TITLE: Velocity in the lift equation
QUESTION [2 upvotes]: What exactly is the nature of the velocity in the lift equation? I have read that it is the velocity of the aircraft, and I think I have read that it is the relative velocity of the air to the wing/airfoil. If I were holding a wing over my head and running at 5 mph into a breeze of 5 mph, would the velocity for the lift equation in this instance be 10 mph?

REPLY [0 votes]: It's not the speed you run, it's the speed of the plane relative to the air.

As shown above, the red arrow is the direction of air movement at the top of the wing. This direction is not perpendicular to the normal line at the top of the wing and is greater than 90 degrees. Therefore, the air tends to leave the wing along the normal line, so the air pressure at the top of the wing decreases.
As shown above, the blue arrow is the movement direction of the air at the bottom of the wing, which is not perpendicular to the normal at the bottom of the wing and is less than 90 degrees. Therefore, the air tends to approach the wing along the normal, so the air pressure at the bottom of the wing increases.
The air pressure at the top of the wing decreases, and the air pressure at the bottom of the wing increases. There is a pressure difference, which leads to the generation of lift.
It can be seen from the cause of lift that the relative velocity of wing and air is necessary.


TITLE: Isn't the category of abelian groups obviously Grothendieck?
QUESTION [1 upvotes]: In Peter Freyd's Abelian Categories, it is mentioned in passing that the category of abelian groups (more generally, $R$-modules for a ring $R$) satisfy the axiom AB5:

For each linearly ordered family $\{S_i\}_I$ in the lattice of subobjects of an object $S$, and $T$ is any subobject of $S$, then we have $$T \cap \bigcup S_i = \bigcup (T \cap S_i).$$

Isn't this completely obvious?
Let $r \in RHS$. Then $r \in LHS$. Conversely if $r \in LHS$ then $r \in RHS$.
(The only non-triviality is that the union of a family of submodules is a module in the first place, and this holds because the family is linearly ordered.)
I am confused because here https://ncatlab.org/nlab/show/Grothendieck+category#Kiersz they link to a 9 page paper that proves that the category of $R$-modules satisfies what they say is an equivalent condition to AB5 ('small filtered colimits are exact').

REPLY [5 votes]: Yes, you are correct.  The formulation of AB5 used on the nlab page you linked is substantially different from Freyd's (though it turns out to be equivalent) and so it takes a bit more work to prove it for modules over a ring.  Note that the 9-page paper you refer to also starts by introducing all the concepts needed to formulate the statement from scratch.  The actual proof of the final result (Proposition 4) is quite short and the important tools it uses from the previous 8 pages are just basic background on how to concretely identify all of the relevant category-theoretic notions in the case of modules over a ring (which is comparable to how you tacitly used the fact that the lattice of subobjects of a module can be identified with its usual lattice of submodules).


TITLE: Is $Pic^0(X)$ of a curve of genus $\geq 1$ over a non-algebraically closed field still non-finitely generated?
QUESTION [3 upvotes]: Qing Liu's "Algebraic Geometry and Arithmetic Curves" page 299 COrollary 7.4.41 gives the following result.
Let $X$ be a smooth, connected, projective curve over an algebraically closed field $k$, of genus $g$. Let $Pic^0(X)$ denote the subgroup of $Pic(X)$ consisting of divisors of degree $0$. Let $n\in \mathbb{Z}$ be non-zero and $Pic^0(X)[n]$ denote the kernel of the multiplication by $n$ map.

If $(n,\text{char} (k))=1$, then $Pic^0(X)[n]\cong (\mathbb{Z}/n\mathbb{Z})^{2g}$;
If $p=\text{char} (k)>0$, then there exists an $0\leq h\leq g$ such that for any $n=p^m$, we have $Pic^0(X)[n]=(\mathbb{Z}/n\mathbb{Z})^h$.

From this it is easy to deduce that if $X$ is a smooth, connected, projective curve over an algebraically closed field $k$, of genus $g\geq 1$, then $Pic^0(X)$ is not a finitely generated abelian group. (This is Exercise 4.9 (d) in page 301 of Qing Liu's book.)
$\textbf{My question}$ is: if the base field $k$ is not algebraically closed, is the above statement still true? I.e. if $X$ is a smooth, geometrically connected, projective curve over a field $k$, of genus $g\geq 1$, then is $Pic^0(X)$ finitely generated?

REPLY [9 votes]: As Will Sawin says, the key phrase here is Mordell-Weil theorem. Also, this isn't really a theorem about Picard groups of curves, it's a theorem about abelian varieties. Here is a fairly general statement:
Theorem (Mordell-Weil-Lang-Neron) Let $K$ be a field that is of finite type over its prime field (where the prime field is either $\mathbb Q$ or $\mathbb F_p$), and let $A/K$ be an abelian variety. Then $A(K)$ is finitely generated.
More generally, let $k$ be any field and let $K/k$ with $K$ of finite type over $k$. Then for any abelian variety $A/K$ there is an abelian variety $B/k$ (possibly  trivial) called the $K/k$-trace of $A$ and an inclusion $i:B\times_kK\hookrightarrow A$. Roughly speaking, $B$ is the largest piece of $A$ that comes from an abelian variety defined over $k$. Then $A(K)/i(B(k))$ is finitely generated.
I believe that this is all proven in Lang's Fundamentals of Diophantine Geometry.
Conversely, if $K$ is not finitely generated over its prime field, then I suspect that there always exists an abelian variety such that $A(K)$ is not finitely generated. (But you probably won't be able to prove it using torsion points.)


TITLE: $f(x)=\alpha e^{-x^2-\beta x}$ Find $\alpha$ and $\beta$ . Expectation is given.
QUESTION [0 upvotes]: Let the probability density function of a random variable $X$ be given
by
$f(x)=\alpha e^{-x^2-\beta x}\ \ \ \ \ \ \infty<x<\infty $
If $E(X)= -\dfrac{1}{2}$ then
$(A)  \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=1$
$(B)  \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=-1$
$(C)  \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=-1$
$(D)  \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=1$

The way I did this question:
$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{\infty}\alpha e^{-(x^2+\beta x +\frac{\beta^2}{4}-\frac{\beta^2}{4})}dx=\int_{-\infty}^{\infty}\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx=1 $
put $y=(x+\frac{\beta}{2})$
$\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\alpha\sqrt{\pi}e^{\frac{\beta^2}{4}}=1\implies \alpha=\dfrac{e^{\frac{-\beta^2}{4}}}{\sqrt{\pi}}$
Now using $E(X)=\int_{-\infty}^{\infty}x\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx$
put $y=(x+\frac{\beta}{2})$
$E(X)=\int_{-\infty}^{\infty}(y-\frac{\beta}{2})\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\underbrace{\alpha e^{\frac{\beta^2}{4}} \int_{-\infty}^{\infty}(y) e^{-(y)^2}dx}_{\text {odd function}}-\frac{\beta}{2}\underbrace{\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{-\beta^2}{4}}}_{\text{pdf}}dx$
$0-\dfrac{\beta}{2}=-\dfrac{1}{2} \implies \fbox {$\beta$ =1}$ and $\fbox{$\alpha=\dfrac{e^{\frac{-1}{4}}}{\sqrt{\pi}}$} $
Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?

REPLY [2 votes]: $f(x)$ is in the form of a normal PDF: $\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-(x-\mu)^2/2\sigma^2)$, which should be equal to $\alpha\exp(-x^2-\beta x)$. Coefficients should match in LHS and RHS, so $2\sigma^2=1$ and $E[X]=\mu=-1/2$. Then, LHS becomes
$$\frac{1}{\sqrt{\pi}}\exp(-x^2+2\mu x - \mu^2)=\alpha\exp(-x^2-\beta x)$$
which gives you $\beta=1$ and $\alpha=\frac{\exp(-1/4)}{\sqrt{\pi}}$.


text:
\begin{document}

\title[Schwarzschild Black Hole Threaded by a Cosmic String]{Renormalized Vacuum Polarization on the Horizon of a Schwarzschild Black Hole Threaded by a Cosmic String}
\author{Adrian C. Ottewill and Peter Taylor}
\address{School of Mathematical Sciences \& Complex and Adaptive Systems Laboratory, University College Dublin, Belfield, Dublin 4, Ireland}
\eads{\mailto{adrian.ottewill@ucd.ie} and \mailto{peter.taylor@ucd.ie}}
\date{\today}
\begin{abstract}
We obtain an analytic expression for the vacuum polarization on the horizon of the Schwarzschild black hole threaded by an infinite cosmic string. This calculation relies on a generalized Heine
identity for Legendre functions which we derive without using specific properties of the Legendre functions themselves.
\end{abstract}
\pacs{04.62.+v, 04.70.Dy, 11.27.+d}

\maketitle

\section{Introduction}
There is a long and successful history of  calculating the renormalized vacuum polarization and stress-energy tensor on spherically-symmetric black holes (see \cite{Candelas:1980zt,Candelas:1984pg,Anderson:1989vg,JensenOttewill, JMO1, JMO2, Anderson:1990jh,Winstanley:2007}), however, in the astrophysically significant case of the Kerr-Newman black hole, such calculations  have proved intractable. This is due to the fact that all the calculations in the spherically symmetric case have relied heavily on specific properties of the Legendre functions (mainly the Legendre Addition Theorem and the Heine Identity) and also the applicability of the Watson-Sommerfeld Formula to obtain a mode-sum expression that is numerically tractable. In the axially symmetric case, however, addition theorems for the angular functions do not exist nor do we have an analog of the Heine identity for the angular functions that arise. We will show, in the framework of the Schwarzschild black hole threaded by a cosmic string, that we can proceed in the axially symmetric case by obtaining useful summation formulae based on the Hadamard structure of the Green function without using specific properties of the mode-functions themselves. In this paper we will
use this approach to obtain an analytic expression for the renormalized vacuum polarization on the horizon of such a space-time.

It should be noted that the result presented here is a generalization of a result given by Davies and Sahni \cite{DaviesSahni} where the authors calculate the vacuum polarization for a restricted and discrete set of values for the azimuthal deficit.

\section{The Green's Function}
 The cosmic string is modeled by introducing an azimuthal deficit parameter, $\alpha$, into the standard Schwarzschild metric. We
may describe this in  coordinates $(t, r, \theta, \tilde{\phi})$, where  $\tilde{\phi}$ is periodic with period  $2\pi \alpha$, so we may take $\tilde{\phi}\in[0,  2\pi \alpha)$, in which the line element is given by
\begin{equation}
\fl
{\rmd}s^2 = -(1-2M/r) {\rmd}t^2 +(1-2M/r)^{-1} {\rmd}r^2 +r^2 {\rmd} \theta^2 +  r^{2} \sin^{2}\theta {\rmd} \tilde{\phi}^{2}.
\end{equation}
For GUT scale cosmic strings $\alpha = 1- 4 \mu$ where $\mu$ is the mass per unit length of the string so we shall assume $0 < \alpha \le 1$.
We can alternatively define a new azimuthal coordinate by
\begin{equation}
\tilde{\phi} = \alpha \phi
\end{equation}
so that  $\phi$ is periodic with period  $2\pi$ and we may take $\phi \in [0, 2 \pi)$. In coordinates $(t, r, \theta, \phi)$,  the line element is given by
\begin{equation}
\fl
{\rmd}s^2 = -(1-2M/r) {\rmd}t^2 +(1-2M/r)^{-1} {\rmd}r^2 +r^2 {\rmd} \theta^2 + \alpha^{2} r^{2} \sin^{2}\theta {\rmd} \phi^{2}.
\end{equation}

