TITLE: Proof of a property of the derivative of 4-velocity
QUESTION [0 upvotes]: In General Relativity, An Introduction for Physicists, in between equations 8.7 and 8.8, the authors make use of the property that the contraction of the 4 velocity gives the speed of light squared ($u_{\nu}u^{\nu} = c^2$), so the partial derivative will be zero. They also claim
$$
\left(\partial_{\mu} u^{\nu}\right) u_{\nu}+u^{\nu}\left(\partial_{\mu} u_{\nu}\right)=2\left(\partial_{\mu} u^{\nu}\right) u_{\nu}=0
$$
where I assume the second part of the sum must be equal to the first (hence the 2). Is this true by raising or lowering the indices with the metric? How would I prove this statement?
Edit:
I can't just contract the $u^{\nu}\left(\partial_{\mu} u_{\nu}\right)$ with the metric without it being part of an equation. Also, it's not necessarily valid that I can contract a metric inside a derivative because it is not true that the derivative of that metric are zero. (I eventually want to generalise this statement to covariant instead of partial derivatives. )

REPLY [1 votes]: In contractions you can lower one index if you bring another one up. This is a really useful property:
\begin{align}A^\mu B_\mu&=A^\mu g_{\mu\nu}B^\nu\\&=A_\nu B^\nu\\&=A_\mu B^\mu\end{align}
Partial derivatives are bit less obvious though. Looking at your second term:
\begin{align}u^\nu(\partial_\mu u_\nu)&=u^\nu(\partial_\mu g_{\nu\lambda}u^\lambda)\\
&=u^\nu g_{\nu\lambda}\partial_\mu u^\lambda+u^\nu u^\lambda\partial_\mu g_{\nu\lambda}\\
&=u_\nu \partial_\mu u^\nu+u^\nu u^\lambda\partial_\mu g_{\nu\lambda}
\end{align}
In the Minkowski metric $\partial_\mu g_{\nu\lambda}=0$ so partials commute with the lowering of indices. In GR this is generally not the case. This last term is often referred to as the non-tensorial part, because it is what prevents the partial derivative from transforming like a tensor. So in GR we generally have
$$(\partial_\mu u^\nu)u_\nu \neq u^\nu(\partial_\mu u_\nu)$$
If you replace the partials by covariant derivatives (like Ben mentioned) you can commute the metric $\nabla_\lambda g_{\mu\nu}=g_{\mu\nu}\nabla_\lambda$ so you would get a similar identity.


TITLE: Semialgebraic Morse-Sard Theorem - why are the critical points of a semi-algebraic map again semi-algebraic?
QUESTION [1 upvotes]: 2.5.12 Exercise (Semi-algebraic Morse-Sard theorem)
Let $f:M\to N$ be a $C^\infty$ semi-algebraic map between semi-algebraic submanifolds of $\Bbb R^n$ and $\Bbb R^m$, respectively. Set $$C = \{x\in M; \operatorname{rank} df_x < \dim N\},$$ $$S=\{y\in N; y=f(x), x\in C\}.$$
Then $S$ is a semi-algebraic set and $\dim S<\dim N$.

I was reading this theorem from Benedetti and Risler book, and I cannot understand why is $C$ a semialgebraic set. Is the rank of a linear transformation a semialgebraic map? I can't see why.

REPLY [1 votes]: The condition that a matrix is of rank $<n$ is that the determinants of all $n\times n$ minors vanish. Since the determinant is a polynomial in the entries of the matrix, this is a collection of polynomial conditions on the entries of the matrix.
As the matrix $df$ has semi-algebraic entries because the derivative of a semi-algebraic function is semi-algebraic (where it exists), the determinant of any minor of $df$ is again semi-algebraic, being a polynomial in semi-algebraic functions. Therefore the condition that $\operatorname{rank} df_x < \dim N$ is semi-algebraic.


TITLE: Is there a way to extract a primary decomposition of $(0)\subset R/I$ given a decomposition of $I\subset R$?
QUESTION [0 upvotes]: Suppose we have a (minimal) primary decomposition of an ideal $I$ in a polynomial ring $R$ over a field; for simplicity assume $I=I_1\cap I_2$ where $I_j$ is $P_j$-primary. Is there a way to extract from this a primary decomposition of $(0)$ in $R/I$? If so, say $(0)=J_1\cap\dots\cap J_k$ where $J_i$ is $Q_i$-primary.
Can I just take the equality $I=I_1\cap I_2$ and replace all generators by their residues mod $I$? If so, how to justify this? How are $Q_i$ and $P_j$ related?

REPLY [1 votes]: Here are three results that together solve your problem:

For $I\subset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $J\supset I \mapsto J/I$. (Correspondence theorem)
If $I\subset J$ are ideals in $R$, then $(R/I)/(I/J)\cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)
An ideal $I\subset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i \cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)\subset R/I$ gives rise bijectively to a primary decomposition of $I\subset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.


TITLE: Determining the 3rd vertex of an Equilateral and Right angled isoceles triangle.
QUESTION [5 upvotes]: I am really having problems solving the following problems:

If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the two vertices on the hypotenuse of a right angled isosceles triangle then the coordinates of the 3rd vertex are?
Similarly-If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the two vertices of an Equilateral Triangle then the coordinates of the 3rd vertex are?

The answer to 1 and 2 respectively are-
$$\left(\frac{x_1+x_2 \pm(y_1-y_2)}{2},\frac{y_1+y_2 \pm(x_1-x_2)}{2}\right)$$
And
$$\left(\frac{x_1+x_2 \pm\sqrt 3(y_1-y_2)}{2},\frac{y_1+y_2 \pm\sqrt 3(x_1-x_2)}{2}\right)$$
How do I solve this problem?
I tried attempting it by using the fact that the triangle is isosceles and right angled in the first case. So I applied pythagoras theorem but failed to get this answer it was a complicated equation. Secondly I tried multiplying the slopes of the legs of the triangle to get -1 but even this attempt failed.
It was similar with the equilateral triangle problem.
I guess if I understand how to do even one easily the second should bit be difficult.
Please help.

REPLY [1 votes]: Consider the fact that if P(x,y) is the unknown point, then when you apply rotation around that point, the point P1(x1, y1) goes to (gets mapped to) P2(x2,y2). From this you can construct equations and solve them to get the x and y. Just lookup the formulas for rotation in the 2D plane. The rotation angles are nice in both cases: 90 and 60 degrees. If you do this, you should get the answers you have posted here.
Say you rotate the point (x1,y1) around the point (a,b) on an angle of $\theta$.
Then the point (x1,y1) gets mapped to (x2, y2) where:
$x_2 = (x_1-a) * cos(\theta) - (y_1-b) * sin(\theta) + a$
$y_2 = (x_1-a) * sin(\theta) + (y_1-b) * cos(\theta) + b$
In your case you don't know the $a$ and $b$, you want to find them.
The $\theta$ is $\pm \pi/2$ and $\pm \pi/3$ respectively.
The $x_1, y_1, x_2, y_2$ - these you know.
Here is how one can solve the case: $\pi/3$. You have 3 more cases to solve.

See also:
Rotation
Rotation matrix


TITLE: Symmetry behind Taylor expansion
QUESTION [1 upvotes]: We consider the function $f:\mathbb R^2 \to \mathbb C.$
$$f(x,y) = 2i \sum_{k=0}^2 \zeta^k \sin\left( \frac{x\bar \zeta^k-y\zeta^k}{2i} \right)$$
where $\zeta= e^{2\pi i/3}$ is the third root of unity
I computed using Mathematica the Taylor expansion at $x=y=0$ up to order $8$
$f(x,y) = \left(-\frac{y^5}{640}\right)+x
\left(3+\frac{y^6}{15360}\right)+x^2 \left(-\frac{3
y}{8}-\frac{y^7}{860160}\right)+x^3
\left(\frac{y^2}{64}+\frac{y^8}{82575360}\right)+x^4
\left(-\frac{y^3}{3072}\right)+x^5
\left(\frac{y^4}{245760}\right)+x^6
\left(-\frac{y^5}{29491200}\right)+x^7
\left(\frac{1}{107520}+\frac{y^6}{4954521600}\right)+x^8
\left(-\frac{y}{3440640}-\frac{y^7}{1109812838400}\right)+O\left(x^9+y^9
\right).$$
Then since the function is anti-symmetric by switching both $x,y$, we only get products of odd order in $x^iy^j$, i.e. $i+j$ is odd.
However, there is another feature that is more subtle for me to explain. For a given power in $x$ we have some $y^i$ and then a term $y^{i+6}$ appearing. Why do we have these jumps of order $6$ in the expansion?
Please let me know if you have any questions.

REPLY [1 votes]: Overview. Your problem is connected to the general problem of constructing analytic functions that possess prescribed symmetries. One trick for creating such functions is to average "any" analytic function over the action of all elements of the  desired symmetry group.  This trick compels the Taylor coefficients  of the symmetrized function to have some special symmetrical patterns.
Details. The crux of the matter is that the derived function $\partial f/\partial x$ is loosely speaking a symmetrization of $h(x,y)=\cos(\frac{x-y}{2i})$ under the action of a finite group generated by powers of $\zeta$.
More precisely you can verify the identity
(1) $f_x= g(x,y)=\sum_{k=0}^2 h( \zeta^{-k} x, \zeta^k y)$
Consider the  implications of the symmetry
(2) $h(-x,-y)=h(x,y)$
on the Taylor expansion
(3) $h(x,y)= \sum_{m,n\geq 0} c_{m,n} x^m y^n$
Deduce that $c_{m,n} =(-1)^{n+m} c_{m,n}$.
That is
(4)  $c_{m,n}=0$ unless $n$ and $m$ have the same parity (either both even, or both odd).
Substitute  (3) into (1) to deduce that
$g(x,y)= \sum_{k=0}^2 h(  \zeta^{-k} x, \zeta^k y)$ satisfies
(5) $g(x,y)= \sum_{m,n\geq 0}  c_{m,n}  x^m y^m (\sum_{k=0}^2 \zeta^{k(n-m)})$.
The rightmost sum is a three-term geometric series consisting of three consecutive powers of  $\zeta^{n-m}$.
That geometric series vanishes unless $n-m \equiv 0 $(mod 3).
Thus the non-vanishing terms in the Taylor expansion of $g(x,y)$ consist only  of those terms for which
(6) $n \equiv m$ (mod 3).
But as noted in (4),  the additional requirement that $n$ and $m$ must have the same parity  forces the condition that $c_{m,n}=0$ unless $n\equiv m $(mod 6).
Finally, this pattern satisfied by $f_x$ naturally leads to the pattern that you noted about $f$ itself. (Just integrate the series for $f_x$ term-wise to obtain $f$.)


TITLE: Does an accelerating mass radiate energy?
QUESTION [1 upvotes]: This question is actually 2 parts

Just like binary star system or the 2 black wholes (which generated the recently discovered gravitational waves) shouldn't the Earth also radiate giving off energy. Reading the Wikipedia article confirmed my belief that Earth should radiate and give off gravitational waves. Now, since the Earth is 'radiating' shouldn't it lose energy (howsoever small it may be) resulting in shrinking of its orbit (howsoever ever small the shrink may be)? Radiating means losing energy after all.
All mass (except few like neutrinos) is made up of charges and accelerating charges radiate. So normal matter, though neutral, must radiate in the sense that each of its individual charge should radiate. Shouldn't it? And there should be a net loss of energy?

REPLY [0 votes]: The earth-sun system absolutely emits gravitational waves, and they will tend to reduce the radius of the earth's orbit. But the effect will be very small (I think negligible) because the accelerations involved are small in relativistic terms.


TITLE: Why is the distance between two circles/spheres that don't intersect minimised at points that are in the line formed by their centers?
QUESTION [5 upvotes]: From GRE 0568

From MathematicsGRE.Com:

I'm guessing the idea applies to circles also?
Is there a way to prove this besides the following non-elegant way?

Form a line between centers $C_1$ and $C_2$
Given a point on circle/sphere 1 $(x_1,y_1)$, minimise

$$f(x_2,y_2) = (x_1-x_2)^2 + (y_1-y_2)^2$$
to get $(x_2^*,y_2^*)$

Minimise

$$g(x_1,y_1) = (x_1-x_2^*)^2 + (y_1-y_2^*)^2$$
to get
$(x_1^*,y_1^*)$

Show that the $(x_2^*,y_2^*)$ and $(x_1^*,y_1^*)$ are on the line.

REPLY [2 votes]: Basically the question in your title is answered by the fact that a straight line is the shortest path between two points. If $P,Q$ are the points on your two spheres (or circles, if you are in a plane), and $C_i$ and $r_i$ are their respective centres and radii, for $i=1,2$, then $C_1-P-Q-C_2$ is a path from $C_1$ to $C_2$ of length $r_1+d(P,Q)+r_2$. Here only the middle term, the distance $d(P,Q)$ from $P$ to $Q$, depends on the choice of these points; the other two terms are constant. The shortest possible path from $C_1$ to $C_2$ is a straight line, and given that $d(C_1,C_2)\geq r_1+r_2$, this path can be obtained by choosing $P$ and $Q$ on the segment $[C_1,C_2]$. Every other choice of $P,Q$ gives a longer path, hence a larger value of $d(P,Q)$.


TITLE: Is it true, that $\sum\limits_{n=1}^{\infty}(e^{\frac{1}{n!e}}-1)=\frac{10}{11+e}$?
QUESTION [1 upvotes]: If
$$\sum\limits_{n=1}^{\infty}(e^{\frac{1}{n!e}}-1)=\frac{10}{11+e}$$
is true, so how can we prove it (if not, how can we came to this approximation)?
If I made some mistakes, sorry for my English.

REPLY [3 votes]: Notice that if $h > 0$, then by the mean value theorem $e^h - 1 < e^h h$. So for any fixed integer $N \geq 1$ we have
\begin{align*}
\sum_{n=1}^{\infty} \left( e^{1/n!e} - 1 \right)
&< \sum_{n=1}^{N-1} \left( e^{1/n!e} - 1 \right) + e^{1/N!e} \sum_{n=N}^{\infty} \frac{1}{n!e} \\
&= \sum_{n=1}^{N-1} \left( e^{1/n!e} - 1 \right) + e^{1/N!e} \left(1 -
 \frac{1}{e} \sum_{n=0}^{N-1} \frac{1}{n!} \right).
\end{align*}
Now with the choice $N = 7$ with aid of numerical computation, we can check that
$$ \sum_{n=1}^{\infty} \left( e^{1/n!e} - 1 \right) < 0.7289542\color{red}{1703628121330\cdots} $$
Comparing this with
$$ \frac{10}{11+e} \approx 0.7289542\color{blue}{6155006217295\cdots} $$
it follows that the sum is strictly smaller than the proposed value.


