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Tagged: linear algebra Problem 146
Let $A$ be an $n \times n$ nilpotent matrix, that is, $A^m=O$ for some positive integer $m$, where $O$ is the $n \times n$ zero matrix.
Prove that $A$ is a singular matrix and also prove that $I-A, I+A$ are both nonsingular matrices, where $I$ is the $n\times n$ identity matrix.Add to solve later
Problem 143
Let $V$ be the vector space over $\R$ consisting of all $n\times n$ real matrices for some fixed integer $n$. Prove or disprove that the following subsets of $V$ are subspaces of $V$.
(a) The set $S$ consisting of all $n\times n$ symmetric matrices. (b) The set $T$ consisting of all $n \times n$ skew-symmetric matrices.
Add to solve later
(c) The set $U$ consisting of all $n\times n$ nonsingular matrices. Problem 142
Let $T:\R^2 \to \R^3$ be a linear transformation such that $T(\mathbf{e}_1)=\mathbf{u}_1$ and $T(\mathbf{e}_2)=\mathbf{u}_2$, where $\mathbf{e}_1=\begin{bmatrix}
1 \\ 0 \end{bmatrix}, \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ are unit vectors of $\R^2$ and \[\mathbf{u}_1= \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{u}_2=\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}.\] Then find $T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right)$. Problem 141
Let $V$ be a vector space over a field $K$. Let $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$ be linearly independent vectors in $V$. Let $U$ be the subspace of $V$ spanned by these vectors, that is, $U=\Span \{\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n\}$.
Let $\mathbf{u}_{n+1}\in V$. Show that $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly independent if and only if $\mathbf{u}_{n+1} \not \in U$. Problem 131
Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
\[V:=\left\{ \quad\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \in \R^4 \quad \middle| \quad x_1-x_2+x_3-x_4=0 \quad\right\}.\] Find a basis of the subspace $V$ and its dimension. Problem 125
Let $S$ be the following subset of the 3-dimensional vector space $\R^3$.
\[S=\left\{ \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, x_1, x_2, x_3 \in \Z \right\}, \] where $\Z$ is the set of all integers. Determine whether $S$ is a subspace of $\R^3$. Problem 121
Let $A$ be an $m \times n$ real matrix. Then the
null space $\calN(A)$ of $A$ is defined by \[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\] That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.
Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.) Read solution Problem 120
Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r$ are linearly dependent $n$-dimensional real vectors.
For any vector $\mathbf{v}_{r+1} \in \R^n$, determine whether the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r, \mathbf{v}_{r+1}$ are linearly independent or linearly dependent.Add to solve later
Problem 119
Let $\mathbf{a}$ and $\mathbf{b}$ be fixed vectors in $\R^3$, and let $W$ be the subset of $\R^3$ defined by
\[W=\{\mathbf{x}\in \R^3 \mid \mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0\}.\]
Prove that the subset $W$ is a subspace of $\R^3$.
Read solution |
In
classical thermodynamics the change in internal energy is defined by the first law as$$\Delta U = q + w$$so that only the difference in $U$ is known; $q$ is the heat absorbed by the 'system' and $w$ the work done on the system.
For example in a closed system (no exchange of matter with environment) we can write for a reversible change\begin{align} \mathrm{d}q &= T\mathrm{d}S \\ \mathrm{d}w &= -p\mathrm{d}V \end{align}and then if the only form of work on a gas is volume change$$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V$$and this is the fundamental equation for a closed system. Thus only difference in internal energy are measurable from thermodynamics, and this follows from the first law. (Even if you integrate this equation from say state $a$ to $b$ the result will be $U(b)-U(a)=\ddot {}$ in other words $\Delta U$.)
Thermodynamics was developed before the nature of matter was known, i.e. it does not depends on matter being formed of atoms and molecules. However, if we use additional knowledge about the nature of molecules then the internal energy (and entropy) can be determined from statistical mechanics.
The internal energy ($U$ not $\Delta U$) of a perfect monoatomic gas is the ensemble average and is $$U=(3/2)NkT$$or in general $U=(N/Z)\Sigma_ j\exp(−\epsilon_j/(kT))$ where $Z$ is the partition function, $k$ Boltzmann constant, $\epsilon_j$ energy of level $j$, and $T$ temperature and $N$ Avogadro's number. The absolute value of the entropy $S$ (for a perfect monoatomic gas) can also be determined and is given by the Sakur-Tetrode equation. |
L # 1
Show that
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Last edited by krassi_holmz (2006-03-09 02:44:53)
IPBLE: Increasing Performance By Lowering Expectations.
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 2
If
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Let
log x = x' log y = y' log z = z'. Then:
x'+y'+z'=0.
Rewriting in terms of x' gives:
IPBLE: Increasing Performance By Lowering Expectations.
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Well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 3
If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)?
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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loga=2logx+3logy
b=logx-logy loga+3b=5logx loga-2b=3logy+2logy=5logy logx/logy=(loga+3b)/(loga-2b). Last edited by krassi_holmz (2006-03-10 20:06:29)
IPBLE: Increasing Performance By Lowering Expectations.
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Very well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 4
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You are not supposed to use a calculator or log tables for L # 4. Try again!
Last edited by JaneFairfax (2009-01-04 23:40:20)
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No, I didn't
I remember
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You still used a calculator / log table in the past to get those figures (or someone else did and showed them to you). I say again:
no calculators or log tables to be used (directly or indirectly) at all!! Last edited by JaneFairfax (2009-01-06 00:30:04)
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log a = 2log x + 3log y
b = log x log y
log a + 3 b = 5log x
loga - 2b = 3logy + 2logy = 5logy
logx / logy = (loga+3b) / (loga-2b)
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Hi ganesh
for L # 1 since log(a)= 1 / log(b), log(a)=1 b a a we have 1/log(abc)+1/log(abc)+1/log(abc)= a b c log(a)+log(b)+log(c)= log(abc)=1 abc abc abc abc Best Regards Riad Zaidan
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Hi ganesh
for L # 2 I think that the following proof is easier: Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t So Log(x)=t(b-c),Log(y)=t(c-a) , Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0 So Log(xyz)=0 so xyz=1 Q.E.D Best Regards Riad Zaidan
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Gentleman,
Thanks for the proofs.
Regards.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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log_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64) - \log_2(4) = 6 - 2 = 4, \,
log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \,
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L # 4
I don't want a method that will rely on defining certain functions, taking derivatives,
noting concavity, etc.
Change of base:
Each side is positive, and multiplying by the positive denominator
keeps whatever direction of the alleged inequality the same direction:
On the right-hand side, the first factor is equal to a positive number less than 1,
while the second factor is equal to a positive number greater than 1. These facts are by inspection combined with the nature of exponents/logarithms.
Because of (log A)B = B(log A) = log(A^B), I may turn this into:
I need to show that
Then
Then 1 (on the left-hand side) will be greater than the value on the
right-hand side, and the truth of the original inequality will be established.
I want to show
Raise a base of 3 to each side:
Each side is positive, and I can square each side:
-----------------------------------------------------------------------------------
Then I want to show that when 2 is raised to a number equal to
(or less than) 1.5, then it is less than 3.
Each side is positive, and I can square each side:
Last edited by reconsideryouranswer (2011-05-27 20:05:01)
Signature line:
I wish a had a more interesting signature line.
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Hi reconsideryouranswer,
This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Hi all,
I saw this post today and saw the probs on log. Well, they are not bad, they are good. But you can also try these problems here by me (Credit: to a book):
http://www.mathisfunforum.com/viewtopic … 93#p399193
Practice makes a man perfect.
There is no substitute to hard work All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
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JaneFairfax, here is a basic proof of L4:
For all real a > 1, y = a^x is a strictly increasing function.
log(base 2)3 versus log(base 3)5
2*log(base 2)3 versus 2*log(base 3)5
log(base 2)9 versus log(base 3)25
2^3 = 8 < 9
2^(> 3) = 9
3^3 = 27 < 25
3^(< 3) = 25
So, the left-hand side is greater than the right-hand side, because
Its logarithm is a larger number.
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The first thing you should realize to understand it is that the $H(Y)$ notation corresponds to the calculation of the information entropy for the states in $Y$, such that $H(Y) = -\sum\limits_{y \in Y} p(y) \log[p(y)]$. With that in mind, we can interpret the expression and equate the terms. You are correct that the three parts correspond to the probability of $0$, $e$, and $1$, where $e$ is the state of erased information. Expanding it we get:
$H((1-\pi)(1-\alpha), \alpha, \pi(1-\alpha)) $$\quad= -(1-\pi)(1-\alpha)\log[(1-\pi)(1-\alpha)] - \alpha\log(\alpha) - \pi(1-\alpha)\log[\pi(1-\alpha)]$
This gives us the full expression for calculating the entropy of the system composed of those three states. Next, as we usually do when proving an equality, we cheat and look at the answer to see what form we need it in. Your last term has all the $\pi$ terms clumped onto one side, and all the $\alpha$ terms clumped onto another (except for where they are also a coefficient). Thus we will organize the log terms in this manner. First we use the log rules to expand the products:
$ \quad= -(1-\pi)(1-\alpha)\log(1-\pi) - (1-\pi)(1-\alpha)\log(1-\alpha) $$\quad\phantom{=} - \alpha\log(\alpha) - \pi(1-\alpha)\log(\pi) - \pi(1-\alpha)\log(1-\alpha)$
Then we sort all the log terms in the order of the final expression:
$ \quad = -\alpha\log(\alpha) - (1-\pi)(1-\alpha)\log(1-\alpha) - \pi(1-\alpha)\log(1-\alpha) $$ \quad\phantom{=} - \pi(1-\alpha)\log(\pi) -(1-\pi)(1-\alpha)\log(1-\pi)$
Then we simplify and factor:
$ \quad = -\alpha\log(\alpha) - (1-\alpha)\log(1-\alpha) + (1-\alpha) \cdot [ -\pi\log(\pi) - (1-\pi)\log(1-\pi) ]$
And this is equivalent to:
$$H(\alpha, 1-\alpha) + (1-\alpha) H(\pi, 1-\pi)$$
This can be interpreted as the left part being the entropy from erasing (probability $\alpha$) or not erasing, and the right part being, in the event it was not erased, the entropy from it being a 1 (probability $\pi$) or a 0. It should also make some intuitive sense that this will be equal to the entropy from the distribution of final states of $0$, $e$ (erased), and $1$. |
(Statistics|Probability|Machine Learning|Data Mining|Data and Knowledge Discovery|Pattern Recognition|Data Science|Data Analysis) Table of Contents 1 - What Is
The terms pattern recognition, machine learning, data mining and knowledge discovery in databases (KDD) are hard to separate, as they largely overlap in their scope.
Machine Learning is the common term for supervised learning methods and originates from artificial intelligence, whereas KDD and data mining have a larger focus on unsupervised methods and stronger connection to business use. Pattern recognition has its origins in engineering, and the term is popular in the context of computer vision.
?? TODO: Splits between proba/stat en machine leraning ?
2 - Articles Related It used to be that we [computer scientists] studied discrete math, now we study probability and statistics I keep saying that the sexy job in the next 10 years will be statisticians.“ In the article, there's a picture of Carrie Grimes, who was a graduate from Stanford Statistics. She was one of the first statisticians hired at Google. My personal statistical paradigm I use statistical models, which are sets of equations involving random variables, with associated distributional assumptions, devised in the context of a question and a body of data concerning some phenomenon, with which tentative answers can be derived, along with measures of uncertainty concerning these answers.
<MATH> \begin{array}{rcl} questions + data & \rightarrow & answers + \text{measures of uncertainty} \\ & \underbrace{}_{\displaystyle \text{equations, distributions}} & \\ \end{array} </MATH> |
I hope this question is not too basic, but I have no experience with partial differential equations and would like to ask for some hints on how to solve the following problems:
The visual idea is to describe the diffusion of some dilute chemical around a spherical sink or a sink at some point.
It would be nice to obtain a time evolution when starting with a uniform density (this is only possible in problems 1) and 5)), but I would already be satisfied with a "nice" steady-state solution.
1) The simplest case would be a diffusion equation in one dimension with a sink at $x=0$, i.e. $$\frac{\partial{\rho}(x)} {\partial{t}} = - \nabla \cdot J(x), \qquad x \in \mathbb{R}$$ with $J(x) = -D \nabla \rho(x)$ for $x \neq 0$ and $J(x) = -D \nabla \rho(x) -k\rho(x)$ for $x = 0$ with $k$ some large depletion constant.
Moreover, I would like to require that $\rho(x) \rightarrow c$ for $|x| \rightarrow \infty$ for some constant $c >0$.
The initial conditions are not so important, let's say $\rho(x;t=0) = c_0$ for all $x \in \mathbb{R}$. Actually, I am not as much interested in the time evolution as in finding a nice steady-state solution, which does not diverge for $x \rightarrow \infty$. Thus, some more appropriate initial conditions could be chosen.
2) Alternatively, One could just require the boundary condition $\rho(x=0;t) = 0$ for all $t$ in the above situation. Then the initial condition needs to be adjusted accordingly.
3) The problem in 3D is the same: to solve the diffusion equation with $x \in \mathbb{R}^3$ and boundary conditions $\rho(x) = 0$ for $x = 0$ and $\rho(x) \rightarrow c$ for $|x| \rightarrow \infty$ with some $c >0$. I guess with spherically symmetric initial conditions, this case is completely equivalent to 2).
4) Now I would like to have a real sphere in $\mathbb{R}^3$ of radius $r >0$, i.e. the boundary conditions $\rho(x) = 0$ for $|x| < r$ and $\rho(x) \rightarrow c$ for $|x| \rightarrow \infty$ with some $c >0$.
5) Maybe 4) is simpler when formulated with a finite depletion rate $k$ as in 1) instead of the boundary condition around $0$. Physically, this is very similar anyway.
Any literature suggestions are also highly appreciated. |
79 0
Hello there,
In scalar algebra, I find solving for variables a useful tool. Say ohms law, I want to find ##R## so:
$$ U=RI \iff R = \frac{U}{I} $$
Can I do something analogous in vector equations? I.e. May I solve for ##\vec{\omega}## in equations using cross or dot products?
$$ \vec{v} = \vec{\omega} \times \vec{r} \iff \vec{\omega} = ? $$
or:
$$ \vec{\alpha} \cdot \vec{\beta} = \gamma \iff \vec{\beta} = ? $$
It would be fantastic if I could solve for vectors in some way.
Hope you are able to help.
Thank you for your time!
Kind regards, Marius
In scalar algebra, I find solving for variables a useful tool. Say ohms law, I want to find ##R## so:
$$
U=RI \iff R = \frac{U}{I}
$$
Can I do something analogous in vector equations? I.e. May I solve for ##\vec{\omega}## in equations using cross or dot products?
$$
\vec{v} = \vec{\omega} \times \vec{r} \iff \vec{\omega} = ?
$$
or:
$$
\vec{\alpha} \cdot \vec{\beta} = \gamma \iff \vec{\beta} = ?
$$
It would be fantastic if I could solve for vectors in some way.
Hope you are able to help.
Thank you for your time!
Kind regards,
Marius |
"tan is sine / cosine" means that to find the value of tangent you find the values of sine and of cosine
at that same value, then you divide the sine and cosine. So, you use$$\tan \frac{11\pi}6 = \frac{{\sin \frac{{11\pi }}{6}}}{{\cos \frac{{11\pi }}{6}}}$$
Similarly,$$\sec \frac{{ - 3\pi }}{4} = \frac{1}{{\cos \frac{{ - 3\pi }}{4}}}$$and $$\cot\frac{{ - 5\pi }}{3} = \frac{{\cos \frac{{ - 5\pi }}{3}}}{{\sin \frac{{ - 5\pi }}{3}}}$$
Since you say you "know the unit circle back and forth" you should be able to find each reference angle, find the sine and cosine for that reference angle, calculate your desired trig ratio, then give that ratio the proper sign given the quadrant of the original angle. Ask if you did not understand that brief summary. Good luck! |
When correlation function has branch cut in momentum space,how to find correlation in coordinate space? For example$$ \tilde {G}(\omega) = \frac{2i}{\omega+(\omega^2-\nu^2)^{1/2}}$$How to get the $G(t)$ usng Fourier transformation ?
t>0 HERE. This problem is from matrix model of Iizuka and Polchinski. They discuss the propagator in the model and find that the propagator $G(t)$ has power law decay behavior if there is branch cut in $\tilde{G}(\omega)$. If there is a pole in the lower half plane for $\tilde{G}(\omega)$, there is an exponential decay in $G(t)$.
When correlation function has branch cut in momentum space,how to find correlation in coordinate space? For example$$ \tilde {G}(\omega) = \frac{2i}{\omega+(\omega^2-\nu^2)^{1/2}}$$How to get the $G(t)$ usng Fourier transformation ?
Cutology
The math books aren't good for this, you need seat-of-the-pants intuition. The quick and dirty physicist answer is that a branch cut is best thought of as a continuum of poles densely spread over a line. You reproduce branch cuts by integrating poles spread over an interval, for example with a constant "residue density" (this is not the standard term for it, see below)
$$ \log\left({x-a\over x-b}\right) = \int_a^b {1\over x-u} du$$
And this is a continuous density of poles with unit residue between a and b, just as you can see from the right hand side, and a function with a cut between a and b on the left hand side. If you make the residue density between a and b some function $\rho(u)$
$$ f(x) = \int_a^b {\rho(u)\over x-u} du $$
you get different functions, but always with a cut inside [a,b] wherever $\rho(u)$ is nonzero. The residue density is not usually called the residue density--- it's called the "cut discontinuity", because if you consider the value of the function f defined by integral just above and just below the real axis somewhere in the interval [a,b], and take the difference between the two values, you get 2\pi i times $\rho$ as the difference. This is because you can deform the two integrals in opposite sense into a small circle, or, if you like, because of the Cauchy-distribution representation of the delta function as:
$$\delta(x) = {1\over 2\pi i} \left({1\over x-i\epsilon} - {1\over x+i\epsilon}\right)$$
Which you can work out explicitly.
To show you how it works in detail, say you want reproduce the square-root function's branch cut, running along the negative real axis, you look for the jump-discontinuity in $\sqrt{x}$ along the negative axis. It goes from $i\sqrt{|x|}$ to $-i\sqrt{x}$, the jump discontinuity is the square root of the distance from the origin. So you write
$$ f(x) = {1\over \pi} \int_{-\infty}^0 {\sqrt{|u|}\over x-u} du $$
And this should reproduce the square-root function. Except this is nonsense, since the integral is divergent! This is still morally true, however, by doing the following manipulation: change u to -u, and split the integral as follows:
$$ {2\over \pi} \int_0^\infty {1\over2\sqrt{u}} {u\over x+u} du $$
Then split ${u\over x+u}$ into $1 - {x\over x+u}$. Discard the 1, because this is an (infinite) x independent constant, and change variables to $\alpha=\sqrt{u}$, and you get
$$ {2\over \pi} \int_0^\infty {x\over x+\alpha^2} d\alpha $$
Now you can see that this evaluates to $\sqrt{x}$ by rescaling by x. So it works, but you have to watch out for divergences.
Polology
When there is a single pole on the negative imaginary axis at position $-ai$, when you integrate with $e^{-i\omega t}$, for t positive, the integration contour can be moved down to the $\mathrm{Im}(\omega)=-ia$ line, just by sliding it down (there are no singularities along the way) and the integral along the new contour is a sum of contributions each with an exponential decay $e^{-at}$. So the dominant exponential decay at infinity is the pole closest to the real axis.
If the pole is the only singularity and with appropriate stuff at negative imaginary infinity, you can do the integral explicitly, but the contour moving is simple and doesn't depend on anything, and shows the exponential decay immediately. Usually you have many singularities, and you only want the leading behavior at infinity.
If there is a pole very close to the real axis with position $a-i\epsilon$, the fourier transform isn't exponentially decaying (or rather it is decaying with a rate $\epsilon$), but oscillating. There can't be any singularities in the positive imaginary $\omega$ halfplane, as this would lead to a blow up in the future. This means that you should imagine all singularities shifted infinitesimally to the negative imaginary half-plane, just like this.
When there is a continuous density of poles, you get a continuous density of decay rates, and if they accumulate arbitrarily close to the real axis, you get a continuous superposition of different decay rates that can reproduces a power law, if the pole-density near the real axis is the appropriate power.
Your thing
In you case, you have a cut starting at $\omega=\nu$ on the real axis, and the cut runs off to infinity somehow along the negative imaginary axis. The discontinuity in the imaginary part along the cut can be found by writing the thing as
$$ {\omega - \sqrt{\omega^2 - \nu^2}}\over \nu^2 $$
and you can see there is a square-root pole-density near $\omega=\nu$. Each pole gives a decy of $e^{-at}$ in real time, so you superpose the decays:
$$ \int_0^\infty \sqrt{a} e^{-at} da \approx t^{-\frac{3}{2}} $$
This is all on top of a general oscillation of $e^{-i\nu t}$ which you get from the fact that this stuff is happening near $\omega=\nu$. This sort of stuff is very useful, and is secret lore for some reason. |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Something I learned recently which really helped me in understanding exchange is that the case of exchanging two electrons is really a special case of enforcing symmetry (or anti-symmetry) in which the permutations of labels in cycles out to length $N$ are used. $N$ is the number of particles.
Just to write it out, the two-electron exchange integral is defined as, $$K_{ab}=\int d\textbf{r}_1d\textbf{r}_2\psi^*_a(\textbf{r}_1)\psi_b(\textbf{r}_1)r_{12}^{-1}\psi_b^*(\textbf{r}_2)\psi_a(\textbf{r}_2)$$ where everything means what it normally means and I took this integral and notation straight out of Szabo and Ostlund.
Now, this is clearly only a permutation of two electrons, and introduces the only correlation present in Hartree-Fock by allowing for exchange of spin-paired electrons.
Shifting gears, I've been reading
Quantum Mechanics and Path Integrals by Richard Feynman and Albert Hibbs. In the chapter on Statistical Mechanics, they talk about liquid helium, which has a peculiar transition around $2K$ in which the heat capacity begins to increase. Feynman was the first to explain that this is due to the exchange interaction, and at this temperature is when permutations of the atom labels larger than just one (i.e. two particle exchange) become important. The following statements from the book are relevant to electrons (as opposed to helium because $\ce{^4He}$ is a boson!).
In a different context, the calculation of partition functions, the authors state:
If we were dealing with Fermi particles, e.g., the isotope of helium which has three nucleons, we would have to include an extra factor of $\pm1$, positive for even permutations and negative for odd permutations. There would also be some extra features which depend upon the spin of the atom in our result.
And more directly about electrons,
Consider the behavior of electrons in a solid metal. The mass of the electron is so much smaller than that of a molecule that the critical temperature is much higher. At room temperature, electrons in a metal are described accurately only by equations which include the exchange effects of these cyclic permutations. From this point of view, room temperature is very cold for electrons.
So, hopefully we have now reached a place where it is clear why I am a bit confused. Namely, to my knowledge, we only ever consider the exchange of two electrons of the same spin in our electronic structure methods. First, is this actually true? That is, are there methods which implement (in principle) $N$-electron integrals to describe all permutations of the labels?
Second, is the comparison I am making even strictly valid? That is, in the case of a partition function, which is what Feynman is talking about, permutations of even length tend to cause the electrons to stay further away from each other. This is the behavior we are used to with exchange. In contrast, for permutations of odd numbers, the contribution to the partition function picks up a minus sign. So, the total partition function for a system of fermions is an alternating sum of positive and negative terms in the path integral formalism. Can I simply substitute partition function for wavefunction and sum for integral and find what would happen in the electronic structure case? That is, will the 3-electron exchange integral (assuming this is real and I'm not mistaken) be a negative number, so that we can picture the wavefunction relaxing a little bit from the so-called "heaps" which form due to exchange?
Finally, if I'm not just delusional about this whole thing, is there evidence that permutations beyond the two labels are important for describing any known chemical processes?
References:
[1]: Szabo, A., & Ostlund, N. S. (2012). Modern quantum chemistry: introduction to advanced electronic structure theory. Courier Corporation.
[2]: Feynman, R. P., Hibbs, A. R., & Styer, D. F. (2010). Quantum mechanics and path integrals. Courier Corporation. |
I've got this simple assignment, to find out the density for a give sphere with a radius = 2cm and the mass 296g. It seems straightforward, but it all got hairy when i've got to a fraction with three elements(
more precisely a fraction divided by a number actually this was wrong, the whole point was that the number is divided by a fraction, and it's different than a fraction being divided by a number.). I tend to solve these by dividing the element on the bottom by 1, and extracting from that 2 fraction division like this :
$$ \frac{a}{\frac{b}{c}} \Rightarrow \frac{\frac{a}{b}}{\frac{c}{1}} \Rightarrow \frac{a}{b} \div \frac{c}{1} => \frac{a}{b} \cdot \frac{1}{c} \Rightarrow \frac {a} {b \cdot c} $$
And it used to work, though for the next example it doesn't seem to, it looks like another technique is used:
$$ \frac{a}{\frac{b}{c}} \Rightarrow a \div \frac{b}{c} \Rightarrow a \cdot \frac{c}{b} \Rightarrow \frac {a \cdot c}{ b} $$
For the example below cleary the second method is used/needed, to get the right response. But i'm confused when to use each, as i've use both before, and both gave correct asnwers(matching with the answers at the end of the book).
$$ v = \frac43\pi r^3 $$
$$ d = \frac mv $$
$$ m = 296g $$
$$ r=2cm $$
$$ v = \frac43\pi 2^3 \Rightarrow \frac{32\pi}{3} $$
$$ d = \frac{m}{v} \Rightarrow \frac{296}{\frac{32\pi}{3}} \Rightarrow \frac {296}{32\pi} \div \frac31 \Rightarrow \frac{296}{32\pi} \cdot \frac{1}{3} \Rightarrow \frac{296}{96\pi} \approx 0.9814\frac{g}{cm^3} $$
$$ d_{expected} = 8.8 \frac{g}{cm^3} $$
I am, clearly, missing something fundamental about the use of these.
Can anyone enlighten me please? Can't quite find a good explanation online. |
The context of this question is that I'm taking a convex optimization class and was studying about Lagrangians. I was looking at an example problem when I experienced some difficulty understanding something.
The optimization problem in question is a simple least-squares solution of linear equations:
$\text{minimize}\quad \ x^Tx$
$\text{subject to}\quad Ax = b$
The Lagrangian of this problem is
$$L(x, \nu) = x^Tx + \nu^T(Ax - b)$$
When we're finding the dual function $g(\nu)$, we use the gradient of the Lagrangian:
$$\nabla_xL(x, \nu) = 2x + A^T\nu=0$$
We can obtain the solution $x = -\frac{1}{2}A^Tv$ from this. Plugging this into the dual function I get:
$$ \begin{align} g(\nu)& = L(-\frac{1}{2}A^T\nu,\ \nu)\\ & = (-\frac{1}{2}A^T\nu)^T(-\frac{1}{2}A^T\nu) + \nu^T(A(-\frac{1}{2}A^T\nu) - b) \\ & = \frac{1}{4}\nu^TAA^T\nu - \frac{1}{2}\nu^TAA^T\nu - \nu^Tb \\ & = -\frac{1}{4}\nu^TAA^T\nu - \nu^Tb\end{align}$$
However, the dual function that's in the textbook is:
$$g(\nu) = -\frac{1}{4}\nu^TAA^T\nu - b^T\nu$$
I have two questions regarding this derivation:
When we find the gradient of the Lagrangian, I initially thought the second term would be $\nu^TA$ because we find the derivative of $\nu^TAx$. This probably stems from my lack of understanding of vector differentiation but why is it $A\nu^T$? This is similar to question 1, but in the derivation of the dual function I got $\nu^Tb$ but apparently the correct term is $b^T\nu$. When we unfold the parentheses, do vector placements usually change? |
In May and Sigurdsson's
Parameterized Homotopy Theory, Proposition 2.2.11, four isomorphisms of functors are given. For a pullback square of base spaces $C=holim(A\overset{f}\to B\overset{j}\leftarrow D)$, with induced maps $C\overset{g}\to D$ and $C\overset{i}\to A$, there are natural isomorphisms:
$$(1)~~j^\ast f_!\cong g_!i^\ast~~~(2)~~f^\ast j_\ast \cong i_\ast g^\ast~~(3)~~f^\ast j_!\cong i_!g^\ast~~~(4)~~ j^*f_\ast\cong g_*i^*$$
Where for a map $f$ of base spaces, $f^\ast$ is the usual pullback of a parameterized
space, $f_\ast$ is its right adjoint, and $f_!$ is its left adjoint.
Do these same natural equivalences hold for parameterized
spectra over a pullback diagram of spaces? In particular, I am interested in equivalence $(3)$ above. And even more specifically, I am interested in comparing the two sides when $A$ above is a contractible space. In other words, $C$ is the homotopy fiber of the map $D\to B$. In that case, for a bundle of sphere spectra over $B$, we have that $i_!g^\ast$ yields the Thom spectrum of the pullback. How then should we think of $f^\ast j_!$ of the same bundle of sphere spectra? |
Chapters
Chapter 2: Relations
Chapter 3: Functions
Chapter 4: Measurement of Angles
Chapter 5: Trigonometric Functions
Chapter 6: Graphs of Trigonometric Functions
Chapter 7: Values of Trigonometric function at sum or difference of angles
Chapter 8: Transformation formulae
Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle
Chapter 10: Sine and cosine formulae and their applications
Chapter 11: Trigonometric equations
Chapter 12: Mathematical Induction
Chapter 13: Complex Numbers
Chapter 14: Quadratic Equations
Chapter 15: Linear Inequations
Chapter 16: Permutations
Chapter 17: Combinations
Chapter 18: Binomial Theorem
Chapter 19: Arithmetic Progression
Chapter 20: Geometric Progression
Chapter 21: Some special series
Chapter 22: Brief review of cartesian system of rectangular co-ordinates
Chapter 23: The straight lines
Chapter 24: The circle
Chapter 25: Parabola
Chapter 26: Ellipse
Chapter 27: Hyperbola
Chapter 28: Introduction to three dimensional coordinate geometry
Chapter 29: Limits
Chapter 30: Derivatives
Chapter 31: Mathematical reasoning
Chapter 32: Statistics
Chapter 33: Probability
RD Sharma Mathematics Class 11 Chapter 11: Trigonometric equations Chapter 11: Trigonometric equations Exercise 11.10 solutions [Pages 21 - 22] Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation:
Find the general solution of the following equation: Find the general solution of the following equation: Find the general solution of the following equation:
Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sin x + \cos x = \sqrt{2}\]
Solve the following equation:
\[\sqrt{3} \cos x + \sin x = 1\]
Solve the following equation:
Solve the following equation:
\[cosec x = 1 + \cot x\]
Solve the following equation:
\[\cot x + \tan x = 2\]
Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]
Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]
Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Solve the following equation:
4sin x cos x + 2 sin x + 2 cos x + 1 = 0
Solve the following equation:
cosx + sin x = cos 2x + sin 2x
Solve the following equation:
sin x tan x – 1 = tan x – sin x
Solve the following equation:
3tanx + cot x = 5 cosec x
Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0
Solve the following equation:
3sin 2x – 5 sin x cos x + 8 cos 2 x = 2
Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]
If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.
Chapter 11: Trigonometric equations solutions [Page 26]
Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].
Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]
Write the general solutions of tan
2 2x = 1.
Write the set of values of a for which the equation
If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.
Write the number of points of intersection of the curves
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
Write the number of points of intersection of the curves
Write the solution set of the equation
Write the number of values of x in [0, 2π] that satisfy the equation \[\sin x - \cos x = \frac{1}{4}\].
If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
Chapter 11: Trigonometric equations solutions [Pages 26 - 28]
The smallest value of x satisfying the equation
\[2\pi/3\]
`pi/3`
`pi/6`
`pi/12`
If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]
\[\pi/3\] \[2\pi/3\] \[4\pi/6\] \[5\pi/12\]
If \[\tan px - \tan qx = 0\], then the values of θ form a series in
AP
GP
HP
none of these
If a is any real number, the number of roots of \[\cot x - \tan x = a\] in the first quadrant is (are).
2
0
1
none of these
The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is
\[x = 2 n\pi \pm \frac{\pi}{6}, n \in Z\] \[x = 2 n\pi \pm \frac{2\pi}{3}, n \in Z\] \[x = n\pi \pm \frac{\pi}{3}, n \in Z\]
none of these
A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval
\[\left( - \pi/4, \pi/4 \right)\] \[\left( \pi/4, 3\pi/4 \right)\] \[\left( 3\pi/4, 5\pi/4 \right)\] \[\left( 5\pi/4, 7\pi/4 \right)\]
The number of solution in [0, π/2] of the equation \[\cos 3x \tan 5x = \sin 7x\] is
5
7
6
none of these
The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\] \[x = n\pi + \left( - 1 \right)^n \frac{\pi}{4} + \frac{\pi}{3}, n \in Z\]
\[x = n\pi + \left( - 1 \right)^n \frac{\pi}{3} + \frac{\pi}{6}, n \in Z\]
\[x = n\pi \pm \frac{\pi}{6}, n \in Z\]
\[x = n\pi \pm \frac{\pi}{3}, n \in Z\]
The smallest positive angle which satisfies the equation
\[\frac{5\pi}{6}\] \[\frac{2\pi}{3}\] \[\frac{\pi}{3}\] \[\frac{\pi}{6}\]
If \[4 \sin^2 x = 1\], then the values of x are
\[2 n\pi \pm \frac{\pi}{3}, n \in Z\]
\[n\pi \pm \frac{\pi}{3}, n \in Z\]
\[n\pi \pm \frac{\pi}{6}, n \in Z\]
\[2 n\pi \pm \frac{\pi}{6}, n \in Z\]
If \[\cot x - \tan x = \sec x\], then, x is equal to
\[2 n\pi + \frac{3\pi}{2}, n \in Z\]
\[n\pi + \left( - 1 \right)^n \frac{\pi}{6}, n \in Z\]
\[n\pi + \frac{\pi}{2}, n \in Z\]
none of these.
