text:
\begin{document}
\paper{Statistical correlations in a Coulomb gas with a test
charge}

\author{Henning Schomerus}
\address{Department of Physics, Lancaster University, Lancaster, LA1 4YB, UK}
\begin{abstract}
A recent paper [Jokela {\em et al.}, arxiv:0806.1491 (2008)] contains a
surmise about an expectation value in a Coulomb gas which interacts with an
additional charge $\xi$ that sits at a fixed position. Here I demonstrate the
validity of the surmised expression and extend it to a certain class of
higher cumulants. The calculation is based on the analogy to statistical
averages in the circular unitary ensemble of random-matrix theory and
exploits properties of orthogonal polynomials on the unit circle.
 \end{abstract}
\pacs{02.50.Sk, 05.20.-y, 05.30.Fk, 05.45.Mt}

\section{Purpose and result} In a recent paper Jokela, J{\"a}rvinen
and Keski-Vakkuri studied $n$-point functions in timelike boundary
Liouville theory via the analogy to a Coulomb gas on a unit circle
\cite{jokela}. In this analogy, $N$ unit charges at position $t_i$
interact with additional charges of integer value $\xi_a$,
situated at position $\tau_a$. To illustrate this technique the
authors of \cite{jokela} considered the canonical expectation
value
\begin{equation}
\langle\cdot\rangle\equiv\frac{1}{Z}
\int\prod_{i=1}^N\frac{dt_i}{2\pi}\prod_{i<j}\left|\rme^{\rmi t_i}-\rme^{\rmi t_j}\right|^2
\prod_{i}\left|\rme^{\rmi \tau}-\rme^{\rmi t_i}\right|^{2\xi} (\cdot)
\end{equation}
(where $Z$ is a normalization factor so that $\langle 1 \rangle=1$) and
surmised that
\begin{equation}
\langle {\rm Re}\, a_1\rangle \equiv\left\langle\sum_{i}\cos(\tau-t_{i})\right\rangle=-\frac{\xi N}{N+\xi}.
\label{eq:surmise}
\end{equation}

In this communication I demonstrate the validity of
(\ref{eq:surmise}), and also compute expectation values of the
more general quantities
\begin{equation}
a_n\equiv \sum_{i_1<i_2<\ldots<i_n}\exp\left(\rmi \sum_{k=1}^n(t_{i_k}-\tau)\right).
\label{eq:an}
\end{equation}
As a result, I find
\begin{equation}
\langle a_n\rangle
=(-1)^n\frac{(N-n+1)^{(\xi)}(n+1)^{(\xi-1)}}{(N+1)^{(\xi)}(1)^{(\xi-1)}}
\quad \forall\,\, n=0,1,2,\ldots,N;\,\, \xi \geq 0,
\label{eq:result}
\end{equation}
where $(x)^{(y)}=\Gamma(x+y)/\Gamma(x)$ is the generalized rising
factorial (Pochhammer symbol).
 In
particular, the validity of (\ref{eq:surmise}) follows from
(\ref{eq:result}) by setting $n=1$.

Expression (\ref{eq:result}) will be obtained by relating the generating
polynomial
\begin{equation} \varphi_{N,\xi}(\lambda)\equiv\sum_{n=0}^N \langle a_n\rangle
(-\lambda)^{N-n}
\label{eq:pol}
 \end{equation}
to a weighted average of the secular polynomial in the circular
unitary ensemble (CUE). This in turn establishes a relation to the
Szeg\H{o} polynomial of a Toeplitz matrix composed of binomial
coefficients. This calculation sidesteps Jack polynomials and
generalized Selberg integrals, which can be used to tackle general
expectation values in  multicomponent Coulomb gases
\cite{forrester1}.

\section{Reformulation in terms of random matrices} The CUE is composed of
$N\times N$ dimensional unitary matrices $U$ distributed according
to the Haar measure. Identify $t_i$ with the eigenphases of such a
matrix. The joint probability distribution is then given by
\cite{mehta}
\begin{equation}
P(\{t_i\}_{i=1}^N)=z
\prod_{i<j}\left|\rme^{\rmi t_i}-\rme^{\rmi t_j}\right|^2,
\end{equation}
where $z$ is again a normalization constant. This expression can also be
written as the product of two Vandermonde determinants $\det V^+\det V^-$
with matrices $V_{lm}^{\sigma}=\rme^{\rmi \sigma(m-1) t_l}$. Furthermore, we
can write
\begin{equation}
\prod_i\left|\rme^{\rmi \tau}-\rme^{\rmi t_i}\right|^{2\xi} =[\det(1-U\rme^{-\rmi \tau})\det(1-U^\dagger \rme^{\rmi \tau})]^{\xi}.
\end{equation}
Finally, the expressions $a_n$ in (\ref{eq:an}) arise as the
expansion coefficients of the secular polynomial
\begin{equation}
\det(U\rme^{-\rmi \tau}-\lambda)=\sum_{n=0}^N a_n(-\lambda)^{N-n} .
\end{equation}
Note that in all these expressions $\tau$ can be shifted to any fixed value
by a uniform shift of all $t_i$'s, which leaves the unitary ensemble
invariant. Therefore the expectation values are independent of $\tau$.
Collecting all results, we have the identity
\begin{equation}
\varphi_{N,\xi}(\lambda)=\frac{\left\langle[\det(1-U)\det(1-U^\dagger)]^{\xi}\det(U-\lambda)\right\rangle_{\rm CUE}}
{\left\langle[\det(1-U)\det(1-U^\dagger)]^{\xi}\right\rangle_{\rm CUE}}.
\label{eq:sec}
\end{equation}
This can be interpreted as a weighted average of the secular polynomial in
the CUE.

