text:
\begin{document}
\title[The values of zeta functions at integers]{The values of zeta functions composed by the Hurwitz and periodic zeta functions at integers}
\author[T.~Nakamura]{Takashi Nakamura}
\address[T.~Nakamura]{Department of Liberal Arts, Faculty of Science and Technology, Tokyo University of Science, 2641 Yamazaki, Noda-shi, Chiba-ken, 278-8510, Japan}
\email{nakamuratakashi@rs.tus.ac.jp}
\urladdr{https://sites.google.com/site/takashinakamurazeta/}
\subjclass[2010]{Primary 11M35, Secondary 11M20, 60G18}
\keywords{Hurwitz zeta function, periodic zeta function, values at integers, zeros, stationary self-similar Gaussian distributions}
\maketitle

\begin{abstract}
For $s \in {\mathbb{C}}$ and $0 < a <1$, let $\zeta (s,a)$ and ${\rm{Li}}_s (e^{2\pi ia})$ be the Hurwitz and periodic zeta functions, respectively. For $0 < a \le 1/2$, put $Z(s,a) := \zeta (s,a) + \zeta (s,1-a)$, $P(s,a) := {\rm{Li}}_s (e^{2\pi ia}) + {\rm{Li}}_s (e^{2\pi i(1-a)})$, $Y(s,a) := \zeta (s,a) - \zeta (s,1-a)$ and $O(s,a):= -i \bigl( {\rm{Li}}_s (e^{2\pi ia}) - {\rm{Li}}_s (e^{2\pi i(1-a)}) \bigr)$.

Let $n \ge 0$ be an integer and $b := r/q$, where $q>r>0$ are coprime integers. In this paper, we prove that the values $Z(-n,b)$, $\pi^{-2n-2} P(2n+2,b)$, $Y(-n,b)$ and $\pi^{-2n-1} O(2n+1,b)$ are rational numbers, in addition, $\pi^{-2n-2} Z(2n+2,b)$, $P(-n,b)$, $\pi^{-2n-1} Y(2n+1,b)$ and $O(-n,b)$ are polynomials of $\cos (2\pi/q)$ and $\sin (2\pi/q)$  with rational coefficients. Furthermore, we show that $Z(-n,a)$, $\pi^{-2n-2} P(2n+2,a)$, $Y(-n,a)$ and $\pi^{-2n-1} O(2n+1,a)$ are polynomials of $0<a<1$ with rational coefficient, in addition, $\pi^{-2n-2} Z(2n+2,a)$, $P(-n,a)$, $\pi^{-2n-1} Y(2n+1,a)$ and $O(-n,a)$ are rational functions of $\exp (2 \pi ia)$ with rational coefficients. Note that the rational numbers, polynomials and rational functions mentioned above are given explicitly.

Moreover, we show that $P(s,a) \equiv 0$ for all $ 0 < a < 1/2$ if and only if $s$ is a negative even integer. We also prove similar assertions for $Z(s,a)$, $Y(s,a)$, $O(s,a)$ and so on. In addition, we prove that the function $Z(s,|a|)$ appears as the spectral density of some stationary self-similar Gaussian distributions.
\end{abstract}

\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction and statement of main results}

\subsection{Special values of the Riemann zeta function}

For a complex variable $s = \sigma +it$, where $\sigma,t \in {\mathbb{R}}$, the Riemann zeta function is defined by
$$
\zeta (s) := \sum_{n=1}^\infty \frac{1}{n^s}, \qquad \sigma >1.
$$
According to the integral representation
$$
\zeta (s) = \frac{e^{-i\pi s} \Gamma (1-s)}{2 \pi i} \int_C \frac{z^{s-1}}{e^z-1} dz,
$$
where the contour $C$ starts at infinity on the positive real axis, encircles the origin once in the positive direction, excluding the points $\pm 2\pi i$, $\pm 4\pi i,\ldots$, and returns to the positive infinity, we can see that $\zeta (s)$ is a meromorphic function with a simple pole at $s=1$ with residue $1$. Moreover, the function $\zeta (s)$ satisfies the functional equation
\begin{equation}\label{eq:RZfe}
\zeta(1-s) =\frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \zeta(s)
\end{equation}
from the integral representation above. The functional equation and integral representation derive the following (see for example \cite[Section 12.12]{Apo} and \cite[Section 2.4]{Tit}). Note that the $n$-th Bernoulli number $B_n$ is defined in Section 2.1.

\begin{thm}
For $n \in {\mathbb{N}}$, we have
$$
\zeta (2n) = (-1)^{n+1} \frac{(2\pi)^{2n}}{(2n)!} B_{2n}.
$$
For every integer $n \ge0$, it holds that
$$
\zeta (-n) = -\frac{B_{n+1}}{n+1}.
$$
\end{thm}

Let $\chi$ be a Dirichlet character and $L(s,\chi)$ be the Dirichlet $L$-function associated to the character $\chi$. And let $\chi$ be a primitive character and $n$ be a natural number. Then it is known that $\pi^{-n} L(n,\chi)$ is written by a Gauss sum and the generalized Bernoulli number if $\chi (-1) = (-1)^n$ (see for example \cite[Theorem 9.6]{AIK}). Furthermore, it is also known that $L(-n,\chi)$ is written by a generalized Bernoulli number when $n$ is a non-negative integer (see \cite[Theorem 9.10]{AIK}). Obviously, these facts are an analogue of Theorem A. In general, there are no such explicit evaluation formulas at integers for automorphic $L$-function, Epstein zeta functions, the prime zeta function and so on. Hence, there are few zeta or $L$-functions of which the values at both positive and negative integers are expressed by $\pi$ and (generalized) Bernoulli numbers.