We shall consider a massless, minimally coupled scalar field, $\varphi$, in the Hartle-Hawking vacuum state.
Since this is a thermal state, it is convenient to work with the Euclidean Green's function, performing a Wick rotation of the
 temporal coordinate $t\rightarrow -i \tau$ and eliminating the conical singularity at $r=2M$ by making $\tau$ periodic with period
$2\pi/\kappa$ where $\kappa=1/(4M)$ is  the surface gravity of the black hole.
The massless, minimally coupled scalar field, satisfies the homogenous wave-equation
\begin{equation}
\Box \varphi(\tau,r,\theta,\phi)=0,
\end{equation}
which can be solved by a separation of variables by writing
\begin{equation}
\varphi(\tau,r,\theta,\phi)\sim \rme^{i n \kappa \tau+i m\phi} P(\theta)R(r)
\end{equation}
where $P(\theta)$ is regular and satisfies
\begin{equation}
\label{eq:legendre}
\Big\{\frac{1}{\sin\theta} \frac{{\rmd}}{{\rmd} \theta}\Big(\sin\theta\frac{{\rmd}}{{\rmd} \theta}\Big) -\frac{m^{2}}{\alpha^{2}\sin^{2}\theta}+\lambda(\lambda+1)\Big\}P(\theta)=0
\end{equation}
while $R(r)$ satisfies
\begin{equation}
\label{eq:radialhomogeneous}
\Big\{\frac{{\rmd}}{{\rmd} r}(r^2-2Mr)\frac{{\rmd}}{{\rmd}r} - \lambda(\lambda+1)-\frac{n^2 \kappa^2 r^4}{r^2-2Mr}\Big\}R(r)= 0.
\end{equation}
 The $\lambda(\lambda+1)$ term arises as the separation constant. The choice of $\lambda$ is arbitrary for $\varphi$ to satisfy the wave equation but requires a specific choice in order for the mode-function $P(\theta)$ to satisfy the boundary conditions of regularity on the poles. In the Schwarzschild case (no cosmic string, $\alpha=1$), regularity on the poles means that $\lambda=l$, i.e the separation constant is $l(l+1)$. In the cosmic string case, the appropriate choice of $\lambda$ that guarantees regularity of the angular functions on the poles is
\begin{equation}
\label{eq:lambda}
\lambda=l-|m|+|m|/\alpha.
\end{equation}

With this choice of $\lambda$, the angular function is the Legendre function of both non-integer order and non-integer degree,\textit{viz.},
\begin{equation}
P(\theta)=P_{\lambda}^{-|m|/\alpha}(\cos\theta).
\end{equation}
It can be shown that these angular functions satisfy the following normalization condition,
\begin{eqnarray}
\label{eq:norm}
&\int_{-1}^{1}P_{l-|m|+|m|/\alpha}^{-|m|/\alpha}(\cos\theta) P_{l'-|m|+|m|/\alpha}^{-|m|/\alpha}(\cos\theta) d(\cos\theta)
=\nonumber\\
&\qquad\qquad\qquad\frac{2}{(2\lambda+1)}\frac{\Gamma(\lambda-|m|/\alpha+1)}{\Gamma(\lambda+|m|/\alpha+1)}\delta_{ll'}.
\end{eqnarray}

The periodicity of the Green's function with respect to $(\tau-\tau')$ and $(\phi-\phi')$ with periodicity $2\pi/\kappa$ and $2\pi$, respectively, combined with Eq.(\ref{eq:norm}) allow us to write the mode-sum expression for the Green's function as

\begin{eqnarray}
\label{eq:greensfn}
\fl
G(x,x')=\frac{\kappa}{8\pi^{2}}\sum_{n=-\infty}^{\infty} {\rme}^{i n \kappa (\tau-\tau')}&\sum_{m=-\infty}^{\infty} {\rme}^{im(\phi-\phi')} \sum_{l=|m|}^{\infty} (2\lambda+1)\frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)}  \nonumber\\
 &\qquad P_{\lambda}^{-|m|/\alpha}(\cos\theta) P_{\lambda}^{-|m|/\alpha}(\cos\theta')\chi_{n\lambda}(r,r'),
\end{eqnarray}

where $\chi_{n\lambda}(r,r')$ satisfies the inhomogeneous equation,
\begin{equation}
\label{eq:radialr}
\fl
\Big\{\frac{{\rmd}}{{\rmd} r}(r^2-2Mr)\frac{{\rmd}}{{\rmd}r} - \lambda(\lambda+1)-\frac{n^2 \kappa^2 r^4}{r^2-2Mr}\Big\}\chi_{n\lambda}(r,r')=-\frac{1}{\alpha}\delta(r-r').\nonumber\\
\end{equation}

It is convenient to write the radial equation in terms of a new radial variable

$\eta=r/M-1$,

 the radial equation then reads
\begin{eqnarray}
\label{eq:radial}
\fl
\Big\{\frac{{\rmd}}{{\rmd}\eta}\Big((\eta^{2}-1)\frac{{\rmd}}{{\rmd}\eta}\Big)-\lambda(\lambda+1)-\frac{n^{2}(1+\eta)^{4}}{16(\eta^{2}-1)}\Big\}\chi_{n\lambda}(\eta,\eta')=-\frac{1}{\alpha M}\delta(\eta-\eta'),
\end{eqnarray}
where we have used the fact that $\kappa=1/4M$. For $n=0$, the two solutions of the homogeneous equation are the Legendre functions of the first and second kind. For $n\ne 0$, the homogeneous equation cannot be solved in terms of known functions and must be solved numerically. We denote the two solutions that are regular on the horizon and infinity (or some outer boundary) by $p_{\lambda}^{|n|}(\eta)$ and $q_{\lambda}^{|n|}(\eta)$, respectively. A near-horizon Frobenius analysis for $n \neq 0$ shows that the indicial exponent is $\pm |n|/2$, and so we have the following asymptotic forms:
\begin{equation}
\label{eq:asymp}

\eqalign{
 p_{\lambda}^{|n|}(\eta)\sim (\eta-1)^{|n|/2} \qquad\qquad &\eta\rightarrow 1, \cr
q_{\lambda}^{|n|}(\eta) \sim (\eta-1)^{-|n|/2} &\eta \rightarrow 1.}

\end{equation}
Defining the normalizations by these asymptotic forms and using the Wronskian conditions one can obtain the appropriate normalization of the Green's function:
 \begin{eqnarray}
 \label{eq:chi}
 \chi_{n\lambda}(\eta,\eta') =
\cases{
\displaystyle{\frac{1}{\alpha M}} P_{\lambda}(\eta_{<})Q_{\lambda}(\eta_{>})&$n=0$, \\
\displaystyle \frac{1}{2|n|\alpha M} p_{\lambda}^{|n|}(\eta_{<}) q_{\lambda}^{|n|}(\eta_{>})\qquad&$n\neq 0$.}
 \end{eqnarray}

\section{Renormalization}
 From the asymptotic forms (\ref{eq:asymp}), it is clear that taking one point on the horizon (with the other outside) means that all the $n\neq 0$ modes vanish. The unrenormalized Green's function then reduces to
\begin{eqnarray}
\label{eq:greensfnhorizon}
\fl
G(x,x')=\frac{1}{32 \pi^{2} M^{2} \alpha}&\sum_{m=-\infty}^{\infty} {\rme}^{im(\phi-\phi')} \sum_{l=|m|}^{\infty} (2 \lambda+1) \frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)} \nonumber\\
& \qquad P^{-|m|/\alpha}_{\lambda}(\cos\theta) P^{-|m|/\alpha}_{\lambda}(\cos\theta') Q_{\lambda}(\eta).
\end{eqnarray}
In the absence of the cosmic string, i.e. $\alpha = 1$, this sum can be done using a combination of the Legendre Addition Theorem and the Heine Identity, yielding a closed form expression~\cite{Candelas:1980zt}, that,
of course, diverges in the coincidence limit.

To renormalize we employ the Christensen-DeWitt\cite{Christensen:1976vb} point-separation method subtracting the geometrical singular
part of the Green's function for small separations in the coordinates. This method relies on the universal Hadamard singularity structure  of the Green's function which ensures that once we have subtracted the geometric singular part, `$U/\sigma + V \ln \sigma$', the coincidence limit of the remainder `$W$' is finite. To calculate $\langle \hat{\varphi}^2 \rangle_{ren}$ in the massless Ricci-flat case, the only subtraction term needed is
\begin{equation}
G_{sing}(x,x') = \frac{1}{4\pi^{2} s^2}
\end{equation}
where $s$ is the geodesic distance between $x$ and $x'$. Having already taken one point on the horizon
we choose to separate radially, placing $x'$ at $2M+\epsilon$; the geodesic distance is then given by
\begin{eqnarray}
\fl
 s(2M,2M+\epsilon)&=&\int_{2 M}^{2M+\epsilon} \frac{\rmd r'}{(1-2M/r')^{1/2}}  =  (2M\epsilon)^{1/2} \bigg[2+1/3(\epsilon/2M) +O(\epsilon^2)\bigg].
 \end{eqnarray}

Note that this behaves like $(\Delta x)^{1/2}$ rather that the usual $\Delta x$ since the metric is singular at
$r=2M$ so $g^{r'r'}=O(\epsilon)$, but this is simply a coordinate singularity and in no way affects the validity of the geometrical subtraction.  The subtraction terms up to $O(\epsilon)$ are
 \begin{equation}
 \label{eq:gdiv}
 G_{sing}  =\frac{1}{32 \pi^{2} M \epsilon} - \frac{1}{192 \pi^{2} M^{2}} + O(\epsilon).
 \end{equation}

Taking the partial coincidence limit $\phi \rightarrow \phi'$, $\theta\rightarrow \theta'$ in (\ref{eq:greensfnhorizon}), subtracting the renormalization terms (\ref{eq:gdiv}) and taking the limit as $\eta \rightarrow 1$ ($\epsilon\rightarrow 0$) gives us the renormalized expression for the vacuum polarization. However, we now face our fundamental challenge as the limit cannot be taken in a meaningful way in the forms given. In order to take the limit, we must either invent a way to write the divergent terms as an appropriate mode-sum and do a mode-by-mode subtraction or alternatively, we must attempt to perform the sum in (\ref{eq:greensfnhorizon}) so that we can write the Green's function in closed form. In \cite{CSHorizon} we follow the former route to obtain numerical values of $\langle \hat{\varphi}^2 \rangle_{ren}$ outside our black hole. However, when possible, the latter is clearly preferred since it can give us a simple closed-form solution for the renormalized vacuum polarization. In the next section, we derive a formula that will allow us to perform the sum (\ref{eq:greensfnhorizon}).

\section{Generalized Heine Identity}
We recall that for real arguments, the Heine Identity for the Legendre Functions is \cite{Erdelyi}
\begin{equation}
\label{eq:heine}
\sum_{l=0}^{\infty} (2 l+1) P_{l}(\Psi)Q_{l}(\zeta)=\frac{1}{(\zeta-\Psi)},
\end{equation}
valid for $|\Psi|<|\zeta|$. Writing $\Psi=\cos\gamma=\cos\theta\cos\theta'+\sin\theta\sin\theta'\cos(\phi-\phi')$, and using the Legendre Addition Theorem,
\begin{equation}
P_{l}(\cos\gamma) = \sum_{m=-l}^{l} \rme^{im (\phi-\phi')} \frac{(l-m)!}{(l+m)!} P_{l}^{m}(\cos\theta)P_{l}^{m}(\cos\theta'),
\end{equation}
we can re-write the Heine Identity as
\begin{equation}
\label{eq:heinenew}
\fl
\sum_{l=0}^{\infty}\sum_{m=-l}^{l} \rme^{im (\phi-\phi')}(2 l+1)\frac{(l-m)!}{(l+m)!} P_{l}^{m}(\cos\theta)P_{l}^{m}(\cos\theta') Q_{l}(\zeta)=\frac{1}{(\zeta-\cos\gamma)}.
\end{equation}
Below we shall generalize this form of the Heine Identity to the non-integer Legendre functions that arise in the mode-sum of a cosmic string space-time.
It is this form rather than a generalization of (\ref{eq:heine}) that is useful to us since the fundamentaly axially-symmetric nature of the Green's function for $\alpha\neq 1$ means that its radial part  will always depend on $m$ (through $\lambda$), and therefore the $m$-sum over the angular functions can never be done independently of the radial part.

In the Appendix, we have derived other identities involving Legendre Functions of non-integer order and/or degree. We have done so since, although they are not directly useful here, they turn out to be useful in other contexts, for example, determining the Hadamard singularity structure of a 3D Green's function on a cosmic string black hole space-time \cite{CSHorizon}.

Returning to the problem at hand, we want to construct a generalization of (\ref{eq:heinenew}). In addition, we would prefer to do this in such a way that the derivation does not rely on specific properties of the Legendre functions themselves, since
ultimately we wish to apply the same techniques in Kerr space-time. There is a very natural way to do this by equating equivalent expressions for the same Green's function on a conveniently chosen space-time.