\section{Random-matrix average}  Statistical properties of the secular
polynomial without the weight factor ($\xi=0$) have been
considered in \cite{haake}. Clearly, $\varphi_{N,0}=(-\lambda)^N$,
so that in this case the attention quickly moves on to higher
moments of the $a_n$. The main technical observation in
\cite{haake} which allows to address the case of finite $\xi$
concerns averages of expressions $g(\{t_i\}_{i=1}^N)$ that are
completely symmetric in all eigenphases. In this situation the
average can be found via
\begin{equation}
\langle g(\{t_i\}_{i=1}^N)\rangle_{\rm CUE}=\int\prod_i\frac{\rmd t_i}{2\pi} g(\{t_l\}_{l=1}^N) \det W,
\label{eq:cuesimp}
\end{equation}
where $W_{lm}=\rme^{\rmi t_m (l-m)}$. Equation (\ref{eq:cuesimp})  is simpler
than the general expression involving the product of two Vandermonde
matrices, since each eigenphase only appears in a single column of $W$.

In the present problem, the numerator in (\ref{eq:sec}) is
represented by the completely symmetric function
\begin{equation}
g_1(\{t_i\}_{i=1}^N)=\prod_{i=1}^N[(\rme^{\rmi t_i}-\lambda)(1-\rme^{\rmi t_i})^\xi(1-\rme^{-\rmi t_i})^\xi],
\end{equation}
while for the denominator we need to consider the similar expression
\begin{equation}
g_2(\{t_i\}_{i=1}^N)=\prod_{i=1}^N[(1-\rme^{\rmi t_i})^\xi(1-\rme^{-\rmi t_i})^\xi].
\end{equation}

Using the multilinearity of the determinant we can now pull each factor into
the $i$th column and perform the integrals. This delivers the representation
\begin{equation}
\varphi_{N,\xi}(\lambda)=\frac{\det(B-\lambda A)}{\det A},
\end{equation}
where the matrices $A_{lm}=(-1)^{l-m}{{2\xi}\choose{\xi+l-m}}$,
$B_{lm}=(-1)^{l-m+1}{{2\xi}\choose{\xi+l-m+1}}$ have entries given
by binomial coefficients. We now exploit the regular structure of
these matrices in two steps.

1) Matrix $B$ contains the same entries as matrix $A$, but shifted
to the left by one column index. In order to exploit this, let us
expand the determinant in the numerator into a sum of determinants
of matrices labeled by $X=(x_m)_{m=1}^N$, where we select each
column either from $A$ ($x_m={\rm A}$) or from $B$ ($x_m={\rm
B}$). [Note that we set these symbols in roman letters.] The
related structure of $A$ and $B$ then entails that $\det X$
vanishes if $X$ contains a subsequence $(x_m,x_{m+1})=({\rm
A},{\rm B})$. Consequently we only need to consider determinants
of matrices $X_n \equiv ({\rm B})_{m=1}^{n}\oplus({\rm
A})_{m=n+1}^N$, associated to sequences that contain $n$ leading
B's and $N-n$ trailing A's. As $A$ is multiplied by $-\lambda$,
$\det X_n$ contributes to order $(-\lambda)^{N-n}$. (Note that
$X_0=A$ and $X_N=B$.)

2) Next, consider the matrix $A_{N+1}$, where the subscript denotes the
dimension, and strike out the first row and the $n+1$st column
($n=0,1,2,\cdots,N$). This takes exactly the form of the matrix $X_{n}$ of
dimension $N$. Therefore, the expressions $(-1)^{n}\det X_{n}$ are the
cofactors of the first row of $A_{N+1}$. These, in turn, are  proportional to
the first column of $A_{N+1}^{-1}$, where the proportionality factor is given
by $\det A_{N+1}$. Consequently, taking care of all alternating signs,
\begin{equation}
\varphi_{N,\xi}(\lambda)= (-1)^N\frac{\det A_{N+1}}{\det A_N} \sum_{n=0}^N(A_{N+1}^{-1})_{1,1+n}\lambda^{N-n}.
\label{eq:aexp}
\end{equation}
Via steps 1) and 2) we have eliminated any reference to the matrix $B$.

