TITLE: Finding the function of the power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$
QUESTION [0 upvotes]: Find the sum function of the following power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$

I actually felt like I had the right idea on this one, my solution is:
Moving one $x$ out of the sum and then differentiating we get
$$\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}=x\sum\limits _{n=1}^{\infty}\frac{x^{2n}}{n}=2x\int\left(\sum\limits _{n=2}^{\infty}x^{2n-1}\right)dx=2x\int\left(\sum\limits _{n=1}^{\infty}x^{2n-1}-x\right)dx$$
Taking the remaining sum, we can split it to get
$$\sum\limits _{n=1}^{\infty}x^{2n-1}=\sum\limits _{n=1}^{\infty}\left(x^{n}+\left(-1\right)^{n-1}x^{n}\right)=\sum\limits _{n=0}^{\infty}x^{n}-\sum\limits _{n=0}^{\infty}\left(-x\right)^{n}$$
now using the geometric sum, and putting it all back, we get that
$$\begin{aligned}\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}= & 2x\int\left(\frac{1}{1-x}-\frac{1}{1+x}-x\right)dx\\
= & 2x\left(-\ln\left(1-x\right)-\ln\left(1+x\right)-\frac{x^{2}}{2}\right)
\end{aligned}$$
But this solution is apparently wrong, and I'm not sure where my mistake is...

REPLY [5 votes]: Using the Taylor series $$\sum_{n\geq1}\frac{x^{n}}{n}=-\log\left(1-x\right),\,\left|x\right|<1
 $$ we get $$\sum_{n\geq1}\frac{x^{2n+1}}{n}=x\sum_{n\geq1}\frac{x^{2n}}{n}=-x\log\left(1-x^{2}\right).$$


TITLE: For line $ax+by+k=0$ which intercepts form a triangle rectangle with area $A$, find $k$
QUESTION [1 upvotes]: I know that the area of a triangle is given by the formula $A=\frac{1}2Bh$ and the intercepts of line $ax+bx+k=0$ are $(B,0)$ and $(0,h)$ which forms a square with area $2A$, but without brute-forcing my way through it, I got no idea how can I find this the analytic way. For example, for line $4x-5y+k=0$ and area $A=2\frac{1}2$, $k$ is $\pm10$, but how I know is?

REPLY [1 votes]: To find the intercepts, plug in $x=0$ and $y=0$ to $ax + by + k = 0$. We see that it has intercepts of $(0, -k/b)$ and $(-k/a, 0)$.
Therefore, the area of this triangle along with $(0,0)$ is $$A = \dfrac{1}{2} \cdot Bh = \dfrac{k^2}{2|ab|}$$.
Therefore, when $a=4, b=5, A = 2.5$, we have $$2.5 =\dfrac{k^2}{2 \cdot 4 \cdot 5} \implies k^2 = 100 \implies k=10,-10$$


TITLE: Is the energy of an electromagnetic wave, in electron volts, an 'average' or a 'peak' value?
QUESTION [0 upvotes]: Since EM waves oscillate, is the energy given in eV an average or maximum value?
In a related question, can a detector (of whatever sort) 'miss' an EM wave if the amplitude is at a minimum when it goes past the detector?

REPLY [1 votes]: With electromagnetic waves, we talk about peak or average power per unit area, but this is not measured in electron-volts. With individual photons, we talk about their energy, which can be measured in electron-volts, but the energy of a photon doesn’t oscillate.


TITLE: Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$
QUESTION [1 upvotes]: I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\frac{\sin(x)}{x}}$ (I have no clue, what to do next with it), but with Taylor and ended up with:
$$\lim_{x\to 0}\frac{\sqrt{1-\frac{x^2}{6}+o(x^2)}+o(\sqrt{1-\frac{x^2}{6}+o(x^2)})}{x^4} $$
which tends to infinity

REPLY [0 votes]: You can reason as follows:
$$\frac{\sin x}x\sim 1-\frac{x^2}{3!}$$ and after taking one minus the square root,
$$\frac{x^2}{12}.$$
Then one minus the cosine yields
$$\frac12\left(\frac{x^2}{12}\right)^2=\frac{x^4}{288}.$$


TITLE: Atiyah-MacDonald, exercise 2.11
QUESTION [29 upvotes]: Let $A$ be a commutative ring with $1$ not equal to $0$.  (The ring A is not necessarily a domain, and is not necessarily Noetherian.)  Assume we have an injective map of free $A$-modules $A^m \to A^n$.  Must we have $m \le n$?
I believe the answer is yes.  For instance, why is there no injective map from  $A^2 \to A^1$?  Say it's represented by a matrix $(a_1, a_2)$.  Then clearly $(a_2, -a_1)$ is in the kernel.  In the $A^{n+1} \to A^{n}$ case, we can look at the $n \times (n+1)$ matrix which represents it; call it $M$.  Let $M_i$ denote the determinant of the matrix obtained by deleting the $i$-th column.  Let $v$ be the vector $(M_1, -M_2, ..., (-1)^nM_{n+1})$.  Then $v$ is in the kernel of our map, because the vector $Mv^T$ has $i$-th component the determinant of the $(n+1) \times (n+1)$ matrix attained from $M$ by repeating the $i$-th row twice.
That almost finishes the proof, except it is possible that $v$ is the zero vector.  I would like to see either this argument finished, or, even better, a nicer proof.
Thank you!

REPLY [12 votes]: Every few years this question re-appears in my life (as it just did this week) and I have to rediscover the Euler Characteristic proof which I half-remember. Let me just write my thoughts down here so I don't ever have to do this again.
Definition. Let $A$ be a non-zero commutative ring. A finite free resolution (or FFR) of an $A$-module $M$ is an exact sequence
$$0\to F_n \to F_{n-1} \to \cdots \to F_1 \to F_0 \to M\to 0$$
where each $F_i$ is a free $A$-module, of rank $r_i$.
If $M$ has an FFR then we define the Euler Characteristic $\chi(M)$ of $M$ to be $\sum_i(-1)^ir_i$. This turns out to be a well-defined invariant of $M$ (i.e. independent of the resolution) because a relatively straightforward induction on $n$ shows that if $G_i$ is another resolution then $F_0\oplus G_1\oplus F_2\oplus\cdots\oplus (F\ \mbox{or}\ G)_n$ is isomorphic to $G_0\oplus F_1\oplus G_2\oplus\cdots$. For more details on this part of the argument see Lemma 4 in section 19 of Matsumura's book "Commutative Ring Theory".
The fundamental fact is:
Theorem. If $M$ has an FFR then $\chi(M)\geq0$.
As a consequence, not only does an injection $A^m \to A^n$ imply $m\leq n$ (apply the theorem to the cokernel), but if $0\to A^m\to A^n \to A^r$ is exact then $m+r\geq n$ and so on and so on. Do the other solutions also generalise to prove this?
The theorem is Theorem 19.7 of Matsumura's "commutative ring theory". Here's a sketch. Localising at a minimal prime one can assume that $A$ is local with nilpotent maximal ideal. It suffices to prove that in this case $M$ is free, as then the exact sequence splits and the Euler Characteristic is just the rank of $M$. It suffices to prove that the cokernel of $F_n\to F_{n-1}$ is free, because then one is done by induction on $n$. The trick now is to replace $F_n$ and $F_{n-1}$ by a minimal free resolution, look at the image of $F_n$ and see that its coordinates must lie in the maximal ideal, and it's then not difficult to concoct a non-zero element of $A$ which kills this finite set of coordinates, and hence $F_n$ is a free module annihilated by a non-zero element of $A$ and must hence be zero.
This last part of the argument feels very similar to Georges Elencwajg's argument above, which reduces to the Noetherian case and uses lengths, but Matsumura explicitly avoids this reduction.