In our case, the most appropriate case to consider is that of  an infinite cosmic string in otherwise flat space-time. Of course, normally in this case one would exploit translational invariance along the
string and work in cylindrical polar coordinates but for our purposes we choose to work in spherical polar coordinates, so
\begin{equation}
{\rmd}s^{2} = {\rmd}\tau^{2} +{\rmd}r^{2} +r^{2} {\rmd}\theta^{2} +\alpha^{2} r^{2} \sin^{2}\theta {\rmd}\phi^{2},
\end{equation}
where $\phi$ is periodic with period  $2\pi$.
The Green's function for a minimally coupled scalar field on this space-time has a closed-form solution,
which can be obtained, for example, by separating in cylindrical polar coordinates~\cite{Smith}
\begin{equation}
\label{eq:gclosed}
G(x,x') = \frac{1}{8\pi^{2}\alpha} \frac{\sinh(\chi/\alpha)}{\rho\rho' \sinh\chi(\cosh(\chi/\alpha)-\cos(\phi-\phi'))},
\end{equation}
where
\begin{equation}
\cosh\chi = \frac{(\tau-\tau')^{2}+r^2 +r'^2-2r r' \cos\theta\cos\theta'}{2\rho\rho'}
\end{equation}
and $\rho=r\sin\theta$ is the cylindrical polar coordinate.

One can also obtain the spherical polar mode-sum expression for this Green's function:

\begin{eqnarray}
\fl
G(x,x') = \frac{1}{4\pi^{2}\alpha} \int_{0}^{\infty}\cos\omega(\tau-\tau') \rmd\omega \sum_{m=-\infty}^{\infty} \rme^{im(\phi-\phi')} \sum_{l=|m|}^{\infty} (2 \lambda+1) \frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)}  \nonumber\\
 P^{-|m|/\alpha}_{\lambda}(\cos\theta) P^{-|m|/\alpha}_{\lambda}(\cos\theta') R_{\omega\lambda}(r,r'),
\end{eqnarray}
where $R_{\omega\lambda}(r,r')$ is the 1D Green's function satisfying
\begin{equation}
\Big[\frac{{\rmd}}{{\rmd} r}\Big(r^{2} \frac{{\rmd}}{{\rmd} r}\Big)-\omega^{2} r^{2} -\lambda(\lambda+1)\Big]R_{\omega\lambda} = -\delta(r-r').
\end{equation}
The solution of this equation can be given in terms of Modified Bessel Functions,
\begin{equation}
R_{\omega\lambda}(r,r') = \frac{I_{\lambda+1/2}(\omega r_{<}) K_{\lambda +1/2}(\omega r_{>})}{(r r')^{1/2}},
\end{equation}
where $I$ and $K$ are the Modified Bessel Functions of the first and second kinds, respectively.
The integral over $\omega$ can now be performed using the identity \cite{gradriz}
\begin{equation}
\int_{0}^{\infty} \cos\omega(\tau-\tau') I_{\lambda+1/2}(\omega r_{<}) K_{\lambda +1/2}(\omega r_{>})\rmd\omega = \frac{1}{2(r r')^{1/2}} Q_{\lambda}(\zeta)
\end{equation}
valid for $\lambda > -1$, where $Q_\lambda$ is the Legendre function of the second kind and
\begin{equation}
\zeta = \frac{(\tau-\tau')^{2} + r^{2} +r'^{2}}{2 r r'}.
\end{equation}
Thus, the spherical polar mode-sum expression for the Green's Function may be written as
\begin{eqnarray}
\fl
G(x,x') = \frac{1}{8 \pi^{2}\alpha r r'} \sum_{m=-\infty}^{\infty} e^{im(\phi-\phi')} \sum_{l=|m|}^{\infty} (2 \lambda+1) \frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)} \nonumber\\
 P^{-|m|/\alpha}_{\lambda}(\cos\theta) P^{-|m|/\alpha}_{\lambda}(\cos\theta') Q_{\lambda}(\zeta).
\end{eqnarray}

We now equate this Green's Function with the equivalent closed form expression (\ref{eq:gclosed}) to obtain the following generalized Heine Identity:
\begin{eqnarray}
\label{eq:generalizedheine}
\fl
\sum_{m=-\infty}^{\infty} \rme^{im(\phi-\phi')} \sum_{l=|m|}^{\infty} (2 \lambda+1) \frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)}P^{-|m|/\alpha}_{\lambda}(\cos\theta) P^{-|m|/\alpha}_{\lambda}(\cos\theta')Q_{\lambda}(\zeta)  \nonumber\\
=\frac{\sinh(\chi/\alpha)}{\sin\theta\sin\theta' \sinh\chi(\cosh(\chi/\alpha)-\cos(\phi-\phi'))}
\end{eqnarray}
where
\begin{equation}
\fl
\cosh\chi = \frac{\zeta-\cos\theta\cos\theta'}{\sin\theta\sin\theta'}\quad \Leftrightarrow \quad
\zeta = \cos \gamma + (\cosh\chi - \cos (\phi-\phi')) \sin\theta\sin\theta'.
\end{equation}

This identity, valid for $\zeta>|\cos \gamma|$, is completely analagous to (\ref{eq:heinenew}) and reduces to it in the $\alpha\rightarrow 1$ limit.

\section{Calculation of $\langle \hat{\varphi}^{2} \rangle_{ren}$}
Returning now to our calculation of the vacuum polarization on the horizon of the Schwarzschild black hole threaded by a cosmic string, we see that our generalized Heine Identity (\ref{eq:generalizedheine}) has precisely the right form to enable us to perform the mode-sum in the Green's function expression (\ref{eq:greensfn}). The Green's function, where one point has been taken to lie on the horizon, may now be written as
\begin{equation}
G(x,x')=\frac{1}{32\pi^{2} M^{2}\alpha}\frac{\sinh(\chi/\alpha)}{\sin\theta\sin\theta' \sinh\chi(\cosh(\chi/\alpha)-\cos(\phi-\phi'))}
\end{equation}
For purely radial separation, we take $\phi'\rightarrow\phi$, $\theta'\rightarrow\theta$, to yield
\begin{equation}
\label{eq:gclosedradial}
G(x,\theta;0,\theta)=\frac{1}{32\pi^{2} M^{2}\alpha}\frac{\sinh(\chi/\alpha)}{\sin^{2}\theta\sinh\chi(\cosh(\chi/\alpha)-1)}
\end{equation}
where we now have
\begin{equation}
\cosh\chi=\frac{\eta-\cos^{2}\theta}{\sin^{2}\theta}=1+ \frac{\epsilon}{M\sin^{2}{\theta}},
\end{equation}
on taking $r'=2M+\epsilon \Leftrightarrow \eta=\epsilon/M+1$.
From Eqs.(\ref{eq:gdiv}) and (\ref{eq:gclosedradial}), the renormalized vacuum polarization is given by
\begin{eqnarray}
\label{eq:phi2limit}
&&\fl\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon} = \lim_{\epsilon \rightarrow 0} \Big[ G(x,0)-G_{sing}(x,0)\Big] \nonumber\\
&&\fl \ = \lim_{\epsilon \rightarrow 0} \Big[ \frac{1}{32\pi^{2} M^{2} \alpha \sin^{2}\theta} \frac{\sinh(\chi/\alpha)}{\sinh\chi(\cosh(\chi/\alpha)-1)}-\frac{1}{32\pi^{2} M \epsilon}-\frac{1}{192\pi^{2} M^{2}}\Big].
\end{eqnarray}
It is now convenient first expand in terms of $\chi$

 \begin{equation}
  \frac{\sinh(\chi/\alpha)}{\sinh\chi(\cosh(\chi/\alpha)-1)}=\frac{2\alpha}{\chi^{2}} +\Big(\frac{1}{6\alpha}-\frac{\alpha}{3}\Big) + O(\chi^{2}).
  \end{equation}

then
  \begin{equation}
  \cosh\chi = 1+\frac{\epsilon}{M \sin^{2}\theta}\quad \Rightarrow \quad \frac{1}{\chi^{2}} = \frac{M\sin^{2}\theta}{2 \epsilon} + \frac{1}{12} +O(\epsilon^{2})
  \end{equation}

so we  obtain
\begin{equation}
\label{eq:gexpansion}
 \frac{\sinh(\chi/\alpha)}{\sinh\chi(\cosh(\chi/\alpha)-1)}=\frac{\alpha M\sin^{2}\theta}{\epsilon} +\frac{1}{6\alpha}(1-\alpha^{2}) + O(\epsilon^{2}).
 \end{equation}
Substituting (\ref{eq:gexpansion}) into (\ref{eq:phi2limit}) and taking the limit, we arrive at the renormalized vacuum polarization on the horizon,
\begin{equation}
\label{eq:phi2ren}
\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon} = \frac{1}{192\pi^{2} M^{2}}\Big(1+\frac{1-\alpha^{2}}{\alpha^{2}\sin^{2}\theta}\Big).
  \end{equation}

  The first term here is simply the Candelas result \cite{Candelas:1980zt}, which, of course we recover when $\alpha \rightarrow 1$, and the second term represents the contribution due to the presence of the cosmic string. The result (\ref{eq:phi2ren}) has in fact been calculated by Davies and Sahni \cite{DaviesSahni} but for a restricted set of $\alpha$ values such that $(1/\alpha)$ is an integer. For this restricted class of values, the cosmic string solution is obtained as a sum of $(1/\alpha)$ images of the solution in the absence of a string. The method of images cannot be applied for general azimuthal deficits, however, so our method presents a generalization and independent derivation of the result of Davies and Sahni.

  We have plotted the vacuum polarization in Figure~\ref{fig:phi2horizon} for a range of values of $\alpha$. We see that the presence of the string increases $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$ everywhere. In the equatorial plane it is increased by a factor of $1/\alpha^2$ from its value in the absence of a string.
As we approach the axis  $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$ diverges in a non-integrable manner as
 \begin{equation}
\langle \varphi^{2} \rangle_{ren}^{horizon} \sim \frac{1}{48\pi^{2} }\frac{(1-\alpha^{2})}{\alpha^{2}} \frac{1}{(2M \sin\theta)^{2}} \qquad \sin\theta \to 0,
\end{equation}
  as one would expect a distance  $(2M \sin\theta)$ from a flat-space cosmic string.
  The region over which the string dominates may be characterised by the range of $\cos \theta$ (near $\theta=0$) at which
  $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$ exceeds twice its value in the equatorial plane, this is given
by
  \begin{equation}
1-(\cos \theta)_2 = 1-\frac{1}{(2-\alpha^2)^{1/2}} = (1-\alpha) + O\bigl((1-\alpha)^2)\bigr) .
\end{equation}

\begin{figure}
\centering
\includegraphics[width=11cm]{Figure1.pdf}
\caption{\emph{ $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$} (in units $M=1$) as a function of $\cos\theta$ for a range of values of $\alpha$. For $\alpha \neq 1$,  $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$  diverges at the poles since there is a curvature singularity there due to the presence of the string. $\langle \hat{\varphi}^{2} \rangle_{ren}^{horizon}$ increases everywhere as $\alpha$ decreases.}
\label{fig:phi2horizon}
\end{figure}

\section{Conclusions}
We have considered  a massless, minimally coupled quantum scalar field on a Schwarzschild black hole threaded by an infinite straight cosmic string. We have obtained a simple analytic expression for the vacuum polarization on the black hole horizon for this field in the Hartle-Hawking vacuum state, which is a generalization of the Candelas \cite{Candelas:1980zt} result. In order to do this, we have derived a very useful summation formula involving non-integer Legendre functions, which we have called a generalized Heine Identity given its analogy with the standard Heine Identity involving Legendre functions of integer degree. The derivation of this formula involved equating equivalent expressions for the same Green's function on a conveniently chosen space-time. Furthermore, the derivation did not rely on specific properties of the Legendre functions. This idea can be used to derive many useful summation formulae and addition theorems for non-integer Legendre functions, some of which are presented in the Appendix.  Elsewhere \cite{CSHorizon}, we have used this idea to obtain an expression for the vacuum polarization on the entire exterior region of the space-time. We believe this framework will prove useful in the other axially symmetric black hole calculations, most importantly, the Kerr-Newman black hole.

