measurable cardinal could also be the least weakly compact cardinal. Despite being left open, there are promising developments that are being undertaken jointly by Gitik, Hamkins, and Schanker, which are aimed at this possibility.
References Schanker, Jason A. Weakly measurable cardinals.MLQ Math Log Q 57(3):266--280, 2011. www DOI bibtex Schanker, Jason A. Weakly measurable cardinals and partial near supercompactness.Ph.D. Thesis, CUNY Graduate Center, 2011. bibtex Mitchell, William J. The Covering Lemma.Handbook of Set Theory , 2001. www bibtex Jech, Thomas J. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex Set Theory. |
Roots of the algebraic equation x3+ x2+ x + 1 = 0 are
With K as a constant, the possible solution for the first order differential equation dydx=e-3x is
The r.m.s value of the current i (t) in the circuit shown below is
The fourier series expansion $\style{font-size:18px}{f\left(t\right)=a_0+\sum\limits_{n=1}^\infty a_n\cos n\omega t+b_n\sin n\omega t}$ of the periodic signal shown below will contain the following nonzero terms
A 4 point starter is used to start and control the speed of a
A three phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current
A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options
The frequency response of a linear system G(jω) is provided in the tubular form below
The gain margin and phase margin of the system are
The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r (t) having a magnitude of 10 and a duration of one second, as shown in the figure is
Consider the following statement
(i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.
A low – pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as
for the circuit shown below
The CORRECT transfer characteristic is
A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.
The proper configuration for realizing switches S1 to S6 is
A point Z has been plotted in the complex plane, as shown in figure below.
The plot of the complex number y=1Z is
The voltage applied to a circuit is 100 2 cos (100πt) volts and the circuit draws a current of 10 2 sin (100πt + π / 4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is
In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3Ω is
Given two continuous time signals x(i)=e-t and y(i)=e-2t which exist for t>0, the convolution z(t)=x(t)*y(t) is
A single phase air core transformer, fed from a rated sinusoidal supply, is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform
A negative sequence relay is commonly used to protect
For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a
This is not the official website of GATE.
It is our sincere effort to help you. |
Wythoff's game is as follows: there are two players $A$ and $B$ ( $A$ being the first player ) and there are $2$ piles of stones. When his turn a player can remove one or more stones from anyone pile or same number of stones from both the piles. A player unable to make a move loses. Also it is assumed that both the players play optimally.
As mentioned in the link the losing configurations $(n_k,m_k)$ ( $n_k$ stones in one pile and $m_k$ stones in another ) are given by $$ n_k = \lfloor( k*\phi) \rfloor$$ $$ m_k = \lfloor( k*\phi^2) \rfloor$$ where $k$ is any natural number ( and $n_k \le m_k$ ) and $\phi=\frac{1+\sqrt{5}}{2}$. For example $\{1,2\}$ ( two stones in one pile and one stone in another ) is a losing position.
Modification to Wythoff's game"
In this modified game instead of ${\it2}$ piles there are ${\it 3}$ piles. And when his turn a player can move one or more stones from anyone pile or same number of stones from any $2$ piles or same number of stones from all the $3$ piles.
How do I compute the losing positions for this modified game efficiently ? The inefficient way of course is to find the grundy number of each configuration $\{a,b,c\}$. This method is inefficient because I want to calculate number of losing positions given that number of stones in each pile can be between $1$ and $1000$. This is a project Euler question ( stone game ). I would appreciate hints only. |
There are several definitions of the second law. No matter which definition is used to describe the second law it will end in a mathematical form. The most common mathematical form is Clausius inequality which state that
\[ \oint \dfrac {\delta Q} { T} \ge 0 \label{thermo:eq:clausius} \tag{11} \] The integration symbol with the circle represent integral of cycle (therefor circle) in with system return to the same condition. If there is no lost, it is referred as a reversible process and the inequality change to equality. \[ \oint \dfrac {\delta Q} { T} = 0 \label{thermo:eq:clausiusE} \tag{12} \] The last integral can go though several states. These states are independent of the path the system goes through. Hence, the integral is independent of the path. This observation leads to the definition of entropy and designated as \(S\) and the derivative of entropy is \[ ds \equiv \left( \dfrac{ \delta Q}{T} \right)_{\hbox{rev}} \label{thermo:eq:engropy} \tag{13} \] Performing integration between two states results in \[ S_2 -S_1 = \int^2_1 \left( \dfrac{ \delta Q}{T} \right)_{\hbox{rev}} = \int^2_1 dS \label{thremo:eq:deltaS} \tag{14} \] One of the conclusions that can be drawn from this analysis is for reversible and adiabatic process \(dS=0\). Thus, the process in which it is reversible and adiabatic, the entropy remains constant and referred to as isentropic process. It can be noted that there is a possibility that a process can be irreversible and the right amount of heat transfer to have zero change entropy change. Thus, the reverse conclusion that zero change of entropy leads to reversible process, isn't correct. For reversible process equation (12) can be written as \[ \delta Q = T\, dS \label{thermo:eq:dQ} \tag{15} \] and the work that the system is doing on the surroundings is \[ \delta W = P\,dV \label{thermo:eq:dW} \tag{16} \] Substituting equations (15) (16) into (10) results in \[ T\,dS = d\,E_U + P\, dV \label{thermo:eq:Tds} \tag{17} \] Even though the derivation of the above equations were done assuming that there is no change of kinetic or potential energy, it still remain valid for all situations. Furthermore, it can be shown that it is valid for reversible and irreversible processes. Enthalpy
It is a common practice to define a new property, which is the combination of already defined properties, the enthalpy of the system.
\[ H = E_U + P\,V \label{thermo:eq:enthalpy} \tag{18} \] The specific enthalpy is enthalpy per unit mass and denoted as, \(h\). Or in a differential form as \[ dH = dE_U + dP\,V + P\,dV \label{thermo:eq:dEnthalpy} \tag{19} \] Combining equations (18) the (17) yields
(one form of) Gibbs Equation
\[
\label{thermo:eq:TdSH} T\,dS = dH -V\,dP \tag{20} \]
For isentropic process, equation (17) is reduced to \(dH=VdP\). The equation (17) in mass unit is
\[ T\,ds = du + P\,dv = dh - \dfrac{dP}{\rho} \label{thermo:eq:Tdsh} \tag{21} \] when the density enters through the relationship of \(\rho=1/v\). Specific Heats
The change of internal energy and enthalpy requires new definitions. The first change of the internal energy and it is defined as the following
Specific Volume Heat
\[
\label{thermo:eq:cv} C_v \equiv \left( \dfrac {\partial E_u }{\partial T} \right) \tag{22} \]
And since the change of the enthalpy involve some kind of boundary work is defined as
Specific Pressure Heat
\[
\label{thermo:eq:cp} C_p \equiv \left( \dfrac {\partial h }{\partial T} \right) tag\{23} \]
The ratio between the specific pressure heat and the specific volume heat is called the ratio of the specific heat and it is denoted as, \(k\).
Specific Heats Ratio
\[
\label{thermo:eq:k} k \equiv \dfrac {C_p}{C_v} \tag{24} \]
For solid, the ratio of the specific heats is almost 1 and therefore the difference between them is almost zero. Commonly the difference for solid is ignored and both are assumed to be the same and therefore referred as \(C\). This approximation less strong for liquid but not by that much and in most cases it applied to the calculations. The ratio the specific heat of gases is larger than one.
Equation of state
Equation of state is a relation between state variables. Normally the relationship of temperature, pressure, and specific volume define the equation of state for gases. The simplest equation of state referred to as ideal gas. And it is defined as
\[ P = \rho\, R\, T \label{thermo:eq:idealGas} \tag{25} \] Application of Avogadro's law, that 'all gases at the same pressures and temperatures have the same number of molecules per unit of volume,' allows the calculation of a "universal gas constant.'' This constant to match the standard units results in \[ \bar{R} = 8.3145 \dfrac{kj} {kmol\; K } \label{thermo:eq:Rbar} \tag{26} \] Thus, the specific gas can be calculate as \[ R = \dfrac{\bar{R}} {M} \label{thermo:eq:R} \tag{27} \] The specific constants for select gas at 300K is provided in table 2.1. Table 2.1 Properties of Various Ideal Gases [300K]
Gas Chemical Formula Molecular Weight \(R\, \left[\dfrac{kj}{Kg K}\right]\) \(C_p\, \left[\dfrac{kj}{Kg K}\right]\) \(C_V\, \left[\dfrac{kj}{Kg K}\right]\) \(k\) Air - 28.970 0.28700 1.0035 0.7165 1.400 Argon \(Ar\) 39.948 0.20813 0.5203 0.3122 1.400 Butane \(C_4H_{10}\) 58.124 0.14304 1.7164 1.5734 1.091 Carbon Dioxide \(CO_2\) 44.01 0.18892 0.8418 0.6529 1.289 Carbon Monoxide \(CO\) 28.01 0.29683 1.0413 0.7445 1.400 Ethane \(C_2H_6\) 30.07 0.27650 1.7662 1.4897 1.186 Ethylene \(C_2H_4\) 28.054 0.29637 1.5482 1.2518 1.237 Helium \(He\) 4.003 2.07703 5.1926 3.1156 1.667 Hydrogen \(H_2\) 2.016 4.12418 14.2091 10.0849 1.409 Methane \(CH_4\) 16.04 0.51835 2.2537 1.7354 1.299 Neon \(Ne\) 20.183 0.41195 1.0299 0.6179 1.667 Nitrogen \(N_2\) 28.013 0.29680 1.0416 0.7448 1.400 Octane \(C_8H_{18}\) 114.230 0.07279 1.7113 1.6385 1.044 Oxygen \(O_2\) 31.999 0.25983 0.9216 0.6618 1.393 Propane \(C_3H_8\) 44.097 0.18855 1.6794 1.4909 1.327 Steam \(H_2O\) 18.015 0.48152 1.8723 1.4108 1.327
From equation (25) of state for perfect gas it follows
\[ d\left(P\,v\right) = R\,dT \label{thermo:eq:stateD} \tag{28} \] For perfect gas \[ dh = dE_u + d(Pv) = dE_u + d(R\,T) = f(T)\hbox{ (only)} \label{thermo:eq:dhIdeal} \tag{29} \] From the definition of enthalpy it follows that \[ d(Pv) = dh - dE_u \label{thermo:eq:defHd} \tag{30} \] Utilizing equation (28) and subsisting into equation (29) and dividing by \(dT\) yields \[ C_p - C_v = R \label{thermo:eq:CpCvR} \tag{31} \] This relationship is valid only for ideal/perfect gases. The ratio of the specific heats can be expressed in several forms as
\(C_v\) to Specific Heats Ratio
\[
\label{thermo:eq:Cv} C_v = \dfrac{R}{k-1} \tag{32} \]
\(C_p\) to Specific Heats Ratio
\[
\label{thermo:eq:Cp} C_p = \dfrac{k\,R}{k-1} \tag{33} \]
The specific heat ratio, \(k\) value ranges from unity to about 1.667. These values depend on the molecular degrees of freedom (more explanation can be obtained in Van Wylen "F. of Classical thermodynamics.'' The values of several gases can be approximated as ideal gas and are provided in Table 2.1. The entropy for ideal gas can be simplified as the following
\[ s_2 - s_1 = \int_1^2 \left(\dfrac{dh}{T}- \dfrac{dP}{\rho\, T}\right) \label{thermo:eq:deltaSidealI} \tag{34} \] Using the identities developed so far one can find that \[ s_2 - s_1 = \int_1^2 C_p \dfrac{dT}{T} - \int_1^2 \dfrac{R\,dP}{P} = C_p \, \ln \dfrac{T_2}{T_1} - R\, \ln \dfrac{P_2}{P_1} \label{thermo:eq:deltaSideal} \tag{35} \] Or using specific heat ratio equation (35) transformed into \[ \dfrac{s_2 - s_1} {R} = \dfrac{k}{ k -1} \,\ln \dfrac{T_2}{T_1} - \ln \dfrac{P_2}{P_1} \label{thermo:eq:deltaSidealK} \tag{36} \] For isentropic process, \(\Delta s=0\), the following is obtained \[ \ln \dfrac{T_2}{T_1} = \ln \left(\dfrac{P_2}{P_1} \right) ^ {\dfrac{k -1 }{k}} \label{thermo:eq:sZero} \tag{37} \] There are several famous identities that results from equation (37) as
Ideal Gas Isentropic Relationships
\[
\label{thermo:eq:famousIdeal} \dfrac{T_2}{T_1} = \left(\dfrac{P_2}{P_1} \right) ^ {\dfrac{k -1 }{k}} = \left(\dfrac{V_1}{V_2} \right) ^ {k -1 } \tag{38} \]
The ideal gas model is a simplified version of the real behavior of real gas. The real gas has a correction factor to account for the deviations from the ideal gas model. This correction factor referred as the compressibility factor and defined as
Z deviation from the Ideal Gas Model
\[
\label{thermo:eq:Z} Z = \dfrac{P\,V}{R\,T} \tag{39} \] Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
Under the auspices of the Computational Complexity Foundation (CCF)
In 1990 Karchmer and Widgerson considered the following communication problem $Bit$: Alice and Bob know a function $f: \{0, 1\}^n \to \{0, 1\}$, Alice receives a point $x \in f^{-1}(1)$, Bob receives $y \in f^{-1}(0)$, and their goal is to find a position $i$ such that $x_i \neq y_i$. Karchmer and Wigderson proved that the minimal size of a boolean formula for the function $f$ equals the size of the smallest communication protocol for the $Bit$ relation. In this paper we consider a model of dag-like communication complexity (instead of classical one where protocols correspond to trees). We prove an analogue of Karchmer-Wigderson Theorem for this model and boolean circuits. We also consider a relation between this model and communication PLS games proposed by Razborov in 1995 and simplify the proof of Razborov's analogue of Karchmer-Wigderson Theorem for PLS games.
We also consider a dag-like analogue of real-valued communication protocols and adapt a lower bound technique for monotone real circuits to prove a lower bound for these protocols.
In 1997 Krajicek suggested an interpolation technique that allows to prove lower bounds on the lengths of resolution proofs and Cutting Plane proofs with small coefficients ($CP^*$). Also in 2016 Krajicek adapted this technique to ``random resolution''. The base of this technique is an application of Razborov's theorem. We use real-valued dag-like communication protocols to generalize the ideas of this technique, which helps us to prove a lower bound on the Cutting Plane proof system ($CP$) and adopt it to ``random $CP$''.
Our notion of dag-like communication games allows us to use an analogue Raz-McKenzie transformation from Pitassi and Goos paper, which yields a lower bound on the real monotone circuit size for the $CSPSAT$ problem.
Minor corrections.
In 1990 Karchmer and Widgerson considered the following communication problem $Bit$: Alice and Bob know a function $f: \{0, 1\}^n \to \{0, 1\}$, Alice receives a point $x \in f^{-1}(1)$, Bob receives $y \in f^{-1}(0)$, and their goal is to find a position $i$ such that $x_i \neq y_i$. Karchmer and Wigderson proved that the minimal size of a boolean formula for the function $f$ equals the size of the smallest communication protocol for the $Bit$ relation. In this paper we consider a model of dag-like communication complexity (instead of classical one where protocols correspond to trees). We prove an analogue of Karchmer-Wigderson Theorem for this model and boolean circuits. We also consider a relation between this model and communication PLS games proposed by Razborov in 1995 and simplify the proof of Razborov's analogue of Karchmer-Wigderson Theorem for PLS games.
We also consider a dag-like analogue of real-valued communication protocols and adapt a lower bound technique for monotone real circuits to prove a lower bound for these protocols.
In 1997 Krajicek suggested an interpolation technique that allows to prove lower bounds on the lengths of resolution proofs and Cutting Plane proofs with small coefficients ($CP^*$). Also in 2016 Krajicek adapted this technique to ``random resolution''. The base of this technique is an application of Razborov's theorem. We use real-valued dag-like communication protocols to generalize the ideas of this technique, which helps us to prove a lower bound on the Cutting Plane proof system ($CP$) and adopt it to ``random $CP$''.
Our notion of dag-like communication games allows us to use an analogue Raz-McKenzie transformation from Pitassi and Goos paper, which yields a lower bound on the real monotone circuit size for the $CSPSAT$ problem. |
Bernoulli Bernoulli Volume 23, Number 1 (2017), 58-88. On the survival probability for a class of subcritical branching processes in random environment Abstract
Let $Z_{n}$ be the number of individuals in a subcritical Branching Process in Random Environment (BPRE) evolving in the environment generated by i.i.d. probability distributions. Let $X$ be the logarithm of the expected offspring size per individual given the environment. Assuming that the density of $X$ has the form
\[p_{X}(x)=x^{-\beta-1}l_{0}(x)e^{-\rho x}\] for some $\beta>2$, a slowly varying function $l_{0}(x)$ and $\rho\in(0,1)$, we find the asymptotic of the survival probability $\mathbb{P}(Z_{n}>0)$ as $n\rightarrow\infty$, prove a Yaglom type conditional limit theorem for the process and describe the conditioned environment. The survival probability decreases exponentially with an additional polynomial term related to the tail of $X$. The proof uses in particular a fine study of a random walk (with negative drift and heavy tails) conditioned to stay positive until time $n$ and to have a small positive value at time $n$, with $n\rightarrow\infty$.