TITLE: Apostol's Calculus - Prove the set is open (inequality)
QUESTION [3 upvotes]: This is from Apostol's Calculus, Vol. II Section 8.3 #3(a).
Prove that $S=\{(x,y,z) \mid z^2 - x^2 - y^2 -1>0\}$ is open.
The only way to prove that a set is open which has been covered so far is to prove that for an arbitrary point $\mathop a\limits^{\small \to}   \in S$ there is an open ball $B(\mathop a\limits^{\small \to}  , r) \subset S$ by finding an explicit $r$ and showing that $\mathop x\limits^{\small \to}   \in B(\mathop a\limits^{\small \to}  , r) \implies \mathop x\limits^{\small \to}   \in S$. Nothing topological (compactness, completeness, etc.) has been discussed. For some reason I just am stuck trying to figure out an $r$ which would work.
Hints > Complete answers

REPLY [3 votes]: Choose arbitrary $(a,b,c)\in S$. Assume without loss of generality that $a,b,c\geq 0$. Put $k=c^2-a^2-b^2-1$.
Then, for any $(a',b',c')\in B((a,b,c),r)$ (with $0<r<c$, $r$ to be specified later -- notice that $c>0$) we have that
$$c'^2-a'^2-b'^2-1\geq (c-r)^2-(a+r)^2-(b+r)^2-1=k-2r(a+b+c)-r^2$$
Thus, we can see that if we choose $r>0$ small enough to have $r^2+2r(a+b+c)<k$ (and $r<c$), $B((a,b,c),r)$ will be contained in $S$.


text:
\begin{document}

\title{Identities of some special mixed-type polynomials }

\author{Dae San Kim}

\address{\noindent Department of Mathematics, Sogang University, Seoul 121-742,
Republic of Korea}

\email{dskim@sogang.ac.kr}

\author{Taekyun Kim}

\address{\noindent Department of Mathematics, Kwangwoon University, Seoul
139-701, Republic of Korea}

\email{tkkim@kw.ac.kr}

\subjclass[2010]{05A19; 11B68; 11B83.}

\keywords{mixed-type polynomial, Bernoulli-Euler, Daehee-Changhee, Cauchy-Daehee, Cauchy-Changhee }
\begin{abstract}
In this paper, we consider various speical mixed-type polynomials which are
related to Bernoulli, Euler, Changhee and Daehee polynomials. From
those polynomials, we derive some interesting and new identities.
\end{abstract}
\maketitle

\global\long\def\Zp{\mathbb{Z}_{p}}

\section{Introduction}

Let $p$ be a fixed odd prime number. Throughout this paper, $\Zp$,
$\mathbb{Q}_{p}$ and $\mathbb{C}_{p}$ will denote the ring of $p$-adic
integers, the field of $p$-adic rational numbers and the completion
of the algebraic closure of $\mathbb{Q}_{p}$. Let $\nu_{p}$ be the
normalized exponential valuation of $\mathbb{C}_{p}$ with $\left|p\right|_{p}=\frac{1}{p}=p^{-\nu_{p}\left(p\right)}$.
Let $UD\left(\Zp\right)$ be the space of uniformly differentiable
functions on $\Zp$.

For $f\in UD\left(\Zp\right)$, the bosonic $p$-adic integral is
given by
\begin{equation}
I_{0}\left(f\right)=\int_{\Zp}f\left(x\right)d\mu_{0}\left(x\right)=\lim_{N\rightarrow\infty}\frac{1}{p^{N}}\sum_{x=0}^{p^{N}-1}f\left(x\right),\quad\left(\mbox{see \cite{key-11}}\right),\label{eq:1}
\end{equation}
and the fermionic $p$-adic integral on $\Zp$ is defined by Kim to
be
\[
I_{-1}\left(f\right)=\int_{\Zp}f\left(x\right)d\mu_{-1}\left(x\right)=\lim_{N\rightarrow\infty}\sum_{x=0}^{p^{N}-1}f\left(x\right)\left(-1\right)^{x},\quad\left(\mbox{see \cite{key-10}}\right).
\]

In \cite{key-6,key-7}, the higher-order Daehee polynomials are defined
by
\begin{equation}
\left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(1+t\right)^{x}=\sum_{n=0}^{\infty}D_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!},\quad\left(r\in\mathbb{N}\right).\label{eq:2}
\end{equation}

When $x=0$, $D_{n}^{\left(r\right)}=D_{n}^{\left(r\right)}\left(0\right)$
are called the Daehee numbers of order $r$.

When $r=1$, $D_{n}^{\left(1\right)}\left(x\right)=D_{n}\left(x\right)$
are called the Daehee polynomials (see \cite{key-6}).

As is known, the Changhee polynomials of order $s$$\left(\in\mathbb{N}\right)$
are defined by the generating function to be
\begin{equation}
\left(\frac{2}{t+2}\right)^{s}\left(1+t\right)^{x}=\sum_{n=0}^{\infty}Ch_{n}^{\left(s\right)}\left(x\right)\frac{t^{n}}{n!},\quad\left(\mbox{see \cite{key-8}}\right).\label{eq:3}
\end{equation}

When $x=0$, $Ch_{n}^{\left(s\right)}=Ch_{n}^{\left(s\right)}\left(0\right)$
are called the Changhee numbers of order $s$.

For $s=1$, $Ch_{n}^{\left(1\right)}\left(x\right)=Ch_{n}\left(x\right)$
are called the Changhee polynomials.

The Bernoulli polynomials of order $r\in\mathbb{N}$ are given by

\begin{equation}
\left(\frac{t}{e^{t}-1}\right)^{r}e^{xt}=\sum_{n=0}^{\infty}B_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!},\quad\left(\mbox{see \cite{key-11,key-12,key-13,key-14,key-15,key-16,key-17,key-18,key-19,key-20,key-21}}\right).\label{eq:4}
\end{equation}

When $x=0$, $B_{n}^{\left(r\right)}=B_{n}^{\left(r\right)}\left(0\right)$
are called the Bernoulli numbers of order $r$.

For $r=1$, $B_{n}^{\left(1\right)}\left(x\right)=B_{n}\left(x\right)$
are called the ordinary Bernoulli polynomials.

We recall that the Euler polynomials of order $r$ are defined by
the generating function to be

\begin{equation}
\left(\frac{2}{e^{t}+1}\right)^{r}e^{xt}=\sum_{n=0}^{\infty}E_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!},\quad\left(\mbox{see \cite{key-1,key-2,key-3,key-4,key-5,key-6,key-7,key-8,key-9,key-10,key-11,key-12}}\right).\label{eq:5}
\end{equation}

When $x=0$, $E_{n}^{\left(r\right)}=E_{n}^{\left(r\right)}\left(0\right)$
are called the Euler numbers of order $r$.

For $r=1$, $E_{n}^{\left(1\right)}\left(x\right)=E_{n}\left(x\right)$
are called the ordinary Euler polynomials.

Finally, the Cauchy polynomials of the first kind of order $r$ are given by
\begin{equation}
\left(\frac{t}{\log\left(1+t\right)}\right)^{r}\left(1+t\right)^{x}=\sum_{n=0}^{\infty}C_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!},\quad\left(\mbox{see \cite{key-3,key-9}}\right).\label{eq:6}
\end{equation}

When $x=0$, $C_{n}^{\left(r\right)}=C_{n}^{\left(r\right)}\left(0\right)$
are called the Cauchy numbers of the first kind of order $r$.

For $r=1$, $C_{n}^{\left(1\right)}\left(x\right)=C_{n}\left(x\right)$
are called the ordinary Cauchy polynomials of the first kind (see \cite{key-3}).

From (\ref{eq:1}) and (\ref{eq:2}), we have
\begin{equation}
I_{0}\left(f_{1}\right)-I_{0}\left(f\right)=f^{\prime}\left(0\right)\label{eq:7}
\end{equation}

and
\begin{equation}
I_{-1}\left(f_{1}\right)=-I_{-1}\left(f\right)+2f\left(0\right),\label{eq:8}
\end{equation}
where $f_{1}\left(x\right)=f\left(x+1\right)$ (see \cite{key-10,key-12}).