\section{Orthogonal polynomials} Matrix $A$ is a Toeplitz matrix,
$A_{lm}=c_{l-m}$. In order to find the explicit expression
(\ref{eq:result}) we now make contact to the theory of orthogonal
polynomials on the unit circle \cite{barry}. Among its many
applications, this theory provides a general expression for the
inverse of any Toeplitz matrix in terms of Szeg\H{o} polynomials
$\psi_N(\lambda)$. For the case of real symmetric coefficients,
the inverse is generated via
\begin{equation}\fl
\frac{\lambda\mu^{N}\psi_N(\lambda)\psi_N(\mu^{-1})-\lambda^N\mu\psi_N(\lambda^{-1})
\psi_N(\mu)}{\lambda-\mu}
=  \frac{\det A_{N+1}}{\det A_N}\sum_{n,m=0}^N(A_{N+1}^{-1})_{m+1,n+1}\lambda^{N-n}\mu^{m}.
\label{eq:aexp1}
\end{equation}
Comparison of this equation with $m=0$ to (\ref{eq:aexp})
immediately leads to the identification of
$(-1)^N\varphi_{N,\xi}(\lambda)$ with the Szeg\H{o} polynomial
$\psi_N(\lambda)$ of degree $N$. These polynomials satisfy
recursion relations which for real symmetric coefficients take the
form \numparts
\begin{eqnarray}
\gamma_{N}=-\frac{1}{\delta_{N-1}} \oint\frac{\rmd\lambda}{2
\pi\rmi} \psi_{N-1}(\lambda) \sum_{n=-\infty}^\infty c_n\lambda^n
,
\\
\psi_{N}(\lambda)=\lambda\psi_{N-1}(\lambda)+\gamma_N\lambda^{N-1}\psi_{N-1}(\lambda^{-1}),
\\[.3cm]
 \delta_N=\delta_{N-1}(1-\gamma_N^2).
\end{eqnarray}
\endnumparts
The initial conditions are $\delta_0=c_{0}$, $\psi_0(\lambda)=1$. The numbers
$\gamma_N$ are known as the Schur or Verblunsky coefficients.

It can now be seen in an explicit if tedious  calculation that the
polynomials \numparts
\begin{eqnarray}
\psi_N(\lambda)=(-1)^N\varphi_{N,\xi}(\lambda)=
\sum_{n=0}^N\frac{(N-n+1)^{(\xi)}(n+1)^{(\xi-1)}}{(N+1)^{(\xi)}(1)^{(\xi-1)}}\lambda^{N-n}
\\=\lambda^N{}_2F_1(-N,\xi;-N-\xi;\lambda^{-1})
\end{eqnarray}
[with coefficients and expansion given in (\ref{eq:result}),
(\ref{eq:pol})] indeed fulfill the Szeg\H{o} recursion generated
by the binomial coefficients $c_n=(-1)^n{{2\xi}\choose \xi-n}$.
The recursion coefficients take the simple form \begin{equation}
\gamma_N=\frac{\xi}{\xi+N},\qquad
\delta_N=\frac{N!(2\xi+1)^{(N)}}{[(\xi+1)^{(N)}]^2}.
\end{equation}
\endnumparts
 This completes the proof of
(\ref{eq:result}), and also entails the validity of (\ref{eq:surmise}).

\ack

The author wishes to thank Niko Jokela, Matti J{\"a}rvinen and
Esko Keski-Vakkuri for useful correspondence regarding reference
\cite{jokela}, and Mitsuhiro Arikawa and Peter Forrester for
useful communications about reference \cite{forrester1}. This work
was supported by the European Commission, Marie Curie Excellence
Grant MEXT-CT-2005-023778 (Nanoelectrophotonics).

\section*{References}


TITLE: Bayes' Rules: The probability of at least one event occurring?
QUESTION [2 upvotes]: There is a 60 percent chance that the event $A$ will occur. If $A$ does not occur, then
  there is a 10 percent chance that $B$ will occur.
(a) What is the probability that at least one of the events $A$ or $B$ occurs?

I am given $P(A)=0.6$, $P(A')=0.4$, $P(B\mid A')=0.1$.
If the probability that at least one of the events occurring is the probability of the union of $A$ and $B$:
$$P(A ∪ B)=P(A)+P(B)-P(A ∩ B)$$
then I must find $P(B): P(B)=P(B\mid A)P(A)+P(B\mid A')P(A')$.
To do that I have to find $P(B|A)$. I can't assume that $P(B\mid A)=1-P(B\mid A')$ right? So $P(B\mid A)=\frac{P(AB)}{P(A)}=\frac{P(A\mid B)P(B)}{P(A)}$, which means I have to find $P(B)$ which I had to find in the first place. It feels like I am caught in a loop.
Where was my error, and how can I solve this problem?