Article information Source Bernoulli, Volume 23, Number 1 (2017), 58-88. Dates Received: December 2013 Revised: March 2015 First available in Project Euclid: 27 September 2016 Permanent link to this document https://projecteuclid.org/euclid.bj/1475001348 Digital Object Identifier doi:10.3150/15-BEJ723 Mathematical Reviews number (MathSciNet) MR3556766 Zentralblatt MATH identifier 1362.60074 Citation
Bansaye, Vincent; Vatutin, Vladimir. On the survival probability for a class of subcritical branching processes in random environment. Bernoulli 23 (2017), no. 1, 58--88. doi:10.3150/15-BEJ723. https://projecteuclid.org/euclid.bj/1475001348 |
Symbols:Greek/Lambda/Left Regular Representation $\lambda_a$
Let $\struct {S, \circ}$ be an algebraic structure.
The mapping $\lambda_a: S \to S$ is defined as: $\forall x \in S: \map {\lambda_a} x = a \circ x$ This is known as the left regular representation of $\struct {S, \circ}$ with respect to $a$. The $\LaTeX$ code for \(\map {\lambda_a} x\) is
\map {\lambda_a} x . |
세미나
Information Center for Mathematical Science
세미나
아주대학교 수학과 세미나
Title Bruhat Interval polytopes Date 2019-02-11 Speaker 박선정(아주대학교) Sponsors 2019-02-11 Host 아주대학교 Place 팔달관 621호 Abstract Given $v,w\in \mathfrak{S}_n$ with $v\leq w$, the Richardson variety~$X^v_w$ is the intersection of the Schubert variety$X_w$ and the opposite Schubert variety $X^v$. A Bruhat interval polytope~$Q_{v,w}$ is the convex hull of all permutation vectors $x = (x(1), x(2), . . . , x(n))$ with $v\leq x\leq w$. It is known that $Q_{v^{-1},w^{-1}}$ is the moment map image of $X^v_w\subset\mathrm{Fl}(\mathbb{C}^n)$. In this talk, we discuss the properties of Bruhat interval polytopes and some open problems. |
TL;DR
I recommend using LIPO. It is provably correct and provably better than pure random search (PRS). It is also extremely simple to implement, and has no hyperparameters. I have not conducted an analysis that compares LIPO to BO, but my expectation is that the simplicity and efficiency of LIPO imply that it will out-perform BO.
(See also: What are some of the disavantage of bayesian hyper parameter optimization?)
Bayesian Optimization
Bayesian Optimization-type methods build Gaussian process surrogate models to explore the parameter space. The main idea is that parameter tuples that are closer together will have similar function values, so the assumption of a co-variance structure among points allows the algorithm to make educated guesses about what best parameter tuple is most worthwhile to try next. This strategy helps to reduce the number of function evaluations; in fact, the motivation of BO methods is to keep the number of function evaluations as low as possible while "using the whole buffalo" to make good guesses about what point to test next. There are different figures of merit (expected improvement, expected quantile improvement, probability of improvement...) which are used to compare points to visit next.
Contrast this to something like a grid search, which will never use any information from its previous function evaluations to inform where to go next.
Incidentally, this is also a powerful
global optimization technique, and as such makes no assumptions about the convexity of the surface. Additionally, if the function is stochastic (say, evaluations have some inherent random noise), this can be directly accounted for in the GP model.
On the other hand, you'll have to fit at least one GP at every iteration (or several, picking the "best", or averaging over alternatives, or fully Bayesian methods). Then, the model is used to make (probably thousands) of predictions, usually in the form of multistart local optimization, with the observation that it's much cheaper to evaluate the GP prediction function than the function under optimization. But even with this computational overhead, it tends to be the case that even nonconvex functions can be optimized with a relatively small number of function calls.
A widely-cited paper on the topic is Jones et al, "Efficient Global Optimization of Expensive Black-Box Functions." But there are many variations on this idea.
Random Search
Even when the cost function is expensive to evaluate, random search can still be useful. Random search is dirt-simple to implement. The only choice for a researcher to make is setting the the
probability $p$ that you want your results to lie in some quantile $q$; the rest proceeds automatically using results from basic probability.
Suppose your quantile is $q = 0.95$ and you want a $p=0.95$ probability that the model results are in top $100\times (1-q)=5$ percent of all hyperparameter tuples. The probability that all $n$ attempted tuples are
not in that window is $q^n = 0.95^n$ (because they are chosen independently at random from the same distribution), so the probability that at least one tuple is in that region is $1 - 0.95^n$. Putting it all together, we have
$$1 - q^n \ge p \implies n \ge \frac{\log(1 - p)}{\log(q)}$$
which in our specific case yields $n \ge 59$.
This result is why most people recommend $n=60$ attempted tuples for random search. It's worth noting that $n=60$ is comparable to the number of experiments required to get good results with Gaussian Process-based methods when there are a moderate number of parameters. Unlike Gaussian Processes, the number of queries tuples does not change with the number of hyperparameters to search over; indeed, for a large number of hyperparameters, a Gaussian process-based method can take many iterations to make headway.
Since you have a probabilistic guarantee of how good the results are, it can be a persuasive tool to convince your boss that it's not necessary to run more experiments.
LIPO and its Variants
This is an exciting arrival which, if it is not
new, is certainly new to me. It proceeds by alternating between placing informed bounds on the function, and sampling from the best bound, and using quadratic approximations. I'm still working through all the details, but I think this is very promising. This is a nice blog write-up, and the paper is Cédric Malherbe and Nicolas Vayatis "Global optimization of Lipschitz functions." |
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$\approx$$\frac{\partial S}{\partial t}\geq 0$Powers of ten -- In video
Feynman's dream
I play trombone, piano, guitar, theremin, some ocarinas, ...
(eating in fact) |
Homework 6 Solutions
p. 226: 1.
$ \mathcal{L}[t^2-2t]= \mathcal{L}[t^2]-2\mathcal{L}[t] $
$ = \frac{2}{s^3}-2\frac{1}{s^2} $
Odd solutions in the back of the book.
p. 226: #2:
$ \mathcal(t^2 - 3)^2 $
$ = (t^2 - 3)(t^2 - 3) = t^4 - 6t^2 + 9 $
So
$ \mathcal{L}[(t^2-3)^2] = \mathcal{L}[t^4]-6\mathcal{L}[t^2]+9\mathcal{L}[1]= $
$ = \frac{4!}{s^5} - 6\frac{2!}{s^3} + \frac{9}{s} $ p. 226: #4:
$ \ sin^2 4 t = \frac {1 - cos2(4t)}{2} $
So the Laplace Transform can be gotten from the table.
p. 226: #23.
$ \mathcal{L}[f(t)]=\int_0^\infty e^{-st}f(t)\ dt=F(s). $
So
$ \mathcal{L}[f(ct)]=\int_0^\infty e^{-st}f(ct)\ dt. $
Make the change of variables
$ \tau=ct $
to get
$ \mathcal{L}[f(ct)]=\int_{\tau=0}^\infty e^{-(s/c)\tau}f(\tau)\ (1/c) d\tau= $
$ \frac{1}{c}F(s/c). $
Even solutions (added by Adam M on Oct 5, please check results):
p. 226: 10.
$ \mathcal{L}[-8sin(0.2t)]=\frac{-1.6}{s^2+0.04} $
p. 226: 12.
$ \mathcal{L}[(t+1)^3]=\frac{6}{s^4}+\frac{6}{s^3}+\frac{3}{s^2}+\frac{1}{s} $
p. 226: 30.
$ \mathcal{L}^{-1}[\frac{2s+16}{s^2-16}]=2cosh(4t)+4sinh(4t) $
(AJ) I have the same solutions for p 226 #10 and #12, but on #30, I factored the denominator and used partial fraction decomposition to get
$ \mathcal{L}^{-1}=-e^{-4t}+3e^{4t} $
AJ, I was able to get your answer and verified that it is correct. However, I am unable to see why my initial answer was wrong. I seperated as follows:
$ \mathcal{L}^{-1}[2*\frac{s}{s^2-4^2}+4*\frac{4}{s^2-4^2}] $
then used (8) and (9) from Table 6.1. Thoughts?
They are actually the exact same thing, so both answers should be correct. This can be proven using:
cosh(bx) = (1/2)*(e^(bx) + e^(-bx))
sinh(bx) = (1/2)*(e^(bx) - e^(-bx))
--Idougla 23:40, 7 October 2010 (UTC)
Thank you!
P. 226: 39. Can somebody post a solution for 39? I must be missing something on this one.
Re-write as:
$ \mathcal{L}^{-1}[\frac{1}{s^2+5}-\frac{1}{s+5}]=\mathcal{L}^{-1}[\frac{1}{\sqrt{5}}*\frac{\sqrt{5}}{s^2+\sqrt{5}^2}-\frac{1}{s-(-5)}] $ |
Didn’t you buy _________ when you went shopping?
Which of the following options is the closest in meaning to the sentence below?
She enjoyed herself immensely at the party
Which one of the following combinations is incorrect?
Based on the given statements, select the most appropriate option to solve the given question.
If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building?
Statements:
(I) Each step is ¾ foot high.
(II) Each step is 1 foot wide.
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16?
Select the alternative meaning of the underlined part of the sentence.
The chain snatchers took to their heels when the police party arrived.
The given statement is followed by some courses of action. Assuming the statement to be true, decide the correct option.
Statement:
There has been a significant drop in the water level in the lakes supplying water to the city.
Course of action:
(I) The water supply authority should impose a partial cut in supply to tackle the situation.
(II) The government should appeal to all the residents through mass media for minimal use of water.
(III) The government should ban the water supply in lower areas.
The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?
The probabilities that a student passes in Mathmatics, Physics and Chemistry are m,p, and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c:
(I) p + m + c = 27/20
(II) p + m + c = 13/20
(III) (p)×(m)×(c) = 1/10
The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.
What is the average of the marks obtained by the class in the examination?
A random variable $X$ has probability density function $f\left(x\right)$ as given below:
f(x) = a+bxfor 0<x<10otherwise
If the expected value EX =2/3, then Pr X<0.5 is __________.
If a continuous function $f\left(x\right)$ does not have a root in the interval [a, b], then which one of the following statements is TRUE?
If the sum of the diagonal elements of a 2 × 2 matrix is -6, then the maximum possible value of determinant of the matrix is ________.
Consider a function f→ = 1r2 r^, where r is the distance from the origin and r^ is the unit vector in the radial direction. The divergence of the function over a sphere of radius R, which includes the origin, is
When the Wheatstone bridge shown is used to find value of resistance RX , the galvanometer G indicates zero current when R1 = 50Ω, R2 =65Ω and R3=100Ω. If R3 is known with ±5% tolerance on its nominal value of $100\mathrm\Omega$, what is the range of RX in Ohms?
A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0-500 A) is ________.
Of the four characteristic given below, which are the major requirements for an instrumentation amplifier?
P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance
In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is ________.
A moving average function is given by $ \style{font-family:'Times New Roman'}{y\left(t\right)=\;\frac1t\int_{t-T}^tu\left(\tau\right)} $. If the input u is a sinusoidal signal of frequency 12THz, then in steady state, the output $y$ will lag $u$ (in degree) by ________.
The impulse response g(t) of a system, G, is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure(b)?
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Analyzing the Viscous and Thermal Damping of a MEMS Micromirror
Micromirrors have two key benefits: low power consumption and low manufacturing costs. For this reason, many industries use micromirrors for a wide range of MEMS applications. To save time and money when designing micromirrors, engineers can accurately account for thermal and viscous damping and analyze device performance via the COMSOL Multiphysics® software.
The Many Applications of Micromirrors
Picture a micromirror as a single string on a guitar. The string is so light and thin that when you pluck it, the surrounding air dampens the string’s motion, bringing it to a standstill.
Micromirrors have a wide variety of potential applications. For instance, these mirrors can be used to control optic elements, an ability that makes them useful in the microscopy and fiber optics fields. Micromirrors are found in scanners, heads-up displays, medical imaging, and more. Additionally, MEMS systems sometimes use integrated scanning micromirror systems for consumer and telecommunications applications.
When developing a micromirror actuator system, engineers need to account for its dynamic vibrating behavior and damping, both of which greatly affect the operation of the device. Simulation provides a way to analyze these factors and accurately predict system performance in a timely and cost-efficient manner.
To perform an advanced MEMS analysis, you can combine features in the Structural Mechanics Module and Acoustics Module, two add-on products to the COMSOL Multiphysics simulation platform. Let’s take a look at frequency-domain (time-harmonic) and transient analyses of a vibrating micromirror.
Performing a Frequency-Domain Analysis of a Vibrating Micromirror
We model an idealized system that consists of a vibrating silicon micromirror — which is 0.5 by 0.5 mm with a thickness of 1 μm — surrounded by air. A key parameter in this model is the penetration depth; i.e., the thickness of the viscous and thermal boundary layers. In these layers, energy dissipates via viscous drag and thermal conduction. The thickness of the viscous and thermal layers is characterized by the following penetration depth scales:
where f is the frequency, \rho is the fluid density, \mu is the dynamic viscosity, \kappa is the coefficient of thermal conduction, C_\textrm{p} is the heat capacity at constant pressure, and \textrm{Pr} is the nondimensional Prandtl number.
For air, when the system is excited at a frequency of 10 kHz (which is typical for this model), the viscous and thermal scales are 22 µm and 18 µm, respectively. These are comparable to the geometric scales, like the mirror thickness, meaning that thermal and viscous losses must be included. Moreover, in real systems, the mirrors may be located near surfaces or in close proximity to each other, creating narrow regions where the damping effects are accentuated.
The frequency-domain analysis provides insight into the frequency response of the system, including the location of the resonance frequencies, Q-factor of the resonance, and damping of the system.
The micromirror model geometry, showing the symmetry plane, fixed constraint, and torquing force components.
In this example, we use three separate interfaces:
The Shellinterface to model the solid micromirror, available in the Structural Mechanics Module The Thermoviscous Acoustics, Frequency Domaininterface to model the air domain around the mirror, available in the Acoustics Module The Pressure Acoustics, Frequency Domaininterface to truncate the computational domain, available in the Acoustics Module
By modeling the detailed thermoviscous acoustics and using the
Thermoviscous Acoustics, Frequency Domain interface, we can explicitly include thermal and viscous damping while solving the full linearized Navier-Stokes, continuity, and energy equations. In doing so, we accomplish one of the main goals for this model: accurately calculating the damping experienced by the mirror.
To set up and combine the three interfaces, we use the
Acoustics-Thermoviscous Acoustics Boundary and Thermoviscous-Acoustics-Structure Boundary multiphysics couplings. We then solve the model using a frequency-domain sweep and an eigenfrequency study. These analyses enable us to study the resonance frequency of the mirror under a torquing load in the frequency domain. Results of the Frequency-Domain Analysis
Let’s take a look at the displacement of the micromirror for a frequency of 10 kHz and when exposed to the torquing force. In this scenario, the displacement mainly occurs at the edges of the device. To view displacement in a different way, we also plot the response at the tip of the micromirror over a range of frequencies.
Micromirror displacement at 10 kHz for phase 0 (left) and the absolute value of the z -component of the displacement field at the micromirror tip (right).
Next, let’s view the acoustic temperature variations (left image below) and acoustic pressure distribution (right image below) in the micromirror for a frequency of 11 kHz. As we can see, the maximum and minimum temperature fluctuations occur opposite to one another and there is an antisymmetric pressure distribution. The temperature fluctuations are closely related to the pressure fluctuations through the equation of state. Note that the temperature fluctuations fall to zero at the surface of the mirror, where an isothermal condition is applied. The temperature gradient near the surface gives rise to the thermal losses.
Temperature fluctuation field within the thermoviscous acoustics domain (left) and the pressure isosurfaces (right).
The two animations below show a dynamic extension of the frequency-domain data using the time-harmonic nature of the solution. Both animations depict the mirror movement in a highly exaggerated manner, with the first one showing an instantaneous velocity magnitude in a cross section and the second showing the acoustic temperature fluctuations. These results indicate that there are high-velocity regions close to the edge of the micromirror. We determine the extent of this region into the air via the scale of the viscous boundary layer (viscous penetration depth). We can also identify the thermal boundary layer or penetration depth using the same method.
Animation of the time-harmonic variation in the local velocity. Animation of the time-harmonic variation in the acoustic temperature fluctuations.