A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is
`(5pi)/3`
\[\frac{4\pi}{3}\]
`(2pi)/3`
\[\frac{\pi}{3}\]
In (0, π), the number of solutions of the equation \[\tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x\] is
7
5
4
2
The number of values of x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]
1
2
3
4
If \[e^{\sin x} - e^{- \sin x} - 4 = 0\], then x =
0
\[\sin^{- 1} \left\{ \log_e \left( 2 - \sqrt{5} \right) \right\}\]
1
none of these
The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.
finite
infinite
one
no
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of
x is \[n \pi + \left( - 1 \right)^n \frac{\pi}{4}, n \in Z\]
\[\left( - 1 \right)^n \frac{\pi}{4} - \frac{\pi}{3}, n \in Z\]
\[n \pi + \frac{\pi}{4} - \frac{\pi}{3}, n \in Z\]
\[n \pi + \left( - 1 \right)^n \frac{\pi}{4} - \frac{\pi}{3}, n \in Z\]
General solution of \[\tan 5 x = \cot 2 x\] is
\[\frac{n \pi}{7} + \frac{\pi}{2}, n \in Z\]
\[x = \frac{n \pi}{7} + \frac{\pi}{3}, n \in Z\] \[x = \frac{n \pi}{7} + \frac{\pi}{14}, n \in Z\] \[x = \frac{n \pi}{7} - \frac{\pi}{14}, n \in Z\]
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
\[\left( - \pi/4, \pi/4 \right)\] \[\left(\pi/4,3 \pi/4 \right)\] \[\left( 3\pi/4, 5\pi/4 \right)\] \[\left( 5\pi/4, 7\pi/4 \right)\]
If \[\cos x = - \frac{1}{2}\] and 0 < x < 2\pi, then the solutions are
\[x = \frac{\pi}{3}, \frac{4\pi}{3}\] \[x = \frac{2\pi}{3}, \frac{4\pi}{3}\] \[x = \frac{2\pi}{3}, \frac{7\pi}{6}\] \[\theta = \frac{2\pi}{3}, \frac{5\pi}{3}\]
The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is
0
5
6
10
Chapter 11: Trigonometric equations RD Sharma Mathematics Class 11 Textbook solutions for Class 11 RD Sharma solutions for Class 11 Mathematics chapter 11 - Trigonometric equations
RD Sharma solutions for Class 11 Maths chapter 11 (Trigonometric equations) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Class 11 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 11 Mathematics chapter 11 Trigonometric equations are Transformation Formulae, Values of Trigonometric Functions at Multiples and Submultiples of an Angle, Sine and Cosine Formulae and Their Applications, 180 Degree Plusminus X Function, 2X Function, 3X Function, Expressing Sin (X±Y) and Cos (X±Y) in Terms of Sinx, Siny, Cosx and Cosy and Their Simple Applications, Concept of Angle, Introduction of Trigonometric Functions, Signs of Trigonometric Functions, Domain and Range of Trigonometric Functions, Trigonometric Functions of Sum and Difference of Two Angles, Trigonometric Equations, Truth of the Identity, Negative Function Or Trigonometric Functions of Negative Angles, 90 Degree Plusminus X Function, Conversion from One Measure to Another, Graphs of Trigonometric Functions.
Using RD Sharma Class 11 solutions Trigonometric equations exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.
Get the free view of chapter 11 Trigonometric equations Class 11 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation |
Problem 676
Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]
Find the rank and nullity of $T$.Add to solve later
Problem 675
The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined by
\[ T(f)(x) = f(0) + f'(0) x + \frac{f^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3.\] Here, $f’, f^{\prime\prime}$ and $f^{\prime \prime \prime}$ denote the first, second, and third derivatives of $f$, respectively.
Prove that $T$ is a linear transformation.Add to solve later
Problem 674
Let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis. Let $T : \mathrm{P}_4 \rightarrow \mathrm{P}_{4}$ be the map defined by, for $f \in \mathrm{P}_4$,
\[ T (f) (x) = f(x) – x – 1.\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_4$.Add to solve later
Problem 673
Let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the
standard basis.
Let $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
\[T (f) (x) = ( x^2 – 2) f(x).\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.Add to solve later
Problem 672
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,
\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.Add to solve later
Problem 669 (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
Add to solve later
(c) Let $A$ be a $4\times 4$ matrix and let \[\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}.\] Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Problem 668
Consider the system of differential equations
\begin{align*} \frac{\mathrm{d} x_1(t)}{\mathrm{d}t} & = 2 x_1(t) -x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_2(t)}{\mathrm{d}t} & = -x_1(t)+2x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_3(t)}{\mathrm{d}t} & = -x_1(t) -x_2(t) +2x_3(t) \end{align*} (a) Express the system in the matrix form. (b) Find the general solution of the system.
Add to solve later
(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$. Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization Problem 667 (a) Find all solutions of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},\] where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.
Add to solve later
(b) Solve the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}\] with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less Problem 665
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less.
(a) Prove that the set $\{ 1 , 1 + x , (1 + x)^2 \}$ is a basis for $\mathbf{P}_2$.
Add to solve later
(b) Write the polynomial $f(x) = 2 + 3x – x^2$ as a linear combination of the basis $\{ 1 , 1+x , (1+x)^2 \}$. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later |
The "projective semion" model was considered in http://arxiv.org/abs/1403.6491 (page 2). It is a symmetry enriched topological (SET) phase. There is one non-trivial anyon, a semion $s$ which induces a phase factor of $\pi$ when going around another semion.The chiral topological order is the same as the $\nu = 1/2$ bosonic fractional quantum Hall state, whose effective field theory is the $K = 2$ Chern-Simons theory: \begin{equation} \mathcal{L} = \frac{2}{4\pi}\epsilon^{\mu\nu\lambda}a_{\mu}\partial_{\nu}a_{\lambda} \end{equation}
The symmetry group for the theory is $G = \mathbb{Z}_2 \times \mathbb{Z}_2$. We label the three non-trivial group elements as $g_x, g_y, g_z$. The symmetry can act on the semion in the following ways:
Each semion carries half charge for all three $\mathbb{Z}_2$ transformations. Moreover the three $\mathbb{Z}_2$ transformations anticommute with each other and can be represented as $g_x = i\sigma_x, g_y = i\sigma_y, g_z = i\sigma_z$.
The semion carries integral charge under two of the three $\mathbb{Z}_2$ transformations, and half charge under the the other $\mathbb{Z}_2$ transformation. There are three variants of this, and the symmetry group can be represented as $g_x = \sigma_x, g_y = \sigma_y, g_z = i\sigma_z$, or $g_x = \sigma_x, g_y = i\sigma_y, g_z = \sigma_z$, or $g_x = i\sigma_x, g_y = \sigma_y, g_z = \sigma_z$.
Symmetry fractionalization in case 1 is anomaly free but is anomalous in case 2, as shown in http://arxiv.org/abs/1403.6491.
I want to write down an effective field theory description to describe symmetry fractionlization pattern in cases 1 and 2 on the semion $a$, and can explicitly see that the field theory I write down for case 1 is anomaly free whereas that for case 2 has an anomaly.
One possible way is to gauge the symmetry $\mathbb{Z}_2 \times \mathbb{Z}_2$, and couple the gauge fields to the semion $a$. The different coupling terms reflect the different ways that the symmetry is represented on the semion. I think this is essentially what Eq.(5) on page 21 of http://arxiv.org/abs/1404.3230 is trying to describe. The action they wrote down is
\begin{equation} \mathcal{L} = \frac{2}{4\pi}\epsilon^{\mu\nu\lambda}a_{\mu}\partial_{\nu}a_{\lambda} + \frac{p_1}{2\pi}\epsilon^{\mu\nu\lambda}a_{\mu}\partial_{\nu}A_{1\lambda} + \frac{p_2}{2\pi}\epsilon^{\mu\nu\lambda}a_{\mu}\partial_{\nu}A_{2\lambda} + \frac{p_3}{\pi^2}\epsilon^{\mu\nu\lambda}a_{\mu}A_{1\nu}A_{2\lambda} \end{equation}
I can understand the second and third terms in this action, which says (with $p_1=p_2=1$) that the semion $a$ carries half symmetry charge under the two generators (say $g_x$ and $g_y$) of $\mathbb{Z}_2\times \mathbb{Z}_2$.
However, I am having trouble understanding the last term in the action, presumably, it means that the semion carries half charge under all three elements $g_x,g_y,g_z$ in $\mathbb{Z}_2\times\mathbb{Z}_2$. If this is correct then setting $p_1=p_2=0, p_3=1$ gives us an effective description of case 1. The theory is anomaly free; whereas setting $p_1=p_2=p_3=1$ gives us an effective description of case 2 (semion $a$ carries half $g_x,g_y,g_z$ charge from the last term, and an additional half $g_x,g_y$ charge from the second and third term), and the theory is anomalous. This is consistent with the claim on page 24 of http://arxiv.org/abs/1404.3230.
Does any people have an idea why the last term in $\mathcal{L}$ says that the semion carries half charge under all three elements $g_x,g_y,g_z$ in $\mathbb{Z}_2\times\mathbb{Z}_2$? |
This question is similar to This Dice Blackjack game - is there Nash equilibrium?, but the game rules are completely different.
This Dice Blackjack is game for two. It is also played with fair six-sided dice. In this game is no bettor or dealer, both players have the same role.
Rules: Any player can roll dice as many times he/she desires while the sum of rolls is lower than 21.Players does not see the sum of rolls of the other player till both finish (they can only guess what other player has).When both players finish, they compare their sums of rolls. Winner:Winner is the one with lower penalization $ =\begin{cases} 21 - \sum,& \text{if } \sum \leq 21\\ 100, & \text{otherwise}\end{cases}$
where $\sum$ is sum of player rolls. If tie, nobody wins. Goal is to achieve as many wins as possible in huge number of games.
Example games:
game - player A stops at 18, player B stops at 16 (4, 4, 4, 4). Player A wins.
(player B was too afraid to risk)
game - player A stops at 19, player B reach 22 (5, 6, 1, 6,
5). Player A wins. (player B start risking but has no luck)
game - player A get to 23, player B stops at 17 (4, 2, 6, 5). Player B wins.
(player B become again careful and this time it payed off) Primary question:Is here the Nash equilibrium? That means existence of optimal strategy, that you should play no matter what the opponent plays? Or, can be proficient to guess the opponent strategy and adjust your strategy according to what he/she is probably playing?
I think it is similar to rock/paper/scissor. As I read somewhere there is "optimal" strategy, but if your opponent strategy is biased, you can get better outcome with adjusted strategy.
If anything is not clear, please comment, I will do my best to improve the question.
Secondary question:What should be name of this game? Blind dice Blackjack? Is there any name convention for game modifications like this one? Please comment if you have idea or mention the idea in your answer for primary question. |
Early in the 17th century, Johannes Kepler established, from actual observations of the positions of the planets in the sky, three laws of planetary motion.
A Mersenne number is an integer of the form $2^p-1$ where $p$ is a prime.
$x$ and $y$ are unequal positive integers. Prove that $xy$ does not divide x^2 +y^2$.
\begin{align*}
1+\cos(x) & = 1 + (1-\sin^2(x))^{1/2} \\ (1+\cos(x))^2 & = \left\{1 + (1-\sin^2(x))^\frac{1}{2}\right\}^2 \\ \end{align*} Let $x=\pi$.
Suppose you have two glasses; one contains water and the other contains the same amount of cordial.
J151 (i) Prove that if $k$ is not a prime then neither is $2^k-1$ |
Following Schumacher and Westmoreland, we address the problem of the capacity of a quantum wiretap channel. We first argue that, in the definition of the so-called “quantum privacy,” Holevo quantities should be used instead of classical mutual informations. The argument actually shows that the security condition in the definition of a code should limit the wiretapper’s Holevo quantity. Then we show that this modified quantum privacy is the optimum achievable rate of secure transmission.
Characteristics of a successive cancellation scheme, widely used in iterative decoding, is investigated. Comparison with another popular method, the minimum mean-square error (MMSE) method, is also provided.
A Steiner quadruple system SQS(v) of order v is a 3-design T (v; 4; 3;λ) with λ = 1. In this paper we describe all nonisomorphic systems SQS(16) that can be obtained by the generalized concatenated construction (GC-construction). These Steiner systems have rank at most 13 over F2. In particular, there is one system SQS(16) of rank 11 (points and planes of the a fine geometry AG(4; 2)), fifteen systems of rank 12, and 4131 systems of rank 13. All these Steiner systems are resolvable.
A new statistical test is proposed for testing the hypothesis H 0 that symbols of an alphabet are generated with equal probabilities against the alternative hypothesis H 1, the negation of H 0. The new method is applied to testing generators of pseudorandom numbers. It is experimentally demonstrated that the method makes it possible to detect deviations from randomness for many generators which “withstand” previously known statistical tests.
A (w, r)-cover-free code is the incidence matrix of a family of sets where no intersection of w members of the family is covered by the union of r others. We obtain a new condition in view of which (w, r)-cover-free codes with a simple structure are optimal. We also introduce (w, r)-cover-free codes with a constraint set.
We establish the properness of some classes of binary block codes with symmetric distance distribution, including Kerdock codes and codes that satisfy the Grey-Rankin bound, as well as the properness of Preparata codes, thus augmenting the list of very few known proper nonlinear codes.
Characteristics of a successive cancellation scheme, widely used in iterative decoding, is investigated. Comparison with another popular method, the minimum mean-square error (MMSE) method, is also provided.
A Steiner quadruple system SQS(v) of order v is a 3-design T (v, 4, 3, ) with = 1. In this paper we describe all nonisomorphic systems SQS(16) that can be obtained by the generalized concatenated construction (GC-construction). These Steiner systems have rank at most 13 over $$\mathbb{F}$$ 2. In particular, there is one system SQS(16) of rank 11 (points and planes of the a fine geometry AG(4, 2)), fifteen systems of rank 12, and 4131 systems of rank 13. All these Steiner systems are resolvable.
Under the assumption that the distribution function of the observation noise is known, both for the case of a predefined observation design and the case where observation designing is possible, we construct estimates of smooth functionals of the regression function, for which lower bounds on mean-square risks of arbitrary estimates of smooth functionals obtained in [1, 2] are asymptotically attained.
Following Schumacher and Westmoreland, we address the problem of the capacity of a quantum wiretap channel. We first argue that, in the definition of the so-called quantum privacy, Holevo quantities should be used instead of classical mutual informations. The argument actually shows that the security condition in the definition of a code should limit the wiretappers Holevo quantity. Then we show that this modified quantum privacy is the optimum achievable rate of secure transmission.
We establish the properness of some classes of binary block codes with symmetric distance distribution, including Kerdock codes and codes that satisfy the Grey-Rankin bound, as well as the properness of Preparata codes, thus augmenting the list of very few known proper nonlinear codes.
In this paper, we address the problem of image denoising using a stochastic differential equation approach. Proposed stochastic dynamics schemes are based on the property of diffusion dynamics to converge to a distribution on global minima of the energy function of the model, under a special cooling schedule (the annealing procedure). To derive algorithms for computer simulations, we consider discrete-time approximations of the stochastic differential equation. We study convergence of the corresponding Markov chains to the diffusion process. We give conditions for the ergodicity of the Euler approximation scheme. In the conclusion, we compare results of computer simulations using the diffusion dynamics algorithms and the standard Metropolis Hasting algorithm. Results are shown on synthetic and real data.
A convolutional code can be used to detect or correct infinite sequences of errors or to correct infinite sequences of erasures. First, erasure correction is shown to be related to error detection, as well as error detection to error correction. Next, the active burst distance is exploited, and various bounds on erasure correction, error detection, and error correction are obtained for convolutional codes. These bounds are illustrated by examples.
A mathematical model of an adaptive random multiple access communication network is investigated. The value of the network critical load is found; in the critical load, asymptotic probability distributions for states of the information transmission channel and for the number of requests in the source of repeated calls are found. It is proved that distributions of the normalized number of requests belong to the class of normal and exponential distributions, and it is shown how conditional normal distributions pass in the limit to the class of exponential ones.
We consider the 2 statistic, destined for testing the symmetry hypothesis, which has the form n2 = n -[Fn(x) + Fn(-x) - 1]2 dFn(x), where Fn(x) is the empirical distribution function. Based on the Laplace method for empirical measures, exact asymptotic (as n ) of the probability Pr{n2 > nv} for 0 < v < 1/3 is found. Constants entering the formula for the exact asymptotic are computed by solving the extreme value problem for the rate function and analyzing the spectrum of the second-order differential equation of the Sturm Liouville type.
An infinite-server queueing system is considered where access of customers to service is controlled by a gate. The gate is open only if all servers are free. Otherwise, customers are put on a queue. Asymptotic behavior of the system in heavy traffic is studied under the assumption that the service time distribution has a power tail.
We consider the ω2 statistic, destined for testing the symmetry hypothesis, which has the form $$\omega _n^{\text{2}} = n\;\int\limits_{ - \infty }^\infty {[F_n (x)\; + F_n ( - x)\; - 1]^2 dF_n (x),}$$ where F n (x) is the empirical distribution function. Based on the Laplace method for empirical measures, exact asymptotic (as n → ∞) of the probability $$P\{ \omega _n^2 > nv\} $$ for 0 < v < 1/3 is found. Constants entering the formula for the exact asymptotic are computed by solving the extreme value problem for the rate function and analyzing the spectrum of the second-order differential equation of the Sturm–Liouville type.
We consider the optical-acoustic tomography problem. In the general case, the problem is to reconstruct a real-valued function with a compact support in the n-dimensional Euclidean space via its spherical integrals, i.e., integrals over all (n - 1)-dimensional spheres centered at points on some (n - 1)-dimensional hypersurface. We deal with the cases n = 2 and n = 3, which are of the most practical interest from the standpoint of possible medical applications. We suggest a new effective method of reconstruction, develop restoration algorithms, and investigate the quality of the algorithms for these cases. The main result of the paper is construction of explicit approximate reconstruction formulas; from the mathematical standpoint, these formulas give the parametrix for the optical-acoustic tomography problem. The formulas constructed is a background for the restoration algorithms. We performed mathematical experiments to investigate the quality of the restoration algorithms using the generally accepted tomography quality criteria. The results obtain lead to the general conclusion: the quality of the restoration algorithms developed for optical-acoustic tomography is only slightly lower then the quality of the convolution and back projection algorithm used in Radon tomography, which is a standard de facto.
The paper considers the problem of estimating a signal with finitely many points of discontinuity from observations against white Gaussian noise. It is shown that, with an appropriate choice of a generator polynomial, an estimation method based on wavelets yields asymptotically minimax (up to a constant) estimates for functions sufficiently smooth outside the discontinuity points. |
Welcome (belatedly) to a new year – I don’t know what life is like for you, but with the present cut-backs, it is getting harder and harder to have
Parabola out on time.
Welcome (belatedly) to a new year – I don’t know what life is like for you, but with the present cut-backs, it is getting harder and harder to have
In a previous article (
Parabola, Volume 32 Number 2), the swing and reverse swing of a cricket ball was discussed.
In the 3-Unit Maths course you are asked to prove (by induction) various formulae such as
$$ 1^2+2^2+3^2+.....+n^2 = \sum _{x=1}^n x^2 = \frac{1}{6}n(n+1)(2n+1),$$
When you began to sketch curves early in high school, you evaluated the “$y$-value” for several “$x$-values”, plotted the resulting points and then joined them up as smoothly as you could.
Q993 Consider \begin{eqnarray*} p(n)&=&a_0+a_1n+a_2n^2+\cdots+a_kn^k\ ,\\ q(n)&=&b_0+b_1\binom{n}{1}+b_2\binom{n}{2}+\cdots+b_k\binom{n}{k}\ , \end{eqnarray*} |
Wave energy converters in coastal structures: verschil tussen versies
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An interesting result is that the maximum average wave power that a point absorber can absorb <math>P_{abs} </math>(W) from the waves does not depend on its dimensions <ref name ="ref4">De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010,
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An interesting result is that the maximum average wave power that a point absorber can absorb <math>P_{abs} </math>(W) from the waves does not depend on its dimensions <ref name ="ref4">De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010, 899–918. </ref>. It is theoretically possible to absorb a lot of energy with only a small buoy. It can be shown that for a body with a vertical axis of symmetry (but otherwise arbitrary geometry) oscillating in heave the capture (or absorption) width <math>L_{max}</math>(m) is as follows <ref name =ref4/>:
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Versie van 3 sep 2012 om 11:57 Introduction
Fig 1: Construction of a coastal structure.
Coastal works along European coasts are composed of very diverse structures. Many coastal structures are ageing and facing problems of stability, sustainability and erosion. Moreover climate change and especially sea level rise represent a new danger for them. Coastal dykes in Europe will indeed be exposed to waves with heights that are greater than the dykes were designed to withstand, in particular all the structures built in shallow water where the depth imposes the maximal amplitude because of wave breaking.
This necessary adaptation will be costly but will provide an opportunity to integrate converters of sustainable energy in the new maritime structures along the coasts and in particular in harbours. This initiative will contribute to the reduction of the greenhouse effect. Produced energy can be directly used for the energy consumption in harbour area and will reduce the carbon footprint of harbours by feeding the docked ships with green energy. Nowadays these ships use their motors to produce electricity power on board even if they are docked. Integration of wave energy converters (WEC) in coastal structures will favour the emergence of the new concept of future harbours with zero emissions.
Inhoud Wave energy and wave energy flux
For regular water waves, the time-mean wave energy density E per unit horizontal area on the water surface (J/m²) is the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy
[1] both contributing half to the time-mean wave energy density E that is proportional to the wave height squared according to linear wave theory [1]:
(1)
[math]E= \frac{1}{8} \rho g H^2[/math]
g is the gravity and [math]H[/math] the wave height of regular water waves. As the waves propagate, their energy is transported. The energy transport velocity is the group velocity. As a result, the time-mean wave energy flux per unit crest length (W/m) perpendicular to the wave propagation direction, is equal to
[1]:
(2)
[math] P= Ec_{g}[/math]
with [math]c_{g}[/math] the group velocity (m/s). Due to the dispersion relation for water waves under the action of gravity, the group velocity depends on the wavelength λ (m), or equivalently, on the wave period T (s). Further, the dispersion relation is a function of the water depth h (m). As a result, the group velocity behaves differently in the limits of deep and shallow water, and at intermediate depths:
[math](\frac{\lambda}{20} \lt h \lt \frac{\lambda}{2})[/math]
Application for wave energy convertersFor regular waves in deep water:
[math]c_{g} = \frac{gT}{4\pi} [/math] and [math]P_{w1} = \frac{\rho g^2}{32 \pi} H^2 T[/math]
The time-mean wave energy flux per unit crest length is used as one of the main criteria to choose a site for wave energy converters.
For real seas, whose waves are random in height, period (and direction), the spectral parameters have to be used. [math]H_{m0} [/math] the spectral estimate of significant wave height is based on zero-order moment of the spectral function as [math]H_{m0} = 4 \sqrt{m_0} [/math] Moreover the wave period is derived as follows
[2].
[math]T_e = \frac{m_{-1}}{m_0} [/math]
where represents the spectral moment of order n. An equation similar to that describing the power of regular waves is then obtained :
[math]P_{w1} = \frac{\rho g^2}{64 \pi} H_{m0}^2 T_e[/math]
If local data are available ([math]H_{m0}^2, T_e [/math]) for a sea state through in-situ wave buoys for example, satellite data or numerical modelling, the last equation giving wave energy flux [math]P_{w1}[/math] gives a first estimation. Averaged over a season or a year, it represents the maximal energetic resource that can be theoretically extracted from wave energy. If the directional spectrum of sea state variance F (f,[math]\theta[/math]) is known with f the wave frequency (Hz) and [math]\theta[/math] the wave direction (rad), a more accurate formulation is used:
[math]P_{w2} = \rho g\int\int c_{g}(f,h)F(f,\theta) dfd \theta[/math]
Fig 2: Time-mean wave energy flux along
West European coasts
[3] .
It can be shown easily that equations (5 and 6) can be reduced to (4) with the hypothesis of regular waves in deep water. The directional spectrum is deduced from directional wave buoys, SAR images or advanced spectral wind-wave models, known as third-generation models, such as WAM, WAVEWATCH III, TOMAWAC or SWAN. These models solve the spectral action balance equation without any a priori restrictions on the spectrum for the evolution of wave growth.
From TOMAWAC model, the near shore wave atlas ANEMOC along the coasts of Europe and France based on the numerical modelling of wave climate over 25 years has been produced
[4]. Using equation (4), the time-mean wave energy flux along West European coasts is obtained (see Fig. 2). This equation (4) still presents some limits like the definition of the bounds of the integration. Moreover, the objective to get data on the wave energy near coastal structures in shallow or intermediate water requires the use of numerical models that are able to represent the physical processes of wave propagation like the refraction, shoaling, dissipation by bottom friction or by wave breaking, interactions with tides and diffraction by islands.
The wave energy flux is therefore calculated usually for water depth superior to 20 m. This maximal energetic resource calculated in deep water will be limited in the coastal zone:
at low tide by wave breaking; at high tide in storm event when the wave height exceeds the maximal operating conditions; by screen effect due to the presence of capes, spits, reefs, islands,...
Technologies
According to the International Energy Agency (IEA), more than hundred systems of wave energy conversion are in development in the world. Among them, many can be integrated in coastal structures. Evaluations based on objective criteria are necessary in order to sort theses systems and to determine the most promising solutions.
Criteria are in particular:
the converter efficiency : the aim is to estimate the energy produced by the converter. The efficiency gives an estimate of the number of kWh that is produced by the machine but not the cost. the converter survivability : the capacity of the converter to survive in extreme conditions. The survivability gives an estimate of the cost considering that the weaker are the extreme efforts in comparison with the mean effort, the smaller is the cost.
Unfortunately, few data are available in literature. In order to determine the characteristics of the different wave energy technologies, it is necessary to class them first in four main families
[3].
An interesting result is that the maximum average wave power that a point absorber can absorb [math]P_{abs} [/math](W) from the waves does not depend on its dimensions
[5]. It is theoretically possible to absorb a lot of energy with only a small buoy. It can be shown that for a body with a vertical axis of symmetry (but otherwise arbitrary geometry) oscillating in heave the capture (or absorption) width [math]L_{max}[/math](m) is as follows [5]:
[math]L_{max} = \frac{P_{abs}}{P_{w}} = \frac{\lambda}{2\pi}[/math] or [math]1 = \frac{P_{abs}}{P_{w}} \frac{2\pi}{\lambda}[/math]
Fig 4: Upper limit of mean wave power
absorption for a heaving point absorber.
where [math]{P_{w}}[/math] is the wave energy flux per unit crest length (W/m). An optimally damped buoy responds however efficiently to a relatively narrow band of wave periods.
Babarit et Hals propose
[6] to derive that upper limit for the mean annual power in irregular waves at some typical locations where one could be interested in putting some wave energy devices. The mean annual power absorption tends to increase linearly with the wave power resource. Overall, one can say that for a typical site whose resource is between 20-30 kW/m, the upper limit of mean wave power absorption is about 1 MW for a heaving WEC with a capture width between 30-50 m.
In order to complete these theoretical results and to describe the efficiency of the WEC in practical situations, the capture width ratio [math]\eta[/math] is also usually introduced. It is defined as the ratio between the absorbed power and the available wave power resource per meter of wave front times a relevant dimension B [m].
[math]\eta = \frac{P_{abs}}{P_{w}B} [/math]
The choice of the dimension B will depend on the working principle of the WEC. Most of the time, it should be chosen as the width of the device, but in some cases another dimension is more relevant. Estimations of this ratio [math]\eta[/math] are given
[6]: 33 % for OWC, 13 % for overtopping devices, 9-29 % for heaving buoys, 20-41 % for pitching devices. For energy converted to electricity, one must take into account moreover the energy losses in other components of the system.
Civil engineering
Never forget that the energy conversion is only a secondary function for the coastal structure. The primary function of the coastal structure is still protection. It is necessary to verify whether integration of WEC modifies performance criteria of overtopping and stability and to assess the consequences for the construction cost.
Integration of WEC in coastal structures will always be easier for a new structure than for an existing one. In the latter case, it requires some knowledge on the existing coastal structures. Solutions differ according to sea state but also to type of structures (rubble mound breakwater, caisson breakwaters with typically vertical sides). Some types of WEC are more appropriate with some types of coastal structures.
Fig 5: Several OWC (Oscillating water column) configurations (by Wavegen – Voith Hydro).
Environmental impact
Wave absorption if it is significant will change hydrodynamics along the structure. If there is mobile bottom in front of the structure, a sand deposit can occur. Ecosystems can also be altered by change of hydrodynamics and but acoustic noise generated by the machines.
Fig 6: Finistere area and locations of
the six sites (google map).
Study case: Finistere area
Finistere area is an interesting study case because it is located in the far west of Brittany peninsula and receives in consequence the largest wave energy flux along the French coasts (see Fig.2). This area with a very ragged coast gathers moreover many commercial ports, fishing ports, yachting ports. The area produces a weak part of its consumption and is located far from electricity power plants. There are therefore needs for renewable energies that are produced locally. This issue is important in particular in islands. The production of electricity by wave energy will have seasonal variations. Wave energy flux is indeed larger in winter than in summer. The consumption has peaks in winter due to heating of buildings but the consumption in summer is also strong due to the arrival of tourists.
Six sites are selected (see figure 7) for a preliminary study of wave energy flux and capacity of integration of wave energy converters. The wave energy flux is expected to be in the range of 1 – 10 kW/m. The length of each breakwater exceeds 200 meters. The wave power along each structure is therefore estimated between 200 kW and 2 MW. Note that there exist much longer coastal structures like for example Cherbourg (France) with a length of 6 kilometres.
(1) Roscoff (300 meters) (2) Molène (200 meters) (3) Le Conquet (200 meters) (4) Esquibien (300 meters) (5) Saint-Guénolé (200 meters) (6) Lesconil (200 meters) Fig.7: Finistere area, the six coastal structures and their length (google map).
Wave power flux along the structure depends on local parameters: bottom depth that fronts the structure toe, the presence of caps, the direction of waves and the orientation of the coastal structure. See figure 8 for the statistics of wave directions measured by a wave buoy located at the Pierres Noires Lighthouse. These measurements show that structures well-oriented to West waves should be chosen in priority. Peaks of consumption occur often with low temperatures in winter coming with winds from East- North-East directions. Structures well-oriented to East waves could therefore be also interesting even if the mean production is weak.
Fig 8: Wave measurements at the Pierres Noires Lighthouse.
Conclusion
Wave energy converters (WEC) in coastal structures can be considered as a land renewable energy. The expected energy can be compared with the energy of land wind farms but not with offshore wind farms whose number and power are much larger. As a land system, the maintenance will be easy. Except the energy production, the advantages of such systems are :
a “zero emission” port industrial tourism test of WEC for future offshore installations.
Acknowledgement
This work is in progress in the frame of the national project EMACOP funded by the French Ministry of Ecology, Sustainable Development and Energy.
See also Waves Wave transformation Groynes Seawall Seawalls and revetments Coastal defense techniques Wave energy converters Shore protection, coast protection and sea defence methods Overtopping resistant dikes
References Mei C.C. (1989) The applied dynamics of ocean surface waves. Advanced series on ocean engineering. World Scientific Publishing Ltd Vicinanza D., Cappietti L., Ferrante V. and Contestabile P. (2011) : Estimation of the wave energy along the Italian offshore, journal of coastal research, special issue 64, pp 613 - 617. Mattarolo G., Benoit M., Lafon F. (2009), Wave energy resource off the French coasts: the ANEMOC database applied to the energy yield evaluation of Wave Energy, 10th European Wave and Tidal Energy Conference Series (EWTEC’2009), Uppsala (Sweden) Benoit M. and Lafon F. (2004) : A nearshore wave atlas along the coasts of France based on the numerical modeling of wave climate over 25 years, 29th International Conference on Coastal Engineering (ICCE’2004), Lisbonne (Portugal), 714-726. De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010, pp. 899–918. Babarit A. and Hals J. (2011) On the maximum and actual capture width ratio of wave energy converters – 11th European Wave and Tidal Energy Conference Series (EWTEC’2011) – Southampton (U-K). |
Equilibrium Law of Chemical Equilibrium and Equilibrium Constants, Henry's Law Law of mass action : (by guldberg and waage) A + B → Products rate ∝ [A] [B] [A], [B] are the active masses of A and B. Rate = k[A] [B] , where k = rate constant Active mass : Molar concentration = \frac{n}{V_{L}} for both gases and solution. For gases active mass can be expressed in partial pressures also. 8.5 gm of NH 3 is present in 500ml vessel Active mass of \left[NH_{3}\right] = \frac{wt}{m.wt}\times \frac{1000}{V_{ml}} =\frac{8.5}{17}\times\frac{1000}{500} = 1 rate = k[A] [B] ⇒ k = \frac{rate}{[A][B]} ⇒ rate ∝ k Equilibrium constant in terms of partial pressure (Kp) : k_{p} = \frac{P_C^c \ P_D^d}{P_A^a \ P_B^b} Relation between k p and k c PV = nRT ⇒ p = \left(\frac{n}{v}\right)RT P = CRT k_{p} = \frac{\left(C_{C}RT\right)^{C}.\left(C_{D}RT\right)^{D}}{\left(C_{A}RT\right)^{A}.\left(C_{B}RT\right)^{B}} k_{p} = k_{c}RT^{\Delta n} Arrhenius equation with equilibrium constant : According to Arrhenius k = A.e^{\frac{-E_{a}}{RT}} A → frequency factor R → Gas constant T → Temperature E a → Activation energy \log k = \log A - \frac{E_{a}}{2.303 \ RT} \log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \ R}\left[\frac{1}{T_{1}} - \frac{1}{T_{2}}\right]
\log\frac{k_{2}}{k_{1}} = \frac{\Delta H}{2.303 \ R}\left[\frac{1}{T_{1}} - \frac{1}{T_{2}}\right] Van't Hoff reaction
ΔH = enthalpy of reaction k 1, k 2 → equilibrium constant Case 1 : If ΔH = 0 logK 2 − logK 1 = 0 : k 1 = k 2 Case 2 : If ΔH > 0 endo k 2 > k 1 Prediction of extent of completion of reaction : k c > 1 forward reaction is favourable k c < 1 backward reaction is favourable A + B \rightleftharpoons C + D k_{c} = \frac{[C] [D]}{[A][B]}
∴ k
C = 10 3 forward favours k c = 10 −3 backward favours If k c = 1 equilibrium Reaction quotient (Q C):
A + B \rightleftharpoons C + D
Q_{C} = \frac{[C] [D]}{[A][B]} and Q_{P} = \frac{P_{C}\times P_{D}}{P_{A}\times P_{B}}
case I : If Q
C= k C, reaction is in equilibrium
Case II : If Q
C< k Cforward reaction
Case III : Q
C> k Cback ward reaction Part1: View the Topic in this Video from 0:11 to 11:26 Part2: View the Topic in this Video from 3:55 to 14:00
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1. For a general reaction, \tt aA + bB \rightleftharpoons cC + dD
Rate of forward reaction \propto \left[A\right]^{a} \left[B\right]^{b} = k_{f}\left[A\right]^{a} \left[B\right]^{b} Rate of backward reaction \propto \left[C\right]^{c} \left[D\right]^{d} = k_{b}\left[C\right]^{c} \left[D\right]^{d} Where, k f and kare rate constants. b
2. At equilibrium,
Rate of forward reaction = Rate of backward reaction k_{f}\left[A\right]^{a} \left[B\right]^{b} = k_{b} \left[C\right]^{c} \left[D\right]^{d} \frac{k_{f}}{k_{b}} = K_{c} = \frac{\left[C\right]^{c} \left[D\right]^{d}}{\left[A\right]^{a} \left[B\right]^{b}} Where, K c is called the equilibrium constant.