\subsection{The Hurwitz and periodic zeta functions}

The Hurwitz zeta function $\zeta (s,a)$ is defined by the series
$$
\zeta (s,a) := \sum_{n=0}^\infty \frac{1}{(n+a)^s}, \qquad \sigma >1, \quad 0<a \le 1.
$$
The function $\zeta (s,a)$ is  meromorphic and has a simple pole at $s=1$ whose residue is $1$ (see for instance \cite[Section 12]{Apo}).
Next, we define the periodic zeta function by
$$
{\rm{Li}}_s (e^{2\pi ia}) := \sum_{n=1}^\infty \frac{e^{2\pi ina}}{n^s}, \qquad \sigma >1, \quad 0<a \le 1
$$
(see for example \cite[Exercise 12.2]{Apo}). The periodic zeta function ${\rm{Li}}_s (e^{2\pi ia})$ with $0<a<1$ is analytically continuable to the whole complex plane since the Dirichlet series of ${\rm{Li}}_s (e^{2\pi ia})$ converges uniformly in each compact subset of the half-plane $\sigma >0$ when $0<a<1$ (see for example \cite[p.~20]{LauGa}). Note that $\zeta (-n,a)$ and ${\rm{Li}}_{-n} (e^{2\pi ia})$ , where $n$ is a non-negative integer, are written by the Bernoulli number and Stirling number of the second kind, respectively (see Lemmas \ref{lemB2} and \ref{eq:SKKY2}). However, there are no such formulas for $\zeta (n,a)$ and ${\rm{Li}}_{n} (e^{2\pi ia})$ , where $n$ is a integer greater than $1$.

For $0 <a \le 1/2$, let
\begin{equation*}
\begin{split}
&Z(s,a) := \zeta (s,a) + \zeta (s,1-a), \qquad P(s,a) := {\rm{Li}}_s (e^{2\pi ia}) + {\rm{Li}}_s (e^{2\pi i(1-a)}), \\
&2Q(s,a):= Z(s,a) + P(s,a) = \zeta (s,a) + \zeta (s,1-a) + {\rm{Li}}_s (e^{2\pi ia}) + {\rm{Li}}_s (e^{2\pi i(1-a)}),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
&Y(s,a) := \zeta (s,a) - \zeta (s,1-a), \qquad O(s,a):= -i \bigl( {\rm{Li}}_s (e^{2\pi ia}) - {\rm{Li}}_s (e^{2\pi i(1-a)}) \bigr) ,\\
&2X(s,a) := Y(s,a) + O(s,a) = \zeta (s,a) - \zeta (s,1-a) -i \bigl( {\rm{Li}}_s (e^{2\pi ia}) - {\rm{Li}}_s (e^{2\pi i(1-a)}) \bigr) .
\end{split}
\end{equation*}
It should be noted that the functions $Y(s,a)$, $O(s,a)$ and $X(s,a)$ are entire (see \cite[Section 3.3]{NPCZ}). We remark that one has $Y(s,1/2) \equiv O(s,1/2) \equiv X(s,1/2) \equiv 0$.

In \cite[Section 1.2]{NPCZ} and \cite[Section 1.1]{NPRCZ}, the following are shown.

\begin{thm}\label{th:m1}
All real zeros of the function $Z(s,a)$ are simple and only at the non-positive even integers if and only if $1/4 \le a \le 1/2$.

Moreover, all real zeros of the function $P(s,a)$ are simple and only at the negative even integers if and only if $1/4 \le a \le 1/2$.
\end{thm}

\begin{thm}\label{th:mQ1}
All real zeros of the quadrilateral zeta function $Q(s,a)$ are simple and only at the negative even integers if and only if $a_0 < a \le 1/2$, where $a_0 = 0.1183751396...$ satisfies $Z(1/2, a_0) = P(1/2,a_0) = Q(1/2, a_0) = 0$.
\end{thm}

\begin{thm}\label{th:OY1}
All real zeros of the functions $Y(s,a)$, $O(s,a)$ or $X(s,a)$ with $0 < a < 1/2$ are simple and only at the negative odd integers.
\end{thm}

It should be emphasised that from the theorems above, the gap between consecutive real zeros of $Y(s,a)$, $O(s,a)$, $X(s,a)$ with $0 < a < 1/2$,  $Z(s,a)$ and $P(s,a)$ with $1/4 \le a \le 1/2$ and $Q(s,a)$ with $a_0 < a \le 1/2$ is always $2$, in other words, the the gaps do not depend on $a$ just like the Riemann zeta function $\zeta (s)$.

\subsection{Main results}

In the present paper, we investigate the values of zeta functions $Z(s,a)$, $P(s,a)$, $Q(s,a)$, $Y(s,a)$, $O(s,a)$ and $X(s,a)$ at integers (see Theorems \ref{th:Ber}, \ref{th:poly}, \ref{th:d1} and \ref{th:d2}). Moreover, we show that the zeta function $Z(s,a)$ is related to some stationary self-similar Gaussian distributions in Proposition \ref{th:apSi1}.