When the problem is formulated in the frequency domain, eigenmodes or eigenfrequencies can also be identified. From the eigenfrequency study (also performed in the model), we can determine the vibrating modes, shown in the animation below (only half the mirror is shown as symmetry applies). Our results show that the fundamental mode is around 10.5 kHz, with higher modes at 13.1 kHz and 39.5 kHz. The complex value of the eigenfrequency is related to the Q-factor of the resonance and thus the damping. (This relationship is discussed in detail in the Vibrating Micromirror model documentation.)
Animation of the first three vibrating modes of the micromirror. Transient Analysis of Viscous and Thermal Damping in a Micromirror
As of version 5.3a of the COMSOL® software, a different take on this example solves for the transient behavior of the micromirror. Using the same geometry, we extend the frequency-domain analysis into a transient analysis. To achieve this, we swap the frequency-domain interfaces with their corresponding transient interfaces and adjust the settings of the transient solver. In the simulation, the micromirror is actuated for a short time and exhibits damped vibrations.
The resulting model includes some of the most advanced air and gas damping mechanisms that COMSOL Multiphysics has to offer. For instance, the
Thermoviscous Acoustics, Transient interface generates the full details for the viscous and thermal damping of the micromirror from the surrounding air.
In addition, by coupling the transient perfectly matched layer capabilities of pressure acoustics to the thermoviscous acoustics domain, we can create efficient nonreflecting boundary conditions (NRBCs) for this model in the time domain.
Results of the Transient Analysis
Let’s start with the displacement results. The 3D results (left image below) visualize the displacement of the micromirror and the pressure distribution at a given time. We also generate a plot (right image below) to illustrate the damped vibrations caused by thermal and viscous losses. The green curve represents the undamped response of the micromirror when the surrounding air is not coupled to the mirror movement. The time-domain simulations make it possible to study transients of the system, like the decay time, and the response of the system to an anharmonic forcing.
Micromirror displacement and pressure distribution (left) and the transient evolution of the mirror displacement (right).
We can also examine the acoustic temperature variations surrounding the micromirror. The isothermal condition at the micromirror surface produces an acoustic thermal boundary layer. As with the frequency-domain example, the highest and lowest temperatures are located opposite to one another.
In addition, by calculating the acoustic velocity variations of the micromirror, we see that a no-slip condition at the micromirror surface results in a viscous boundary layer.
Acoustic temperature variations (left) as well as acoustic velocity variations for the x -component (center) and z -component (right). Next Steps
These examples demonstrate that we can analyze micromirrors using advanced modeling features available in the Acoustics Module in combination with the Structural Mechanics Module. For more details on modeling micromirrors, check out the tutorials below.
Comments (1) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
The man who is now Municipal Commissioner worked as ____________________.
Find the odd one in the following group of words.
mock, deride, praise, jeer
Pick the odd one from the following options.
In a quadratic function, the value of the product of the roots $ (\alpha,\beta) $ is 4. Find the value of
αn+βnα-n+β-n
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A function $ y(t) $, such that $ y(0)=1 $ and $ y(1)=3e^{-1} $,is a solution of the differential equation d2ydt2+2dydt+y=0. Then $ y(2) $ is
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$\oint_c\frac{2z+5}{\left(z-{\displaystyle\frac12}\right)\left(z^2-4z+5\right)}dz$
over the contour z=1, taken in the anti-clockwise direction, would be
The transfer function of a system is YSRS=SS+2 The steady state output $ y(t) $ is A cos2t+φ for the input cos2t The values of A and φ respectvely are
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Which one of the following can be another eigenvector of A?
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$ y\lbrack n\rbrack=(1+(-1)^n)x\lbrack n\rbrack $
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Consider $g(t)=\left\{\begin{array}{lc}t-\left\lfloor t\right\rfloor,&t\geq0\\t-\left\lceil t\right\rceil,&otherwise\end{array}\right.$, where $ t\in\mathbb{R}. $
Here, $\left\lfloor t\right\rfloor$ represent the largest integer less than or equal to t and $\left\lfloor t\right\rfloor$ denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic of the Fourier series represent g(t) is__________
Let $I=\;c\iint_R\;xy^2\;dxdy$, where R is the region shown in the figure and c = 6 × 10-4. The value of I equals _________. (Give the answer up to two decimal places.)
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The equivalent resistance between the terminal A and B is__________$ \Omega$.
A three-phase, 50 Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draw current at unity power factor (UPF). The synchronous reactance of the motor is 30 $\Omega$ per phase. The load angle is 30°. The power delivered to the motor in kW is________. (Give the answer up to two decimal phase).
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The definition of a Hamiltonian system I am working with is a triple $(X,\omega, H)$ where $(X,\omega)$ is a symplectic manifold and $H\in C^\infty(X)$ is the Hamiltonian function.I am wondering if ...
Let's take the pendulum equation $\ddot{x} = -\sin x$. Here $x \in \mathbb{T}^{1}$. Now rewrite it as a coupled first order system $$\dot{y} = -\sin x, \quad \dot{x}=y.$$Intuitively we know that $y$ ...
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I have already used
\sum symbol for another reason, I needed an additional symbol and I have used
\Xi, but the problem is I cannot put the super/sub scripts directly under or above like
\sum or
\prod.
\displaystyle does not work here. Does anyone have any solution? Alternatively one may try to answer different but related question here
\documentclass{article}\usepackage{amsmath}\begin{document}\begin{equation}{W}^k=\Big(\Xi_{s=1}^{S}\prod_{i=1}^{I}{}\Big)\end{equation}\end{document} |
2019-10-14 17:21
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Aabaz Member From: Paris, France Registered: 2008-05-11 Posts: 36 Latex capability
Hi,
I would like to know if someone know a way of giving users the ability to type scientific equations with a latex-like editor inside fluxbb ?
The only requirement is that it got to work only with php because of the hosting restrictions.
I know of PHPMathPublisher wich does it but i wanted the input of your community, just in case i missed a good alternative.
I also post this in feature requests because i would find it a nice feature for fluxbb, or maybe an official extension as i heard the fluxxbb developement could go that way.
Thanks for your help.
Aabaz
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Jacobson New member Registered: 2012-09-28 Posts: 3 Re: Latex capability
Hi,
I would like to know if someone know a way of giving users the ability to type scientific equations with a latex-like editor inside fluxbb ?
The only requirement is that it got to work only with php because of the hosting restrictions.
I know of PHPMathPublisher wich does it but i wanted the input of your community, just in case i missed a good alternative.
I also post this in feature requests because i would find it a nice feature for fluxbb, or maybe an official extension as i heard the fluxxbb developement could go that way.
Thanks for your help.
Aabaz
I am interested in this aswell. Any update(s) on this?
Also, does anyone know what this board is using for the math part? - mathisfunforum.com/viewtopic.php?id=14873
Edit
Found this after doing some searching and I believe it works ok, if anyone is interested..
mathjax.org/
Last edited by Jacobson (2012-09-28 18:56:11)
Offline
mathsbeauty Member Registered: 2013-11-04 Posts: 5 Re: Latex capability
How do you included mathajax in a forum created by fluxbb? Waiting for your reply
Offline
Askelon Member From: Bretagne − France Registered: 2010-06-09 Posts: 202 Website Re: Latex capability
Edit the header.php at the root of your forum, find line 147-148:
// JavaScript tricks for IE6 and olderecho '<!--[if lte IE 6]><script type="text/javascript" src="style/imports/minmax.js"></script><![endif]-->'."\n";
And replace it by:
// JavaScript tricks for IE6 and olderecho '<!--[if lte IE 6]><script type="text/javascript" src="style/imports/minmax.js"></script><![endif]-->'."\n";// MathJaxecho '<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"></script>'."\n";
Save and upload. Now you can test it, simply post a new post:
\[\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.\]
It should render a nice equation.
Offline
mathsbeauty Member Registered: 2013-11-04 Posts: 5 Re: Latex capability
Thanks for your reply. It is not working. I did exactly as you mentioned. Where may be the problem?
Offline
mathsbeauty Member Registered: 2013-11-04 Posts: 5 Re: Latex capability
Sorry, It worked. It takes time to display after posting. Thanks for all your help.
Last edited by mathsbeauty (2013-11-05 04:37:33)
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Aabaz Member From: Paris, France Registered: 2008-05-11 Posts: 36 Re: Latex capability
Coming back more than 4 years later oO
Thanks a lot for suggesting MathJax, Jacobson, looks like what I was looking for exactly.
Hope it will not impact performance too much.
Offline
adaur Developer From: France Registered: 2010-01-07 Posts: 843 Website Re: Latex capability
It will affect user's performances, not your server's .
FeatherBB - A simple and lightweight new generation forum system
Based on FluxBB, written in PHP, using Slim Framework for a proper OOP-MVC architecture.
Offline |
Can someone help me?
How can I prove that exists a number $k \in \mathbb R$ that $$A = \{x + k;\ x \in \text{Cantor set} \} \subset\text{ Irrationals}\;?$$
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
Let $C$ be the middle-thirds Cantor set. If $x\in\Bbb R$ and $q\in\Bbb Q\cap(x+C)$, then there is a $y\in C$ such that $q=x+y$, and hence $x=q-y\in q-C$. If $\Bbb Q\cap(x+C)\ne\varnothing$ for every $x\in\Bbb R$, then
$$\Bbb R=\bigcup_{q\in\Bbb Q}(q-C)\;.$$
Verify that each of the sets $q-C$ for $q\in\Bbb Q$ is a closed, nowhere dense subset of $\Bbb R$, and then apply the Baire category theorem to get a contradiction.
If you read Spanish, you may want to look at the book
Wilman Brito,
El Teorema de Categoría de Baire y aplicaciones, Consejo de publicaciones, Universidad de los Andes, Mérida, Venezuela, 2011.
I found this book through this MO question. What follows comes from this reference.
Section 1.10 is about the Cantor set and the result you are asking about, that was first noted by Ludwig Scheeffer, in 1884, with the argument in Brian's answer. The reference is
Ludwig Scheeffer.
Zur Theorie der stetigen Funktionen einer reellen Veränderlichen, Acta Math. 5 (1), (1884), 49–82. MR1554648.
(The article is behind a paywall.) The proof can be seen in English in
The theory of sets of points, by William Henry Young and Grace Chisholm Young, recently (2006) republished as part of the Cambridge Library Collection. (See Theorem 18 in Chapter IV, and $\S67$ in Chapter V.)
More precisely, Scheeffer shows that if $E$ is countable and $N$ is nowhere dense, then there is a dense set $D$ of points $d$ such that $(E+d)\cap N$ is empty. Note that Brian's answer actually shows a bit more (for $E=\mathbb Q$ and $N=C$, the Cantor set), namely, that the set of $x$ such that $C+x$ consists entirely of irrational numbers, is comeager. This is a particular case of a 1954 result by Frederick Bagemihl, see
Bagemihl showed that if $F$ is meager and $N$ is countable, then there is a comeager set $G$ such that $(x+N)\cap F=\emptyset$ for all $x\in G$. This is an improvement of Scheeffer's result, replacing the assumption that $N$ is nowhere dense with the weaker requirement that it is meager, and the conclusion that $D$ is dense with the stronger requirement that it is comeager.
In 1981, in
it is proved that there is a comeager set $G\subseteq[0,1]$ such that $(0,1)\cap C_\alpha$ consists only of irrational numbers, for all $\alpha\in G$. Here, for $0<\alpha\le 1$, $C_\alpha$ is defined as the Cantor set, starting with $[0,1]$ and at stage $n>0$ removing the middle interval of length $\alpha/3^n$ from each remaining interval. (So $C_1=C$ is the usual Cantor set.)
Returning to the question, note that all the results above are established via applications of the Baire category theorem. For explicit examples of reals $x$ such that $x+C$ consists solely of irrational numbers, see this MO question. In particular, any $x$ such that every finite sequence of $0$s, $1$s, and $2$s appears as a substring of the ternary expansion of $x$ has this property.
We had a similar problem in one of our Real Analysis assignments. It was a generalization of the given problem.
Let $E \subset \mathbb{R}$ be a set with Lebesgue outer measure $0$. Then there exists $h \in \mathbb{R}$ such that $E+h \subseteq \mathbb{Q}^c$.
We can prove this using the identity Brian derived.
If $\Bbb Q\cap(E+h)\ne\varnothing$ for every $h\in\Bbb R$, then
$$\Bbb R=\bigcup_{h\in\Bbb Q}(h-E)\;.$$
$$m(\Bbb R) \leq \sum_{h \in \Bbb Q}{} m^*(h-E)=0$$
which gives us a contradiction. So, $\Bbb Q\cap(E+h)=\varnothing$ for some $h \in \Bbb R$. |
Hitchhiker Orbit Decay
Hitchhiker, the first server sky experiment, will deploy from a GTO transfer stage into a highly elliptical orbit, using a high-ISP electric engine like the Accion electrospray thruster. This orbit is purposely designed to decay quickly during high solar activity years, so it will not leave debris in orbit in the long term.
The orbit will decay in two regimes:
1 Highly elliptical to near-circular by perigee drag in the denser atmosphere close to Earth. Perigee around 6,800 kilometers radius, apogee around 42,000 kilometers radius 2 Rapidly decaying near-circular orbit, descending quickly through the ISS orbital radius (6792 to 6772 km on 20161004 ) to reentry.
Decaying Near-Circular Orbit
This is symmetrical and easier to solve. Presume the thinsat (or group of thinsats) has a "perpendicular" area-to-mass ratio of A/M = 5 m
2/kg edge on, but that the thinsat is tumbling randomly, for an average A/M of 2.5 m 2/kg. Assume the coefficient of drag C_D is 2.
For an orbit radius of r , the orbit velocity is \sqrt{ \mu / r } and the air density is \rho( r ) . The drag force is 0.5 C_D A \rho v^2 , and the drag power is dE/dt = 0.5 C_D A \rho ( r ) v^3 . The circular orbit energy is E = M ( v^2 / 2 - \mu / r ) ~ = ~ - M \mu / 2 r , so dE/dr = M \mu / 2 r^2 and dr/dE = 2 r^2 / M \mu Thus:
{ \Large { dr \over dt } = } { \LARGE { \left( dr \over dE \right) \left( dE \over dt \right) = \left( { 2 r^2 } \over { M \mu } \right) { \large ( 0.5 C_D A \rho( r ) v^3 ) = } \left( { C_D A \rho( r ) } \over { M \mu } \right) \left( { \large r^2 } \left( \mu \over r \right)^{3/2} \right) } } { \Large = C_D (A/M) \rho( r ) \sqrt{ \mu r } }
At 414 km equatorial altitude ( 6792 km radius ), the average air density is 1.4e-12 kg/m
3. \mu = 3.9860044e14 m 3/s 2 amd r = 6.792e6 m, so the orbit descent rate is about 0.36 m/s, or 1.3 kilometers per hour. At solar max, the air density is about 3.1e-12 kg/m 3, so the orbit descent rate is about 0.8 m/s or 2.8 km/hr . To minimize the ISS collision rate, we should end the experiment during a solar maximum. |
Under the auspices of the Computational Complexity Foundation (CCF)
There are three different types of nondeterminism in quantum communication: i) $\nqp$-communication, ii) $\qma$-communication, and iii) $\qcma$-communication. In this \redout{paper} we show that multiparty $\nqp$-communication can be exponentially stronger than $\qcma$-communication. This also implies an exponential separation with respect to classical multiparty nondeterministic communication complexity. We argue that there exists ... more >>>
The threshold degree of a Boolean function $f$ is the minimum degree of
a real polynomial $p$ that represents $f$ in sign: $f(x)\equiv\mathrm{sgn}\; p(x)$. In a seminal 1969 monograph, Minsky and Papert constructed a polynomial-size constant-depth $\{\wedge,\vee\}$-circuit in $n$ variables with threshold degree $\Omega(n^{1/3}).$ This bound underlies ... more >>>
The threshold degree of a Boolean function $f$ is the minimum degree of
a real polynomial $p$ that represents $f$ in sign: $f(x)\equiv\mathrm{sgn}\; p(x)$. Introduced in the seminal work of Minsky and Papert (1969), this notion is central to some of the strongest algorithmic and complexity-theoretic results for more >>>
In both query and communication complexity, we give separations between the class NISZK, containing those problems with non-interactive statistical zero knowledge proof systems, and the class UPP, containing those problems with randomized algorithms with unbounded error. These results significantly improve on earlier query separations of Vereschagin [Ver95] and Aaronson [Aar12] ... more >>>
The threshold degree of a Boolean function $f\colon\{0,1\}^n\to\{0,1\}$ is the minimum degree of a real polynomial $p$ that represents $f$ in sign: $\mathrm{sgn}\; p(x)=(-1)^{f(x)}.$ A related notion is sign-rank, defined for a Boolean matrix $F=[F_{ij}]$ as the minimum rank of a real matrix $M$ with $\mathrm{sgn}\; M_{ij}=(-1)^{F_{ij}}$. Determining the maximum ... more >>> |
DrivAer: grille shutters
In this post we continue to explore road vehicle aerodynamics with aid of the DrivAer model. This time, considering the benefit of adding grille shutters to improve aerodynamic drag.