REPLY [3 votes]: $$\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B\cap A')=\mathbb P(A)+\mathbb P(B\mid A')\mathbb P(A')$$


TITLE: Creating a function F(x) such that all outputs exist between -5 and 5
QUESTION [3 upvotes]: Take this sample graph illustration:

As $X$ approaches negative infinity the output approaches $-5$
As $X$ approaches positive infinity the output approaches $5$
From what I recall this would be leveraging $\log, \ln$ or $e$ but I'm failing to remember the specific principles involved to come up with this function.
Other things that would be nice for this function is that it accelerates very quickly from the origin and tapers off which I tried to illustrate.
Bring this back to the real world and not just conceptual, realistic values of $X$ will primarily exist between $-10,10$ and much less frequently $-20,20$ and $-50,50$. My goal is to produce scoring algorithm with constrained limits on the output of the score.
The simpler the function, the better.

This function seems to be very close to what I'm looking for:
$5*\frac{x}{1+|x|}$
The graph it produces is:

At $F(5)$ and $F(-5)$ the graph is approximately 4 and -4.
How can I stretch out this function that $F(10)$ and $F(-10)$ are roughly 4 and -4 instead?
Able to answer this question myself, replace $X$ with $0.5X$

REPLY [2 votes]: You have many alternatives. Have a look at this picture:

You can choose any one of these functions and just multiply it by the max value you want to obtain (you wrote about $5$). If you replace $x$ with $kx$ you are able to tune how steep the curve is in the middle. You can find this functions and even a few other examples on Wikipedia's entry about sigmoid functions.
FOLLOW UP: you can tune the purple function to get what you need in two different ways: $y=5\frac{x/2}{1+|x/2|}$ and $y=5\frac{x}{5/2+|x|}$.

REPLY [2 votes]: It looks like you want a sigmoid function.
$$y = 5\left(\frac{2}{1 + e^{-kx}} - 1\right)$$
Increasing k will give a faster change through the origin. This one will go between $-5$ and $5$, replace the $5$ out front for a different range.

REPLY [2 votes]: The simplest is $$f(x)=\frac{10}{1+e^{-x}}-5$$
It is a simple logistic function subtracted by 0.5 multiplied by $10$.

REPLY [1 votes]: What about this :$$f(x)=\frac{10}{\pi} \tan^{-1}x$$


TITLE: Show that radius of rolling clay cylinder decreases at a rate proportional to the radius
QUESTION [0 upvotes]: Question:
Consider a rolling clay cylinder. It keeps being cylindrical and its volume remains the same. However, the length increases proportionally to the length. Show that the radius decreases proportionally to the radius.
Attempted solution:
So at time $t_{0}$ it has a certain length h and a certain radius r. Once it starts rolling, the rate at which the length increases at any given time is proportional to the length at that time. Now our job is to show that the rate at which the radius decreases is proportional to the radius at any given time.
It seems like an odd question and I am by far much worse for applied questions like this than other types of questions.
From the question, it is obvious that:
$$\frac{dh}{dt} = ch$$
where h is the length and c is a constant. The volume of a cylinder is also pretty obvious:
$$V = r^2 \pi h$$
Implicit differentiation of this second equation (using the product rule and treating r(t) and h(t) as two functions) gives:
$$0 = r^2 \frac{dh}{dt} + 2r \frac{dr}{dt} h$$
Putting in the first equation and solving for the rate of change in r gives:
$$\frac{dr}{dt} = \frac{-r^2 \frac{dh}{dt}}{2rh} = \frac{-r^2c}{2r} = c \frac{-r^2}{2r}$$
Now I have shown that the rate at which the radius changes is proportional to some negative expression involving $r$ and several other things. But this isn't quite what is being asked. The question is to show how the decrease is proportional to $r$. Here I am, perhaps naively, expecting the answer to come out as $cr$.
What crucial part am I missing?

REPLY [1 votes]: You're essentially done. $-cr^2/2r = (-c/2)r$. So $dr/dt$ is proportional to $r$ with proportionality constant $-c/2$.


TITLE: Bessel identities
QUESTION [1 upvotes]: Please help me prove the following identity
$$
a(J_1(a)Y_0(a)-J_0(a)Y_1(a))=\frac{2}{\pi}
$$
for any $a$.
$J$ and $Y$ are bessel functions of the first and second kind respectively.

Thank you.