3. For a gaseous reaction, \tt aA + bB \rightleftharpoons cC + dD
K_{p} = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b}
4.
Relation between K c and Kp
K_{p} = K_{c}\left[RT\right]^{\Delta n_{g}}
Where, Δn
g= moles of products − moles of reactants (gaseous only)
5. For a general reaction, \tt aA + bB \rightleftharpoons cC + dD
Which is not at equilibrium, Q_{c} = \frac{\left[C\right]^{c}\left[D\right]^{d}}{\left[A\right]^{a}\left[B\right]^{b}} |
This feat is notorious for its poor wording. The “+100%” phrasing is completely unique within D&D 3.5e as far as I know, for example. Ultimately, I can’t imagine any other interpretation here than adding again the number subtracted from your attack rolls, and it does have the nice feature of specifying the “normal” damage from Power Attack which means that features like the frenzied berserker’s supreme power attack that already give one-handed weapons 2:1 returns don’t get doubled to 4:1, but instead go to the 3:1 you would normally expect from D&D’s multiplication rules.
But then there is the line you haven’t quoted:
If you use this tactic with a two-handed weapon, you instead triple the extra damage from Power Attack.
No bizarre “+100%” in sight! But also we have lost the useful reference to “normal” and now it is multiplying “the extra damage from Power Attack,” whatever that is for you. This is going to get us in trouble, you can just tell already.
So you are tripling the extra damage—not tripling the penalty applied. The problem here, well the first problem here, is that “the extra damage from Power Attack” is “twice the number subtracted from your attack rolls” when attacking two-handed. Worse, since “the extra damage from Power Attack” is calculated as twice the penalty, but isn’t itself subject to any multiplier, arguably the repeated-multiplication rules don’t apply, and that gets you a 2×3=6 rather than 1+(2−1)+(3−1)=4. So instead of 2:1 returns on Power Attack, you get
6:1 returns on Power Attack. Or maybe you get 5:1; it’s impossible to say since it’s worded so poorly. Plus, ya know, I suspect what they meant to do was give you 3:1 returns, but of course they didn’t say that.
And that would combine quite nicely with, say, the supreme power attack feature of the frenzied berserker, who was getting 4:1 returns to begin with. Now they’re arguably getting 8:1.
On top of those issues, this is
only the Power Attack bonus damage. The result is added to the rest of your damage, and that gets you your full damage... which might be multiplied again, e.g. with valorous. This effectively multiplies your multiplier, which is exactly what the multiplication rules try to avoid, but since two different things are being multiplied, the multiplication rules don’t actually come into play.
So for the example: 2d6+1 damage from the weapon itself, +6 for Strength, and the −6 attack penalty for maximum Power Attack results in double that for +12 damage from Power Attack without Leap Attack. Thus 2d6+19 is the baseline for all interpretations, and
valorous doubles that for 4d6+38.
With the 6:1 returns, we are instead looking at Power Attack bonus of +36 (six times the penalty, triple “the extra damage from Power Attack” which would have been +12). Using 5:1 brings that down to +30, which is somewhat better, but not, ya know, great, when what they probably meant was +18. Note that +36 is nearly what
valorous was giving the entire attack before. Now with valorous, we’re looking at a total of 4d6+ 66—of which, 52 comes from Power Attack.
It may not be a bad idea to try to eliminate the multiplication of a multiplier here through houserule, but note that the Power Attack bonus damage isn’t the only case of this: the bonus damage due to Strength also has a multiplier, +1½×, which is
also being doubled by valorous. This, unlike Leap Attack, has strong precedent in the rules. The “fix” would be to apply the multiplication rule individually to all sources of damage, like so:
\begin{array}{r}2 \times ( && 2\text{d}6 && +1 && +1\tfrac{1}{2}\times 4 && +3\times 2\times 6 & ) \\= && 2\times 2\text{d}6 && + 2\times 1 && + 2\times 1\frac{1}{2}\times 4 && + 2\times 3\times 2\times 6 \\= & [1 \\ && +\left(2-1\right) \\ & ] & \times 2\text{d}6 & +[1 \\ && && +\left(2-1\right) \\ && & ] & \times 1 & +[1 \\ && && && +\left(2-1\right) \\ && && && +\left(1\frac{1}{2}-1\right) \\ && && & ] & \times 4 & +[1 \\ && && && && +\left(2-1\right) \\ && && && && +\left(3-1\right) \\ && && && && +\left(2-1\right) \\ && && && & ] & \times 6 \\= && 2\times 2\text{d}6 && +2\times 1 && +2\frac{1}{2}\times 4 && +5\times 6 \\= && 4\text{d}6 && +2 && +10 && +30 \\= && && && && 4\text{d}6+42 \\\end{array}
But this is very-definitely a houserule, and I’m not convinced that it
is good (I mean, good luck calculating that for every attack!), even though it “enforces” the idea that you’re not supposed to get to mulitply multipliers. |
The shared secret generated by the Diffie–Hellman key exchange is a random element of the subgroup of the multiplicative group modulo $p$ generated by $g$.
In particular, for $g$ and $p$ chosen as specified in RFC 2631 section 2.2, i.e. so that $p = jq+1$, where $q$ and $p$ are both prime, $j$ is a small number (often 2, making $p$ as safe prime) and $g$ generates the order $q$ subgroup modulo $p$, this subgroup contains $q$ elements and the shared secret therefore (assuming that the private keys are appropriately generated, of course) has $\log_2 q$ bits of entropy.
The recommended way to extract this entropy into a useful form is to hash the shared secret or, better yet, feed it into a key derivation function like HKDF (RFC 5869).What you're asking is how bad is it if we simply take the first (or last) 256 bits of the binary encoding of the shared secret and use them as an AES-256 key?
My answer, a bit surprisingly, would be "
not too bad." We may not get quite 256 bits of entropy out of the secret that way, but we ought to get at least 247 bits or so (assuming that the secret is long enough to have that many bits in the first place, of course), which is still plenty enough for any practical purposes.
Specifically, let's assume that $p$ is a safe prime (i.e. $j=2$), and that the secret is padded to the minimum number of bytes needed to encode $p$, as specified in RFC 2631. Thus, the padded secret will be a bitstring with $n = 8 \cdot \lceil (\log_2 (p+1)) /8 \rceil \le \log_2 p + 8$ bits, and will contain $$m = \log_2 q = \log_2((p-1)/2) = \log_2(p-1)-1 \approx \log_2 p - 1$$ bits of entropy. When we truncate the secret to 256 bits, we throw away some of this entropy, but at most only as many bits as we remove from the length. Thus, the truncated secret will contain at least $m - (n - 256) = 256 - (n - m) \ge 256 - 9 = 247$ bits of entropy.
Now, in the code you refer to, the secret is apparently
not padded to a fixed number of bytes, but rather has leading zero bytes removed. However, at least assuming that $q \gg 2^{256}$, this still should not make much difference. (In fact, in some cases it might even slightly increase the entropy left after truncation.)
To see why, let $k = 8 \cdot \lceil (\log_2(z+1))/8 \rceil$ be the number of bits needed to encode the secret $z$. First, we'll observe that ${\rm Pr}[k < 256] = {\rm Pr}[z < 2^{256}] \le 2^{256} / q = 2^{256-\log_2 q}$. For, say, $q > 2^{336}$, this probability is negligible (less than $2^{-80}$). Thus, for large enough $q$, we may safely assume $k \ge 256$.
Now, for each possible length $k$, the secret $z$ lies in the interval $2^{k-8} \le z < 2^k$. Using the same argument as above, we may show that, for any given $k$, $z$ must have at least $k - 9$ bits of entropy, and that truncating it to 256 bits must leave at least $256 - 9 = 247$ bits of entropy. Since we'll have at least 247 bits of entropy regardless of $k$ (except in negligible cases), we'll have at least 247 bits of entropy overall.
(For $j > 2$, the constant $9$ above becomes $8 + \log_2 j$ instead, but that makes little difference unless $j$ is huge, which it normally shouldn't be.)
All that said, none of the above should be taken as an argument against using a proper KDF to convert Diffie–Hellman secrets into key material. You definitely
should do that, if only because a) the standards say so, b) it's still more secure and c) it provides other benefits such as salting.Still, it does appear that simply using the raw truncated D–H shared secret as an encryption key may not be quite as bad as one might assume just based on reading the RFCs. |
This question already has an answer here:
Is it possible to derive the Coulomb's law using the principles of quantum electrodynamics? How?
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This question already has an answer here:
Is it possible to derive the Coulomb's law using the principles of quantum electrodynamics? How?
There are multiple ways to interpret Coulomb's law in quantum electrodynamics (QED). Interestingly, they don't lead to quite the same conclusion (but there is no inconsistency because they are not defined in the same way).
The most commonly used way (that ACuriousMind refers to in his comment) consists in relating the notion of classical potential with that of the scattering cross section between two charges. The reason why this is the most commonly used method is purely field-related in the sense that (high-energy) particle physicists are the ones who use the most QED and what they measure in fact is cross sections; so that makes sense.
In this scattering interpretation of the Coulomb's law, the interaction is related to the S matrix expression in terms of Feynman diagrams. The first diagram gives the classical Coulomb's law which is independent of the Planck constant $h$ while higher order terms contain quantum corrections that vanish as $h \rightarrow 0$. An example for scalar QED can be found here (page 12).
There is however another way to interpret the Coulomb's law that is linked to the field of low energy physics (out of which we get, atoms, molecules and so forth).
In this interpretation (more adapted to low energy physics), the Coulomb potential is the QED ground state energy of a system of two point charges $q_1$ and $q_2$ pinned at two locations separated by a distance $r_{12}$.
If we write first the hamiltonian of the free Electro Magnetic (EM) field, we have:
\begin{equation} \hat{H}_{R} = \sum_{i = 1,2} \sum_{\mathbf{k}} \hbar \omega(\mathbf{k}) \:a^{\dagger}_{i,\mathbf{k}} a_{i,\mathbf{k}} \end{equation} where $i$ stands for the two possible transverse polarizations and $\mathbf{k}$ is the wave vector of the photon.
With this definition of the free EM energy, we have that the vacuum energy for the free EM field is
\begin{equation} \langle 0| \: \hat{H}_{R} \: |0 \rangle = 0 \end{equation}
Now, when we introduce the two pinned charges, we need to account for the coupling between them and the EM field. The hamiltonian of the whole system is then:
\begin{equation} \hat{H}_{tot} = \hat{H}_{R} + \hat{H}_{coupling} \end{equation}
where
\begin{equation} \hat{H}_{coupling} = \int d^3r \: A_{\mu}\: j^{\mu} \end{equation}
where $\{A_{\mu}\}$ is the 4-potential of the EM field and $\{j^{\mu}\} \equiv (c\rho, \mathbf{j})$ is the 4-current density. In our case of pinned charges we have that $\{j^{\mu} \} = (c[q_1\delta(\mathbf{r}-\mathbf{r}_1)+q_2\delta(\mathbf{r}-\mathbf{r}_2)], \mathbf{0})$.
In a covariant formulation of QED, the field $A_s = A_0$ can be written in term of scalar photons (which have some unusual properties):
\begin{equation} A_s(\mathbf{r}) = \int d^3k \sqrt{\frac{\hbar}{2\varepsilon_0 \omega (2\pi)^3}}(a_s(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{r}} + a^{\dagger}_s(\mathbf{k})e^{-i\mathbf{k}\cdot \mathbf{r}}) \end{equation} which leads to \begin{equation} \hat{H}_{coupling} = \int d^3k c\sqrt{\frac{\hbar}{2\varepsilon_0 \omega (2 \pi)^3}}\left[a_s(\mathbf{k})(q_1 e^{i\mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i\mathbf{k}\cdot \mathbf{r}_2}) + a^{\dagger}_s(\mathbf{k})(q_1 e^{-i\mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{-i\mathbf{k}\cdot \mathbf{r}_2}) \right] \end{equation}
which are not observable as they do not contribute to the radiation field energy $\hat{H}_R$.
Because there are charges present in the system, the ground state energy is now different from zero. We can try to estimate the shift $\Delta E$ by which the vacuum energy has been changed by using perturbation theory up to the first non zero term:
\begin{equation} \Delta E = \langle 0| \hat{H}_{coupling} | 0 \rangle + \sum_{n \neq 0} \frac{|\langle n|\hat{H}_{coupling}| 0 \rangle |^2 }{- E_n} + .... \end{equation}
as it is quite obvious, the first term is zero by definition of the vacuum of the free field. The second term is only non zero for Fock states where there is a scalar photon produced. We thus get:
\begin{equation} \Delta E \approx \int d^3k \frac{|\langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle |^2}{-\hbar \omega} \end{equation}
We now have that:
\begin{equation} \langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle = c\sqrt{\frac{\hbar}{2 \varepsilon_0 \omega(\mathbf{k})}}(q_1 e^{i \mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i \mathbf{k}\cdot \mathbf{r}_2}) \end{equation}
This is now where the weirdness of the scalar photons shows up (in this very sketchy calculations). When we compute the square norm of the matrix element, it turns out that it is negative for scalar photons (this trick enables to make more troublesome thing disappear and long story short that's how it works). We thus have that:
\begin{equation} |\langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle |^2 = - \frac{c^2 \hbar}{2 \varepsilon_0 \omega(\mathbf{k})} |(q_1 e^{i \mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i \mathbf{k}\cdot \mathbf{r}_2})|^2 \end{equation}
Upon replacing in the expression for $\Delta E$, one finds that:
\begin{equation} \Delta E \approx \epsilon_1 + \epsilon_2 + V_{coul} \end{equation}
where
\begin{equation} \epsilon = \int d^3k \frac{q^2}{2 \varepsilon_0 k^2} \end{equation}is the self interaction energy of the charges with themselves (can be interpreted as the emission and absorption of a scalar photon by the same charge)
and
\begin{equation} V_{coul} = \int d^3k \frac{q_1 q_2 e^{i\mathbf{k}\cdot(\mathbf{r}_1-\mathbf{r}_2)}}{2\varepsilon_0 k^2} = \frac{q_1q_2}{4\pi \varepsilon_0 |\mathbf{r}_1-\mathbf{r}_2|} \end{equation}is the Coulomb interaction potential.
Note that although for simplicity I have only used the perturbation expansion up to second order, it turns out that in fact the exact EM ground state with pinned charges at fixed locations is
exactly equal to the $\Delta E$ we have calculated. To make the point clear, there is no such thing as quantum correction to the Coulomb interaction in this setup as it is an exact result from QED. In this view, quantum corrections arise when one introduces a dynamic field theory for the charged particles as well; hence the differences with the scattering picture.
For more on this picture, I recommend greatly this book from which the above summary is heavily inspired (although written in a simpler fashion). |
I'm sorry if this is a duplicate in any way. Throughout I use cycle notation and write maps $m:X\to Y$ on the right of their arguments (e.g. $xm=y$ for $m(x)=y$).
This is Exercise 1.7 of Howie's
Fundamentals of Semigroup Theory. The Details:
Definition 1:Let $S$ be a semigroup and $A$ be a subset of $S$. The subsemigroup of $S$ generated by $A$is given by the intersection of all the subsemigroups of $S$ containing $A$, denoted $\langle A\rangle$.
Definition 2:The full transformation semigroup$\mathscr{T}_n$ on the set $N=\{1, \dots,n\}$ is given by all maps $\alpha:N\to N$, together with composition of transformations.
Definition 3:For $i\ne j$ in $N$, let $\lvert\lvert i\, j \rvert\rvert$ denote the map $\phi\in\mathscr{T}_n$ for which $$i\phi =j,\quad x\phi=x\quad (x\ne i).$$
Let $\pi=\lvert\lvert 1\, 2 \rvert\rvert$.
Lemma 1:The following hold: $$\begin{align} (1\, i)\circ\pi\circ (1\, i)&=\lvert\lvert i\, 2 \rvert\rvert\quad (i\ge 3), \\ (2\, j)\circ\pi\circ (2\, j)&=\lvert\lvert 1\, j \rvert\rvert\quad (i\ge 3), \\ (i\, j)\circ\lvert\lvert i\, j \rvert\rvert\circ (i\, j)&=\lvert\lvert j\, i \rvert\rvert\quad (i, j\ge 1, i\ne j). \end{align}$$ Proof: Just plug & chug.$\square$
Lemma 2:Let $\varphi\in\mathscr{T}_n$ with $\lvert\operatorname{im}\varphi\rvert=r\le n-1$. Let $i\ne j$ such that $i\varphi=j\varphi$ and let $z\in N\setminus\operatorname{im}\varphi.$ Then $$\varphi=\lvert\lvert i\, j \rvert\rvert\circ\hat{\varphi},$$ where $$i\hat{\varphi}=z,\quad k\hat{\varphi}=k\varphi\quad (k\ne i).$$ Proof: This is again just a matter of plug & chug: the maps agree on $N$.$\square$ The Question:
Let's get a possible typo out of the way first.
A subquestion:Does $$(1\, i)\circ(2\, j)\circ\lvert\lvert i\, j \rvert\rvert\circ (2\, j)\circ (1\, i)=\lvert\lvert i\, j \rvert\rvert$$ for $i,j\ge 3, i\ne j$?
Perhaps I'm being stupid but I can't get the LHS to agree with the RHS on $1$ (as $1\mapsto i\mapsto i \mapsto j\mapsto 2\mapsto 2$ but I need $1\lvert\lvert i\, j \rvert\rvert=1$). The correct version of this result is considered as part of
Lemma 1.
Let $\tau=(1\, 2), \zeta=(1\, 2\, \dots \, n)$.
Deduce (from
Lemma 1and Lemma 2) that $\mathscr{T}_n=\langle\zeta, \tau, \pi\rangle$. My Attempt:
I'm not sure where to start. I did the previous question of Howie's book (showing the
symmetric group $\mathscr{S}_n=\langle\tau, \zeta\rangle$) without any trouble. That might be of some use here. I can see how Lemma 1 might be made to fit $\zeta, \tau$ and $\pi$, making elements of $\mathscr{T}_n$.
Please help :) |
Class to display expressions in a format that Sympy can parse. Will throw an exception if a Cadabra Ex object cannot be understood by Sympy. Can also convert expressions back to Cadabra notation. This class contains the printing logic that you see in action when you call the
__sympy__ method on an Ex object.
#include <DisplaySympy.hh>
virtual bool needs_brackets (Ex::iterator it) override Determine if a node needs extra brackets around it. More...
void print_multiplier (std::ostream &, Ex::iterator) Output the expression to a sympy-readable form. More... void print_opening_bracket (std::ostream &, str_node::bracket_t) void print_closing_bracket (std::ostream &, str_node::bracket_t) void print_parent_rel (std::ostream &, str_node::parent_rel_t, bool first) void print_children (std::ostream &, Ex::iterator, int skip=0) virtual void dispatch (std::ostream &, Ex::iterator) override For every object encountered, dispatch will figure out the most appropriate way to convert it into a LaTeX expression. More... void print_productlike (std::ostream &, Ex::iterator, const std::string &inbetween) Printing members for various standard constructions, e.g. More... void print_sumlike (std::ostream &, Ex::iterator) void print_fraclike (std::ostream &, Ex::iterator) void print_commalike (std::ostream &, Ex::iterator) void print_arrowlike (std::ostream &, Ex::iterator) void print_powlike (std::ostream &, Ex::iterator) void print_intlike (std::ostream &, Ex::iterator) void print_equalitylike (std::ostream &, Ex::iterator) void print_components (std::ostream &, Ex::iterator) void print_partial (std::ostream &str, Ex::iterator it) void print_matrix (std::ostream &str, Ex::iterator it) void print_other (std::ostream &str, Ex::iterator it) bool children_have_brackets (Ex::iterator ch) const
std::map< std::string, std::string > symmap Map from Cadabra symbols to Sympy symbols. More... std::multimap< std::string, std::string > regex_map std::map< nset_t::iterator, Ex, nset_it_less > depsyms Map from symbols which have had dependencies added to an expression containing these dependencies. More... ◆ DisplaySympy()
DisplaySympy::DisplaySympy ( const Kernel & kernel, const Ex & e )
◆ children_have_brackets()
bool DisplaySympy::children_have_brackets ( Ex::iterator ch ) const private
◆ dispatch()
void DisplaySympy::dispatch ( std::ostream & str, Ex::iterator it ) overrideprivatevirtual
For every object encountered, dispatch will figure out the most appropriate way to convert it into a LaTeX expression.
This may be done by simply looking at the object's name (e.g.
\prod will print as a product) but may also involve looking up properties and deciding on the best course of action based on the attached properties.
Implements cadabra::DisplayBase.
◆ import()
void DisplaySympy::import ( Ex & ex )
Rewrite the output of sympy back into a notation used by Cadabra.
This in particular involves converting 'sin' and friends to
\sin and so on, as well as converting all the greek symbols. Currently only acts node-by-node, does not do anything complicated with trees.
◆ needs_brackets()
bool DisplaySympy::needs_brackets ( Ex::iterator it ) overrideprotectedvirtual
Determine if a node needs extra brackets around it.
Uses context from the parent node if necessary. Has to be implemented in a derived class, because the answer depends on the printing method (e.g.
(a+b)/c needs brackets when printed like this, but does not need brackets when printed as
\frac{a+b}{c}).
Implements cadabra::DisplayBase.
◆ preparse_import()
std::string DisplaySympy::preparse_import ( const std::string & in )
◆ print_arrowlike()
void DisplaySympy::print_arrowlike ( std::ostream & str, Ex::iterator it ) private
◆ print_children()
void DisplaySympy::print_children ( std::ostream & str, Ex::iterator it, int skip =
0
) private
◆ print_closing_bracket() ◆ print_commalike()
void DisplaySympy::print_commalike ( std::ostream & str, Ex::iterator it ) private
◆ print_components()
void DisplaySympy::print_components ( std::ostream & str, Ex::iterator it ) private
◆ print_equalitylike()
void DisplaySympy::print_equalitylike ( std::ostream & str, Ex::iterator it ) private
◆ print_fraclike()
void DisplaySympy::print_fraclike ( std::ostream & str, Ex::iterator it ) private
◆ print_intlike()
void DisplaySympy::print_intlike ( std::ostream & str, Ex::iterator it ) private
◆ print_matrix()
void DisplaySympy::print_matrix ( std::ostream & str, Ex::iterator it ) private
◆ print_multiplier()
void DisplaySympy::print_multiplier ( std::ostream & str, Ex::iterator it ) private
Output the expression to a sympy-readable form.
For symbols which cannot be parsed by sympy, this can convert to an alternative, Such rewrites can then be undone by the 'import' member above.
◆ print_opening_bracket() ◆ print_other()
void DisplaySympy::print_other ( std::ostream & str, Ex::iterator it ) private
◆ print_parent_rel() ◆ print_partial()
void DisplaySympy::print_partial ( std::ostream & str, Ex::iterator it ) private
◆ print_powlike()
void DisplaySympy::print_powlike ( std::ostream & str, Ex::iterator it ) private
◆ print_productlike()
void DisplaySympy::print_productlike ( std::ostream & str, Ex::iterator it, const std::string & inbetween ) private
Printing members for various standard constructions, e.g.
print as a list, or as a decorated symbol with super/subscripts etc. The names reflect the structure of the output, not necessarily the meaning or name of the object that is being printed.
◆ print_sumlike()
void DisplaySympy::print_sumlike ( std::ostream & str, Ex::iterator it ) private
◆ depsyms
std::map<nset_t::iterator, Ex, nset_it_less> cadabra::DisplaySympy::depsyms private
Map from symbols which have had dependencies added to an expression containing these dependencies.
◆ regex_map
std::multimap<std::string, std::string> cadabra::DisplaySympy::regex_map private
◆ symmap
std::map<std::string, std::string> cadabra::DisplaySympy::symmap private
Map from Cadabra symbols to Sympy symbols.
The documentation for this class was generated from the following files: |
Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form$$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$into the int...
[Some time travel comments] Since in the previous paragraph, we have explained how travelling to the future will not necessary result in you to arrive in the future that is resulted as if you have never time travelled (via twin paradox), what is the reason that the past you travelled back, has to be the past you learnt from historical records :?
@0ßelö7 Well, I'd omit the explanation of the notation on the slide itself, and since there seems to be two pairs of formulae, I'd just put one of the two and then say that there's another one with suitable substitutions.
I mean, "Hey, I bet you've always wondered how to prove X - here it is" is interesting. "Hey, you know that statement everyone knows how to prove but doesn't bother to write down? Here is the proof written down" significantly less so
Sorry I have a quick question: For questions like this physics.stackexchange.com/questions/356260/… where the accepted answer clearly does not answer the original question what is the best thing to do; downvote, flag or just leave it?
So this question says express $u^0$ in terms of $u^j$ where $u$ is the four-velocity and I get what $u^0$ and $u^j$ are but I'm a bit confused how to go about this one? I thought maybe using the space-time interval and evaluating for $\frac{dt}{d\tau}$ but it's not workin out for me... :/ Anyone give me a quickie starter please? :p
Although a physics question, this is still important to chemistry. The delocalized electric field is related to the force (and therefore the repulsive potential) between two electrons. This in turn is what we need to solve the Schrödinger Equation to describe molecules. Short answer: You can calculate the expectation value of the corresponding operator, which comes close to the mentioned superposition. — Feodoran13 hours ago
If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?
@0ßelö7 I just looked back at chat and noticed Phase's question, I wasn't purposefully ignoring you - do you want me to look over it? Because I don't think I'll gain much personally from reading the slides.
Maybe it's just me having not really done much with Eigenbases but I don't recognise where I "put it in terms of M's eigenbasis". I just wrote it down for some vector v, rather than a space that contains all of the vectors v
If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?
Honey, I Shrunk the Kids is a 1989 American comic science fiction film. The directorial debut of Joe Johnston and produced by Walt Disney Pictures, it tells the story of an inventor who accidentally shrinks his and his neighbor's kids to a quarter of an inch with his electromagnetic shrinking machine and throws them out into the backyard with the trash, where they must venture into their backyard to return home while fending off insects and other obstacles.Rick Moranis stars as Wayne Szalinski, the inventor who accidentally shrinks his children, Amy (Amy O'Neill) and Nick (Robert Oliveri). Marcia... |
I'm trying to design a circuit with an operational amplifier (OPA625). The input signal is amplified and sent to an ADC. In order to estimate the input noise, I'm using the following equation from an Analog Devices tutorial:
$$ v_{n,\text{rms}} (F_L, F_C) = v_{\text{nw}} \cdot \sqrt{F_C} \cdot \sqrt{\displaystyle\int_{F_L}^{F_C} \frac{1}{f} \mathrm{d} f} $$
$$ v_{n,\text{rms}} (F_L, F_C) = v_{\text{nw}} \cdot \sqrt{F_C \ln{\dfrac{F_C}{F_L}}} $$
\$v_{\text{n,rms}}\$ — input noise RMS
\$F_L, F_H\$ — frequency bandwidth
\$v_{\text{nw}}\$ — voltage noise density in the white noise area
It's quite easy to calculate the noise for the specified bandwidth. But what happens when \$F_L\$ is close to 0? In the Analog Devices' tutorial I mentioned earlier, \$0.1\text{ Hz}\$ is used as a lower frequency limit, but how close it should be to 0? How do I estimate the noise for DC input? |
Hey guys, I'm concerned with bounding the following sum of gauss sums from above $$\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)},$$ where $p$ runs through the primes $\leq x$, $\chi$ runs through the multiplicative characters modulo $p$ and $e\left(\frac{a}{p}\right)=\exp\left(\frac{2\pi ia}{p}\right)$. By using orthogonality relations of characters one gets $$\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}=(p-1)\sum_{a=1}^{p-1}~{e\left(\frac{a}{p}\right)\frac{p-1}{ord_pa}},$$ where $ord_pa$ denotes the multiplicative order of $a$ modulo $p$. The right side can be bounded trivially by $$(p-1)\sum_{a=1}^{p-1}~{\frac{p-1}{ord_pa}}=(p-1)^2\sum_{d\mid p-1}{\frac{\varphi(d)}{d}},$$ $\varphi(d)$ denoting Euler's totient function. Using $\varphi(n)\leq n$ one gets the estimate $$\left|\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}\right|\leq\sum_{p\leq x}{\tau(p-1)},$$ where $\tau(n)$ is the number of divisors of $n$. The latter sum can be shown to be asymptotically equivalent to a positive constant times $x$. I would like to know if there is a way to show that the sum is $o(x)$.
There should be a bunch of cancellation. Here is an idea. You need to relate your sums $\sum_a e(a/p)/ord_p a$ to the sums $\sum_{ord_p a | m} e(a/p)/m$. Now, if $mr = p-1$,
$\sum_{ord_p a | m} e(a/p) = (1/r)\sum_{n=1}^{p-1} e(n^r/p) = O(p^{1/2})$
by the Weil bound. This will deal with the elements of large order, I believe. There is work to do, but this should get you going.
This is a comment rather than an answer, but it is too long. Let $g$ be some generator of the multiplicative group. Then
$$\sum_{a=1}^{p-1}e\left(\frac{a}{p}\right)\frac{p-1}{ord_{p}a}=\sum_{k=1}^{p-1}e\left(\frac{g^{k}}{p}\right)\gcd,\left(p-1,k\right).$$
Rearranging yields $$\sum_{d|p-1} \phi(d) \sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right) $$ so that the entire sum is $$\sum_{p\leq x}\frac{1}{p-1}\sum_{d|p-1}\phi(d)\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right).$$
My hope in posting this is that there are existing bounds on sums of the form $\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right)$. It might be strange to deal with, as it is a sum over elements chosen for their multiplicative properties. Essentially, we would need a theorem regarding how these multiplicative elements are distributed among the residue classes, and that it cannot be "too far from uniform". |
Two elementary electric charges \(e\) interact. From the Coulomb force \(F(r)= e^2/ 4 \pi \epsilon_0 r^2\) between them, which decreases as \(1/r^2\), the fine structure constant is defined as:
\(\alpha= \frac{F(r) \ r^2 }{\hbar c}\). This gives \(\alpha = \frac{e^2}{ 4 \pi \epsilon_0 \hbar c}\) , which is Sommerfeld's formula in SI units. The value of \(\alpha\) is measured to be about 1/137.04 and describes the strength of the Coulomb force, or, equivalently, the strength of electromagnetism.
Here, \(c\) is the speed of light, and \(\hbar\) is Planck's constant, and \(r\) is the distance between the charges. Now, a researcher proposes the following argument in two steps:
1. The Coulomb force \(F(r) \) between two elementary charges, i.e. the force due to exchanging *virtual* photons, is surely larger than than the force between two *neutral* (microscopic) *black holes* that appears when they exchange *thermal* photons.
Both forces are electromagnetic in origin, and we neglect gravity effects completely, by assuming extremely small bodies throughout, in the microscopic regime, with negligible masses and/or negligible gravitational interaction. We also assume that the charges are of the same sign, and that the forces are thus repulsive. In particular, the thermal photons from one hot particle push against the other particle and thus induce a force.
2. This last force can be calculated. It is given as
\(F_{BHTP}= \frac{P_{BH}}{c} \frac{A_{BH}/ 2}{ 4 \pi \ r^2 }\)
where \(P_{BH}\) is the power of the black hole Bekenstein-Hawking radiation, \(A_{BH}\) the surface of the black hole, and \(r\) the distance between the two black holes (which is assumed much larger than the black holes, so that space can be approximated as flat).
If we insert the formulae for \(P_{BH}\) and \(A_{BH}\) from the wikipedia (article on Hawking radiation), we get
\(\alpha > \frac{F_{BHTP} \ r^2}{ \hbar c} = \frac{1}{ 7680 \pi} = \frac{1}{ 24127,...}\)
Since alpha needs to be smaller than 1 for many reasons (a simple one: probabilities are always smaller than one), we have deduced:
\(1 > \alpha > \frac{1}{ 24127,...}\)
which is not a great result, but still a result that is better than nothing.
Is this argument correct? Or is there a mistake? In particular, is it correct to assume that
F_(Coulomb) > F _ (neutral black holes)
in the case of negligible gravitation? The algebra of the calculations is correct, but is the argument correct? I have never seen a limit of alpha, ever, and this one, even though it is a rough one, is the first I have ever seen deduced from physical formulae. The issue is really whether the force comparison is valid also for microscopic hot bodies and microscopic black holes. What do you think? The force sequence indirectly assumes that microscopic particles might be "hot"- so it is questionable. But still intriguing.
All this is not my own idea, nor of a friend of mine: I found it via the wikipedia entry on the fine structure constant in this paper by John P. Lestone: https://arxiv.org/abs/physics/0703151 I simplified the argument in the paper, took away all misleading issues, and corrected the mistakes (there are two factors of 4 pi lost in the paper). The idea is already ten years old but I only found it a few days ago.
---
A few comments:
A. The expression for \(F_{BHTP}\) might not be an equality, because the effective surface might be different and not exactly given by \(A_{BH}/2\).
B. This reasoning does for electrodynamics the same that Verlinde did for gravity: both assume that a macroscopic force is in fact due to many microscopic degrees of freedom.
C. Also the strength of gravity is conventionally defined using the above expression \(\alpha_G= \frac{F(r) \ r^2 }{\hbar c}\) |
Let
$(\Omega_i,\mathcal A_i)$ be a measurable space $\emptyset\ne A_i\subseteq\Omega_i$
Are we able to show that $\left.\mathcal A_1\otimes\mathcal A_2\right|_{A_1\times A_2}=\left.\mathcal A_1\right|_{A_1}\otimes\left.\mathcal A_2\right|_{A_2}$?