In this paper, we consider several special polynomials which are
derived from the bosonoic or fermionic $p$-adic integral on $\Zp$.

Finally, we give some relation or identities of those polynomials.

\section{Some special mixed-type polynomials }

In this section, we assume that $t\in\mathbb{C}_{p}$ with $\left|t\right|_{p}<p^{-\frac{1}{p-1}}$.
From (\ref{eq:7}), we can derive the following equation :
\begin{align}
 & \int_{\Zp}\cdots\int_{\Zp}\left(1+t\right)^{\left(x_{1}+\cdots+x_{r}+x\right)}d\mu_{0}\left(x_{1}\right)\cdots d\mu_{0}\left(x_{r}\right)\label{eq:9}\\
= & \left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(1+t\right)^{x}=\left(\frac{\log\left(1+t\right)}{e^{\log\left(1+t\right)}-1}\right)^{r}e^{x\log(1+t)}\nonumber \\
=& \sum_{m=0}^{\infty}B_{m}^{(r)}(x)\frac{(\log(1+t))^{m}}{m!}=\sum_{m=0}^{\infty}B_{m}^{(r)}(x)\sum_{n=m}^{\infty}S_1(n,m)\frac{t^{n}}{n!}\nonumber\\
= & \sum_{n=0}^{\infty}\left(\sum_{m=0}^{n}B_{m}^{\left(r\right)}\left(x\right)S_{1}\left(n,m\right)\right)\frac{t^{m}}{n!},\nonumber
\end{align}
and
\begin{equation}
\left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(1+t\right)^{x}=\sum_{n=0}^{\infty}D_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!}.\label{eq:10}
\end{equation}

Therefore, by (\ref{eq:9}) and (\ref{eq:10}), we obtain the following
equation :
\begin{align}
 & \int_{\Zp}\cdots\int_{\Zp}\dbinom{x_{1}+\cdots+x_{r}+x}{n}d\mu_{0}\left(x_{1}\right)\cdots d\mu_{0}\left(x_{r}\right)\label{eq:11}\\
= & \frac{D_{n}^{\left(r\right)}\left(x\right)}{n!}=\frac{1}{n!}\sum_{m=0}^{n}B_{m}^{\left(r\right)}\left(x\right)S_{1}\left(n,m\right)\nonumber
\end{align}
where $S_{1}\left(n,m\right)$ is the Stirling number of the first
kind.

From (\ref{eq:8}), we have
\begin{align}
 & \int_{\Zp}\cdots\int_{\Zp}\left(1+t\right)^{x_{1}+\cdots+x_{r}+x}d\mu_{-1}\left(x_{1}\right)\cdots d\mu_{-1}\left(x_{r}\right)\label{eq:12}\\
= & \left(\frac{2}{t+2}\right)^{r}\left(1+t\right)^{x}=\left(\frac{2}{e^{\log\left(1+t\right)}+1}\right)^{r}e^{x\log\left(1+t\right)}\nonumber \\
= & \sum_{m=0}^{\infty}E_{m}^{\left(r\right)}\left(x\right)\frac{\left(\log\left(1+t\right)\right)^{m}}{m!}=\sum_{m=0}^{\infty}E_{m}^{\left(r\right)}\left(x\right)\sum_{n=m}^{\infty}S_{1}\left(n,m\right)\frac{t^{n}}{n!}\nonumber \\
= & \sum_{n=0}^{\infty}\left\{ \sum_{m=0}^{n}E_{m}^{\left(r\right)}\left(x\right)S_{1}\left(n,m\right)\right\} \frac{t^{n}}{n!}\nonumber
\end{align}
and
\begin{equation}
\left(\frac{2}{t+2}\right)^{r}\left(1+t\right)^{x}=\sum_{n=0}^{\infty}Ch_{n}^{\left(r\right)}\left(x\right)\frac{t^{n}}{n!}.\label{eq:13}
\end{equation}

From (\ref{eq:12}) and (\ref{eq:13})
\begin{align}
 & \int_{\Zp}\cdots\int_{\Zp}\dbinom{x_{1}+\cdots+x_{r}+x}{n}d\mu_{-1}\left(x_{1}\right)\cdots d\mu_{-1}\left(x_{r}\right)\label{eq:14}\\
= & \frac{Ch_{n}^{\left(r\right)}\left(x\right)}{n!}=\frac{1}{n!}\sum_{m=0}^{n}E_{m}^{\left(r\right)}\left(x\right)S_{1}\left(n,m\right).\nonumber
\end{align}

Note that
\begin{align}
\left(1+t\right)^{x} & =\left(\frac{t}{\log\left(1+t\right)}\right)^{r}\left(1+t\right)^{x}\left(\frac{\log\left(1+t\right)}{t}\right)^{r}\label{eq:15}\\
 & =\left(\sum_{l=0}^{\infty}C_{l}^{\left(r\right)}\left(x\right)\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}D_{m}^{\left(r\right)}\frac{t^{m}}{m!}\right)\nonumber \\
 & =\sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}\dbinom{n}{l}C_{l}^{\left(r\right)}\left(x\right)D_{n-l}^{\left(r\right)}\right)\frac{t^{n}}{n!}\nonumber
\end{align}
and
\begin{equation}
\left(1+t\right)^{x}=\sum_{n=0}^{\infty}\left(x\right)_{n}\frac{t^{n}}{n!}.\label{eq:16}
\end{equation}

From (\ref{eq:15}) and (\ref{eq:16}), we have
\begin{align}
\left(x\right)_{n} & =\sum_{l=0}^{n}\dbinom{n}{l}C_{l}^{\left(r\right)}\left(x\right)D_{n-l}^{\left(r\right)}\label{eq:17}\\
 & =\sum_{l=0}^{n}\dbinom{n}{l}D_{n-l}^{\left(r\right)}\left(x\right)C_{l}^{\left(r\right)}.\nonumber
\end{align}

That is,
\[
\dbinom{x}{n}=\frac{1}{n!}\sum_{l=0}^{n}\dbinom{n}{l}C_{l}^{\left(r\right)}\left(x\right)D_{n-l}^{\left(r\right)}.
\]

Let us consider the Bernoulli-Euler mixed-type polynomials of order $(r,s)$ as follows
:
\begin{equation}
BE_{n}^{\left(r,s\right)}\left(x\right)=\int_{\Zp}\cdots\int_{\Zp}E_{n}^{\left(s\right)}\left(x+y_{1}+\cdots+y_{r}\right)d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{r}\right).\label{eq:18}
\end{equation}

Then, we can find the generating function of $BE_{n}^{\left(r,s\right)}\left(x\right)$
as follows :
\begin{align}
 & \sum_{n=0}^{\infty}BE_{n}^{\left(r,s\right)}\left(x\right)\frac{t^{n}}{n!}\label{eq:19}\\
= & \int_{\Zp}\cdots\int_{\Zp}\sum_{n=0}^{\infty}E_{n}^{\left(s\right)}\left(x+y_{1}+\cdots+y_{r}\right)\frac{t^{n}}{n!}d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{r}\right)\nonumber \\
= & \left(\frac{2}{e^{t}+1}\right)^{s}\int_{\Zp}\cdots\int_{\Zp}e^{\left(x+y_{1}+\cdots+y_{r}\right)t}d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{r}\right)\nonumber \\
= & \left(\frac{2}{e^{t}+1}\right)^{s}\left(\frac{t}{e^{t}-1}\right)^{r}e^{xt}.\nonumber
\end{align}

Note that
\begin{align}
\left(\frac{2}{e^{t}+1}\right)^{s}\left(\frac{t}{e^{t}-1}\right)^{r}e^{xt} & =\left(\sum_{l=0}^{\infty}E_{l}^{\left(s\right)}\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}B_{m}^{\left(r\right)}\left(x\right)\frac{t^{m}}{m!}\right)\label{eq:20}\\
 & =\sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}\dbinom{n}{l}E_{l}^{\left(s\right)}B_{n-l}^{\left(r\right)}\left(x\right)\right)\frac{t^{n}}{n!}.\nonumber
\end{align}