REPLY [4 votes]: This is equivalent to computing the Wronskian of $J_0$ and $Y_0$, since $J'_0 = -J_1$ and $Y'_0 = -Y_1$.
$$
W(x) =
\begin{vmatrix}
J_0 & Y_0 \\\
J'_0 & Y'_0
\end{vmatrix} =
\begin{vmatrix}
J_0 & Y_0 \\\
-J_1 & -Y_1
\end{vmatrix} =
J_1 Y_0 - J_0 Y_1.
$$
But the zeroth-order Bessel equation is:
$$
\frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} + y = 0.
$$
By Abel's formula, $W(a) = \frac{W(1)}{a}$. You now have to compute any value of $W$.


TITLE: If $A$ and $B$ are two events and is $S$ is the sample space, does $\overline{(A \cap B)} \cup A = S$?
QUESTION [0 upvotes]: If $A$ and $B$ are two events and is $S$ is the sample space, does $\overline{(A \cap B)} \cup A = S$?
I get that $\overline{(A \cap B)} \cup A  =  (\overline {A} \cup \overline{B}) \cup A = \overline{A} \cup  (\overline{B} \cup A) = \overline{A} \cup A = S?$ Is this right?

REPLY [1 votes]: Your argument is correct .
Please edit the title so it matches your proof.


TITLE: How to prove that $\dim \mathrm{Spec}~A = \dim \mathrm{Spec}~ A_\mathfrak{p} + \dim \mathrm{Spec}~A/\mathfrak{p}$.
QUESTION [4 upvotes]: $\DeclareMathOperator{Spec}{Spec}$
$\newcommand{p}{\mathfrak{p}}$
I was trying to solve the following exercise.

Let $A$ be a finitely generated $k$-algebra over an infinite field $k$. Assume that
  $A$ is a domain, thus $\Spec A$ is irreducible. Let $\p \subset A$ be a prime
  ideal and let $A_\p$ be the localization of $A$ at $\p$. Show that $\dim \Spec A =
\dim \Spec A_\p + \dim \Spec A/\p$.

I'm not very sure how to proceed. Is it possible to proceed somehow by using Noether Normalization Theorem and that in this setting we have $\dim \Spec A = \mathrm{tr.deg}_k A$? To be clear, this has been asked in a somewhat different form previously, but if  possible, I wonder if one can furnish a proof using the transcendence degree and the normalization lemma?

REPLY [2 votes]: You can indeed use the transcendence degree:

Proposition: Let $A$ be a finitely generated $k$-algebra and suppose $A$ is a domain. Suppose $\mathfrak p_0\subsetneqq \mathfrak p_1\subsetneqq \dots \subsetneqq \mathfrak p_r$ is a chain of prime ideals and denote by $d$ the transcendence degree of $K=\mathrm{Frac}(A)$. Then $r\leq d$ with equality if and only if the chain is a maximal one.

Proof: By Noether Normalization, $A$ is module finite over $P=k[t_1,\dots,t_\nu]$ such that $\mathfrak p_i\cap P=\langle t_1,\dots,t_{h_i}\rangle$ for suitable $h_i$. If $L$ is the fraction field of $P$, then $\nu=\mathrm{tr.deg}_k(L)$. But $P\hookrightarrow A$ is an integral extension and module finite, hence $K/L$ is algebraic and therefore $\nu=d$. By the incomparability property of integral extensions we find $h_i<h_{i+1}$ for every $i$, hence $r\leq h_r$. But $h_r\leq \nu=d$, hence $r\leq d$. If $r=d$, then $r$ is maximal as we just showed that no chain can be longer. Conversely, assume $r$ maximal. Then $\mathfrak p_0=\langle 0\rangle$, so $h_0=0$. Furthermore $\mathfrak p_r$ is maximal, hence $\mathfrak p_r\cap P$ must be maximal too. Hence $h_r=\nu$. Now suppose there is $i$ such that $h_i+1<h_{i+1}$. Then $(\mathfrak p_i\cap P)\subsetneqq\langle t_1,\dots,t_{h_i+1}\rangle \subsetneqq (\mathfrak p_{i+1}\cap P)$. But $P/(\mathfrak p_i\cap P)$ is equal to $k[t_{h_i+1},\dots,t_\nu]$, which is a normal ring. As furthermore the extension $P/(\mathfrak p_i\cap P)\hookrightarrow A/{\mathfrak p_i}$ must be integral as $P\subset A$ is, Going-down yields a prime $\mathfrak p$ between $\mathfrak p_i$ and $\mathfrak p_{i+1}$ such that $\mathfrak p\cap P=\langle t_1,\dots,t_{h_i+1}\rangle$. Then $\mathfrak p_i\subsetneqq \mathfrak p\subsetneqq \mathfrak p_{i+1}$ which contradicts the maximality assumption of $r$. Hence $h_i+1=h_{i+1}$ for all $i$, which implies $r=h_r=\nu$.
Now you can prove your formula quite easily: Pick a maximal chain $\mathfrak p_0\subsetneqq \dots\subsetneqq \mathfrak p\subsetneqq\dots\subsetneqq \mathfrak p_n$ in $A$ (which is possible by the above proposition and has length $\dim (A)$), then $\mathfrak p_0A_{\mathfrak p}\subsetneqq \dots\subsetneqq \mathfrak pA_{\mathfrak p}$ is a maximal chain in $A_{\mathfrak p}$ and $\mathfrak p/\mathfrak p\subsetneqq \dots\subsetneqq \mathfrak p_n/\mathfrak p$ is a maximal chain in $A/\mathfrak p$ and since every such chain in $A$ can be constructed from two chains in $A_{\mathfrak p}$ and $A/\mathfrak p$, respectively, we see that the lengths of both chains must be the dimension of both rings.