The active grille shutters know when the engine, brakes and other components need air and automatically open the air vents. If no further air intake is required, the shutters close improving aerodynamics and reducing fuel consumption.
The following video demonstrates the system’s operating principle.
Computational domain
For this case I re-used the same meshDict as for the external aerodynamics case, just adding additional refinements for the underhood area. Specially around the grid area in order to better capture the small features; which ends up with 9.4 million cells.
For the boundary conditions I’m keeping the same settings as for the external aerodynamics cases, with the exception of adding a porous region.
Porous region
The porosity properties are added to a certain region of the mesh, the heat exchanger. This region can be defined during the meshing process, or afterwards using the setSet command-line utility.
The most common approach in OpenFOAM is to define the Darcy-Forchheimer coefficients (alternatives include the power law and fixed coefficients). Those two coefficients are added to the sink term of the Navier-Stokes equations, which looks like
$$ S_i = -\left( \mu D_{ij} + \frac{1}{2}\rho |u_{kk}| F_{ij} \right) u_i $$
where $D_{ij}$ is the resistance of the medium (viscous loss term) and $F_{ij}$ is the inertial resistance (inertial loss term). In OpenFOAM we define the coefficients as two vectors,
d and f which are then added to the diagonal of the tensors $D_{ij}$ and $F_{ij}$, respectively. Here you can see a sample of the fvOption dictionary used to define the porosity source term of my simulations.
As I didn’t find experimental values for the pressure drop in the heat exchangers defined in the DrivAer model, I re-used the same inertial and viscous resistance as in the formula student aerodynamics workshop, that is,
d d [ 0 -2 0 0 0 0 0 ] ( 18463.81 -1000 -1000 ); f f [ 0 -1 0 0 0 0 0 ] ( 1.09 -1000 -1000 );
The code will automatically multiply negative values with the largest component of the vector and switch the sign to a positive value.
Results
As expected, considering the underhood in the simulation has increased the aerodynamic drag of the vehicle. Introducing the engine bay in the simulation pumped up drag in 13 counts (+5.25 %), while closing the upper grille improved the latter value in about 11 counts (-4.21 %).
Of course, by opening up the grilles, the stagnating region at the front of the vehicle is reduced and so is the pressure drag at the front of the vehicle. That’s why the orange line has a lower peak. With the active grille shutters we have a higher peak at the front; because on the front side of the shutters the flow stagnates —just as in the reference model—, and on the other side we have the engine bay low pressure region, which combined makes the resistance in the front to be greater. We can appreciate this in the following pictures.
If we consider only the results of simplified models like this one, we might be inclined towards installing this kind of devices on any car; but actually one must also consider heat exchanger performance. Obviously, when the grille shutters are closed there’s no fresh air going through the radiators; but when fully open, real systems include complex mechanisms and elements and reconfigure the flow path inside the engine bay usually restricting the amount of flow available for cooling. So, if the base design is already performing at the lower target boundary in terms of cooling performance, adding grille shutters to improve a few counts aerodynamic drag might not be the best option.
References
m.bmw.co.uk. (2017).
Lightweight and Aerodynamics | BMW EfficientDynamics | BMW UK. [online] Available at: https://m.bmw.co.uk/en_gb/discover-bmw/technology/efficientdynamics/lightweight-aerodynamics [Accessed 8 Dec. 2017].
Elvar, H. (2009).
Porous Media in OpenFOAM. Göteborg: Chalmers University of Technology.
Schütz, T. (Ed.). (2015).
Aerodynamics of road vehicles (5th ed.). Warrendale, Pennsylvania, USA: SAE International. |
The complex number x+iy has two alternative forms, the polar form r(cos(theta)+i sin(theta)) and the exponential form r.e^(i theta), where r is the positive square root of x^2 + y^2 and theta is arctan(y/x). Both polar and exponential form are often abbreviated to r,angle(theta). (Apologies for the text by the way, I can't be bothered fiddling around with LaTex.)
The angle theta is multivalued and strictly should be written theta + 2k.pi, where k is any integer, positive or negative. In most cases the 2k.pi is omitted, but, if the complex number is being raised to a fractional power, it must be included.
Consider e^{i(theta+2k.pi)} raised to the power one third. Using the usual rule for indices this becomes e^{i(theta/3 + 2k.pi/3)} and, as you can see, the angle takes on different values for different values of k. In fact, ignoring the multivaluedness, there will be just three different angles, theta/3, theta/3+2pi/3 and theta/3+4pi/3. The next one theta/3+6pi/3 = theta/3+2pi is a repeat of theta/3. Other values of k simply repeat one of these three angles.
So, there will be three different cube roots of a number, and this applies to any number, not just complex ones. For example, there are three cube roots of 8, 2 angle 0 = 2, 2 angle 2pi/3 = 2(-1/2 + i 3^(1/2)/2) = -1 + i.3^(1/2) and 2 angle 4pi/3 = -1 - i.3^(1/2).
Similarly there will be two square roots (which everyone is used to for positive numbers at least), four fourth roots, five fifth roots and so on.
I've been outside cleaning my car Melody, a seagull had made a mess (and you wouldn't believe the extent of the mess) of the roof and windscreen.
As to LaTex I'm certainly not an expert, I just tend to muddle my way through. Muddling my way through this one, the best I can come up with is
\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\ \end{array} \frac{3x+h}{x+h}.
That produces
$$\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\ \end{array} \frac{3x+h}{x+h}$$
I've tried to make the h arrow 0 smaller, but without success.
The notation is sloppy isn't it ?
Is it (-1)^4 divided by 3, (and why should the 4 come first, that only happens because the fraction is written on a single line, why not (-1)^(1/3) raised to the power 4 ?), or is it (-1)^(4/3) ?
Enclosing the 4/3 in brackets easily removes the ambiguity, and as this hasn't been done, my interpretion would be the same as Melody's.
Speaking of Melody, was it yesterday that she asked about
$$e^{i\pi}=-1$$ ?
In which case,
$$(-1)^{4/3}=(e^{i\pi})^{4/3}=e^{4i\pi/3}$$
$$=\cos(4\pi/3)+i\sin(4\pi/3)$$
$$=-\frac{1}{2}-i\frac{\sqrt{3}}{2}.$$
There will of course be two other complex 'answers' one of which is the real number 1. |
The Rise of Artificial Intelligence
Over the past decade, artificial intelligence (AI) has achieved remarkable success in many fields such as healthcare, automotive, and marketing. The capabilities of sophisticated, autonomous decision systems driven by AI keep evolving and moving from lab to reality. Many of these systems are
black-box which means we don’t really understand how they work and why they reach such decisions. The Dangers of Black-Box
As black-box decision systems have come into greater use, they have also come under greater criticism. One of the main concerns is that it is dangerous to rely on black-box decisions without knowing the way they are made. Here is an example of why they can be dangerous.
Risk-assessment tools have been widely used in the federal and state courts to facilitate and improve judges’ decisions in the criminal justice processes. They provide defendants’ future criminal risk based on socio-economic status, family background, and other factors. In May 2016, ProPublica claimed that one of the most widely used risk-assessment tools, COMPAS, was biased against black defendants while being more generous to white defendants [link]. Northpointe, a for-profit company that provides the software, disputed the analysis but refused to disclose the software’s decision mechanism. So it is not possible for either stakeholders or the public to see what might be actually creating the disparity.
The Need for Interpretability
It is dangerous to rely on black-box decisions without knowing the way they are made. Here, we raise a question: How can we possibly go about resolving this concern? Explaining how a black-box decision system works or why it reaches such decisions helps to decide whether or not to follow its decisions. The need for interpretability is especially urgent in fields where black-box decisions can be life-changing and have significant consequences, such as disease diagnosis, criminal justice, and self-driving cars.
What is a desired explanation?
What makes a ‘good’ explanation for a black-box? Assume that you give a black-box predictive model an image of an apple. You open the black-box and explain why it believes this is indeed an apple on the image. Simply saying that
“it is red, so this is an apple” is not sufficient to justify your thought, but you should also avoid redundant explanation. It is important to give enough information concisely in explaining a black-box decision system. In other words, explanations should be brief but comprehensive. Variational Information Bottleneck for Interpretation
How can we take into account both briefness and comprehensiveness for explaining a black-box? Our work uses an information theoretic perspective to quantify the idea of briefness and comprehensiveness.
The information bottleneck principle (Tishby et al., 2000) provides an appealing information theoretic view for learning supervised models by defining what we mean by a
‘good’ representation. The principle says that the optimal model transmits as much information as possible from its input to its output through a compressed representation called the information bottleneck. And the information bottleneck is a good representation that is maximally informative about the output while compressive about a given input. Recently, Shwartz-Ziv et al. (2017) and Tishby et al. (2015) showed that the principle also applies to deep neural networks and each layer of the networks can work as an information bottleneck.
We adopt the information bottleneck principle as a criterion for finding a
‘good’ explanation. In the information theoretic view, we define a brief but comprehensive explanation as maximally informative about the black-box decision while compressive about a given input. In other words, the explanation should maximally compress the mutual information regarding an input while preserving as much as possible mutual information regarding its output. Variational Information Bottleneck for Interpretation
We introduce the
variational information bottleneck for interpretation (VIBI), a system-agnostic information bottleneck model that provides a brief but comprehensive explanation for every single decision made by a black-box.
VIBI is composed of two parts:
explainer and approximator, each of which is modeled by a deep neural network. Using the information bottleneck principle, VIBI learns an explainer that favors brief explanations while enforcing that the explanations alone suffice for an accurate approximation to the black-box. See the following illustration for an illustration of VIBI.
For each instance, the explainer returns a probability whether a chunk of features, called a
cognitive chunk, will be selected as an explanation or not. Cognitive chunk is defined as a group of raw features that work as a unit to be explained and whose identity is recognizable to a human, such as a word, phrase, sentence or a group of pixels. The selected chunks act as an information bottleneck that is maximally compressed about input and informative about the decision made by a black-box system on that input.
Now, we formulate the following optimization problem inspired by the information bottleneck principle to learn the explainer and approximator:
$$ p(\mathbf{z} | \mathbf{x}) = \mathrm{argmax}_{p(\mathbf{z} | \mathbf{x}), p(\mathbf{y} | \mathbf{t})} ~~\mathrm{I} ( \mathbf{t}, \mathbf{y} ) – \beta~\mathrm{I} ( \mathbf{x}, \mathbf{t} )$$ where \( \mathrm{I} ( \mathbf{t}, \mathbf{y} ) \) represents the sufficiency of information retained for explaining the black-box output \( \mathbf{y} \), \(-\mathrm{I} ( \mathbf{x}, \mathbf{t} ) \) represents the briefness of the explanation \( \mathbf{t} \), and \( \beta \) is a Lagrange multiplier representing a trade-off between the two. Challenges in Learning VIBI
The current form of information bottleneck objective is intractable due to the mutual informations and the non-differentiable sample \( \mathbf{z} \). We address these challenges as follows.
Variational Approximation to Information Bottleneck Objective
The mutual informations \( \mathrm{I} ( \mathbf{t}, \mathbf{y} ) \) and \( \mathrm{I} ( \mathbf{x}, \mathbf{t} ) \) are computationally expensive to quantify (Tishby et al., 2000; Chechik et al., 2005). In order to reduce the computational burden, we use a variational approximation to our information bottleneck objective: $$\mathrm{I} ( \mathbf{t}, \mathbf{y} )~-~\beta~\mathrm{I} ( \mathbf{x}, \mathbf{t} )
\geq \mathbb{E}_{\mathbf{y} \sim p(\mathbf{x})} \mathbb{E}_{\mathbf{y} | \mathbf{x} \sim p(\mathbf{y} | \mathbf{x})} \mathbb{E}_{\mathbf{t} | \mathbf{x} \sim p(\mathbf{t} | \mathbf{x})} \left[ \log q(\mathbf{y} | \mathbf{t}) \right] ~-~\beta~\mathbb{E}_{\mathbf{x}\sim p(\mathbf{x})} \mathrm{KL} (p(\mathbf{z}| \mathbf{x}), r(\mathbf{z})) $$
Now, we can integrate the Kullback-Leibler divergence \( \mathrm{KL} (p(\mathbf{z}| \mathbf{x}), r(\mathbf{z})) \) analytically with proper choices of \( r(\mathbf{z}) \) and \( p(\mathbf{z}|\mathbf{x}) \). We also use the empirical data distribution to approximate \( p(\mathbf{x}, \mathbf{y}) = p(\mathbf{x})p(\mathbf{y}|\mathbf{x}) \).
Continuous Relaxation and Re-parameterization
We use the generalized Gumbel-softmax trick (Jang et al., 2017; Chen et al., 2018), which approximates the non-differentiable categorical subset sampling with Gumbel-softmax samples that are differentiable. This trick allows using standard backpropagation to compute the gradients of the parameters via reparameterization.
Result
VIBI provides instance-specific keywords to explain an LSTM sentiment prediction model using Large Movie Review Dataset, IMDB.
The keywords such as “waste,” and “horrible,” are selected for the negative-predicted movie review, while keywords such as “most fascinating,” explain the model’s positive-predicted movie review. Also, we could see that the LSTM sentiment prediction model makes a wrong prediction for a negative review because the review includes several positive words such as ‘enjoyable’ and ‘exciting’.
VIBI also provides instance-specific key patches containing \( 4 \times 4 \) pixels to explain a CNN digit recognition model using the MNIST image dataset.
The first two examples show that the CNN recognizes digits using both shapes and angles. In the first example, the CNN characterizes ‘1’s by straightly aligned patches along with the activated regions although ‘1’s in the left and right panels are written at different angles. Contrary to the first example, the second example shows that the CNN recognizes the difference between ‘9’ and ‘6’ by their differences in angles. The last two examples show that the CNN catches a difference of ‘7’s from ‘1’s by patches located on the activated horizontal line on ‘7’ (see the cyan circle) and recognizes ‘8’s by two patches on the top of the digits and another two patches at the bottom circle.
VIBI Explains Black-box Models Better
We assume that a better explanation allows humans to better infer the black-box output given the explanation. Therefore, we asked humans to infer the output of the black-box system (Positive/Negative/Neutral) given five keywords as an explanation generated by VIBI and other competing methods (Saliency, LIME, and L2X). Each method was evaluated by the human intelligences on Amazon Mechanical Turk who are awarded the Masters Qualification (i.e. high-performance workers who have demonstrated excellence across a wide range of tasks). We also evaluated the interpretability for the CNN digit recognition model using MNIST. We asked humans to directly score the explanation on a 0 to 5 scale (0 for no explanation, 1-4 for insufficient or redundant explanation and 5 for concise explanation). Each method was evaluated by 16 graduate students at the School of Computer Science, Carnegie Mellon University who have taken at least one graduate-level machine learning class.
VIBI Approximates Black-box Models Better
We assessed fidelity of the approximator by prediction performance with respect to the black- box output. We introduce two types of formalized metrics to quantitatively evaluate the fidelity:
approximator fidelity and rationale fidelity.
Approximator fidelity implies the ability of the approximator to imitate the behaviour of a black-box. As shown above, VIBI and L2X outperform the others in approximating the black-box models. However, it does not mean both approximators are same in fidelity. See below.
Rationale fidelity implies how much the selected chunks contribute to the approximator fidelity. As shown above, the selected chunks of VIBI account for more approximator fidelity than L2X. Note that L2X is a special case of VIBI having the information bottleneck trade-off parameter \( \beta = 0 \) (i.e. not using the compressiveness constraint \( −\mathrm{I} ( \mathbf{x}, \mathbf{t} ) \)). Therefore, compressing information through the explainer achieves not only conciseness of explanation but also better fidelity of explanation to a black-box.
Note that the number of cognitive chunks to be selected, \( k \), should be given in advance. It also impacts conciseness of the actual total explanation and should be chosen carefully. In our analysis, we choose \( k \) as the minimum number that exceeds a certain fidelity.
Where Should I Look to Learn More?
Further details can be found here. The code is publicly available here.
DISCLAIMER: All opinions expressed in this post are those of the author and do not represent the views of CMU. |
I am encountering a problem concerning Reproducing Kernel Hilbert Spaces (
RKHS) in the context of machine learning using Support Vector Machines ( SVMs).
With refernece to this paper [Olivier Chapelle, 2006], Section 3, I will try to be brief and focused on my problem, thus I may avoid giving rigorous description of what I am using below.