From (\ref{eq:19}) and (\ref{eq:20}), we have
\begin{equation}
BE_{n}^{\left(r,s\right)}\left(x\right)=\sum_{l=0}^{n}\dbinom{n}{l}E_{l}^{\left(s\right)}B_{n-l}^{\left(r\right)}\left(x\right).\label{eq:21}
\end{equation}

By replacing $t$ by $\log\left(1+t\right)$, we get
\begin{align}
 & \sum_{n=0}^{\infty}BE_{n}^{\left(r,s\right)}\left(x\right)\frac{\left(\log\left(1+t\right)\right)^{n}}{n!}\label{eq:22}\\
= & \left(\frac{2}{t+2}\right)^{s}\left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(1+t\right)^{x}\nonumber \\
= & \left(\sum_{l=0}^{\infty}Ch_{l}^{\left(s\right)}\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}D_{m}^{\left(r\right)}\left(x\right)\frac{t^{m}}{m!}\right)\nonumber \\
= & \sum_{n=0}^{\infty}\left\{ \sum_{m=0}^{n}\dbinom{n}{m}D_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(s\right)}\right\} \frac{t^{n}}{n!},\nonumber
\end{align}
and
\begin{align}
\sum_{m=0}^{\infty}BE_{m}^{\left(r,s\right)}\left(x\right)\frac{\left(\log\left(1+t\right)\right)^{m}}{m!} & =\sum_{m=0}^{\infty}BE_{m}^{\left(r,s\right)}\left(x\right)\sum_{n=m}^{\infty}S_{1}\left(n,m\right)\frac{t^{n}}{n!}\label{eq:23}\\
 & =\sum_{n=0}^{\infty}\left\{ \sum_{m=0}^{n}BE_{m}^{\left(r,s\right)}\left(x\right)S_{1}\left(n,m\right)\right\} \frac{t^{n}}{n!}.\nonumber
\end{align}

Therefore, by (\ref{eq:22}) and (\ref{eq:23}), we obtain the following
equation :
\begin{equation}
\sum_{m=0}^{n}\dbinom{n}{m}D_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(s\right)}=\sum_{m=0}^{n}BE_{m}^{\left(r,s\right)}\left(x\right)S_{1}\left(n,m\right).\label{eq:24}
\end{equation}

Let us consider the Daehee-Changhee mixed-type polynomials of order $(r,s)$ as follows
:

\begin{equation}
DC_{n}^{\left(r,s\right)}\left(x\right)=\int_{\Zp}\cdots\int_{\Zp}D_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{r}\right),\label{eq:25}
\end{equation}
where $n\ge0$.

From (\ref{eq:25}), we can derive the generating function of $DC_{n}^{\left(r,s\right)}\left(x\right)$
as follows :
\begin{align}
 & \sum_{n=0}^{\infty}DC_{n}^{\left(r,s\right)}\left(x\right)\frac{t^{n}}{n!}\label{eq:26}\\
= & \int_{\Zp}\cdots\int_{\Zp}\sum_{n=0}^{\infty}D_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)\frac{t^{n}}{n!}d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{s}\right)\nonumber \\
= & \left(\frac{\log\left(1+t\right)}{t}\right)^{r}\int_{\Zp}\cdots\int_{\Zp}\left(1+t\right)^{x+y_{1}+\cdots+y_{s}}d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{s}\right)\nonumber \\
= & \left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(\frac{2}{t+2}\right)^{s}\left(1+t\right)^{x}.\nonumber
\end{align}

We observe that
\begin{align}
\left(\frac{2}{t+2}\right)^{s}\left(\frac{\log\left(1+t\right)}{t}\right)^{r}\left(1+t\right)^{x} & =\left(\sum_{l=0}^{\infty}Ch_{l}^{\left(s\right)}\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}D_{m}^{\left(r\right)}\left(x\right)\frac{t^{m}}{m!}\right)\label{eq:27}\\
 & =\sum_{n=0}^{\infty}\left\{ \sum_{m=0}^{n}\dbinom{n}{m}D_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(s\right)}\right\} \frac{t^{n}}{n!}.\nonumber
\end{align}

From (\ref{eq:26}) and (\ref{eq:27}), we have
\begin{equation}
DC_{n}^{\left(r,s\right)}\left(x\right)=\sum_{m=0}^{n}\dbinom{n}{m}D_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(r\right)},\label{eq:28}
\end{equation}
 where $n\ge0$, $r,s\in\mathbb{N}$.

Now, we define the Cauchy-Daehee mixed-type polynomials of order $(r,s)$ as follows
:
\begin{equation}
CD_{n}^{\left(r,s\right)}\left(x\right)=\int_{\Zp}\cdots\int_{\Zp}C_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{r}\right).\label{eq:29}
\end{equation}

From (\ref{eq:29}), we can derive the generating function of $CD_{n}^{\left(r,s\right)}\left(x\right)$
as follows :
\begin{align}
 & \sum_{n=0}^{\infty}CD_{n}^{\left(r,s\right)}\left(x\right)\frac{t^{n}}{n!}\label{eq:30}\\
= & \int_{\Zp}\cdots\int_{\Zp}\sum_{n=0}^{\infty}C_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)\frac{t^{n}}{n!}d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{s}\right)\nonumber \\
= & \left(\frac{t}{\log\left(1+t\right)}\right)^{r}\int_{\Zp}\cdots\int_{\Zp}\left(1+t\right)^{x+y_{1}+\cdots+y_{s}}d\mu_{0}\left(y_{1}\right)\cdots d\mu_{0}\left(y_{s}\right)\nonumber \\
= & \left(\frac{t}{\log\left(1+t\right)}\right)^{r}\left(\frac{\log\left(1+t\right)}{t}\right)^{s}\left(1+t\right)^{x}.\nonumber \\
= & \begin{cases}
\sum_{n=0}^{\infty}C_{n}^{\left(r-s\right)}\left(x\right)\frac{t^{n}}{n!} & \mbox{if }r>s\\
\sum_{n=0}^{\infty}D_{n}^{\left(s-r\right)}\left(x\right)\frac{t^{n}}{n!} & \mbox{if }r<s\\
\sum_{n=0}^{\infty}\left(x\right)_{n}\frac{t^{n}}{n!} & \mbox{if }r=s.
\end{cases}\nonumber
\end{align}

Thus, by (\ref{eq:30}), we get
\begin{equation}
CD_{n}^{\left(r,s\right)}\left(x\right)=\begin{cases}
C_{n}^{\left(r-s\right)}\left(x\right) & \mbox{if }r>s\\
D_{n}^{\left(s-r\right)}\left(x\right) & \mbox{if }r<s\\
\left(x\right)_{n} & \mbox{if }r=s
\end{cases}\label{eq:31}
\end{equation}
where $n\ge0$.

By replacing $t$ by $e^{t}-1$ in (\ref{eq:26}), we get
\begin{align}
\sum_{n=0}^{\infty}DC_{n}^{\left(r,s\right)}\left(x\right)\frac{\left(e^{t}-1\right)^{n}}{n!} & =\left(\frac{t}{e^{t}-1}\right)^{r}e^{xt}\left(\frac{2}{e^{t}+1}\right)^{s}\label{eq:32}\\
 & =\left(\sum_{n=0}^{\infty}B_{l}^{\left(r\right)}\left(x\right)\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}E_{m}^{\left(s\right)}\frac{t^{m}}{m!}\right)\nonumber \\
 & =\sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}\dbinom{n}{l}B_{l}^{\left(r\right)}\left(x\right)E_{n-l}\right)\frac{t^{n}}{n!},\nonumber
\end{align}
and
\begin{align}
\sum_{m=0}^{\infty}DC_{n}^{\left(r,s\right)}\left(x\right)\frac{\left(e^{t}-1\right)^{m}}{m!} & =\sum_{m=0}^{\infty}DC_{m}^{\left(r,s\right)}\left(x\right)\sum_{n=m}^{\infty}S_{2}\left(n,m\right)\frac{t^{n}}{n!}\label{eq:33}\\
 & =\sum_{n=0}^{\infty}\left(\sum_{m=0}^{n}DC_{m}^{\left(r,s\right)}\left(x\right)S_{2}\left(n,m\right)\right)\frac{t^{n}}{n!}.\nonumber
\end{align}

Therefore, by (\ref{eq:32}) and (\ref{eq:33}), we get
\begin{equation}
\sum_{m=0}^{n}DC_{m}^{\left(r,s\right)}\left(x\right)S_{2}\left(m,n\right)=\sum_{l=0}^{n}\dbinom{n}{l}B_{l}^{\left(r\right)}\left(x\right)E_{n-l},\label{eq:34}
\end{equation}
where $S_{2}\left(n,m\right)$ is the Stirling number of the second
kind.