Above $\left.\mathcal B\right|_B:=\left\{B\cap B':B'\in\mathcal B\right\}$ denotes the trace $\sigma$-algebra of $B$ in $\mathcal B$ and $\mathcal B_1\otimes\mathcal B_2$ the product $\sigma$-algebra of $\mathcal B_1$ and $\mathcal B_2$.
I'm pretty sure that the statement is correct, but I didn't find an approach to show it. |
The Annals of Probability Ann. Probab. Volume 21, Number 3 (1993), 1671-1690. Strong Large Deviation and Local Limit Theorems Abstract
Most large deviation results give asymptotic expressions for $\log P(Y_n \geq y_n)$, where the event $\{Y_n \geq y_n\}$ is a large deviation event, that is, $P(Y_n \geq y_n)$ goes to 0 exponentially fast. We refer to such results as weak large deviation results. In this paper we obtain strong large deviation results for arbitrary random variables $\{Y_n\}$, that is, we obtain asymptotic expressions for $P(Y_n \geq y_n)$, where $\{Y_n \geq y_n\}$ is a large deviation event. These strong large deviation results are obtained for lattice valued and nonlattice valued random variables and require some conditions on their moment generating functions. These results strengthen existing results which apply mainly to sums of independent and identically distributed random variables. Since $Y_n$ may not possess a probability density function, we consider the function $q_n(y; b_n,S) = \lbrack(b_n/\mu(S))P(b_n(Y_n - y) \in S)\rbrack$, where $b_n \rightarrow \infty, \mu$ is the Lebesgue measure on $R$, and $S$ is a measurable subset of $R$ such that $0 < \mu(S) < \infty$. The function $q_n(y; b_n,S)$ is the p.d.f. of $Y_n + Z_n$, where $Z_n$ is uniform on $-S/b_n$, and will be called the pseudodensity function of $Y_n$. By a local limit theorem we mean the convergence of $q_n(y_n; b_n,S)$ as $n \rightarrow \infty$ and $y_n \rightarrow y^\ast$. In this paper we obtain local limit theorems for arbitrary random variables based on easily verifiable conditions on their characteristic functions. These local limit theorems play a major role in the proofs of the strong large deviation results of this paper. We illustrate these results with two typical applications.
Article information Source Ann. Probab., Volume 21, Number 3 (1993), 1671-1690. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176989136 Digital Object Identifier doi:10.1214/aop/1176989136 Mathematical Reviews number (MathSciNet) MR1235434 Zentralblatt MATH identifier 0786.60026 JSTOR links.jstor.org Citation
Chaganty, Narasinga Rao; Sethuraman, Jayaram. Strong Large Deviation and Local Limit Theorems. Ann. Probab. 21 (1993), no. 3, 1671--1690. doi:10.1214/aop/1176989136. https://projecteuclid.org/euclid.aop/1176989136 |
Toda system and interior clustering line concentration for a singularly perturbed Neumann problem in two dimensional domain
1.
Department of Mathematics, Chinese University of Hong Kong, Shatin, New Territories, Hong Kong
2.
Department of Mathematics, Shenzhen University, Shenzhen, China
non-degenerateand non-minimalwith respect to the curve length. For any given integer $N\ge 2$ and for small $\varepsilon$ away from certain critical numbers, we construct a solution exhibiting $N$ interior layers at mutual distances $O(\varepsilon|\ln\varepsilon|)$ whose center of mass collapse onto $\Gamma$ at speed $O(\varepsilon^{1+\mu})$ for small positive constant $\mu$ as $\varepsilon\to 0$. Asymptotic location of these layers is governed by a Toda system. Mathematics Subject Classification:Primary: 35B25, 35J60; Secondary: 35K99, 92C4. Citation:Juncheng Wei, Jun Yang. Toda system and interior clustering line concentration for a singularly perturbed Neumann problem in two dimensional domain. Discrete & Continuous Dynamical Systems - A, 2008, 22 (3) : 465-508. doi: 10.3934/dcds.2008.22.465
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2018 Impact Factor: 1.143
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After 4 months of work, it's finally done.
why the heck did this take this long
Photo is edited from Wikipedia. Thanks to Faminepulse for helping with the ionization effect.
I don't get it
would you read it/explain it to me ples?
M = |M_{1}|e^{i\theta_{1}}e^{i\phi_{1}} + |M_{2}|e^{i\theta_{2}}e^{i\phi_{2}}\tilde{M} = |M_{1}|e^{i\theta_{1}}e^{-i\phi_{1}} + |M_{2}|e^{i\theta_{2}}e^{-i\phi_{2}}
…
would you read it/explain it to me pleeeees?
Matter is being converted every two weeks to anti-matter, and back again. Apparently, the Doctor who was 'annihilated' was retroactively removed from existence. But it seems to have been kinda hap-hazard, as memories of him readily exist, and you can refer to him. It's just, everything he ever did personally has been undone on some level.
This is worse than him just being
entirely removed retroactively. If he were entirely removed, reality would've changed to fit his absence. But it hasn't. There are gaps forming in reality and paradoxes building up over time because it's happening to others as well.
My take from this is that he wasn't retroactively removed, but rather everything he directly created was converted out-of-phase just like he was. So, ink in anything he wrote, bytes in any files he created, etc. And then poof, and it's all gone.
It looks like you've done your research really well, and I know enough about antimatter for the twist near the end to make sense and be kind of a cool hook, but…most of the text of the article is so in-depth that I just glaze over while reading it. Unfortunately, I can't think of a good way to fix this (and retain the article's integrity.)
Upvoted despite begrudgingly reading through a physics lesson. The hook was worth it, and I think I learned something.
Impressive.
Not nearly as good as 1718, partly because I imagine a background in particle physics is what you need for this one to truly work (the gradual realization of the punchline yourself, if you will) but it's still a great article.
I'd say, don't change a thing. This is and will be a niche article, but it's a very good one.
Actually, if it helps, I know next to jack about particle physics and I was still starting to get it by the time the addendum handed it to me. It helped that I knew about matter/antimatter asymmetry, though.
SCP Wiki Administrator | Earth: We're all in this together.
It's a cool thought, a YK-Scenario happens bi-weekly. But I feel that the last note was
kind-of hanging of the edge of "believable" and "un-believable". I may be missing something, since I don't understand a lot of physics. Writing is good, tone, spelling and the rest is fantastic, so is the idea, but the last note felt a bit abrupt.
Sorry, but no-vote. May be liable to change.
650mSv
Thank you for knowing the difference between dose and exposure
Thank you for knowing the difference between dose and exposure Thank you for knowing the difference between dose and exposure Thank you for knowing the difference between dose and exposure
[REDACTED, SEE ADDENDUM 2123-ALPHA, LEVEL 4 CLEARANCE REQUIRED]
This should be a semicolon after REDACTED tho
In fact, though that would cut it, I'd change it to
[REDACTED; SEE ADDENDUM 2123-ALPHA (LEVEL 4 CLEARANCE REQUIRED)]
There are multiple spots where you forgot to replace their SCP Designation placeholder (instead of 2123, it says "y"). However, I like the detail you put in with the collapsibles. Adds to my immersion. +1 |
Define a sequence $a_0,a_1,a_2,...$ by setting $a_0=1$, $a_1=2$, $a_n=(a_{n-1}+a_{n-2})/2$ for $n \ge 2$.
A) Find the radius of convergence of the series $\sum_{n=0}^{\infty}a_nz^n$.
B) Find an explicit formula for the function f(z) defined by the series of part A.
My approach for A
Using the definition of ratio test, I obtain $$\frac{|a_{n-1}+a_{n-2}|}{|1+a_{n-1}|}$$ So I didn't know what to do next as $n \to \infty$ and went ahead to plug in some numbers $n=2,3..$. I get $1.5$ then it decreases to $1.2222...$ and it keeps decreasing so I'm guessing that the radius of convergence will be $R=1.5$.
My approach for B
Computing the first five terms of $a_n$,
I get $$a_0=1$$
$$a_1=2$$
$$a_2=1.5$$
$$a_3=1.75$$
$$a_4=1.625$$ So that $$f(z)=1+2z+1.75z^2+1.75z^2+1.625z^4+...$$
Am I doing this correctly? |
Just an update: with help from Dan Shepherd and Greg Kuperberg, I now understand how to solve this problem in deterministic, classical polynomial time,
even when the degree $d$ is as large as the field characteristic or larger (the case that originally interested me, and that David Speyer's excellent answer doesn't treat). As a consequence, I now know that (alas) no exponential quantum speedup is possible for this problem.
For concreteness, let's assume we're working with a degree-$d$ polynomial $p:\mathbb{F}_2^n \rightarrow \mathbb{F}_2$, whose coefficients are given to us explicitly. Our goal is to find the subspace of "secret directions" $s \in \mathbb{F}_2^n$ such that $p(x)=p(x\oplus s)$ for all $x\in \mathbb{F}_2^n$. If $\operatorname{char}(\mathbb{F})$ were large enough, then (as David's answer discusses) we would simply use Gaussian elimination to find those $s$'s for which the formal partial derivative, $\frac{\partial p}{\partial s}$, is the identically-$0$ polynomial. Our problem is that formal partial derivatives are no longer "semantically sane" when $d \ge \operatorname{char}(\mathbb{F})$. For example, over $\mathbb{F}_2$, we can have a formal polynomial like $x^3 + x$ that's constant even though its derivative $3x^2+1=x+1$ is nonzero, or conversely, a formal polynomial like $x^2$ that's non-constant even though its derivative is $2x=0$. Likewise, one can construct examples of polynomials $p:\mathbb{F}_2^3 \rightarrow \mathbb{F}_2$ that have a "secret direction" even though their three partial derivatives
$\frac{\partial p}{\partial x},\frac{\partial p}{\partial y},\frac{\partial p}{\partial z}$
are linearly independent (e.g., $p(x,y,z)=xy+yz+xz+x+y+z$); or conversely, that have no "secret direction" even though their three partial derivatives are linearly dependent (e.g., $p(x,y,z)=xy+yz+xz$).
But we can solve this problem with the help of two new ideas. The first idea is to look at the "finite differences,"
$\frac{dp}{ds}(x) := p(x)+p(x\oplus s)$,
instead of formal partial derivatives. Finite differences are "semantically sane" over $\mathbb{F}_2$; in particular, the problem we're trying to solve is simply to find the subspace of $s$'s such that $\frac{dp}{ds}(x)$ is the identically-$0$ polynomial.
Now, the problem with finite differences is that they don't behave simply under linear combinations: in particular, it's not true in general that
$\frac{dp}{d(s+t)}(x) = \frac{dp}{ds}(x) + \frac{dp}{dt}(x)$.
But this brings us to the second new idea: namely,
the above equation is true "to leading order." In more detail, $\frac{dp}{ds}(x)$ can be expanded out as a polynomial in both $s$ and $x$, which has total degree at most $d$, and degree at most $d-1$ in $x$ (since the degree-$d$ terms cancel). Now, suppose we look only at the terms of $\frac{dp}{ds}(x)$ that have degree $d-1$ in $x$, and not those terms that involve products of $d-2$ of the $x_i$'s or fewer. Then those terms will depend linearly on $s$, since the total degree can never exceed $d$. Thus, letting $q_s(x)$ be the degree-$(d-1)$ component of $\frac{dp}{ds}(x)$, we do indeed have
$q_{s+t}(x) = q_s(x) + q_t(x).$
Now let $S$ be the set of all $s\in \mathbb{F}_2^n$ on which $q_s(x)$ is the identically-$0$ polynomial---or in other words, on which
$\frac{dp}{ds}(x)$ has degree at most $d-2$, considered as a polynomial in $x$. Then what we learn from the above is that $S$ is a subspace of $\mathbb{F}_2^n$. Moreover, it's a subspace that can be found in polynomial time, by simply using Gaussian elimination to find those $s$'s for which $q_s(x)$ is the identically-$0$ polynomial.
Now, once we've found this degree-$(d-2)$ subspace $S$, we can then simply apply the same algorithm inductively, in order to find the subspace $S' \le S$ of directions $s$ such that $\frac{dp}{ds}(x)$ has degree at most $d-3$, considered as a polynomial in $x$. Once again, the trick is to restrict attention to the "leading-order," degree-$(d-2)$ components of the polynomials $\frac{dp}{ds}(x)$ (for $s\in S$), and then use Gaussian elimination to find the subspace of directions $s$ on which those degree-$(d-2)$ components vanish. (Once again, we take advantage of the fact that these degree-$(d-2)$ components behave linearly as functions of $s$, even though the rest of the polynomials don't.)
Next we find the subspace $S'' \le S'$ of directions such that $\frac{dp}{ds}(x)$ has degree at most $d-4$, and so on, until we're down to the subspace on which $\frac{dp}{ds}(x)$ has degree $0$, and then we're done. Or rather, we're
almost done: we still need to find the subspace on which $\frac{dp}{ds}(x)$ is the identically-$0$ polynomial, rather than the identically-$1$ polynomial. But that last part is trivial. |
Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$
Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water
does turn into vapour from liquid over time without any apparent interference.
Your math is correct but you left out a very important symbol from your equations. There is a big difference between $\Delta G$ and $\Delta G^\circ$. Only $\Delta G^\circ$ means the Gibbs energy change under
standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor.
Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$. Whether something is spontaneous under other conditions is determined by $\Delta G$. To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions.
Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas. But the partial pressure of water in our atmosphere is usually much lower than this. Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$. If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity. That would translate to an infinitely
negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction.
Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water. So water evaporation is still spontaneous.
Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation. And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set! |
Consider $u \in C_1^2(\Omega \times [0,T]), \Omega\subset\mathbb{R}^n$ as a solution of the problem
$ u_t - \Delta u = f, \text{ in } \Omega \times (0, T]$,
$u = 0, \text{ on } \partial\Omega \times [0,T]$,
$u(x,0) = g(x) , x \in \Omega$.
Then the so called "energy inequality"
$\int_\Omega |u(x,t)|^2 dx + \int_0^t \int_\Omega |\nabla u(x,\tau)|^2 dxd\tau \leq c \int_0^t \int_\Omega |f(x,\tau)|^2 dxd\tau + \int_\Omega |g(x)|^2 dx$
holds. I have never heard of such an energy inequality and I´m looking for a proof but I have not found one yet. |
Let $M$ be (say smooth) manifold. From the short exact sequence of groups $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0$ (where the first map is multiplication by $2$) one obtain long exact sequence in cohomology. In particular one obtains the connecting map $\beta:H^2(M,\mathbb{Z}_2) \to H^3(M,\mathbb{Z})$. This map is called the Bockstein homomorphism. Define $W_3(M):=\beta(w_2(M))$ where $w_2(M)$ is the second Stiefel-Whitney class. The class $W_3(M)$ is called the
third integral Stiefel-Whitney class. On the other hand there is another class in $H^3(M,\mathbb{Z})$ which at the first sight has nothing to do with $W_3(M)$: it is called Dixmier-Douady class and is defined in terms of bundles of simple $C^*$-algebras.
It turns out that these two classes coincide: this is proved in this paper by Plymen-see Theorem 2.8. However the proof relies on another result. The author gives precise reference:
Marry P. ,,
Varietes spinorielles. Geometrie riemannienne en dimension 4'', Seminaire Arthur Besse, CEDIC, Paris 1981
however I was unable to find it (even if I could, unfortunately I don't speak French). So
I would like to understand why $W_3(M)=\delta(M)$, in particular understand the last two lines of the case ''i)'' in the proof of Theorem 2.8 in the above paper.
EDIT: The relevant bundle for defining $\delta(M)$ is the (even part of) the complex Clifford bundle of the tangent bundle. Recall that the complex Clifford algebras are isomorphic to either $M_{2^n}(\mathbb{C})$ or to $M_{2^n}(\mathbb{C}) \oplus M_{2^n}(\mathbb{C})$ thus the even part of Clifford algebra is always a simple algebra. |
In his celebrated paper "Conjugate Coding" (written around 1970), Stephen Wiesner proposed a scheme for quantum money that is unconditionally impossible to counterfeit, assuming that the issuing bank has access to a giant table of random numbers, and that banknotes can be brought back to the bank for verification. In Wiesner's scheme, each banknote consists of a classical "serial number" $s$, together with a quantum money state $|\psi_s\rangle$ consisting of $n$ unentangled qubits, each one either
$$|0\rangle,\ |1\rangle,\ |+\rangle=(|0\rangle+|1\rangle)/\sqrt{2},\ \text{or}\ |-\rangle=(|0\rangle-|1\rangle)/\sqrt{2}.$$
The bank remembers a classical description of $|\psi_s\rangle$ for every $s$. And therefore, when $|\psi_s\rangle$ is brought back to the bank for verification, the bank can measure each qubit of $|\psi_s\rangle$ in the correct basis (either $\{|0\rangle,|1\rangle\}$ or ${|+\rangle,|-\rangle}$), and check that it gets the correct outcomes.
On the other hand, because of the uncertainty relation (or alternatively, the No-Cloning Theorem), it's "intuitively obvious" that, if a counterfeiter who
doesn't know the correct bases tries to copy $|\psi_s\rangle$, then the probability that both of the counterfeiter's output states pass the bank's verification test can be at most $c^n$, for some constant $c<1$. Furthermore, this should be true regardless of what strategy the counterfeiter uses, consistent with quantum mechanics (e.g., even if the counterfeiter uses fancy entangled measurements on $|\psi_s\rangle$).
However, while writing a paper about other quantum money schemes, my coauthor and I realized that we'd never seen a rigorous proof of the above claim anywhere, or an explicit upper bound on $c$: neither in Wiesner's original paper nor in any later one.
So,
has such a proof (with an upper bound on $c$) been published? If not, then can one derive such a proof in a more-or-less straightforward way from (say) approximate versions of the No-Cloning Theorem, or results about the security of the BB84 quantum key distribution scheme?
Update: In light of the discussion with Joe Fitzsimons below, I should clarify that I'm looking for more than just a reduction from the security of BB84. Rather, I'm looking for an explicit upper bound on the probability of successful counterfeiting (i.e., on $c$)---and ideally, also some understanding of what the optimal counterfeiting strategy looks like. I.e., does the optimal strategy simply measure each qubit of $|\psi_s\rangle$ independently, say in the basis
$$\{ \cos(\pi/8)|0\rangle+\sin(\pi/8)|1\rangle, \sin(\pi/8)|0\rangle-\cos(\pi/8)|1\rangle \}?$$
Or is there an entangled counterfeiting strategy that does better?
Update 2: Right now, the best counterfeiting strategies that I know are (a) the strategy above, and (b) the strategy that simply measures each qubit in the $\{|0\rangle,|1\rangle\}$ basis and "hopes for the best." Interestingly, both of these strategies turn out to achieve a success probability of (5/8) n. So, my conjecture of the moment is that (5/8) n might be the right answer. In any case, the fact that 5/8 is a lower bound on c rules out any security argument for Wiesner's scheme that's "too" simple (for example, any argument to the effect that there's nothing nontrivial that a counterfeiter can do, and therefore the right answer is c=1/2).
This post has been migrated from (A51.SE)
Update 3: Nope, the right answer is (3/4) n! See the discussion thread below Abel Molina's answer. |
On small oscillations of mechanical systems with time-dependent kinetic and potential energy
1.
Dipartimento di Matematica Pura e Applicata, Università de L'Aquila, I-67100 L'Aquila
2.
Bolyai Institute, University of Szeged, Aradi vértanúk tere 1, H-6720 Szeged, Hungary
$\sum$
n k=1$(a_{ik}(t)\ddot q_k+c_{ik}(t)q_k)=0,
(i=1,2,\ldots,n).$(*)
A nontrivial solution $q_1^0,\ldots ,q_n^0$ is called
small
if
$\lim _{t\to \infty}q_k(t)=0 (k=1,2,\ldots n).
It is known that in the scalar case ($n=1$, $a_{11}(t)\equiv 1$,
$c_{11}(t)=:c(t)$) there exists a small solution if $c$ is increasing and
it tends to infinity as $t\to \infty$.
Sufficient conditions for the existence of a small solution of the general system (*) are given in the case when coefficients $a_{ik}$, $c_{ik}$ are step functions. The method of proofs is based upon a transformation reducing the ODE (*) to a discrete dynamical system. The results are illustrated by the examples of the coupled harmonic oscillator and the double pendulum. Mathematics Subject Classification:Primary:70J25, 34D05; Secondary: 39A1. Citation:Nicola Guglielmi, László Hatvani. On small oscillations of mechanical systems with time-dependent kinetic and potential energy. Discrete & Continuous Dynamical Systems - A, 2008, 20 (4) : 911-926. doi: 10.3934/dcds.2008.20.911
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Perturbation solution of the coupled Stokes-Darcy problem.
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Rigorous description of macroscopic wave packets in infinite periodic
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Isn't orbital velocity proportional to $1/\sqrt r$?
As Emilio Pisanty mentioned in his answer, that's the speed of an object on a parabolic trajectory. That requires a fairly high velocity. The orbital speed drops to $1/\sqrt{2r}$ for an object in a circular orbit. The general rule, for all kinds of orbits in the two body problem is given by the
vis viva equation,$$v^2 = G(M_1+M_2)\left(\frac2r-\frac1a\right)$$where $v$ is the relative velocity between the two gravitating objects, $M_1$ and $M_2$ are their masses, $r$ is the distance between the objects, and $a$ is the semi-major axis length of the conic section (circles, ellipses, parabolae, and hyperbolae). Note that the semi-major axis length is infinite for a parabola and negative for a hyperbola. In the case of a tiny mass such as the Parker Space Probe and a large mass such as the Sun, this reduces to$$v^2 = \mu_\odot\left(\frac2r-\frac1a\right)$$where $\mu_\odot \equiv GM_\odot$ is the Sun's standard gravitational parameter.
The
vis viva equation appears so often in orbital mechanics that it's worthwhile remembering it.
After multiple gravity assists from Venus, the Parker Space Probe's perihelion distance will be
h=3.83 million miles above the surface of the Sun, or $r_\odot+h$ from the center of the Sun. It's aphelion distance will be a bit above $R_v$, Venus's orbital radius. The semimajor axis length of the orbit is thus $(R_v+r_s+h)/2$, and the velocity at perihelion is$$v_p = \sqrt{\mu_\odot \left(\frac2{r_\odot+h} - \frac2{R_v+r_\odot+h}\right)}= \sqrt{\frac{2\mu_\odot}{R_v+r_\odot+h}\frac{R_v}{r_\odot+h}}$$
One could go through the tedium of looking up the Sun's radius ($695700\, \text{km}$) and gravitational parameter ($132712440018\,\mathrm{km}^3/\mathrm{s}^2$), Venus's orbital radius ($108208930\,\text{km}$), and converting 3.83 million miles to something sensible ($6164000\,\mathrm{km}$), arrive at an answer of 190.8 km/s with the help of a nice online calculator.
But why bother? The linked online calculator is rather smart. It can do that tedious stuff. Just tell it what you want. |
Answer
.93 meters
Work Step by Step
We use the known equation to obtain: $ h = \frac{p-p_{atm}}{\rho g} = \frac{(1)(101.3\times 10^3)-(.91)(101.3\times10^3)}{1000\times9.81}=.93\ m$
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have
24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback. |
A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals
Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ...
So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.
Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.
Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.
Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.
Well, we do know what the eigenvalues are...
The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.
Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker
"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.
@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)
Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism
@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.
Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again
O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)
For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism |
This question already has an answer here:
At school I asked the question and I kept wondering "Can fractions or decimals be odd or even?"
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This question already has an answer here:
At school I asked the question and I kept wondering "Can fractions or decimals be odd or even?"
There is a natural way to define "oddness" for fractions. For integer $x$, let $\nu_{2}(x)$ denote the number of $2$'s that divide $x$. For example, $\nu_2(6)=1, \nu_2(4)=2, \nu_2(12)=2, \nu_2(1)=0, \nu_2(7)=0$. We can leave $\nu_2(0)$ undefined, or set it equal to $\infty$, as you like.
We can extend this to fractions via $$\nu_2\left(\frac{m}{n}\right)=\nu_2(m)-\nu_2(n)$$
This satisfies the lovely relation $$\nu_2(xy)=\nu_2(x)+\nu_2(y)$$ which holds even when $x,y$ are fractions.
With this tool in hand, we can define a number $x$ as "odd" if $\nu_2(x)=0$. The product of two odd numbers is odd, while the product of an odd number and a non-odd number is non-odd. However there is no natural definition of "even" numbers. We could take $\nu_2(x)\neq 0$ (but then the product of two even numbers might be odd), or $\nu_2(x)>0$ (but then we need a third term for $\nu_2(x)<0$).
See also a more comprehensive answer here.
No, odd-ness and even-ness is defined only for Integers.
For more info: Parity
No. Parity (whether a number is even or odd) only applies to integers.
Parity does not apply to non-integer numbers.
A non-integer number is neither even nor odd.
Parity applies to integers and also functions. So I wouldn't say that parity only applies to integers.
For instance, if $f(x)=x^n$ and $n$ is an integer, then the parity of $n$ is the parity of the function. |
EDIT: Here is a related post which concern quadratic vector fields rather than Van der pol equation. In this linked post we see that the convexity of limit cycle play a crucial role. On the other hand the unique limit cycle of Van der pol equation is convex. So there is a Riemanian metric on $\mathbb{R}^2 \setminus C$ such that all solutions of the Van der pol equation are geodesics. Here $C$ is the algebraic curve $yP-xQ=0$ where $P,Q$ are the components of the Van der pol equation. Moreover the limit cycle of the vander pol equation do not intersect this algebraic curve $C$.
The classical Van der Pol equation is the following vector field on $\mathbb{R}^{2}$:
\begin{equation}\cases{\dot{x}=y-(x^{3}-x)\\ \dot{y}=-x}\end{equation}
This equation defines a foliation on $\mathbb{R}^{2}-\{ 0\}$. It is well known that this vector field has a unique limit cycle(isolated closed leaf) in the (punctured) plane.
I search for a geometric proof for a particular case of this fact. In fact I search for an
alternative proof of the fact that this system has
at most one limit cycle.
Here is my question:
Question:
Is there a Riemannian metric on $\mathbb{R}^{2}-\{0\}$ with the following two properties?:
The Gaussian curvature is nonzero at all points of $\mathbb{R}^{2}-\{0\}$.
Each leaf of the corresponding foliation of $\mathbb{R}^{2}-\{0\}$ is a geodesic.
Obviously from the Gauss Bonnet theorem we conclude that existence of such metric implies that there are no two distinct simple closed geodesics on $\mathbb{R}^2\setminus \{0\}$, otherwise we glue two copy of the annular region surrounded by closed geodesics along the boundary then we obtain a torus with non zero curvature.(So this gives us an alternative proof for having at most one limit cycle for the Van der pol equation)
For a related question see Conformal changes of metric and geodesics
My initial motivation for this question goes back to more than 15 years ago, when I was reading a statement in the book of De Carmo, differential geometry of curves and surface, who wrote that:
A topological cylinder in $\mathbb{R}^{3}$ whose curvature is negative, can have at most one closed geodesic.
After this, I asked my supervisor for a possible relation between limit cycles and Riemannian metrics. As a response to my question, he introduced me a very interesting paper by Romanovski entitled "Limit cycles and complex geometry"
Note 1:For the moment we forget "negative curvature".We just search for a metric compatible to the Van der pol foliation. In this regard, one can see that for every metric on $\mathbb{R}^2 \setminus \{0\}$, with the property that all solutions of the Van der pol equations are (non parametrized) geodesics, then either the metric is not complete or the punctured plane does not possess a polynomial convex function or an strictly convex function. This is a consequence of Proposition 2.1 of this paper and also the following fact. Note 2: What is the answer if we replace the Vander pol vector field by an arbitrary foliation of $\mathbb{R}^{2}\setminus \{0\}$ with a unique compact leaf? Remark: The initial motivation is mentioned in page 3, item 5 of this arxiv note. |
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about
Resolve, the replacement for Vengeance.
Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec.
In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts.
Mathemagic
Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$:
$$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$
where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds.
The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this:
My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve.
Once you’re in combat and taking damage, the second term makes a contribution:
I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage:
$$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$
The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve.
The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns.
The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve.
However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation.
You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth.
Comparing to Vengeance
Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading.
Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss.
Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was.
Celestalon’s post also outlines a few other significant differences:
No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs
The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had
some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not.
However, Resolve ramps up a
lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds.
To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason.
As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented).
The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having
any ramp-up mechanism.
Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists.
This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage:
You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly.
One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown.
The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction:
It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve.
I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference:
Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords.
If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta.
Is It Fixed Yet?
If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years.
But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around.
It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve.
While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices.
All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the
removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve. |
Let $(a_n)$ be a bounded sequence. Let $c \in \mathbb{R}$ and suppose that $c \geq 0$. Prove then that $$ \lim_{n \to \infty} \sup (ca_n) = c \cdot \lim_{n \to \infty} \sup a_n. $$
Attempt at proof: I managed to prove one inequality. Here is my reasoning:
$(a_n)$ is a bounded sequence. For every $n \in \mathbb{N}$ the subset $A_n = \left\{ a_m \mid m \geq n \right\}$ is then a bounded subset of $\mathbb{R}$. Hence it has a supremum. Let $y_n = \sup A_n$. Now consider the set $$B_n = \left\{ c a_m \mid m \geq n \right\}. $$ Let $ m \geq n$ be arbitrary. Because $a_m \in A_n$, we have that $a_m \leq y_n$. Because $c \geq 0$, it also holds that $ca_m \leq c y_n$. But $ca_m \in B_n$. Since $m$ was chosen arbitrarily, this proves that $c y_n$ is an upper bound for $B_n$. But $\sup(B_n)$ is the least upper bound. Hence it follows that $\sup(B_n) \leq c y_n$. By taking the limit it now follows that $$ \lim_{n \to \infty} \sup (ca_n) \leq c \cdot \lim_{n \to \infty} \sup a_n.$$ But I don't know how to prove the other inequality. I tried to assume that $$ \lim_{n \to \infty} \sup (ca_n) < c \cdot \lim_{n \to \infty} \sup a_n $$ and then to derive a contradiction, but I was not able to.
Any help please to proof the other inequality? |
Probabilists often work with Polish spaces, though it is not always very clear where this assumption is needed.
Question: What can go wrong when doing probability on non-Polish spaces?
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One simple thing that can go wrong is purely related to the size of the space (Polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology:
Let $X$ be a measurable space with $|X| > 2^{\aleph_0}$. Then the diagonal in $X^2$ is not measurable.
We'll prove this by way of a theorem:
Let $U \subseteq X^2$ be measurable. $U$ can be written as a union of at most $2^{\aleph_0}$ subsets of the form $A \times B$.
Proof: First note that we can find some countable collection $(A_i)_{i\ge 0}$ of subsets of $X$, such that $U \subseteq \sigma(\{A_i \times A_j:i,j\ge 0\})$, where $\sigma(\cdot)$ denotes the $\sigma$-algebra generated by the given subsets (proof: The set of $V$ such that we can find such $A_i$ is a $\sigma$-algebra containing the basis sets).
For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.
Consider all subsets of $X^2$ which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a $\sigma$-algebra and obviously contains all the $A_i \times A_j$, so contains $U$.
But now we're done. There are at most $2^{\aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ subsets of the form $A \times B$.
QED
Corollary: The diagonal is not measurable.
Evidently the diagonal cannot be written as a union of at most $2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.
Separability is a key technical property used to avoid measure-theoretic difficulties for processes with uncountable index sets. The general problem is that measures are only countably additive and $\sigma$-algebras are closed under only countably many primitive set operations. In a variety of scenarios, uncountable collections of measure zero events can bite you; separability ensures you can use a countable sequence as a proxy for the entire process without losing probabilistic content. Here are two examples.
Weak convergence: the classical theory of weak convergence utilizes Borel-measurable maps. When dealing with some function-valued random elements, such as cadlag functions endowed with the supremum norm, Borel-measurability fails to hold. See the motivation for
Weak Convergence and Empirical Processes. The $J1$ topology is basically a hack which ensures the function space is separable and thereby avoids measurability issues. The parallel theory of weak convergence described in the book embraces non-measurability.
Existence of stochastic processes with nice properties: a key property of Brownian motion is continuity of the sample paths. Continuity, however, is a property involving uncountably many indices. The existence of a continuous version of a process can be ensured with separable modifications. See this lecture and the one that follows.
Metrizability allows us to introduce concepts such as convergence in probability. Completeness (the Cauchy convergence kind, not the null subsets kind) makes it easier to conduct analysis.
There's already been some good responses, but I think it's worth adding a very simple example showing what can go wrong if you don't use Polish spaces.
Consider $\mathbb{R}$ under its usual topology, and let X be a non-Lebesgue measurable set. e.g., a Vitali set. Using the subspace topology on X, the diagonal $D\subseteq\mathbb{R}\times X$, $D=\{(x,x)\colon x\in X\}$ is Borel Measurable. However, its projection onto $\mathbb{R}$ is X, which is not Lebesgue measurable. Problems like this are avoided by keeping to Polish spaces. A measurable function between Polish spaces always take Borel sets to analytic sets which are, at least, universally measurable.
The space X in this example is a separable metrizable metric space, whereas Polish spaces are separable completely metrizable spaces. So things can go badly wrong if just the completeness requirement is dropped.
Below is a copy of an answer I gave here https://stats.stackexchange.com/questions/2932/metric-spaces-and-the-support-of-a-random-variable/20769#20769
Here are some technical conveniences of separable metric spaces
(a) If $X$ and $X'$ take values in a separable metric space $(E,d)$ then the event $\{X=X'\}$ is measurable, and this allows to define random variables in the elegant way: a random variable is the equivalence class of $X$ for the "almost surely equals" relation (note that the normed vector space $L^p$ is a set of equivalence class)
(b) The distance $d(X,X')$ between the two $E$-valued r.v. $X, X'$ is measurable; in passing this allows to define the space $L^0$ of random variables equipped with the topology of convergence in probability
(c) Simple r.v. (those taking only finitely many values) are dense in $L^0$
And some techical conveniences of complete separable (Polish) metric spaces :
(d) Existence of the conditional law of a Polish-valued r.v.
(e) Given a morphism between probability spaces, a Polish-valued r.v. on the first probability space always has a copy in the second one
(f)
Doob-Dynkin functional representation: if $Y$ is a Polish-valued r.v. measurable w.r.t. the $\sigma$-field $\sigma(X)$ generated by a random element $X$ in any measurable space, then $Y = h(X)$ for some measurable function $h$.