Let the following optimization problem: $$ \displaystyle \min_{\mathbf{w},b}\: \lVert\mathbf{w}\rVert^2 + C\sum_{i=1}^{n}L(y_i, \mathbf{w}\cdot\mathbf{x}_i+b), $$ where $L(y,t)=\operatorname{max}(0,1-yt)$ is a loss function, the so-called "hinge loss". Trying to introduce kernels, in order to consider non-linear SVMs, the author reformulates the aforementioned optimization problem, looking for a function in a RKHS, $\mathcal{H}$, that minimizes the following functional: $$ F[f]=\lambda\lVert f \rVert^2_\mathcal{H} + \sum_{i=1}^{n}L(y_i, f(\mathbf{x}_i)+b). $$ I understand the following of his work; my question is the following: What if I had some other loss function (not the hinge-loss above), which is not expressed solely by the inner product $\mathbf{w}\cdot\mathbf{x}_i$, which -if I understand correctly- is "replaced" by $f(\mathbf{x}_i)$, but instead I had some loss function of the form: $$ \mathbf{w}\cdot\mathbf{x}_i+b+\sqrt{\mathbf{w}^TA\mathbf{w}}, $$ where $A$ is a positive-definite symmetric matrix? I mean, is there any way of expressing the above quadratic form ($\sqrt{\mathbf{w}^TA\mathbf{w}}$) using the function $f$, such that I can express my optimization problem in the context of RKHS?
On the other hand, the theory suggests that, whatever the loss-function, $L$, is, the solution of the above reformulated problem would be in the form: $$ f(\mathbf{x})=\sum_{i=1}^{n}\alpha_ik(\mathbf{x}_i, \mathbf{x}), $$ where $k$ is the kernel associated with the adopted RKHS. Have I understood that correctly? The solution would be the above even if my loss function included terms like $\sqrt{\mathbf{w}^TA\mathbf{w}}$? |
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Woodin cardinals (named after W. Hugh Woodin) are a generalization of the notion of strong cardinals and have been used to calibrate the exact proof-theoretic strength of the axiom of determinacy. They can also be seen as weakenings of Shelah cardinals, defined below. Their exact definition has several equivalent but different characterizations, each of which is somewhat technical in nature. Nevertheless, an inner model theory encapsulating infinitely many Woodin cardinals and slightly beyond has been developed. Definition and some properties
We first introduce the concept of
$\gamma$-strongness for $A$: an ordinal $\kappa$ is $\gamma$-strong for $A$ (or $\gamma$-$A$-strong) if there exists a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$ such that $V_{\kappa+\gamma}\subseteq M$ and $A\cap V_{\kappa+\gamma} = j(A)\cap V_{\kappa+\gamma}$. Intuitively, $j$ preserves the part of $A$ that is in $V_{\kappa+\gamma}$. We say that a cardinal $\kappa$ is <$\delta$-$A$-strong if it is $\gamma$-$A$-strong for all $\gamma<\delta$.
We also introduce
Woodin-ness in $\delta$: for an infinite ordinal $\delta$, a set $X\subseteq\delta$ is Woodin in $\delta$ if for every function $f:\delta\to\delta$, there is an ordinal $\kappa\in X$ with $\{f(\beta):\beta<\kappa\}\subseteq\kappa$ ($\kappa$ is closed under $f$), there exists a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$ such that $V_{j(f)(\kappa)}\subseteq M$.
An inaccessible cardinal $\delta$ is
Woodin if any of the following (equivalent) characterizations holds [1]: For any set $A\subseteq V_\delta$, there exists a $\kappa<\delta$ that is <$\delta$-$A$-strong. For any set $A\subseteq V_\delta$, the set $S=\{\kappa<\delta:\kappa$ is <$\delta$-$A$-strong$\}$ is stationary in $\delta$. The set $F=\{X\subseteq \delta:\delta\setminus X$ is not Woodin in $\delta$$\}$ is a proper filter, the Woodin filter over $\delta$. For every function $f:\delta\to\delta$ there exists $\kappa<\delta$ such that $\{f(\beta):\beta\in\kappa\}\subseteq\kappa$ (that is, $\kappa$ is closed under $f$) and there exists a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$ such that $V_{j(f)(\kappa)}\subseteq M$.
Let $\delta$ be Woodin, $F$ be the Woodin filter over $\delta$, and $S=\{\kappa<\delta:\kappa$ is <$\delta$-$A$-strong$\}$. Then $F$ is normal and $S\in F$. [1] This implies every Woodin cardinal is Mahlo and preceeded by a stationary set of measurable cardinals, in fact of <$\delta$-strong cardinals. However, the least Woodin cardinal is not weakly compact as it is not $\Pi^1_1$-indescribable.
Woodin cardinals are weaker consistency-wise then superstrong cardinals. In fact, every superstrong is preceeded by a stationary set of Woodin cardinals. On the other hand the existence of a Woodin is much stronger than the existence of a proper class of strong cardinals.
The existence of a Woodin cardinal implies the consistency of $\text{ZFC}$ + "the nonstationary ideal on $\omega_1$ is $\omega_2$-saturated". Huge cardinals were first invented to prove the consistency of the existence of a $\omega_2$-saturated $\sigma$-ideal on $\omega_1$, but turned out to be stronger than required, as a Woodin is enough.
Shelah cardinals
Shelah cardinals were introduced by Shelah and Woodin as a weakening of the necessary hypothesis required to show several regularity properties of sets of reals hold in the model $L(\mathbb{R})$ (e.g., every set of reals is Lebesgue measurable and has the property of Baire, etc...). In slightly more detail, Woodin had established that the axiom of determinacy (a hypothesis known to imply regularity properties for sets of reals) holds in $L(\mathbb{R})$ assuming the existence of a nontrivial elementary embedding $j:L(V_{\lambda+1})\to L(V_{\lambda+1})$ with critical point $<\lambda$. This axiom, a rank-into-rank axiom, is known to be very strong and its use was first weakened to that of the existence of a supercompact cardinal. Following the work of Foreman, Magidor and Shelah on saturated ideals on $\omega_1$, Woodin and Shelah subsequently isolated the two large cardinal hypotheses which bear their name and turn out to be sufficient to establish the regularity properties of sets of reals mentioned above.
Shelah cardinals were the first cardinals to be devised by Woodin and Shelah. A cardinal $\delta$ is
Shelah if for every function $f:\delta\to\delta$ there exists a nontrivial elementary embedding $j:V\to M$ with critical point $\delta$ such that $V_{j(f)(\delta)}\subseteq M$. Every Shelah is Woodin, but not every Woodin is Shelah: indeed, Shelah cardinals are always measurable and in fact strong, while Woodins are usually not. However, just like Woodins, Shelah cardinals are weaker consistency-wise than superstrong cardinals.
A related notion is
Shelah-for-supercompactness, where the closure condition $V_{j(f)(\delta)}\subseteq M$ is replaced by $M^{j(f)(\delta)}\subseteq M$, a much stronger condition. The difference between Shelah and Shelah-for-supercompactness cardinals is essentially the same as the difference between strong and supercompact cardinals, or between superstrong and huge cardinals. Also, just like every Shelah is preceeded by a stationary set of strong cardinals, every Shelah-for-supercompactness cardinal is preceeded by a stationary set of supercompact cardinals.
Much weaker, consistent with $V=L$ variant: A cardinal $κ$ is
virtually Shelah for supercompactness iff for every function $f : κ → κ$ there are $λ > κ$ and $\bar{λ}< κ$ such that in a set-forcing extension there is an elementary embedding $j : V_{\bar{λ}}→ V_{λ}$ with $j(\mathrm{crit}(j)) = κ$, $\bar{λ} ≥ f(\mathrm{crit}(j))$ and $f ∈ \mathrm{ran}(j)$. If $κ$ is virtually Shelah for supercompactness, then $V_κ$ is a model of proper class many virtually $C^{(n)}$-extendible cardinals for every $n < ω$ and if κ is 2-iterable, then $V_κ$ is a model of proper class many virtually Shelah for supercompactness cardinals.[2] Woodin for strong compactness
(from [3] unless otherwise noted)
A cardinal $δ$ is
Woodin for strong compactness (or Woodinised strongly compact) iff for every $A ⊆ δ$ there is $κ < δ$ which is $<δ$-strongly compact for $A$.
This definition is obviously analogous to one of the characterisations of Woodin and
Woodin-for-supercompactness (Perlmutter proved that [4] it is equivalent to Vopěnkaness) cardinals.
Results:
Woodin for strong compactness cardinal $δ$ is an inaccessible limits of $<δ$-strongly compact cardinals. $κ$ is Woodin and there are unboundedly many $<δ$-supercompact cardinals below $δ$, then $δ$ is Woodin for strong compactness. The existence of a Woodin for strong compactness cardinal is at least as strong as a proper class of strongly compact cardinals and at most as strong as a Woodin limit of supercompact cardinals (which lies below an extendible cardinal). Woodin cardinals and determinacy See also: axiom of determinacy, axiom of projective determinacy
Woodin cardinals are linked to different forms of the axiom of determinacy [1][5][6]:
$\text{ZF+AD}$, $\text{ZFC+AD}^{L(\mathbb{R})}$, ZFC+"the non-stationary ideal over $\omega_1$ is $\omega_1$-dense" and $\text{ZFC}$+"there exists infinitely many Woodin cardinals" are equiconsistent. Under $\text{ZF+AD}$, the model $\text{HOD}^{L(\mathbb{R})}$ satisfies $\text{ZFC}$+"$\Theta^{L(\mathbb{R})}$ is a Woodin cardinal". [6] gives many generalizations of this result. If there exists infinitely many Woodin cardinals with a measurable above them all, then $\text{AD}^{L(\mathbb{R})}$. If there assumtion that there is a measurable above those Woodins is removed, one still has projective determinacy. In fact projective determinacy is equivalent to "for every $n<\omega$, there is a fine-structural, countably iterable inner model $M$ such that $M$ satisfies $\text{ZFC}$+"there exists $n$ Woodin cardinals". For every $n$, if there exists $n$ Woodin cardinals with a measurable above them all, then all $\mathbf{\Sigma}^1_{n+1}$ sets are determined. $\mathbf{\Pi}^1_2$-determinacy is equivalent to "for every $x\in\mathbb{R}$, there is a countable ordinal $\delta$ such that $\delta$ is a Woodin cardinal in some inner model of $\text{ZFC}$ containing $x$. $\mathbf{\Delta}^1_2$-determinacy is equivalent to "for every $x\in\mathbb{R}$, there is an inner model M such that $x\in M$ and $M$ satisfies ZFC+"there is a Woodin cardinal". $\text{ZFC}$ + lightface $\Delta^1_2$-determinacy implies that there many $x$ such that $\text{HOD}^{L[x]}$ satisfies $\text{ZFC}$+"$\omega_2^{L[x]}$ is a Woodin cardinal". $\text{Z}_2+\Delta^1_2$-determinacy is conjectured to be equiconsistent with $\text{ZFC}$+"$\text{Ord}$ is Woodin", where "$\text{Ord}$ is Woodin" is expressed as an axiom scheme and $\text{Z}_2$ is second-order arithmetic. $\text{Z}_3+\Delta^1_2$-determinacy is provably equiconsistent with $\text{NBG}$+"$\text{Ord}$ is Woodin" where $\text{NBG}$ is Von Neumann–Bernays–Gödel set theory and $\text{Z}_3$ is third-order arithmetic. Role in $\Omega$-logic Stationary tower forcing
References Main library Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Gitman, Victoria and Shindler, Ralf. Virtual large cardinals. www bibtex Dimopoulos, Stamatis. Woodin for strong compactness cardinals. The Journal of Symbolic Logic 84(1):301–319, 2019. arχiv DOI bibtex Perlmutter, Norman. The large cardinals between supercompact and almost-huge. , 2010. arχiv bibtex Larson, Paul B. A brief history of determinacy. , 2013. www bibtex Koellner, Peter and Woodin, W Hugh. Chapter 23: Large cardinals from Determinacy. Handbook of Set Theory , 2010. www bibtex |
I am a TeX/LaTeX noob, but I am learning rapidly. I am about to attempt to draw the following figure using LaTeX/TeX/TikZ I already have a, (I think) good working knowledge of how to draw circles, braces, lines, segments, and nodes, etc.
The last time I drew one though, I noticed that I spent a lot of time fine tuning parameters, making sure that lines intersected circles at the proper points, making sure lines were parallel, etc.
I need to draw the following figure, so I wanted first inquire if there is a fast/easy way, or anything I should keep in mind, when creating parallel lines (the red ones), as per the following figure:
The main time sinks I foresee are going to be the following:
Making sure the length of the purple lines are exactly correct, so as not to overshoot the red ones.
Making sure the red lines are all parallel to each other.
Making sure than I can draw nodes/dots where lines intersect each other and/or the circle.
Is there any particular way(s) that is recommended to make sure that the above points do not waste too much time? I would appreciate any advice/examples.
Thanks!
P.S. I am making good use of TikZ as well.
EDIT:
I am including a minimal example of my 'skill' set. As you can see, I painstakingly usually just have to fine tune co-ordinates, for where things intersect, etc. I also do not know how to not-waste-time if I wanted to make parallel lines, as shown in the red lines above. Here is my example code:
\documentclass[journal]{IEEEtran}\usepackage{graphicx}\usepackage{float}\usepackage{amsmath}\usepackage{amsfonts}\usepackage{tikz}\usetikzlibrary{matrix}\usetikzlibrary{decorations.pathreplacing}\usepackage{xspace} \usepackage{float}\usepackage{capt-of}\usepackage{cases}\begin{document}\newcommand{\var}{1.5}\begin{tikzpicture}%\begin{tikzpicture} \draw [help lines] (-4,-4) grid (4,4); % Draw the axes \draw [->,black, ] (-4,0) -- (4,0) ; \draw [->,black] (0,-4) -- (0,4) ; % Draw the circle \path [draw, ultra thick, black] (0,0) circle (3); %Radial Lines \draw[black](0:0)--(45:3); \draw[black](0:0)--(135:3); \draw[black](0:0)--(225:3); \draw[black](0:0)--(315:3); %Wavefront normal \draw[blue](0:0)--(60:3); \draw[blue](0:0)--(240:3); \draw[blue](0,1) arc (90:60:1); \draw[blue] (0,0.95)arc(90:60:0.95); \node[] at (75:1.2) {$\alpha$}; \draw[blue] (0.7,1.2)arc(90:30:0.37); %\node[] at (50:2) {$\frac{pi}{4}-\alpha$}; \draw[-latex](1.5,3)node[right]{$\frac{\pi}{4}-\alpha$} to[out=180,in=90] (50:1.5); %Points \path[fill, black](45:3) circle(0.1); \node[] at (45:3.3){1}; \path[fill, black](135:3) circle(0.1); \node[] at (133:3.3){4}; \path[fill,black](225:3) circle(0.1); \node[] at (225:3.3){3}; \path[fill, black](315:3) circle(0.1); \node[] at (313:3.3){2}; %Wavefronts \draw[blue, ultra thick](-.5,3.6347)--(3.5,1.3252); \draw[blue, ultra thick](-2.5,2.34)--(3.5,-1.1242); \draw[blue, ultra thick](-3.2,0.9509)--(2.8,-2.5132); \draw[blue, ultra thick](-3.2,-1.4985)--(0.2,-3.4615); %Right angle signs \draw[red](60:2.8)--(62:2.8)--(62:2.9); \draw[red](60:.68)--(68:.69)--(67:.79); % Done\end{tikzpicture}\end{document}
The above code is now giving me this: |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
I really don't understand what he said about why $\alpha$ is not an embedding... No worry about my knowledge on topology. Can anyone help me “translate” it to the common language that's easy to understand?
The set $C=\alpha\bigl((-3,0)\bigr)$ has two topologies:
the topology it inherits from the usual topology in $\mathbb{R}^2$: a set $A\subset C$ is open if there is an open subset $A^\star$ of $\mathbb{R}^2$ such that $A=A^\star\cap C$. the topology it gets from $(-3,0)$: a set $A\subset C$ is open if there is an open subcet $A^\star$ of $(-3,0)$ such that $A=\alpha(A^\star)$.
Then, Do Carmo explains
why these two topologies are distinct: the second one is locally connected, whereas the first one is not. |
Suppose I have a linear system
$$ \left\lbrack \begin{array}{cc} M_1& S\\ S^{\mathrm{T}}& M_2 \end{array} \right\rbrack \left\lbrack \begin{array}{c} X\\ Y\end{array} \right\rbrack= \left\lbrack \begin{array}{c} F\\G\end{array}\right\rbrack, $$ where $M_1\in \mathrm{R}^{n\times n}$ is symmetric positive definite, $M_2\in \mathrm{R}^{m\times m}$ is symmetric positive semi-definite. $S$ is a full-rank matrix of compatible size. $X,Y,F,G$ are all vectors of appropriate size.
What factors are relevant in deciding between using GMRES on the big system and using a Schurr complement technique?
Factors that I would guess are relevant are how expensive $M_1$ is to invert, the condition number of $S$, and whether $M_2$ is the zero matrix.
Any links to the literature are appreciated.