When $a =r/q$ is a rational number, we have the following as an analogue of the fact $\zeta (2n) \in {\mathbb{Q}} \pi ^{2n}$ proved by Theorem A. Note that the explicit evaluation formulas for the special vales below are given in Section 2.2.
\begin{theorem}\label{th:Ber}
Let $n$ be a non-negative integer and $q>r>0$ be coprime integers. Then
$$
Z(-n, r/q), \quad Y(-n, r/q), \quad \pi^{-2n-2} P(2n+2, r/q), \quad \pi^{-2n-1} O(2n+1, r/q)
$$
are rational numbers. Moreover,
\begin{equation*}
\begin{split}
&\pi^{-2n-2} Z(2n+2, r/q) , \quad P(-n, r/q), \quad \pi^{-2n-1} Y(2n+1, r/q), \quad O(-n, r/q), \\
&\pi^{-2n-2} Q(2n+2, r/q), \quad Q(-n, r/q), \quad \pi^{-2n-1} X(2n+1, r/q), \quad X(-n, r/q)
\end{split}
\end{equation*}
are elements of the polynomial ring ${\mathbb{Q}} [\cos (2\pi/q), \sin (2\pi/q)]$.
\end{theorem}

Next we prove the following when $0 < a < 1/2$ is irrational. It should be emphasised that the polynomials and rational functions in the theorem below are given explicitly in Section 3.2.
\begin{theorem}\label{th:poly}
Let $n$ be a non-negative integer. Then
$$
Z(-n,a), \quad \pi^{-2n-2}P(2n+2,a),  \quad Y(-n,a) , \quad \pi^{-2n-1} O(2n+1,a)
$$
are polynomials with rational coefficients of $0 < a < 1/2$. Furthermore,
\begin{equation*}
\begin{split}
&\pi^{-2n-2} Z(2n+2,a), \quad P(-n,a), \quad \pi^{-2n-1} Y(2n+1,a) \quad O(-n,a) \\
&\pi^{-2n-2} Q(2n+2, a), \quad Q(-n, a), \quad \pi^{-2n-1} X(2n+1, a), \quad X(-n, a)
\end{split}
\end{equation*}
are rational functions with rational coefficients of $\exp (2 \pi ia)$.
\end{theorem}

We can see that $P(s,a)$ identically vanishes for all $0<a<1/2$ if $s$ is a negative even integer by the functional equation of $P(s,a)$ (see Theorem B and Lemma \ref{lem:fe1}). The next theorem implies that $P(s,a) \equiv 0$ for all $0<a<1/2$ only if $s$ is a negative even integer.

\begin{theorem}\label{th:d1}
Let $s \ne 1$. Then we have
$$
Z(s,a) \equiv 0 \quad \mbox{ for all } \quad 0 < a < 1/2
$$
if and only if $s$ is a non-positive even integer. Furthermore it holds that
$$
Q(s,a) \equiv 0 \quad \mbox{ for all } \quad 0 < a < 1/2
$$
if and only if $s$ is a negative even integer.

Next let $s \in {\mathbb{C}}$. Then one has
$$
P(s,a) \equiv 0 \quad \mbox{ for all } \quad 0 < a < 1/2
$$
if and only if $s$ is a negative even integer. Moreover, we have
$$
Y(s,a) \equiv 0 \quad \mbox{ for all } \quad 0 < a < 1/2
$$
if and only if $s$ is an odd negative integer. The same statement holds for the zeta functions $O(s,a)$ and $X(s,a)$.
\end{theorem}

On the other hand, we have the following for $\zeta(s,a)$ and ${\rm{Li}}_s (e^{2\pi ia})$.
\begin{theorem}\label{th:d2}
For any $1 \ne s \in {\mathbb{C}}$, there exists $0 < a < 1/2$ such that
$$
\zeta (s,a) \ne 0.
$$
For any $s \in {\mathbb{C}}$, there is $0 < a < 1/2$ such that
$$
{\rm{Li}}_s (e^{2\pi ia}) \ne 0.
$$
\end{theorem}

Moreover, we have the following proposition which implies that $Z(s,|a|)$ appears as the spectral density of some stationary self-similar Gaussian distributions (for details see Appendix or \cite[Section 1]{Sinai}).
\begin{proposition}\label{th:apSi1}
Let $P$ be a one-dimensional stationary Gaussian distribution on $X$ with ${\mathbb{E}} x_l =0$. Then the distribution $P$ is an s.s.d, if and only if its spectral density $\rho_\lambda(\alpha)$ has the form
$$
\rho_\lambda(\alpha):= C | e^{2\pi i \alpha} -1|^2 Z(\lambda+1, |\alpha|),
\qquad -1/2 \le \alpha \le 1/2,
$$
where $C>0$ is a constant.
\end{proposition}

In Section 2 and 3, we prove Theorems \ref{th:Ber} and \ref{th:poly}, respectively. We prove Theorems \ref{th:d1} and \ref{th:d2} in Section 4. In Section 5, we prove Proposition \ref{th:apSi1}.

\section{Proof of Theorem \ref{th:Ber}}

\subsection{Bernoulli polynomials and functional equations}

We denote by $B_n (t)$ the Bernoulli polynomial of order $n$ defined as
$$
\frac{ze^{tz}}{e^z-1} = \sum_{n=0}^{\infty} B_n (t) \frac{z^n}{n!} .
$$
The first few are:
$$
B_0 (t) =1, \qquad B_1 (t) = t - \frac{1}{2}, \qquad B_2 (t) = t^2 - t + \frac{1}{6},
$$
$$
B_3 (t) = t^3 - \frac{3}{2} t^2 + \frac{1}{2} t, \qquad B_4 (t) = t^4 - 2t^3 + t^2 - \frac{1}{30}  .
$$
The following equation is well-known (see for example \cite[Exercise 12.11]{Apo}).
\begin{equation}\label{eq:ab1-a}
B_n (1-a) = (-1)^n B_n (a), \qquad n \ge 0.
\end{equation}
And we define the $n$-th Bernoulli number $B_n$ by
$$
B_n : = B_n(1).
$$
The following are well-known (see for instance \cite[Theorems 12.19 and 12.13]{Apo}).
\begin{lemma}\label{lemB1}
If $k \in {\mathbb{N}}$ and $0 < a < 1$, one has
$$
B_{2k} (a) = (-1)^{k+1} \frac{2(2k)!}{(2\pi)^{2k}} \sum_{m=1}^\infty \frac{\cos 2\pi ma}{m^{2k}}, \quad
B_{2k-1} (a) = (-1)^k \frac{2(2k-1)!}{(2\pi)^{2k-1}} \sum_{m=1}^\infty \frac{\sin 2\pi ma}{m^{2k-1}}.
$$
\end{lemma}
\begin{lemma}\label{lemB2}
For every integer $n \ge0$, it holds that
$$
\zeta (-n,a) = - \frac{B_{n+1}(a)}{n+1}.
$$
\end{lemma}