TITLE: finding the pdf and expected value of first order statistic
QUESTION [0 upvotes]: Let $Y_1,Y_2,...,Y_n$ be  independent  and  identically  distributed  with  the  following  probability density function $f(y) =4(1−y)^3$ for $y$ between 0 and 1
(a)  Find the probability density function of $Y_{(1)}=\min(Y_1,...,Y_n)$
(c)  Find the expected value of $Y_{(1)}$
I did part a and got that the pdf of the first order statistic is $$f_{Y_{(1)}}(y)=4n(1-y)^3 \cdot [1+(1-y)^4]^{n-1}$$
I tried to find the expected value of this by taking the integral from 0 to 1 of the pdf of the first statistic times y. This integral got way to complicated so I am wondering if there is a way to find the expected value of this first order statistic differently.

REPLY [0 votes]: For $0\leq y<1$ you should have :
$$\begin{split}f_{Y_{(1)}}(y) &= n~f(y)~\mathsf P(Y > y)^{n-1}\\[1ex] &=4n~(1-y)^3\left(\int_y^1 4(1-t)^3\mathsf d t\right)^{n-1}\\[1ex]&= 4n~(1-y)^3(1-y)^{4n-4}\\[1ex]&= 4n~(1-y)^{4n-1}\\[3ex]\therefore\quad\mathsf E(Y_{(1)})&=4n\int_0^1 y~(1-y)^{4n-1}~\mathsf d y\end{split}$$
Which is not too difficult.


TITLE: What is the indefinite integral of the error function times a gaussian?
QUESTION [3 upvotes]: Wikipedia lists
$$\int\limits_{-\infty}^{\infty} \operatorname{erf}(ax+b) \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( \frac{-(x-\mu)^2}{2 \sigma^2}\right) \, dx \\ = \operatorname{erf}\left(\frac{a\mu+b}{\sqrt{1+2a^2\sigma^2}}\right)$$
How is this obtained? Is there an indefinite solution to this integral?

REPLY [1 votes]: There's an elegant solution based on @fGDu94's hint using probability theory. (I'm sure I've used this argument in a previous question of which this one might be a duplicate, but I can't find it.) Let's first restate the claimed result with $\Phi$ instead of $\operatorname{erf}$ as$$\int_{\Bbb R}(2\Phi((ax+b)\sqrt{2})-1)\frac{d}{dx}\Phi\left(\frac{x-\mu}{\sigma}\right)dx=2\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg)-1,$$or equivalently as$$\int_{\Bbb R}\Phi((ax+b)\sqrt{2})\frac{d}{dx}\Phi\left(\frac{x-\mu}{\sigma}\right)dx=\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg).$$The left-hand side is the mean of $\Phi(Y)$ with $Y:=(aX+b)\sqrt{2}$ for $X\sim N(\mu,\,\sigma^2)$ so that $Y\sim N((a\mu+b)\sqrt{2},\,2a^2\sigma^2)$. But in terms of $Z\sim N(0,\,1)$ independent of $Y$ so that $Y-Z\sim N((a\mu+b)\sqrt{2},\,1+2a^2\sigma^2)$,$$\Phi(Y)=P(Z\le Y)=P(Y-Z\ge 0)=\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg).$$


TITLE: bounding the sum of the entries of the inverse of a 0-1 matrix away from 1?
QUESTION [1 upvotes]: Let $A\in\mathbb{R}^{n\times n}$ be an invertible 0-1 matrix. Is it possible that the sum $a:=\sum_{i,j}(A^{-1})_{ij}$ of entries of $A^{-1}$ is not equal to 1, but exponentially close (w.r.t. $n$) to 1?
More precisely, the conjecture is that if $a\geq 1$ then either $a=1$ or $|a|\geq 1+p(n)/q(n)$, for $p$ and $q$ fixed polynomials (some other not too fast growing functions might do, too).
Remark. The condition $a\geq 1$ is not redundant, as there exist matrices $A$ with $a<1$, e.g. $$A=\left(\begin{array}{rrrr}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1
\end{array}\right), \quad A^{-1}=\left(\begin{array}{rrrr}
-1 & -1 & 0 & 1 \\
-1 & -1 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{array}\right), \quad a=\sum_{i,j} (A^{-1})_{ij}=0.
$$
An (admittedly very shaky) evidence for this is that it is true if the row sums $r_i$ of $A$ are all equal to each other. Indeed, let $\overline 1$ denote the all-1 vector. Then $a=\sum_{i=1}^n x_i$ for $x$ satisfying $Ax=\overline 1$, and
$$
\overline{1}^\top Ax=\sum_i r_i x_i=\overline{1}^\top \overline{1}=n,\quad \text{implying }\quad a\geq \frac{n}{n-1}
$$

REPLY [2 votes]: Apparently, this question had already been answered in the negative here.


text:
\begin{document}

\title{Operator methods applied to special functions}

\author{H. Moya-Cessa and F. Soto-Eguibar}
\affiliation{Instituto Nacional de Astrof\'isica \'Optica y Electr\'onica\\
Calle Luis Enrique Erro No. 1, Sta. Ma. Tonantzintla, Pue. CP 72840, Mexico}

\begin{abstract}
Based on operator algebras commonly used in quantum
mechanics some properties of special functions such as Hermite  and Laguerre polynomials and Bessel functions are derived.
\end{abstract}
\maketitle
\section{Introduction}
Based on some techniques that are common to quantum mechanics, we give some examples on how some series of special functions may be added.
For this, we use some theorems and lemmas that are not usually known when one studies special functions, such as the Baker-Hausdorff formula and the Hadamard lemma.
We developed further preliminary results presented in  \cite{arXive}.

\section{Hermite polynomials}
The generating function for the Hermite polynomials is
\begin{equation}
e^{-\alpha ^{2}+2\alpha x}=\sum\limits_{n=0}^{\infty}H_{n}(x)\frac{\alpha^{n}}{n!}.
\end{equation}

The Hermite polynomials may be obtained from Rodrigues' formula \cite{A}
\begin{equation}\label{1q}
H_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}.
\end{equation}
From the recurrence relations
\begin{equation}
H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x),
\end{equation}
and
\begin{equation}
\frac{dH_n(x)}{dx}=2nH_{n-1}(x),
\end{equation}
we can generate all the Hermite polynomials. \\

 From the above recurrence relations, we can also prove that, if we define the functions
\begin{equation}\label{fooa}
\psi_n(x)=\frac{\pi^{-1/4}}{\sqrt{2^{n}n!}}\mathrm{e}^{-x^2/2}H_n(x),
\end{equation}
then
\begin{equation}
A^{\dagger}\psi_n(x)\equiv\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right)\psi_n(x)=\sqrt{n+1}\psi_{n+1}(x)
\end{equation}
and
\begin{equation}
A\psi_n(x)\equiv\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right)\psi_n(x)=\sqrt{n}\psi_{n-1}(x).
\end{equation}
The functions (\ref{fooa}) constitute a complete orthonormal set for the space of square integrable functions, then we can expand any function in that space as
\begin{equation}\label{2.9}
f(x)=\sum_{n=0}^{\infty}c_n \psi_n(x),
\end{equation}
where
\begin{equation}
c_n =\int_{-\infty}^{\infty}dxf(x)\psi_n(x).
\end{equation}

Hermite polynomials are also solutions of the second order ordinary differential equation
\begin{equation}
y''-2xy'+2ny=0.
\end{equation}

Let us define the differential operator $p$ as
\begin{equation}\label{p}
p=-i\frac{d}{dx},
\end{equation}
then we have that
\begin{equation}
\frac{d^{n}}{dx^{n}}=\left( -\frac{p}{i}\right) ^{n},
\end{equation}
and we can rewrite (\ref{1q}) in the form
\begin{equation}\label{herop}
H_{n}(x)=(-i)^{n}e^{x^{2}}p^{n}e^{-x^{2}}=(-i)^{n}\left(e^{x^{2}}pe^{-x^{2}}\right) ^{n}.
\end{equation}
The operator inside the parenthesis above has the form
\begin{equation}
e^{\xi A}Be^{-\xi A},
\end{equation}
for which we will use the following lemma \cite{L,Soto}.
\begin{quote}
\textbf{Hadamard lemma:} Given two linear operators $A$ and $B$ then
\begin{eqnarray}   \label{bch0}
e^{\xi A}Be^{-\xi A}=B+\xi\left[ A,B\right]+\frac{\xi^{2}}{2!}\left[ A,\left[A,B\right]\right]+\frac{\xi^{3}}{3!}\left[A,\left[A,\left[A,B\right]\right]\right]+\cdots,
\end{eqnarray}
where $[A,B]\equiv AB-BA$ is the commutator of operators $A$ and $B$. \\
\end{quote}

This allows us  to obtain an expression for
\begin{equation}
H_{n}(x)=(-i)^{n}\left(e^{x^{2}}pe^{-x^{2}}\right) ^{n},
\end{equation}
formula (\ref{herop}) developed above.\\
We identify
\begin{equation}
\xi =1,\qquad A=x^{2},\qquad B=p,\nonumber\\
\end{equation}
in equation  (\ref{bch0}), so that
\begin{equation}\label{formop}
e^{x^{2}}pe^{-x^{2}}
=p+1\left[ x^{2},p\right]+\frac{1}{2!}\left[ x^{2},\left[x^{2},p\right]\right]+\frac{1}{3!}\left[x^{2},\left[x^{2},\left[x^{2},p\right]\right]\right]+\cdots
\end{equation}
To calculate the first commutator $\left[ x^{2},p\right]$, we use the general property
\begin{equation}\label{p1c}
[AB,C]=A[B,C]+[A,C]B
\end{equation}
of commutators, and that
\begin{equation}
[x,p]f(x)\equiv
-i(x\frac{d}{dx}-\frac{d}{dx}x)f(x)=-xf'(x)+xf'(x)+if(x)=if(x);
\end{equation}
i.e. $[x,p]=i$, to get the commutation relation
\begin{equation}
\left[ x^{2},p\right] =2ix.
\end{equation}
It is obvious that all the other commutators in (\ref{formop}) are zero, and we finally get
\begin{equation}\label{Hermite}
H_{n}(x)=(-i)^{n}\left( p+2ix\right)^{n}1 .
\end{equation}
This last expression can be used to obtain the generating function. We have,
\begin{equation}\label{2q}
\sum_{n=0}^{\infty }H_{n}(x)\frac{\alpha ^{n}}{n!}
=\sum_{n=0}^{\infty }\frac{\alpha ^{n}}{n!}(-i)^{n}\left(
p+2ix\right) ^{n}1=e^{-i\alpha \left( p+2ix\right) } 1.
\end{equation}
We obtained the exponential of the sum of two quantities that do not commute.
The above exponential can be factorized in the product of
exponentials via the Baker-Hausdorff formula
\subsection{ Baker-Hausdorff formula}
\begin{quote}
\textbf{Baker-Hausdorff formula:} \textit{Given two operators} $A$
\textit{and} $B$ \textit{that obey}
\begin{equation}
\left[ \left[ A,B\right] ,A\right] =\left[ \left[ A,B\right],B\right] =0,
\end{equation}
\textit{then}
\begin{equation}\label{bhformula}
e^{A+B}=e^{-\frac{1}{2}\left[ A,B\right]}e^{A}e^{B}.
\end{equation}
\end{quote}

Then we  identify
\begin{equation}
A=2\alpha x,
\end{equation}
\begin{equation}
B=-i\alpha p,
\end{equation}
and then
\begin{equation}
\left[ A,B\right] =-2i\alpha ^{2}\left[ x,p\right] =2\alpha ^{2},
\end{equation}
such that
\begin{equation}
\sum_{n=0}^{\infty }H_{n}(x)\frac{\alpha ^{n}}{n!}=e^{-\alpha^{2}}e^{2\alpha x}e^{-i\alpha p}1.
\end{equation}
Using now the obvious fact that $e^{-i\alpha p}1=1$, we finally obtain
\begin{equation}
\sum_{n=0}^{\infty }H_{n}(x)\frac{\alpha ^{n}}{n!}=e^{-\alpha ^{2}+2\alpha x},
\end{equation}
that is the generating function for Hermite polynomials. \\