TITLE: How to write $\sin(x_1).\sin(x_2)$ as an integral?
QUESTION [2 upvotes]: We know that $\sin(x)$ is the same as writing $\int_{0}^3 \cos(x)dx$.
However, I am interested if there is a similar way to write $\sin(3). \sin(4) $ as an integral.
One way I think is $$\sin(3).\sin(4) \Longrightarrow  \int_{0}^3 \cos(x)dx.\int_{0}^4 \cos(y)dy $$However, it is not so satisfying.Please help me to know some other way to write $\sin(3).\sin(4)$ (or, in general, $\sin(x_1).\sin(x_2)$) as an integral. Thanks for your help.

REPLY [5 votes]: \begin{align}
\sin(x_1)\sin(x_2) &= \frac12 \left( \cos (x_1-x_2) - \cos(x_1+x_2\right)\\
&=\frac12 \int_{x_1+x_2}^{x_1-x_2} -\sin(t) \, dt \\
&=\frac12 \int_{x_1-x_2}^{x_1+x_2} \sin(t) \, dt \\
\end{align}

REPLY [2 votes]: We have:
$$\sin x_1 \sin x_2 = \frac12(\cos (x_1-x_2) - \cos (x_1+x_2))$$
Hence we have:
$$\int_{x_1-x_2}^{x_1+x_2}\frac{\sin x}2 dx = \frac12(\cos (x_1-x_2) - \cos (x_1+x_2))=\sin x_1 \sin x_2$$


TITLE: Chebyshev inequality, lower bound on $P(X \ge 200)$
QUESTION [1 upvotes]: We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X \ge 200) \ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X \ge 200) \le t$ and consequently, $P(X < 200) \ge t$ but not $P(X \ge 200) \ge t$.
Could you help me with that $p$ and the estimation?
Thank you!

REPLY [0 votes]: Various points:

As you say in your comment, the expected value of each throw is $\frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $\frac{105}{36}$ rather than $\frac{21}{36}$
So with $100$ throws the expected value of the sum is $\frac{2100}{6} = 350$ while the variance of the sum is  $\frac{10500}{36} \approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $\frac{350-200}{17.08} \approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $\frac{350-199}{17.08} \approx 8.842$ below the mean to deal with your $P(X \ge 200)=P(X \gt 199)$ point
One version of Chebyshev's inequality is $\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$ and you can rewrite this as $\Pr(|X-\mu|\lt k\sigma) \geq 1-\frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-\frac1{8.842^2}\approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $\Pr(X-\mu \geq k\sigma) \leq \frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $\Pr(\mu - X \geq k\sigma) \leq \frac{1}{1+k^2}$ or $\Pr(\mu - X \lt k\sigma) \geq 1-\frac{1}{1+k^2}$ and here this would give about $1-\frac1{1+8.842^2}\approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here.  In fact $P(X \lt 200) \approx 2.5234\times 10^{-20}$ so $P(X \ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest


TITLE: Annihilator of a quotient module over a local ring
QUESTION [2 upvotes]: Assume $R$ is a commutative local ring, $I$ is a proper ideal in $R$ and $M$ is a finitely generated $R$-module. Is it true that $\operatorname{Ann}(M/IM)=I+\operatorname{Ann}(M)$?

(Note that $\supseteq$ is clear.)
If it's true I would like (a hint for) an elementary proof (preferably avoiding any homological algebra).

REPLY [1 votes]: Typically only a power of $Ann(M/IM)$ will be contained in $I+Ann (M)$. For a simple example, take the `universal' example. Let $R$ be the power series over a field in three variables $x,y,z$. Consider the module $M$ given as the quotient of the free module $Re_1\oplus Re_2$ by $xe_1+ye_2, ze_1+xe_2$. You can easily check that $Ann(M)$ is the ideal generated by $x^2-yz$, and if $I=(y,z)$, then $Ann(M/IM)=(x,y,z)$.