\section*{Appendix}
In this Appendix, we shall derive some interesting summation formulae for non-integer Legendre functions based on considering equivalent forms of the 3D Green's function on flat space threaded by a cosmic string
\begin{equation}
\rmd s^{2}=\rmd r^{2}+r^{2} \rmd\theta^{2} + \alpha^{2} r^{2}\sin^{2}\theta \rmd\phi^{2}.
\end{equation}
The 3D Green's function for Laplace's equation satisfying
\begin{equation}
\nabla^2 G_3(\mathbf{x},\mathbf{x'}) = -{g_3}(\mathbf{x})^{-1/2} \delta(\mathbf{x}-\mathbf{x'})
\end{equation}
can be written in spherical polar coordinates as the mode sum:
\begin{eqnarray}
\label{eq:app1}
\fl
G_3(\mathbf{x},\mathbf{x'})=\frac{1}{4\pi\alpha}&\sum_{m=-\infty}^{\infty} \rme^{im (\phi-\phi')}\sum_{l=|m|}^{\infty}\frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)}\nonumber\\ &\qquad P^{-|m|/\alpha}_{\lambda}(\cos\theta)P^{-|m|/\alpha}_{\lambda}(\cos\theta')(r_{<})^{\lambda} (r_{>})^{-(\lambda+1)} .
\end{eqnarray}
Note that this corresponds to the $n=0$ term of the 4D Green function in the corresponding ultrastatic ($g_{tt}=-1$) space-time, but that the
corresponding statement is not true in the merely static Schwarzschild cosmic string space-time.
We can also write this same Green's function in several other forms and in several coordinate systems, for example,
performing the mode decomposition in cylindrical polar coordinates, $\rho=r \sin\theta$ and $z=r\cos\theta$, we have
\begin{eqnarray}
\label{eq:app2}
\fl
G_3(\mathbf{x},\mathbf{x'})=\frac{1}{2 \pi^{2} \alpha} \sum_{m=-\infty}^{\infty} \rme^{im(\phi-\phi')}\int_{0}^{\infty} \cos k(z-z') I_{|m|/\alpha}(k\rho_{<}) K_{|m|/\alpha}(k\rho_{>}) \rmd k  .
\end{eqnarray}
The integral over $k$ may further be performed to yield
\begin{eqnarray}
\label{eq:app3}
\fl
G_3(\mathbf{x},\mathbf{x'})=\frac{1}{4 \pi^{2}\alpha (\rho \rho')^{1/2}} \sum_{m=-\infty}^{\infty} e^{im (\phi-\phi')} Q_{|m|/\alpha-1/2}\left(\frac{(z-z')^{2}+\rho^{2}+\rho'^{2}}{2 \rho \rho'}\right) .
\end{eqnarray}
Another form of this Green's function is obtained from General Axi-Symmetric Potential Theory~\cite{Weinstein, Linet1977}:
\begin{eqnarray}
\label{eq:app4}
\fl
G_3(\mathbf{x},\mathbf{x'})=\frac{1}{4 \pi^{2} \alpha}& \sum_{m=-\infty}^{\infty}\rme^{im(\phi-\phi')}\nonumber\\
 & \int_{0}^{\pi} \frac{(\rho \rho')^{|m|/\alpha} \sin^{2|m|/\alpha} \Psi }{[(z-z')^{2}+\rho^{2}+\rho'^{2}-2 \rho \rho' \cos\Psi]^{(|m|/\alpha+1/2)}} \rmd\Psi.
\end{eqnarray}
These expressions are valid for all $\alpha \in \mathbb{R}_+$; equating any one of these expressions with another gives us a formula for the functions involved. Some of these results are well known and can be found in volumes such as  Gradsteyn \& Rhyzik~\cite{gradriz}, nevertheless, we have non-obvious, important results, for example,  equating (\ref{eq:app1}) with (\ref{eq:app3}) and taking $r\rightarrow r'$, we obtain
\begin{eqnarray}
\label{eq:app5}
\fl \sum_{l=|m|}^{\infty}
\frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)}&& P^{-|m|/\alpha}_{\lambda}(\cos\theta)P^{-|m|/\alpha}_{\lambda}(\cos\theta')=\nonumber\\
&&
\frac{1}{\pi (\sin\theta\sin\theta')^{1/2}} Q_{|m|/\alpha-1/2}\left(\frac{1-\cos\theta\cos\theta'}{\sin\theta\sin\theta'}\right).
\end{eqnarray}

There are also a set of formulae that hold for a more restrictive set of $\alpha$ values. Linet \cite{LinetCosmicString1987} has shown that, when $\alpha > 1/2$, the Green's function may be written as
\begin{equation}
G_3(\mathbf{x},\mathbf{x'}) = \frac{1}{4 \pi \sigma}+\frac{1}{8\pi^{2}\alpha} \int_{0}^{\infty} \frac{1}{R} F_{\alpha}(u,\phi-\phi') \rmd u
\end{equation}
where
\begin{eqnarray}
\sigma&=[\rho^{2}+\rho'^{2}+(z-z')^{2}-2\rho\rho' \cos\alpha(\phi-\phi')]^{1/2} \nonumber\\
R&=[\rho^{2}+\rho'^{2}+(z-z')^{2}+2\rho\rho'\cosh u]^{1/2}
\end{eqnarray}
and
\begin{eqnarray}
\fl
F_{\alpha}(u,\Psi) = \frac{\sin(\Psi-\pi/\alpha)}{\cosh(u/\alpha)-\cos(\Psi-\pi/\alpha)}-\frac{\sin(\Psi+\pi/\alpha)}{\cosh(u/\alpha)-\cos(\Psi+\pi/\alpha)}.
\end{eqnarray}
Equating this to (\ref{eq:app1}) for $\alpha>1/2$ and taking $r\rightarrow r'$, we find
\begin{eqnarray}
\fl
& \frac{1}{\alpha}\sum_{m=-\infty}^{\infty}e^{im (\phi-\phi')}\sum_{l=|m|}^{\infty}\frac{\Gamma(\lambda+|m|/\alpha+1)}{\Gamma(\lambda-|m|/\alpha+1)} P^{-|m|/\alpha}_{\lambda}(\cos\theta)P^{-|m|/\alpha}_{\lambda}(\cos\theta')\nonumber\\
\fl &\qquad =\frac{1}{ [2(1-\cos\alpha\gamma)]^{1/2}}+\frac{1}{2\pi\alpha}\frac{1}{(2 \sin\theta \sin\theta')^{1/2}} \int_{0}^{\infty}\frac{F_{\alpha}(u,\phi-\phi')}{[\cosh\xi+\cosh u]^{1/2}} \rmd u
\end{eqnarray}
where
\begin{eqnarray}
\cos\alpha\gamma&= \cos\theta\cos\theta'+\sin\theta\sin\theta'\cos\alpha(\phi-\phi') \nonumber\\
\cosh\xi &= \frac{1-\cos\theta\cos\theta'}{\sin\theta\sin\theta'}.
\end{eqnarray}
This formula has the benefit of relating a 3D mode-sum expression to an expression that isolates the 3D Hadamard singularity structure
(the first term on the RHS) together with a regular term that depends on the boundary conditions. We have used this result to prove the regularity of the mode-sum that arises in the calculation of the vacuum polarization on the exterior region of the spacetime \cite{CSHorizon}.

Another form of the Green's function can be given in terms of the half-integer Legendre functions by considering the Green's function in toroidal coordinates, which are related to the cartesian coordinates by
\begin{eqnarray}
\fl x=\frac{\sinh\mu \cos\phi}{\cosh\mu-\cos\eta} \qquad y=\frac{\sinh\mu\sin\phi}{\cosh\mu-\cos\eta}\qquad  z=\frac{\sin\eta}{\cosh\mu-\cos\eta},
 \end{eqnarray}
 where the ranges of the coordinates are
$ 0\le\mu <\infty$, $0\le\eta < 2 \pi$,  $0\le\phi < 2 \pi$.
The mode-sum form of the Green's function obtained by separating in these coordinates is
\begin{eqnarray}
\fl
&G_3(\mathbf{x},\mathbf{x'})= \frac{1}{4\pi^2 \alpha}\sum_{m=-\infty}^{\infty}\rme^{im(\phi-\phi')}\sum_{n=-\infty}^{\infty}e^{i n(\eta-\eta')}[(\cosh\mu-\cos\eta)(\cosh\mu'-\cos\eta')]^{1/2} \nonumber\\
\fl&\qquad
   \rme^{|m|/\alpha \pi i}\frac{\Gamma(n+|m|/\alpha+1/2)}{\Gamma(n-|m|/\alpha+1/2)}P_{n-1/2}^{-|m|/\alpha}(\cosh\mu_{<}) Q_{n-1/2}^{-|m|/\alpha}(\cosh\mu_{>}) .
\end{eqnarray}
Equating this to (\ref{eq:app3}) gives us the following addition theorem
\begin{eqnarray}
\fl
&\sum_{n=-\infty}^{\infty}\rme^{i n(\eta-\eta')}e^{|m|/\alpha \pi i}\frac{\Gamma(n+|m|/\alpha+1/2)}{\Gamma(n-|m|/\alpha+1/2)}P_{n-1/2}^{-|m|/\alpha}(\cosh\mu_{<}) Q_{n-1/2}^{-|m|/\alpha}(\cosh\mu_{>}) \nonumber\\
\fl&\qquad
=\frac{Q_{|m|/\alpha-1/2}(\chi)}{[\sinh\mu\sinh\mu'(\cosh\mu-\cos\eta)(\cosh\mu'-\cos\eta')]^{1/2}}.
\end{eqnarray}

The final coordinate system we shall investigate is prolate spheroidal coordinates which may trivially extended to the oblate spheroidal case. The relationship between prolate spheroidal coordinates and cartesian coordinates is
given by
\begin{equation}
\fl
x=\sinh\sigma\sin\theta\cos\phi \qquad y=\sinh\sigma\sin\theta\sin\phi  \qquad z=\cosh\sigma\cos\theta.
\end{equation}
where the ranges of the coordinates are $ 0\le\sigma <\infty$, $0\le\theta\le \pi$,  $0\le\phi
< 2\pi$.
The mode form of the Green's function obtained by separating in these coordinates is
\begin{eqnarray}
\fl
& G_3(\mathbf{x},\mathbf{x'}) = \frac{1}{4\pi\alpha} \sum_{m=-\infty}^{\infty} e^{i m (\phi-\phi')}\sum_{l=|m|}^{\infty}(2\lambda+1)e^{|m|/\alpha \pi i}\frac{\Gamma(\lambda+|m|/\alpha+1)^{2}}{\Gamma(\lambda-|m|/\alpha+1)^{2}}  \nonumber\\
\fl & \qquad P_{\lambda}^{-|m|/\alpha}(\cos\theta)P_{\lambda}^{-|m|/\alpha}(\cos\theta')P_{\lambda}^{-|m|/\alpha}(\cosh\sigma_{<})Q_{\lambda}^{-|m|/\alpha}(\cosh\sigma_{>})
\end{eqnarray}
We shall write down only one possible summation formula here, by equating the Green's function expression above to (\ref{eq:app3}), we obtain a summation formula for a product of four Legendre functions:
\begin{eqnarray}
\fl &
\sum_{l=|m|}^{\infty}(2\lambda+1)e^{|m|/\alpha \pi i}\frac{\Gamma(\lambda+|m|/\alpha+1)^{2}}{\Gamma(\lambda-|m|/\alpha+1)^{2}}  P_{\lambda}^{-|m|/\alpha}(\cos\theta)P_{\lambda}^{-|m|/\alpha}(\cos\theta')  \nonumber\\
\fl & \qquad P_{\lambda}^{-|m|/\alpha}(\cosh\sigma_{<})Q_{\lambda}^{-|m|/\alpha}(\cosh\sigma_{>})
=\frac{1}{\pi\alpha} \frac{Q_{|m|/\alpha-1/2}(\chi)}{\sqrt{\sinh\sigma\sinh\sigma'\sin\theta\sin\theta'}}
\end{eqnarray}
where
\begin{equation}
\fl
\chi=\frac{\cosh^2\sigma+\cosh^2\sigma'-\sin^2\theta-\sin^2\theta'-2\cosh\sigma\cosh\sigma'\cos\theta\cos\theta'}{2\sinh\sigma\sinh\sigma'\sin\theta\sin\theta'}
\end{equation}
There are many more summation and addition formulae that can be derived in this way. Those presented here were either of particular use to us elsewhere or are to the best of our knowledge previously unpublished.

\section*{Acknowledgements}

PT is supported by the Irish Research Council for Science, Engineering and Technology, funded by the National
Development Plan.