\subsection{Series of even Hermite polynomials}
In order to show the power of the operator methods, we calculate now the value of the following even Hermite polynomials series,
\begin{equation}
F(t)=\sum\limits_{n=0}^{\infty }\frac{t^{n}}{n!}H_{2n}\left( x\right).
\end{equation}
 From (\ref{Hermite}), we get
\begin{equation}
H_{2n}\left( x\right) =\left( -1\right) ^{n}\left( p+2ix\right)^{2n}1.
\end{equation}
Therefore,
\begin{eqnarray}    \nonumber
F(t)&=&\sum\limits_{n=0}^{\infty }\frac{t^{n}}{n!}H_{2n}\left( x\right)
=\sum\limits_{n=0}^{\infty }\frac{t^{n}}{n!}\left( -1\right)
^{n}\left( p+2ix\right) ^{2n}1 \\
&=& \sum\limits_{n=0}^{\infty
}\frac{t^{n}}{n!}\left( -1\right) ^{n}\left[ \left( p+2ix\right)
^{2}\right] ^{n}1=\exp \left[ -t\left( p+2ix\right) ^{2}\right] 1.
 \end{eqnarray}
Developing the power in the exponential above we get
\begin{equation}\label{211000}
F\left(t\right)=\sum\limits_{n=0}^{\infty}\frac{t^{n}}{n!}H_{2n}\left(x\right)=\exp\left\{-t\left[p^{2}-4x^{2}+2i\left(xp+px\right)\right]\right\}1.
\end{equation}
The operators in the exponential in this last expression do not satisfy the conditions of the Baker-Hausdorff formula, so we need another method to understand the action of the full operator that appears in the right side of expression (\ref{211000}). What we do is to propose the ansatz,
\begin{equation}\label{ansatz}
{F\left(t\right)=\exp\left[f\left( t\right)x^{2}\right]\exp\left[g\left(t\right)\left(xp+px\right)\right]\exp\left[h\left(t\right)p^{2}\right]1},
\end{equation}
where $f(t), g(t)$ and $h(t)$ are functions to be  determined.
Deriving this expression with respect to $t$, and dropping the explicit dependence of $f(t), g(t)$ and $h(t)$ on $t$,
\begin{eqnarray}\label{44}\nonumber
&&\frac{dF\left(t\right)}{dt}= \\
&=&\frac{df}{dt}x^{2}F\left(t\right)+\frac{dg}{dt}\exp\left(fx^{2}\right)\left( xp+px\right)\exp\left[g\left( xp+px\right) \right] \exp\left(hp^{2}\right)1\nonumber\\
&+&\frac{dh}{dt}\exp\left(fx^{2}\right)\exp\left[g\left(xp+px\right)\right]p^{2}\exp\left(hp^{2}\right)1.
\end{eqnarray}
Introducing an "smart" 1 in the second and third term, we get
\begin{eqnarray}  \label{44}\nonumber
\frac{dF\left(t\right)}{dt}&=&\frac{df}{dt}x^{2}F\left(t\right)+\frac{dg}{dt}e^{fx^{2}}\left( xp+px\right)e^{-fx^{2}}F(t)  \\
&+&\frac{dh}{dt}e^{fx^{2}} \exp \left[ g\left( xp+px\right) \right] p^{2}\exp\left[ -g\left( xp+px\right) \right] e^{-fx^{2}}F(t).
\end{eqnarray}
We work then with the operator in the second term; we have to use the very useful expression
\begin{eqnarray}\nonumber
e^{\xi A}Be^{-\xi A}=B+\xi \left[ A,B\right] +\frac{\xi ^{2}}{2!}\left[ A,
\left[ A,B\right] \right] +\frac{\xi ^{3}}{3!}\left[ A,\left[ A,\left[ A,B\right] \right] \right] +\cdots,
\end{eqnarray}
and obtain
\begin{eqnarray} \label{48}
&&\exp \left( fx^{2}\right) \left( xp+px\right) \exp \left(-fx^{2}\right)=xp+px+f\left[ x^{2},xp+px\right] \\
&+&\frac{f^{2}}{2!}\left[ x^{2},\left[ x^{2},xp+px\right] \right]+\frac{f^{3}}{3!}\left[ x^{2},\left[ x^{2},\left[x^{2},xp+px\right] \right] \right]+\cdots.
\end{eqnarray}
The first commutator that appears in the above expression is easily calculated,
\begin{equation}
[x^{2},xp+px] =4ix^{2},
\end{equation}
and so all the others commutators are zero. Substituting back in (\ref{48}), we get
\begin{equation}
\exp \left( fx^{2}\right) \left( xp+px\right) \exp \left(-fx^{2}\right) =xp+px+4ifx^{2}.
\end{equation}
We analyze now the third operator in expression (\ref{44}), i.e.
\begin{equation}
\exp(-fx^{2})\exp \left[ g\left( xp+px\right) \right] p^{2}\exp\left[ -g\left( xp+px\right) \right]\exp(-fx^{2}).
\end{equation}
We study first only a part of it, $\exp \left[ g\left( xp+px\right) \right] p^{2}\exp \left[-g\left(xp+px\right) \right]$. Using again (\ref{bch0}),
\begin{eqnarray}\nonumber
&&\exp \left[ g\left( xp+px\right) \right] p^{2}\exp \left[-g\left(xp+px\right) \right] = \\
&&p^{2}+g\left[ xp+px,p^{2}\right] +\frac{g^{2}}{2!}\left[ xp+px,\left[ xp+px,p^{2}\right] \right] \\
&+&\frac{g^{3}}{3!}\left[ xp+px,\left[ xp+px,\left[xp+px,p^{2}\right] \right] \right] +\cdots.
\end{eqnarray}
Calculating the first commutators,
\begin{equation}
\left[ xp+px,p^{2}\right] =4ip^{2},
\end{equation}
\begin{equation}
\left[ xp+px,\left[ xp+px,p^{2}\right]\right] =-16p^{2},
\end{equation}
\begin{equation}
\left[ xp+px,\left[xp+px,\left[ xp+px,p^{2}\right] \right] \right] =-64ip^{2},
\end{equation}
and so on. It is clear that
\begin{equation}
\exp \left[ g\left( xp+px\right) \right] p^{2}\exp \left[-g\left(
xp+px\right) \right] =p^{2}\sum\limits_{j=0}^{\infty }\left( 4i\right) ^{j}
\frac{g^{j}}{j!}=p^{2}\exp \left( 4ig\right).
\end{equation}
We proceed now to complete the study of the third operator in expression (\ref{44}). Until now we have
\begin{eqnarray}\nonumber
&&\exp(-fx^{2})\exp \left[ g\left( xp+px\right) \right] p^{2}\exp\left[ -g\left( xp+px\right) \right]\exp(-fx^{2})= \\
&&\exp(-fx^{2}) p^{2}\exp \left( 4ig\right) \exp(-fx^{2}).
 \end{eqnarray}
We use once more formula (\ref{bch0}), to write
\begin{eqnarray}\nonumber
&&\exp \left( fx^{2}\right) p^{2}\exp \left( -fx^{2}\right)= \\
&&p^{2}+f\left[x^{2},p^{2}\right] +\frac{f^{2}}{2!}\left[ x^{2},\left[ x^{2},p^{2}\right]
\right] +\frac{f^{3}}{3!}\left[ x^{2},\left[ x^{2},\left[x^{2},p^{2}\right] \right] \right] + \cdots
 \end{eqnarray}
The first commutator gives
\begin{equation}
\left[ x^{2},p^{2}\right] =-2+4ipx,
\end{equation}
the second one gives
\begin{equation}
\left[x^{2},\left[ x^{2},p^{2}\right] \right] =-8x^{2},
\end{equation}
and the third one
\begin{equation}
\left[ x^{2},\left[ x^{2},\left[ x^{2},p^{2}\right] \right]\right] =0;
\end{equation}
such that all the other commutators are zero, and
\begin{eqnarray}\nonumber
&&\exp \left( fx^{2}\right) p^{2}\exp \left( -fx^{2}\right)= \\
&&p^{2}+f\left( -2+4ipx\right) +\frac{f^{2}}{2!}\left(-8x^{2}\right)
=p^{2}+2if\left( xp+px\right) -4f^{2}x^{2}.
  \end{eqnarray}
Finally, we can write a reduced expression for the derivative of the original series $F(t)$,
\begin{eqnarray}
&&\frac{dF\left( t\right) }{dt}=
\allowbreak \Big\{ \frac{df}{dt}x^{2}+\frac{dg}{dt}\left( xp+px+4ifx^{2}\right) + \\
&&\exp \left( 4ig\right)\frac{dh}{dt}\left[ p^{2}+2if\left( xp+px\right)-4f^{2}x^{2}\right] \Big\} F(t)
   \end{eqnarray}
and rearranging terms
\begin{eqnarray}\nonumber
&&\frac{dF\left( t\right) }{dt}=  \\ \nonumber
&&\allowbreak \Big\{ \left[ \frac{df}{dt}+4if\frac{dg}{dt}-4f^{2}\exp \left( 4ig\right) \frac{dh}{dt}\right] x^{2}+ \\
&&\left[ \frac{dg}{dt}+2if\exp \left( 4ig\right) \frac{dh}{dt}\right] \left(xp+px\right) +\exp \left( 4ig\right) \frac{dh}{dt}p^{2}\Big\}F(t).
\end{eqnarray}
We get back to the original expression for the series
\begin{equation}
F\left( t\right) =\exp \left\{ -t\left[ p^{2}-4x^{2}+2i\left( xp+px\right)\right] \right\} 1
\end{equation}
and taken the derivative with respect to $t$,
\begin{eqnarray}\nonumber
\frac{dF\left( t\right) }{dt}&=&-\left[ p^{2}-4x^{2}+2i\left( xp+px\right)\right] \exp \left\{ -t\left[ p^{2}-4x^{2}+2i\left( xp+px\right)\right] \right\} 1 \\
&=&\left[ -p^{2}+4x^{2}-2i\left( xp+px\right)\right] F.
  \end{eqnarray}
Comparing now both expressions, we get the system of differential equations
\begin{eqnarray}\label{212000}
  \frac{df}{dt}+4if\frac{dg}{dt}-4f^{2}\exp \left( 4ig\right) \frac{dh}{dt} &=& 4, \nonumber \\
  \frac{dg}{dt}+2if\exp \left( 4ig\right) \frac{dh}{dt} &=& -2i, \\
  \exp \left( 4ig\right) \frac{dh}{dt} &=& -1. \nonumber
\end{eqnarray}
The initial conditions that we must set on these equations  and from the ansatz (\ref{ansatz}) are that for $t=0$ the operator in the right side of (\ref{211000}) is the identity, so  we must impose $f(0)=g(0)=h(0)=0$.
The solutions then are simply
\begin{equation}f=\frac{4t}{4t+1},\end{equation}
\begin{equation}g=-\frac{i}{2}\ln \left( 4t+1\right),\end{equation}
\begin{equation}h=-\frac{t}{4t+1}.\end{equation}

We calculate now explicitly
\begin{equation}
F\left( t\right) =\exp \left( fx^{2}\right) \exp \left[ g\left(xp+px\right) \right] \exp \left( hp^{2}\right) 1.
\end{equation}
Remembering  the definition of $p$ (expression \ref{p}) it is very easy to see that
\begin{equation}   \exp \left( hp^{2}\right) 1=1   \end{equation}
and then
\begin{equation}
F\left( t\right) =\exp \left( fx^{2}\right) \exp \left[ g\left(xp+px\right) \right] 1.
\end{equation}
Using now the $\left[ x,p\right] =i$, we write
\begin{equation}
\exp \left[ g\left( xp+px\right) \right] 1=\exp \left[ g\left( 2xp-i\right)\right]
1=\exp \left( -ig\right) \exp \left( 2gxp\right) 1
\end{equation}
and it is also clear that
\begin{eqnarray}
\exp \left( 2gxp\right) 1&&=\left[ 1+gxp+\frac{g^{2}}{2}\left(xp\right) ^{2}+...\right] 1= \\
&&1+gxp1+\frac{g^{2}}{2}\left(xp\right) ^{2}1+...=1\\
 \end{eqnarray}
and also that
\begin{equation}
\exp \left[ g\left( xp+px\right) \right] 1=\exp \left( -ig\right).
\end{equation}
We then have
\begin{equation}
F\left( t\right) =\exp \left( fx^{2}-ig\right).
\end{equation}
Substituting the functions $f$ y $g$,
\begin{equation}
F\left( t\right) =\exp \left[ \left( \frac{4t}{4t+1}\right) x^{2}-\frac{1}{2}\left[
\ln \left( 4t+1\right) \right] \right] =\frac{1}{\sqrt{4t+1}}\exp \left( \frac{4tx^{2}}{4t+1}\right),
\end{equation}
and finally we get the formula we were looking for
\begin{equation} \label{finsumherpar}
\sum\limits_{n=0}^{\infty }\frac{t^{n}}{n!}H_{2n}\left( x\right) =\frac{1}{\sqrt{4t+1}}\exp \left( \frac{4tx^{2}}{4t+1}\right).
\end{equation}
\\

\subsection{Addition formula\index{Hermite polynomials!addition formula}}
We want to apply the form obtained in (\ref{Hermite}), $H_{n}(x)=(-i)^{n}\left( p+2ix\right)^{n}1$, to evaluate the quantity
\begin{equation}\nonumber
H_n(x+y).
\end{equation}
We write it as
\begin{equation}\label{addin}
H_n(x+y)=(-i)^n{[-i\frac{d}{d(x+y)}+2i(x+y)]^n},
\end{equation}
by using the chain rule we have
\begin{equation}
\frac{d}{d(x+y)}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right),
\end{equation}
so that we may re-express (\ref{addin}) in the form
\begin{equation}
H_n(x+y)=\left(\frac{-i}{\sqrt{2}}\right)^n(-i\frac{\partial}{\partial\sqrt{2}x}+2i\sqrt{2}x-i\frac{\partial}{\partial \sqrt{2}y}+2i\sqrt{2}y)^{n}.
\end{equation}
By defining
\begin{equation}
p_X=-i\frac{\partial}{\partial X}, \qquad p_Y=-i\frac{\partial}{\partial Y}
\end{equation}
with $X=\sqrt{2}x$ and $Y=\sqrt{2}y$, we obtain
\begin{eqnarray}
H_n(x+y)=\frac{1}{2^{n/2}}\sum_{k=0}^{n}
\left(
\begin{array}{c}
n\\k
\end{array}
\right)
(-i)^k{(p_X+2iX)^k}(-i)^{n-k} (p_Y+2iY)^{n-k},
\end{eqnarray}
that by using $H_{j}(x)=(-i)^{n}(p+2ix)^{j}1$ adds to
\begin{equation}
H_n(x+y)=\frac{1}{2^{n/2}}\sum_{k=0}^{n}\left(
\begin{array}{c}
n\\k
\end{array}
\right)H_k(\sqrt{2}x)H_{n-k}(\sqrt{2}y),
\end{equation}
which is the addition formula we were looking for.
\section{Associated Laguerre polynomials}
The generating function for the associated Laguerre polynomials\index{associated Laguerre polynomials!generating function} is
\begin{equation}\label{lag1}
\sum_{n=0}^\infty L_n^\alpha(x) t^n = \frac{1}{{(1-t)}^{\alpha+1}} \exp{\left( \frac {-xt}{1-t} \right ) },\qquad |t|<1 .
\end{equation}

The associated Laguerre polynomials may be obtained from the corresponding Rodrigues' formula\index{associated Laguerre polynomials!Rodrigues' formula}
\begin{equation}\label{lag2}
L_{n}^{\alpha }\left( x\right) =\frac{1}{n!}x^{-\alpha}e^{x}\frac{d^{n}}{dx^{n}}\left( e^{-x}x^{n+\alpha }\right).
\end{equation}

The associated Laguerre polynomials satisfy several recurrence relations\index{associated Laguerre polynomials!recurrence relations}. One very useful, when extracting properties of the wave functions of the hydrogen atom, is
\begin{equation}\label{lagrr}
(n+1)L_{n+1}^{\alpha}(x)=(2n+\alpha+1-x)L_{n}^{\alpha}(x)-(n+\alpha)L_{n-1}^{\alpha}(x).
\end{equation}