TITLE: Does EMR emanate in all directions? Does the wave expand like a sphere surface whose radius grows at $c$?
QUESTION [3 upvotes]: I made a commitment to understand special relativity.
Right now I'd like to understand  Electro-Magnetic Radiation.  What would it "look like"if I could see it emanate.  From a book (Simply Einstein) I'm told any movement of an electric charge (e.g. an electron) can start an electromagnetic wave.  Now that is cool.  So, finally, my question - once a wave is started does it emanate in all directions (a rock thrown in a lake goes in all directions in 2 dimensional lake surface, so would shaking an election start an emf wave that expands as an ever-growing surface of a sphere.  That is, a sphere whose radius is growing at the speed of light?  If so, where is the photon on this sphere?

REPLY [2 votes]: You are right; when an electric charge is forced to vibrate in all directions, it gives off electromagnetic radiation in all directions. You can think of this as consisting of a large number of photons being sprayed in all directions. That radiation then moves away through space as an expanding spherical wavefront, traveling at the speed of light.


TITLE: Solving $xdy+(y-xe^x)dx=0$ from GRE 8767.
QUESTION [2 upvotes]: I'm a little confused about GRE 8767 problem #40.

Let $y=f(x)$ be a solution of the differential equation $xdy+(y-xe^x)dx=0$ such that $y=0$ when $x=1$. What is the value of $f(2)$?

I see that this is an exact differential since $g(x,y)=xy+e^x-xe^x+C$ is such that $g_x=y-xe^x$ and $g_y=x$. How can I use this to find $f(2)$? I tried writing
$$
g(2,f(2))=2f(2)+e^2-2e^2+C=\cdots
$$
but I'm not sure what follows. Thanks. The final answer is $e^2/2$ by the way.

REPLY [2 votes]: Re-write the ODE by multiplying through by $dx$ to give $x \, y'(x) + y(x) - xe^x = 0$. This ODE can be solved very simply. We want to solve $xy' + y = xe^x$ which is the same as $(xy)' = xe^x$. Integrating both sides with respect to $x$ gives $xy = (x-1)e^x + k$ and so:
$y(x) = \frac{x-1}{x}e^x + \frac{k}{x} \, ,$
where $k$ is an arbitrary constant. If $y(1) = 0$ then $k = 0.$ The specific solution is
$y(x) = \frac{x-1}{x}e^x \, , $
and so $y(2) = \frac{1}{2}e^2.$

REPLY [1 votes]: You have arrived at the conclusion that for a solution $y=f(x)$ we have:
$$
   g(x, f(x)) = C
$$
Initial condition $f(1) = 0$ determines $C=0$, hence
$$
     0 = g(2, f(2)) = 2 f(2) + \mathrm{e}^2 - 2 \mathrm{e}^2 = 2 f(2) - \mathrm{e}^2
$$
Hence $f(2) = \frac{1}{2} \mathrm{e}^2$


TITLE: Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
QUESTION [1 upvotes]: I am trying to evaluate the integral:
$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dx$
with sgn$(x)$ the sign function and $a$ positive real. Naively applying the residue theorem and pushing the contour around the pole $+i/\sqrt{a}$, I obtain:
$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dx=-i ~\frac{2\pi i}{a}~{\rm sgn}\left(i/\sqrt{a}\right)^2~\left(\frac{i}{\sqrt{a}}\right) \frac{\sqrt{a}}{2i}~e^{-1/\sqrt{a}}=-\frac{\pi}{a}~e^{-1/\sqrt{a}}$
where I used: ${\rm sgn}\left(i/\sqrt{a}\right)=i~{\rm sgn}\left(1/\sqrt{a}\right)=i$.
However, I know the correct result (and Mathematica agrees) should be just
$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dx=\frac{\pi}{c}~e^{-1/\sqrt{a}}$
Why do I pick up the extra minus sign? Am I not applying the residue theorem correctly?
Note: this integral is a special case ($n=3$) of the integral I'm actually interested in, namely $-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^{n-1} ~x^{n-2}~ e^{i x}}{1+ax^2} dx$

REPLY [1 votes]: Assume $a \gt 0$.  Ignore the $\operatorname{sgn}^2{x}$, as that is identically $1$ almost everywhere, except at $x=0$ (which can be ignored as it is measure zero).  Thus you want to find
$$-i \int_{-\infty}^{\infty} dx \frac{x \, e^{i x}}{1+a x^2}$$
Then consider the integral
$$-i \oint_C dz \frac{z \, e^{i z}}{1+a z^2}$$
where $C$ is a semicircle of radius $R$ in the upper half plane.  The contour integral is equal to
$$-i \int_{-\infty}^{\infty} dx \frac{x \, e^{i x}}{1+a x^2} + R^2 \int_0^{\pi} d\theta\, e^{i 2 \theta} \frac{e^{i R e^{i \theta}}}{1+a R^2 e^{i 2 \theta}}$$
The integral about the semicircle has a magnitude bounded by
$$\frac{2}{a} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{\pi}{a R}$$
as $R \to \infty$.  The contour integral is also equal to, by the residue theorem, $i 2 \pi$ times the residue of the pole at $z=i/\sqrt{a}$.  Then we have
$$-i \int_{-\infty}^{\infty} dx \frac{x \, e^{i x}}{1+a x^2} = i 2 \pi (-i) \frac{ e^{-1/\sqrt{a}}}{2 a} = \frac{\pi}{a} \, e^{-1/\sqrt{a}}$$