\section*{References}

\bibliographystyle{unsrt}

config: arxiv


TITLE: What is the Jacobian-determinant of this problem?
QUESTION [1 upvotes]: Calculate the area of the walls of the solid given by $x^2+y^2=1$, $z=0$ and $z=1+xy$. The walls mean the area between $z=0$, and $z=1+xy$.
I know that I can solve this problem by calculating the area of the surface by treating the walls as a toilet paper, unfold the toilet paper on a flat surface,  and do a double integral over the toilet paper with the function$f(x,y)$ inside of the double inegral set to 1.
$\int _0^{2\pi}\:\int _0^{1+cos\left(\theta \right)\cdot \:sin\left(\theta \right)}\:1\:dz\:d\theta$
For this problem though I am more interested in using the formula given by:
$\int \:\int _A\left|\frac{\partial r}{\partial u}\times\frac{\partial r}{\partial v}\left(u,v\right)\:\right|du\:dv$
And I am unsure of the parametrization of this problem and then what the jacobian-determinant becomes.
I think the correct parametrization is:
$r\left(\theta ,z\right)=cos\left(\theta \right)i+sin\left(\theta \right)j+\left(z\right)k\:$
Choosing $r(\theta,z)$ as above $\left|\frac{\partial \:r}{\partial \:u}\times \frac{\partial \:r}{\partial \:v}\left(u,v\right)\:\right|$ becomes 1. I generally have a problem with these things when the functions aren't given on the form $z=..$
But then, the question becomes, if that is the correct parametrization, what is the jacobian-determinant?

REPLY [1 votes]: Everything you say is correct. You have the parametrization
$${\bf r}(\theta,z)=(\cos\theta,\sin\theta,z)\ ,\tag{1}$$
giving
$$|{\bf r}_\theta\times{\bf r}_z|\equiv1\ ,$$
hence the surface element on $S$ is given by ${\rm d}\omega={\rm d}(\theta,z)$. There is no Jacobian determinant here, but a scaling factor from the measure ${\rm d}(\theta,z)$ in the parameter plane to the surface measure ${\rm d}\omega$ in $3$-space. This scaling factor happens to be $1$ in the case at hand. You end up with the integral of your first attempt, obtained using "geometric intuition". Given the parametrization $(1)$ the condition  $0\leq z\leq 1+x y$ translates into $0\leq z\leq1+\cos\theta\sin\theta$.


TITLE: Convergence Proof!
QUESTION [0 upvotes]: The truth is, I don't even know where to begin with this question... Can someone please offer their assistance?

REPLY [1 votes]: By the definition of the infimum $\inf S$ we have
$$\forall \epsilon>0\quad \exists x\in S\ \text{such that}\ \inf S\leq x<\inf S+\epsilon$$
so take $\epsilon=\frac 1 n, \ n\in\mathbb{N}^*$ we find $x_n\in S$ such that
$\inf S\leq x_n<\inf S+\frac 1 n$ and passing to the limit in the last inequalities gives $\displaystyle\lim_{n\to\infty}x_n=\inf S$.


TITLE: Variance of 1/X
QUESTION [0 upvotes]: Given a random variable $X$ that is exponentially distributed, what is the variance of $\frac{1}{X}$, i.e. $\operatorname{Var}\frac{1}{X}?$
Since $\operatorname{Var} X = \frac{1}{\lambda^2}$, is it $\operatorname{Var}\frac{1}{X}=\lambda^2$?

REPLY [1 votes]: If $X\sim \mbox{Exp}(\lambda)$,  then $X^{-1}$ has distribution function $F(z)=e^{-\lambda/z}$. In particular it's expected value does not exist and hence neither does it's variance.


TITLE: Normal Distribution sample mean and population mean?
QUESTION [1 upvotes]: Assume that house prices in an area are normally distributed with a standard deviation of \$ 20,000. A random sample of 16 houses is taken. What is the probability that the sample mean differs from the population mean by more than \$ 5,000.

I can write two equations as follows
For case I - $N(\mu_1, 20000)$
For case II - $N\left(\mu_2, \sqrt{\frac{20000^2}{16}}\right)$
But in both cases $\mu_1 = \mu_2$ right? Can anyone give me a tip?

REPLY [1 votes]: The arithmetic average of $K$ equally normal-distributed quantities has the same mean $\mu$; but the standard deviation decreases by the factor of $\sqrt{K}$. In this case, $K=16$ so the standard deviation of the mean will be $20,000/\sqrt{16} = 5,000$ dollars. It just happens to be the value equal to the deviation in the last, main question you asked.
So the question is what's the probability that the quantity (average of 16 prices) differs from the mean by more than 1 standard deviations. This value is well-known numerically. The probability that the quantity is "within one sigma" is 68.3% or so. The probability that the deviation exceeds one standard deviation is 37.3% and that's the answer to your question.


TITLE: Erdos distance problem n=12
QUESTION [15 upvotes]: The recent paper On the Erdos distinct distance problem in the plane
Authors: Larry Guth, Nets Hawk Katz prodded me to get a non-trivial example.  Here is what I cannot find:  an example of 12 distinct points in the plane with only 5 different distances between points.  The regular 12-polygon has 6 different lengths but I cannot do better. http://oeis.org/A186704 implies that there is one>

REPLY [5 votes]: Thought it might be nice to show the set placed on the hexagonal lattice.

Here’s a Desmos link if you wish to play around with such point configurations.


TITLE: How to find the curl, divergence, gradient and laplacian of functions without calculating the individual indices
QUESTION [1 upvotes]: I know the definitions of the $\nabla$ operators but I don't understand how to find something like this:
Given the wave equation $\vec E = \vec a \sin(\vec k \cdot \vec r - \omega t)$
find $\vec \nabla (\vec \nabla \cdot \vec E)$
I could start calculating it all in components but that seems a bit unnecessary since the $\nabla$ operator is comparable to taking the derivative to $\vec r$. I just don't understand what the difference between $\vec \nabla$, $\vec \nabla\cdot$, $\vec \nabla\times$ and $\vec\nabla^2$ would be.

REPLY [1 votes]: Note that if $a$ is a constant vector,
$$ \nabla \cdot (af) = \partial_i (a_i f) = a_i \partial_i f = (a \cdot \nabla)f = a \cdot (\nabla f), $$
using summation convention. You also then have that if $f$ is a function of $k \cdot r$, then
$$ (\nabla f(k \cdot r))_i = \partial_i f(k_j r_j) = k_j (\partial_i r_j) f'(k_j r_j) = k_j \delta_{ij} f'(k_j r_j) = (k f'(k \cdot r))_i, $$
using the chain rule. These two identities should make it much easier for you to compute $\nabla(\nabla \cdot E)$.


TITLE: Matching numbers and colors
QUESTION [1 upvotes]: I'm sorry I can't ask this problem in a more formal way, the issue being that I'm not even sure what type of math this involves; I think it actually translate to a graph-coloring problem?
FWIW, it's about figuring how many races I need so I can distribute them among guilds/factions for a game.

I have the number 1-10, separated between even and odd numbers.
Each number needs to have at least three colors.
No more than two odd or two even numbers can share a given color, but colors can be shared otherwise between odd and even numbers.
Each number has a unique set of colors.
This include not sharing all colors with a number that can be a greater number of colors.

Given these parameters, how many colors am I going to need? Does that minimum changes if I allow...

For a number that can only be 2 color?
For a color that can be shared by 4 numbers?
for both of the above?

REPLY [1 votes]: As I understand it, each color can appear at most 4 times: on two odd teams, and two even teams. Since there are at least 30 slots to fill, we certainly need at least $30/4=7.5$ colors. Indeed, 8 will suffice.
Say the colors are ABCDEFGH. One (uninspired) arrangement satisfying the requirements is:

ABC
ABD
ABE
ABF
FGH
EGH
DGH
CGH
CDE
DEF

Perhaps you can arrange them in a more varied manner.
If the teams only require two colors each, then five colors suffice:

AB
AC
AE
AD
BC
BD
CD
BE
DE
CE


TITLE: Prove that $\mathbb{Z}^{\mathbb{N}}$ is not isomorphic to $\mathbb{Z}^{\oplus \mathbb{N}}$
QUESTION [1 upvotes]: I am not quite sure how to show the following:

Let $\mathbb{Z}^{\oplus \mathbb{N}}$ be set of functions $f:\mathbb N \to \mathbb Z$ such that $f(i) \ne 0$ for finitely many $i\in \mathbb N$. Let $\mathbb Z^{\mathbb N}$ be set of functions from $\mathbb N \to\mathbb Z$. Show that $\mathbb{Z}^{\mathbb{N}}$ is not isomorphic to $\mathbb{Z}^{\oplus \mathbb{N}}$.

So far, I know that $\mathbb{Z}^{\oplus \mathbb{N}}$ is generated by $f_i:\mathbb N \to \mathbb Z$ for $i \in \mathbb N$ where $f_i(x)=1$ if $x=i$ and $0$ otherwise. In other words, every $f\in \mathbb{Z}^{\oplus \mathbb{N}}$ can be represented as a finite sum
$\sum m_i f_i$ where $m_i\in \mathbb Z$. So the question above is
equivalent to asking that whether $\mathbb Z^{\mathbb N}$ is generated
by a countable set of functions $f:\mathbb N \to \mathbb Z$.
Any help is appreciated. Thanks!

REPLY [5 votes]: Isomorphisms are defined to be bijections. So if two algebraic structures are isomorphic their corresponding carrier sets have the same cardinality.
Starting with
1) $\text{#}( \mathbb{Z}^{\oplus \mathbb{N}}) = \aleph _{0}$.
2) $\aleph _{0} \lt \text{#}(2^{\Bbb N}) = \mathfrak {c}$
use celtschk's hint

Construct an injection from $ 2^{\Bbb N}$ into $\mathbb{Z}^{\mathbb{N}}$ to show that $\mathfrak {c} \le \text{#}(\mathbb{Z}^{\mathbb{N}})$.

Note that by using a forgetful functor, the problem of analyzing algebraic morphisms melts away.

The OP can show that $\mathbb{Z}^{\oplus \mathbb{N}}$ is countable by using the following technical machinery
(see Wikipedia: Countable set):
Proposition: If $A_n$ is a countable set for each $n \in \Bbb N$ then the union of all $A_n$ is also countable.
and defining for each $n \in \Bbb N$ the set
$$ F_n = \bigr\{\sum_{k=0}^n m_k f_k \, | \, -n \le m_k \le n \bigr\}$$


TITLE: $Z(G)$ is not maximal subgroup of G
QUESTION [2 upvotes]: How to prove that $Z(G)$ is not a maximal subgroup of G, where G is an arbitrary group?
Thanks in advance.

REPLY [5 votes]: The hint given in the comments is probably the easiest way to do this. Here's an idea for a different solution.
Hint 1: If a maximal subgroup $M$ is normal in $G$, then $G/M$ is cyclic.
Hint 2: If $G/Z(G)$ is cyclic, then $G$ is abelian.


TITLE: In how many of the integer numbers between 0 and 10,000 does the digit 3 appear to the left of 4
QUESTION [18 upvotes]: In how many of the integer numbers between $0$ and $10\,000$ does the digit $3$ appear some place to the left of the digit $4$? This would include, for example, the numbers $34$, $374$, $4384$ and $3874$, but would not include $27$, $43$, $3650$ or $4333$.
So I am thinking you do $$8\cdot9\cdot1\cdot1+1\cdot9\cdot1+8\cdot1\cdot9\cdot1+1\cdot9\cdot9\cdot1 = 234$$
But I feel like that is too low.

REPLY [1 votes]: As David shows, the formula for this with $d$ digits is
$$10^d - (d+9)9^{d-1}$$
We can prove this without the recurrence by showing that there are $9^d + d9^{d-1}$ ($=(d+9)9^{d-1}$) numbers where $3$ does not appear before $4$.
There are $9^d$ $d$-digit string with no $4$ at all, giving us $9^d$.
Every other $d$-digit string has at least one $4$.  Suppose the rightmost occurrence of $4$ is at digit $i \in [1, d]$, then every number before it has $9$ choices (anything but $3$, by assumption) and every number after it has $9$ choices (anything but $4$).  This gives us our $d9^{d-1}$.
Adding the two together gives us all numbers where $3$ is not before $4$, and since there are $10^d$ digits, the answer above is shown to be correct.


TITLE: In how many ways can we select $5$ children from this group so that Mary and Jane are always in the selection?
QUESTION [0 upvotes]: In a group of $10$ children no two kids have the same name. We know that Mary and Jane are among these children. In how many ways can we select $5$ children from this group so that Mary and Jane are always in the selection?
So I started by looking at all the possible outcomes
$$\frac{n!}{r!(n-r)!}  \implies  \frac{10!}{5!(10-5)!} = 6048$$
from here I get stuck and I don't know how to proceed.
Thanks for your help in advance.