Edit
I am asking this question because I need to solve the following system.
$$ \left\lbrack \begin{array}{cccc} M_1& S &0 &\Gamma^{\mathrm{T}} \\ -S^{\mathrm{T}}& 0 &\Xi^{\mathrm{T}}& 0\\ 0 & \Xi & 0 &0 \\ \Gamma&0&0&0 \end{array} \right\rbrack \left\lbrack \begin{array}{c} x_1\\x_2\\x_3\\x_4 \end{array} \right\rbrack= \left\lbrack \begin{array}{c} f_1\\f_2\\f_3\\f_4 \end{array} \right\rbrack $$
Right now, I solve this system using the Schurr complement method but it requires three iterative CG solves. I am wondering if it would be better to just "brute force it" and use GMRES. |
What is the difference between the three terms below?
percentile quantile quartile
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0 quartile = 0 quantile = 0 percentile
1 quartile = 0.25 quantile = 25 percentile
2 quartile = .5 quantile = 50 percentile (median)
3 quartile = .75 quantile = 75 percentile
4 quartile = 1 quantile = 100 percentile
Percentiles go from $0$ to $100$.
Quartiles go from $1$ to $4$ (or $0$ to $4$).
Quantiles can go from anything to anything.
Percentiles and quartiles are examples of quantiles.
In order to define these terms rigorously,it is helpful to first define the
quantile functionwhich is also known as the inverse cumulative distribution function.Recall that for a random variable $X$,the cumulative distribution function $F_X$ is defined by the equation$$F_X(x) := \Pr(X \le x).$$The quantile function is defined by the equation$$Q(p)\,=\,\inf\left\{ x\in \mathbb{R} : p \le F(x) \right\}.$$
Now that we have got these definitions out of the way, we can define the terms:
percentile: a measure used in statistics indicatingthe value below which a given percentage of observationsin a group of observations fall.
Example: the 20th percentile of $X$ is the value $Q_X(0.20)$
quantile: values taken from regular intervalsof the quantile function of a random variable.For instance, for some integer $k \geq 2$,the $k$-quartiles are defined as the valuesi.e. $Q_X(j/k)$ for $j = 1, 2, \ldots, k - 1$.
Example: the 5-quantiles of $X$ are the values $Q_X(0.2), Q_X(0.4), Q_X(0.6), Q_X(0.8)$
It may be helpful for you to work out an example of what these definitions mean when say $X \sim U[0,100]$, i.e. $X$ is uniformly distributed from 0 to 100.
References from Wikipedia:
From wiki page: https://en.wikipedia.org/wiki/Quantile
Some q-quantiles have special names:
The only 2-quantile is called the median The 3-quantiles are called tertiles or terciles → T The 4-quantiles are called quartiles → Q The 5-quantiles are called quintiles → QU The 6-quantiles are called sextiles → S The 8-quantiles are called octiles → O (as added by @NickCox - now on wiki page also) The 10-quantiles are called deciles → D The 12-quantiles are called duodeciles → Dd The 20-quantiles are called vigintiles → V The 100-quantiles are called percentiles → P The 1000-quantiles are called permilles → Pr
The difference between
quantile,
quartile and
percentile becomes obvious.
Percentile : The percent of population which lies below that value Quantile : The cut points dividing the range of probability distribution into continuous intervals with equal probability There are q-1 of q quantiles one of each k satisfying 0 < k < q Quartile : Quartile is a special case of quantile, quartiles cut the data set into four equal parts i.e. q=4 for quantiles so we have First quartile Q1, second quartile Q2(Median) and third quartile Q3
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Bauer, D., Broersma, H. J., van den Heuvel, J., Kahl, N. and Schmeichel, E. (2013)
Toughness and vertex degrees. Journal of Graph Theory, 72 (2). pp. 209-219. ISSN 0364-9024 Abstract
We study theorems giving sufficient conditions on the vertex degrees of a graph $G$ to guarantee $G$ is $t$-tough. We first give a best monotone theorem when $t\ge1$, but then show that for any integer $k\ge1$, a best monotone theorem for $t=\frac1k\le 1$ requires at least $f(k)\cdot|V(G)|$ nonredundant conditions, where $f(k)$ grows superpolynomially as $k\rightarrow\infty$. When $t<1$, we give an additional, simple theorem for $G$ to be $t$-tough, in terms of its vertex degrees.
Item Type: Article Official URL: http://onlinelibrary.wiley.com/ Additional Information: © 2013 John Wiley & Sons, Inc. Divisions: Mathematics Subjects: Q Science > QA Mathematics Sets: Departments > Mathematics Date Deposited: 09 Apr 2010 13:16 Last Modified: 20 Feb 2019 10:23 Projects: EP/F064551/1 Funders: Engineering and Physical Sciences Research Council URI: http://eprints.lse.ac.uk/id/eprint/27680 Actions (login required)
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Mass
The big issue here lies in determining the mass of these black holes. On the one hand, the traditional primordial black holes you mention - formed by density perturbations - occupy a relatively low-mass regime. Constraints from a variety of observations indicate a peak in the mass distribution at $\sim10^{17}\text{ kg}$ depending on the conditions at the time of formation, as well as the behavior of inflation. These black holes should still be around today, even if Hawking radiation had dissipated them.
Figure 3, Primordial Black Holes in the Inflationary Universe, Masahiro Kawasaki. A plot of the primordial black hole mass distribution, with various constraints shaded out. Notice the peaks. I should note that new constraints are continuously being added; see, for example, a recent press release from the Subaru Telescope.
On the other hand, it has been suggested that black holes of a few tens of solar masses could make up the dark matter we observe - a variant of the Massive Compact Halo Object (MACHO) hypothesis, and primordial clouds of $10^{4-5}M_{\odot}$ could have collapsed to form black holes. If we take both populations into account as "primordial", we have an enormous mass range to consider.
At the lower end, we see black holes with Schwarzschild radii of a few tenths of a nanometer. At the upper end, we find Schwarzschild radii of 46 Earth radii. To have a radius of, say, 10% of Earths, we have to limit ourselves to about $200M_{\odot}$, possible from the collapse of a low-mass Population III star. However, this is still going to pose problems for your planet, and much of its interior could be accreted on short timescales. Therefore, I suggest staying with the classic primordial black hole formed via density fluctuations, with a much nicer mass of $10^{17}\text{ kg}$. After all, we don't want to swallow the planet!
Surface effects
At this point, we can run through the calculations pretty quickly. Using a Newtonian approximation (which I think is justified this far from the black hole), we find that the surface gravity is$$g=\frac{GM_{\text{BH}}}{R_{\oplus}^2}\approx1.67\times10^{-8}g_{\text{Earth}}$$where $g_{\text{Earth}}$ is the surface gravity on Earth; our result is so small that it wouldn't be felt at the surface. On the other hand, deep inside the planet, there
might be problems, as the black hole would accrete matter. However, as might happen with a black hole at the center of the Sun, the radiation pressure induced could stop a collapse. So this might be stable over reasonable timescales. For the $200M_{\odot}$ black hole, of course, we see that the surface gravity is $6.66\times10^7g_{\text{Earth}}$. That's way too high.
Finally, the time dilation on the surface would be$$t_{\text{surf}}=t_0\sqrt{1-\frac{r_s}{R_{\oplus}}}\approx t_0$$because the term inside the square root is approximately 1. In other words, there's negligible time dilation. For the $200M_{\odot}$ black hole, however, $t_{\text{surf}}=0.96t_0$ - not terribly significant, but not a huge amount, either.
To be honest, on the surface of the planet, there shouldn't be major effects from time dilation or the extra mass of the black hole. Of course, don't take this as an endorsement of the scheme; you've still got a black hole at the center of your planet, and that rarely turns out well. |
I have a (circular) segment of known height and known chord length. Is is possible to determine the radius of the circle?
Any help much appreciated.
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We can apply the
Intersecting Chords Theorem.
You chord length is the length $UV$ and the segment height is the length $PX$.
The intersecting chords theorem tells us that $XP \times XQ = XU \times XV$.
Let $\ell = UV$ and $h=XP$. It follows that $UX = XV = \tfrac{1}{2}\ell$. The ICT then tells us that
$$\tfrac{1}{2}\ell \times \tfrac{1}{2}\ell = h \times XQ \, ,$$ i.e. $XQ = \tfrac{1}{4h}\ell^2$. The diameter $PQ=PX+XQ$ and $$PX + XQ = h + \frac{\ell^2}{4h}=\frac{4h^2+\ell^2}{4h}$$
The radius is then one half of this, i.e.
$$CQ = \frac{4h^2+\ell^2}{8h} \, . $$
Here is the solution using trigonometry
Consider given the height $h=DE$ and the half length $a=DB$. The triangle
ADB produces the following two equations
$$ a = r \sin \theta \\ r-h = r \cos \theta $$
where $r$ is the unknown radius, and $\theta$ is the included half angle (denoted on point
A above)
If these two equations are squared and added together they make
$$ a^2 + (r-h)^2 = (r \sin\theta)^2 + (r \cos \theta)^2 = r^2 $$
The solution to this is $$ \boxed{ r = \dfrac{a^2+h^2}{2 h} } $$
This problem can be solved as follows. There are two knowns and two unknowns. The two knowns are the chord length $UV$ ($l$) and the chord height $XP$ ($h$). The two unknowns are the radius ($PC$ or $r$) and $XC$ which is part of the radius. Lets call this $q$. I used the drawing and symbols from the first answer to the problem.
We can write two distinct equations using the two knowns and two unknowns, so we can solve this problem:
(1) $h + q = r$
(2) Using the Pythagorean theorem, $(\frac{l}{2})^2 + q^2 = r^2$. To see this draw a line from $C$ to $V$ (this distance is $r$ of course). Since $PC$ bisects $UV$, the angle $CXV$ must be a right angle, so the Pythagorean theorem applies
By rearranging (1) as $q = r - h$, substitute $r - h$ for $q$ in equation (2)
Without going through all the math, you can solve equation (2) for $r$
The solution is: $$r = \dfrac{\left(\frac{l}{2}\right)^2 + h^2}{2h}$$
or
$$r = \dfrac{\frac{l^2}{4} + h^2}{2h}$$
Multiply numerator and denominator by $4$ and you get the same answer as previously submitted
Any two cords that intersect within the same circle will create two equal area rectangles if you multiply the two segments of the same cord together, the segments created by the intersection of the other second cord.
Given one cord and creating a second imaginary cord that intersects the given cord at the mid point and also intersects the circles center, we can find the diameter of the circle. The height of the circular segment becomes one of the segments of the second imaginary cord. We can solve the second segment by dividing the square of the given two segments by the height of the circular segment. The height of the circular segment is one of the segments of our imaginary created cord. If we add them both together they create the diameter length of the circle.
(1/2 cord)^2 / circular segment height, equals the diameter if you add the height of the circular segment to it. If you want the radius just divide the diameter by 2.
Sincerely,
William McCormick
I have been exploring the design of certain church doorways from the mid 1700's and this problem came up. I tried to think of how they might have solved this problem back then and came up with the following solution:
Draw a long vertical line. Measure and mark h from the top. Draw a horizontal line equal to l centered on that point. Take your compass with the sharp end moving down the vertical line and expand it until it scribes an arc that includes the top of the horizontal line and the end of the vertical lines. Measure that length and you get the radius.
Math is a powerful interpretive language to describe certain things, but it is not those things. It's interesting to speculate how the masons of those days solved these complex problems with little or no knowledge of the calculations that we would use today.
I would suggest that you read a book on telescope mirror making. The telescope maker's approximate formula for the saggita of a mirror is s = r^2/2R, where r is the radius of the mirror surface, R is the big radius (half the focal length). That means R = r^2/2s. If the mirror is a parabola instead of a sphere, this formula is exact.
For an exact formula, use the Pythagorean theorem to get (after some algebra) R=(r^2+s^2)/2s.
In real world telescope mirror making, no mirror comes out exactly spherical, not even for John Dobson (PBUH), so the approximation is good enough.
Contact me richard1941@gmail.com if you need help building a Foucault tester or interpreting the test results. I have plenty of glass from the estate of Bill Kelly waiting to made into mirrors. You would honor a great inventor.
Let $h$ be the segment height, $2c$ the chord and $r$ radius. For Pythagoras: $$r^2=c^2+(r-h)^2\rightarrow r^2=c^2+r^2-2hr+h^2\rightarrow h^2-2rh+c^2=0\rightarrow$$ $$2rh=h^2+c^2\rightarrow r=\dfrac{1}{2h}(h^2+c^2)$$
Consider one half of the segment. Slant side length = $\sqrt{c^2+h^2}$.
Note similarity of this right triangle of smaller sides $(c,h)$ at left with main triangle diameter as hypotenuse $2r$
Ratio of corresponding sides should be same:
$$\dfrac{h}{\sqrt{c^2+h^2}}=\dfrac{\sqrt{c^2+h^2}}{2r} \rightarrow 2rh ={c^2+h^2} $$
Note that the result is true either if $h$ is the major segment height or the minor segment height.
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I am reading a proof of the result: a square matrix $A$ is invertible if and only if $\lambda = 0$ is not an eigenvalue of $A$.
Proof goes as follows: Assume that $A$ is an $n\times n$ matrix and observe that $\lambda = 0$ is the solution of the characteristic equation of $A$ given by
$det (\lambda I - A) = \lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.
On setting $\lambda = 0$ in above equation, we get $det(-A) = c_n$ or $(-1)^n det(A) = c_n$ Now it follows that $det(A) = 0$ if and ony if $a_n \neq 0$.
I am curious to know that if I want to write the characteristic equation of $A$ as $det( A - \lambda I)$ then what would be the general form of characteristic equation in terms of polynomial in $\lambda$.
Can I still write the characteristic equation of $A$ as
$det( A - \lambda I)$= $\lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.
Please clarify my doubt. As writing in this form changes the fact that the constant term of characteristic polynomial is $(-1)^n det(A)$, where $n$ is the order of the matrix.
Thank you |
If a body is at rest at Earth's surface, can we say that its kinetic energy $E_{\textrm{kin}}=0$, and its potential energy $E_{\textrm{pot}}=0$ also? Because its velocity $v$ and height above ground $h$ are zero.
Yes, absolutely.
Potential energy is not a measurable physical quantity. What can be measured are
differences in potential energy. So, if you compare the potential energies of a given mass at the Earth's surface and $100 \, \mathrm{m}$ above the surface, you cannot choose their difference, because that is governed by the laws of physics. However, you can freely choose to assign any point in space a specific potential energy (for the mass $m$). You could define that $m$'s potential energy at the surface is $0$, or $1234 \, \mathrm{J}$, or any other value.
Kinetic energy of a point particle of mass $m$ is defined by $\frac{mv^2}{2}$, where $v$ is the velocity. "The velocity with respect to what," you may ask. The choice is, again, yours to make; for different frames of references, you get different speeds and kinetic energies, and different notions of "rest". However, as long as $m$ is not accelerating (and $m\ne 0$), there is always a reference frame in which $v=0$ and thus $E_{\textrm{kin}}=0$.
So yes, for a body at rest (with regard to an inertial frame) at the surface (or any other point), you could very well claim that it has zero potential and kinetic energy.
It all depends on where you have set your coordinate system. If it is on the earths surface then yes but if you set it say on the sun then no.
The kinetic energy of a particle is dependent on the reference frame, so if a particle is at rest in a particular reference frame, the KE is zero. If someone moving with respect to the particle calculates the KE based on the reference frame in whichthey are at rest they will say $K \neq 0$.
If you are using $$U_g = mgh$$ for gravitational potential energy, then the reference position is arbitrary and $h$ is a small deviation from that reference position. If someone else chooses a different reference position, say 2 meters above your reference position, they will say $U_g \neq 0.$
What's important for the gravitational potential energy in most simple problems is the
change in $U_g$, not the actual value.
Always specify the reference frame being used and the reference height.
Also note that for large changes in vertical position one should use $$U_g = \frac{Gm_1m_2}{r_{12}}$$ which rightly recognizes that the gravitational potential energy rightly involves 2 masses. $r_{12}$ is the distance between the centers of the masses. For the simplified $mgh$ form, $m=m_2$ and $g=\frac{GM_{earth}}{r}$
Yes of course. A stable body has potential energy... as the definition itself says that "the energy of a body at rest is called potential energy". So as the stable body which is of course in rest will contain potential energy.
Kinetic energy is due to motion and potential energy is due to height. There is no motion and height. So,there is only inertia in the body.its kinetic and potential is zero
protected by Emilio Pisanty Aug 2 '17 at 16:10
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Precisely the title of the question.
I have encountered these terms in two areas: conjugate gradient method, and adaptive finite elements.
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Preasymptotic refers to the following concept: A priori error estimates only say that as the mesh size $h\rightarrow 0$, we have that (for example) $\|e\|\le Ch^2$. But this is an asymptotic statement: it is not an equality that holds for all $h$ but one typically only sees the quadratic decay whenever $h$ is small enough. In other words, in numerical experiments, if you plot the error as a function of the mesh size in a log-log plot, you get a nice straight line with slope 2 for small enough mesh sizes. But, for large mesh sizes, the data points are more scattered and do not follow the exact relationship. This, the range of large $h$ where the decay has not yet set in, is called the preasymptotic range (range as in "a part of the possible range of mesh sizes $h$").