We know by Ulam's theorem that a Borel measure on a Polish space is necessarily tight. If we just assume that the metric space is separable, we have that each Borel probability measure on $X$ is tight if and only if $X$ is universally measurable (that is, given a probability measure $\mu$ on the metric completion $\widehat X$, there are two measurable subsets $S_1$ and $S_2$ of $\widehat X$ such that $S_1\subset X\subset S_2$ and $\mu(S_1)=\mu(S_2)$. So a probability measure is not necessarily tight (take $S\subset [0,1]$ of inner Lebesgue measure $0$ and outer measure $1$), see Dudley's book
Real Analysis and Probability.
An other issue related to tightness. We know by Prokhorov theorem that if $(X,d)$ is Polish and if for all sequence of Borel probability measures $\{\mu_n\}$ we can extract a subsequence which converges in law, then $\{\mu_n\}$ is necessarily uniformly tight. It may be not true if we remove the assumption of "Polishness". And it may be problematic when we want results as "$\mu_n\to \mu$ in law if and only if there is uniform tightness and convergence of finite dimensional laws."
Google "image measure catastrophe" with quotation marks.
It can also be useful to have the set of Borel probability measures on $X$ (with weak* convergence, a.k.a. convergence in law) to be metrizable, for instance to be able to treat the convergences sequentially. For this you need the space $X$ to be separable and metrizable (see Lévy-Prohorov metric).
Fun fact: you can find a non-separable Banach space and a Gaussian probability measure on it which gives measure $0$ to
every ball of radius $1$. (In particular your intuition about notions like the "support" of a measure goes pretty badly wrong.)
Consider i.i.d. $\xi_n$ and take as your norm $|\xi|^2 = \sup_{k\ge 0}2^{-k}\sum_{n=1}^{2^{k}} |\xi_n|^2$. This is almost surely finite by Borel-Cantelli and almost surely at least $1$ by the law of large numbers. The fact that it gives measure $0$ to every ball of radius $1$ is left as an exercise. This norm isn't even very exotic: if you interpret the $\xi_n$'s as Fourier coefficients, then $B$ is really just the Besov space $B^{1/2}_{2,\infty}$.
As far as I remember, the projection of a measurable set may fail to be measurable, so something very natural may become not an event. Besides, constructing conditional probabilities as measures on sections becomes problematic. Perhaps, there are more reasons but these two are already good enough. |
We learned about the class of regular languages $\mathrm{REG}$. It is characterised by any one concept among regular expressions, finite automata and left-linear grammars, so it is easy to show that a ...
There are many techniques to prove that a language is not context-free, but how do I prove that a language is context-free?What techniques are there to prove this? Obviously, one way is to exhibit ...
Is there an algorithm/systematic procedure to test whether a language is context-free?In other words, given a language specified in algebraic form (think of something like $L=\{a^n b^n a^n : n \in \...
I have the following language$\qquad \{0^i 1^j 2^k \mid 0 \leq i \leq j \leq k\}$I am trying to determine which Chomsky language class it fits into. I can see how it could be made using a context-...
The Chomsky-Schutzenberger representation theorem states that a language $L$ is context-free iff there is a homomorphism $h$, a regular language $R$, and a paired alphabet $\Sigma = T \cup \overline{T}...
Hypothetically, let's say you are using the pumping lemma for either regular or context free languages. Now using either, you come across a case that remains true despite pumping it. In this situation,...
I've came up with a result while reading some automata books, that Turing machines appear to be more powerful than pushdown automata. Since the tape of a Turing machine can always be made to behave ...
So I have a problem that I'm looking over for an exam that is coming up in my Theory of Computation class. I've had a lot of problems with the pumping lemma, so I was wondering if I might be able to ... |
Perhaps this completely basic illustration helps for the understanding. If it is not needed, I can delete that answer.
The little theorem of Fermat says, that for all primes $p$ and bases $b$ where $\gcd(b,p)=1$ we have that $b^{p-1} \equiv 1 \pmod p $ . So this suggests, that we could test some $n$ for primeness with this very easy to compute relation. But unfortunately, not only primes $p$ satisfy this, but also some composites $n$ - at least this depends also on the selection of the base $b$ .
I've made a little program to show this for some $n$ and some $b$ in a spreadsheet-like display:
base\ n -->
| \
V \ 2 3 4 5 6 7 8 9 10 11 12
-------+----------------------------------------------------
2 | 0 1 0 1 2 1 0 4 2 1 8
3 1 0 3 1 3 1 3 0 3 1 3
4 0 1 0 1 4 1 0 7 4 1 4
5 1 1 1 0 5 1 5 7 5 1 5
6 0 0 0 1 0 1 0 0 6 1 0
7 1 1 3 1 1 0 7 4 7 1 7
8 0 1 0 1 2 1 0 1 8 1 8
9 1 0 1 1 3 1 1 0 9 1 9
10 0 1 0 0 4 1 0 1 0 1 4
11 1 1 3 1 5 1 3 4 1 0 11
12 0 0 0 1 0 1 0 0 2 1 0
13 1 1 1 1 1 1 5 7 3 1 1
14 0 1 0 1 2 0 0 7 4 1 8
15 1 0 3 0 3 1 7 0 5 1 3
16 0 1 0 1 4 1 0 4 6 1 4
17 1 1 1 1 5 1 1 1 7 1 5
18 0 0 0 1 0 1 0 0 8 1 0
19 1 1 3 1 1 1 3 1 9 1 7
20 0 1 0 0 2 1 0 4 0 1 8
21 1 0 1 1 3 0 5 0 1 1 9
22 0 1 0 1 4 1 0 7 2 0 4
The entries are always $ b^{n-1} \pmod n$ and we see for instance at the $n$ which are primes $p=3$ or $p=5$ and so on, that indeed for all bases $b$ that do not contain $p$ the result is indeed $1$. At composite $n$ this is not so; in most cases the entries are different from $1$ and thus the little Fermat would work in this cases. For instance, for $n=4,6,8,10,12 $ there occur $1$ - but only at bases $b=k\cdot n+1$ meaning $(k \cdot n+1)^{n-1} \equiv 1 \pmod n$
But it becomes a bit more complicated for larger $n$.
base\ n -->
| \
V \ 7 9 11 13 15 17 19 21 23 25 27
------------------------------------------------------------
2 1 4 1 1 4 1 1 4 1 16 13
3 1 0 1 1 9 1 1 9 1 6 0
4 1 7 1 1 1 1 1 16 1 6 7
5 1 7 1 1 10 1 1 4 1 0 16
6 1 0 1 1 6 1 1 15 1 21 0
7 0 4 1 1 4 1 1 7 1 1 4
8 1 1 1 1 4 1 1 1 1 21 10
9 1 0 1 1 6 1 1 18 1 11 0
10 1 1 1 1 10 1 1 16 1 0 19
11 1 4 0 1 1 1 1 16 1 16 22
12 1 0 1 1 9 1 1 18 1 11 0
13 1 7 1 0 4 1 1 1 1 11 25
14 0 7 1 1 1 1 1 7 1 16 25
15 1 0 1 1 0 1 1 15 1 0 0
16 1 4 1 1 1 1 1 4 1 11 22
17 1 1 1 1 4 0 1 16 1 21 19
18 1 0 1 1 9 1 1 9 1 1 0
19 1 1 1 1 1 1 0 4 1 21 10
20 1 4 1 1 10 1 1 1 1 0 4
21 0 0 1 1 6 1 1 0 1 6 0
22 1 7 0 1 4 1 1 1 1 6 16
We see, that regularly simple composite numbers like $n=9,15,21,25$ have $1$ at some small bases, so using that bases for the little Fermat prime-test would sign false positives. However, the base $b=2$ works nice so far: it perfectly separates between primes and composites. The same is true for $b=3$.
But still we get in troubles. Let's look around $n=341$
| 337 339 341 343 345 347 349 351 353 355 357
-----|-------------------------------------------------------
2| 1 4 1 162 31 1 1 121 1 229 67
3| 1 9 56 337 96 1 1 243 1 294 30
4| 1 16 1 176 271 1 1 250 1 256 205
5| 1 25 67 141 220 1 1 259 1 270 319
6| 1 36 56 57 216 1 1 270 1 231 225
7| 1 49 56 0 301 1 1 166 1 129 259
8| 1 64 1 43 121 1 1 64 1 49 169
For the primes around $n=341$ things are ok, and also for the composites around $n$ when using base $b=2$. Only, $n=341$ gives a false positive. So this is called a "pseudoprime (to base 2)" Here base $b=3$ would still detect correctly the compositeness. But let's see some more base-2-pseudoprimes:
| 341 561 645 1105 1387 1729 1905 2047 2465 2701 2821
-----|-------------------------------------------------------
2| 1 1 1 1 1 1 1 1 1 1 1
3| 56 375 36 1 875 1 276 1013 1 1 1
4| 1 1 1 1 1 1 1 1 1 1 1
5| 67 1 595 885 1122 1 400 622 1480 2554 1
6| 56 375 36 1 875 1 276 1013 1 1 1
7| 56 1 436 1 1122 742 1561 1013 1 2554 2016
8| 1 1 1 1 1 1 1 1 1 1 1
9| 67 375 6 1 1 1 1881 622 1 1 1
10| 67 1 595 885 1122 1 400 622 1480 2554 1
11| 253 154 1 1 1141 1 226 1 1 2554 1
12| 56 375 36 1 875 1 276 1013 1 1 1
We see, that a couple of them are correctly signalled by base-3, some still not and a really hard beast is $n=1729$ which needs tests up to $b=7$ to detect the compositeness.
The introduction of the concept of the Carmichael numbers is now to denote that numbers $n$ which are pseudoprime to all bases $b$ which are coprime to $n$ itself - so they are in a sense "stronger" pseudoprimes than that for only a single base. For instance, $n=561$ is a strong pseudoprime for bases $ b=2,4,5,7,8,10,...$ and only bases which happen to have a common factor with $n$ indicate correctly compositeness.
Other answers have indicated some properties of that Carmichael-numbers, such like squarefreeness, number of primefactors $\ge 3$ and some more - that is what you actually have asked for (and can find in the other answers).
Additional remark, in which way "Carmichael-numbers" is an interesting concept: stronger prime-test-procedures than the simple little Fermat on base $b=2$ use a couple of bases first for pseudoprime-tests before they enter more complicated and time-consumptive procedures (and it is somehow a kind of art to define the optimal set of bases for mutually correcting pseudoprime-tests up to some lower bound $n \le N$ which costs least and are still perfect indicators) |
For the dynamics of open quantum systems, the Kraus operators $K_\kappa$ can be derived from the unitary orbit $U(t)\rho U(t)^\dagger$ for $\rho=\rho_S\otimes\rho_E$ of the composite system given by the time-propagators $U(t)=\exp(-\mathrm{i} tH)$ via tracing over the environmental states $|\mu\rangle$: $$K_\kappa(t)=K_{\mu\nu}(t):=\langle\mu|U(t)|\nu\rangle$$
I wanted to derive an expression for $K_\kappa$ directly in terms of the interaction Hamiltonian $H_I=\sum_\alpha S_\alpha\otimes E_\alpha$ for the whole Hamiltonian $H=H_S\otimes I_E+I_S\otimes H_E+H_I$ (assuming time-independent $H$). So I wrote $U(t)$ as the exponential $\sum_k (-\mathrm{i} tH)^k/k!$ and employed binomial expansions for the powers of sums of Hamiltonians to obtain
$$K_{\mu\nu}(t)=\sum_{\alpha,k,j\leq k,i\leq j} (-\mathrm{i} t)^k/k!{k\choose j}{j\choose i}\langle\mu|H_S^iI_S^{j-i}S_\alpha^{k-j}\otimes I_E^iH_E^{j-i} E_\alpha^{k-j}|\nu\rangle$$
How can this be simplified to get a clean formula only depending on $H_S$ and $S_\alpha$? It should be possible, as the Kraus operators only act on the reduced Hilbert space alone. In addition, how would one proceed for time-dependent $H(t)$ and their time-ordered products? |
In
classical thermodynamics the change in internal energy is defined by the first law as$$\Delta U = q + w$$so that only the difference in $U$ is known; $q$ is the heat absorbed by the 'system' and $w$ the work done on the system.
For example in a closed system (no exchange of matter with environment) we can write for a reversible change\begin{align} \mathrm{d}q &= T\mathrm{d}S \\ \mathrm{d}w &= -p\mathrm{d}V \end{align}and then if the only form of work on a gas is volume change$$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V$$and this is the fundamental equation for a closed system. Thus only difference in internal energy are measurable from thermodynamics, and this follows from the first law. (Even if you integrate this equation from say state $a$ to $b$ the result will be $U(b)-U(a)=\ddot {}$ in other words $\Delta U$.)
Thermodynamics was developed before the nature of matter was known, i.e. it does not depends on matter being formed of atoms and molecules. However, if we use additional knowledge about the nature of molecules then the internal energy (and entropy) can be determined from statistical mechanics.
The internal energy ($U$ not $\Delta U$) of a perfect monoatomic gas is the ensemble average and is $$U=(3/2)NkT$$or in general $U=(N/Z)\Sigma_ j\exp(−\epsilon_j/(kT))$ where $Z$ is the partition function, $k$ Boltzmann constant, $\epsilon_j$ energy of level $j$, and $T$ temperature and $N$ Avogadro's number. The absolute value of the entropy $S$ (for a perfect monoatomic gas) can also be determined and is given by the Sakur-Tetrode equation. |
I'm trying to explain to someone learning elementary physics (16 year old) that linear momentum and energy are conserved independently in a 2-body collision. I'm not a professional physicist and haven't tried to explain this stuff for years, and I can't think of any convincing elementary argument to show that this is the case. Does anyone know of an elementary approach to this? (i.e. one that does
not contain the expressions "Lagrangian" and "Noether's Theorem".)
I'm trying to explain to someone learning elementary physics (16 year old) that linear momentum and energy are conserved independently in a 2-body collision. I'm not a professional physicist and haven't tried to explain this stuff for years, and I can't think of any convincing elementary argument to show that this is the case. Does anyone know of an elementary approach to this? (i.e. one that does
Give him an example of inelastic collisions and explain why momentum is conserved but kinetic energy is not. If you explain the reasoning, (all forces are internal hence momentum is conserved) and that there are losses so KE of the system is lost in heat/sound/other forms of energy.. he should get the idea that the two are different beasts. At such a level it is best to illustrate with counterexamples.
Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you
can prove him wrong using conservation of energy. Therefore conservation of energy has implications that conservation of momentum does not.
Suppose we have two identical billiard balls, one traveling east and one traveling west, with the same velocities and they collide. Suppose someone claims that both balls will now travel west, at the same velocity. You can't prove him wrong using conservation of energy, but you
can prove him wrong using conservation of momentum. Therefore conservation of momentum has implications that conservation of energy does not.
For a particle in a 1D external time-dependent field there are no energy and momentum conservation laws, yet there are two independent conserved quantities. It is because the differential equation is of the second order and it is accompanied with two independent initial data - the initial position and initial velocity. See an example
here.
All the following explanation is for elementary students.First ensure that he/she understands that momentum is a vector quantity and energy is scalar. Also s/he might be knowing Newton's 2nd law of motion. Momentum of a system is unique in given direction. For a given system if net external force (on the system) is zero then net momentum of the system
does not change. He can argue that so what energy will also not change. Yes, but we can have situation where we apply force on a body and its momentum is changed but energy is not. (A force applied perpendicular to the direction of motion all times). I hope this sets the sail for discussion.
Consider for simplicity a non-relativistic collision between two point particles of
same masses in their center-of-mass frame. From total momentum conservation, we know that the center-of-mass (COM) frame is an inertial frame. Moreover, if particle $1$ at some instance $t$ has position ${\bf r}_1$ and velocity ${\bf v}_1$ (relative to the COM frame), then particle $2$ is completely dictated to have opposite position ${\bf r}_2=-{\bf r}_1$ and opposite velocity ${\bf v}_2=-{\bf v}_1$. So from the COM perspective, the two-particle system is completely determined by knowing the state of particle $1$ alone.
Up until now, we have only used momentum conservation, and it doesn't matter whether the collision is elastic, partially elastic, or inelastic. The above is true regardless.
Now let us investigate a collision at initial and final instances $t_i$ and $t_f$ well before and well after the collision takes place. Note that we have already completely extracted all the information in the momentum conservation law to conclude that whatever the particle $1$ does, the particle $2$ would do the opposite. There are no more information available. In particular, momentum conservation gives us no clue about how initial and final velocity of particle $1$ are related.
Finally, let us restrict to an elastic collision. The kinetic energy conservation is in this context the
independent statement that the initial and final speed $v_{1i}=v_{1f}$ of particle $1$ are equal (still measured relative to the COM frame).
Momentum and energy are both different depending on what I compare the motion of an object with. If I'm in a train, I have no momentum or kinetic energy relative to the train. Relative to the fields outside the train, however, I have lots of momentum and energy. If I jump off the train, I will come to a stop relative to the fields, so the momentum and energy relative to the fields have to go somewhere. On the other hand, I will then be moving relative to the train, so the momentum and energy to make that happen also have to come from somewhere.
The difference between the momentum and the energy comes from the fact that the forces that change my speed have to act both for a certain amount of time
and over a certain distance.
Suppose the force that slows me down is constant, and that the force doesn't make me spin or break me up into pieces. This is the sort of wild idealization that gets Physics a bad name with 16 year olds, but it's a first approximation from which we can go on to a second approximation that's better, and no-one has yet thought up a better first approximation. If the train travels twice as fast, the force has to act for
twice as much time to bring me to stop, that's the change of momentum, but the force has to act for four times the distance to bring me to a stop, that's the change of energy.
This gets very tricky, because someone in an airplane that's moving really fast sees the force acting for the same amount of time as someone standing in the field sees the force acting for, but the person in the airplane sees the force acting for much more distance, because when the force started I was right next to the airplane, say, but when the force ended I was a
long way behind. So, the change of energy from the point of view of the person in the airplane was much bigger than someone in the fields thinks it was, even though everyone agrees that the change of momentum was the same.
To switch to a different analogy, the energy is important because it determines how far it takes me to stop a car using the brakes, so it determines whether I hit the brick wall that I suddenly see in front of me. The momentum determines how much time it takes to come to a stop, but I can't immediately think of a really graphic situation when that's important.
One can construct different situations endlessly. It can be done in equations, of course, but you'll have to decide whether that's appropriate. I'll be interested if there's anything about this Answer that you think could be made clearer. It certainly isn't complete.
Welcome from an Englishman in the USA.
EDIT: Of course overnight I realize that I mention
conservation not once. From the point of view of the above it's enough to note that both can be understood to be because of Newton's third law, which is, from Wikipedia, "The mutual forces of action and reaction between two bodies are equal, opposite and collinear". As a result, we can say that the energy added to an object is taken away from the other object, and the same for momentum. The independence of the two conservation laws is essentially just because the two quantities are independent.
I've decided to add a few simplified equations, $$Force = Mass \times Acceleration,$$ $$Kinetic\ Energy = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Distance\ each\ Force\ is\ Applied\ For,$$ $$Momentum = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Length\ of\ Time\ each\ Force\ is\ Applied\ For,$$ or, as vector equations, almost certainly beyond what you need, $$\underline{F}=m\underline{a}, \qquad E=\int \underline{F}(t,\underline{s}(t))\cdot\frac{d\underline{s}(t)}{dt}dt, \qquad \underline{P}=\underline{F}(t,\underline{s}(t))dt.$$
Really, this stuff should be left to specialist educators, the best of whom will take time not only to create new ways to explain ideas, but also to study how well different strategies of explanation work for different kinds of student, but I've always been interested occasionally to put myself in this mindset. It's always humbling to discover how much creativity is needed to do it well.
As often, trawling around in Wikipedia, starting from the page on Newton's laws that I mention above, will render up some gems amongst the too-much-information for the purposes of your Question. I particularly like the tail-end comment that "Conservation of energy was discovered nearly two centuries after Newton's lifetime, the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infra-red light."
Consider elastic collision of a small mass with a large mass (say, fast bullet with (initially) non-moving heavy metall ball). I guess you are considering Newtonian mechanics, so there are following options here:
a) bullet stops after collision, and heavy ball starts moving. If you will require that energy is conserved than you will see that momentum is not conserved, and vise versa. That means that in elastic collision the bullet
cannot stop and transfer all its energy ( or all its momentum) to the heavy ball because in this case energy and momentum cannot be conserved. both
b) Heavy ball remains not moving after collision, while the bullet is moving with the same velocity in arbitrary direction. Obviously, the energy is conserved, but momentum is not.
Hence, we can imagine the number of outcomes where only one quantity is conserved. But nature leaves us only one choice out of many because
both energy and momentum need to be conserved. There are many scenarios where only momentum (or only energy) is conserved, but if we require that both are conserved, only one scenario remains possible.
It is also interesting to consider the situation where the momentum is
seemingly not conserved. The simplest case is when a man (or woman) stays on the floor and at some moment of time start moving (walking). The initial momentum of a man (women) is zero, and the final momentum is not. What happens to conservation of momentum? Here it is important to consider a concept of closed system, because only in closed system the momentum is conserved. In this case the closed system includes the Earth. So when we start walking we move the Earth! :-) edited
Conservation of energy is almost self-evident. Conservation of linear momentum comes from the fact the in an isolated system of particles there is no preferred location. To try to make a point that the two are related seems that would invoke some really convoluted arguments. One deals with invariance of energy and the other with invariance of location.
Check out Noether's Theorem.
Conservation laws are best explained as consequences of space and time symmetries: Because space is the same in every direction we see momentum as beng conserved, because time is uniform we see energy as being conserved. http://en.wikipedia.org/wiki/Noether's_theorem
Note that energy is not conserved in relativistic situations but momentum is - this is related to the nature of relativity.
The 16 year old is on to something but no one realizes it because physicists and those who teach physics assume that kinetic energy is a valid scientific principle. This opening sentence will naturally solicite laughter and other reactions but take a moment anyway to consider the following analogy. It looks at what happens when you define energy or force with respect to distance. Energy (kinetic energy) is related to force acting through a distance. Imagine a passenger sitting in the rear seat of a car and he hands his cell phone to the driver. If the car is stationary, that phone travels about 3 feet with respect to the road. If the car is moving, the phone might travel 10, 20, 30 or more feet. The time it takes to hand the phone does not change; it is a constant. Now think about force, it causes things to accelerate. When a body accelerates, it changes its velocity/speed. As this occurs, that body will travel a certain distance and take TIME to do so. First, take wind resistance and other unrelated things out of the mix. Accelerate a body from 0 mph to 10 mph; this takes time and the force will act through a distance. Accelerate the same body from 10 mph to 20 mph using the same amount of force. The time to change that body's speed by 10 mph does not change; the distance it travels during that act will. In short, there is something wrong with the kinetic energy formula. Do not do what everyone does and assume that because something has been around for a long time that it is correct. |
At some stage of our school life, most of us would have sent or received coded messages from our classmates.
General mathematical principle: To investigate a complicated situation apply simplifying transformations, investigate the simplified situation, and try to transfer this information back to the original situation.
I have been doing some work on the $n$'th prime number, but was impeded for some time because of the lack of prime numbers.
This game is a contest between two gladiators in an arena consisting of a 20 $\times$ 20 grid.
Determine all ordered triples $(a,b,c)$ of natural numbers, $a\geq b\geq c$, such that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$ Prove that you have found all solutions J241 By inserting brackets in $$1\div 2 \div 3 \div 4 \div 5 \div 6 \div 7 \div 8 \div 9 ,$$ the value of the expression can be made equal to $7/10$. How? |
Dear Readers,
this is my last message to you as Editor of Parabola. Dear Readers, this is my first message to you as Editor of Parabola.
Dear Readers,
When learning the intuition behind definite integration, calculus students often learn how to find the area under a curve by using a Riemann sum.
Ever thought of batting in cricket as a life and death struggle against hostile forces? It always seemed that way when I batted anyway.
Well you might be more accurate than you think in looking at it that way.
An enclosure of length $1$ unit is constructed around two adjoining walls of unlimited length. It is made of $n \geq 2$ straight sections, referred to as an
$n$-enclosure, designed so as to maximise the enclosed area $A_n(\omega)$, where $\omega \leq \pi$ is the angle formed by the walls. Competition Winners - Senior DivisionSeyoon Ragavan Knox Grammar School 1st prize Competition Winners - Junior DivisionRichard Gong Sydney Grammar School 1st prize Junior Division - Problems and Solutions Solutions by Denis Potapov, UNSW Australia. Problem 1 Q1481 Prove that if the denominator $q$ of a fraction $p/q$ is the number consisting of $n$ digits, all equal to $9$, and if $p$ is less than $q$, then $p/q$ can be written as a repeating decimal in which the repeating part has length $n$ and contains the digits of $p$, preceded by a sufficien |
I came across Shimura (1971) notes about cosets representatives of the congruence subgroups $ \Gamma_0(N) $. He firstly proves that its index in the modular group $\Gamma$ is
\begin{equation} [\Gamma : \Gamma_0(N)]=N \cdot \prod_{p|N} (1+p^{-1} ) \end{equation}
Then he comes up with a sets of cosets representatives for $\Gamma_0(N)$ in $\Gamma$ made in this way: we first choose pairs $(c,d)$ of positive integers such that
\begin{equation} (c,d)=1, \qquad d|N, \qquad 0 < c \le N/d \end{equation}
then for each pairs we fix integers $a,b$ such that $ad-bc=1$. Our list of cosets representatives is made of the matrices with such entries.
However, let us take for example $N=12$ when we know the index is 24 and thus this is the cardinality of the set of cosets representatives. Using the rule above, I only find 22 cosets representatives, namely the ones corresponding to the following $(c,d)$ pairs: $$(1,1),(2,1),\dots,(12,1),(1,2),(3,2),(5,2),(1,3),(2,3),(4,3),(1,4),(3,4),(1,6),(1,12).$$
I also tried to run SAGE and it gives me 24 cosets representatives but they seem redundant, for example $[[1, 0] [2, 1]]$ and $[[1, 2][2, 5]]$ are listed as different cosets representatives, but
$$\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}$$
and thus it seems to me that these 2 matrices belong in fact to the same coset.
Something is clearly wrong, I hope you can help me. |
Applications of the Integrals Area under Simple Curves The area region bounded by the y = f(x), the x - axis and the lines x = a, x = b is =\begin{vmatrix}\int_{a}^{b} y \ dx \end{vmatrix}=\begin{vmatrix}\int_{a}^{b} f(x) \ dx \end{vmatrix} The area region bounded by the curve y = f(x) the y-axis and the lines y = c, y = d is =\begin{vmatrix}\int_{c}^{d} x \ dy \end{vmatrix}=\begin{vmatrix}\int_{c}^{d} f^{-1}(y) \ dy \end{vmatrix} The area region bounded by y = f(x), y = g(x) the lines x = a, x = b is \begin{vmatrix}\int_{a}^{b} f(x)-g(x) \ dx \end{vmatrix} The area region bounded by y = f(x) and y = g(x) and the lines y = c, y = d is \begin{vmatrix}\int_{c}^{d} f^{-1}(y)-g^{-1}(y) \ dy \end{vmatrix} If y = f(x) and y = g(x) and these curves intersect at P(α 1, β 1), Q (α 2, β 2), then the area between curves is \begin{vmatrix}\int_{\alpha_{1}}^{\alpha_{2}} f(x)-g(x) \ dx \end{vmatrix} on x-axis \begin{vmatrix}\int_{\beta_{1}}^{\beta_{2}} f^{-1}(y)-g^{-1}(y) \ dy\end{vmatrix} on y-axis The area of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 is πab sq.units The area of the circle x 2+ y 2= a 2is π a 2sq.units. The area region bounded by the curve y = sin ax (or) cos ax and x-axis is \frac{2}{a} sq.units by one arc. The area region bounded by the curve y = sin ax (or) y = cos ax and x-axis in [0, nπ] is \frac{2n}{a} sq.units Part1: View the Topic in this video From 14:04 To 48:16 Part2: View the Topic in this video From 04:50 To 12:40
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1. The area of the region bounded by the curve y = f(x), x-axis and the lines x = a and x = b (b > a) is given by the formula: Area = \int_{a}^{b} y \ dx=\int_{a}^{b} f(x)\ dx.
2. The area of the region bounded by the curve x = Φ(y), y-axis and the lines y = c, y = d is given by the formula: Area = \int_{c}^{d} x \ dy=\int_{c}^{d} \phi(y)\ dy. |
Statistics - Piecewise polynomials Table of Contents 1 - About
Instead of having a
single polynomial over the whole domain of the variable, we fit different polynomials in each region (partition) instead of different constants.
The polynomials are fitted locally and has different level of continuity (smooth) throughout the whole range.
2 - Articles Related 3 - Formula
<MATH> y_i = \left\{ \begin{matrix} \beta_{10} + \beta_{11} x_i + \beta_{12} x_i^2 + \dots + \beta_{1n} x_i^n & \text{if } x_i < \href{knot}{knot} \\ \beta_{20} + \beta_{21} x_i + \beta_{22} x_i^2 + \dots + \beta_{2n} x_i^n & \text{if } x_i \geq \href{knot}{knot} \\ \end{matrix} \right. </MATH>
4 - Continuity
The amount of continuity you can enforce, the order of continuity, is one less than the degree of the polynomial.
Splines are the maximum amount of continuity
5 - Example 5.1 - Non Continuous piecewise cubic Piecewise cubic polynomial in two regions. The knot is at 50.
They're two different cubic polynomials that just fit to the data with a break in the middle.
The left one's fit to the data on the left. The right one's fit to the data on the right.
To suppress the break, it's better to add constraints to the polynomials, for example, continuity.
5.2 - Continuous piecewise cubic
Two cubic polynomials (one in the left and one in the right) forced to be continuous at the knot.
In some situations, continuity alone is not enough (Notice the little kink in the middle) to have a smooth line. Continuity can then be enforced on the derivatives. See below.
5.3 - Cubic Spline
Cubic splines is a cubic polynomial in the left and the right and continuous at the knot.
There's the following constraints:
The first derivatives continues at the knot, and the second derivatives continues at the knot.
We couldn't make the third derivative continuous, because then it would just be a global cubic polynomial. It's believed that a cubic spline, the discontinuity which exists in the third derivative is not detectable by the human eye.
The degree of continuity is 2 because it's a third degree polynomial. |
This paper is a classic, and all of it is highly original, and it's deductions are mostly accurate up to section VII in their context. But here, I would like to focus only on the inaccuracies, as this is the thing that helps with historical material.
The detailed assumptions about the strong interactions in this paper aren't at all right, since Wilson was writing well before QCD. But there are also a few accuracy problems in the context of pure 1960s physics. These are outweighed by the great originality of the paper's main method. The paper's virtue is not that the physical hypotheses are right, nor that the results of Wilson's model are particularly enlightening in the special case of the strong interactions, but that the OPE produces a formalism to analyse arbitrary strongly coupled field theory fixed points, a language for talking about the algebra of operators in any quantum field theory without paradoxes. The results he gives are examples of the power of the OPE to produce results, it's just that the results themselves are not physically right inside QCD.
The strong interactions, as understood in the 1960s, consisted of families of hadrons of mass <5GeV, interacting by scattering at energies well below the scale where the quark and gluon substructure becomes manifest. It was known that not all these particles could be quanta of elementary quantum fields. The Regge trajectory idea was developed by Regge and Mandelstam, and shown to fit the experimental data starting with Chew and Frautschi. The empirical Regge law said that the mesons form families of spin and mass so that
m 2 is proportional to J, with a universal proportionality constant. The assumption that this behavior continued made predictions for the scaling laws for the near-beam scattering at higher energy, and Regge theory fits worked to predict the small-angle scattering at higher energies and fit the total cross sections (Regge theory still works as well as ever for this, although people largely stopped talking about it). The trajectories meant that there were going to be infinitely many strongly interacting particles of ever higher mass and spin, extrapolating the m 2 vs. J line of the known ones.
The Regge behavior was emphasized by Chew and collaborators, who worked on S-matrix theory. These folks believed that field theory was hopeless for the strong interactions, and that it is best to formulate a new kind of theory from scratch, without using the concept of local fields at all, only using experimental data to infer relativistic amplitudes, and dispersion relations and unitarity. When supplemented with the hypotheses of linear Regge trajectories in a narrow-resonance approximation, and Dolen-Horn-Schmidt duality, this idea leads to string theory.
Separate from the Regge trajectory story, which linked meson resonances of increasing mass and spin, the mesons and baryons of one particular spin and parity clustered into groups according to their isospin, hypercharge, and strangeness, each of them making the simplest SU(3) multiplets which were all broken to lowest order in a universal mass-breaking pattern, which was parametrized by introducing a diagonal SU(3) matrix diag($m_s$,$m_d$,$m_u$) as a noninvariant spurion to contribute to the mass matrix for the hadrons. This SU(3) breaking term was identified as the quark-mass matrix, which was appreciated long before the quarks themselves were understood. The interpretation of the mass-relations in SU(3) flavor multiplets was that the difference in quark-masses (whatever quarks were exactly) was responsible for splitting the hadrons in an SU(3) multiplet apart in mass, and the fundamental underlying strong interaction theory, ignoring quark masses, was SU(3) symmetric (see, e.g., "The Eightfold Way").
The SU(3) relations meant that the strong interactions were pictured as split in two--- one part was the universal SU(3) symmetric super-strong interactions, describing high energies where the quark masses were negligible--- this part today we would identify as the theory of QCD at zero-quark-mass. The other part of the theory was what were called the medium-strong interactions, which produced or included the quark masses as perturbation over the SU(3) symmetric super-strong interactions. Now we also know that the quark masses are the whole story regarding the medium strong interactions, at least in the fundamental Lagrangian.
But further, it was known, from the Nambu Jona-Lasinio model and the sigma model, that the true (flavor) symmetry group of the strong-stong interactions must be an $SU(2)\times SU(2)$ symmetry and that it was fundamentally breaking down to Isospin, since the pions provided three Goldstone bosons for a broken SU(2). This was generalized to broken flavor $SU(3)\times SU(3)$. The formalism for analyzing the symmetries was provided by Gell-Mann, who proposed that the currents for these symmetries were still useful to consider as local fields, even though the underlying theory might not be a field theory. He suggested to use the local current commutators to make predictions, because the form of these local commutators would be determined by the symmetry of the theory. Then you could interpret the current commutation relations, which were determined by group theory, by considering the unbroken symmetry generators to be integrals of the unbroken currents (these are the "conserved" currents), while the Goldstone boson fields were identified with the broken symmetry currents (these are the "partially conserved" currents).