We will use the operator method outlined above for the Hermite polynomials, to derive the usual explicit expression for the associated Laguerre polynomials. We rewrite expression (\ref{lag2}) as
\begin{equation}
L_{n}^{\alpha }\left( x\right) =\frac{1}{n!}x^{-\alpha}e^{x}\left(ip\right) ^{n}e^{-x}x^{n+\alpha },
\end{equation}
where again the operator $p=-id/dx$, defined in (\ref{p}), is used. \\
We notice that $e^{x}\left(ip\right) ^{n}e^{-x}=[e^{x}\left(ip\right) e^{-x}]^n$ and that, using (\ref{bch0}), $e^xpe^{-x}=(p+i)$, so
\begin{eqnarray*}
L_n^{\alpha}(x)&=&\frac{i^{n}}{n!}x^{-\alpha }\left( p+i\right) ^{n}x^{n+\alpha },
\end{eqnarray*}
or writing explicitly the operator $p$,
\begin{equation}\label{Lag}
L_n^{\alpha}(x)=\frac{1}{n!}x^{-\alpha}\left(\frac{d}{dx}-1\right)^nx^{n+\alpha}.
\end{equation}
Using the binomial expansion,
\begin{equation}
L_n^{\alpha}(x)= \frac{1}{n!}x^{-\alpha}\sum_{m=0}^n\left(
\begin{array}{c}
n\\m
\end{array}
\right)(-1)^{n-m}\frac{d^m}{dx^m}x^{n+\alpha},
\end{equation}
and because
\begin{equation}
\frac{d^m}{dx^m}x^{n+\alpha}=\frac{(n+\alpha)!}{(n+\alpha-m)!}x^{n+\alpha-m},
\end{equation}
we obtain the usual form for associated Laguerre polynomials,
\begin{equation}
L_n^{\alpha}(x)=\sum_{k=0}^n
\left(
\begin{array}{c}
n+\alpha\\n-k
\end{array}
\right)(-1)^k\frac{x^k}{k!}.
\end{equation}
\\

\section{Bessel functions of the first kind of integer order}
Bessel functions of the first kind of integer order, $J_n(x)$, are solutions of the Bessel differential equation\index{Bessel functions of the first kind!differential equation}
\begin{equation} \label{bessel10}
x^2y''+xy'+(x^2-n^2)y=0,
\end{equation}
where $n$ is an integer.
They may be obtained from the generating function\index{Bessel functions of the first kind!generating function}
\begin{equation} \label{bessel20}
\exp\Bigg[{\frac{x}{2}\left(t-\frac{1}{t}\right)}\Bigg]=\sum_{n=-\infty}^{\infty}t^nJ_n(x),
\end{equation}
and also from the following recurrence relations\index{Bessel functions of the first kind!recurrence relations}
\begin{equation}\label{bessel30}
\frac{2n}{x}J_{n}(x)=J_{n-1}(x)+J_{n+1}(x).
\end{equation}

Bessel functions of the first kind of integer order may be written as
\begin{equation} \label{bessel40}
J_n(x)=\sum_{m=0}^{\infty}\frac{(-1)^mx^{2m+n}}{2^{2m+n}m!(m+n)!},
\end{equation}
and also the following integral representation is very useful
\begin{equation}\label{bessel50}
J_{n}\left( x\right) =\frac{1}{2\pi }\int_{-\pi }^{\pi }e^{-i(n\tau-x\sin\tau)}d\tau.
\end{equation}
Some other important relations for the Bessel functions of the first kind are the Jacobi-Anger expansions\index{Bessel functions of the first kind!Jacobi-Anger expansions}:
\begin{equation}\label{ja1}
    e^{ix\cos y}=\sum_{n=-\infty}^{\infty}i^nJ_n(x)e^{iny}
\end{equation}
and
\begin{equation}\label{ja2}
    e^{ix\sin y}=\sum_{n=-\infty}^{\infty}J_n(x)e^{iny}.
\end{equation}

\subsection{Addition formula\index{Bessel functions of the first kind!addition formula}}
Using the operator methods developed in previous sections, we will obtain here the addition formula for the Bessel functions of the first kind of integer order.
\newline
First, we will derive the following expression for any "well behaved" function $f$,
\begin{equation}\label{af1}
 f(x+y)= e^{iyp_x}f(x)e^{-iyp_x}1,
\end{equation}
where $p_x=-id/dx$ is the operator introduced in Section 1, expression (\ref{p}).
Because $e^{-iyp_x}1=1$, developing the $f$ function in a Taylor series (we call $c_n$ to the coefficients in the expansion) and using the linearity of the $e^{iyp_x}$ operator,
\begin{equation}
    e^{iyp_x}f(x)e^{-iyp_x}1=e^{iyp_x}f(x)=e^{iyp_x}\sum_{k=0}^{\infty}c_{k}x^k=\sum_{k=0}^{\infty}c_{k}e^{iyp_x}x^k.
\end{equation}
Now
\begin{equation}
e^{iyp_x}x^k=\sum_{l=0}^{\infty}\frac{(iy)^l}{l!}(-i)^l\frac{d^l}{dx^l}x^k=\sum_{l=0}^{k}(y)^l
\left(
\begin{array}{c}
k\\l
\end{array}
\right)
 x^{k-l}=(x+y)^k,
\end{equation}
then
\begin{equation}
    e^{iyp_x}f(x)e^{-iyp_x}1=\sum_{k=0}^{\infty}c_{k}(x+y)^k=f(x+y),
\end{equation}
as we wanted to prove. \\
Now consider the Bessel function $J_n$ evaluated at $x+y$. From expression (\ref{af1}) we have
\begin{equation}
J_n(x+y)= e^{iyp_x}J_n(x)e^{-iyp_x}1,
\end{equation}
because $e^{-iyp_x}1=1$, and developing the first exponential in Taylor series, we obtain
\begin{equation}\label{forad}
J_n(x+y)= \sum_{m=0}^{\infty}\frac{y^m}{m!}\frac{d^m}{dx^m}J_n(x).
\end{equation}
To calculate the \textit{m}-derivative of $J_{n}$, we use the integral representation (\ref{bessel50}) to write
\begin{equation}
   \frac{d^m}{dx^m}J_n(x) = i^m \frac{1}{2\pi }\int_{-\pi }^{\pi } \sin^m\tau e^{-i(n\tau-x\sin\tau)}d\tau,
\end{equation}
substituting $\sin\tau=(e^{i\tau}-e^{-i\tau})/2i$, and using the binomial expansion,
\begin{eqnarray}\nonumber
   \frac{d^m}{dx^m}J_n(x) &=&
    \frac{1}{2^m} \dfrac{1}{2\pi } \sum_{k=0}^{m} (-1)^k
\left(
\begin{array}{c}
m\\k
\end{array}
\right) \int_{-\pi }^{\pi } e^{i(m-k)\tau} e^{-ik\tau} e^{-i(n\tau-x\sin\tau)}d\tau \\
    &=&\frac{1}{2^m} \frac{1}{2\pi } \sum_{k=0}^{m} (-1)^k \left(
\begin{array}{c}
m\\k
\end{array}
\right) \int_{-\pi }^{\pi }  e^{-i\left[(n-m+2k)\tau-x\sin\tau\right]}d\tau,
\end{eqnarray}
and therefore, using again the integral representation (\ref{bessel50}), we obtain
\begin{equation}
\frac{d^m}{dx^m}J_n(x)=\frac{1}{2^m}\sum_{k=0}^{m}(-1)^k \left(
\begin{array}{c}
m\\k
\end{array}
\right)J_{n-m+2k}(x).
\end{equation}
Substituting this last expression in equation (\ref{forad}), we obtain (we have taken the sum up to infinite as we add only zeros)
\begin{equation}
J_n(x+y)= \sum_{m=0}^{\infty}\frac{y^m}{m!}
\frac{1}{2^m}\sum_{k=0}^{\infty}(-1)^k  \left(
\begin{array}{c}
m\\k
\end{array}
\right)J_{n-m+2k}(x).
\end{equation}
We now change the order of summation and start the second sum at $m=k$ (because from $m<k$ all the terms are zero)
\begin{equation}
J_n(x+y)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}
\sum_{m=k}^{\infty}\frac{y^m}{2^m(m-k)!} J_{n-m+2k}(x).
\end{equation}
We do now $j=m-2k$ and obtain
\begin{equation}
J_n(x+y)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}
\sum_{j=-k}^{\infty}\frac{y^{j+2k}}{2^{j+2k}(j+k)!} J_{n-j}(x),
\end{equation}
take the second sum from minus infinite, and exchange the order of the sums
\begin{equation}
J_n(x+y)=\sum_{j=-\infty}^{\infty}J_{n-j}(x)\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}
\frac{y^{j+2k}}{2^{j+2k}(m+k)!}=
\sum_{j=-\infty}^{\infty}J_{n-j}(x)J_j(y).
\end{equation}
The final expression
\begin{equation}
J_n(x+y)=\sum_{k=-\infty}^{\infty}J_{n-k}(x)J_k(y)
\end{equation}
is known as the {\it addition formula} for the Bessel functions of the first kind of integer order.
\section{Conclusions}
We have shown how to apply some of the formalism of operator theory to some special functions, namely, Hermite and Laguerre polynomials and Bessel functions.
We have applied them to obtain  some known series of functions, such as the addition formula for Bessel functions, and some, to our knowledge, not known series, such as the
sum of even Hermite polynomials.

config: arxiv


text:
\begin{document}

\title{ADDENDUM \\ Details of the Derivation of the Probability of Coverage for the Relaying Scheme (Section IV in the paper: ``A Poisson Line Process based Framework for Determining the Needed RSU Density and Relaying Hops in Vehicular Networks'')}

\author{Hussein~A.~Ammar, Abdel-karim Ajami
		and~Hassan~Artail\\% <-this
		Department of Electrical and Computer Engineering\\
		American University of Beirut, Beirut 1107 2020\\
		e-mail: \{haa141, aa377, hartail\}@aub.edu.lb
\thanks{This document is an addendum for Section IV of the paper~\cite{VehicularRelayingJournal} accepted to be published at the IEEE Transactions on Wireless Communications.}

}

\maketitle

\makeatletter
\def\tagform@#1{\maketag@@@{\normalsize(#1)\@@italiccorr}}
\makeatother

\section{Abstract}
This paper develops a framework to study multi-hop relaying in a vehicular network consisting of vehicles and Road Side Units (RSUs), and the effect of this relaying on the network coverage and the communication delay. We use a stochastic geometry model that consists of a combination of Poisson Line Process (PLP) and 1D Poisson Point Process (PPP) to reliably characterize the vehicular network layout and the locations of the vehicles and the RSUs. Using this model, we analyze the effect of the different network parameters on the coverage provided by the RSUs to the vehicles. Then, we investigate how the uncovered vehicles can receive their intended packets by relaying them through multiple hops that form connected paths to the RSUs. We also analyze the delay introduced to packet delivery due to multi-hop relaying. Namely, we present results that illustrate the coverage gains achieved through multi-hop relaying and the delays induced. Such results could be used by network planners and operators to decide on the different configurations and operational parameters of the vehicular network to suit particular scenarios and objectives.

\section{System Model}
We consider a vehicular network that consists of RSUs and vehicles communicating on a typical geographical area containing roads. Since in most areas, the roads' layout is a set of straight and randomly orientated lines, we model these roads' locations as a motion-invariant Poisson Line Process (PLP) $\mathrm{\Phi}_l$~\cite{chiu2013stochastic} with a line density $\rho$. The lines of PLP can be represented by two parameters. The first one is the distance $y_n\in[0,\infty)$ for $n=\{1,2,3,\text{etc}\}$, which represents the perpendicular projection (or simply the distance) from the origin $o$ to  the $n^{th}$ nearest line $L_n$. The second parameter is the angle $\theta_n\in [0, 2\pi)$ between the $x$-axis and $y_n$ measured in a counterclockwise direction (check Figure~\ref{fig:NetworkModel1Zoomed}). When the angles of the lines have a uniform probability measure on $(0, 2\pi]$, the PLP is motion-invariant. We denote the density of the equivalent Poisson Point Processes (PPP) of $\mathrm{\Phi}_l$ on a 2D space $\mathcal{S}=[0,2\pi) \times [0,\infty)$ as $\lambda_l=\frac{\rho}{\pi}$.
On each road $L_n$, i.e., on each line from the PLP, we use two independent 1D PPP $\mathrm{\Phi}_{{\rm ru}_n}$ and $\mathrm{\Phi}_{{\rm v}_n}$ with respective densities $\lambda_{{\rm ru}}$ and $\lambda_{\rm v}$ to model the RSUs and the vehicles locations, respectively.
The RSUs are the service providers for the vehicles, i.e., they are the transmitters, and they are connected to each other through backhaul links~\cite{7973038}. Although, we are studying the downlink performance, it is worth mentioning that due to the channel reciprocity and the considered Tx parameters for the RSUs and the vehicles, the uplink performance is the same as the downlink. If we assume that the vehicles are transmitting according to a probability $p_1$, we obtain on each line $n$ a 1D PPP $\mathrm{\Phi}_{{\rm vt}_n}$ with a density $p_1\lambda_{\rm v}$ representing the locations of the vehicles currently transmitting, which follows from the independent thinning on each $\mathrm{\Phi}_{{\rm v}_n}$~\cite{haenggi2012stochastic}. Also, consistent with the literature (e.g.~\cite{6042301}) we assume that the sent signals experience Rayleigh fading with mean $1/\mu$.

\begin{figure}[H]
	\vspace{-2em}
	\centering
	\includegraphics[width=0.5\textwidth]{Figures/ZoomedNetworkView}
	\caption{Zoomed view of the vehicular network.}
	\label{fig:NetworkModel1Zoomed}

\end{figure}

\par We study the average network performance at a typical vehicle (observation point) $v_{\rm typ}$ found on the origin $o$, and served by the nearest RSU. According to Slivnyak's Theorem~\cite{haenggi2012stochastic} and the stationarity of the considered model, we can translate the model so that the typical vehicle falls on the origin of the network. The translated model can be treated as the superposition of the PLP with an additional line ($L_0$) passing through the center $\mathrm{\Phi}_{l_0}=\mathrm{\Phi}_l\cup L_0$, and then adding an additional vehicle $v_{\rm typ}$ (the observation point) to the 1D PPP $\mathrm{\Phi}_{{\rm vt}_0}$ falling on this $L_0$~\cite{6260478}\cite{chiu2013stochastic}. We note that by adding $L_0$, we have automatically added $\mathrm{\Phi}_{{\rm ru}_0}$, which is reasonable, else we would be assuming that $v_{\rm typ}$ cannot connect to RSUs on the same road.
A representation of this model used to create our vehicular network is shown in Figure~\ref{fig:NetworkModel1Zoomed} with a zoomed view of the origin.