The reason this preasymptotic range exists is that the exact error relationship is a nonlinear function of the mesh size. For example, if the exact error were of the form $\|e\|=c_1h^2 + c_2h^3 + c_3h^4$, then for small enough $h$, all but the first term are essentially irrelevant. However, for large $h$, the other terms may contribute.
The meanings of those terms depend on context.
Superconvergence is usually used to mean you are converging faster than the "optimal" rate, and occasionally this sort of weirdly fast convergence can be proven rigorously. One example in DG is that hyperbolic problems generally have an optimal estimate that decays like $h^{n+1/2}$ but some recent superconvergence papers shows that you can actually recover $h^{2n+1}$ in some limited cases.
Preasymptotic means: some numerical results you have demonstrated don't fit the analysis for a small parameter choice, but past some cut-off point it does. Sometimes I have seen authors use this to explain why their numerical results don't fit their analysis.
Pre-asymptoticity in conjugate gradient method: Asymptotic behavior sometimes assumes certain conditions. For example, given $h$ the meshsize or similar, and $h\to 0$:$$\|f_h - g_h \| \leq c h.$$but what if $$\|f_h - g_h \| \leq c k h^2$$if we assume $kh = O(1)$ then $f_h$ is asymptotic to $g_h$, this is asympototic analysis. If we do not assume that condition, for example $k$ may be increasing wave number, then this is pre-asymptotic analysis.
In preconditioned conjugate gradient(PCG) method, the asymptotic convergence rate for large sparse SPD matrices depends on the spectral condition number of the PCG iteration matrix. Normally we just suppose this condition number is fixed, the
pre-asymptotic analysis for conjugate gradient method also takes into account the behavior of this condition number in the analysis of the convergence rate. Superconvergence in adaptive finite elements: suppose $u$ be the true solution to the usual $H^1_0(\Omega)$ variational problem$$\int_{\Omega} \nabla u \cdot \nabla v = \int_{\Omega} f\,v$$and $u_h$ be the finite elements solution. Adaptive finite elements method(AFEM) essentially want to use an estimator $\eta$ to approximate the error $\|\nabla(u-u_h)\|$ locally and globally, such that we could perform the adaptive mesh refinement.
The superconvergence says something like this, for a
postprocessed computable quantity $\sigma$:$$\|\nabla u - \sigma\| \leq c h^{\alpha} \|\nabla u - \nabla u_h\|.$$As you can see, when the meshsize is small, $\|\nabla u - \sigma\|$ is much smaller than the true error, convergent at a much faster rate than the approximation error $\|\nabla u - \nabla u_h\|$, and this is what superconvergence means in AFEM.
Now simply by triangle inequality: $$ \Big| \|\nabla u - \sigma\| - \|\sigma- \nabla u_h\| \Big|\leq \|\nabla u - \nabla u_h\| \leq \|\nabla u - \sigma\| + \|\sigma- \nabla u_h\| $$ Since $\|\nabla u - \sigma\|$ is negligible when the meshsize is small, whenever superconvergence is true, we could use the quantity $\|\sigma- \nabla u_h\|$ to be a computable error estimator to estimate the non-computable approximation error $\|\nabla u - \nabla u_h\|$.
Superconvergent refers to the concept that sometimes convergence is faster than one would usually expect. As an example, we know that for the Laplace equation, pointwise convergence happens as $\|u-u_h\|_{L_\infty} \le Ch^2 |\log h|$ when using linear elements. But, on sufficiently regular meshes, it can be shown that at certain points (on quadrilateral cells the $2\times 2$ Gauss integration points) the error is actually of order $h^3$. In other words, at some points of the mesh, the error is significantly smaller than at all the others.
As hinted at in one of the other answers to this question, this can be exploited. If we know, for example, that the error is smaller at certain points than others, then we should interpolate a higher order polynomial through these points and we can hope to get a more accurate solution than what the finite element method provided us initially. This process is called a "recovery procedure". It can be used to get a more accurate solution, or a better approximation of the gradient. In either case, the difference between the original finite element solution and the "recovered" one (or their respective gradients) can be used to approximate the error between the original finite element solution and the exact solution. |
Intuitively, it makes little sense that the magnetic field running
through the area enclosed by the loop of wire would induce a voltage
It's just Stokes'
theorem which isn't particularly intuitive. First, recall this point relationship from Maxwell's equations:
$$\nabla \times \vec E = -\frac{\partial\vec B}{\partial t}$$
There is no area involved here; this equation relates the curl of the electric field
at a point to the time rate of change of the magnetic field at the same point.
Now, integrate to find the flux of the vector fields through a surface $\Sigma$:
$$\iint_\Sigma \left(\nabla \times \vec E\right) \cdot \vec n\,\mathrm{d}\Sigma = -\iint_\Sigma \frac{\partial\vec B}{\partial t} \cdot \vec n\,\mathrm{d}\Sigma$$
Now, apply Stokes' theorem on the left hand side:
$$\iint_\Sigma \left(\nabla \times \vec E\right) \cdot \vec n\,\mathrm{d}\Sigma = \oint_{\partial\Sigma} \vec E \cdot \mathrm{d}\vec l$$
By Stokes' theorem, the line integral of the electric field along a closed contour equals the surface integral of the curl of the electric field through a surface bounded by the contour. And so
$$ \oint_{\partial\Sigma} \vec E \cdot \mathrm{d}\vec l = -\iint_\Sigma \frac{\partial\vec B}{\partial t} \cdot \vec n\,\mathrm{d}\Sigma$$ |
$i\uparrow\uparrow t$ Setup
All of the relevant patterns here seem to involve only nonnegative inputs, so define $f:[0,\infty)\to \mathbb C$ by $f\left(t\right)=\begin{cases}i^{t} & \text{ if }t\in[0,1)\\i^{f\left(t-1\right)} & \text{ if }t>1\end{cases}$. To get a sense for the graph of $f$, we can use different colors for $[0,1)$, $[1,2)$,...
Graph
Here is a graph of $f(t)$ for $t\in[0,9)$:
It starts at $1$, then follows an indigo quarter-circle to $i$, then an orange sort of vertical sigmoid to $e^-\pi/2$, and then a green sigmoid up towards what appears to be a point on the initial quarter-circle, and spiraling inwards with colors like red, purple, brown, blue, yellow, and then pink.
For notational convenience, set $p=\dfrac{\pi}{2}$ and $g\left(t\right)=i^{t}=\left(e^{ip}\right)^{t}=e^{ipt}=\exp\left(ipt\right)$. This makes $f\left(t\right)=\begin{cases}g\left(t\right) & \text{ if }t\in[0,1)\\g\left(f\left(t-1\right)\right) & \text{ if }t>1\end{cases}$.
Do these actually connect? Yes
Note that $f$ is continuous (i.e. the graph is connected) since $1=g\left(0\right)$. For example, $f\left(2\right)=g\left(g\left(g\left(2-2\right)\right)\right)=g\left(g\left(2-1\right)\right)={\displaystyle \lim_{t\to2^{-}}}f\left(t\right)$. Therefore, there no breaks in the graph at $i$, $e^{-p}$, etc.
There is another sequence of apparent connections "in the middle", with the first being where the third arc (green) meets the original quarter-circle (indigo) at $\exp\left(ipe^{-p}\right)$. For this particular one, note that $f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The others are simply the result of applying $g$ to both sides: For example, since $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$ and there's an intersection around $f\left(4-\varepsilon\right)=g\left(f\left(3-\varepsilon\right)\right)\approx g\left(f\left(e^{-p}\right)\right)=f\left(1+e^{-p}\right)$ as well, etc.
"the branches connect perpendicularly" True.
Now we will show that the intersections in the diagram that look perpendicular all really are. To start, let's look at two particular ones.
Intersection at $i$
On $\left[0,1\right]$, we have $f\left(t\right)=g\left(t\right)$, which draws a quarter-circle in the complex plane, and it's horizontal at $i$ ($t=1$). On $\left[1,2\right]$, we have $f\left(t\right)=g\left(g\left(t-1\right)\right)$. Taking the derivative of this and evaluating at $t=1$ (to take the limit of the derivative of $f$ as $t$ approaches $1$ from above), we get $\boxed{-ip^{2}}$, so $f$ is moving vertically for $t$ just above $1$. As vertical is perpendicular to horizontal, we do have a right angle at $i$.
Intersection at $\exp\left(ipe^{-p}\right)$
Note that $f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The derivative of $g\left(g\left(g\left(t-2\right)\right)\right)$ as $t$ approaches $3$ is $p^{3}e^{-p}\sin\left(p\left(1+e^{-p}\right)\right)-i\left(p^{3}e^{-p}\cos\left(p\left(1+e^{-p}\right)\right)\right)$. Since $\cos\left(p+x\right)=-\sin x$ and $\sin\left(p+x\right)=\cos x$, this simplifies to $\boxed{p^{3}e^{-p}\left(\cos\left(pe^{-p}\right)+i\sin\left(pe^{-p}\right)\right)}$.
And the derivative of $f\left(t\right)=g\left(t\right)$ at $t=e^{-p}$ is $-p\sin\left(pe^{-p}\right)+i\left(p\cos\left(pe^{-p}\right)\right)=\boxed{p\left(-\sin\left(pe^{-p}\right)+i\cos\left(pe^{-p}\right)\right)}$.
These two complex numbers are perpendicular as vectors, so this is a right angle as well.
The other intersections
To get all of the other intersections, note that $g\left(t\right)$ has a complex derivative of $ipe^{ipt}\ne0$, so it is conformal (see, for example, this MSE question), meaning that $g$ preserves the local angles in the diagram. Because of the recursive definition of $f$, this means that the two intersections considered above propagate along the diagram. For example, since there's a perpendicular intersection around $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$, there's a perpendicular intersection around $f\left(4-\varepsilon\right)\approx f\left(1+e^{-p}\right)$.
"Are these shapes similar to one another?" No.
I'm not certain how to interpret this question, but everything that comes to mind has the answer "no". For example, the first part of the graph of f is a quarter circle, but the fourth part (red) is certainly not.
$z\uparrow\uparrow t$
I think these questions about $z\uparrow\uparrow t$, which were added significantly after the original posting, deserve their own separate Math(s) StackExchange question. That said, here are some observations that are a bit too long for a comment.
When $|z|=1$, $\theta_0=\pi/2$? Probably.
Suppose we started with $g\left(t\right)=\exp\left(i\theta t\right)$ instead of $g\left(t\right)=\exp\left(i\frac{\pi}{2}t\right)$? Then $f\left(3\right)=g\left(g\left(g\left(1\right)\right)\right)$ is at $\exp\left(i\theta e^{i\theta e^{i\theta}}\right)$ $=\exp\left(i\theta e^{i\theta\left(\cos\theta+i\sin\theta\right)}\right)$ $=\exp\left(i\theta e^{i\theta\cos\theta-\theta\sin\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}e^{i\theta\cos\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}\left(\cos\left(\theta\cos\theta\right)+i\sin\left(\theta\cos\theta\right)\right)\right)$ $=\exp\left(\theta e^{-\theta\sin\theta}\left(i\cos\left(\theta\cos\theta\right)-\sin\left(\theta\cos\theta\right)\right)\right)$. So in absolute value, this is $\exp\left(-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\right)$. To determine if $f\left(3\right)$ is in the unit disc, we need to check that $h\left(\theta\right)=-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\le0$. Certainly $h\left(0\right)=h\left(\frac{\pi}{2}\right)=0$, and $h$ is negative in between, so when $\theta$ is slightly more than $\frac{\pi}{2}$, $f\left(3\right)$ leaves the unit disc. The next greatest zero of $h$ is at $\frac{3\pi}{2}$ (and there is another at approximately $5.34$).
However, a similar analysis shows that $f\left(2\right)$ leaves the unit disc for $\theta\in\left(\pi,2\pi\right)$ (the corresponding $h\left(\theta\right)$ is just $-\theta\sin\theta$). Therefore, $f\left(2\right)$ and $f\left(3\right)$ are both in the unit disc (considering $\theta\in[0,2\pi)$) only for $\theta\in\left[0,\frac{\pi}{2}\right]$.
I would consider this evidence suggestive that that would be the "stable range". If you were interested in $\theta\in\left[-\pi,\pi\right]$ instead, note that both $h$'s are even functions, and so we get a conjectured "stable range" of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, which seems to agree with numerical experiment. |
Given a custom pdf $f_x(x)$, I'm trying to find it's transformation $f_y(y)$ where $$y=x^2$$ and $$f_x(x)=30*x^2 (1 - x)^2, 0<x<1$$ I tried to using the following commands:
y = x^2PDF[TransformedDistribution[y, x \[Distributed] ProbabilityDistribution[30*x^2 (1 - x)^2, {x, 0, 1}]]]
However, the answer comes up in terms of x, not y:
$\text{Function}\left[x, \begin{array}{cc} \{ & \begin{array}{cc} 15. \left(\sqrt{x}-1.\right)^2 \sqrt{x} & 0.\leq x\leq 1.\land 0.<\sqrt{x}<1. \\ 0. & \text{True} \\ \end{array} \\ \end{array} ,\text{Listable}\right]$
Could you please point me in the right direction? |
I) Let us just for fun generalize OP's question to $n$ spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic $(-,+,\ldots,+)$ notation with $c=1$, where $\mu,\nu\in\{0,\ldots,n-1\}$ denote spacetime indices, while $i,j \in\{1,\ldots,n-1\}$ denote spatial indices. Maxwell eqs. are the following.
Source-free Bianchi identities:$${\rm d}F~=~0 \qquad\qquad \Leftrightarrow \qquad\qquad
\sum_{\rm cycl.~\mu,\nu,\lambda} d_{\lambda} F_{\mu\nu} ~=~0, \qquad\qquad
F~:=~\frac{1}{2} F_{\mu\nu}~ {\rm d}x^{\mu} \wedge {\rm d}x^{\nu}.$$ Here $$\left(\begin{array}{c} n \cr 3\end{array}\right) {\rm~Bianchi~identities} ~=~ \left(\begin{array}{c} n-1 \cr 3\end{array}\right) {\rm~constraints}~+~ \left(\begin{array}{c} n-1 \cr 2\end{array}\right) {\rm~dynamical~eqs.} $$$$~=~ ({\rm No~magnetic~monopole~eqs.})~+~ ({\rm Faraday's~law}). $$
Maxwell eqs. with source terms:$$ d_{\mu}F^{\mu\nu}~=~-j^{\nu} .$$ Here $$n {\rm~source~eqs.}~=~1 {\rm~constraint} ~+~ (n-1) {\rm~dynamical~eqs.}$$$$~=~({\rm Gauss'~law}) ~+~ ({\rm Ampere's~law~with~displacement~term}).$$
We have used the terminology that a
dynamical eq. contains time derivatives, while a constraint does not. So the number of dynamical eqs. is
$$ \left(\begin{array}{c} n-1 \cr 2\end{array}\right)~+~(n-1)~=~ \left(\begin{array}{c} n \cr 2\end{array}\right),$$
which precisely matches
$${\rm the~number~} \left(\begin{array}{c} n \cr 2\end{array}\right) {\rm~of~}
F_{\mu\nu} {\rm~fields}$$$$~=~\left(\begin{array}{c} n-1 \cr 2\end{array}\right){~\rm magnetic~fields~} F_{ij} ~+~(n-1) {\rm~electric~fields~}F_{i0} .$$
Maxwell eqs. with source terms imply the continuity eq.
$$ d_{\nu}j^{\nu} ~=~-d_{\nu}d_{\mu}F^{\mu\nu}~=~0,\qquad\qquad F^{\mu\nu}~=~-F^{\nu\mu},$$
so one must demand that the background sources $j^{\nu}$ obey the continuity eq.
For consistency, the time derivative of each of the constraints should vanish. In the case of the no-magnetic-monopole-eqs., this follows from Faraday's law. In the case of Gauss' law, this follows from the modified Ampere's law and the continuity eq.
II) The previous section (I) made the counting in terms of the $\left(\begin{array}{c} n \cr 2\end{array}\right)$ field strengths $F_{\mu\nu}$. In terms of the $n$ gauge potentials $A_{\mu}$, the counting goes as follows. The Bianchi identities are now trivially satisfied,
$$F~=~{\rm d}A\qquad\qquad A~:=~A_{\mu}~ {\rm d}x^{\mu}. $$
There are still the $n$ Maxwell eqs. with source terms
$$ (\Box\delta^{\mu}_{\nu}-d^{\mu}d_{\nu})A^{\nu}~=~-j^{\mu} , \qquad\qquad \Box~:=~d_{\mu}d^{\mu}. $$
There is a single gauge d.o.f. because of gauge symmetry $A \to A + {\rm d}\Lambda$ and $F \to F$. If one gauge-fixes using the Lorenz gauge condition
$$d_{\mu}A^{\mu}~=~0, $$
the Maxwell eqs. become $n$ decoupled wave equations
$$ \Box A^{\mu}(x)~=~-j^{\mu}(x). $$
By a spatial Fourier transformation, these become decoupled linear second-order ODEs with constant coefficients,
$$ (d^2_t+\vec{k}^2) \hat{A}^{\mu}(t;\vec{k})~=~\hat{j}^{\mu}(t;\vec{k}) , $$
which, starting from some initial time $t_0$, may be solved for all times $t$, cf. OP's question. [One should check that the solution
$$\hat{A}^{\mu}(t;\vec{k})
~=~\int {\rm d} t^{\prime} ~G(t-t^{\prime};\vec{k})~\hat{j}^{\mu}(t^{\prime};\vec{k}), \qquad\qquad
(d^2_t+\vec{k}^2)G(t-t^{\prime};\vec{k})~=~\delta(t-t^{\prime}),$$
satisfies the Lorenz gauge condition. This follows from the continuity eq.]