Next we quote the functional equations for $\zeta (s,a)$ and ${\rm{Li}}_s (e^{2\pi ia})$ (see \cite[Theorem 12.6 and Exercise 12.2]{Apo}) and $Z(s,a)$, $P(s,a)$, $Y(s,a)$, $Q(s,a)$ and $X(s,a)$ (see \cite[Sections 3.3 and 4.2]{NPCZ}).
\begin{lemma}\label{lem:fe1}
It holds that
\begin{equation*}
\begin{split}
&\zeta (1-s, a) =
\frac{\Gamma (s)}{(2\pi)^s} \Bigl( e^{-\pi is/2} {\rm{Li}}_s (e^{2\pi ia}) + e^{\pi is/2} {\rm{Li}}_s (e^{2\pi i(1-a)}) \Bigr),\\
&\,\,\, {\rm{Li}}_{1-s} (e^{2\pi ia}) =
\frac{\Gamma (s)}{(2\pi )^s} \Bigl( e^{\pi is/2} \zeta (s,a) + e^{-\pi is/2} \zeta (s,1-a) \Bigr),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
&Z(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) P(s,a), \qquad
P(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) Z(s,a), \\
&Y(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) O(s,a), \qquad
O(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) Y(s,a), \\
&Q(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) Q(s,a), \qquad
X(1-s,a) = \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) X(s,a).
\end{split}
\end{equation*}
\end{lemma}

\subsection{Proof of Theorem \ref{th:Ber}}

In this subsection, we give explicit evaluation formulas for $Z(-n,b)$, $P(2n+2,b)$, $Y(-n,b)$, $O(2n+1,b)$, $Z(2n+2,b)$, $P(-n,b)$, $Y(2n+1,b)$ and $O(-n,b)$, where $n$ is a non-positive integer, $q>r>0$ are coprime integers and $b := r/q$, which prove Theorem \ref{th:Ber}.

The next well-known formula plays an important role in the proof of Theorem \ref{th:Ber}.
\begin{lemma}\label{lem:invmul}
Let $r,q \in {\mathbb{N}}$ be coprime and $q>r>0$. The one has
\begin{equation}\label{eq:mul1}
{\rm{Li}}_s (e^{2\pi irm/q}) = q^{-s} \sum_{m=1}^q e^{2\pi irm/q} \zeta(s,m/q), \qquad s \in {\mathbb{C}}.
\end{equation}
\end{lemma}
\begin{proof}
For readers convenience, we write the proof. Let $|z|=1$ and $\sigma >1$. Then it is easy to see that
$$
\sum_{l=1}^\infty \frac{z^l}{l^s} = \sum_{m=1}^q \sum_{l=0}^\infty \frac{z^{ql+m}}{(ql+m)^s} =
q^{-s} \sum_{m=1}^q z^m \sum_{l=0}^\infty \frac{z^{ql}}{(l+m/q)^s}.
$$
By putting $z = e^{2\pi ir/q}$, we have (\ref{eq:mul1}).
\end{proof}

\begin{proposition}\label{pro:raZYPOn}
Let $r,q \in {\mathbb{N}}$ be coprime and $q>r>0$. Then, for $n \in {\mathbb{N}}$, we have
\begin{equation*}
\begin{split}
&Z(-n,r/q) = \frac{(-1)^n-1}{n+1} B_{n+1}(r/q), \qquad Y(-n,r/q) = \frac{(-1)^{n+1}-1}{n+1} B_{n+1}(r/q),\\
&\qquad \qquad P(-n,r/q) = -\frac{2 q^n}{n+1} \sum_{m=1}^q \cos (2\pi rm/q) B_{n+1} (m/q), \\
&\qquad \qquad O(-n,r/q) = -\frac{2 q^n}{n+1} \sum_{m=1}^q \sin (2\pi rm/q) B_{n+1} (m/q) .
\end{split}
\end{equation*}
\end{proposition}
\begin{proof}
By using Lemma \ref{lemB2}, we have
$$
Z(-n,r/q) = \zeta (-n,r/q) + \zeta(-n,1-r/q) = - \frac{B_{n+1}(r/q)}{n+1} - \frac{B_{n+1}(1-r/q)}{n+1}.
$$
Hence we obtain the first equation of Proposition \ref{pro:raZYPOn} from (\ref{eq:ab1-a}). Similarly, we have
$$
Y(-n,r/q) = \zeta (-n,r/q) - \zeta(-n,1-r/q) = - \frac{B_{n+1}(r/q)}{n+1} + \frac{B_{n+1}(1-r/q)}{n+1}
$$
which implies the second equation of Proposition \ref{pro:raZYPOn}. From (\ref{eq:mul1}), it holds that
\begin{equation*}
\begin{split}
&\quad q^s P(s,r/q) = q^s \Bigl( {\rm{Li}}_s (e^{2\pi irm/q}) + {\rm{Li}}_s (e^{2\pi i(q-r)m/q}) \Bigr) \\
&= \sum_{m=1}^q e^{2\pi irm/q} \zeta(s,m/q) + \sum_{m=1}^q e^{-2\pi irm/q} \zeta(s,m/q)
= 2 \sum_{m=1}^q \cos (2\pi rm/q) \zeta (s,m/q) .
\end{split}
\end{equation*}
By (\ref{eq:mul1}), we similarly obtain
\begin{equation*}
\begin{split}
&\quad i q^s O(\sigma,r/q) = i q^s \Bigl( {\rm{Li}}_s (e^{2\pi irm/q}) - {\rm{Li}}_s (e^{2\pi i(q-r)m/q}) \Bigr) \\
&= \sum_{m=1}^q e^{2\pi irm/q} \zeta(s,m/q) - \sum_{m=1}^q e^{-2\pi irm/q} \zeta(s,m/q)  =
2 i \sum_{m=1}^q \sin (2\pi rm/q) \zeta (s,m/q) .
\end{split}
\end{equation*}
Hence we have the third and fourth formulas of Proposition \ref{pro:raZYPOn} from Lemma \ref{lemB2}.
\end{proof}