\section{Relaying Scheme - (Section IV in manuscript)}
We study the scenario when the typical vehicle $v_{\rm typ}$ is not directly covered by its nearest RSU, hence its signal is relayed through another vehicle $v_{{\rm rel}_1}$ found at distance $r_1$, which we denote as Scenario B. In Figure~\ref{fig:figureRelaying1}, we show an example of this scenario.
\begin{figure}[H]

	\centering
	\includegraphics[width=0.45\textwidth]{Figures/RelayingFigure4_v2}
	\caption{Typical vehicle covered by RSU through a relaying vehicle at distance $r_1$.}
	\label{fig:figureRelaying1}

\end{figure}
For this scenario to be successful, $v_{\rm typ}$ must be covered by $v_{{\rm rel}_1}$ given that it is not already covered directly by the serving RSU (its own nearest RSU), and $v_{{\rm rel}_1}$ must be covered by its nearest RSU. The two constraints that render the one-hop relaying scheme useful, are listed below and are shown in Figure~\ref{fig:figureRelaying1}:
\begin{itemize}
	\item $r_{\rm b1}>r_1$: will be accomplished by imposing a lower bound on the variable distance $r_{\rm b1}$.
	\item $r_0<r_{\rm b1}$: will be accomplished by imposing an upper bound on the variable distance $r_0$.
\end{itemize}

\subsection{Summary of the Approach}
We define the Signal to Interference and Noise Ratio (SINR) at the relaying vehicle when it is served by its nearest RSU and the SINR at the typical vehicle when it is served by the relaying vehicle from distance $r_1$. We formulate the probability of coverage for the typical vehicle through one-hop relaying. The formula for this coverage uses the joint Laplace Transform (LT) of interference, which are defined in Lemma 1 in the manuscript and appear naturally in the expression of the coverage because of the dependence of the SINR condition for the typical vehicle before and after relaying, where the typical vehicle will get its signal relayed only when it cannot be directly covered by its nearest RSU. Hence, by using Bayes rule, the joint LT expressions appear.

\subsection{Details of Deriving the Probability of Coverage through One-hop Relaying}
The probability $p_{\rm c}^{{\rm B,rel}_1}$ for a typical vehicle $v_{\rm typ}$ to be covered by an RSU through one-hop relaying is equivalent to the probability that its SINR, received from a relaying vehicle at distance $r_1$, is greater than an SINR threshold $T$; this means that ${\rm SINR}_{\rm B}(r_1) > T$, given that this $v_{\rm typ}$ is not directly covered by the nearest RSU, i.e., ${\rm SINR}_{\rm A}(r_{\rm b1}) < T$. Moreover, the relaying vehicle should be covered by its own nearest RSU, i.e., ${\rm SINR}_{{\rm rel}}(r_0) > T$. Hence we can write the following
\begingroup
\allowdisplaybreaks
\begin{align}
p_{\rm c}^{{\rm B,rel}_1}=\mathbb{P}\left[{\rm SINR}_{\rm B}(r_1)>T|{\rm SINR}_{\rm A}(r_{\rm b1})<T\right]\times\mathbb{P}\left[{\rm SINR}_{{\rm rel}}(r_0)>T\right]
\end{align}
By using Bayes rule for the first term, we express Equation (41) in the manuscript, by writing:
\begin{align}
p_{\rm c}^{{\rm B,rel}_1}&
= \frac{\mathbb{P}\left[{\rm SINR}_{\rm B}(r_1)>T,{\rm SINR}_{\rm A}(r_{\rm b1})<T\right]}{\mathbb{P}\left[{\rm SINR}_{\rm A}(r_{\rm b1})<T\right]} \times\mathbb{P}\left[{\rm SINR}_{{\rm rel}}(r_0)>T\right]
\nonumber \\
&=
\frac{{\xi_1}_{(r_1)}}{{\xi_3}_{(r_1)}} \times {\xi_2}_{(r_1)}
\end{align}

\par Now, we describe how the three elements of Equation (2) are derived. We can therefore think of this section as a detailed version of Appendix C in the manuscript.
\newline
\par Since the small-scale fading experienced by the signal powers is Rayleigh fading, the fading power is exponentially distributed. Hence, we can write the first term ${\xi_1}_{(r_1)}$ as follows \cite[Appendix~A]{6047548}
\begin{flalign}
&{\xi_1}_{(r_1)}=
\resizebox{0.86\columnwidth}{!}
{$\displaystyle
	\mathbb{E}_{r_{\rm b1}}\Bigg[e^{\left(-\frac{\mu T r_1^\eta N}{\nu} - \mu\frac{\kappa Tr_1^\eta}{\nu} I_{{\rm ru}} - \mu Tr_1^\eta I_{{\rm vt}}\right)}\Bigg(1-\bigg(p_{\rm c1}^\text{A}\left( r_{\rm b1},s_7,s_8\right)+\sum_{n=1}^{\infty}\int_{0}^{\infty}p_{\rm c2}^\text{A}\left( r_{\rm b1},s_7,s_8,y_n\right)
	\mathop{}\! \mathrm{d}{y_{n}}\bigg)\Bigg)\Bigg]
	$}
\nonumber\\
&
=
\resizebox{0.86\columnwidth}{!}
{$\displaystyle
	\mathbb{E}_{r_{\rm b1}}\Bigg[e^{\left(-\frac{\mu T r_1^\eta N}{\nu}\right)}L_{I_{{\rm ru}}}\left(s_5\right)L_{I_{{\rm vt}}}\left(s_6\right)\Bigg(1-\bigg(p_{\rm c1}^\text{A}\left( r_{\rm b1},s_7,s_8\right)+\sum_{n=1}^{\infty}\int_{0}^{\infty}p_{\rm c2}^\text{A}\left( r_{\rm b1},s_7,s_8,y_n\right)
	\mathop{}\! \mathrm{d}{y_{n}}\bigg)\Bigg)\Bigg]
	$}
\nonumber&&
\end{flalign}
where $s_5=\frac{\kappa\mu Tr_1^\eta}{\nu}$, $s_6=\mu Tr_1^\eta$, $s_7=\mu Tr_{\rm b1}^\eta$, and $s_8=\frac{\nu\mu Tr_{\rm b1}^\eta}{\kappa}$. Hence
\begin{flalign}

&{\xi_1}_{(r_1)}
=
\nonumber \\
&
\resizebox{0.97\columnwidth}{!}
{$\displaystyle
	\mathbb{E}_{r_{\rm b1}}\Bigg[e^{\left(-\frac{\mu T r_1^\eta N}{\nu}\right)}L_{I_{{\rm ru}}}\left(s_5\right)L_{I_{{\rm vt}}}\left(s_6\right)-p_{\rm c1}^\text{AB}\left( r_{\rm b1},s_7,s_5,s_8,s_6\right)-\sum_{n=1}^{\infty}\int_{0}^{\infty}p_{\rm c2}^\text{AB}\left( r_{\rm b1},s_7,s_5,s_8,s_6,y_n\right)
	\mathop{}\! \mathrm{d}{y_n}\Bigg]
	$}
\nonumber\\

&=\Bigg[e^{\left(-\frac{\mu T r_1^\eta N}{\nu}\right)}L_{I_{{\rm ru}}}\left(s_5\right)L_{I_{{\rm vt}}}\left(s_6\right)-\int_{r_1}^{\infty}p_{\rm c1}^\text{AB}\left( r_{\rm b1},s_7,s_5,s_8,s_6\right) f_{R}(r_{\rm b1}|\varepsilon_0)
\mathop{}\! \mathrm{d}{r_{\rm b1}}\nonumber\\
&
\quad
-\sum_{n=1}^{\infty}\int_{0}^{\infty}\int_{m(y_n)}^{\infty}p_{\rm c2}^\text{AB}\left( r_{\rm b1},s_7,s_5,s_8,s_6,y_n\right) f_{R}(r_{\rm b1}|\varepsilon_n,y_n)
\mathop{}\! \mathrm{d}{r_{\rm b1}} \mathop{}\! \mathrm{d}{y_n}\Bigg]

\end{flalign}
As can be seen this term involves expectation over the variable $r_{\rm b1}$ with the known Probability Density Functions (PDFs) $f_{R}(r_{\rm b1}|\varepsilon_0)$ and $f_{R}(r_{\rm b1}|\varepsilon_n,y_n)$. Also, we have
\begin{align}\label{eq:pc1AB}
p_{\rm c1}^\text{AB}\left( r_{\rm b1},s_5,s_7,s_6,s_8\right)=&\mathbb{P}\left[\varepsilon_0\right]e^{\left(-\mu TN\left(\frac{r_1^\eta}{\nu}+\frac{r_{\rm b1}^\eta}{\kappa}\right)\right)} L_{\{I_{0},I_{0}\}}\left(\{s_5,s_7\}|\varepsilon_0\right)L_{\{I_{3},I_{3}\}}\left(\{s_5,s_7\}|\varepsilon_0\right)\nonumber\\
&\times L_{\{I_{4},I_{4}\}}\left(\{s_5,s_7\}|\varepsilon_0\right)L_{\{I_{{\rm vt}},I_{{\rm vt}}\}}\left(\{s_6,s_8\}\right)
\end{align}
\vspace{-1.5em}
\begin{align}\label{eq:pc2AB}
&p_{\rm c2}^\text{AB}\left( r_{\rm b1},s_5,s_7,s_6,s_8,y_n\right)=
\resizebox{0.7\columnwidth}{!}
{$\displaystyle
	\mathbb{P}\left[\varepsilon_n|Y_n\right]e^{\left(-\mu TN\left(\frac{r_1^\eta}{\nu}+\frac{r_{\rm b1}^\eta}{\kappa}\right)\right)} L_{\{I_{0},I_{0}\}}\left(\{s_5,s_7\}|\varepsilon_n,y_n\right)L_{\{I_{1},I_{1}\}}\left(\{s_5,s_7\}|\varepsilon_n,y_n\right)
	$}
\nonumber\\
&\ \ \times
\resizebox{0.95\columnwidth}{!}
{$\displaystyle
	L_{\{I_{2},I_{2}\}}\left(\{s_5,s_7\}|\varepsilon_n,y_n\right)L_{\{I_{3},I_{3}\}}\left(\{s_5,s_7\}|\varepsilon_n,y_n\right) L_{\{I_{4},I_{4}\}}\left(\{s_5,s_7\}|\varepsilon_n,y_n\right)L_{\{I_{{\rm vt}},I_{{\rm vt}}\}}\left(\{s_6,s_8\}\right)f_{Y_n}(y_n)
	$}
\end{align}
The $L_{\{I_{x},I_{y}\}}\left(\{s_5,s_7\}\right)$ are the joint Laplace Transform (LT) terms of $I_x$ and $I_y$ evaluated at $(s_5,s_7)$ and will be discussed later. These expressions appear in the coverage because of the dependence of the SINR condition for the typical vehicle before and after relaying, where the typical vehicle will get its signal relaying only when it cannot be directly covered by its nearest RSU.
\newline
\par The second term ${\xi_2}_{(r_1)}$ can be derived as follows.
\begin{align}\label{xi_2_main}
&{\xi_2}_{(r_1)}=\mathbb{E}_{r_0,r_{\rm b1}}\bigg[p_{\rm c1}^\text{A}\left( r_0,s_3,s_4\right)+\sum_{n=1}^{\infty}\int_{0}^{r_{\rm b1}}p_{\rm c2}^\text{A}\left( r_0,s_3,s_4,y_q\right)
\mathop{}\! \mathrm{d}{y_q}\bigg]
\end{align}
When applying the expectation over the distances $r_0$ and $r_{\rm b1}$, equation \eqref{xi_2_main} can be divided into four cases:
\begin{itemize}
	\item Case 1: Both the relaying and the typical vehicles have their nearest RSU on the same road.
	\item Case 2: The relaying vehicle has its nearest RSU on the same road, and the typical vehicle has its nearest RSU on the $n^\text{th}$ nearest road.
	\item Case 3: The relaying vehicle has its nearest RSU on the $n^\text{th}$ nearest road, and the typical vehicle has its nearest RSU on the same road.
	\item Case 4: The relaying vehicle has its nearest RSU on the $n^\text{th}$ nearest road, and the typical vehicle has its nearest RSU on the $q^\text{th}$ nearest road.
\end{itemize}
We can account for these cases using the following.
\begin{align}
&{\xi_2}_{(r_1)} =\nonumber \\
&
\resizebox{0.97\columnwidth}{!}
{$\displaystyle
	\mathbb{E}_{r_{\rm b1}}\Bigg[\boldsymbol{\Bigg[}\int_{0}^{r_{\rm b1}}p_{\rm c1}^\text{A}\left( r_0,s_3,s_4\right)f_{R}(r_0|\varepsilon_0)
	\mathop{}\! \mathrm{d}{r_0}\boldsymbol{\Bigg]} +\boldsymbol{\Bigg[}\sum_{n=1}^{\infty}\int_{0}^{r_{\rm b1}}\int_{y_q}^{r_{\rm b1}}p_{\rm c2}^\text{A}\left( r_0,s_3,s_4,y_q\right)f_{R}(r_0|\varepsilon_q,y_q)
	\mathop{}\! \mathrm{d}{r_0} \mathop{}\! \mathrm{d}{y_q}\boldsymbol{\Bigg]}\Bigg]
	$}
\nonumber\\

&=\boldsymbol{\Bigg[}
\underbrace{
	\int_{r_1}^{\infty}\bigg(\int_{0}^{r_{\rm b1}}p_{\rm c1}^\text{A}\left( r_0,s_3,s_4\right)f_{R}(r_0|\varepsilon_0)
	\mathop{}\! \mathrm{d}{r_0}\bigg)\mathbb{P}\left[\varepsilon_0\right]f_{R}(r_{\rm b1}|\varepsilon_0)
	\mathop{}\! \mathrm{d}{r_{\rm b1}}
}_{\text{case 1}}
\nonumber\\
&\quad
+
\resizebox{0.945\columnwidth}{!}
{$\displaystyle
	\underbrace{
		\sum_{n=1}^{\infty}\int_{0}^{\infty}\int_{m(y_n)}^{\infty}\bigg(\int_{0}^{r_{\rm b1}}p_{\rm c1}^\text{A}\left( r_0,s_3,s_4\right)f_{R}(r_0|\varepsilon_0)
		\mathop{}\! \mathrm{d}{r_0}\bigg)