Finally, we consider the Cauchy-Changhee mixed-type polynomials of order $(r,s)$ as
follows :
\begin{equation}
CC_{n}^{\left(r,s\right)}\left(x\right)=\int_{\Zp}\cdots\int_{\Zp}C_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{s}\right),\label{eq:35}
\end{equation}
where $n\ge0$.

By (\ref{eq:35}), we see that the generating function of $CC_{n}^{\left(r,s\right)}\left(x\right)$
are given by
\begin{align}
 & \sum_{n=0}^{\infty}CC_{n}^{\left(r,s\right)}\left(x\right)\frac{t^{n}}{n!}\label{eq:36}\\
= & \int_{\Zp}\cdots\int_{\Zp}\sum_{n=0}^{\infty}C_{n}^{\left(r\right)}\left(x+y_{1}+\cdots+y_{s}\right)\frac{t^{n}}{n!}d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{s}\right)\nonumber \\
= & \left(\frac{t}{\log\left(1+t\right)}\right)^{r}\int_{\Zp}\cdots\int_{\Zp}\left(1+t\right)^{x+y_{1}+\cdots+y_{s}}d\mu_{-1}\left(y_{1}\right)\cdots d\mu_{-1}\left(y_{s}\right)\nonumber \\
= & \left(\frac{t}{\log\left(1+t\right)}\right)^{r}\left(\frac{2}{t+2}\right)^{s}\left(1+t\right)^{x}\nonumber \\
= & \left(\sum_{m=0}^{\infty}C_{m}^{\left(r\right)}\left(x\right)\frac{t^{m}}{m!}\right)\left(\sum_{l=0}^{\infty}Ch_{l}^{\left(s\right)}\frac{t^{l}}{l!}\right)\nonumber \\
= & \sum_{n=0}^{\infty}\left\{ \sum_{m=0}^{n}\dbinom{n}{m}C_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(s\right)}\right\} \frac{t^{n}}{n!}.\nonumber
\end{align}

Thus, by (\ref{eq:36}), we get
\begin{equation}
CC_{n}^{\left(r,s\right)}\left(x\right)=\sum_{m=0}^{n}\dbinom{n}{m}C_{m}^{\left(r\right)}\left(x\right)Ch_{n-m}^{\left(s\right)}.\label{eq:37}
\end{equation}

By replacing $t$ by $e^{t}-1$, we get
\begin{align}
\sum_{n=0}^{\infty}CC_{n}^{\left(r,s\right)}\left(x\right)\frac{\left(e^{t}-1\right)^{n}}{n!} & =\left(\frac{e^{t}-1}{t}\right)^{r}\left(\frac{2}{e^{t}+1}\right)^{s}e^{xt}\label{eq:38}\\
 & =\left(\sum_{l=0}^{\infty}\frac{S_{2}\left(l+r,l\right)}{\binom{l+r}{r}}\frac{t^{l}}{l!}\right)\left(\sum_{m=0}^{\infty}E_{m}^{\left(s\right)}\left(x\right)\frac{t^{m}}{m!}\right)\nonumber \\
 & =\sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}\frac{S_{2}\left(l+r,l\right)E_{n-l}^{\left(s\right)}\left(x\right)}{\binom{l+r}{l}}\dbinom{n}{l}\right)\frac{t^{n}}{n!},\nonumber
\end{align}
and
\begin{align}
\sum_{l=0}^{\infty}CC_{l}^{\left(r,s\right)}\left(x\right)\frac{\left(e^{t}-1\right)^{l}}{l!} & =\sum_{l=0}^{\infty}CC_{l}^{\left(r,s\right)}\left(x\right)\sum_{n=l}^{\infty}S_{2}\left(n,l\right)\frac{t^{n}}{n!}\label{eq:39}\\
 & =\sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}CC_{l}^{\left(r,s\right)}\left(x\right)S_{2}\left(n,l\right)\right)\frac{t^{n}}{n!}.\nonumber
\end{align}

Therefore, by (\ref{eq:38}) and (\ref{eq:39}), we obtain the following
identities.

\[
\sum_{l=0}^{n}CC_{l}^{\left(r,s\right)}\left(x\right)S_{2}\left(n,l\right)=\sum_{l=0}^{n}\frac{\binom{n}{l}}{\binom{l+r}{l}}S_{2}\left(l+r,l\right)E_{n-l}^{\left(s\right)}\left(x\right),
\]
where $n\ge0$.

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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

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  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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config: arxiv


TITLE: Conceptual understanding of the effects of time shift on the coefficients of a Fourier Series.
QUESTION [1 upvotes]: I have a function of a triangle wave that goes from a minimum value of 0 volts to a maximum value of V. It has a period of T, and is a maximum at t = 0. I'm trying to understand how a time shift $\tau$ would affect the coefficients $a_n$ and $b_n$ of the Fourier series.
Without a time shift I can write:
$$a_n = \frac 2T \int_{-\frac T2}^0 f(t) \cos(\frac {2\pi nt}{T})\,dt\ + \frac 2T \int_{0}^{\frac T2}g(t) \cos(\frac {2\pi nt}{T})\,dt$$
But once a time shift is introduced I have:
$$a_n(\tau) = \frac 2T \int_{-\frac T2}^0 f(t +\tau) \cos(\frac {2\pi nt}{T})\,dt\ + \frac 2T \int_{0}^{\frac T2} g(t+ \tau) \cos(\frac {2\pi nt}{T})\,dt$$
My assumption is that this adds an extra few terms to the oringinal $a_n$ such that $a_n(\tau) = a_n + C$, where C depends on n. Is this statement accurate?
Thinking of the coefficients as a projection of $f(t)$ onto the vector $\cos(n\omega t)$, I don't understand why shifting the graph would change the value of this projection for a given point on the function. Why is this?
Maybe I'm thinking of this all wrong or making wrong assumptions, any help would be appreciated!