It was a very important idea, and the primitive precursor to the OPE. The OPE extends the idea of a current algebra to cover all local fields in a local field multiplication algebra, rather than just currents (although it works for currents too). Current algebra today is most often just identified with some particular universal coefficients of singular parts of the current-current OPE.
The quark masses set a length-scale, and the symmetry breaking set a scale also. At the time, it was not clear whether, at high energies, the strong-strong interactions would have a separate independent scale inside, or whether they would be scale invariant. Wilson's fundamental starting point in this paper is the idea, which he attributes to Kastrup and Mack, that the strong-strong interactions are scale free, they are a scale invariant theory with the only scales coming from spontaneous symmetry breaking, which also produces the quark mass matrix effects. This is false in QCD. The strong interactions are not scale invariant except at extremely high energies where they are free, and they go to this limit slowly, by logarithms, not by power-laws, as in the case of scale invariance broken by masses or condensates.
But Wilson further makes the then radical assumption that this theory will
not be a pure S-matrix bootstrap-type string theory, but a scale invariant quantum field theory, with local operators at every spacetime point and multiplication laws for these fields. What makes this paper radical is that the field theory is not assumed to be a traditional weakly coupled field theory, rather a completely different kind of field theory which is always strongly coupled, a theory at a scale invariant RG limit. He renounces a traditional perturbative Lagrangian description and introduces the OPE for the fields, and makes an assumption on the scale dimension for the fields. He calls the scale invariant super-strong interaction the "Hadronic skeleton theory". It is defined as the hypothetical ultraviolet fixed point of the strong interactions.
This assumption is not correct for the strong interactions. In QCD, the theory at zero quark mass has its own intrinsic scale, $\Lambda_{QCD}$, which is determined by the renormalization group running of the QCD coupling, and is separate and independent of the quark masses, which just happen to have the same order of magnitude in nature by an unexplained coincidence (The quark masses come from the Higgs scale while the QCD scale is independent, but they match up to a few orders of magnitude).
The "hadronic skeleton theory" in modern QCD would just be the ultraviolet fixed point, which is the free theory of quarks and gluons. This is ultimately a perturbative theory. In other theories, like Banks-Zaks theories, where you introduce enough flavors of quarks and colors to make a weakly coupled scale invariant limit, you can have scale-invariant infrared limits, and then you can imagine a theory which is scale invariant forever, with only a little bit of breaking. But these are not QCD.
The theory Wilson is proposing and examining here is an entirely different possibility from the then mainstream bootstrap approach, and Wilson devotes the paper to showing that the operator products subsume and extend the current algebra approach, extending it to the case where the fields have arbitrary scale dimensions. This was a type of field theory which had not been considered in the 1950s, but it was clear that there was at least one example, because the 2 dimensional Thirring model provided one. Wilson would later give the canonical example, the Wilson Fisher fixed point in 3d, with the scaling dimensions the Ising model anomalous dimensions. Wilson had the courage to imagine and propose that such a thing is happening in 4 dimensions and in the strong interactions.
In historical reminiscences, Wilson describes his thinking about the OPE by considering the momentum-indexed variables field theory divided into concentric sectors of momentum $\lambda a^k <|k|<\lambda a^{k-1}$ for some $|a|<1$. Then he considered integrating out the momenta in consecutive shells, from outside in, producing a discrete version of the renormalization group flow. He realized that operators which were non-coincident before integrating out would have to become coinciding after integration. The history is discussed in reminiscences of Michael Peskin which may be found in this video and the associated historical arxiv paper: http://www.physics.cornell.edu/events-2/ken-wilson-symposium/ken-wilson-symposium-videos/michael-peskin-slac-ken-wilson-solving-the-strong-interactions/. This idea is a momentum space version of block-spinning, which was developed, I believe earlier and independently, by Kadanoff and Migdal in statistical physics. But Wilson was studying statistical physics at the time, and perhaps he was influenced by Kadanoff style block spinning. I don't know.
This idea of operator product expansions is as basic as Taylor series in calculus, or Ito calculus for Brownian walks. The main point is that the coefficients are only diverging locally, there are only finitely many singular operators in the product of any two operators, and the scaling behavior of the coefficients are determined by scaling laws for the fields, and you can do calculus on the operator products expansion and the local operators without fear of contradiction or paradox. Wilson introduces these main points correctly, and these main points are used in all subsequent work on the OPE.
The OPE is also important sociologically in rehabilitating field theory, as the confusions with quantum field theory were caused by people proving "theorems", like the Sutherland-Veltman theorem, and then others then showing by explicit calculation that the theorem is false. The failures of theorems intoduced a whole zoo of diseases--- Schwinger terms, anomalies, and so on. All of these are subsumed into the analysis of the singular terms in the OPE, and when you have a systematic calculus for these, you aren't going to be surprised anymore by false theorems. In principle, you can compute the OPE coefficients, define the composite operators appearing in the expansion, and check if your differential-algebraic identities still hold with the singular coefficients of the OPE. This makes it instrumental in demonstrating that quantum field theory actually was a well defined thing.
But because Wilson's physical strong interaction picture is off, the physical deductions regarding the specific family of models are incorrect, the material in section VII (applications), and the specific predictions regarding strong interaction behavior are mostly obsolete and can be disregarded. The analysis of the SU(2)xSU(2) symmetry breaking is recapitulating current algebra, in a context where Wilson doesn't know the dimensions of the pion field anymore, as he is imagining it is something other than a normal weakly interacting field theory. In real life QCD, the pion field can be taken to be the divergence of the quark axial current $\partial_\mu \psi^i \gamma^5 \gamma^\mu \psi^i$ , and the dimension at short distances is exactly four by free-field dimensional analysis. The paper assumes that the rho will become massless in a hadronic skeleton theory, as all scales collapse. This is not what happens in zero quark mass QCD, where the rho mass doesn't vanish, as the chiral condensate doesn't go away. Neither does the Baryon mass vanish, as it is determined by $\Lambda_{QCD}$
In section VIIA, Wilson discusses some sum rules of Weinberg. I didn't review this yet.
In section VIIB, To be reviewed.
In section VII C,D: The paper notes the existence of an axial anomaly, and attempts to apply the OPE to calculate it. It formulates the calculation in OPE language, but doesn't carry it out. It is certainly possible to do in a specific field theory model with a broken axial current, but Wilson finds it sufficient to pinpoint the exact place where the Veltman-Sutherland theorem fails in differentiating the product of the current and photon field without differentiating the singular coefficients to find the extra finite residue. This calculation I think is important, because the OPE makes all the identities of quantum field theory error-free and systematic.
In section VII E, Wilson proposes ideas for weak interactions, which were known to be coupled to the same axial currents which are spontaneously broken in the strong interaction. This was what allowed the PCAC relation between neutron decay (a weak process) and the pion-nucleon coupling constant (strong process). The pion is a goldstone boson of the same current which is involved in the weak interaction. Wilson proposes that the different scaling laws for the different fields in the skeleton theory are responsible for the enhancement or decay of certain hadronic weak processes over others. Since the skeleton theory is not interacting, this is not what is going on. |
Tagged: normal subgroup If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group |
Here are the functions:
a) $\displaystyle\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$
If I used L'Hopital's rule the limit is $2$
b) $\displaystyle\lim _{x\to 0}\:\frac{e^x\cdot \:\sin\left(x\right)-x\cdot \left(1+x\right)}{x^3}$
here $\dfrac{1}{3}$
c) $\displaystyle\lim _{x\to 0}\left(\frac{\ln\left(\sin\left(3 x\right)\right)}{\ln\left(\sin\left(7x\right)\right)}\right)$
and here $1$
but the problem is that I am not allowed to use L'Hopital's rule, can you give me ideas for another type of approaches?
UPDATE:
I apologize, I see there is some discussion and confusion among people, which obviously goes beyond my functions, but still I wanted to explain that I have been missing a lots of lectures recently due to illness and last week I got $0$ points for using L'hopital because we have not learnt it, so my guess was that we are not allowed this time either, but I just talked to my tutor and he told me that just in the last lecture, they introduced L'hopital rule to us so I am free to use it. I'm very sorry. |
Not really. First, note that each functional $\psi $ on $C_b (X) $ restricts to a bounded functional on $C_0 (X) $, so that there is a (unique!) measure (regular, complex valued) with $\psi (f) =\int f d\mu $ for $ f \in C_0$.
The problem is that even though both sides of the equality make sense for $f \in C_b $, it is not necessarily true for all such $f $. As an example, consider the (closed) subspace$$L := \{f : \Bbb {R} \to \Bbb {C} \,:\, \lim_{|x|\to \infty} f (x) \text { exists}\},$$and define a bounded functional on $L $ by $\psi (f) =\lim_{|x|\to\infty} f (x) $. Since $\psi $ vanishes on $C_0$, the measure from above is $\mu =0$, although $\psi (x\mapsto 1) =1 \neq 0$.
Essentially, the problem is that $C_b / C_0$ is quite large in general.
Note though that the theorem is true if $X $ is compact, simply because in this case, $C_0 = C_b $.
EDIT: Related to your additional question: One can identify the set of functionals on $C_b (X)$ with the set of regular complex measures on a suitable compactification of $X$. Indeed, if $\beta X$ denotes the Stone–Čech compactification of $X$, then each $f \in C_b (X)$ extends uniquely to a map $\tilde{f} \in C(\beta X) = C_0 (\beta X)$ (note that $f(X)$ is a bounded subset of $\Bbb{C}$, and thus contained in a compact set; in fact, this shows $\| \tilde{f} \|_\sup = \|f \|_\sup$).
Let $\iota : X \to \beta X$ be the embedding of $X$ into $\beta X$ and note that $\iota$ is a homeomorphism of $X$ onto its range in $\beta X$, since $X$ is locally compact, and thus Tychonoff. Now, if we set$$V := \{ \tilde{f} \, : \, f \in C_b (X)\},$$then $\varphi : V \to \Bbb{C}, \tilde{f} \mapsto \psi(f)$ is well-defined and bounded, and by the Hahn-Banach theorem, we can extend $\varphi$ to a bounded functional on all of $C(\beta X)$. Thus, there is a regular complex measure $\nu$ on $\beta X$ satisfying$$\psi(f)=\varphi (\tilde{f}) = \int \tilde{f} d \nu \quad \forall f \in C_b (X).$$Note though that we integrate $\tilde{f}$, and not $f$ itself.
Conversely, each complex measure $\nu$ on $\beta X$ also induces a linear functional on $C_b (X)$ by setting $\psi(f) := \int \tilde{f} d \nu$.
In fact, $\nu$ is uniquely determined by $\psi$: It is uniquely determined once one knows $\int g d\nu$ for all $g \in C(\beta X)$. But in fact, we have $V = C(\beta X)$, since if $g \in C(\beta X)$ is arbitrary, then $f := g \circ \iota \in C_b (X)$, and $g = \tilde{f}$ is the unique extension of $g$ to a continuous map on $\beta X$. |
In Don Knuth's famous series of books,
The Art of Computer Programming, section 2.3.1, he describes an algorithm to traverse binary tree in inorder, making use of an auxiliary stack:
T1[Initialize.] Set stack $\rm A$ empty and set the link variable $\rm P\gets T$
T2[$\rm P=\Lambda$?] If $\rm P=\Lambda$, go to step T4.
T3[Stack$\rm \;\Leftarrow P$] (Now $\rm P$ points to a nonempty binary tree that is to be traversed.) push the value of $\rm P$ onto stack $\rm A$, then set $\rm P\gets LLINK(P)$
T4[$\rm P\Leftarrow Stack$] If stack $\rm A$ is empty, the algorithm terminates; otherwise pop the top of $\rm A$ to $\rm P$.
T5[Visit $\rm P$] Visit $\rm NODE(P)$. Then set $\rm P\gets RLINK(P)$ and return to step T2.
We can plot a flow chart of the algorithm. In the succeeding paragraph, he gives a
formal proof of the algorithm:
Starting at step T2 with $\rm P$ a pointer to a binary tree of $n$ nodes and with the stack $\rm A$ containing $\rm A[1]\dotsc A[m]$ for some $m\ge 0$, the procedure of steps T2-T5 will traverse the binary tree in question, in inorder, and will then arrive at step T4 with stack $\rm A$ returned to its original value $\rm A[1]\dotsc A[m]$.
However, as far as I know, such a formal proof is quite different from the general method described in section 1.2.1:
for each box in the flow chart, that if an assertion attached to any arrow leading into the box is true before the operation in that box is performed, then all of the assertions on relevant arrows leading away from the box are true after the operation.
In fact, such a method is somewhat equivalent to Hoare logic, which is used to formally check the validity of algorithms.
Can we turn the statement mentioned to prove the traversing algorithm into a schema of Hoare logic, or the assertion-attachment of a flow chart?
Thanks! |
Problem 15
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$. (b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.
(
University of California, Berkeley) Problem 12
Let $A$ be an $n \times n$ real matrix. Prove the followings.
(a) The matrix $AA^{\trans}$ is a symmetric matrix. (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. (c) The matrix $AA^{\trans}$ is non-negative definite.
(An $n\times n$ matrix $B$ is called
non-negative definite if for any $n$ dimensional vector $\mathbf{x}$, we have $\mathbf{x}^{\trans}B \mathbf{x} \geq 0$.)
Add to solve later
(d) All the eigenvalues of $AA^{\trans}$ is non-negative. Problem 11
An $n\times n$ matrix $A$ is called
nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. Prove the followings. (a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
Add to solve later
(b) The matrix $A$ is nilpotent if and only if $A^n=O$. Read solution Problem 9
Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$
Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.
Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
Read solution Problem 5
Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Let $\mathbf{0}_n$ and $\mathbf{0}_m$ be zero vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively. Show that $T(\mathbf{0}_n)=\mathbf{0}_m$.
(
The Ohio State University Linear Algebra Exam)
Add to solve later
Problem 3
Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.
Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.
In particular, the commutator subgroup $[G, G]$ is a normal subgroup of $G$Add to solve later |
Ray Optics and Optical Instruments Total Internal Reflection Critical angle is the angle of incidence in denser for which angle of refraction in rarer medium is 90° \tt \mu = \frac{1}{\sin C} \tt \sin C = \frac{\mu_{rarer}}{\mu_{denser}} = \frac{V_{denser}}{V_{rarer}} = \frac{\lambda_{denser}}{\lambda_{rarer}} C Red> C Violet(C = critical angle) When angle of incidence in the denser medium is greater than critical angle light ray bounces back to the same medium and is called Total internal reflection Field of vision of fish under water at depth h is r = h tan c = \tt \frac{h}{\sqrt{\mu^{2} - 1}} (r = radius of circle view) Optical fibre works on the principle of T I R Launching angle for optical fibre is \tt \sin i_{L} = \sqrt{\mu^{2}_{core} - \mu_{cladding}} Optical fibres are used in communication and Laproscope, Cudoscope Optical fibres are flexible, light weight, and non corrosive. View this Topic in this video from 1:37
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Total internal reflection
\mu = \frac{1}{\sin C} = cosec \ C where μ → Rarerμ Denser |
Answer
a) $P_t = \frac{1}{2}\rho \omega v^2 r^2+P_a + \rho h_0 g$ b) $h_t = h_0+\frac{\omega^2 r^2}{2g}$
Work Step by Step
a) In this problem, there are three sources of pressure at the bottom of the pan. For starters, there is the atmospheric pressure, which we will call $P_a$. In addition, there is the pressure due to the water above the bottom of the pan. We know that pressure is force per unit area, so we find the gravitational force per unit area to be: $P=\rho h_0 g$ (For there is $h_0$ meters of water above the point.) Finally, there is pressure due to centripetal acceleration, which we find is: $P = \frac{1}{2}\rho v^2$ $P = \frac{1}{2}\rho \omega^2 r^2$ Adding all of these gives: $P_t = \frac{1}{2}\rho \omega^2 r^2+P_a + \rho h_0 g$ b) We know that the height will occur where the centripetal force and gravitational force cancel. Thus, we find: $\rho h g=\frac{1}{2}\rho \omega^2 r^2$ $h =\frac{\omega^2 r^2}{2g}$ We add this to the initial height to find: $h_t = h_0+\frac{\omega^2 r^2}{2g}$ |
I am working on this question on my statics assignment:
Approximately at point B there are two forces, 20kN and 50kN.So far, I have only calculated global equilibrium and got the following values: $D_y=(\frac{80}{7})kN$ $D_x=(\frac{20}{\frac{\sqrt{13}}{2}})kN$ $A_y\approx89.094 kN$ My question here is, when I'm drawing the shear force diagram do I sum up the 20kN force and the 50kN force into one resultant and graph it? Or, do I draw them separately. I need help.
I am working on this question on my statics assignment:
To answer your question, you just need to break the inclined force into its component X and Y forces:
$$\begin{alignat}{4} f_x &= \dfrac{20}{\sqrt{1^2+1.5^2}}&&=11.09\text{ kN} \\ f_y &= -1.5f_x&&=-16.64\text{ kN} \\ \end{alignat}$$
and then add the vertical component to the 50 kN force.
That being said, your reactions (other than $D_x$) are incorrect:
$$\begin{align} \sum M_A &= -66.64\times2 - 6\times4\times\left(2+\dfrac{4}{2}\right) + 200 + 9D_y - 2\times2\times\left(9 + \dfrac{2}{2}\right) - 200 = 0 \\ \therefore D_y &= 29.92\text{ kN}\\ \sum F_y &= A_y + 29.92 - 66.64 - 6\times4 - 2\times2 = 0 \\ \therefore A_y &= 64.72\text{ kN} \end{align}$$
You can then draw the diagram as normal, remembering to add the inclined force's vertical component to the vertical force at B:
Diagram obtained with Ftool, a free 2D frame analysis program. |
Speaker
Prof. Tobias Frederico (Instituto Tecnologico de Aeronautica)
Description
We will discuss recent results for the formulation of charged three-body $B$ decays in charmless channels, like $KKK$, $\pi\pi$, $KK\pi$, and $\pi\pi\pi$, introducing the final state interaction in the decay amplitude with relation to the CPT constraint, while CP violation is allowed. We consider the s-wave interaction between the mesons, coupling between different decay channels and a resonance. The p-wave interaction in the resonant states, as the formation of the $\rho$-meson in the $\pi\pi$ channel is considered within the general formulation of the three-body decay channel. In this case the CP violation has contributions from the interference between different mechanisms, like e.g. interference from s and p-wave amplitudes. In particular, we will present preliminary results for the CP asymmetry with dependence on $\cos(\theta)$ in charmless charged channels, as revealed by the recent data from LHCb. We will also discuss briefly a relativistic three-body formalism for the final sate interaction based on the projection of the inhomogeneous Bethe-Salpeter equation onto the light-front with application to the D decay in $K\pi\pi$ channel.
Primary author
Prof. Tobias Frederico (Instituto Tecnologico de Aeronautica) |
No, the converse of Cayley-Hamilton is not true for $n\times n$ matrices with $n\gt 1$; in particular, it fails for $2\times 2$ matrices.
For a simple counterexample, notice that if $p(A)=0$, then for every multiple $q(x)$ of $p(x)$ you also have $q(A)=0$; so you would want to amend the converse to say "if $p(A)=0$, then $p(a)$ is a
multiple of the characteristic polynomial of $A$". But even that amended version is false
However, the only failure in the $2\times 2$ matrix case are the scalar multiples of the identity. If $A=\lambda I$, then $p(x)=x-\lambda$ satisfies $p(A)=0$, but the characteristic polynomial is $(x-\lambda)^2$, not $p(x)$.
For bigger matrices, there are other situations where even this weakened converse fails.
The concept that captures the "converse" of Cayley-Hamilton is the minimal polynommial of the matrix, which is the monic polynomial $p(x)$ of
smallest degree such that $p(A)=0$. It is then easy to show (using the division algorithm) that if $q(x)$ is any polynomial for which $q(A)=0$, then $p(x)|q(x)$. (Be careful to justify that if $m(x)=r(x)s(x)$, then $m(A)=r(A)s(A)$; this is not immediate because matrix multiplication is not in general commutative!) So we have:
Theorem. Let $A$ be an $n\times n$ matrix over $\mathbf{F}$, and let $\mu(x)$ be the minimal polynomial of $A$. If $p(x)\in \mathbf{F}[x]$ is any polynomial such that $p(A)=0$, then $\mu(x)$ divides $p(x)$.
The Cayley-Hamilton Theorem shows that the characteristic polynomial is always a multiple of the minimal polynomial. In fact, one can prove that every irreducible factor of the characteristic polynomial must divide the minimal polynomial. Thus, for a $2\times 2$ matrix, if the characteristic polynomial splits and has distinct roots, then the characteristic and minimal polynomial are equal. If the characteristic polynomial is irreducible quadratic and we are working over $\mathbb{R}$, then again the minimal and characteristic polynomials are equal. But if the characteristic polynomial is of the form $(x-a)^2$, then the minimal polynomial is either $(x-a)$ (when the matrix equals $aI$), or $(x-a)^2$ (when the matrix is not diagonalizable).
As for solving this problem: if $\lambda$ is an eigenvalue of $A$, and $A$ is invertible, then $\lambda\neq 0$, and $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$: for if $\mathbf{x}\neq\mathbf{0}$ is such that $A\mathbf{x}=\lambda\mathbf{x}$, then multiplying both sides by $A^{-1}$ we get $\mathbf{x} = A^{-1}(\lambda \mathbf{x}) = \lambda A^{-1}\mathbf{x}$. Dividing through by $\lambda$ shows $\mathbf{x}$ is an eigenvector of $A^{-1}$ corresponding to $\frac{1}{\lambda}$.
Since $A=A^{-1}$, that means that if $\lambda_1,\lambda_2$ are the eigenvalues of $A$, then $\lambda_1 = \frac{1}{\lambda_1}$ and $\lambda_2=\frac{1}{\lambda_1}$; thus, each eigenvalue is either $1$ or $-1$.
If the matrix is diagonalizable, then we cannot have both equal to $1$ (since then $A=I$), and they cannot both be equal to $-1$ (since $A\neq -I$), so one eigenvalue is $1$ and the other is $-1$. Since the trace of a square matrix equals the sum of its eigenvalues, the sum of the eigenvalues is $0$.
Why is $A$ diagonalizable? If it has two distinct eigenvalues, $1$ and $-1$, then there is nothing to do; we know it is diagonalizable. If it has a repeated eigenvalue, say $1$, but $A-I$ is not the zero matrix, pick $\mathbf{x}\in \mathbb{R}^2$ such that $A\mathbf{x}\neq \mathbf{x}$; then $$\mathbf{0}=(A-I)^2\mathbf{x} = (A^2-2A + I)\mathbf{x} = (2I-2A)\mathbf{x}$$by the Cayley Hamilton Theorem. But that means that $2(A-I)\mathbf{x}=\mathbf{0}$, contradicting our choice of $\mathbf{x}$. Thus, $A-I=0$, so $A=I$ and $A$ is diagonalizable. A similar argument shows that if $-1$ is the only eigenvalue, then $A+I=0$. .(Hiding behind that paragraph is the fact that if the minimal polynomial is squarefree and splits, then the matrix is diagonalizable; since $p(x)=x^2-1=(x-1)(x+1)$ is a multiple of the minimal polynomial, the matrix must be diagonalizable).
So this completes the proof that the trace must be $0$, given that $A\neq I$ and $A\neq -I$. |
There is a simple extension of trapezoid rule to 2d. In 1d, if you have the interval $[0,1]$, the rule takes the linear interpolant$$ l(x) = f(0)(1-x) + f(1)x, $$and integrates it to get$$ \int_0^1 l(x) = \tfrac12 f(0) + \tfrac12 f(1). $$To integrate over a large interval, you then sum up the contributions of the individual small intervals to get the usual trapezoid formula:$$ h \big(\tfrac12 f(x_0) + f(x_1) + \cdots + f(x_{n-1}) + \tfrac12 f(x_n) \big). $$
In 2d, on the square $[0,1]^2$, you can interpolate the function linearly on the bottom edge $y=0$, and separately on the top edge $y=1$, and then interpolate linearly between the two linear interpolants:$$ l_0(x) = f(0,0)(1-x)+f(1,0)x, \qquad l_1(x) = f(0,1)(1-x)+f(1,1)x, $$$$ l(x,y) = l_0(x)(1-y) + l_1(x)y. $$The resulting interpolant is a linear combination of elementary interpolants on the square $[0,1]^2$, with coefficients being the function values, and the elementary interpolants look like this:
where the elementary interpolants are the coefficients of the function values $f(x,y)$ in the interpolant constructed from them:$$ l(x,y) = f(0,0)(1-x)(1-y) + f(1,0)x(1-y) + f(0,1) (1-x)y + f(1,1)xy, $$so of the four functions $xy$, $x(1-y)$, $(1-x)y$ and $(1-x)(1-y)$.
The integral of $l(x,y)$ will be$$ \int_{[0,1]^2}l(x,y)\,dx\,dy = \tfrac14\big( f(0,0)+f(0,1)+f(1,0)+f(1,1) \big). $$
Summing up these contributions over small intervals would then give$$ \frac{h^2}{4}\sum_{j,k} w_{j,k} f(x_j, y_k) $$where $w_{j,k}$ is an integer that counts how many squares the point $(x_j,y_k)$ belongs to: one in the corners, two on the edges, four in the middle. |
I have a normalized energy eigenfunction for the ground state of Hydrogen which is $$ \Psi(r) = \frac{1}{\sqrt{\pi a_0^3}}\exp\left(-\frac{r}{a_o}\right), $$ where $a_o$ is the Bohr radius,
I have been told that the expectation value of the electrons momentum, $ \left\langle\hat{\vec{p}}\right\rangle =\vec{0}$. I want to know why and I think I'm missing something obvious because my current understanding is not very great.
I'm thinking it has something to do with the electron being in a bound state and the orbit being spherically symmetric but I still think there should be a positive expectation value for momentum.
Could somebody throw me a hint here? I tried to search for answers but couldn't find. |
Tagged: group action Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 455
Let $G$ be a finite group.
The centralizer of an element $a$ of $G$ is defined to be \[C_G(a)=\{g\in G \mid ga=ag\}.\]
A
conjugacy class is a set of the form \[\Cl(a)=\{bab^{-1} \mid b\in G\}\] for some $a\in G$. (a)Prove that the centralizer of an element of $a$ in $G$ is a subgroup of the group $G$.
Add to solve later
(b) Prove that the order (the number of elements) of every conjugacy class in $G$ divides the order of the group $G$. Problem 108
Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number.
Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, the $n$-dimensional vector space over $\F_p$. Therefore, $G_n$ acts on $\F_p^n$.
Let $e_n \in \F_p^n$ be the vector $(1,0, \dots,0)$.
(The so-called first standard basis vector in $\F_p^n$.)
Find the size of the $G_n$-orbit of $e_n$, and show that $\Stab_{G_n}(e_n)$ has order $|G_{n-1}|\cdot p^{n-1}$.
Conclude by induction that
\[|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).\] |
While reading a paper Nominal Unification from a Higher-Order Perspective, in the abstract I noticed something feel bit confusing,
Nominal Logic is an extension of first-order logic with equality, name-binding, name-swapping, and freshness of names. Contrarily to higher-order logic, bound variables are treated as atoms, and only free variables are proper unknowns in nominal unification. This allows “variable capture”, breaking a fundamental principle of lambda-calculus.
I know a bit about nominal logic, but as I know it respects $\alpha$-conversion, such as $\lambda x.x \approx_\alpha \lambda y.y$. but I could not understand above statement.
I understand the notion of variable capture which leads to a wrong result during the substitution.
So does variable capture not lead to an incorrect result in nominal logic?
what does it mean saying breaking "variable capture" here?
Can anyone explain with examples? |
Answer
(a) The speed of the transverse waves in the string is $1350~m/s$ (b) The tension is $45.6~N$ (c) The frequency in the air is $450.0~Hz$ The wavelength in the air is $0.756~m$
Work Step by Step
(a) We can find the speed of the transverse waves in the string: $v = \lambda~f$ $v = 2L~f$ $v = (2)(1.50~m)(450.0~Hz)$ $v = 1350~m/s$ The speed of the transverse waves in the string is $1350~m/s$ (b) We can find the tension: $\sqrt{\frac{F}{\mu}} = v$ $\frac{F}{\mu} = v^2$ $F = \mu~v^2$ $F = (25.0\times 10^{-6}~m)(1350~m/s)^2$ $F = 45.6~N$ The tension is $45.6~N$ (c) The frequency in the air is the same as the frequency in the string. Therefore, the frequency in the air is $450.0~Hz$ We can find the wavelength in the air: $\lambda = \frac{v}{f}$ $\lambda = \frac{340~m/s}{450.0~Hz}$ $\lambda = 0.756~m$ The wavelength in the air is $0.756~m$ |
When investigating regular languages, regular expressions are obviously a useful characterisation, not least because they are amenable to nice inductions. On the other hand ambiguity can get in the way of some proofs.
Every regular language is recognized by an unambiguous context-free grammar (take a deterministic automaton which recognises it, and make a production $R \rightarrow tS$ for every edge $R \stackrel{t}{\rightarrow} S$ in the DFA, and $R \rightarrow \epsilon$ for every accepting state $R$).
On the other hand, the natural "grammar" for a regular language is its regular expression. Can
these be made unambiguous?
To be precise, let's define a parse for a regular expression (this is I think a natural definition, but not one I've seen named before).
$x$ is an $x$-parse of $x$, if $x$ is a symbol or $x=\varepsilon$ $(y, 0)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R$-parse of $x$ Similarly, $(y,1)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R'$-parse of $x$ $(y_1, y_2)$ is an $RR'$-parse for $x_1x_2$, if $y_i$ is an $R$-parse for $x_i$ for $i=1,2$ $[]$ is an $R^*$-parse for $\varepsilon$ $[y_1, y_2, \dots, y_n]$ is an $R^*$-parse of $x_1x_2\cdots x_n$, if $y_i$ is an $R$-parse for $x_i$ for $1 \le i \le n$
In short, the parses of a string tell us
how a regular expression matches a string if it does.
A regular expression $R$ is
unambiguous if, for every $x \in L(R)$, there is only one $R$-parse of $x$. Given a regular expression, is there an unambiguous regular expression which matches the same language? |
The
Friedmann–Lemaître–Robertson–Walker ( FLRW) Howard P. Robertson and Arthur Geoffrey Walker — may be named (e.g., Friedmann–Robertson–Walker ( FRW) or Robertson–Walker ( RW) or Friedmann–Lemaître ( FL)). This model is sometimes called the Standard Model of modern cosmology. [4] It was developed independently by the named authors in the 1920s and 1930s.
Contents General metric 1 Reduced-circumference polar coordinates 1.1 Hyperspherical coordinates 1.2 Cartesian coordinates 1.3 Solutions 2 Interpretation 2.1 Cosmological constant 2.2 Newtonian interpretation 2.3 Name and history 3 Einstein's radius of the universe 4 Evidence 5 References and notes 6 Further reading 7 General metric
The FLRW metric starts with the assumption of homogeneity and isotropy of space. It also assumes that the spatial component of the metric can be time-dependent. The generic metric which meets these conditions is
- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}\mathbf{\Sigma}^2
where \mathbf{\Sigma} ranges over a 3-dimensional space of uniform curvature, that is, elliptical space, Euclidean space, or hyperbolic space. It is normally written as a function of three spatial coordinates, but there are several conventions for doing so, detailed below. \mathrm{d}\mathbf{\Sigma} does not depend on
t — all of the time dependence is in the function a( t), known as the "scale factor". Reduced-circumference polar coordinates
In reduced-circumference polar coordinates the spatial metric has the form
\mathrm{d}\mathbf{\Sigma}^2 = \frac{\mathrm{d}r^2}{1-k r^2} + r^2 \mathrm{d}\mathbf{\Omega}^2, \quad \text{where } \mathrm{d}\mathbf{\Omega}^2 = \mathrm{d}\theta^2 + \sin^2 \theta \, \mathrm{d}\phi^2.
k is a constant representing the curvature of the space. There are two common unit conventions: k may be taken to have units of length −2, in which case r has units of length and a( t) is unitless. k is then the Gaussian curvature of the space at the time when a( t) = 1. r is sometimes called the reduced circumference because it is equal to the measured circumference of a circle (at that value of r), centered at the origin, divided by 2π (like the r of Schwarzschild coordinates). Where appropriate, a( t) is often chosen to equal 1 in the present cosmological era, so that \mathrm{d}\mathbf{\Sigma} measures comoving distance. Alternatively, k may be taken to belong to the set {−1,0,+1} (for negative, zero, and positive curvature respectively). Then r is unitless and a( t) has units of length. When k = ±1, a( t) is the radius of curvature of the space, and may also be written R( t).
A disadvantage of reduced circumference coordinates is that they cover only half of the 3-sphere in the case of positive curvature—circumferences beyond that point begin to decrease, leading to degeneracy. (This is not a problem if space is elliptical, i.e. a 3-sphere with opposite points identified.)
Hyperspherical coordinates
In
hyperspherical or curvature-normalized coordinates the coordinate r is proportional to radial distance; this gives \mathrm{d}\mathbf{\Sigma}^2 = \mathrm{d}r^2 + S_k(r)^2 \, \mathrm{d}\mathbf{\Omega}^2
where \mathrm{d}\mathbf{\Omega} is as before and
S_k(r) = \begin{cases} \sqrt{k}^{\,-1} \sin (r \sqrt{k}), &k > 0 \\ r, &k = 0 \\ \sqrt{|k|}^{\,-1} \sinh (r \sqrt{|k|}), &k < 0. \end{cases}
As before, there are two common unit conventions:
k may be taken to have units of length −2, in which case r has units of length and a( t ) is unitless. k is then the Gaussian curvature of the space at the time when a( t ) = 1. Where appropriate, a( t ) is often chosen to equal 1 in the present cosmological era, so that \mathrm{d}\mathbf{\Sigma} measures comoving distance. Alternatively, as before, k may be taken to belong to the set {−1,0,+1} (for negative, zero, and positive curvature respectively). Then r is unitless and a( t ) has units of length. When k = ±1, a( t ) is the radius of curvature of the space, and may also be written R( t ). Note that, when k = +1, r is essentially a third angle along with θ and φ. The letter χ may be used instead of r.