		\mathbb{P}\left[\varepsilon_n|Y_n\right]f_{R}(r_{\rm b1}|\varepsilon_n,y_n)f_{Y_n}(y_n)
		\mathop{}\! \mathrm{d}{r_{\rm b1}} \mathop{}\! \mathrm{d}{y_n}
	}_{\text{case 2}}
	\boldsymbol{\Bigg]}
	$}
\nonumber\\
&\quad+
\boldsymbol{\Bigg[}
\underbrace{
	\int_{r_1}^{\infty}\bigg(\sum_{n=1}^{\infty}\int_{0}^{r_{\rm b 1}}\int_{y_q}^{r_{\rm b1}}p_{\rm c2}^\text{A}\left( r_0,s_3,s_4,y_q\right) f_{R}(r_0|\varepsilon_q,y_q)
	\mathop{}\! \mathrm{d}{r_0}
	\mathop{}\! \mathrm{d}{y_q}\bigg)\mathbb{P}\left[\varepsilon_0\right]f_{R}(r_{\rm b1}|\varepsilon_0)
	\mathop{}\! \mathrm{d}{r_{\rm b1}}
}_{\text{case 3}}
\nonumber\\
&\quad
+
\underbrace{
	\sum_{q=1}^{\infty}\int_{0}^{\infty}\int_{m(y_n)}^{\infty}\bigg(\sum_{n=1}^{\infty}\int_{0}^{r_{\rm b 1}}\int_{y_q}^{r_{\rm b1}}  p_{\rm c2}^\text{A}\left( r_0,s_3,s_4,y_q\right)f_{R}(r_0|\varepsilon_q,y_q)
	\mathop{}\! \mathrm{d}{r_0}
	\mathop{}\! \mathrm{d}{y_q}\bigg)\mathbb{P}\left[\varepsilon_n|Y_n\right]
}_{\text{case 4}}
\nonumber\\
&\quad \quad
\underbrace{
	\times
	f_{R}(r_{\rm b1}|\varepsilon_n,y_n) f_{Y_n}(y_n)
	\mathop{}\! \mathrm{d}{r_{\rm b1}}\mathop{}\! \mathrm{d}{y_n}
}_{\text{case 4 (continued)}}
\boldsymbol{\Bigg]}
\end{align}
where $m(y_n)=\max\{r_1,y_n\}$ is the maximum of $r_1$ and $y_n$.
The expressions for $p_{\rm c1}^\text{A}$, $p_{\rm c2}^\text{A}$, $\mathbb{P}\left[\varepsilon_0\right]$, $\mathbb{P}\left[\varepsilon_n|Y_n\right]$ $f_{Y_n}(y_n)$ are found in the manuscript. Also, more discussion is found at the end of Appendix C and in Theorem 1.
\newline
\par As for ${\xi_3}_{(r_1)}$, it is the same probability of outage derived in Scenario A (direct connection to the nearest RSU from the concerned vehicle).
\begin{align}
{\xi_3}_{(r_1)}&=
\resizebox{0.9\columnwidth}{!}
{$\displaystyle
	1-\Bigg[\int_{r_1}^{\infty}p_{\rm c1}^\text{A}\left( r_{\rm b1},s_7,s_8\right)f_{R}(r_{\rm b1}|\varepsilon_0)
	\mathop{}\! \mathrm{d}{r_{\rm b1}}+\sum_{n=1}^{\infty}\int_{0}^{\infty}\int_{m(y_n)}^{\infty}p_{\rm c2}^\text{A}\left( r_{\rm b1},s_7,s_8,y_n\right)f_{R}(r_{\rm b1}|\varepsilon_n,y_n)
	\mathop{}\! \mathrm{d}{r_{\rm b1}}
	\mathop{}\! \mathrm{d}{y_{n}}\Bigg]
	$}
\end{align}

\endgroup
Now, regarding the joint Laplace transforms (LTs) for the interference found in the equations \eqref{eq:pc1AB} and \eqref{eq:pc2AB} (Lemma 1, equations (32)-(38) in the manuscript) and which depend on the distance $r_{\rm b1}$, there are many works in the literature that show how these LTs can be derived, e.g.,~\cite{6047548} and~\cite{6220221}.

For the case of $L_{\{I_{0},I_{0}\}}\left(\{s_5, s_7\}|\varepsilon_0\right)$, it is derived as follows

\begin{align}
&L_{\{I_{0},I_{0}\}}\left(\{s_5, s_7\}|\varepsilon_0\right)=\mathbb{E}_{g,g',\mathrm{\Phi_{{\rm ru}_0}}}\left[\exp\left(-s_5\sum_{j\in\mathrm{\Phi_{{\rm ru}_0}}}g'_{0,j}R_{0,j}^{-\eta} -s_7\sum_{j\in\mathrm{\Phi_{{\rm ru}_0}}\backslash b(o,r_{\rm b1})}g_{0,j}R_{0,j}^{-\eta} \right)\right]\nonumber\\
&=\mathbb{E}_{g,g',\mathrm{\Phi_{{\rm ru}_0}}}\left[\prod_{j\in\mathrm{\Phi_{{\rm ru}_0}}\backslash b(o,r_{\rm b1})}\exp\left(-s_7g_{0,j}R_{0,j}^{-\eta}\right)\prod_{j\in\mathrm{\Phi_{{\rm ru}_0}}}\exp\left(-s_5g'_{0,j}R_{0,j}^{-\eta}\right)\right]\nonumber\\
&=
\resizebox{0.97\columnwidth}{!}
{$\displaystyle
	\mathbb{E}_{\mathrm{\Phi_{{\rm ru}_0}}}\left[\prod_{j\in\mathrm{\Phi_{ru_0}}\backslash b(o,r_{\rm b1})}\frac{\mu}{\mu+s_7R_j^{-\eta}}\prod_{j\in\mathrm{\Phi_{{\rm ru}_0}}}\frac{\mu}{\mu+s_5R_j^{-\eta}}\right]\stackrel{(a)}{=}\exp\Bigg[-2\lambda_{{\rm ru}}\int_{r_{\rm b1}}^{\infty}\left(1-\frac{\mu}{\mu+s_7x^{-\eta}}\frac{\mu}{\mu+s_5x^{-\eta}}\right)
	\mathop{}\! \mathrm{d}x\Bigg]
	$}
\end{align}
where $r_{\rm b1}>r_1$ and (a) follow from the Probability Generating Functional (PGFL) of the 1D PPP~\cite{haenggi2012stochastic}. The interference from $\mathrm{\Phi_{ru_0}}$ is when the relaying vehicle is transmitting to $v_{\rm typ}$, while the interference from $\mathrm{\Phi_{{\rm ru}_0}}\backslash b(o,r_{\rm b1})$ is when the nearest RSU (w.r.t. to $v_{\rm typ}$) would be transmitting to $v_{\rm typ}$. Additionally, although the term that contains $s_5$ does not contain an exclusion region for the interference (because when the relaying vehicle is sending the data to $v_{\rm typ}$, all RSUs can be an interfering source), we do not add additional terms for the PGFL integral, i.e., $\exp\Bigg[-2\lambda_{{\rm ru}}\bigg(\int_{r_{\rm b1}}^{\infty}\left(1-\frac{\mu}{\mu+s_7x^{-\eta}}\frac{\mu}{\mu+s_5x^{-\eta}}\right)
\mathop{}\! \mathrm{d}x+B\bigg)\Bigg]$, because by definition we already know that the interfering RSUs
are further than distance $r_{\rm b1}$, and they are already included in the integral.

For the other LTs we follow the same derivations done in~\cite[Lemma 12, 13, 14 and 15]{8340239}, which use the main properties of the 1D PPP and the PLP~\cite{haenggi2012stochastic} and include conditioning on the serving distance $r_{\rm b1}$, the distance $y_n$ from the typical vehicle to the roads, and the event $\epsilon_n$. These derivations are however done with replacing the terms $\frac{s_1}{\mu\left(x^2+y^2\right)^\frac{\eta}{2}+s_1}$ in the LTs from Scenario A in the manuscript by
\begin{align}
\zeta_2\left(x,y_n,s_5,s_7\right)=1-\frac{\mu}{\mu+s_5\left(x^2+{y_n}^2\right)^\frac{-\eta}{2}} \left( \frac{\mu}{\mu+s_7\left(x^2+{y_n}^2\right)^\frac{-\eta}{2}} \right)
\end{align}
The readers can refer to~\cite{8340239} for the detailed derivations mentioned above.

\subsection{Final Notes}
We do not impose on the nearest RSU to be the same for the typical vehicle $v_{\rm typ}$ and for the relaying vehicle $v_{{\rm rel}}$, thus making $r_0$ and $r_{\rm b1}$ independent. Note that by using the properties of the homogeneous Poisson process, the numbers of points $N$ in two disjoint regions $R_1$ and $R_2$ are independent~\cite{haenggi2012stochastic}, and we can directly relate the number of points found in a region to the serving distance through the relation
\begin{align}
\mathbb{P}(\text{dist}(u,\Phi) \le r) = \mathbb{P}(N(b(u,r))>0)
\end{align}
where $\text{dist}(u,\Phi) \triangleq \min\{x \in \Phi: \|x - u\|\}$, with $u \in \mathbb{R}^2$, is the distance from point $u$ to the nearest point in the Poisson process $\Phi$, and $N(b(u,r))$ is the number of points in the circular region of center $u$ and radius $r$.
\par If the nearest RSU to $v_{\rm typ}$ cannot provide the coverage condition, then $v_{\rm typ}$ may get its signal relayed via another vehicle $v_{{\rm rel}}$ that is covered by its corresponding nearest RSU. Hence, it follows that the conditions for the distances $r_0$ and $r_{\rm b1}$ may be from different RSUs. As a result, we can apply the expectations over these distances independently, again, because we do not require that the nearest RSU to the two vehicles to be the same. Our postulation makes the system more flexible and the analysis more tractable as compared to a condition that states that the serving and nearest-to-relay RSUs should be the same. By more flexible we mean that with our design choice, we enable another RSU to send the typical vehicle the intended message. This is a valid argument since the RSUs are considered to be connected, and moreover, via high capacity backhaul links, as stated in several works, like~\cite{7973038}. Hence if the serving RSU cannot deliver the message to $v_{\rm typ}$, it can forward it via the backhaul link to the nearest RSU to $v_{\rm rel}$ that can relay the message to the $v_{\rm typ}$. Not to mention the caching capabilities at the RSUs for the vehicles data, and the fact that data generated by some common services, like map-updates, is likely to be available on all RSUs.

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi

\footnotesize
\bibliography{Model_References}
\bibliographystyle{unsrt}

\end{document}


TITLE: why moon has a spectra, as it doesn' t have any energy source in its interior?
QUESTION [1 upvotes]: Moon doesn't have any source of energy, what does causes the spectra of moon to form? Is the spectrum seen is due to the sun.

REPLY [2 votes]: Because the moon has a temperature, it emits "black body radiation', which is electromagnetic radiation with a spectrum characteristic of the temperature.  The composition and microstructure of the surface can modify that spectrum.
Also, as you guessed, sunlight reflects off the moon's surface.  Sunlight has its own spectrum, which is modified when reflected, due to the composition and microstructure of the moon's surface.


TITLE: Computing Euler Characteristics of Line Bundles on the Hilbert Scheme of n points
QUESTION [2 upvotes]: Let $S^{[n]}$ be the Hilbert scheme of $n$ points on a smooth projective surface (actually, right now I am particularly interested in del Pezzo surfaces). Let $B$ be the exceptional divisor of the Hilbert-Chow morphism $S^{[n]}  \rightarrow \operatorname{Sym}^n S$. Let $L$ be a divisor on $S$ and $\tilde L$ the corresponding "symmeterized" divisor on $S^{[n]}$, i.e. the set of $D \in S^{[n]}$ such that $\operatorname{Supp} D \cap \operatorname{Supp} L \neq 0$.
I would like to have a formula for $\chi(\tilde L+\frac{r}{2}B)$ in terms of $r$, $n$, and the invariants of $S$ and $L$. (We have a conjecture for del Pezzo surfaces by doing a different computation that we expect to match.)
In, e.g. [Li, Qin, and Wang], they say that there is an algorithm to compute the cup product of any two cohomology classes on $S^{[n]}$ for an arbitrary $S$. Furthermore, Boissiere has a paper about finding "universal formulas" for chern classes of tangent bundles of $S^{[n]}$.
If I understand the chern classes of the the tangent bundle, maybe I could understand the todd class, and assuming I could convert $\tilde L + \frac{r}{2}B$ into generators suitable for the algorithm mentioned by L,Q,W, maybe I could use GRR to compute the desired Euler characteristic.
I've been reading abstracts and introductions of these and surrounding papers, but I'm having a hard time getting a feel for what they can do. I don't mind putting in some time in to learn some new stuff, but I'd' like to be sure I'm going in the right direction first.
Note
I previously asked this question for the special case $n=2$, and people pointed out in the comments that one could simply use a nice description of $S^{[2]}$ to do this computation, and indeed this turned out to be the case. But now I want to do this for arbitrary $n$.
Question
Is the kind of stuff referenced above the right way to go? If so, do you have any recommendations for where to start reading? Or is there a different strategy that looks better?

REPLY [1 votes]: Here is the paper to look at.
http://arxiv.org/abs/math/9904095


TITLE: What is the Euler characteristic of a Hilbert scheme of points of a singular algebraic curve?
QUESTION [17 upvotes]: Let $X$ be a smooth surface of genus $g$ and $S^nX$ its n-symmetrical product (that is, the quotient of $X \times ... \times X$ by the symmetric group $S_n$).  There is a well known, cool formula computing the Euler characteristic of all these n-symmetrical products:
$$\sum_{d \geq 0} \ \chi \left(X^{[d]} \right)q^d \ \ = \ \ (1-q)^{- \chi(X)}$$
It is known that $S^nX \cong X^{[n]}$, the Hilbert scheme of 0-subschemes of length n over $X$.  Hence, the previous formula also computes the Euler characteristic of these spaces.
What about for singular surfaces?  More precisely, if $X$ is a singular complex algebraic curve, do you know how to compute the Euler characteristic of its n-symmetrical powers $S^nX$? More importantly: what is the Euler characteristic of $X^{[n]}$, the Hilbert scheme of 0-schemes of length n over $X$?
I guess it is too much to hope for a formula as neat as the one given for the smooth case.  Examples, formulas for a few cases or general behaviour (e.g. if for large n, $\chi\left(X^{[n]}\right) = 0)$ are all very welcome!