REPLY [1 votes]: Suppose that $f(t)$ is a periodic function with period $T$.
The fourier series of $f$ is,
$$ f(t) = a_0 + \sum_{n=1}^\infty \Big[a_n \cos(n\omega t) + b_n \sin(n\omega t)\Big]$$
Where
$$ a_0 = \frac1T \int_0^T f(t) \mathrm{d}t$$
$$ a_n = \frac2T \int_0^T f(t) \cos(n\omega t)\mathrm{d}t$$
$$ b_n = \frac2T \int_0^T f(t) \sin(n\omega t)\mathrm{d}t$$
$$ \omega = \frac{2\pi}{T}$$

Now lets study the effect of a shift in $t$ by an amount $\tau$ on the Fourier coefficients of $f$. We could study this by analyzing the integral expressions for the coefficients, but it is easier to study it by manipulating the Fourier series term by term.
$$ f(t+\tau) = a_0 + \sum_{n=1}^\infty \Big[a_n \cos\Big(n\omega (t+\tau)\Big) + b_n \sin\Big(n\omega (t+\tau)\Big)\Big]$$
$$ = a_0 + \sum_{n=1}^\infty \Big[a_n \cos(n\omega t+n\omega\tau) + b_n \sin(n\omega t+ n\omega\tau)\Big]$$
$$ = a_0 + \sum_{n=1}^\infty \Big[a_n \Big(\cos(n\omega t)\cos(n\omega\tau) - \sin(n\omega t)\sin(n \omega \tau) \Big) + b_n \Big( \sin(n\omega t)\cos(n\omega\tau) + \cos(n\omega t) \sin(n\omega\tau)\Big)\Big]$$
Collecting the coefficients of $\cos(n\omega t)$ and $\sin(n\omega t)$ together we get,
$$ = a_0 + \sum_{n=1}^\infty \Big[\Big(\color{blue}{ a_n \cos(n\omega\tau) + b_n \sin(n\omega\tau)} \Big) \cos(n\omega t) + \Big(\color{blue}{b_n \cos(n\omega\tau)-a_n \sin(n\omega\tau) }\Big) \sin(n\omega t) \Big]$$
The blue expressions are the new coefficients, lets call them $a_n'$ and $b_n'$
$$ = a_0 + \sum_{n=1}^\infty \Big[\color{blue}{ a_n'}  \cos(n\omega t) + \color{blue}{b_n'}\sin(n\omega t) \Big]$$
Note that the coefficient $a_0$ remains unchanged.
The transformation rule is,
$$ a_0 \rightarrow a_0$$
$$ a_n \rightarrow a_n \cos(n\omega\tau) + b_n \sin(n\omega\tau) \qquad (n\geq 1)$$
$$ b_n \rightarrow b_n \cos(n\omega\tau)-a_n \sin(n\omega\tau)$$

Note that these coefficients transform exactly as if they were the $x$ and $y$ components of a vector being rotated in the $xy$ plane by an angle $n\omega\tau$.


TITLE: Is $2^{\sqrt{2}} +3^{\sqrt{3}}$ rational?
QUESTION [8 upvotes]: In an exercise found in Lang's A Complete Course in Calculus, it's mentioned that it is unknown if $2^{\sqrt{2}} +3^{\sqrt{3}}$ is rational. Has this problem been solved since then?

REPLY [3 votes]: We know that both $2^{\sqrt2}$ and $3^{\sqrt3}$ are transcendental $($see the Gelfond-Schneider theorem for more information$)$, but proving that their sum $($or any other non-trivial linear combination of the two$)$ is also transcendental lies beyond current knowledge. The same goes for $e+\pi$ and others.


TITLE: Proving $\sin^2 (\pi x) - \frac{1}{x}$ is not eventually positive
QUESTION [0 upvotes]: $f:\mathbb{R}^+ \rightarrow\mathbb{R}$ is eventually positive if
$$
\exists M\in\mathbb{R}^+ \ \forall x\in\mathbb{R}^+ \ \ (x \geq M) \rightarrow (f(x)>0)
$$
Prove that $f(x) = \sin ^2 (\pi x) - \frac{1}{x}$ is not eventually positive.
I found the negation to be
$$\forall M >0 \ \exists x >0 \ \ (x\geq M)\wedge (f(x)\leq 0) $$
I always have trouble doing these type of questions.
Can someone give advice or some intuition in thinking about what choice to make?
In particular, I'm unsure of where to start. In limit proofs I think usually we'd start at the end, so I'd start with
$$f(x) = \sin^2 (\pi x) - \frac{1}{x} \leq 0
$$
which seems to be true for "small" enough $x$ values (by sketching the two graphs up) but not sure what to do now.

REPLY [3 votes]: Suppose that it is, then for some $M>0$, $\sin^{2}(\pi x)-1/x>0$ for all $x\geq M$. Choose some integer $N\geq M$, then $\sin^{2}(\pi N)-1/N=-1/N<0$.


TITLE: Why study CM abelian varieties?
QUESTION [14 upvotes]: I know that abelian varieties of CM type have central importance in algebraic geomtry and number theory. There are many conjectures and concepts related to them like Andre-Oort, Coleman conjecture, Shimura varieties, etc... . But why CM abelian varieties came into consideration in first place? Why did mathematicians begin to study CM abelian varieties?
I am in particular intereseted to know the relation to "special functions".

REPLY [9 votes]: (An elaboration of Timo Keller's answer.)
One of the main sources of historical interest in CM abelian varieties is the pursuit of Kronecker's Jugendtraum and Hilbert's 12th problem. If $K$ is an imaginary quadratic field, then one can always find an elliptic curve $E$ with CM by the full ring of integers $\mathfrak o_K$ of $K$, and then it turns out that, (i) $K(j(E))$ is the Hilbert class field of $K$, and (ii) all the abelian extensions of $K$ can be constructed by adjoining to $K(j(E))$ values of a certain "modular function" (the Weber function $h$) evaluated at torsion points of $E$. For the precise statement and further elaboration, pick up any book that deals with the CM theory of elliptic curves, e.g., Silverman's second book on elliptic curves.
What about more general number fields $K$: can we explicitly construct their abelian extensions in a similar fashion? The answer to this general question is wide open. But, anyhow, one possible approach is to try to think about how to generalize the above picture, in which case we ask: what should replace the field $K$, the elliptic curve $E$, and the special functions $j$ and $h$?
Hecke gave the first attempt at answering these questions. In his thesis he takes up the case of a real quadratic $K$ and uses a certain "Hilbert modular function", but doesn't get far. Later on he tries his hand at dealing with certain biquadratic extensions, and constructs unramified abelian extensions of them by adjoining singular values of two-variable Hilbert modular functions. Implicitly, the geometric object taking the place of $E$ here is a certain abelian surface with CM. Next comes Shimura who takes this one step further and looks at a general CM number field $K$. His $E$ and special functions are then CM abelian varieties and certain "Siegel modular functions". He uses this setup to construct certain class fields, except not over $K$, but over a certain other field $K^*$, the so-called "reflex field". (In the case of elliptic curves, we have $K=K^*$, and we get all the class fields of $K$.) So, unfortunately, unlike in the case of elliptic curves and imaginary quadratic fields, the higher-dimensional theory is more intricate doesn't give as complete an answer. For more details, you can start with Shimura's Automorphic functions and number theory (LMN 1968).

REPLY [8 votes]: Why did mathematicians begin to study CM abelian varieties?
Because they could. The study of CM abelian varieties arguably starts with Fagnano's work on the length of the lemniscate. In 1799, Gauss linked the length of the lemniscate to the arithmetic-geometric mean and in the entries of Gauss's diary from 1814, including famously (but not restricted to) the last one, he studied some properties of biquadratic reciprocity and their links with the division of the lemniscate which he understood to derive from properties of Punktgitter (lattices of points) in the complex planes, i.e CM elliptic curves.
As you see, there is a continuous path from traditional questions in geometry arguably going back to Antiquity (Fagnano's motivation was to rectify the ellipse) to accounts of complex multiplication quite close to the modern version, and the latter was known (at least to Gauss) in the early XIXth century. Legendre, Jacobi, Abel, Galois all knew about CM from one perspective or another.

REPLY [7 votes]: CM elliptic curves and their torsion points are used to explicitly construct ray class fields of imaginary quadratic number fields. E.g., the $j$-invariant generates the Hilbert class field.


TITLE: Why does $\tfrac{a}{b} = \tfrac{c}{d} = \tfrac{e}{f} \rightarrow \tfrac{a}{b} = \tfrac{c-e}{d-f}$
QUESTION [1 upvotes]: Why is it that
$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} \rightarrow \frac{a}{b} = \frac{c-e}{d-f}$$
It is used in the proof of Ceva's theorem. Thank you.

REPLY [0 votes]: $\begin{array}{rrrl}{\bf Hint}
&& d\, x\!\!\! &=&\!\!\! c\\
&& f\, x \!\!\!&=&\!\!\! e\\
\Rightarrow &&(d\!-\!f)\, x\!\!\! &=&\!\!\! c\!-\!e\ \ \text{ by subtracting the prior equations.}\end{array}$
Geometrically: if vectors $(d,c),\,(f,e)$ have the same slope, then so too does their difference.