TITLE: Continuity of $f$ and the existence of $f'(x)$ given $\lim_{x\rightarrow 0} f'(x)$ exists.
QUESTION [0 upvotes]: Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0,$ $f'(x)$ exists. If $\lim_{x\rightarrow 0} f'(x)$ exists, does it follow that $f'(0)$ exists? Prove or disprove.
Intuitively, it seems to me that it does does. That $f'(x)$ would not exist implies a point discontinuity, since $\lim_{x\rightarrow 0} f'(x)$ exists. Given that $f$ is continuous, $f$ is defined at $x.$ However, I'm not sure how to go about proving this (if I'm correct). Any hints? Thanks :)

REPLY [2 votes]: Yes, you are right! Just use Lagrange Mean Value theorem.
$$f'(0)=\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\dfrac{f'(\xi)(x-0)}{x-0}
=\lim_{x\to 0}f'(\xi)=\lim_{\xi\to 0}f'(\xi)=\lim_{x\to 0}f'(x),$$
here $\xi$ is between $0$ and $x$, and  $x\to 0$ implies $\xi\to 0.$


TITLE: $ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $
QUESTION [2 upvotes]: How do I show that, for $ 0 < x < \dfrac{\pi}{4} $ (first quadrant), the inequality $ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $ is valid?
I've tried Bernoulli's, but it took me to a false inequality (though all restrictions were respected). I actually thought of using Jensen's, but I don't know where to begin.

REPLY [4 votes]: Use:
$$\cos x > 1/\sqrt{2},\quad \sin x > x - x^3/6>0$$
So that we have:
$$(1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} >
1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6}$$
Now, we can use the series expansion of $(1+a)^x$:
$$(1+a)^x = 1+ \log(1+a)x+\frac{1}{2}\log^2(1+a) x^2 + \ldots =
\sum_{n=0}^\infty \frac{1}{n!}\log^n(1+a) x^n$$
Since all members of the series are positive for $a>0$, we can write:
$$1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6} >
2+\log(1+1/\sqrt2)x>\frac{x}{2}+2>3x$$
With the last being true since $\pi/4<4/5$, the point where $3x=2+x/2$.


TITLE: Simple way to write set of elements that are in either set but not both
QUESTION [1 upvotes]: Consider two sets $P$ and $P$'. I am trying to find the simplest notation denoting all elements that are in either $P$ or $P'$ but not both. I think this is $(P\cup P')\setminus (P\cap P')$. Is there any symbol in set notation already reserved for this operation?

REPLY [0 votes]: Along with what the other person said, in Boolean Algebra this is an XOR and is denoted by
$$
A \oplus B
$$


TITLE: To select 4 members of a math club from 12
QUESTION [1 upvotes]: A teacher must select four members of the math club to participate in an upcoming competition. How many ways are there for her to make her selection if the club has 12 members?

My answer : So we have to pick 4 people from a 12-members club. It should be : $12×11×10×9$ but apparently this is not the good answer "Incorrect She is selecting a 4-subset from the 12 kids in the club so it's $\binom{12}4$ or $\binom{12}8$"
Anyone can explain it to me?

REPLY [1 votes]: Since order does not matter in the selection, and there are $4!$ ways to permute a selection, $12×11×10×9$ must be divided by $4!$ to arrive at the correct answer, and this is equivalent to $\binom{12}4$.


TITLE: Given $u'(t)\leq au(t)$, $u(0)\leq b$. Find an upper bound on $u(t)$.
QUESTION [0 upvotes]: Let $a$ and $b$ be positive constants, and let $u(t)$ be a differentiable function on $[0,\infty)$ satisfying the inequality
$u'(t)\leq au(t)$, $u(0)\leq b$.
Find an upper bound on $u(t)$ and prove that it is the best possible.
My attempt: Using separation of variables, we can see that $u\leq Ce^{at}$ for some constant $C$. I'm not sure how this could possibly bound $u(t)$ given that $a>0$. Am I making a mistake?
Any help appreciated!
Side note: I realize there is a similar question asked, but the answers don't seem to actually answer the question, at least not as far as I can understand.

REPLY [1 votes]: Use the appropriate integrating factor and integrate
$$
(e^{-at}u(t))'=e^{-at}(u'-au)\le 0
$$
from $0$ to $t$.