REPLY [16 votes]: For singular plane curves, there is a conjectural formula (due to Alexei Oblomkov and myself) in terms of the HOMFLY polynomial of the links of the singularities.  For curves whose singularities are torus knots, i.e. like x^a = y^b for a,b relatively prime, and for a few more singularities, the conjecture has been established.  See this preprint.
Edit: More recently I have given a different characterization of these numbers in terms of multiplicities of certain strata in the versal deformation of the singular curve.


TITLE: Analyze the logical forms of the following statements?
QUESTION [0 upvotes]: Analyze the logical forms of the following statements:
3 is a common divisor of 6, 9, and 15.
Is my solution correct?
Let f(x) stand for 3 is a common divisor of x
f(6) && f(9) && f(15)
Thanks for any help!

REPLY [1 votes]: "Let $f(x)$ stand for 3 is a common divisor of $x$" is unhappy twice over.
First, you must mean "3 is a divisor of $x$" -- 3 can't be a common divisor of just one number: you are a common devisor of some numbers, plural!!
Second, $f$ is standardly a function expression, and what is defined here isn't a function but a property, the property $x$ has when divisible by $3$. So you need a predicate expression.
So you could use the one-place predicate $P$, where $Px$ stands for 3 is a divisor of $x$, and then the original statement can be rendered $P(6) \land P(9) \land P(15)$ -- with bracketing if you want to be fussy.
Or depending on context, you might prefer to use the two-place predicate $R$ where $R(x,y)$ means that $x$ divides $y$, and then render the original statement $R(3,6) \land R(3,9) \land R(3,15)$.
There's no context-free right answer about which of those gives "the" logical form (there is no such thing).


TITLE: Finding SVD of a linear operator (in matrix form)
QUESTION [0 upvotes]: The linear operator $T\in \mathcal{\mathbb{R}^2}$ defined by $T(x,y)=(2y,x)$ has singular value decomposition (SVD) $$T(x,y) = 2\langle (x,y), (0,1)\rangle (1,0)+1\langle (x,y),(1,0)\rangle (0,1).$$
However I would like to consider a "natural" way to express this fact in terms of matrices. I.e., something of the form $A=U\Sigma V^*$, where $U,V$ are unitary and $\Sigma$ is diagonal with singular values of $T$ on the diagonal.
This suggests that I take $\Sigma=\left(\begin{array}{cc} 2 & 0\\ 0 & 1 \end{array}\right)$,
 $V=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right),$ and $U=I_2$. The product gives $$U\Sigma V^*=\left(\begin{array}{cc} 0 & 1\\ 2 & 0 \end{array}\right).$$
Of all matrix representations of $T$, why is the right hand side the matrix of $T$ in the standard basis? There are two bases involved (the columns of $U$ and the columns of $V$), so I don't see why the matrix on the right hand side only takes one of these bases into account.

REPLY [2 votes]: I don't understand your question too well so it might be that this doesn't provide an answer. But let my try to address what I believe confuses you.
What you are doing with the matrices $U,V$ is you change the basis of your vector space $\mathbb R^2$ on both sides. Let $A$ denote a two-dimensional vector space. Then you are considering a linear map $f:A\to A$ and you represent it by a matrix $T$ which we interpret as linear map $T:\mathbb R^2\to\mathbb R^2$. This representation as matrix involves a choice of basis for $A$ on the left and on the right.
So what you are actually doing with singular vector decomposition is that you take a basis $B$ on the left $A$ and another one, say $B'$ on the right $A$ and write down, how $T$ looks with respect to these bases. Check the wikipedia article on base change for further details.
For your specific example, $B=\{(1,0)^t,(0,1)^t\}$ and $B'=\{(0,1)^t,(1,0)^t\}$. It can happen that one of these bases is the standard basis. But you are always taking a basis on the left and one on the right because otherwise you wouldn't be able to speak about a matrix.


TITLE: Sixth and eighth moments of a complex Gaussian random vector
QUESTION [0 upvotes]: Considering the random vector $\textbf{h} = [h_{1}, h_{2}, \ldots, h_{M}]^{T} \sim \mathcal{CN}(\textbf{0}_{M},d\textbf{I}_{M \times M})$, what would be the expectation of
$\mathbb{E}[\| \textbf{h} \|^{6}]$ and $\mathbb{E}[\| \textbf{h} \|^{8}]$ ?

REPLY [1 votes]: Note: the following treatment assumes $d$ generalises the variance in a univariate $N(0,\,\sigma^2)$ distribution. I'm also not an expert on the complex Gaussian distribution, so I'll "flatten" the variable to a $2M$-dimensional real multivariate Gaussian. (This allows us to drop the $||$ signs around even powers.) There might be some powers-of-$2$ errors, but hopefully you can fix the sketch below.
Let $a=\frac{1}{2d}$, so unitarity is the condition $\int d^{2M}\mathbf{h}\exp -ah^2=(\pi/a)^{M}$. Differentiating both sides $k$ times with respect to $a$, and dividing by the original equation, gives $$\mathbf{E}h^{2k}=(M)^{(k)} a^{-k}=(M)^{(k)} (2d)^{k},$$where $(x)^{(k)}:=\prod_{i=0}^{k-1}(x+i)$ is the rising Pochhammer symbol. This technique for evaluating even moments is well-known for the case $M=1/2$ [sic], say with $d=1$ for brevity. For example, if $H\sim N(0,\,1)$ then $$\mathbb{E}H^2=\frac{1}{2}2=1,\,\mathbb{E}H^4=\frac{1}{2}\frac{3}{2}2^2=3,$$and similarly $\mathbb{E}H^6=15,\,\mathbb{E}H^8=105$. In theory, the $6$th & $8$th moments are respectively $8d^3 M(M+1)(M+2),\,16d^4 M(M+1)(M+2)(M+3)$.


TITLE: Is it scalar product given by mapping
QUESTION [0 upvotes]: Firstly, sorry for poor definition of the following example, I translated it from my native language,hope you can understand
So i need help with this:
Mapping is given $\langle , \rangle:R_{2}[x] \times R_{2}[x] \rightarrow R$ by:
$$\langle a+bx+cx^2, \alpha+\beta x+\gamma x^2 \rangle = a\alpha+2b\beta+2c\gamma-a\beta-b\alpha-b\gamma-c\beta$$
Check if scalar product of space $R_{2}[x]$ is given by mapping $\langle , \rangle$
So as far as I know to check if this is a scalar product we need to check next:
1.$\langle x,x \rangle\geq 0 , \forall x\in E$, E is real vector space
$\langle x,x \rangle = 0\Leftrightarrow x=0$
2.$\langle x,y \rangle = \langle y,x \rangle$
3.$\langle \lambda x,y \rangle = \lambda\langle x,y \rangle$, $\lambda \in R$
4.$\langle x+y,z \rangle = \langle x,z \rangle + \langle y,z \rangle$
I know how to do this for simple examples like when $x$ is given only with $x=\lbrace \alpha_{1},\alpha_{2},\alpha_{3}\rbrace$,but for something like this above i got confused

REPLY [1 votes]: Your usage of x and y in the definiton is confusing because x also appears in the polynomial. Your x and y are now polynomials of degree 2, so your real vectorspace now is the vectorspace of polynomials with degree 2. For a better understanding write the definitons with polynomials $p$, $q$ and $p=a+bx+cx^2$, $q=\alpha + \beta x + \gamma x^2$.
Then you have got to Show for example $(p, p) \geq 0$ for All polynomials $p \in R_2 [x] $.
Example:
Let $p \in R_2[x]$, so $p=a +bx + cx^2$ for some $a,b, c \in R$.
Then $0 = (p, p) =a^2 + 2b^2 + 2c^2 - 2ab - 2bc =(a-b)^2 + (b-c)^2 + c^2$, which is equivalent to $(a-b) ^2 =0$, $(b-c)^2=0$ and $c^2=0$ (since for any real number $r$ it holds $r^2 \geq 0$). Now you can go on with equivalences until you get to $p=0$. But notice that $p$ is a polynomial, so all coefficients $a, b, c$ have to be 0.


TITLE: Need help on this geometry problem about angles
QUESTION [0 upvotes]: A certain trapezoid has these properties: Its diagonals are congruent and perpendicular to each other, and its longer base lengths is equal to the length of a diagonal. What is the sum of the degree measures of the two acute angles of this trapezoid?
I looked all the properties and am not able to get there. Any hint is greatly appreciated. I want to solve this with any hints.

REPLY [3 votes]: Because AC = AB, then triangle ABC is isosceles. Hence $45 + x = y$ where y = 180 - 90 - x.
Solve $45 + x = 180 - 90 - x$ for $x$ to determine the two equal base angles.


TITLE: Car horn in the cold
QUESTION [0 upvotes]: I live in Chicago, where it is EXTREMELY cold right now. A couple of mornings ago it was -20C in the morning, and as I was driving through a parking lot a car pulled out of a space right in front of me. I swerved and hit the horn hard to warn the other driver. Fortunately I managed to avoid a collision, however, my car horn did not make a sound. Nothing, nada.
As I was driving to work, after the engine and car had warmed up, all of a sudden the horn started working again.
The fact that it worked tells me there was not a connection problem or such. The only variable was the temperature, but I can't understand why warm air would vibrate, and cold air wouldn't.
Can anyone offer an explanation?

REPLY [2 votes]: Car horn switches are not hermetically sealed or moisture free (unlike some relays found in cars - experience). Assuming your switch/button horn activator had previously fallen below dew point and moisture had condensed on it's contact surfaces and subsequently frozen - ice is not a good conductor of electricity and thus the horn didn't work. Horn switches are normally of the direct push type pressing contacts together on one axis.
Indicators (turn signal) switches are normally of the sliding sort and thus to some extent self cleaning, this is also true of the Tail and headlight switch in my car, perhaps in yours too, thus they are more likely to work first time.
As an experiment, the next time the morning is frosty, you could try pressing the horn repeatedly and hard, this should transfer enough heat energy to de-frost on the contacts and make it work. If this is not effective then an assumption about the diaphragm of the horn being less flexible at frosty temperatures, or the voltage supplied not being sufficient - because you had just started the car (using more than a few amps, starter motors are very power hungry), turned on the lights, rear window demister circuit etc. and the voltage in the battery/alternator circuit had not recovered to one sufficient to make it work.
These are all speculative to some extent, without making specific measurements regarding resistances and voltages, and moisture levels the previous day and of course temperature; it is difficult to do anything other than speculate.


TITLE: Proof of Number Theoretic Function $\sigma$
QUESTION [0 upvotes]: If $N$ is a positive integer then,$$\sum\limits_{n=1}^{N}\sigma(n)=\sum\limits_{n=1}^{N}n\lfloor\dfrac{N}{n}\rfloor$$, where $\lfloor.\rfloor$ denotes greatest integer function and $\sigma(n)$ denotes sum of divisors of $n$ (positive).

I can't think for it, please give very elementary proof here.

REPLY [0 votes]: Let's use induction on $N$. For $N=1$ the equality becomes $1=1$.
For $n=1,\ldots, N$, $n\lfloor N/n\rfloor$ is the greatest multiple of $n$ that is not greater than $N$.
Then, assuming that the equality holds, let's try to prove that
$$\sum_{n=1}^{N+1}\sigma(n)=\sum_{n=1}^{N+1}n\left \lfloor \frac {N+1}n\right\rfloor$$
LHS has increased $\sigma(N+1)$. RHS is the sum of the greatest multiples of $n$ that are not greater than $N+1$. That is, the difference between this sum and the former is that now we are summing $N+1$ when we were summing $N+1-n$, and that happens when $n$ is a divisor of $N+1$. So the difference is also $\sigma(N+1)$.
In symbols,
$$\begin{align}
\sum_{n=1}^{N+1}&n\left \lfloor \frac {N+1}n\right\rfloor-\sum_{n=1}^{N}n\left \lfloor \frac {N}n\right\rfloor\\
&=\sum_{n=1}^{N+1}n\left \lfloor \frac {N+1}n\right\rfloor-\sum_{n=1}^{N+1}n\left \lfloor \frac {N}n\right\rfloor\\
&=\sum_{n=1}^{N+1}n\left(\left \lfloor \frac {N+1}n\right\rfloor-\left \lfloor \frac {N}n\right\rfloor\right)\\
\end{align}$$
The difference in parentheses is $1$ when $n\mid N+1$ and $0$ otherwise. Then the sum is $\sigma(N+1)$.


text:
\captionsetup[figure]{textfont=it,font=normalsize}
\begin{figure*}[tp]
\centering

\captionsetup[subfigure]{textfont=it,font=small}

\begin{subfigure}[b]{0.665\columnwidth}
\centering
\includegraphics[width=1\columnwidth]{Figs/initializers_gaussianMeasurementsMatrix_medianavgMethod_isComplex1_10trials_n16.pdf}
\caption{Gaussian measurements, $N=16$.}
\label{fig:Gaussian16_MN}
\end{subfigure}

\hfill
\begin{subfigure}[b]{0.665\columnwidth}
\centering
\includegraphics[width=1\columnwidth]{Figs/initializers_gaussianMeasurementsMatrix_medianavgMethod_isComplex1_10trials_n256.pdf}
\caption{Gaussian measurements, $N=256$.}
\label{fig:Gaussian256_MN}
\end{subfigure}

\hfill
\begin{subfigure}[b]{0.665\columnwidth}
\centering
\includegraphics[width=1\columnwidth]{Figs/initializers_transmissionMatrixMeasurementsMatrix_medianavgMethod_isComplex1_10trials_n256.pdf}
\caption{Transmission measurements, $N=256$.}
\label{fig:TM256_MN}
\end{subfigure}

\caption{Comparison of normalized MSE (N-MSE) as a function of the oversampling ratio $\delta=M/N$ for complex-valued phase retrieval with different spectral initializers and with different measurement matrices. The proposed LSPEs perform well on low-dimensional problems, for structured measurement ensembles, or at high oversampling ratios~$\delta$.}

\vspace{-0.2cm}
\label{fig:comparison}
\end{figure*}

\section{Numerical Results}
\label{sec:results}
We now compare the performance of our LSPEs against existing spectral initializers proposed for phase retrieval on synthetic and real image data. All our results use the spectral initializers and experimental setups provided by PhasePack~\cite{chandra2017phasepack}.