III) It is interesting to derive the complete solution $\tilde{A}^{\mu}(k)$ in $k^{\nu}$-momentum space
without gauge-fixing. The Fourier-transformed Maxwell eqs. read
$$M^{\mu}{}_{\nu}~\tilde{A}^{\nu}(k)~=~\tilde{j}^{\mu}(k), \qquad\qquad M^{\mu}{}_{\nu}~:=~k^2\delta^{\mu}_{\nu} -k^{\mu}k_{\nu}. $$
To proceed one must analyze the matrix $M^{\mu}{}_{\nu}$ for fixed $k^{\lambda}$. There are three cases.
Constant mode $k^{\mu}=0$. Then the matrix $M^{\mu}{}_{\nu}=0$ vanishes identically. Maxwell eqs. are only possible to satisfy if $\tilde{j}^{\mu}(k=0)=0$ is zero. The gauge potential $\tilde{A}_{\mu}(k=0)$ is not restricted at all by Maxwell eqs., i.e., there is a full $n$-parameter solution. Massive case $k^2\neq 0$. The matrix $M^{\mu}{}_{\nu}$ is diagonalizable with eigenvalue $k^2$ (with multiplicity $n-1$), and eigenvalue $0$ (with multiplicity $1$). The latter corresponds to a pure gauge mode $\tilde{A}^{\mu}~\propto~k^{\mu}$. The complete solution is a $1$-parameter solution of the form$$\tilde{A}^{\mu}(k)
~=~\frac{\tilde{j}^{\mu}(k)}{k^2}~+~ik^{\mu}\tilde{\Lambda}(k).$$Apart from the source term, this is pure gauge. Massless case $k^2=0$ and $k^{\mu}\neq 0$. The matrix $M^{\mu}{}_{\nu}$ is not diagonalizable. There is only eigenvalue $0$ (with multiplicity $n-1$).Maxwell eqs. are only possible to satisfy if the source $\tilde{j}^{\mu}(k)=\tilde{f}(k)k^{\mu}$ is proportional to $k^{\mu}$ with some proportionality factor $\tilde{f}(k)$. In that case Maxwell eqs. become$$ -k_{\mu}\tilde{A}^{\mu}(k)~=~\tilde{f}(k). $$Let us introduce an $\eta$-dual vector$^1$ $$k^{\mu}_{\eta}~:=~(-k^0,\vec{k})\qquad {\rm for}\qquad
k^{\mu}~=~(k^0,\vec{k}).$$ Note that $$k_{\mu}~k^{\mu}_{\eta}~=~(k^0)^2+\vec{k}^2$$ is just the Euclidean distance square in $k^{\mu}$-momentum space. The complete solution is an $(n-1)$-parameter solution of the form$$\tilde{A}^{\mu}(k)
~=~-\frac{k^{\mu}_{\eta}}{k_{\nu}~k^{\nu}_{\eta}}\tilde{f}(k)
~+~ik^{\mu}\tilde{\Lambda}(k)~+~\tilde{A}^{\mu}_{T}(k).$$The term proportional to $k_{\mu}$ is pure gauge. Here $\tilde{A}^{\mu}_{T}(k)$ denote $n-2$ transversal modes, $$k_{\mu}~\tilde{A}^{\mu}_{T}(k)~=~0, \qquad\qquad
k_{\mu}^{\eta}~\tilde{A}^{\mu}_{T}(k)~=~0. $$ The $n-2$ transversal modes $\tilde{A}^{\mu}_{T}$ are the only propagating physical d.o.f. (electromagnetic waves, photon field).
--
$^1$ Longitudinal and timelike polarizations are in the massless case proportional to $k^{\mu}\pm k^{\mu}_{\eta}$, respectively. |
Symbols:Greek/Pi/Probability Generating Function
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$\map {\Pi_X} s$
Let $p_X$ be the probability mass function for $X$.
The probability generating function for $X$, denoted $\map {\Pi_X} s$, is the formal power series defined by: $\displaystyle \map {\Pi_X} s := \sum_{n \mathop = 0}^\infty \map {p_X} n s^n \in \R \left[\left[{s}\right]\right]$ The $\LaTeX$ code for \(\map {\Pi_X} s\) is
\map {\Pi_X} s . |
I am currently looking into post-processing of simulation data using Paraview. I would like to compute certain integrals of field quantities. As an example, consider the following surface integral of an electric field in a simple situation:
$$\iint_{S}^{} \vec{E}\cdot\vec{\omega}=\Phi $$
The surface $S$ is shown in the following picture:
I can define a surface in ParaView and I can compute the surface normals $\vec{w}$ (white arrows in he picture). In order to compute the integral in question, I guess I would have to use a calculator to compute $\vec{E}\cdot\vec{\omega}$ and then use the "Integrate Variables" Filter.
My problem is, that I can only add the calculator to the "case0001.vtu tree" or the "plane tree" in the pipeline browser such that it's impossible to select both the vectors "electric fields" and "normals" in the calculator to compute the integrand $\vec{E}\cdot\vec{\omega}$.
Does anyone know how I could use a calculator to compute the integrand in Paraview? |
This question relates to the topic of Persistent Homology, a branch of (applied) algebraic topology.
In the paper (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.116.2471&rep=rep1&type=pdf, page 6), the "$p$-persistent $k$th homology group" is defined as: $$H_k^{i,p}=Z_k^i/(B_k^{i+p}\cap Z_k^i)$$
While I understand the notation, and also understand the alternative interpretation as the homomorphism $\eta_k^{i,p}: H_k^i\to H_k^{i+p}$ that maps a homology class into the one that contains it, I don't really grasp the intuitive meaning behind it.
For homology group, for example, intuitively we understand that the $n$th homology group counts the number of $n$-dimensional "holes". What then, does the persistent group actually mean? (After reading many papers, I roughly get the idea that it measures the holes that "persist" but my understanding remains vague)
Q2) For the next question, what does the persistent module (defined by Carlsson in page 6 of the same paper) actually mean and how is it related to the persistent group? Again, I vaguely get the idea that the "persistent homology of a filtered complex is standard homology of a graded module over a polynomial ring", but fail to grasp why is this important/useful.
Thanks for any explanation! (Any reference will be greatly appreciated too.) |
HP 17bII+ Silver solver
09-14-2018, 09:20 PM
Post: #1
HP 17bII+ Silver solver
I'm very satisfied with my HP 17bII+ Silver, which I find very powerful, but also nice and not looking too complicated (no [f] and [g] functions on keys like the HP 12c or HP 35s that I used to use at work every day, but a really complete calculator except for trigs and complex numbers calculations).
So the 17BII+ is my new everyday calculator. I don't come back to arguments where a good (HP) calculator is a perfect complement or subsitute to Excel.
I chose the 17BII+ after having carefully studied the programs I use in my day to day work:
- price and costs calculations: can be modeled in the solver, with 4 or 5 equations and share vars
- margin calculations: built-in functions
- time value of money: built-in functions
- time functions: built-in functions
For the few moments I need trigs or complex calculations, I always have Free42 or a real 35s / 15c not really far from me.
After having rebuilt my work environment in the (so powerful) solver, I started to study the behavior of the solver, looking at loops (Σ function), conditional branchings (IF function), menu selection (S function), and the Get and Let functions (G(), L()).
I googled a lot and found an interesting pdf file about the Solvers of the 19B, 17B, 17BII and - according to the author, the 17BII+ Silver.
The file is here : http://www.mh-aerotools.de/hp/documents/...ET-LET.pdf
I tried a few equations in the "Using New and Old Values" chapter, pages 4 and following.
There I found lots of differences with my actual 17BII+ Silver, which I would like to share here with you.
For instance, the following equation found page 4:
Code:
In the next example found page 5:
Code:
Then the equation :
Code:
I don't understand the first 2 cases. In the first one, A should be set to -B, not B, if not using the "old" value of A. In the second one, the solver does not use the "old" value, but it does not also solve the equation, as there is no defined solution.
In the last case, I understand that the equation is evaluated twice before finding an answer. So the old value is used there, but not the way I could expect.
I finally found one - and only one - way to use iterations in the solver, with the equation :
Code:
Note that neither A=G(A)+1 or A=1+G(A) or G(A)+2=A works.
I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Comments are welcomed.
Regards,
Thibault
09-14-2018, 10:14 PM (This post was last modified: 09-14-2018 10:25 PM by rprosperi.)
Post: #2
RE: HP 17bII+ Silver solver
The 17BII+ (Silver Edition) is an excellent machine, in fact it has the best keyboard of any machine made today by HP, but the solver does have a bug, and there are also a few other smaller issues making the solver slightly inferior to the 17B/17BII/19B/19BII/27S version.
See these 3 articles for details:
http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=242551
http://www.hpmuseum.org/forum/thread-657...l#pid58685
http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=134189
Overall these are not dramatic issues, and once you understand the solver bug, you likely can create equations that can avoid the issue.
I have most HP machines but the 17BII and 17BII+ are the ones I use most often for real work (vs. playing, exploring or following along interesting threads here).
Edit: added 3rd link
--Bob Prosperi
09-15-2018, 12:05 AM
Post: #3
RE: HP 17bII+ Silver solver
(09-14-2018 09:20 PM)pinkman Wrote: I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Thibault, when Kinpo built the 17bii+ calculator years ago (both the gold one and the silver one), they basically goofed the solver implementation. This has been discussed at length in the HPMuseum forum. The solvers on the 17b and 17bii work fine, just as you would expect them to. If you plan on making significant use of the solver and its incredible capabilities, forget the + and get an original 17b or 17bii. You won't be sorry.
Also, get the manual (it's on the Museum DVD) Technical Applications for the HP-27s and HP-19b. It applies to the 17b as well.
The Sigma function also works fine on the 17b and 17bii.
09-15-2018, 04:36 AM
Post: #4
RE: HP 17bII+ Silver solver
Thanks to both of you for the details, links and advice.
I've read the threads carefully, I did not find them by myself first. It's the end of the night now, I'll try to make few testing later.
09-15-2018, 11:55 PM (This post was last modified: 09-15-2018 11:58 PM by rprosperi.)
Post: #5
RE: HP 17bII+ Silver solver
If you want to really explore the capabilities of the awesome Pioneer Solver, and confidently try to push it without worrying about using L() this way or that, I agree with Don, buy a 17BII and use that for the Solver stuff, but continue to use the 17BII+ for every day stuff.
Here's a very nice 17BII for only $25 (shipping included, in US) and you can even find them cheaper if you're willing to wait:
https://www.ebay.com/itm/HP-17Bll-Financ...3252533550
The 17BII is bug-free for solver use, while the 17BII+ has a much better LCD, readable in a wider range of lighting & use conditions.
In case you haven't seen this yet, here's an example of what can be done with the solver:
http://www.hpmuseum.org/forum/thread-2630.html
--Bob Prosperi
09-16-2018, 09:59 PM
Post: #6
RE: HP 17bII+ Silver solver
Well I'll try to find one, even if I'm not in the US.
I also want to continue using my actual 17bII+ at work, as the solver is powerful enough for me (for the moment), and it looks really good.
Don and you Bob have done a lot to help understand what the solver can do, that's pretty good stuff.
Regards,
Thibault
09-16-2018, 10:23 PM
Post: #7
RE: HP 17bII+ Silver solver
Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
Of all the various tools built-in to the various calculator models I've explored, the Pioneer Solver is easily the one that most exceeds it's initial apparent capability. This awesome tool must have been incredibly well-tested by the QA team. The fact that the sheer size (and audacity!) of Gerson's Trig formulas still return amazingly accurate results says more about the underlying design and code than any comments I could add.
Enjoy exploring it, and when you've mastered it (or at least tamed it a bit), come back here and share some interesting Solver formulas. There are numerous folks here that enjoy entering the formulas and running some test cases. (well, I suppose "enjoy" is not really the right word about entering the formulas - I guess feel good about accomplishing it successfully is more accurate).
When I see some of these long Solver equations, it reminds of TECO commands back in the PDP-11 days.
--Bob Prosperi
09-16-2018, 11:05 PM
Post: #8
RE: HP 17bII+ Silver solver
If you feel like entering long formulas you can solve the 8-queens problem.
):0)
Cheers
Thomas
09-17-2018, 06:30 AM (This post was last modified: 09-18-2018 09:47 AM by Don Shepherd.)
Post: #9
RE: HP 17bII+ Silver solver
How about a nifty, elegant, simple number base conversion Solver equation for the 17b/17bii, courtesy Thomas Klemm:
BC:ANS=N+(FROM-TO)\(\times \Sigma\)(I:0:LOG(N)\(\div\)LOG(TO):1:L(N:IDIV(N:TO))\(\times\)FROM^I)
Note: either FROM or TO must be 10 unless you are doing HEX conversions
09-17-2018, 07:49 AM
Post: #10
RE: HP 17bII+ Silver solver
09-17-2018, 09:17 AM (This post was last modified: 09-17-2018 10:28 AM by Don Shepherd.)
Post: #11
RE: HP 17bII+ Silver solver
(09-17-2018 07:49 AM)Thomas Klemm Wrote:
Wow, I didn't realize that Thomas. Stunning.
Don
09-17-2018, 10:24 AM
Post: #12
RE: HP 17bII+ Silver solver
As has already been written, unfortunately the solver in this calculator is flawed.
Part of the problems comes from the fact that it evaluates the equation twice which leads to incrementing by two instead of one etc. But there seem to be more quirks.
This is too bad, as the solver is a very capable tool (with G() and L()), while still easy and versatile to use (with the menu buttons).
Accomplishing something similar (solve for one variable today, for another tomorrow) with modern tools like Excel is more complicated.
Martin
09-17-2018, 12:48 PM
Post: #13
RE: HP 17bII+ Silver solver
(09-16-2018 10:23 PM)rprosperi Wrote: Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
An obvious correction is in order here:
Don, Gerson and Thomas Klemm are the real Pioneer Solver Masters...
No disrespect intended by the omission. A review of past posts of significant solver formulas quickly reveals just how often all three of these three guys were the authors.
--Bob Prosperi
09-28-2018, 01:33 PM
Post: #14
RE: HP 17bII+ Silver solver
Be sure I have great respect of everyone mentioned above
Thanks for the advices, my new old 17bii is now in my hands ($29 on French TAS), I'm just waiting for the next rainy Sunday for pushing the solver to it's limits (mmh, to MY limits I guess).
09-28-2018, 07:47 PM (This post was last modified: 09-28-2018 07:48 PM by Jlouis.)
Post: #15
RE: HP 17bII+ Silver solver
Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
TIA
09-29-2018, 01:37 AM
Post: #16
RE: HP 17bII+ Silver solver
(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Check out the document by Martin Hepperle in this thread which lists the functions available for each calculator. The document is an excellent introduction to the Solver application.
Solver GET-LET
~Mark
Who decides?
09-29-2018, 02:25 AM
Post: #17
RE: HP 17bII+ Silver solver
(09-29-2018 01:37 AM)mfleming Wrote:(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Thanks mfleming, I should have read the beginning of the manual. It is the same solver for the calculators of my question. I prefer to use the 19bII due the dedicated alpha keyboard, but the 27s is a pleasure to use too.
Cheers
09-29-2018, 11:13 AM
Post: #18
RE: HP 17bII+ Silver solver
09-29-2018, 11:53 AM (This post was last modified: 09-29-2018 05:12 PM by Don Shepherd.)
Post: #19
RE: HP 17bII+ Silver solver
(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?Here is a 3-page cross-reference I created a few years ago of Solver functions present in the 19bii, 27s, and 17bii.
solver 1.PDF (Size: 1.08 MB / Downloads: 33)
solver 2.PDF (Size: 968.77 KB / Downloads: 21)
solver 3.PDF (Size: 494.75 KB / Downloads: 19)
09-29-2018, 01:36 PM (This post was last modified: 09-29-2018 01:49 PM by Jlouis.)
Post: #20
RE: HP 17bII+ Silver solver
(09-29-2018 11:53 AM)Don Shepherd Wrote:
Really thanks Dom for taking time to make these documents.
This is what makes MoHC a fantastic community.
Cheers
JL
Edited: Looks like the 19BII is a little more complete than the 27s and that the 17BII is the weaker one.
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