Though it is usually defined piecewise as above,
S is an analytic function of both k and r. It can also be written as a power series S_k(r) = \sum_{n=0}^\infty \frac{(-1)^n k^n r^{2n+1}}{(2n+1)!} = r - \frac{k r^3}{6} + \frac{k^2 r^5}{120} - \cdots
or as
S_k(r) = r \; \mathrm{sinc} \, (r \sqrt{k})
where sinc is the unnormalized sinc function and \sqrt{k} is one of the imaginary, zero or real square roots of
k. These definitions are valid for all k. Cartesian coordinates
When
k = 0 one may write simply \mathrm{d}\mathbf{\Sigma}^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2.
This can be extended to
k ≠ 0 by defining x = r \cos \theta \,, y = r \sin \theta \cos \phi \,, and z = r \sin \theta \sin \phi \,,
where
r is one of the radial coordinates defined above, but this is rare. Solutions
Einstein's field equations are not used in deriving the general form for the metric: it follows from the geometric properties of homogeneity and isotropy. However, determining the time evolution of a(t) does require Einstein's field equations together with a way of calculating the density, \rho (t), such as a cosmological equation of state.
This metric has an analytic solution to Einstein's field equations G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu} giving the Friedmann equations when the energy-momentum tensor is similarly assumed to be isotropic and homogeneous. The resulting equations are:
[5] \left(\frac{\dot a}{a}\right)^{2} + \frac{kc^{2}}{a^2} - \frac{\Lambda c^{2}}{3} = \frac{8\pi G}{3}\rho 2\frac{\ddot a}{a} + \left(\frac{\dot a}{a}\right)^{2} + \frac{kc^{2}}{a^2} - \Lambda c^{2} = -\frac{8\pi G}{c^{2}} p.
These equations are the basis of the standard big bang cosmological model including the current ΛCDM model. Because the FLRW model assumes homogeneity, some popular accounts mistakenly assert that the big bang model cannot account for the observed lumpiness of the universe. In a strictly FLRW model, there are no clusters of galaxies, stars or people, since these are objects much denser than a typical part of the universe. Nonetheless, the FLRW model is used as a first approximation for the evolution of the real, lumpy universe because it is simple to calculate, and models which calculate the lumpiness in the universe are added onto the FLRW models as extensions. Most cosmologists agree that the observable universe is well approximated by an
almost FLRW model, i.e., a model which follows the FLRW metric apart from primordial density fluctuations. As of 2003, the theoretical implications of the various extensions to the FLRW model appear to be well understood, and the goal is to make these consistent with observations from COBE and WMAP. Interpretation
The pair of equations given above is equivalent to the following pair of equations
{\dot \rho} = - 3 \frac{\dot a}{a}\left(\rho+\frac{p}{c^{2}}\right) \frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}
with k, the spatial curvature index, serving as a constant of integration for the first equation.
The first equation can be derived also from thermodynamical considerations and is equivalent to the first law of thermodynamics, assuming the expansion of the universe is an adiabatic process (which is implicitly assumed in the derivation of the Friedmann–Lemaître–Robertson–Walker metric).
The second equation states that both the energy density and the pressure cause the expansion rate of the universe {\dot a} to decrease, i.e., both cause a deceleration in the expansion of the universe. This is a consequence of gravitation, with pressure playing a similar role to that of energy (or mass) density, according to the principles of general relativity. The cosmological constant, on the other hand, causes an acceleration in the expansion of the universe.
Cosmological constant
The cosmological constant term can be omitted if we make the following replacements
\rho \rightarrow \rho + \frac{\Lambda c^{2}}{8 \pi G} p \rightarrow p - \frac{\Lambda c^{4}}{8 \pi G}.
Therefore the cosmological constant can be interpreted as arising from a form of energy which has negative pressure, equal in magnitude to its (positive) energy density:
p = - \rho c^2. \,
Such form of energy—a generalization of the notion of a cosmological constant—is known as dark energy.
In fact, in order to get a term which causes an acceleration of the universe expansion, it is enough to have a scalar field which satisfies
p < - \frac {\rho c^2} {3}. \,
Such a field is sometimes called quintessence.
Newtonian interpretation
The Friedmann equations are equivalent to this pair of equations:
- a^3 {\dot \rho} = 3 a^2 {\dot a} \rho + \frac{3 a^2 p {\dot a}}{c^2} \, \frac{{\dot a}^2}{2} - \frac{G \frac{4 \pi a^3}{3} \rho}{a} = - \frac{k c^2}{2} \,.
The first equation says that the decrease in the mass contained in a fixed cube (whose side is momentarily
a) is the amount which leaves through the sides due to the expansion of the universe plus the mass equivalent of the work done by pressure against the material being expelled. This is the conservation of mass-energy (first law of thermodynamics) contained within a part of the universe.
The second equation says that the kinetic energy (seen from the origin) of a particle of unit mass moving with the expansion plus its (negative) gravitational potential energy (relative to the mass contained in the sphere of matter closer to the origin) is equal to a constant related to the curvature of the universe. In other words, the energy (relative to the origin) of a co-moving particle in free-fall is conserved. General relativity merely adds a connection between the spatial curvature of the universe and the energy of such a particle: positive total energy implies negative curvature and negative total energy implies positive curvature.
The cosmological constant term is assumed to be treated as dark energy and thus merged into the density and pressure terms.
During the Planck epoch, one cannot neglect quantum effects. So they may cause a deviation from the Friedmann equations.
Name and history
The main results of the FLRW model were first derived by the Soviet mathematician Alexander Friedmann in 1922 and 1924. Although his work was published in the prestigious physics journal Zeitschrift für Physik, it remained relatively unnoticed by his contemporaries. Friedmann was in direct communication with Albert Einstein, who, on behalf of Zeitschrift für Physik, acted as the scientific referee of Friedmann's work. Eventually Einstein acknowledged the correctness of Friedmann's calculations, but failed to appreciate the physical significance of Friedmann's predictions.
Friedmann died in 1925. In 1927, Catholic University of Leuven, arrived independently at similar results as Friedmann had and published them in Annals of the Scientific Society of Brussels. In the face of the observational evidence for the expansion of the universe obtained by Edwin Hubble in the late 1920s, Lemaître's results were noticed in particular by Arthur Eddington, and in 1930–31 his paper was translated into English and published in the Monthly Notices of the Royal Astronomical Society.
Howard P. Robertson from the US and Arthur Geoffrey Walker from the UK explored the problem further during the 1930s. In 1935 Robertson and Walker rigorously proved that the FLRW metric is the only one on a spacetime that is spatially homogeneous and isotropic (as noted above, this is a geometric result and is not tied specifically to the equations of general relativity, which were always assumed by Friedmann and Lemaître).
Because the dynamics of the FLRW model were derived by Friedmann and Lemaître, the latter two names are often omitted by scientists outside the US. Conversely, US physicists often refer to it as simply "Robertson–Walker". The full four-name title is the most democratic and it is frequently used. Often the "Robertson–Walker"
metric, so-called since they proved its generic properties, is distinguished from the dynamical "Friedmann-Lemaître" models, specific solutions for a( t) which assume that the only contributions to stress-energy are cold matter ("dust"), radiation, and a cosmological constant. Einstein's radius of the universe
Einstein's radius of the Universe is the radius of curvature of space of Einstein's universe, a long-abandoned static model that was supposed to represent our universe in idealized form. Putting \dot{a} = \ddot{a} = 0
in the Friedmann equation, the radius of curvature of space of this universe (Einstein's radius) is
R_E=c/\sqrt {4\pi G\rho},
where c is the speed of light, G is the Newtonian gravitational constant, and \rho is the density of space of this universe. The numerical value of Einstein's radius is of the order of 10
10 light years. Evidence
By combining the observation data from some experiments such as WMAP and Planck with theoretical results of Ehlers–Geren–Sachs theorem and its generalization,
[6] astrophysicists now agree that the universe is almost homogeneous and isotropic (when averaged over a very large scale) and thus nearly a FLRW spacetime. References and notes ^ For an early reference, see Robertson (1935); Robertson assumes multiple connectedness in the positive curvature case and says that "we are still free to restore" simple connectedness. ^ M. Lachieze-Rey; J.-P. Luminet (1995), "Cosmic Topology", ^ G. F. R. Ellis; H. van Elst (1999). "Cosmological models (Cargèse lectures 1998)". In Marc Lachièze-Rey. Theoretical and Observational Cosmology. NATO Science Series C 541. pp. 1–116. ^ L. Bergström, A. Goobar (2006), Cosmology and Particle Astrophysics (2nd ed.), ^ P. Ojeda and H. Rosu (2006), "Supersymmetry of FRW barotropic cosmologies", ^ See pp. 351ff. in Hawking, Stephen W.; . Further reading
English trans. in 'General Relativity and Gravitation' 1999 vol.31, 31– Harrison, E. R. (1967), "Classification of uniform cosmological models", Monthly Notices of the Royal Astronomical Society 137: 69–79, d'Inverno, Ray (1992), Introducing Einstein's Relativity, Oxford: Oxford University Press, (See Chapter 23 for a particularly clear and concise introduction to the FLRW models.).
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The classic Hurwitz theorem for rational approximations (in simplest form; the constant can of course be improved) gives infinitely many approximations $\frac mn$ to an irrational $\alpha$ with $|\frac mn-\alpha|\lt\frac1{n^2}$. Just recently, in trying to answer a question related to rational approximation of $\pi$ I tripped over a limitation of this theorem: it tells us nothing about the specific $m,n$ of an approximation. I'm interested in $n$ particularly, and wondering if there are any 'Dirichlet-style' results that say that for any irrational $\alpha$ and for any $a, d$ we can get good approximations (in the sense above) with $n\equiv a\pmod d$. Is this a known result?
The answer is no. Take $\alpha=\sqrt{2}$ and note that if $|\sqrt{2}-m/n|\le 1/n^2$ then we have $0<|2n^2-m^2| \le (\sqrt{2}n+m)/n \le 3$. Now suppose we want $n\equiv 4\pmod p$ say. Then we must have that $32-m^2 \equiv b \pmod p$ for some $|b|\le 3$. But we can find a prime $p$ for which the numbers $29$ to $35$ are all quadratic non-residues $\pmod p$ (for example, choose $p$ so that $2, 7, 11, 29, 31$ are all non-residues $\pmod p$, and $3, 5, 17$ are residues). Thus there are no good approximations to $\sqrt{2}$ with $n\equiv 4\pmod p$ for such a prime $p$. One can clearly vary this argument a fair bit.
The idea I suggested in the comments works. Let $\phi$ denote the golden ratio
$$ \frac{\sqrt{5} + 1}{2} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cdots}}}$$
Because there are no coefficients of the continued fraction greater than $1$, no terms of the Farey sequence fall between the $n$th and $(n+1)$st convergents until the $(n+2)$nd convergent. This means any other fraction $\frac{p}{q}$ with $F_{n+1} \lt q \le F_{n+2}$ is separated from $\phi$ by at least $\frac{1}{q F_{n+1}} \gt \frac{1}{q^2}$. So, the only reduced good approximations are convergents, ratios of Fibonacci numbers.
Since $|\phi - \frac{F_{n+1}}{F_n}| \approx \frac{1}{\sqrt{5}F_n^2} \gt \frac{1}{(2F_n)^2}$, only reduced fractions can be good approximations.
The Fibonacci sequence does not hit every arithmetic progression. $F_{17} = 1597$ is prime, and the Fibonacci numbers repeat mod $1597$ with period $68$. Anything not hit in that period, such as $4,6,7,9... \mod 1597$, can't be the congruence class of the denominator of a good approximation to $\phi$. Similarly, no good approximation has a denominator that is $4 \mod 233$.
A slight extension works for many other quadratic irrationals. For example,
$$\sqrt{10} = 3+\cfrac{1}{6+\cfrac{1}{6+\cfrac{1}{6+\cdots}}} = [3; \bar 6].$$
The reduced good approximations with denominators between $q_n$ and $q_{n+1}$ must fall between $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$, so the candidate reduced good approximations are $3, 4, [3;2], [3;3], ... [3; 6, 6, ...,6, a],... $ where $1\le a \le 6$. Many of these are too far away to be good approximations. The set of denominators smaller than $q_{n+1}$ of reduced good approximations has $6$ elements for each convergent, and these are linear combinations of convergents, e.g., the denominator of $[3;6,6,...,6,4]$ is $4q_n + q_{n-1}$.
I chose $281$ because it is a factor of a denominator of an early convergent. Mod $281$, the denominators of convergents repeat with period $20$, and hit only the $9$ classes $0,\pm1, \pm6, \pm37, \pm53 \mod 281$. Linear combinations $aq_{n+1}+q_n$ with $1\le a \le 6$ cover $45$ congruence classes. The convergents satisfy $|\sqrt{10} - \frac{p_n}{q_n}| \approx \frac{1}{2\sqrt{10}q_n^2}$ and $4 \lt 2\sqrt{10} \lt 9$, so we also have to consider unreduced approximations from doubling both the numerator and denominator of convergents, e.g., $|\sqrt{10}-\frac{38}{12}| \lt \frac{1}{144}$. That adds only $4$ more congruence classes. There are no good approximations to $\sqrt{10}$ with denominators congruent to $8,9,10,11,14, ... \mod 281$.
Just to make sure that it's not hidden in the comments for future readers, I want to point out that the paper "On the Approximation of Irrational Numbers With Rationals Restricted By Congruence Relations" actually gives a positive answer to a slight extension of my question (where rather than just having $\left|\frac mn-\alpha\right|\lt \frac1{n^2}$ one has that the difference is bounded for $\frac{C}{n^2}$ for some $C$ possibly dependent on the congruence parameters), which was actually sufficient to show what I was originally after. The precise statement is:
For any irrational $\xi$, any $s\geq 1$, and integers $a, b$, there are infinitely many integers $m, n$ satisfying $\displaystyle\left|\xi-\frac{m}{n}\right|\lt\frac{2s^2}{n^2}$, $m\equiv a\pmod s$, $n\equiv b\pmod s$. |
I'll try to make a couple of remarks. it'll be easier for me to write only a bit at a time.
The metric category of metric spaces and metric maps (i.e. Lipschitz with constant $1$, i.e. never stretching) has a certain rigidity to it, so that the idea of a metric base seems attractive. However, I feel that metric dimension makes sense only for a somewhat limited class of, rather then for all, metric spaces. (Possibly, more than one of several competing definitions can make sense).
Consider (see the
above): Question
Property G A subset $B$ of a metric space $(M,d)$ is called a metric g-set for $M$ if and only if $$ \forall_{b \in B}\ d(x,b)=d(y,b)\ \ \implies\ \ x = y $$
Every such set $B$ was called a
metric basis above. Let $B$ be one, and let $\ B\subseteq C\subseteq M.\ $ Then $$ \forall_{c \in C}\ d(x,c)=d(y,c)\ \ \implies\ \ x = y $$
Thus, $\ C\ $ has the metric g-set property too. We see that declaring
Property G as a definition of metric basis goes against the spirit of basis in general. Instead, Property G corresponds to the general notion of a . generating set
It took quite a long time to arrive at the topological definition of a curve and of the topological dimension (from Cantor's curves in $\ \mathbb R^2,\ $ to Brouwers-Urysohn-Menger, then to the algebraic topological dimensions, etc.).
It took also considerable time to arrive at a general definition of basis which ultimately led to matroids.
In this thread the topic is rather something like the metric variations of the independent sets and of a basis. Perhaps one should eye matroids for a comparison. These days things go much faster than in the past but the process of obtaining one or more of such sound metric notions may still take some tries and patience.
Thus let's examine the property of a basis, and more generally, of independent sets for matroids: let $X$ be a matroid; then
if $\ A\subseteq X\ $ is an independent set, and $\ b\in X,\ $ then there exists $\ a\in A\ $ such that $\ (A\setminus\{a\})\cup\{b\}\ $ is independent;
if $\ A\subseteq X\ $ is an independent set (resp. basis), and $\ b\in X\ $ depends on $\ A\ $ (resp. $\ b\in X\ $ is arbitrary), then there exists $\ a\in A\ $ such that $\ (A\setminus\{a\})\cup\{b\}\ $ generates the same set as $\ A\ $ (resp. $\ (A\setminus\{a\})\cup\{b\}\ $ is a basis).
Now let's look at some metric spaces.
EXAMPLE 1 Let $\ X:=\{0\ 1\ 2\ 3\},\ $ and let $\ d\ $ be a metric (thus, symmetric, etc) in $\ X,\ $ such that:\begin{eqnarray} d(0\ n) &:=& 2-\frac 1n&\qquad \mbox{for }\ \ n=1\ 2\ 3\\ d(k\ n) &:=& 2&\qquad \mbox{for every}\ \ 0<k<n\le3\end{eqnarray}We see that there is a $1$-element basis $\ \{0\}\ $ in $\ (X\ d)\ $ in the sense of while there is no other $1$-element Question basis. However, in the spirit of matroids (see the above properties 1. and 2.), every $1$-element subset of $X$ should be a basis. Well, this is not so.
ALSO: Every $2$-element set $\ X\subset\{0\ 1\ 2\}\ $ is a minimal G-set ( basis). These three $2$-element sets and $\ \{0\}\ $ are the only minimal G-sets; thus, there are four of them altogether.
Turning to continuous spaces will not help:
EXAMPLE 2 Let $\ X:=[1;\infty)\ $ be a closed half-line, and let $\ d\ $ be a metric (thus, symmetric, etc) in $\ X,\ $ such that:\begin{eqnarray} d(1\ x) &:=& 2-\frac 1x&\qquad \mbox{for every}\ \ x > 1\\ d(x\ y) &:=& 2&\qquad \mbox{for every}\ \ 1<x<y\end{eqnarray}We see that there is a $1$-element basis $\ \{1\}\ $ in $\ (X\ d)\ $ in the sense of while there is no other $1$-element Question basis. Again, in the spirit of matroids, every $1$-element subset of $X$ should be a basis.
Thus, continuity didn't help.
ALSO: In addition to $\ \{0\},\ $ the only other minimal G-sets are sets $\ X\subset[0;\infty)\ $ such that $\ |[0;\infty)\setminus X|=1$.
CONCLUSION If ( a big ) we had to continue with the definition of if basis from then we would have to restrict the class of the metric spaces under consideration. The first candidate which comes to mind would be the class of transitive space (i.e. such that for every two points there is an isometry of the space onto itself, which sends one of the points onto the other one). Question
(I might continue later on) |
I would like to evaluate the following summation of Clebsch-Gordan and Wigner 6-j symbols in closed form:
$$\sum_{l,m} C_{l_2,m_2,l_1,m_1}^{l,m} C_{\lambda_2,\mu_2,\lambda_1,\mu_1}^{l,m} \left\{ \begin{array}{ccc} l & l_2 & l_1 \\ n/2 & n/2 & n/2 \end{array}\right\} \left\{ \begin{array}{ccc} l & \lambda_2 & \lambda_1 \\ n/2 & n/2 & n/2 \end{array}\right\}$$
with $n \in \left[0,\infty\right)$, $l,l_1,l_2,\lambda_1,\lambda_2 \in \left[0,n\right]$, $m \in \left[-l,l\right]$, $m_1 \in \left[-l_1,l_1\right]$, $m_2 \in \left[-l_2,l_2\right]$, $\mu_1 \in \left[-\lambda_1,\lambda_1\right]$ and $\mu_2 \in \left[-\lambda_2,\lambda_2\right]$. All indices are integers and n must be also even.
I have been using Varshalovich's Book, but can't find any identities that have been useful to simplify this. I am hoping that the result is something like $\delta_{l_2,\lambda_2}\delta_{m_2,\mu_2}\delta_{l_1,\lambda_1}\delta_{m_1,\mu_1}$, but I'm not sure that that will be the case. Any ideas of how to evaluate this? |
The cpr package provides several tools for finding parsimonious B-spline regression models via control polygon reduction. This vignette is an overview of the tools provided by the package.
This vignette is not complete. The details will be filled in soon. As for now, this is a placeholder for many topics to come. A Journal of Statistical Software paper is under development. That paper will become the foundation for this vignette.
A brief overview of the Control Polygon Reduction and Control Net Reduction method for finding parsimonious B-spline regression models are presented first. The later sections of this vignette provide extended detail on the additional tools provided by the cpr package.
# Needed packages to run the examples in this vignettelibrary(cpr)library(splines)
Coming Soon A full description of the CPR method, including the background on B-splines and assessing knot influence will be provided soon.
The quick overview: for a uni-variable B-spline regression model \[\boldsymbol{y}= f(\boldsymbol{x}) + \boldsymbol{Z}_{f} \boldsymbol{\beta} + \boldsymbol{Z}_{r} \boldsymbol{b} + \epsilon\] where \(\boldsymbol{y}\) is a function of one variable \(x\) with other (optional) fixed effects \(\boldsymbol{Z}_{f} \beta\) and (optional) random effects \(\boldsymbol{Z}_{r} \boldsymbol{b},\) we are interested in modeling the function \(f\left( \boldsymbol{x} \right)\) with B-splines.
The B-spline function is \[ f \left(\boldsymbol{x}\right) \approx \boldsymbol{B}_{k, \boldsymbol{\xi}} \left( \boldsymbol{x} \right) \boldsymbol{\theta}\] were \(\boldsymbol{B}\) is the basis matrix defined by the polynomial order \(k\) and knot sequence \(\boldsymbol{\xi},\) and the regression coefficients \(\boldsymbol{\theta}.\)
The Control Polygon Reduction (CPR) approach to finding parsimonious regression models with a good quality of fit is to start with a high cardinal knot sequence, assess the relative influence of each knot on the spline function, remove the least influential knot, refit the regression model, reassess the relative influence of each knot, and so forth until all knots knots have been removed. The selection of a regression model is made by looking at the regression models of sequentially larger knot sequences until it is determined that the use of an additional knot does not provide any meaningful improvements to the regression model.
There are only four steps that the end user needs to take to use the CPR algorithm within the cpr package.
cp.
cpr.
Here is an example.
# Construct the initial control polygoninitial_cp <- cp(log10(pdg) ~ bsplines(day, df = 54), data = spdg)# Run CPRcpr_run <- cpr(initial_cp)
There are two types of diagnostic plots, 1) sequential control polygons, and 2) root mean squared error (RMSE) by model index.
In the plot below, there is no major differences in the shape of the control polygon between model index 4, 5, and 6. As such, we would conclude that model index 4 is sufficient for modeling the data.
# sequential control polygonsplot(cpr_run, color = TRUE)
Further, in the plot below, we see that there is very little improvement in the RMSE from model index 4 to 5, and beyond. As with the plot above, we conclude that model index 4 is sufficient for modeling the data.
# RMSE by model indexplot(cpr_run, type = 'rmse', to = 10)
Extract the model. To save memory when running CPR, the default is to omit the regression model objects from the control polygons with in the
cpr_run object. To get the regression model back, you can extract the knot locations and build the model yourself, or update the
selected_cp object to retain the fit.
selected_cp <- cpr_run[[4]]selected_cp$iknots## [1] -0.04429203 -0.02992955 0.06601307
selected_cp <- update(selected_cp, keep_fit = TRUE)selected_fit <- selected_cp$fitcoef_matrix <- summary(selected_fit)$coefdimnames(coef_matrix)[[1]] <- paste("Vertex", 1:7)coef_matrix## Estimate Std. Error t value Pr(>|t|)## Vertex 1 -0.04067426 0.009447938 -4.305094 1.675568e-05## Vertex 2 -0.52353915 0.020364344 -25.708619 7.278010e-144## Vertex 3 -0.52379368 0.018812577 -27.842740 5.154301e-168## Vertex 4 -0.20824755 0.008756483 -23.782101 1.252151e-123## Vertex 5 0.83530262 0.019968306 41.831421 0.000000e+00## Vertex 6 0.90521730 0.019961035 45.349217 0.000000e+00## Vertex 7 0.05860973 0.009570407 6.124058 9.259636e-10
If you wanted to run this analysis with a better modeling approach, e.g., using mixed effect models handle the multiple observations within a subject, you can specify the regression approach via the
method argument in the
cp call.
library(lme4)initial_lmer_cp <- cp(log10(pdg) ~ bsplines(day, df = 54) + (1 | id), data = spdg, method = lmer)
Coming Soon Control Net reduction is the natural extension of CPR from uni-variable functions to multi-variable functions. Details on this work will be added to this vignette soon.
The CPR and CNR algorithms rely on B-splines, control polygons, tensor products of B-splines, and control nets. The following sections provide additional detail on the tools within the cpr package.
Base R includes the splines package and the
splines::bs call for building B-splines. The cpr package provides the
cpr::bsplines call as there are certain default behaviors and additional required meta-data storage requirements needed for the CPR method that the
splines::bs call does not provide. In this section we compare
splines::bs and
cpr::bsplines so that end users can translate between the two functions.
splines::bs
cpr::bsplines
Arguments
x
x
df
df
knots
iknots
degree = 3
order = 4L
Boundary.knots = range(x)
bknots = range(x)
intercept = FALSE
– Attributes
dim
dim
degree
order
knots
iknots
Boundary.knots
bknots
intercept
– –
xi
–
xi_star
class
class
A major difference between the two functions is related to the
intercept argument of
bs. By default,
bs will omit the first column of the basis whereas
bsplines will return the whole basis. The omission of the first column of the basis generated by
bs allows for additive
bs calls to be used on the right-hand-side of a regression formula and generate a full rank design matrix. If additive
bsplines calls, or additive
bs with
intercept = TRUE, are on the right-hand-side of the regression equation the resulting design matrix will be rank deficient. This is a result of the B-splines being a partition of unity. As the CPR algorithm is based on having the whole basis, the
bsplines function is provided to make it easy to work with the whole basis without having to remember to use non-default settings in
bs.
Both functions use
x, a numeric vector, as the first argument. The degrees of freedom,
df, argument is slightly different between the two functions; the return object from
bs depends on the values of
df and
intercept whereas
bsplines always returns the full basis and thus
df will be equal the number of columns of the returned basis.
bs uses the polynomial
degree whereas
bsplines uses the polynomial
order (order = degree + 1) to define the splines. The default for both
bs and
bsplines is to generate cubic B-splines.
Specifying the internal knots and boundary knots are the same between the two functions save the name of the arguments. For both
bs and
bsplines only the degrees of freedom or the internal knots need to be specified. If the end user specifies both, the specified knots take precedence. If only the degrees of freedom are specified then
bs will generate internal knots via a call equivalent to
stats::quantile(x, probs = seq(0, 1, length = length(knots) + 2L)[-c(1, length(knots) + 2L)].
The default behavior for
bsplines is nearly the same, only that a call to `trimmed_quantile`` is made.
bmat <- bsplines(x = seq(0, 6, length = 500), iknots = c(1.0, 1.5, 2.3, 4.0, 4.5))plot(bmat)
cp
build_tensor and of B-splines
btensor
cn
print(sessionInfo(), local = FALSE)## R version 3.3.2 (2016-10-31)## Platform: x86_64-pc-linux-gnu (64-bit)## Running under: Debian GNU/Linux 8 (jessie)## ## attached base packages:## [1] splines stats graphics grDevices utils datasets methods ## [8] base ## ## other attached packages:## [1] cpr_0.2.3 knitr_1.15.1## ## loaded via a namespace (and not attached):## [1] Rcpp_0.12.9 magrittr_1.5 MASS_7.3-45 munsell_0.4.3 ## [5] colorspace_1.3-2 lattice_0.20-34 R6_2.2.0 minqa_1.2.4 ## [9] plyr_1.8.4 stringr_1.2.0 dplyr_0.5.0 tools_3.3.2 ## [13] grid_3.3.2 gtable_0.2.0 nlme_3.1-128 DBI_0.5-1 ## [17] htmltools_0.3.5 lazyeval_0.2.0 yaml_2.1.14 lme4_1.1-12 ## [21] rprojroot_1.2 digest_0.6.12 assertthat_0.1 tibble_1.2 ## [25] Matrix_1.2-7.1 tidyr_0.6.1 ggplot2_2.2.1 nloptr_1.0.4 ## [29] evaluate_0.10 rmarkdown_1.3 labeling_0.3 stringi_1.1.2 ## [33] scales_0.4.1 backports_1.0.5 |
Let T be a compact operator on an infinite dimensional hilbert space H. I am proving the theorem which says that $Tx=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,x_{n}\rangle y_{n}$ where ($x_{n}$) is an orthonormal sequence consisting of the eigenvectors of $|T|=(T^*T)^{0.5}$, (${\lambda}_{n}$) is the corresponding sequence of eigenvalues, and ($y_{n}$) is an orthonormal sequence of, each $y_{n}$ is an eigenvector of $TT^*$.
I have two questions. Is ($y_{n}$) the full sequence of eigenvectors of $TT^*$? I would have thought the answer was yes because the square root of an operator has the same eigenvectors as the operator and an operator has the same 'number' of eigenvectors as it's adjoint.
Secondly, what makes this decomposition so unsatisfying compared to the decomposition of compact normal operators?
This question wasn't answered on stack exchange so I thought it might be more appropriate here. Thanks |
Background
To use permanence, Lotka-Volterra dynamics have to be assumed, because it is only in this case that a sufficient condition for permanence is known:
\[
\dot{x}_{i} = x_i \cdot f_i(x) = x_i \cdot (r_i + (A \cdot x)_i) \quad \forall i = 1, \ldots, n. \]
Such a dynamical system is permanent if two conditions hold (Hofbauer and Sigmund 1988,
The theory of evolution and dynamical systems p. 98). (1) It is dissipative; this is true for Lotka-Volterra systems where all the basal species are self-limiting and heterotrophs cannot survive without their prey. (2) That: \[ P(x) = \prod_i x_{i}^{p_i} \] for some \( p_i>0 \) is an average Lyapunov function.
In dissipative Lotka-Volterra systems, the test for permanence reduces to a linear programming problem (Jansen 1987,
J. Math. Biol.; Law & Morton 1996, Ecology).
An alternative way of describing the system, pursued in an unpublished paper by Richard Law and R. Daniel Morton, is in terms of the convex hull \( C \) and the set \( D \) where the densities of all species are non-increasing. The convex hull of the boundary equilibria is the smallest convex set of which every boundary equilibrium is a member. If this set is disjoint from \( D \), then the system is permanent. So to test for permanence, one can find a hyperplane passing through the interior equilibrium that has a greater value than that of the hyperplane passing through every boundary equilibrium. One can also measure the ‘strength’ of permanence by \( d \), which is the shortest distance from \( C \) to \( D \), which is always the shortest distance from \( C \) to the interior equilibrium \( \hat{x} \) (Figure 1).
Example
You can download the complete example for Octave from my website: mortongen.m.
Consider the four-species system shown in Figure 2 with the parameter values below, adapted from an unpublished paper by Richard Law and R. Daniel Morton, with:
\[ A = \begin{bmatrix} -0.000610 & -0.000207 & -0.052365 & -0.001789 \\ -0.000400 & -0.000139 & -0.001260 & -0.064501 \\ 0.005740 & 0.000033 & 0 & -0.096816 \\ 0.000010 & 0.000151 & 0.000399 & 0 \\ \end{bmatrix} \] and \[ r^{T} = \begin{bmatrix} 0.220737 & 0.164163 & -0.087756 & -0.096189 \end{bmatrix}. \]
It has an interior equilibrium point:
\[ \hat{x} = \begin{bmatrix} 28.230 & 631.560 & 1.356 & 0.983 \end{bmatrix} \]
The first step is to find all subsystems \( b = (0,0,0),(1,0,0), \ldots, (1,2,0),(1,3,0),(1,4,0),(2,3,0), \ldots, (2,3,4) \), and then identify which of these have a positive boundary equilibrium \( \hat{x}^{b} \).
If we had all subsystems in
subStore, then equilibria would be evaluated by:
for subCnt = 2:noSubs; % Acquire list of species absent and present present = subStore(subCnt,:); present(find(present == 0)) = []; absent = 1:nospp; absent(present) = []; % Evaluate boundary equilibrium ANow = A; ANow(absent,:) = []; ANow(:,absent) = []; rNow = r; rNow(absent) = []; if rank(ANow) == size(ANow,1); xSub = ANow-rNow; % etc.
For our example system, the positive boundary equilibria are:
\[ \hat{x}^{(1)} = \begin{bmatrix} 362 & 0 & 0 & 0 \end{bmatrix}\\ \hat{x}^{(2)} = \begin{bmatrix} 0 & 1181 & 0 & 0 \end{bmatrix}\\ \hat{x}^{(1,3)} = \begin{bmatrix} 15.3 & 0 & 4 & 0 \end{bmatrix}\\ \hat{x}^{(2,4)} = \begin{bmatrix} 0 & 637 & 0 & 1.2 \end{bmatrix}\\ \hat{x}^{(1,2,4)} = \begin{bmatrix} 148 & 627 & 0 & 0.3 \end{bmatrix} \]
The transversal eigenvalues at the boundary equilibria
\[ f(\hat{x}^{}) = r + A\hat{x}^{b} \] are found with code like: rowofG = A*(xStore(subCnt,:)')+r;
and are:
\[ f(\hat{x}^{()}) = \begin{bmatrix} 0.22074 & 0.16416 & -0.08776 & -0.09619\ \end{bmatrix}\\ f(\hat{x}^{(1)}) = \begin{bmatrix} 0.00000 & 0.01942 & 1.98934 & -0.09257\ \end{bmatrix}\\ f(\hat{x}^{(2)}) = \begin{bmatrix} -0.02374 & 0.00000 & -0.04878 & 0.08215\ \end{bmatrix}\\ f(\hat{x}^{(1,3)}) = \begin{bmatrix} 0.00000 & 0.15296 & 0.00000 & -0.09443\ \end{bmatrix}\\ f(\hat{x}^{(2,4)}) = \begin{bmatrix} 0.08678 & 0.00000 & -0.18024 & 0.00000\ \end{bmatrix}\\ f(\hat{x}^{(1,2,4)}) = \begin{bmatrix} 0.00000 & 0.00000 & 0.75719 & 0.00000\ \end{bmatrix}\\ \]
The linear programming problem is then to minimise \( z \) subject to:
\[ \sum p_i \cdot f(\hat{x}^{b_1}) + z \geq 0 \\ \sum p_i \cdot f(\hat{x}^{b_2}) + z \geq 0 \\ \vdots \\ p_i \geq 0 \ \] where each positive equilibrium point on the boundary \( \hat{x}^{b_1},\hat{x}^{b_2}, \ldots \) gives rise to a constraint, and \( p_i \) is part of the average Lyapunov function. So if one can find a solution with \( z < 0 \), then one can be certain that \( P(x) \) is an average Lyapunov function.