\subsection{Impact of Measurement Ensemble}
\label{sec:resultsensemble}

We start by comparing the normalized MSE (N-MSE) defined as~\cite{chandra2017phasepack}
\begin{align*}
\textit{N-MSE} = \frac{\min_{\alpha\in\setH} \|\bmx-\alpha\hat\bmx\|^2}{\|\bmx\|^2}
\end{align*}
for a range of spectral initializers  on different measurement ensembles. Specifically, we focus on the complex-valued case and consider (i) an i.i.d.\ Gaussian measurement matrix with signal dimension $N=16$, (ii) an i.i.d.\ Gaussian measurement matrix with $N=256$, and (iii) the structured ``transmission matrix'' used for image recovery through multiple scattering media as detailed in~\cite{metzler2017coherent}.

We vary the oversampling ratio $\delta=M/N$ and compare the N-MSE of the proposed complex-valued LSPEs, LSPE-$\complexset$ (\fref{esti:phaseinitC}) and LSPE-Exp (\fref{esti:phaseinitExp} with $\gamma=0.001$), to the following spectral initializers:

the original spectral initializer~\cite{netrapalli2013phase,candes2015phase} called ``spectral,''

truncated spectral initializer \cite{chen2015solving} called ``truncated,''

weighted spectral initializer \cite{wang2017solving} called ``weighted,''

amplitude spectral initializer \cite{wang2016solving} called ``amplitude,''

orthogonal spectral initializer \cite{chen2015phase} called ``orthogonal,''

and the asymptotically-optimal spectral initializer~\cite{mondelli2017fundamental} called ``optimal.''

For the following synthetic experiments, we generate the signals to be recovered according to \fref{asms:asms2} and \fref{asms:asms3} for \mbox{LSPE-$\complexset$} and LSPE-Exp, respectively.

\fref{fig:Gaussian16_MN} shows that the proposed LSPEs significantly outperform all existing spectral initializers for small problem dimensions with Gaussian measurements; this improvement is even more pronounced for large oversampling ratios.

The reason is that since we randomly generate a low-dimensional sensing matrix, the system will exhibit strong correlations among the measurements that can be exploited by LSPEs.

For larger dimensions with Gaussian measurements, we see in \fref{fig:Gaussian256_MN} that the proposed LSPEs do not provide an advantage over other methods. In fact, only LSPE-Exp  is able to perform as well as the orthogonal spectral initializer, which achieves the best performance in this scenario.

This behavior can be attributed to the facts that (i) for large random matrices there is no particular correlation structure among the measurements to exploit and (ii) ignoring measurements associated to large values in $y_m$ is increasingly important.

For structured measurements, as it is the case for the transmission matrix from~\cite{metzler2017coherent}, we see in \fref{fig:TM256_MN} that LSPEs significantly outperform existing methods that are designed for random measurement ensembles.

In this scenario, exponential preprocessing does not improve performance since correlations in the transmission matrix are dominating the performance.

\subsection{S-MSE Expressions and Approximation Error}

We now validate our theoretical S-MSE expressions in \fref{thm:MSEofLMSPE} and \fref{prop:SIapproxerror}, and confirm the accuracy of the EER bound given in \fref{cor:LMSPEapproxerror}.

In the following experiment, we set $M=8N$ and vary the dimension~$N$ from $8$ to $64$.

For each pair $(M,N)$, we randomly generate one instance of an i.i.d.\ circularly symmetric complex Gaussian measurement matrix and average the different errors (S-MSE and EER) over $10,000$ Monte-Carlo trials. We consider a noiseless setting and  assume identity preprocessing, i.e., $\setT(y)=y$. The signal vectors are generated according to an i.i.d.\  circularly complex Gaussian random vector.

From \fref{fig:comparison}, we see that our analytical S-MSE expressions for the LSPE-$\complexset$ and spectral initializers match their empirical values. We furthermore see that the empirical EER is only about 6\,dB to 10\,dB lower than our non-asymptotic upper bound given in \fref{cor:LMSPEapproxerror}.

\setlength{\textfloatsep}{15pt}
\captionsetup[figure]{textfont=it,font=normalsize}
\begin{figure}[t]
\centering

\includegraphics[width=0.8\columnwidth]{Figs/Validateerror_FixedGaussianMeasurementsMatrix_meanavgMethod_isComplex1_10000trials_MoverN8_maxN64_rng3.pdf}

\caption{Comparison of the analytical and empirical spectral MSE (S-MSE) and estimation error (EER) for LSPEs and spectral initializers (SI) at oversampling ratio $\delta=8$. Our analytical expressions in \fref{thm:MSEofLMSPE} and \fref{prop:SIapproxerror} match the empirical S-MSE; the upper bound in \fref{cor:LMSPEapproxerror} accurately characterizes the empirical EER.}

\vspace{-0.2cm}
\label{fig:comparison}
\end{figure}

\subsection{Real-World Image Recovery}

We finally illustrate the efficacy of LSPEs in a more realistic scenario. In particular, we show results for a real image reconstruction task by using LSPEs and spectral initializers only, i.e., we are \emph{not} using any additional phase retrieval algorithm.

Our goal is to recover a $16\times16$-pixel and a $40\times40$-pixel image that was captured through a multiple scattering media using the deterministic and highly-structured transmission matrix as detailed in~\cite{metzler2017coherent}.

We compare the proposed LSPEs to the same set of spectral initializers as in \fref{sec:resultsensemble}.

The signal priors are as in \fref{asms:asms2} (LSPE-$\complexset$) and \fref{asms:asms3} (LSPE-Exp).

Figures \ref{fig:TM16} and \ref{fig:TM40} show the recovered images along with the N-MSE values. The proposed LSPEs (often significantly) outperform all spectral initializers in terms of visual quality as well as the N-MSE. This result confirms the observations made in \fref{fig:TM256_MN} that LSPEs outperform existing spectral initializers for structured measurement matrices.

We note that exponential preprocessing for LSPEs does not noticeably improve the N-MSE (over LSPE-$\mathbb{C}$) in this setting since correlations in the transmission measurement matrix are dominating the recovery performance.

\newcommand{\figurescale}{0.6}

\captionsetup[figure]{textfont=it,font=normalsize}
\begin{figure}[t]
\centering

\captionsetup[sub]{textfont=it,font=footnotesize,justification=centering}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-original.png}
\caption{original \phantom{asasdasdfasdf}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-amplitudespectral-real.png}
\caption{amplitude \mbox{$\textit{N-MSE}=0.4927$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-optimal-real.png}
\caption{optimal \mbox{$\textit{N-MSE}=0.4833$}}
\end{subfigure} \\[0.3cm]

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-orthogonal-real.png}
\caption{orthogonal \mbox{$\textit{N-MSE}=0.6850$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-spectral-real.png}
\caption{spectral \mbox{$\textit{N-MSE}=0.4764 $}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-truncatedspectral-real.png}
\caption{truncated \mbox{$\textit{N-MSE}=0.4764$}}
\end{subfigure}  \\[0.3cm]

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-weightedspectral-real.png}
\caption{weighted  \mbox{$\textit{N-MSE}=0.4797$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-initlmspe_c-real.png}
\caption{LSPE-$\mathbb{C}$ \mbox{$\textit{N-MSE}=0.3377$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM16-5-initlmspe_c_exp-real.png}
\caption{LSPE-Exp \mbox{$\textit{N-MSE}=0.2928$}}
\end{subfigure}

\caption{Recovery of a  $16 \times 16$ image from with $ M=5  N$  measurements captured through a scattering medium without the use of a phase retrieval algorithm. LSPEs outperform all initializers for structured measurements.}

\label{fig:TM16}
\end{figure}

\captionsetup[figure]{textfont=it,font=normalsize}
\begin{figure}[t]
\centering

\captionsetup[sub]{textfont=it,font=footnotesize,justification=centering}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-original.png}
\caption{original \phantom{asasdasdfasdf}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-amplitudespectral-real.png}
\caption{amplitude \mbox{$\textit{N-MSE}=0.7010$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-optimal-real.png}
\caption{optimal \mbox{$\textit{N-MSE}=0.5849 $}}
\end{subfigure} \\[0.3cm]

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-orthogonal-real.png}
\caption{orthogonal \mbox{$\textit{N-MSE}=0.7028$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-spectral-real.png}
\caption{spectral \mbox{$\textit{N-MSE}=0.7016$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-truncatedspectral-real.png}
\caption{truncated \mbox{$\textit{N-MSE}= 0.7020$}}
\end{subfigure}  \\[0.3cm]

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-weightedspectral-real.png}
\caption{weighted  \mbox{$\textit{N-MSE}=0.7013$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-initlmspe_c_fast-real.png}
\caption{LSPE-$\mathbb{C}$ \mbox{$\textit{N-MSE}=0.4920$}}
\end{subfigure}

\begin{subfigure}[b]{0.32\columnwidth}
\centering
\includegraphics[width=\figurescale\columnwidth]{Figs/TM40-1-initlmspe_c_exp_fast-real.png}
\caption{LSPE-Exp \mbox{$\textit{N-MSE}=0.4896$}}
\end{subfigure}

\caption{Recovery of a  $40 \times 40$ image from with $ M=10  N$  measurements captured through a scattering medium without the use of a phase retrieval algorithm. LSPEs outperform all initializers for structured measurements.}

\label{fig:TM40}
\end{figure}


TITLE: validity of proof (100 green eyes riddle)
QUESTION [4 upvotes]: Here is the 100 green eyes riddle:
Imagine an island where 100 people,all perfect logicians,are imprisoned by a mad dictator.There's no escape,except for one strange rule. Any prisoner can approach the guards at night and ask to leave. If they have green eyes, they'll be released. If not, they'll be tossed into the volcano. As it happens, all 100 prisoners have green eyes, but they've lived there since birth, and the dictator has ensured they can't learn their own eye color. There are no reflective surfaces, all water is in opaque containers, and most importantly, they're not allowed to communicate among themselves. Though they do see each other during each morning's head count. Nevertheless, they all know no one would
ever risk trying to leave without absolute certainty of success.
On a morning, they are provided the information that "At least one of you has green eyes."
causing that on the hundredth morning after the sentence all the prisoners are gone.
The riddle and solution were available here:
https://www.youtube.com/watch?v=98TQv5IAtY8
In their proof, they consider that having the information that "At least one of you has green eyes."in case of 2 green-eyes people causes them to leave on the second day. This same element being used to prove the same conclusion for 3 green-eyed people on the third day etc.. which leads to conclusion that on the 100th day, all the 100 green eyed people have left.
My question is:
Is it valid to consider "a group of 2 green eyed people would escape on the second day", since actually the sentence "At least one of you has green eyes.", was only pronounced in front of the group of 100 ?

REPLY [0 votes]: I will present a solution to the problem, but first let us make a few assumptions. Assume it is common knowledge on the island that every islander is a perfect reasoner. I.e., if something can be proven, the islander themselves can prove it. Also assume the Guru makes the announcement on day $0$. The Guru never lies, and all the islanders know this. Additionally, if an islander discovers their eye color, they leave the next morning. So someone who discovers their eye color on day $i$ leaves the morning of day $i+1$. Additionally, let us take this as a more general problem:
Given the conditions you describe, but with $n$ people with green eyes on the island, on what day will the first person leave and on what day will the last person leave?. For simplicity, call such a scenario an $n$-green scenario. Observe that an $n$-green scenario is characterized by the fact that any green eyed sees only $n-1$ green eyed people. Note that the number of blue-eyed is irrelevant. Set it at your favorite number; mine is 6.
I will give a proof of the claim that an $n$-green scenario will terminate on day $n$, which I am using as short hand to mean the first and last green-eyed leaves on day $n$.
Now we proceed by induction. For the base case we only need to show that a $1$-green scenario terminates on day $1$. This is sort of trivial. On day $0$, the Guru announces they see a green-eyed. Since the green-eyed looks around and sees a sea of blue-eyed, they are forced to conclude that they themselves are green-eyed.
Okay, now for the inductive step. We may assume that we have a proof that an $n$-green scenario terminates on day $n$$^*$. Now we wish to prove that an $n+1$-green scenario terminates on day $n+1$. So assume that the Guru makes their announcement in front of $n+1$ green eyed people. Any given green-eyed, call them Bob, will look around and see only $n$ other green eyed people. For the sake of deduction, let Bob suppose that he himself is not green eyed. I.e., Bob thinks to himself, "if I'm not green eyed, what follows?" So Bob essentially assumes that the other green-eyed are in an $n$-green scenario. More explicitly, Bob will assume that any other green-eyed only sees $n-1$ green eyed. But since their is a proof that an $n$-green scenario terminates on day $n$, Bob will deduce that all the green eyed will leave on day $n$. But, since Bob's assumption is fallacious and each green-eyed sees $n$, not $n-1$ green-eyed, nobody leaves on day $n$. Bob then concludes that his assumption was false, so Bob concludes he must have green eyes. Thus, Bob will leave the next day, day $n+1$. But since our selection of $Bob$ was arbitrary, each and every green eyed will leave on day $n+1$.
This completes the induction. Thus, we have shown that an $n$-green scenario terminates on day $n$. Set $n=100$, et viola.
(*) this is, for our purposes, the same as assuming that the claim holds for $n$. This is just the type theoretic induction principle. Since the assumption is that the islanders are prefect reasoners, not omniscient, the computational perspective fits better within the framework.


TITLE: Are elements of a sample i.i.d. realizations of the same random variable or realizations of different i.i.d. random variables?
QUESTION [3 upvotes]: If there is some sample $X^n=(x_1,x_2,\dots,x_n)$, do we consider the elements of this sample $x_i$ independent and identically-distributed realizations of the same random variable $X$ or are they all realizations of different independent and identically distributed random variables $X_1,X_2,...,X_n$ (observation $x_1$ is the realization of a random variable $X_1$, observation $x_2$ is the realization of a random variable $X_2$ etc.)?
I hope this makes sense.

REPLY [2 votes]: You may think about $(x_1,\ldots,x_n)$ as a realization of $n$ independent copies of $X$. Basically, there is a probability space $(\Omega,\mathcal{F},\mathsf{P})$ in the background so that $(x_1,\ldots,x_n)=(X_1(\omega)\ldots,X_n(\omega))$ for some $\omega\in\Omega$, which is chosen randomly according to $\mathsf{P}$. Then the statement "independent and identically-distributed realizations of the same random variable" doesn't make sense. Although, sometimes $(x_1,\ldots,x_n)$ is referred to as a random sample from a particular distribution (e.g. $F_X$).