REPLY [1 votes]: If Mary and Jane are always in the selection, then $3$ more children have to be selected from the remaining $8$ children.  Number of ways of selecting $3$ children from $8$ children is
$$= \frac{8!}{(8-3)! \cdot 3!} = 56$$


text:
\section{Conclusion}
\label{sec:conclusion}

We proposed unbiased first-order bilevel optimization (UFO-BLO) -- a modification of first-order bilevel optimization (FO-BLO) which incorporates unbiased gradient estimation at negligible cost (same memory and expected time complexity). UFO-BLO with a SGD-based outer loop is guaranteed to converge to a stationary point of the BLO problem while having a strictly better $O(1)$ memory complexity than the naive BLO approach. We demonstrate a rich family of BLO problems where FO-BLO ends up arbitrarily far away from the stationary point.

\section{Acknowledgements}

Adrian Weller acknowledges support from the David MacKay Newton research fellowship at Darwin College, The Alan Turing Institute under EPSRC grant EP/N510129/1 and U/B/000074, and the Leverhulme Trust via CFI.


TITLE: Gravitational lensing of gravity
QUESTION [8 upvotes]: I have read these questions:
Do gravitational lenses work on gravitational waves?
Are gravitational waves effected by the curvature of space time (gravitational lensing)?
And it made me curious.
I understand that gravitons and gravitational waves are not the same thing.
But this question here:
Gravitons unaffected by gravitational lensing?
It is talking about LIGO experiment and that we detect gravitons by that experiment. And it says that so gravitons do cause gravity and bend spacetime.
So from this question it is not understandable whether LIGO experiment is detecting gravitational waves or gravitons and whether gravitons themselves cause bending of spacetime. I would think they do not, because it is stress energy that bends spacetime.
So it is not understandable whether gravitons are mediating the bending of spacetime, in which case the Sun would shadow most of these effects (depending on whether gravitons are effected by gravitational lensing).
Or is it that it is stress energy that bends spacetime and it does not matter whether the Sun shadows these effects (the Sun is inbetween a far away star and the Earth) or not, the bending of spacetime (from a far away star) still reaches us regardless whether the Sun is there or not.
Imagine there is a star at point A far away from Earth, where Earth is point B, and inbetween the Sun.
We know as per GR that there is gravitational lensing (of photons) because of the Sun's gravitational field. The Sun's stress energy bends spacetime and the photons coming from the far away star are traveling along a bent path.
None of the questions says anything about gravitons following a geodesic or gravitons being affected by bent spacetime.
They did not say anything about the Earth being affected by the gravitational effects of a far away star that would be behind the Sun.
In this case, when a far away star is behind the Sun, I do not know whether the far away star's gravitational effects are still reaching Earth or does the Sun shadow these effects?
I did not find anything whether the Earth itself would be shadowed (gravitationally) by the Sun as an object. And I did not find anything whether gravitons from this far away star would still reach us (like photons do).
Questions:

Do gravitons travel along geodesics, and do they travel along bent spacetime? Is my graph correct with the graviton's path being bent like a photon's path?
If a far away star is behind the Sun, but we still see it (photons still reach us from there) because of gravitational lensing, then do gravitons (and other effects of gravity) reach us from that far away star too?

REPLY [2 votes]: We still don't know if gravitons really exist. We need a proper theory of Quantum Gravity for that. Of course,  in a sense, no particles really exist, they're just a convenient way to model the excitations of quantum fields.
Individual gravitons are far too weak to be detectable with current technology, LIGO can only detect big gravitational waves. It's like the difference between being able to detect a big water wave vs detecting the motion of single water molecules. Or the difference between being able to detect strong light vs single photons.
Yes, gravitons (if they exist) follow geodesics, just like everything else does, so they would be subject to gravitational lensing. One big difference with photons is that gravity isn't blocked by anything, so stars are transparent to gravitons. I don't know how black holes fit into the picture.
Another important difference to photons (which has been mentioned in the comments) is that gravitons carry gravitational charge, purely because they carry energy. So in a way they're more like gluons, the carriers of the strong nuclear interaction, rather than being like photons. However, there's only one type of gravitational charge, unlike the various colors associated with gluons.


TITLE: Multivariable limit studying discontinuity at $x = y$ line
QUESTION [1 upvotes]: This exercise is asking me to study the continuity of the following function:
$$f(x, y) = \frac{\ln (1+x^4-y^4)}{(x^2-y^2)^2}$$
I check the domain of the function: $$1+x^4-y^4 > 0$$ and the important part is that we have to exclude those points from the $\;y = x\;$ and $\;y = -x\;$ lines. For $\;y = x\;$ I followed this approach:
$$\begin{align}\lim_{y \to x^+}\frac{\ln (1+x^4-y^4)}{(x^2-y^2)^2} \approx \\\lim_{y \to x^+}\frac{x^4-y^4}{(x^2-y^2)^2} =
\\\lim_{y \to x^+}\frac{(x^2+y^2)(x^2-y^2)}{(x^2-y^2)^2} =
\\\lim_{y \to x^+}\frac{x^2+y^2}{x^2-y^2} = -\infty\end{align}$$
Also:
$$\lim_{y \to x^-}\frac{x^2+y^2}{x^2-y^2} = +\infty$$
Same would work with $\;y = -x.\;$ I wonder if this is correct as a friend told me I can't do this because either one of the variables must be constant. Anyway, I'm looking for a way to show that the function goes to infinity when $\;(x, y)\;$ goes near the $\;y = x\;$ and $\;y = -x.$
Another approach I've come with is to cut, for example, the 3d graph with the plane $y = x,$ orthogonal to the $y = -x$ one, and then I would find the intersection, and therefore a single variable graph which would go to infinity at... $x = 0?$ The problem with this approach is that I have really no idea how to do it. It seems like I can't just substitute $y = x$ in the function, whatever that means. I have been looking for a similar question in the forum but all I've found were questions regarding single point discontinuities, and my question involves straight lines. Any other approach would be really helpful. This was my first question here, I hope I was clear enough. Thank you.

REPLY [0 votes]: After having thought about this for a while, I havent't quite figured it out, however I have a suggestion. If you factor $(x^2 - y^2)^2$ as $(x+y)^2 (x-y)^2$ then you can rewrite the original expression and have one factor that does not evaluate to $0$ when taking the limit as $x$ approaches $y$. Now you are left with an indeterminate limit, so you can use L'Hopital's rule after this. All of this should yield an expression you can continue to manipulate, since you have gotten rid of the squared terms inside the paranthesis in the numerator in the original expression. This is as far as I've tried to work the problem. It seems promising though, so it might be worth a shot. I tried plotting the graph and it looks a little weird, so you might not have a well defined limit at $x=y$, meaning you should consider the limit when $x$ goes to $y$ from above and when $x$ goes to $y$ from below.
Hope this gets you somewhere if you decide to try it out! Best of luck.


TITLE: What is the specific meaning of "Fourier frequency" (as opposed to simply "frequency")?
QUESTION [6 upvotes]: I've noticed that many journal articles (in optics) use the phrase "Fourier frequency" to describe, well, the frequency of something.

Google scholar search for "Fourier frequency".

Example:

To compare with previous sensitivity curves of a single LISA
  Michelson interferometer, we construct the SNRs as a function of
  Fourier frequency for sinusoidal waves from sources uniformly
  distributed on the celestial sphere.

This phrase seems redundant to me.  Does the adjective "Fourier" add anything?  Is it trying to make a distinction of audio frequencies from optical frequencies?  Or temporal frequencies from spatial frequencies?

REPLY [0 votes]: In many cases people do seem to say "Fourier frequency" when they mean "frequency".
However, when dealing with data defined only on discrete time points the phrase "Fourier frequency" has important meaning.
Consider a sequence of $N$ values $\{ x_n \}$ where $n \in \{1, 2, \ldots N \}$.
This situation comes up all the time if we have a physical signal $x(t)$ which we sample at times $t_n = n \Delta t$, in which case of course $x_n = x(t=n\Delta t)$.
We can analyze the frequency content of this signal by computing its discrete Fourier transform (DFT)$^{a}$
$$ X_k \equiv \sum_{n=1}^N x_n \exp \left( -i 2 \pi n k / N \right) \, . $$
This sum expresses the sequence $\{ x_n \}$ as a sum of sinusoidal signals $\exp(i 2 \pi n k / N)$.
The frequencies of these sinusoids (i.e. the number of radians of oscillation per time step) are
$$\omega_k \equiv 2 \pi k / N \, .$$
These frequencies are some times called "Fourier frequencies".
As noted in Emilio's and Magpie's answers, it can be important to distinguish the notion of a frequency component of a signal from that signal's true underlying frequency.
However, in my experience as an experimental physicist who does a fair amount of signal processing, the phrase "Fourier frequency" is not used in that way.
In fact the usage quoted in the original question is pretty foreign to my ears/eyes, and it seems awkward since the word "frequency" would serve just fine.
I have only seen and used "Fourier frequency" in the way described here and in Oleg's answer.
$[a]$: Many people refer to the discrete Fourier transform as the "FFT". In fact, FFT stands for "fast Fourier transform", which is a particular algorithm for computing the discrete Fourier transform.


TITLE: How can I resolve a matrix of 13 rows and 40 columns?
QUESTION [0 upvotes]: I need to resolve an undeterminated matrix that corresponds to an Flux balance Analysis applied on Glicolysis pathway, I have 19 metabolites in the rows, and 40 fluxes in the columns. I want to know about its null space, rank, and space dimensions...
Some programs? or any methods? Some ideas?

REPLY [1 votes]: You need to be a little more specific about what you want to achieve but here's something you could get started with:
Fire up Octave.
Relevant Functions:
a = rand(19,40) %Just to get a random matrix to work with.
null(a) %Null space
rank(a) %Rank
rref(a) %Row Reduced Echelon Form

Anything else you need?


TITLE: About the decomposition of a rank 2 tensor into its irreducible components
QUESTION [2 upvotes]: A rank 2 tensor $T_{ij}$ of 3D rotation group $SO(3)$ is a reducible representation. It has the decomposition $9=5+3+1$ where 5 is the symmetric traceless tensor, 3 is the vector and 1 is the scalar. When the rotation group acts on a vector, the representative matrix is a $3\times 3$ orthogonal matrix with unit determinant.
After we make this decomposition into irreducible tensors by block diagonalization of a $9\times 9$ matrix into a $5\times 5$, $3\times 3$ and $1\times 1$ blocks, I have some questions.

What kind of matrices are the $9\times 9$ square matrices which act on the $9\times 1$ general tensor $T_{ij}$?
What kind of matrices are the $5\times 5$ square matrices which act on the $5\times 1$ symmetric traceless tensor?

When I ask what kind I am asking if they are also orthogonal and having unit determinant. I guess not. In that case, what kind they are. I am quite sure the $3\times 3$ & $1\times 1$ matrices that act on $3$ and $1$ are respectively orthogonal with determinant one and the number 1 respectively?

REPLY [2 votes]: Take the $3\times 3$ rotation matrix $R(\Omega)$ and tensor it with itself: $R(\Omega)\otimes R(\Omega)$ will be a $9\times 9$ matrix acting on the components of the $9\times 1$ $T_{ij}$.
As you have guessed, $R(\Omega)\otimes R(\Omega)$ is reducible.  The $5\times 5$ block contains the Wigner D-matrices for $L=2$, i.e the $D^2_{MM'}(\Omega)$ functions.  The $3\times 3$ block will contain $D^{1}_{mm'}(\Omega)$ matrices and the $1\times 1$ block is the scalar rep with $D^0=1$.  All these are representations of $SO(3)$ so are unitary with determinant $+1$.


text:
\section{Details of the Gibbs algorithm} \label{sec:details algo}

We detail the algorithm used to sample from the posterior distribution $(\overline{\lambda}, A, \gamma)| \Nb$ in the Poisson process context, in the most complete case (with a hierarchical level on $\gamma$).  In case where $\gamma$ is set to a fixed value, then the corresponding part in the algorithm is removed.