The next proposition is proved by Proposition \ref{pro:raZYPOn} above and the functional equations in Lemma \ref{lem:fe1}.
\begin{proposition}\label{pro:raZYPOp}
Let $r,q \in {\mathbb{N}}$ be coprime. Then, for $n \in {\mathbb{N}}$, we have
\begin{equation*}
\begin{split}
&\qquad \qquad \qquad \quad Z(2n,r/q) = (-1)^{n+1}  q^{2n-1} \frac{(2\pi)^{2n}}{(2n)!} \sum_{m=1}^q \cos (2\pi rm/q) B_{2n} (m/q), \\
&\qquad \qquad \quad Y(2n-1,r/q) = (-1)^{n} q^{2n-2} \frac{(2\pi)^{2n-1}}{(2n-1)!} \sum_{m=1}^q \sin (2\pi rm/q) B_{2n-1} (m/q), \\
&P(2n,r/q) = (-1)^{n+1} \frac{(2\pi)^{2n}}{(2n)!} B_{2n} (r/q), \qquad
O(2n-1,r/q) = (-1)^n \frac{(2\pi)^{2n-1}}{(2n-1)!} B_{2n-1}(r/q).
\end{split}
\end{equation*}
\end{proposition}
\begin{proof}
From the functional equation of $P(1-s,a)$ in Lemma \ref{lem:fe1}, we have
$$
P(1-2n,a) = \frac{2(2n-1)!}{(2\pi)^{2n}} \cos (\pi n) Z(2n,a) = (-1)^{n} \frac{2(2n-1)!}{(2\pi)^{2n}} Z(2n,a) .
$$
Thus we obtain the first formula of this proposition from
$$
P(1-2n,r/q) = -\frac{2 q^{2n-1}}{2n} \sum_{m=1}^q \cos (2\pi rm/q) B_{2n} (m/q)
$$
which is proved by Proposition \ref{pro:raZYPOn}. Similarly, one has
$$
O (2-2n,a) = \frac{2(2n-2)!}{(2\pi)^{n-1}} \sin \Bigl( \frac{2n-1}{2} \pi \Bigr) Y(2n-1, a)
$$
by the functional equation of $O(1-s,a)$ in Lemma \ref{lem:fe1}. Hence, we have the second equation of Proposition \ref{pro:raZYPOp} and
$$
O (2-2n,a) = -\frac{2 q^{2n-2}}{2n-1} \sum_{m=1}^q \sin (2\pi rm/q) B_{2n-1} (m/q)
$$
derived from Proposition \ref{pro:raZYPOn}.

By the definition of $P(s,a)$, it holds that
\begin{equation}\label{eq:PB1}
P(2n, a) = {\rm{Li}}_{2n} (e^{2\pi ia}) + {\rm{Li}}_{2n} (e^{2\pi i(1-a)}) = 2 \sum_{m=1}^\infty \frac{\cos 2\pi ma}{m^{2n}} .
\end{equation}
In addition, we have
\begin{equation}\label{eq:OB1}
O(2n-1, a) = \frac{1}{i} \Bigl( {\rm{Li}}_{2n-1} (e^{2\pi ia}) - {\rm{Li}}_{2n-1} (e^{2\pi i(1-a)}) \Bigr)
= 2 \sum_{m=1}^\infty \frac{\sin 2\pi ma}{m^{2n-1}}
\end{equation}
from the definition of $O(s,a)$. Hence, the third and fourth equations in this proposition are prove by (\ref{eq:PB1}), (\ref{eq:OB1}) and Lemma \ref{lemB1}.
\end{proof}

We can immediately show the following by the propositions above and the definitions of $Q(s,a)$ and $X(s,a)$.
\begin{corollary}\label{cor:QXn}
Let $r,q \in {\mathbb{N}}$ be coprime. Then, for $n \in {\mathbb{N}}$, one has
\begin{equation*}
\begin{split}
2Q(-n,r/q) &= \frac{(-1)^n-1}{n+1} B_{n+1}(r/q) - \frac{2 q^n}{n+1} \sum_{m=1}^q \cos (2\pi rm/q) B_{n+1} (m/q), \\
2X(-n,r/q) &= \frac{(-1)^{n+1}-1}{n+1} B_{n+1}(r/q) - \frac{2 q^n}{n+1} \sum_{m=1}^q \sin (2\pi rm/q) B_{n+1} (m/q).
\end{split}
\end{equation*}
\end{corollary}

\begin{corollary}\label{cor:QXp}
Let $r,q \in {\mathbb{N}}$ be coprime. Then, for $n \in {\mathbb{N}}$, one has
\begin{equation*}
\begin{split}
2Q(2n,r/q) &= (-1)^{n+1} \frac{(2\pi)^{2n}}{(2n)!} \biggl( B_{2n} (r/q) + q^{2n-1} \sum_{m=1}^q \cos (2\pi rm/q) B_{2n} (m/q) \biggr),  \\
2X(2n-1,r/q) &= (-1)^n \frac{(2\pi)^{2n-1}}{(2n-1)!} \biggl( B_{2n-1}(r/q) + q^{2n-2} \sum_{m=1}^q \sin (2\pi rm/q) B_{2n-1} (m/q) \biggr) .
\end{split}
\end{equation*}
\end{corollary}