An introduction on how to code linear programming problems in Octave can be found in a previous post. The vector \( p \) for the Lyapunov function, as found using Jansen’s linear program method, is
\[ p = \begin{bmatrix} 0.835219 & 1 & 0.077317 & 0.999918 \end{bmatrix}. \]
Hofbauer & Sigmund (1988, p. 176-177) states that the convex hull \( C \) can be separated from \( D \) by a hyperplane with a “separating functional” \( pA \), where \( p \) is the same \( p \) used in the Lyapunov function. This separating functional is the vector normal to the hyperplane.
\[ \mathbf{n} = pA \\ = \begin{bmatrix} -4.5568\times10^{-4} & -1.5835\times10^{-4} & -4.4597\times10^{-2} & -7.3481\times10^{-2} \end{bmatrix} \\ = \begin{bmatrix} 0.0062014 & 0.0021550 & 0.6069248 & 1 \end{bmatrix}. \]
The equation for the hyperplane is
\[ 0 = n_1x_1 + n_2x_2 + n_3x_3 + x_4 + b. \]
If one wants to go straight to finding \( d \), then substitute in the boundary steady state for each positive subsystem, reducing the value of \( b \) until the hyperplane is above or equal to every boundary steady state.
So for subsystem \( (1) \),
\[ \hat{x}^{(1)} = \begin{bmatrix} 361.86393 & 0 & 0 & 0 \end{bmatrix}, \] which gives \( b = -2.2441 \)
Then subsystem \( (2) \) is next, with
\[ \hat{x}^{(2)} = \begin{bmatrix} 0 & 1181 & 0 & 0 \end{bmatrix}. \] Substituting into the hyperplane equation gives \[ 0.0021550 \times 1181 -2.2441 = 0.30105 \] which implies that this boundary steady state is above the hyperplane, therefore it is used to find a revised value of \( b \), \( b =-2.5451 \)
This process is continued, checking that all other boundary steady
states are below the plane, and revising \( b \) if they are not, which gives the final value of \( b =-2.5451 \).
Now the distance from the interior equilibrium to the plane can be
found. The interior equilibrium is \[ \hat{x} = \begin{bmatrix} 28.23008 & 631.55967 & 1.35636 & 0.98255 \end{bmatrix} \] and so the perpendicular distance from the hyperplane to \( \hat{x} \) is \[ d = (\mathbf{n}’ \hat{x} + b)/|\mathbf{n}| \\ d = \frac{(0.0062014 \times 28.23008 + 0.0021550 \times 631.55967 + 0.6069248 \times 1.35636 + 0.98255 -2.5451)}{1.1698} \\ d = 0.68108, \]
Alternatively \( d \) can be calculated for each boundary steady state, and the smallest taken as the final value. If
xStore stores the values of all the boundary equilibria, then:
d = (n'*xs-xStore(posSubs,:)*n)/norm(n); disp('d = ') disp(d) mind = min(d); disp('smallest distance = ') disp(mind)
—
You can download the complete example for Octave from my website: mortongen.m. |
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Change to browse by: References & Citations Bookmark(what is this?) Physics > Plasma Physics Title: Magnetic reconnection with null and X-points
(Submitted on 18 Jul 2019 (v1), last revised 8 Oct 2019 (this version, v3))
Abstract: Null and X-points are not themselves directly important to magnetic reconnection because distinguishable field lines do not approach them closely. Even in a collision-free plasma, magnetic field lines that approach each other on a scale $c/\omega_{pe}$ become indistinguishable during an evolution. What is important is the different regions of space that can be explored by magnetic field lines that pass in the vicinity of null and X-points. Traditional reconnection theories made the assumption that the reconnected magnetic flux must be dissipated or diffused by an electric field. This assumption is false in three dimensional systems because an ideal evolution can cause magnetic field lines that cover a large volume to approach each other within the indistinguishability scale $c/\omega_{pe}$. When the electron collision time $\tau_{ei}$ is short compared to the evolution time of the magnetic field $\tau_{ev}$, the importance of $c/\omega_{pe}$ is replaced by the resistive time scale $\tau_\eta=(\eta/\mu_0)L^2$ with $L$ the system scale. The magnetic Reynolds number is $R_m\equiv\tau_\eta/\tau_{ev}$ is enormous in many reconnection problems of interest. Magnetic flux diffusion implies the current density required for reconnection to compete with evolution scales as $R_m$ while flux mixing implies the required current density to compete scales as $\ln R_m$. Submission historyFrom: Allen Boozer [view email] [v1]Thu, 18 Jul 2019 14:09:22 GMT (412kb,D) [v2]Tue, 23 Jul 2019 23:49:50 GMT (412kb,D) [v3]Tue, 8 Oct 2019 12:03:10 GMT (892kb,D) |
I'm trying to implement a normalized cross-correlation algorithm but I don't get what in fact is this measure. What confuses is the wikipedia definition:
$\frac{1}{n} \sum \frac{(f(x,y)- \overline{f})(t(x,y)- \overline{t}) }{\sigma_{f}\sigma_{t}}$
Which result is an scalar (AFAIK)
But then adds other way to measure it:
$\left \langle \frac{F}{\left \| F \right \|},\frac{T}{\left \| T \right \|} \right \rangle $
Where $F$ and $T$ are normalized vectors and $\left \langle . , . \right \rangle$ is the inner product. But the output will be a vector, isn't? Isn't supposed to give me a scalar as well? Am I getting something wrong?
The idea is to implement this formula and use it with matrices with same dimensions.
Thank you |
Suppose that $f$ is a continuous real-valued function on the interval $[0,1]$. Show that $$\int_0^1x^2f(x)\,dx=\frac{1}{3}f(\xi)$$ for some $\xi\in[0,1]$.
I guess that MVT is involved, but I don't know how to apply it.
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$f $ is continuous at $[0,1]$ $$ \implies $$
$$\exists m,M \in \mathbb R \;: \;\;f ([0,1])=[m,M] $$ $$\implies $$ $$\forall x\in [0,1]\;\;mx^2\leq x^2f (x)\leq Mx^2$$ $$\implies $$
$$\frac {m}{3}\leq \int_0^1x^2f (x)dx\leq \frac {M}{3} $$ (because $\int_0^1x^2dx=\frac {1}{3} $.)
$$\implies $$
$$m\leq 3\int_0^1x^2f (x)dx\leq M $$
$$\implies $$
$$\exists \xi \in [0,1]\;\;:\; 3\int_0^1x^2f (x)dx=f (\xi) $$ you can finish.
The second mean value theorem for integrals states that
$$\int_a^b f(x)g(x)dx = f(c)\int_a^b g(x)dx $$ For some $c \in (a,b)$
Apply this theorem with $f(x) = f(x)$ and $g(x) = x^2$ |
In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable.
Problem 91
Show that the matrix $A=\begin{bmatrix}1 & \alpha\\0& 1\end{bmatrix}$, where $\alpha$ is an element of a field $F$ of characteristic $p>0$ satisfies $A^p=I$ and the matrix is not diagonalizable over $F$ if $\alpha \neq 0$. Add to solve later
Thus, over a field of characteristic $p>0$ the condition $A^p=1$ dose not always imply that $A$ is diagonalizable.
Proof.
By induction, it is straightforward to see that\[A^m=\begin{bmatrix}1 & m\alpha\\0& 1\end{bmatrix}\]for any positive integer $m$.Thus\[A^p=\begin{bmatrix}1 & p\alpha\\0& 1\end{bmatrix}=I\]since $p\alpha=0$ in the field $F$.
Since the eigenvalues of $A$ is $1$, if $A$ is diagonalizable, then there exists an invertible matrix $P$ such that $P^{-1}AP=I$, or $AP=P$.Let $P=\begin{bmatrix}a & b\\c& d\end{bmatrix}$. Then we have\[AP=\begin{bmatrix}a+\alpha c & b+\alpha d\\c& d\end{bmatrix}\]and this is equal to $P$, hence\begin{align*}a+\alpha c=a \text{ and } b+\alpha d=b \\\end{align*}Thus\begin{align*}\alpha c=0 \text{ and } \alpha d=0 \\\end{align*}If $\alpha \neq 0$, then $c=d=0$ but this implies that the matrix $P$ is non-invertible, a contradiction.Therefore we must have $\alpha=0$.
Each Element in a Finite Field is the Sum of Two SquaresLet $F$ be a finite field.Prove that each element in the field $F$ is the sum of two squares in $F$.Proof.Let $x$ be an element in $F$. We want to show that there exists $a, b\in F$ such that\[x=a^2+b^2.\]Since $F$ is a finite field, the characteristic $p$ of the field […]
How to Diagonalize a Matrix. Step by Step Explanation.In this post, we explain how to diagonalize a matrix if it is diagonalizable.As an example, we solve the following problem.Diagonalize the matrix\[A=\begin{bmatrix}4 & -3 & -3 \\3 &-2 &-3 \\-1 & 1 & 2\end{bmatrix}\]by finding a nonsingular […]
Quiz 13 (Part 1) Diagonalize a MatrixLet\[A=\begin{bmatrix}2 & -1 & -1 \\-1 &2 &-1 \\-1 & -1 & 2\end{bmatrix}.\]Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that […]
A Matrix Similar to a Diagonalizable Matrix is Also DiagonalizableLet $A, B$ be matrices. Show that if $A$ is diagonalizable and if $B$ is similar to $A$, then $B$ is diagonalizable.Definitions/Hint.Recall the relevant definitions.Two matrices $A$ and $B$ are similar if there exists a nonsingular (invertible) matrix $S$ such […] |
Attwood, Rhianne Elizabeth, Goodwin, S. P. and Whitworth, Anthony Peter 2007. Adaptive smoothing lengths in SPH. Astronomy & Astrophysics 464 (2) , pp. 447-450. 10.1051/0004-6361:20066606
PDF - Published Version
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Abstract
Context.There is a need to improve the fidelity of SPH simulations of self-gravitating gas dynamics. Aims.We remind users of SPH that, if smoothing lengths are adjusted so as to keep the number of neighbours, ${\cal N}$, in the range ${\cal N}_{{\rm NEIB}}\pm\Delta{\cal N}_{{\rm NEIB}}$, the tolerance, $\Delta{\cal N}_{{\rm NEIB}}$, should be set to zero, as first noted by Nelson & Papaloizou. We point out that this is a very straightforward and computationally inexpensive constraint to implement. Methods.We demonstrate this by simulating acoustic oscillations of a self-gravitating isentropic monatomic gas-sphere (cf. Lucy), using ${\cal N}_{{\rm TOT}}\sim6000$ particles and ${\cal N}_{{\rm NEIB}}=50$. Results.We show that there is a marked reduction in the rates of numerical dissipation and diffusion as $\Delta{\cal N}_{{\rm NEIB}}$ is reduced from 10 to zero. Moreover this reduction incurs a very small computational overhead. Conclusions.We propose that this should become a standard test for codes used in simulating star formation. It is a highly relevant test, because pressure waves generated by the switch from approximate isothermality to approximate adiabaticity play a critical role in the fragmentation of collapsing prestellar cores. Since many SPH simulations in the literature use ${\cal N}_{{\rm NEIB}}=50$ and $\Delta{\cal N}_{{\rm NEIB}}\geq10$, their results must be viewed with caution.
Item Type: Article Date Type: Publication Status: Published Schools: Physics and Astronomy Subjects: Q Science > QB Astronomy Uncontrolled Keywords: hydrodynamics -- methods ; numerical -- stars ; oscillations Additional Information: Pdf uploaded in accordance with publisher's policy at http://www.sherpa.ac.uk/romeo/issn/0004-6361/ (accessed 16/04/2014) Publisher: EDP Sciences ISSN: 0004-6361 Date of First Compliant Deposit: 30 March 2016 Last Modified: 04 Jun 2017 04:55 URI: http://orca.cf.ac.uk/id/eprint/46418 Citation Data
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Think about equalities: it is certainly true that if $x=1$, then $x^n=1$; but it is
not true that if $x\neq 1$ then $x^n\neq 1$ (e.g., $x=-1$, $n$ even).
The same is true with congruences:
congruences modulo $n$ may be added and multiplied, just like equalities: if $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$, then $a+c\equiv b+d\pmod{n}$ (adding the two congruences) and $ac\equiv bd\pmod{n}$ (multiplying the two congruences). Same as with equalities. This immediately gives you that if $a\equiv 1 \pmod{n}$, then multiplying the congruence by itself $m$ times you will get $a^m \equiv 1^m=1\pmod{n}$, and in particular $a^n\equiv 1\pmod{n}$.
However, we cannot just add and multiple
non-equalities: that is, if $a\neq b$ and $c\neq d$, then it does not follow that $a+c\neq b+d$ or that $ac\neq bd$ (I'll leave it to you to find easy counterexamples). Note well: these are not inequalities of the form $a\lt b$, but merely of the form "not equal to". Likewise with congruences. Just because $a\not\equiv 1 \pmod{n}$, it does not follow that $a^2\not\equiv 1^2\pmod{n}$, or $a^n\not\equiv 1\pmod{n}$ in general.
It
does follow for $n$ prime, but that's not because of the properties of congruences, but rather because of Fermat's Little Theorem: because Fermat's Little Theorem tells you that $a^p\equiv a\pmod{p}$ when $p$ is a prime, so you can conclude that if $a\not\equiv 1\pmod{p}$, then $a^p \equiv a\not\equiv 1\pmod{p}$. But this is a particular property of primes, not of congruences.
Added. Going from the comments, some situations where it is true and some where it is false:
If $\gcd(a,n)\gt 1$, then of course $a\not\equiv 1\pmod{n}$, and likewise $a^n\not\equiv 1\pmod{n}$; this is the "trivial case", one might say.
If $\gcd(a,n)=1$, then of course we have $a^{\varphi(n)}\equiv 1 \pmod{n}$ by Euler's Theorem; thus, $a^n\equiv a^{\varphi(n)}a^{n-\varphi(n)}\equiv a^{n-\varphi(n)}\pmod{n}$. If $\gcd(\varphi(n),n-\varphi(n))=\gcd(\varphi(n),n)=1$, then from $a^{n-\varphi(n)}\equiv 1\pmod{n}$ you can conclude that $a\equiv 1\pmod{n}$, since no element, other than $1$, of the multiplicative group modulo $n$ has order dividing $n-\varphi(n)$ (since they all have order dividing $\varphi(n)$).
However, if $\gcd(\varphi(n),)\gt 1$, then we can always pick a prime $p$ that divides $\gcd(\varphi(n),n)$, and find an element $a$ of the multiplicative group modulo $n$ that has order exactly $p$. That will mean that $a\not\equiv 1\pmod{n}$, but $a^p\equiv 1\pmod{n}$. Then we will have $a^{n} = a^{\varphi(n)}a^{n-\varphi(n)} = (a^{p})^k\equiv 1\pmod{n}$, where $kp = n-\varphi(n)$, so in this case there will always be a counterexample.
In short, the implication $$a\not\equiv 1\pmod{n}\Longrightarrow a^n\not\equiv 1\pmod{n}$$ holds if and only if $\gcd(n,a)\gt 1$ or $\gcd(a,n)=\gcd(\varphi(n),n)=1$, where $\varphi(n)$ is Euler's phi function. |
I'm trying to understand how the parameters are estimated in ARIMA modeling/Box Jenkins (BJ). Unfortunately none of the books that I have encountered describes the estimation procedure such as Log-Likelihood estimation procedure in detail. I found the website/teaching material that was very helpful. Following is the equation from the source referenced above. $$ LL(\theta)=-\frac{n}{2}\log(2\pi) - \frac{n}{2}\log(\sigma^2) - \sum\limits_{t=1}^n\frac{e_t^2}{2\sigma^2} $$
I want to learn the ARIMA/BJ estimation by doing it myself. So I used $R$ to write a code for estimating ARMA by hand. Below is what I did in $R$,
I simulated ARMA (1,1) Wrote the above equation as a function Used the simulated data and the optim function to estimate AR and MA parameters. I also ran the ARIMA in the stats package and compared the ARMAparameters from what I did by hand. Below is the comparison:
**Below are my questions:
Why is there a slight difference between the estimated and calculated variables ? Does ARIMA function in R backcasts or does the estimation procedure differently than what is outlined below in my code? I have assigned e1 or error at observation 1 as 0, is this correct ? Also is there a way to estimate confidence bounds of forecasts using the hessian of the optimization ?
Thanks so much for your help as always.
Below is the code:
## Load Packageslibrary(stats)library(forecast)set.seed(456)## Simulate Arimay <- arima.sim(n = 250, list(ar = 0.3, ma = 0.7), mean = 5)plot(y)## Optimize Log-Likelihood for ARIMAn = length(y) ## Count the number of observationse = rep(1, n) ## Initialize elogl <- function(mx){ g <- numeric mx <- matrix(mx, ncol = 4) mu <- mx[,1] ## Constant Term sigma <- mx[,2] rho <- mx[,3] ## AR coeff theta <- mx[,4] ## MA coeff e[1] = 0 ## Since e1 = 0 for (t in (2 : n)){ e[t] = y[t] - mu - rho*y[t-1] - theta*e[t-1] } ## Maximize Log-Likelihood Function g1 <- (-((n)/2)*log(2*pi) - ((n)/2)*log(sigma^2+0.000000001) - (1/2)*(1/(sigma^2+0.000000001))*e%*%e) ##note: multiplying Log-Likelihood by "-1" in order to maximize in the optimization ## This is done becuase Optim function in R can only minimize, "X"ing by -1 we can maximize ## also "+"ing by 0.000000001 sigma^2 to avoid divisible by 0 g <- -1 * g1 return(g)}## Optimize Log-Likelihoodarimopt <- optim(par=c(10,0.6,0.3,0.5), fn=logl, gr = NULL, method = c("L-BFGS-B"),control = list(), hessian = T)arimopt############# Output Results###############ar1_calculated = arimopt$par[3]ma1_calculated = arimopt$par[4]sigmasq_calculated = (arimopt$par[2])^2logl_calculated = arimopt$valar1_calculatedma1_calculatedsigmasq_calculatedlogl_calculated############# Estimate Using Arima###############est <- arima(y,order=c(1,0,1))est |
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Another doubt in induction, please help me out.
Show that 2⋅7n+3⋅5n−52\cdot7^{n} + 3\cdot5^{n} -52⋅7n+3⋅5n−5 is divisible by 24.
Note by Swapnil Das 4 years ago
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Step 111: when n=1n=1n=1, we have 2⋅7n+3⋅5n−5=2⋅7+3⋅5−5=242 \cdot 7^{n}+3 \cdot 5^{n}-5=2 \cdot 7+3 \cdot 5 -5=242⋅7n+3⋅5n−5=2⋅7+3⋅5−5=24 which is divisible by 242424.
Step 222: Assume this to be true for n=mn=mn=m. Then 2⋅7m+3⋅5m−52 \cdot 7^{m}+3 \cdot 5^{m}-52⋅7m+3⋅5m−5 is divisible by 242424.
Step 333: Now, 2⋅7m+1+3⋅5m+1−52 \cdot 7^{m+1}+3 \cdot 5^{m+1}-52⋅7m+1+3⋅5m+1−5
=2⋅7m×7+3⋅5m×5−5=2 \cdot 7^{m} \times 7+ 3 \cdot 5^{m} \times 5-5=2⋅7m×7+3⋅5m×5−5
=14⋅7m+15⋅5m−5=14 \cdot 7^{m}+15 \cdot 5^{m}-5=14⋅7m+15⋅5m−5
=2⋅7m+3⋅5m−5+12⋅7m+12⋅5m=2 \cdot 7^{m}+3 \cdot 5^{m}-5+12 \cdot 7^{m}+12 \cdot 5^{m}=2⋅7m+3⋅5m−5+12⋅7m+12⋅5m
Since 2⋅7m+3⋅5m−52 \cdot 7^{m}+3 \cdot 5^{m}-52⋅7m+3⋅5m−5 is divisible by 242424, we have only to prove that 12⋅7m+12⋅5m12 \cdot 7^{m}+12 \cdot 5^{m}12⋅7m+12⋅5m is divisible by 242424.
12⋅7m+12⋅5m=12(7m+5m)12 \cdot 7^{m}+12 \cdot 5^{m}=12(7^{m}+5^{m})12⋅7m+12⋅5m=12(7m+5m) which is divisible by 242424 since 7m7^{m}7m and 5m5^{m}5m are both odd numbers and their sum is an even number.
Hence 2⋅7n+3⋅5n−52 \cdot 7^{n}+3 \cdot 5^{n}-52⋅7n+3⋅5n−5 is divisible by 242424 for all integers n>0n>0n>0.
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Excellent, thank you genius!
Your Welcome :)
Hint: Show that it is true for n=1. Then show that if the statement is true for any arbitrary positive integer m, then it is true for m+1.
That's an obvious hint lol xD
Can you show the m+1 step? This is where I'm getting stuck at.
Only tool (with me ):MATHEMATICAL INDUCTION
I'm surprised that you don't know Induction. You should read the wiki and be familiar with it. This is a standard problem in divisibility by induction.
Lol, I assumed powers of 555 to be even by mistake, and thus couldn't get the result!
This is true only for n equals 1. It is not true for numbers bigger than 1.
@ Mohamed Sultan - You are wrong. Go back and try it for n = 2, 3, and 4, for instance.
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It's been a while since I've done any comprehensive work with structural dynamics, so let me know if I get anything wrong here.
I have a horizontal beam that is clamped on one end and free on the other. It has the following parameters:
length $L$ elastic modulus $E$ moment of inertia $I$ mass $M$
There are no loads on this beam except for its own gravity, which can be modeled by a uniformly distributed load:
$$\psi = \frac{Mg}{L}$$
Using standard beam theory, we can calculate the deflection at points $x$ along the beam, as well as the maximum deflection $\delta_{max}$
$$\delta = \frac{\psi x^2}{24EI}(x^2 + 6L^2 - 4Lx)$$ $$\delta_{max} = \frac{\psi L^4}{8EI}$$
I have a physical constraint that $\delta_{max}$ can be no more than $\epsilon$, so I propose the following:
Spin the beam about the wall it's attached to at angular velocity $\omega$, which creates an axial centrifugal force. This should lower the value of $\delta_{max}$, and for all $\epsilon > 0$, I should be able to find $\omega$ meeting this constraint.
What I'm looking to derive is some function $\delta_{max}(\omega; \psi, L, E, I)$
My question is how can I incorporate the axial force? Each slice of beam $dm$ has centrifugal force equal to $\omega^2 x dm$ where $x$ is the distance from the clamp. I don't immediately see where to proceed from here.
Any help is super appreciated! ^^ |
Consider the infinite product
$$\prod_{n=1}^{\infty} (1-x^n).$$
Prove that the above infinite product is absolutely convergent for $0 \leq x < 1.$
I have considered the sequence of partial products $\{F_m(x) \}$ where $$F_m (x) = \prod_{n=1}^{m} (1-x^n).$$ Then I have observed that $F_{m+1} (x) = (1 - x^{m+1}) F_m (x).$ If $0 \leq x <1$ then $F_{m+1} (x) \leq F_m (x).$ So the sequence of partial products is monotone decreasing and bounded below by $0.$ So for every $0 \leq x < 1$ the sequence $\{F_m(x) \}$ is monotone decreasing and bounded below by $0$ and therefore it is convergent. Since for $0 \leq x < 1$ the sequence $\{F_m (x) \}$ is a sequence of positive terms (as each term is the product of finitely many positive terms) so we can conclude that the sequence $\{F_m (x) \}$ is absolutely convergent for each $0 \leq x < 1.$ Therefore for each $0 \leq x <1$ the infinite product $\prod_{n=1}^{\infty} (1-x^n)$ is absolutely convergent as well.
Is the above argument ok? Please verify it.
Thank you very much for your valuable time. |
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An apsis (Greek ἁψίς, gen. ἁψίδος), plural apsides (/ˈæpsɨdiːz/; Greek: ἁψίδες), is the point of greatest or least distance of a body from one of the foci of its elliptical orbit. In modern celestial mechanics this focus is also the center of attraction, which is usually the center of mass of the system. Historically, in geocentric systems, apsides were measured from the center of the Earth.
The point of closest approach (the point at which two bodies are the closest) is called the periapsis or pericentre, from Greek περί, peri, around, and κέντρον, kentron, centre. The point of farthest excursion is called the apoapsis, apocentre or apapsis (ἀπ(ό), ap(ó), "from") (the last of these terms, although etymologically more correct, is much less used). A straight line drawn through the periapsis and apoapsis is the line of apsides. This is the major axis of the ellipse, the line through the longest part of the ellipse.
The suffix apsis is the most generic term, used to describe all orbits independent of the particular central body. A family of derivative terms are more commonly used to identify the particular central body. Geocentric orbits (around the Earth), the most common, use perigee /ˈpɛrɨdʒiː/ and apogee /ˈæpɵdʒiː/ (from Greek γῆ, gê, "earth"). Heliocentric (solar) orbits use perihelion /ˌpɛrɨˈhiːliən/ (Greek meaning, "near the Sun") and aphelion /əˈfiːliən/ or /æpˈhiːliən/(Greek meaning, "away from the Sun", ἥλιος, hēlios, "sun"). During the Apollo program, the terms pericynthion and apocynthion (referencing Cynthia, an alternative name for the Greek moon goddess Artemis) were used when referring to the Moon.[1]
In orbital mechanics, the apsis technically refers to the distance measured between the centers of mass of the central and orbiting body. However, in the case of spacecraft, the family of terms are commonly used to describe the orbital altitude of the spacecraft from the surface of the central body (assuming a constant, standard reference radius.)
These formulae characterize the periapsis and apoapsis of an orbit:
while, in accordance with Kepler's laws of planetary motion (based on the conservation of angular momentum) and the conservation of energy, these two quantities are constant for a given orbit:
where:
Note that for conversion from heights above the surface to distances between an orbit and its primary, the radius of the central body has to be added, and conversely.
The arithmetic mean of the two limiting distances is the length of the semi-major axis a.The geometric mean of the two distances is the length of the semi-minor axis b.
The geometric mean of the two limiting speeds is \sqrt{-2\epsilon}=\sqrt{\mu/a} which is the speed of a body in a circular orbit whose radius is a.
The words "pericenter" and "apocenter" are occasionally seen, although periapsis/apoapsis are preferred in technical usage.
Various related terms are used for other celestial objects. The '-gee', '-helion' and '-astron' and '-galacticon' forms are frequently used in the astronomical literature, while the other listed forms are occasionally used, although '-saturnium' has very rarely been used in the last 50 years. The '-gee' form is commonly (although incorrectly) used as a generic 'closest approach to planet' term instead of specifically applying to the Earth. The term peri/apomelasma (from the Greek root) was used by physicist Geoffrey A. Landis in 1998 before peri/aponigricon (from the Latin) appeared in the scientific literature in 2002.[2]
Because "peri" and "apo" are Greek, it is considered by some purists[4] more correct to use the Greek form for the body, giving forms such as '-zene' for Jupiter (Zeus) and '-krone' for Saturn. The daunting prospect of having to maintain a different suffix for every orbitable body in the Solar System (and beyond) is the main reason that the generic '-apsis' has become almost universal, with the exception, of course, being the Sun and Earth.
For the orbit of the Earth around the Sun, the time of apsis is often expressed in terms of a time relative to seasons, since this determines the contribution of the elliptical orbit to seasonal variations. The variation of the seasons is primarily controlled by the annual cycle of the elevation angle of the Sun, which is a result of the tilt of the axis of the Earth measured from the plane of the ecliptic. The Earth's eccentricity and other orbital elements are not constant, but vary slowly due to the perturbing effects of the planets and other objects in the solar system.
Currently, the Earth reaches perihelion in early January, approximately 14 days after the December Solstice. At perihelion, the Earth's center is about 0.98329 astronomical units (AU) or 147,098,070 kilometers (about 91,402,500 miles) from the Sun's center.
The Earth reaches aphelion currently in early July, approximately 14 days after the June Solstice. The aphelion distance between the Earth's and Sun's centers is currently about 1.01671 AU or 152,097,700 kilometers (94,509,100 mi).
On a very long time scale, the dates of the perihelion and of the aphelion progress through the seasons, and they make one complete cycle in 22,000 to 26,000 years. There is a corresponding movement of the position of the stars as seen from Earth that is called the apsidal precession. (This is closely related to the precession of the axis.)
Astronomers commonly express the timing of perihelion relative to the vernal equinox not in terms of days and hours, but rather as an angle of orbital displacement, the so-called longitude of the periapsis. For the orbit of the Earth, this is called the longitude of perihelion, and in 2000 was about 282.895 degrees. By the year 2010, this had advanced by a small fraction of a degree to about 283.067 degrees.[6]
The dates and times of the perihelions and aphelions for several past and future years are listed in the following table:[7]
The following table shows the distances of the planets and dwarf planets from the Sun at their perihelion and aphelion.[8]
The following chart shows the range of distances of the planets, dwarf planets and Halley's Comet from the Sun.
Template:Distance from Sun using EasyTimeline
The images below show the perihelion (green dot) and aphelion (red dot) points of the inner and outer planets.
The perihelion and aphelion points of the inner planets of the Solar System
The perihelion and aphelion points of the outer planets of the Solar System
Apollo program, Nasa, Earth, Apollo 11, International Space Station
Apollo program, Kennedy Space Center, Nasa, Soviet Union, Apollo 11
Apollo program, Moon, Apollo 11, Nasa, Kennedy Space Center
Apollo program, Moon, Nasa, Apollo 15, Kennedy Space Center
Apollo program, Moon, Nasa, Kennedy Space Center, Apollo 16 |
Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc. |
Integrals Integration as an Inverse Process of Differentiation Integration is the inverse process of differentiation. INDEFINITE INTEGRALS \int x^{n} dx = \frac{x^{n+1}}{n+1} + c \ \ , n \neq -1 \int x \ dx = \frac{x^{2}}{2} + c \int x^{2} dx = \frac{x^{3}}{3} + c \int x^{3} dx = \frac{x^{4}}{4} + c \int \sqrt{x} dx = \frac{2}{3} x^{3/2} + c \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + c \int \frac{1}{x} dx = \log \mid x \mid + c \int \frac{1}{x^{2}} dx = \frac{-1}{x}+ c \int \frac{1}{x^{3}} dx = \frac{-1}{2x^{2}}+ c \int \log x dx = x \log x - x + c
Rules of Integration:
(i) \frac{d}{dx}\left\{\int f(x)dx\right\}=f(x)
(ii) ∫ k · f(x) dx = k ∫ f(x) dx
(iii) ∫{f
1(x) ± f 2(x) ± f 3(x) ± ...... ± f n(x)} dx = ∫ f 1(x) dx ± ∫ f 2(x) dx ± ∫ f 3(x) dx ± ....... ± ∫ f n(x) dx Part1: View the Topic in this video From 16:50 To 52:18 Part2: View the Topic in this video From 00:50 To 45:10
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1. \int x^{n} dx = \frac{x^{n+1}}{n+1}+C, n \neq -1
2. ∫ dx = x + C
3. ∫ cos x dx = sin x + C
4. ∫ sin x dx = −cos x + C
5. ∫ sec
2 x dx = tan x + C
6. ∫ cosec
2 x dx = −cot x + C
7. ∫ sec x tan x dx = sec x + C
8. ∫ cosec x cot x dx = −cosec x + C
9. \int \frac{dx}{\sqrt{1-x^{2}}}= \sin^{-1} x + C
10. \int \frac{dx}{\sqrt{1-x^{2}}}= -\cos^{-1} x + C
11. \int \frac{dx}{1+x^{2}}= \tan^{-1} x + C
12. \int \frac{dx}{1+x^{2}}= -\cot^{-1} x + C
13. \int a^{x} dx = \frac{a^{x}}{\log a} + c
14. \int e^{x} dx = e^{x} + c 15. \int xe^{x} dx = xe^{x} - e^{x} + c
16. \int \frac{1}{\sqrt{1-x^{2}}} dx = \sin^{-1} + c = - \cos^{-1} x + c
17. \int \frac{1}{1+x^{2}} dx = \tan^{-1} x + c = - \cot^{-1} x + c
18. \int \frac{1}{|x| \sqrt{x^{2} - 1}} dx = \sec^{-1} (x) + c = - cosec^{-1} x + c
19.
Algebric properties of integrals→ \int \left(f \pm g \right) \ (x) \ dx = \int f(x) dx \pm \int g (x) \ dx
→ \int k f(x) \ dx = k \int f(x) \ dx
→ \int \left(k f + lg \right) \ (x) \ dx = k \int f(x) dx + l \int g (x) \ dx |
Charles C.Pinter - Set theory
Let $a,b,c$ be any cardinal numbers.
Give a counterexample to the rule: $$a+b = a+c \implies b=c$$
Does there exist a counterexample?
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$$\begin{align}&1.\quad\aleph_0=\aleph_0+0=\aleph_0+1\\&2.\quad 0\neq 1\end{align}$$
Where $\aleph_0$ is the cardinality of countably infinite sets, e.g. the non-negative integers, $\mathbb N$.
Here's one: $|\Bbb N| + |\{1\}| =|\Bbb N| + |\{2, 3\}| $, but $|\{1\}| \ne|\{2,3\}|$.
You will probably learn later that for any infinite cardinal $a$ the equality $$a+a=a\cdot a=a$$ holds. This implies that for any infinite cardinal $a$ you have a counterexample $a+a=a+0$.
More generally, if $b$, $c$ are infinite cardinals then $$b+c=b\cdot c=\max\{b,c\}.$$
The proof of these general result is not that simple, it requires axiom of choice. See e.g. the following questions: About a paper of Zermelo and How to prove that from "Every infinite cardinal satisfies $a^2=a$" we can prove that $b+c=bc$ for any two infinite cardinals $b,c$?
However, showing the above result for some special cases, like $a=\aleph_0$ or $a=2^{\aleph_0}$ is not that difficult and it might be a useful exercise for someone learning basics of set theory and cardinal arithmetic. |
In the Hamiltonian theory of gravity, see for example comment after equation (5) in "Quantum gravity with a positive cosmological constant", Lee Smolin or end of page 13 in "Background Independent Quantum Gravity: A Status Report" Ashtekar, Lewandowski.
the conjugate momenta to the connection is a vector density of weight 1. I don't know why it is a 'vector density of weight 1'.
The definition that I am aware of density is:
given a vector space $V$ of dimension $n$, a $k$-density on $V$ is a function $\mu$ from $\wedge^n V^*\setminus \{0\}$ to the reals such that
$\mu (t\alpha) = |t|^k \mu(\alpha)$, $\alpha \in \wedge^n V^*\setminus \{0\}$
Is this what should I understand as vector density? |
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