TITLE: Picard group and jacobian of a singular curve
QUESTION [3 upvotes]: I found somewhere written that the jacobian of a reducible curve is the product of the jacobians, which seems quite reasonable to me. Where can I find some proof of it and a description of the Picard group? What can we say if we admit singularities in general?

REPLY [3 votes]: This is a long comment.
Definition. A projective nodal curve $C$ is said to be of compact type if one of the following equivalent conditions holds:

Its dual graph is a tree.
Its Jacobian is compact.
Every node is disconnecting, i.e. $C\setminus \{P\}$ is disconnected for every node $P$.
Every connected component of $\textrm{Pic}^d(C)$ is proper.

To address one of the comments, I would like to say: life is "much easier" when the curve is of compact type (e.g. your union $C=E\cup F$ is).
The nice issue about curves $C=X_1\cup\dots\cup X_r$ of compact type is that, for all $d\geq 0$,
$$\textrm{Pic}^d(C)=\coprod_{d_1+\dots+d_r=d}\textrm{Pic}^{d_1}(X_1)\times \dots\times \textrm{Pic}^{d_r}(X_r).\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$
In particular, $$\textrm{Pic}^0(C)=\prod_{i=1}^r\textrm{Pic}^0(X_i).$$
In other words, specifying a line bundle of degree $d$ on a curve of compact type boils down to fixing a "multidegree" $(d_1,\dots,d_r)$ and to choosing a line bundle of degree $d_i$ on each component $X_i$. This is really special to curves of compact type (For instance, a union of two elliptic curves meeting in two points is not so well behaved).

Two more words on the compact type condition: Why is this really a nice condition to ask when dealing with line bundles?
Suppose $C$ is a projective nodal curve, appearing as special fiber $f^{-1}(0)$ of a one-parameter family of curves $f:X\to B$, where $X$ is a smooth surface. Suppose that over the locus $U=B\setminus \{0\}$ the family is smooth, and we have a family of linear series $L_U$ on $X_U=X|_{f^{-1}(U)}$.
$$Question. \textrm{Does $L_U$ extend to a linear series on the whole of }X\textrm{?}$$
It seems like there is no problem: as $X$ is smooth, divisors extend, so line bundle extend. No assumption on $C$ is needed! However, if for some reason we wanted to base change the family, the total space of the new family would not be smooth any longer, in general. Here is where an assumption on $C$ is needed: if $C$ is of compact type, then, even after base change, line bundles always extend. (Maybe not uniquely, as the Picard scheme $\pi:\textrm{Pic}(X/B)\to B$ is not proper, because of $(\star)$).


TITLE: Maximum matching in bipartite graph
QUESTION [0 upvotes]: Prove that for every bipartite graph G with bipartition {A, B}, the size of the
maximum matching in G equals
|A| − δ
 where
$$ δ = max_{S⊆A} (|S| − |N_G(S)|) $$

REPLY [0 votes]: It is trivial that the size of a maximum matching cannot be greather than $|A| − δ$. We will then construct a matching of size $|A| − δ$ to complete the proof. let $S\subseteq A$ be a set of maximum size satisfiying $|S| - |N(S)| = δ$, let $T=N(S)$ and $G' = S \cup T$.
$T$ has a saturating matching in $G'$ otherwise due to hall's condition there exists a subset of T with less neighbors than its size, then omitting its neighbors from A would lead to a subset of A with less neighbors than $δ$ allows.
all that is left is to show $G-G'$ has a matching saturating its A part. if for some $S' \subseteq (G-G')\cap A: |N(S')| \leq |S'|$, then $S \cup S'$ form a set satisfiying $|S \cup S'| - N(S \cup S') \leq δ$, contradicting our choice of $S$.
thus $(G-G')\cap A$ satisfies hall's condition of a saturating matching. this matching combined with our matching saturating $T$ is a matching of size $|A| − δ$ in $G$.


TITLE: Measurability when conditioning on a sub-sigma field
QUESTION [0 upvotes]: I'm currently going through a book on measure-theoretic probability. In it, we defined conditional expectation:
Let X belong to $M^+(Omega, F)$ and P be a probability measure on F. For each sub-sigma-field G of F, the conditional expectation$ P(X|G)$ is the random variable $X_G$ in $M^+(X,G)$ for which $P(gX) = P(gX_G)$.
However, there is a small disclaimer: be careful to check that $X_G$ is G-measurable. I'm confused about how this can be a problem, since isn't $X_G\in M^+(X,G)$? Doesn't this by definition make it $G-$measurable?

REPLY [0 votes]: They are just saying that if you want to prove that $Y$ is $P(X|G)$ you have to verify two facts: $Y \in M^{+}(X,G)$ and the condition $P(gX)=P(gY)$. A common mistake it to ignore the first part.


TITLE: How much does it cost every time your car stops for a traffic light?
QUESTION [3 upvotes]: Every time we have to stop the car, it is costing us extra money, because we have to then accelerate to full speed again. I would like to know how much.
In order to simplify the situation, we can make these assumptions.

we disregard cost of car maintenance (brake pads wearing out etc.). This is focused entirely on fuel economy.
A car is always traveling at 60km/h when it needs to stop
The car's fuel economy is 8L/100kms when traveling at 60km
The cost of fuel is $1.25 per litre.
The car weighs 1000kg

I will be happy with either a definitive answer, an explanation of how to arrive at the answer, or an explanation of why there is not enough information in the question to come up with the answer.

REPLY [9 votes]: Here is a very basic estimation:
The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$
The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate the car to a given speed which would cost you 5 cents.

REPLY [0 votes]: I believe more information is needed to formulate an answer. Either you'd need the potential energy per liter of the gas and the necessary information to determine how much energy is lost in breaking (in which case, assuming no other loss of energy for simplicity's sake, you can work it out based on pure energy considerations) or you need more detail regarding the acceleration and deceleration phases of motion (in which case you can use forces and work).


TITLE: solution to $\text{div}(f)=0$
QUESTION [0 upvotes]: What is the most compact representation of the general solution to
$$
\text{div}(f)=\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+\frac{\partial f_3}{\partial x_3}=0
$$
for a $f:\mathbb{R}^3\to\mathbb{R}$ that goes to zero exponentially as $|\vec{x}|\to\infty$.
My first approach was the Helmholtz theorem with a potential that satisfies the Laplace equation.
Edit: Using the Helmholtz theorem,  $f=-\nabla \Phi + \nabla\times f$,   one can write
$$
\nabla\cdot f = -\nabla^2 \Phi + \nabla\cdot \nabla\times f=-\Delta \Phi =0
$$

REPLY [2 votes]: The Poincar'e Lemma, as in Spivak's Calculus on Manifolds, includes
the fact that for smooth functions ( $x$ and $f(x)$ in $R^3$, domain
of $f$ a star-shaped set) you have
$$
 f(x) = {\rm curl\,}\Big(
 \int_0^1 f(tx)\times tx\,dt
                    \Big)
 +\int_0^1 t^2x{\rm \, div\,}f(tx)\,dt.
$$
(I have changed the notation to express it for vector fields; Spivak
has it for differential forms.)
When the divergence of $f$ is zero in a star-shaped set there are many
functions $g$ such that $f = {\rm curl\,} g$; the first integral
above is one of them.
It looks like we can't guarantee the exponential decay:
Suppose $|f(x)| \le ce^{-|x|}$. Then
$$
 \bigg|
  \int_0^1 f(tx)\times tx\,dt
 \bigg|
  \le
  \int_0^1 ce^{-t|x|}t|x|\,dt
$$
$$
 =\bigg[
 -c\big(t+\frac{1}{|x|}\big)e^{-t|x|}
 \bigg]_0^1
 =
 c\Big(-\big(1+\frac{1}{|x|}\big)e^{-|x|}+\frac{1}{|x|}
 \Big).
$$


TITLE: Elimination of $'t'$ in a locus problem
QUESTION [2 upvotes]: Okay , so I need to find a relation between $x$ and $y$ using
$$x=\dfrac{3+3t^2}{t^2-1} \qquad and \qquad y=\dfrac{3t}{t^2-1}$$
I tried by solving for $t$ from $y$ , but that ended up in a quadratic in $t$ .Now solving for $t$ and then putting $t$ in expression of $x$ further complicates the result.
Isn't there another , less tedious  way ? Any help would be appreciated !