\begin{proof}[Proof of Theorem \ref{th:Ber}]
We can prove Theorem \ref{th:Ber} from Propositions \ref{pro:raZYPOn} and \ref{pro:raZYPOp}, Corollary \ref{cor:QXn} and \ref{cor:QXp} and de Moivre's identity
$$
\cos n\theta + i \sin n \theta = (\cos \theta + i \sin \theta)^n, \qquad n \in {\mathbb{N}}, \quad \theta \in {\mathbb{R}},
$$
and fact that Bernoulli polynomials are polynomials with rational coefficients.
\end{proof}

\section{Proof of Theorem \ref{th:poly}}

\subsection{Generalized Euler polynomials}

For $0 < a < 1$, we define the generalized Euler polynomial $E_{c,n} (t)$ by
$$
\frac{(1+c) e^{tz}}{e^z+c} = \sum_{n=0}^\infty E_{c,n}(t) \frac{z^n}{n!}, \qquad c := -\exp (2\pi ia).
$$
The polynomial $E_{c,n} (t)$ above is introduced in \cite[Section 4.1]{Shi}. Note that similar polynomials are defined by Apostol \cite{ApoL} and Frobenius \cite{Fro}. For simplicity, we put $b := - (1+c)^{-1}$. Then we have (see \cite[Section 4.1]{Shi})
$$
E_{c,n} (t) = t^n + b \sum_{k=0}^{n-1} \binom{n}{k} E_{c,n} (t), \qquad
\frac{d}{dt} E_{c,n} (t) = n E_{c,n-1} (t), \qquad n>0,
$$
$$
E_{c,n} (t+1) + cE_{c,n} (t) = (1+c)t^n, \qquad E_{1,2n+1}(1/2) =0,
$$
$$
E_{c,n} (1-t) = (-1)^n E_{c^{-1}\!, \!\; n} (t), \qquad E_{c^{-1}\!, \!\; n} (0) = (-1)^{n+1} c E_{c,n} (0).
$$
For instance, one has
\begin{equation*}\begin{split}
&\quad E_{c,0} (t) = 1, \qquad E_{c,1} (t) = t+b, \qquad E_{c,2} = t^2+2bt+2b^2+b, \\
&E_{c,3} (t) = t^3 + 3bt^2 + (6b^2+3b)t + 6b^3 + 6b^2 +b, \qquad b := - (1+c)^{-1}.
\end{split}\end{equation*}
When $n \in {\mathbb{N}}$ and $0<a<1$, we define $F_n (a)$ by
$$
F_n (a) := \sum_{l \in {\mathbb{Z}}} \frac{1}{(l+a)^{n+1}} =
\sum_{l = 0}^\infty \frac{1}{(l+a)^{n+1}} + (-1)^{n+1} \sum_{l = 0}^\infty \frac{1}{(l+1-a)^{n+1}}.
$$
We have the following by $E_{c,n} (0) = (1+c^{-1}) n! (2\pi i)^{-n-1} F_n (a)$ proved in \cite[Theorem 4.2]{Shi}.
\begin{lemma}\label{lem:P1}
For $n \in {\mathbb{N}}$, it holds that
\begin{equation}\label{eq:SKKY1}
F_n (a) = \frac{(2\pi i)^{n+1} E_{c,n} (0)}{n! (1+c^{-1})}, \qquad c := -\exp (2\pi ia).
\end{equation}
\end{lemma}

By using Yamamoto's formula (see \cite[Proposition 3.2]{Yam} or \cite[p.~17]{KKY}) and the functional equation of ${\rm{Li}}_{1-s} (e^{2\pi ia})$ (see Lemma \ref{lem:fe1}), we have the following.
\begin{lemma}\label{lem:P2}
For every integer $n \ge0$, it holds that
\begin{equation}\label{eq:SKKY2}
{\rm{Li}}_{-n} (e^{2\pi ia}) =
\sum_{r=0}^n \frac{r! (-c)^{r} S(n,r)}{(1+c)^{r+1}} = \frac{n! F_n (1-a)}{(2\pi i)^{n+1}} = \frac{E_{c^{-1}\!, \!\; n} (0)}{1+c} ,
\end{equation}
where $c := -\exp (2\pi ia)$ and $S(n,r)$ is the Stirling numbers of the second kind which is defined as $r! S(n,r) := \sum_{m=1}^r (-1)^{r-m} \binom{r}{m} m^n$.
\end{lemma}
\begin{proof}
From the Yamamoto formula proved in \cite[Proposition 3.2]{Yam}, we have
$$
{\rm{Li}}_{-n} (e^{2\pi ia}) = \sum_{r=0}^n  \frac{r! S(n,r) ( e^{2\pi i a})^r}{(1-e^{2\pi i a})^{r+1}} =
\sum_{r=0}^n \frac{r! S(n,r) (-c)^{r}}{(1+c)^{r+1}} .
$$
Hence we obtain the first equal sign of (\ref{eq:SKKY2}). By putting $s=n+1$ in the functional equation of ${\rm{Li}}_{1-s} (e^{2\pi ia})$ (see Lemma \ref{lem:fe1}), we have
\begin{equation*}\begin{split}
{\rm{Li}}_{-n} (e^{2\pi ia}) &= \frac{n!}{(2\pi)^{n+1}} \Bigl( i^{n+1} \zeta (n+1,a) + (-i)^{n+1} \zeta (n+1,1-a) \Bigr) \\
&= \frac{(i)^{n+1} n!}{(2\pi)^{n+1}} F_n (a) = \frac{n!}{(2\pi i)^{n+1}} F_n (1-a)
\end{split}\end{equation*}
which implies the second equal sign of (\ref{eq:SKKY2}). We obtain the third equal sign of (\ref{eq:SKKY2}) from Lemma \ref{lem:P1}.
\end{proof}