\noindent As a standard Gibbs algorithm, the simulation is decomposed into  three steps:
$$[1] \quad  \overline{\lambda} | A, \gamma, \Nb \quad \quad [2] \quad    A | \overline{\lambda}, \gamma, \Nb   \quad\quad [3]  \quad \gamma | A, \overline{\lambda}, \Nb. $$
where $N$ is the observed Poisson process over $[0,T]$,  namely a number of jumps $N(T)$ and jump instants $(T_1,\dots, T_{N(T)})$.
\noindent In order to avoid an artificial truncation in $\overline{\lambda}$, we use the slice sampler strategy proposed by \cite{fall2012}.
\noindent More precisely, we consider  the stick breaking representation of $\overline{\lambda}$. Let $c_i$ be the affectation variable of data $W_i$. The DPM model is written as:
$$W_i  | c_i, \btheta^* \sim  h_{\theta^{\star}_{c_i}}, \quad  P(c_i = k) = w_k,  \forall k \in \mathbb{N}^{*}\quad   (w_k)_{k \in \mathbb{N}^{\star}} \sim  Stick(A),  \quad (\theta^{*}_{k})_{k \in \mathbb{N}^{\star}} \sim_{i.i.d} G_{\gamma}.$$
The  slice sampler strategy consists in introducing a latent variable $u_i$ such that the joint distribution of $(W_i,u_i)$ is
$
 p(W_i,u_i|\bw,\btheta^{*}) = \sum_{k=1}^\infty w_k h_{\theta^{*}_{k}}(W_i) \frac{1}{\xi_k}\ind_{[0,\xi_k]}(u_i)$ with  $\xi_k = \min(w_k,\zeta)$, which  can be reformulated as:
 \begin{equation}\label{eq:u x jointe2}
 p(W_i,u_i |\bw,\btheta^*)=  \frac{1}{\zeta}\ind_{[0,\zeta]}(u_i)  \sum_{k=1,w_k >\zeta}^\infty w_k h_{\theta^{*}_{k}}(W_i) +   \sum_{k=1,  u_i \leq w_k \leq \zeta}^\infty h_{\theta^{*}_{k}}(W_i)\ind_{[0,w_k]}(u_i)
 \end{equation}
 $(w_k)_{k \geq 1} $ verifying  $\sum_{k \geq 1} w_k=1$  (implying $\lim_{k \rightarrow \infty} w_k =0$), the cardinal of $\{k, w_k > \varepsilon\}$ is finite for   every $\varepsilon >0$,  and  the sum in  (\ref{eq:u x jointe2}) is finite.

 \begin{remark}
 Note that the number of non-null terms in (\ref{eq:u x jointe2})  is rigorously  dependent of  the observation index $i$ (denoted $K_i^{*}$). But in the algorithm we will deal with the maximum of the $K^{*}_i$:  $$K^{*} = \max\{K_i^{*},i=1\dots N(T)\}$$
 \end{remark}

\noindent The Gibbs algorithm with the Slice sampler strategy now takes into account the latent variable $\bu=(u_1,\dots,u_{N(T)})$ which is sampled conjointly with $\overline{\lambda}$, resulting into the following steps:
$$[1^{\star}] \quad  \overline{\lambda},\bu | A, \gamma, \Nb \quad \quad [2^{\star}] \quad    A | \overline{\lambda}, \bu,\gamma, \Nb   \quad\quad [3^{\star}]  \quad \gamma | A, \overline{\lambda}, \bu,\Nb. $$
We now detail steps $[1^{\star}]$ , $[2^{\star}]$ and $[3^{\star}]$.

\subsection{Details of the algorithm}

\noindent \emph{Initialisation} The Gibbs algorithms are initialized on $ A^{(0)}=10$ ($0$ referring to the iteration number of the Gibbs algorithm). We set:
$K^{(0)}=  N_T$. for $k=1\dots K$, $(\theta^*)^{(0)}_k \sim G_{\gamma}$.
 When the hierarchical approach is considered on $\gamma$ ($\gamma \sim \Gamma(a_\gamma,b_\gamma)$), we initialize $\gamma$ on its prior mean value: $\gamma^{(0)}= \frac{a_{\gamma}}{b_{\gamma}}$.

\paragraph{$[1^{\star}]$ Sampling from  $ \overline{\lambda},\bu | A, \gamma, \Nb$}$\;$ \\

\noindent Step $[1^{\star}]$ of the Gibbs algorithm is  decomposed into $5$ steps which are detailed below. Let  $\bu = (u_1,\dots,u_{N(T)})$, $\bc=(c_1,\dots,c_{N(T)})$, $\btheta^* = (\theta_1,\dots,\theta_{K^{\star}})$, $\bw = (w_1,\dots,w_{K^{\star}})$ --$\bc$, $\btheta^*$ and $\bw$ representing $\overline{\lambda}$-- be the current object. We denote by $K_{N(T)}$ the number of non-empty classes:
$$ K_{N(T)} = \# \{k\in \{1\dots K^{\star}\} | \exists j \in \{1,\dots,N(T)\} \mbox{ such that }  c_j = k\} $$
$\btheta^*$ and $\bw$ are ordered such that the elements indexed from $K_{N(T)}+1$ to $K^{\star}$ correspond to empty classes.
$\bu$, $\bc$, $\btheta^*$ and $\bw$ are iteratively sampled as follows:

 \begin{enumerate}
\item[[$1^{\star}$.a]]  First we sample $\bu |  \bw,\btheta^*,\bc,\Nb,A, \gamma$ using the following identities:
\begin{eqnarray*}
p(\bu | \bw,\btheta^*,\bc,\Nb,A, \gamma) &\propto& p(\bu,\Nb |  \bw,\btheta^*,\bc) =  p(\bu,\Nb |   \btheta^*,\bc) = \prod_{i=1}^{N(T)} p(u_i,W_i | c_i,\theta_{c_i}^{\star}) \\
&=&   \prod_{i=1}^{N(T)} h_{\theta^{\star}_{c_i}}(W_i) \frac{1}{\xi_{c_i}}\ind_{[0,\xi_{c_i}]}(u_i) \propto \prod_{i=1}^{N(T)}   \frac{1}{\xi_{c_i}}\ind_{[0,\xi_{c_i}]}(u_i) \\
 \end{eqnarray*}
 where $\xi_{c_i} =\min(w_{c_i},\zeta)$.  So for every $i=1\dots N(T)$, $u_i \sim \mathcal{U}_{[0,\min(w_{c_i},\zeta)]}$.

\item[[$1^{\star}$.b]]  Secondly, we sample the weights of the empty classes $(w_{k})_{k\geq K_{N(T)}+1} | \Nb,\bw,\bc,\btheta^*,A, \gamma$. Without the slice sampler strategy, there is an infinite number to sample. But, thanks to the slice sampling, we only need to sample a finite number $K^{\star}$.  The weights of the empty classes follow the prior distribution (stick breaking).  For $k>K_{N(T)}$,
\begin{eqnarray*}
v_k &\sim& \mathcal{B}(1,A)\\
w_k &=& v_k r_{k-1}\\
r_k &=& r_{k-1} (1-v_k)
\end{eqnarray*}
As explained in \cite{fall2012}, we know that we have to represent all the components such that their weights $w_k > u_i$ for all the $u_i$. Assume that we have sampled $w_1,\dots, w_k$, then the weights of the following components can not exceed the rest $1- \sum_{j=1}^k w_j = r_{k}$. So if $r_k$ is such that $r_k < u_i$, for all $i=1\dots n$, i.e. if $r_k < u^{\star}$ with $u^{\star}= \min\{u_1,\dots,u_{N(T)}\}$ we are sure that there is no ``interesting component" after that,  ``interesting" meaning that they won't appear in joint the distribution of $(\Nb,\bu)$. We can stop and get
$$ K^{\star} = \min \{k, r_k < u^{\star}\}.$$
 In the end we have sampled $(w_{K_{N(T)}+1},\dots,w_{K^{\star}})$.

\item[[$1^{\star}$.c]]   Sample the parameters of the empty classes, $(\theta^*_{K_{N(T)}+1},\dots,\theta^*_{K^{\star}})$
 $$\forall k=K_{N(T)}+1, \dots, K^{\star}, \quad \theta^*_{k} \sim_{i.i.d} G_{\gamma}  $$

\item[[$1^{\star}$.d]]  Sample the index $\bc = (c_1,\dots,c_{N(T)}) | \Nb,\bu,\btheta^*,\bw$, i.e. affect the observations to the classes  $\{1, \dots,K^{\star}\}$. Note that we will get new empty classes. The affectations are done using the following probabilities:
\begin{eqnarray*}
p(\bc | \btheta^*,\bw,\bu,\Nb) &\propto&p(\bc, \btheta^*,\bw,\bu,\Nb)=  p(\Nb,\bu, \bc|  \btheta^*,\bw) p(\bw) p(\btheta^*) \\
&\propto& \prod_{i=1}^{ N(T)}  p(W_i,u_i, c_i|  \btheta^*,\bw)
\end{eqnarray*}
So, the $c_i$ are independent and,  for $i=1\dots N(T)$, for $k=1\dots, K^{\star}$
\begin{eqnarray*}
p(c_i = k |  \btheta^*,\bw,u_i,W_i) = w_{i,k} \propto  h_{\theta^{*}_{k}}(W_i) \frac{w_k}{\min(\zeta,w_{k}) }\ind_{ \{k | u_i< \min(\zeta,w_{k})\}}(k)
\end{eqnarray*}

We obtain a new $K_{N(T)}$, which is the number of non-empty classes. We re-arrange the weights and the parameters by order of appearance in this  affectation.

\item[[$1^{\star}$.e]]   Update $(w_1,\dots,w_{K_{N(T)}})$ and  $(\theta_1,\dots,\theta_{K_{N(T)}})$ for the non-empty classes.

\begin{eqnarray}\label{loi theta}
 p(\theta_k | \bu,\Nb,\bw,\bc) &\propto& G_{\gamma}(\theta_k) \prod_{i=1, c_i=k}^n h_{\theta_k}(W_i), \quad \forall k=1\dots K_{N(T)}\\
w_1,\dots,w_{K_{N(T)}},r_{K_{N(T)}} &\sim& \mbox{Dir}(n_1,\dots,n_{K_{N(T)}},A) \nonumber
\end{eqnarray}
where $n_k = \#\{i | c_i=k\}$ and $r_{K_{N(T)}} = 1-\sum_{k=1}^{K_{N(T)}} w_k$.

Note that when $G_{\gamma}$ is the inverse of the translated inverse exponential distribution, $ p(\theta_k | \bu,\Nb,\bw,\bc)$ given in equation (\ref{loi theta}) is:
\begin{equation}\label{loi theta2}
p(\theta_k | \bu,\Nb,\bw,\bc) \propto  \frac{1}{\left(\frac{1}{\theta_k}-\frac{1}{T}\right)^{a-1}} e^{-\frac{\gamma}{\frac{1}{\theta_k}-\frac{1}{T}}}\frac{1}{\theta_k^{n_k}}\ind_{\left[\frac{1}{\max_{i | c_i=k} W_i} , + \infty \right[}(\theta_k)
\end{equation}
Its simulation can not be performed directly and we  resort to an accept-reject procedure to simulate exactly under this distribution.
\end{enumerate}

\begin{remark}
The accept-reject procedure we propose is detailed and discussed here after. Note that $(\theta_k)$ changes of dimension at each iteration of the algorithm. As a consequence, a Metropolis-Hastings procedure can not be considered easily, since it could jeopardize the theoretical and practical convergence properties of the algorithm.
\end{remark}

\paragraph{[$2^{\star}$.]  Sampling from  $A | \bw,\btheta^*,\bc, \bu, \gamma, \Nb$ }$\;$ \\

\noindent    Let  $K_{N(T)}$ be the current number of non-empty classes. \cite{west92} proves that under the prior distribution $A \sim \Gamma(a_\alpha, b_\alpha)$,
we have
\begin{equation}
A | x,K_{N(T)} \sim \pi_x \Gamma(a_{A}+K_{N(T)},b_{A}-\log(x))+(1-  \pi_x) \Gamma(a_{A}+K_{N(T)}-1,b_{A}-\log(x))
\end{equation}

where
\begin{eqnarray*}
x| A,K _{N(T)} &\sim& \mathcal{B}(A+1,N(T))\\
\frac{\pi_x}{1-\pi_x} &=& \frac{a_{A}+K_{N(T)}-1}{n(b_A-\log(x))}
\end{eqnarray*}
Note that the generation of the new value of $A$ relies on the current value of $A$.