REPLY [0 votes]: Hint...
Consider that $$\frac xy=\frac 1t+t$$ and $$\frac 3y=t-\frac 1t$$ so add these equations and you can get $t$ which can then be substituted


TITLE: Show that each variance-covariance-matrix is symmetric and nonnegative definite
QUESTION [0 upvotes]: As the title already says I have to show that for $Z=(Z_1,...,Z_n)^T$ it is
    $$
y^T\text{Cov}(Z)y\geqslant 0~\forall~y\in\mathbb{R}^n.
$$

In a book I read
$$
y^T\text{Cov}(Z)y=\text{Var}(y^TZ)\geqslant 0
$$
but that's not clear to me, because I would have said that
$$
y^T\text{Cov}(Z)y=\text{Cov}(y^TZ)
$$
(and that can be negative).
So in other words, I do not see why
$$
\text{Cov}(y^TZ)=\text{Var}(y^TZ).
$$

Symmetric is clear because of $\text{Cov}(Z_i,Z_j)=\text{Cov}(Z_j,Z_i)$.

REPLY [1 votes]: Indeed $y^T\text{Cov}(Z)y=\text{Var}(y^TZ)$ (variance, not covariance) since $y^TZ$ is a random matrix of size $1\times1$, that is, a real-valued random variable. Thus, $y^T\text{Cov}(Z)y$ is a nonnegative real number for every vector $y$, which proves that the symmetric matrix $\text{Cov}(Z)$ of size $n\times n$ is nonnegative definite.


TITLE: Are uncertainties higher than measured values realistic?
QUESTION [3 upvotes]: Whenever I measure a positive quantity (e.g. a volume) there is some uncertainty related to the measurement. The uncertainty will usually be quite low, e.g. lower than 10%, depending on the equipment. However, I have recently seen uncertainties (due to extrapolation) larger than the measurements, which seems counter-intuitive since the quantity is positive.
So my questions are:

Do uncertainties larger than the measurements make sense?
Or would it be more sensible to "enforce" an uncertainty (cut-off) no higher than the measurement?

(The word "measurement" might be poorly chosen in this context if we are including extrapolation.)

REPLY [1 votes]: Something which has an exponential distribution (so positive) with expected value $\mu$ also has standard deviation $\mu$  - there are other distributions on positive values where the standard deviation can be many times the size of the expected value
By a vague recollection from the normal distribution, you might naively think there could be roughly a $95\%$ chance that a observation of this would be in the range $\mu \pm 2 \mu$. This is a wrong approach in general, but in this special case you would turn out to be correct; more precisely this is the probability of an observation below $3\mu$, which is $1-e^{-3} \approx 0.9502$
The mistake, if any, is thinking that the interval of uncertainty should always be symmetric about the central value


TITLE: Calculus book for computer science students
QUESTION [3 upvotes]: I'm going to teach calculus I and II to undergraduate computer science students and I would like to know if someone here knows some book or site with easy calculus applications in computer science.
Thanks

REPLY [0 votes]: I think The Calculus 7 by Louis Leithold would serve as a good textbook for your undergraduate computer science students.  I used this book in teaching students with the same major a few years back, and all of them found it to be detailed and challenging -- worthy to be read thoroughly.


TITLE: "strongly mixing" action on dimers?
QUESTION [5 upvotes]: In Local Statistics of Lattice Dimers we study a nice familiar object, domino tilings in the plane extending out to infinity.

His paper is going to discuss the frequency of various "motifs" in tiligns of infinite dimer regions and I am hoping to verify all such frequences lie in $\mathbb{Q}[\tfrac{1}{\pi}]$.
However, I am thrown off by the following phrase:

It is known that the translation-action of $\mathbb{Z}^2$ on dominos or lozenges is topologically mixing

I didn't even know where was a translation action on domino tilings.  Kenyon proceeds to quantify this "mixing" as
$$ \mu(U_{T_1}\cap U_{v+T_2})  =  \mu(U_{T_1})\,\mu(U_{v+T_2})  +  O(\tfrac{1}{|v^2|}) $$
Apparently the translation $T_v: \mathbb{Z}^2 \to \mathbb{Z}^2$ induce a translation on the set of infinite dimer tilings on the plane $T_v: X \to X$.
This mixing result says any two motifs $T_1, T_2$ are relatively independent of each other.  Then a tiling is expected to have a motif $T_1$ in one area with probability $\mu(U_{T_1})$ and $T_2$ with probability   $\mu(U_{T_2})$.  The odds of having both is close to $\mu(U_{T_1})\mu(U_{T_2}) $

Intuitively, I would have just embedded both regions $T_1$ and $T_2$ in a tilings of the square and computed the frequences of both motifs $T_1$ and $v + T_2$ occurring in random tilings as the size of the square got large.
However, the more they try to get precise, the more confused I get.

Translation is not mixing in the plane, and yet it is mixing on the space of domino tilings.  How does that make sense?
What was the important of identifying the unique measure $\mu$ of maximal entropy? and showing $\mu$ was the same as that of uniform random tiling for large squares with entropy ?
$$ H = \frac{1}{k^2} \log  |\#\{ \text{ tilings of }\square_{k \times k}\}|$$

REPLY [3 votes]: Let's start with Ian's example of $\{ 0 , 1 \}^{\mathbb{Z}}$. I'll write a point of $\{ 0,1 \}^{\mathbb{Z}}$ as a doubly infinite sequence $(x_i)$. The topology on $\{ 0,1 \}^{\mathbb{Z}}$ has a basis of open sets of the form $\Omega(L,R, (a_L, \ldots, a_R)) := \{ (x_i) : (x_L, x_{L+1}, \ldots, x_R) = (a_L, a_{L+1}, \ldots, a_R) \}$ for some integers $L \leq R$ and some fixed bit sequence $(a_L, \ldots, a_R)$. The measure of this open set is $1/2^{R-L+1}$. Intuitively, we are talking about independent random coin flips.
Saying that translation $T$ is strongly mixing means that, for any $(L,R, (a_L, \ldots, a_R))$ and $(L', R', (a_{L'}, \ldots, a_{R'}))$, the probability that $x \in \Omega(L,R, (a_L, \ldots, a_R)) \cap T^N \Omega(L',R', (a_L', \ldots, a_R'))$ approaches $2^{-(R-L+1)} 2^{-(R'-L'+1)}$ as $N \to \infty$. Sure enough, as soon as $N$ is large enough that the intervals $[L,R]$ and $[L'+N, R'+N]$ don't overlap, the bits in those intervals become independent.
Note that it is exactly the fact that translation is NOT mixing on $\mathbb{Z}$ that allows it to be mixing on $\{ 0,1 \}^{\mathbb{Z}}$. If translation were mixing on $\mathbb{Z}$, then arbitrary translates of $[L',R']$ would continue to intersect  $[L',R']$, so the bit strings would be correlated on these intervals and the probabilities wouldn't become independent.
Dimers are more complicated, because the condition that the region outside two planar patches is tileable imposes constraints even when the regions don't overlap. But it is intuitively plausible that this correlation decays as the regions become further separated, and the equation you quote makes this precise.

Your stated goal is true, and is pointed out by Kenyon immediately after Theorem 1. He expresses the frequency of motif with $k$ dimers as a $k \times k$ determinant, each entry of which is in $\mathbb{Q} + \mathbb{Q} (1/\pi)$.