\subsection{Proof of Theorem \ref{th:poly}}

In this subsection, we give explicit evaluation formulas for $Z(-n,a)$, $P(2n+2,a)$, $Y(-n,a)$, $O(2n+1,a)$, $Z(2n+2,a)$, $P(-n,a)$, $Y(2n+1,a)$, $O(-n,a)$, $Q(2n+2,a)$, $Q(-n,a)$, $X(2n+1,a)$ and $X(-n,a)$, where $n$ is a non-positive integer, which prove Theorem \ref{th:poly}.
\begin{proposition}\label{Pro:Z1}
For $n \in {\mathbb{N}}$, we have
$$
Z(2n, a) = \frac{(2\pi i)^{2n} E_{c,2n-1} (0)}{(1+c^{-1}) (2n-1)!}  =
\frac{c (2\pi i)^{2n}}{(2n-1)!} \sum_{r=0}^{2n-1} \frac{(-1)^r r! S(2n-1,r)}{(1+c)^{r+1}} .
$$
For every integer $n \ge0$, it holds that
$$
Z (-n,a) = - \frac{B_{n+1}(a)+ B_{n+1}(1-a)}{n+1} = \frac{(-1)^n-1}{n+1} B_{n+1} (a).
$$
\end{proposition}
\begin{proof}
Obviously, we have
$$
Z(2n,a) = F_{2n-1} (a).
$$
Hence, the first formula is proved by (\ref{eq:SKKY1}), (\ref{eq:SKKY2}) and
\begin{equation}\label{eq:ES1}
\frac{E_{c,n} (0)}{1+c^{-1}} = \sum_{r=0}^n \frac{r! (-c^{-1})^r S(n,r)}{(1+c^{-1})^{r+1}} =
c \sum_{r=0}^n \frac{r! (-1)^r S(n,r)}{(c+1)^{r+1}}.
\end{equation}
We obtain the second formula from (\ref{eq:ab1-a}) and Lemma \ref{lemB2}.
\end{proof}

\begin{proposition}\label{Pro:P1}
For $n \in {\mathbb{N}}$, we have
$$
P(2n, a) = (-1)^{n+1} \frac{(2\pi)^{2n}}{(2n)!} B_{2n} (a).
$$
For every integer $n \ge0$, it holds that
$$
P (-n,a) = \frac{1-(-1)^n}{1+c^{-1}} E_{c,n} (0) = \frac{1-(-1)^n}{c^{-1}+c^{-2}} \sum_{r=0}^n \frac{(-1)^r r! S(n,r)}{(1+c)^{r+1}}.
$$
\end{proposition}
\begin{proof}
We have the first formula from (\ref{eq:PB1}) and Lemma \ref{lemB1}. The second formula is shown by (\ref{eq:ES1}), the definition of $P(s,a)$, Lemma \ref{lem:P2}, the formula
$$
P (-n,a) = \frac{E_{c^{-1}\!, \!\; n} (0)}{1+c} + \frac{E_{c,n} (0)}{1+c^{-1}}
$$
and the equation $E_{c^{-1},n} (0) = (-1)^{n+1} c E_{c,n} (0)$ (see Section 2.3 or \cite[(4.3g)]{Shi}).
\end{proof}

\begin{proposition}\label{Pro:Y1}
For $n \in {\mathbb{N}}$, we have
$$
Y(2n-1, a) = \frac{(2\pi i)^{2n-1} E_{c,2n-2} (0)}{(1+c^{-1}) (2n-2)!} =
\frac{c(2\pi i)^{2n-1}}{(2n-2)!} \sum_{r=0}^{2n-2} \frac{(-1)^r r! S(2n-2,r)}{(1+c)^{r+1}} .
$$
For every integer $n \ge0$, it holds that
$$
Y (-n,a) = - \frac{B_{n+1}(a)- B_{n+1}(1-a)}{n+1} = \frac{(-1)^{n-1}-1}{n+1} B_{n+1} (a).
$$
\end{proposition}
\begin{proof}
The first formula is shown by (\ref{eq:ES1}), Lemma \ref{lem:P1} and
$$
Y(2n-1,a) = F_{2n-2} (a)
$$
if $n \ge 2$. The case $n=1$ is shown by
$$
\lim_{s \to 1} Y(s,a) = \sum_{n=0}^\infty \biggl( \frac{1}{n+a} - \frac{1}{n+1-a} \biggr) =\psi(1-a) - \psi (a) = \pi \cot \pi a,
$$
where $\psi (a)$ the digamma function. We have the second formula of this proposition from (\ref{eq:ab1-a}) and Lemma \ref{lemB2} again.
\end{proof}

\begin{proposition}\label{Pro:O1}
For $n \in {\mathbb{N}}$, we have
$$
O(2n-1, a) = (-1)^n \frac{(2\pi)^{2n-1}}{(2n-1)!} B_{2n-1}(a) .
$$
For every integer $n \ge0$, it holds that
$$
O (-n,a) = \frac{1+(-1)^n}{i(1+c^{-1})} E_{c,n} (0) = \frac{1+(-1)^n}{i(c^{-1}+c^{-2})} \sum_{r=0}^n \frac{(-1)^r r! S(n,r)}{(1+c)^{r+1}}.
$$
\end{proposition}
\begin{proof}
We have the first formula of this proposition from (\ref{eq:OB1}) and Lemma \ref{lemB1}. We obtain the second formula by Lemma \ref{eq:SKKY2}, the equations
$$
O (-n,a) = \frac{1}{i} \biggl( \frac{E_{c^{-1}\!, \!\; n} (0)}{1+c} - \frac{E_{c,n} (0)}{1+c^{-1}} \biggr)
$$
and $E_{c^{-1},n} (0) = (-1)^{n+1} c E_{c,n} (0)$ again.
\end{proof}