\paragraph{$[3^{\star}.]$  Sampling from  $\gamma | \bw,\btheta^*,\bc, \bu,A, \Nb$}  If a hierarchical level is set on $\gamma$,  we have to sample
$$ \gamma | \btheta^*, \bw,\bu,\Nb,\bc$$
Using  the previous conditional distributions, we have:
\begin{eqnarray*}
p( \gamma | \btheta^*, \bw,\bu,\Nb,\bc ) &\propto& p(\bu,\Nb,\btheta^*,\bw,\bc|\gamma)\pi(\gamma) =   p(\bu,\Nb| \btheta^*, \bc )p(\bc | \bw) p(\btheta^*|\gamma) \pi(\gamma) =\pi(\gamma) \prod_{k=1}^{K^{\star}} G_{\gamma}(\theta_k)\\
&\propto& \gamma^{a_\gamma-1} e^{-b_\gamma \gamma}   \prod_{k=1}^{K^{\star}}  \gamma^a e^{-\gamma /(\frac{1}{\theta^{\star}_k}-\frac{1}{T})} =   \gamma^{a_\gamma +  a K^{\star}-1} e^{-\gamma\left(b_\gamma + \sum_{k=1}^{K^{\star}} \frac{1}{(\frac{1}{\theta^{\star}_k} - \frac{1}{T})} \right)}
\end{eqnarray*}
where $K^{\star}$ is the total number of classes used to represent $\overline{\lambda}$. Finally, we get:

\begin{equation}\label{eq:postgamma}
\gamma | \btheta^*, \bw,\bu,\Nb,\bc \sim \Gamma\left(   a_\gamma +  a K^{\star}, b_\gamma + \sum_{k=1}^{K^{\star}} \frac{1}{(\frac{1}{\theta^{\star}_k} - \frac{1}{T})} \right)
\end{equation}


TITLE: Why the perimeter of a ring is $2\pi(R+r)$ and not $2\pi(R-r)$ ? where $R$ is the bigger circle radius and $r$ is smaller circle radius
QUESTION [0 upvotes]: Why the perimeter of a ring is $2\pi(R+r)$ and not $2\pi(R-r)$ ? where $R$ is the bigger circle radius and $r$ is smaller circle radius

REPLY [0 votes]: The basic reason is that the outer radius of the ring is $R+r$.  In order to make this easier to discuss, pretend that we are talking about a metal ring that you are wearing on your finger.  To get to the outer edge of the metal, you have to first travel from the center of your finger to the middle of the metal part ($R$) and then from the middle of the metal part to the outer edge of the metal part ($r$).  So the outer edge of the ring is a circle with radius $R+r$, so its circumference is $2\pi(R+r)$.
I have added this diagram for clarity.  As noted in the comments below, I think that you and the author of the result that you are talking about have different ideas about what the "bigger circle" and "smaller circle" are, and possibly about what the "perimeter" means in this case.  This is my interpretation.  I have marked the perimeter as $p$, as well as marking $r$ and $R$.


TITLE: How do I find the minimum price p of various items given these constraints?
QUESTION [3 upvotes]: I have six food items $n_1,n_2,...,n_6$. I would like to purchase a combination of these such that the price is minimized, while there are constraints on the total carbs, protein, and fat contents:
$C_{total} = 2100 ≤ 33n_1+16n_2+29n_3+23n_4+7n_5+89n_6$
$P_{total} = 2100 ≤ 4n_1+7n_2+17n_3+1n_4+27n_5+47n_6$
$F_{total} = 470 ≤ 3n_1+0n_2+12n_3+0n_4+32n_5+3n_6$
$p_{total} = 8.95n_1+2.50n_2+2.65n_3+4.30n_4+6.82n_5+7.08n_6$
Variables must be integers; carbs/fat/protein measured in grams, price in USD.
How would I go about finding the minimum price?
I have thought about using matrices, Lagrange multipliers, vector calculus, but I am overall confused as to what the proper method would be and would like a pointer in the right direction.

REPLY [0 votes]: Lagrange constrained works. I get n3=30.78, n6=33.55. The rest are zero. Fat and protein boundaries are tight. Price of 319.08
I'm curious what the foods actually are...


TITLE: Finding the partial derivative
QUESTION [0 upvotes]: I'm new to partial derivatives, and have read about the topic and understood how to solve simple examples.
I came across the following example, and didn't actually understand its solution:

Can you kindly guide me on how to find the partial derivative for such example?
Thanks.

REPLY [1 votes]: Letting $\Psi = \text{Error}_{(m,b)}$ to simplify the notation, for each $k\in \Bbb N, 1 \le k \le N$,
$$\begin{align}{\partial \Psi\over \partial y_k} &= {\partial \over \partial y_k}\left( {1\over N} \sum\limits_{i=1}^n(y_i - (mx_i + b))^2\right)\\
&= {1\over N} \sum\limits_{i=1}^n{\partial \over \partial y_k}(y_i - (mx_i + b))^2\\
&={1\over N}\left ( 0 + ... + 0 + {\partial \over \partial y_k}(y_k - (mx_k + b))^2 + 0 + ... + 0 \right )\\
&= {1\over N}2(y_k - (mx_k + b)){\partial \over \partial y_k}(y_k - (mx_k + b))\\
&= {2\over N}(y_k - (mx_k + b))
\end{align}$$
Since if $i \ne k$, then $y_i$ and $y_k$ are independent variables. The equations for partials in $x_k$ proceed in the same fashion, and are almost identical (except for a factor of $-m$).


TITLE: injectivity is a local property over noetherian rings
QUESTION [1 upvotes]: Let $R$ be a commutative noetherian ring, and let $M$ be an $R$-module. How I can show that if any localization $M_p$ at a prime ideal $p$ of the ring $R$ is injective over $R_p$, then $M$ is injective?

REPLY [6 votes]: Let $A \to B$ be an injection of $R$-modules.  We want to show that $\mathrm{Hom}(B,M) \to \mathrm{Hom}(A,M)$ is surjective.  It suffices to check this locally (since localization is an exact functor and the cokernel of the map is zero iff it's zero locally).  So let $p$ be a prime, and consider the localized map $\mathrm{Hom}_R(B,M) \otimes R_p \to \mathrm{Hom}_R(A,M) \otimes R_p$.  Now, for $B$ and $A$ finitely presented over $R$, this is $\mathrm{Hom}_{R_p}(B_p,M_p) \to \mathrm{Hom}_{R_p}(A_p,M_p)$.  Again, localization is exact so $A_p \to B_p$ is injective, and $M_p$ is injective by hypothesis, so this map is surjective.
What's left is to show that we only need to check injectivity against maps of finitely presented $R$-modules; since $R$ is noetherian, 'finitely presented' is equivalent to 'finitely generated.'  In fact, there's a thing called Baer's injectivity criterion that says you only have to check injectivity against inclusions $I \to R$ for $I$ an ideal!  So we're done.


TITLE: Limit of argmin
QUESTION [0 upvotes]: Fix $t \in [0,1]$ and $n \in \mathbb{N}$. Consider, for $r \in \mathbb{N}$, the ratio $\frac{i}{n+m}$ where $i \in I=\{0,1,...,n+r\}$. How can i prove that
$$\lim_\limits{r \to \infty}\frac{i(r)}{n+r}=t$$
where $i(r)=\underset{i \in I}{\mathrm{argmin}}(\lvert t-\frac{i}{n+r}\rvert)$.

REPLY [0 votes]: By construction
$$ \Big| t - \frac{i(r)}{n+r} \Big| = \min_{i \in I} \Big| t - \frac{i}{n+r} \Big| \leq \Big| t - \frac{\lfloor t (n+r) \rfloor }{n+r} \Big| \leq \frac 1 {n+r} \underset{r \to \infty}{\longrightarrow} 0 $$


TITLE: Simplifying and Factoring Formulae
QUESTION [1 upvotes]: I'm am doing an 8th grade math text book, and I came across this simple problem:
$8l^3 - 36l^2m + 54lm^2 - 27m^3$ simplifies to?
I immediately got to know that it is $(2l - 3m)^3$ ,
but how do you Factor it (step by step)?
Basically how do you factor expressions which has more than 3 or 4 terms.
Please help.
The variables here and l[l for last] and m. (I don't know how to type symbols, italics { it would be nice if someone includes that also })

REPLY [1 votes]: For classwork, there will always be a simple answer.  A slightly less "out of the blue" solution is to take the cube roots of the first and last terms, put the proper sign between, and see if it works.  Here you win.  A second approach is like the rational root theorem-take $l$ as a constant and find the list of roots in $m$.  Now for each root $r$, divide by $m-r$ and see if it comes out.  You can do it from the other end as well.  If it works, you are down to a quadratic, and the quadratic formula or factoring techniques will suffice.


TITLE: Representing a vector in rotated frame
QUESTION [0 upvotes]: If $R \in SO(3) $ is a rotation matrix defined from body fixed frame to the inertial frame such that $e_1, e_2, e_3$ are the basis of inertial frame and $b_1, b_2, b_3$ are the basis of body frame, then what would be the relation between $b_i$ and $e_i$.
According to me, it should be:- $$e_i = Rb_i$$
But consistently across several papers I have come across the relation:-  $$b_i = Re_i$$ while describing the quad-rotor dynamics as:- $$ m\dot{v} = mge_3 - fRe_3$$ where $f$ is the magnitude of force by propellers (whose direction is $-b_3$).
One such paper being - Geometric Tracking Control of a Quadrotor UAV on SE(3) for Extreme Maneuverability (Equation - 3).
What is it that I am interpreting wrong? Shouldn't $R$ take a vector from body frame, $i.e. b_3$ to inertial frame $i.e. e_3$.

REPLY [2 votes]: Rotations have an ambiguity, and that is whether they are alias or alibi (also called passive or active). The latter rotates a vector to a new position while the former expresses the same vector in different coordinate system. Check to see which you are using, and which the papers are using.


TITLE: I'm using the formula below and want to find the length of γ
QUESTION [0 upvotes]: I want to find the length of the smooth curve $γ(t)$ from $t = 0$ to $t = \pi/4$
I'm using the formula for γ:
$$l(γ) = \int_a^b|\gamma'(t)|\,dt$$
apologies I can't work out how to format that correctly.
I have $ \gamma'(t) $
I am now aware this is incorrect but am struggling to amend it myself
I'm not sure how to apply this into the formula without it just integrating back into its original form?
Any help or identification of areas where I've made a mistake would be really useful.

REPLY [1 votes]: Note that if $$\gamma(t)=e^{it}+t(\sin(t)-i\cos(t))=e^{it}+t\sin(t)-it\cos(t)$$
so, you have that $$\gamma'(t)=\color{blue}{ie^{it}}+\sin(t)+t\cos(t) +it\sin(t)-i\cos(t). \quad [1]$$
Now, since that for all $t\in \mathbb{R}$, we have that $$e^{it}=\cos(t)+i\sin(t)$$
The, you can re-write $[1]$ as $$\gamma'(t)=\sin(t)+t\cos(t)\color{blue}{-\sin(t)}+it\sin(t)-i\cos(t)\color{blue}{+i\cos(t)}.$$
Using that for all $z\in \mathbb{C}$ we have that $$|z|=\sqrt{\operatorname{Re}(z)^{2}+\operatorname{Im}(z)^{2}}$$
So, you have that \begin{eqnarray*}
|\gamma'(t)|&=&\sqrt{(\sin(t)+t\cos(t)-\sin(t))^{2}+(t\sin(t)-\cos(t)+\cos(t))^{2}}\\
&=&\sqrt{t^{2}\cos^{2}(t)+t^{2}\sin^{2}(t)}\\
&=&\sqrt{t^{2}(\cos^{2}(t)+\sin^{2}(t))}\\
&=&\sqrt{t^{2}}, \quad \text{since that} \quad \forall x\in \mathbb{R}: \quad \cos^{2}(x)+\sin^{2}(x)=1.
\end{eqnarray*}
Finally, you need calculate $$\ell(\gamma(t))=\int_{0}^{\frac{\pi}{2}}|\gamma'(t)|dt.$$


TITLE: What is Photoproduction
QUESTION [6 upvotes]: I wonder what photoproduction means in the context of pion decay and vector meson dominance? What is the reaction formula, Feynman diagram for such a photoproduction thing? Is it simply a reaction where the final state is a bunch of photons?

REPLY [6 votes]: Photoproduction is a process where something is produced by the interaction of a high energy photon. Something like
$$ \gamma + p \to p + \pi^0 \,. $$
Experimentally it is useful because the electromagnetic vertex is well understood, and photon taggers allow the selection of events with incident photons with well know energy and momentum.


TITLE: $\frac{x}{1+x^2}$ is injective on $[1,\infty)$
QUESTION [0 upvotes]: Prove $\frac{x}{1+x^2}$ is injective on $[1,\infty)$. I'm having a huge brain-fart... I'm assuming $\frac{x}{1+x^2}=\frac{y}{1+y^2}$, yada yada, $\implies \frac{1}{x}+x=\frac{1}{y}+y$... but I can't get past this step!!!

REPLY [1 votes]: if $1 \leq x < y,$  then $y-x > 0 $  and $xy - 1 > 0$ and
$$  \frac{x}{1+x^2} - \frac{y}{1 + y^2}  = \frac{(y-x)(xy-1)}{(1+x^2)(1+y^2)} >0$$