By the propositions above and definitions of $Q(s,a)$ and $X(s,a)$, we have the following.
\begin{corollary}\label{cor:Q1}
For $n \in {\mathbb{N}}$, we have
$$
2Q(2n, a) = \frac{(2\pi i)^{2n} E_{c,2n-1} (0)}{(1+c^{-1}) (2n-1)!} - (-1)^n \frac{(2\pi)^{2n}}{(2n)!} B_{2n} (a).
$$
For every integer $n \ge0$, it holds that
$$
2Q(-n,a) = \frac{1-(-1)^n}{1+c^{-1}} E_{c,n} (0) - \frac{1-(-1)^n}{n+1} B_{n+1} (a).
$$
\end{corollary}
\begin{corollary}\label{cor:X1}
For $n \in {\mathbb{N}}$, we have
$$
2X(2n-1, a) = \frac{(2\pi i)^{2n-1} E_{c,2n-2} (0)}{(1+c^{-1}) (2n-2)!} + (-1)^n \frac{(2\pi)^{2n-1}}{(2n-1)!} B_{2n-1}(a) .
$$
For every integer $n \ge0$, it holds that
$$
2X(-n,a) = \frac{1+(-1)^n}{i(1+c^{-1})} E_{c,n} (0) - \frac{1+(-1)^n}{n+1} B_{n+1} (a) .
$$
\end{corollary}

\section{Proofs of Theorems \ref{th:d1}{} and \ref{th:d2}}

\subsection{Proof of Theorem \ref{th:d1}}

Recall the Hermite formula
$$
\zeta (s,a) = \frac{a^{-s}}{2} + \frac{a^{1-s}}{s-1} + 2
\int_0^\infty \frac{\sin (s \arctan (x/a))}{(x^2+a^2)^{s/2} (e^{2\pi x}-1)} dx,
$$
where the integral involved in the formula above converges for all $s \in {\mathbb{C}}$ (see for example \cite[Section 13.2]{WW}). On the other hand, the following equation is well-known:
\begin{equation}\label{eq:1+a}
\zeta (s,a) = a^{-s} + \zeta (s,1+a).
\end{equation}

\begin{proof}[Proof of Theorem \ref{th:d1} for $Z(s,a)$ and $P(s,a)$]
From (\ref{eq:1+a}), for all $1 \ne s \in {\mathbb{C}}$ with $\sigma >0$, it holds that
$$
|Z(s,a)| \ge a^{-\sigma} - (1-a)^{-\sigma} - |\zeta (s,1+a)| - |\zeta (s,2-a)|.
$$
This inequality and the Hermite formula imply
\begin{equation}\label{asy:Z}
|Z(s,a)| \to \infty, \qquad a \to + 0.
\end{equation}
Hence for any $1 \ne s \in {\mathbb{C}}$ with $\sigma >0$, there is $0 < a < 1/2$ such that $Z(s,a) \ne 0$.
Next let $\sigma > 1$. Then we have
\begin{equation*}
\begin{split}
&\int_0^1 Z(1-s,a)^2 da = \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2 \int_0^1 P(s,a)^2 da \\
= & \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2
\int_0^1 \sum_{m,n=1}^\infty \frac{\cos 2 \pi ma \cos 2 \pi na}{m^s n^s} da =
 \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2 \frac{\zeta (2s)}{2}
\end{split}
\end{equation*}
from the functional equation of $Z(1-s,a)$ in Lemma \ref{lem:fe1} and the formula
$$
2 \cos \alpha \cos \beta = \cos (\alpha + \beta) + \cos (\alpha - \beta), \qquad \alpha, \beta \in {\mathbb{R}}.
$$
According to (\ref{eq:1+a}) and the Hermite formula, the integral $\int_0^1 Z(1-s,a)^2 da$ converges when $\sigma > 1/2$
since one has
$$
\int_0^1 Z(1-s,a)^2 da \ll_s \int_0^1 a^{2s-2} da = O_s (1).
$$
It is well-known that $\zeta (2s)$ does not vanish when $\sigma > 1/2$ by the Euler product of the Riemann zeta function. Therefore, for any $s \in {\mathbb{C}}$ with $\sigma < 1/2$ and $-s \not \in 2 {\mathbb{N}} \cup \{ 0 \}$, there exists $0 < a < 1/2$ such that $Z(s,a) \ne 0$ from
\begin{equation*}
\begin{split}
0 \ne & \int_0^1 Z(1-s,a)^2 da = \biggl( \int_0^{1/2} + \int_{1/2}^1 \biggr) Z(1-s,a)^2 da \\ = &
\int_0^{1/2} Z(1-s,a)^2 da + \int_0^{1/2}  Z(1-s,1-a)^2 da = 2 \int_0^{1/2} Z(1-s,a)^2 da
\end{split}
\end{equation*}
which is shown by $Z(s,a) = Z (s,1-a) = \zeta (s,a) + \zeta (s,1-a)$. Thus, we have $Z(s,a) \equiv 0$ for all $0 < a < 1/2$ if and only if $s$ is a non-positive even integer.

When $\sigma >1$, we have
$$
2 \int_0^{1/2} P(s,a)^2 da = \int_0^1 P(s,a)^2 da = \int_0^1 \sum_{m,n=1}^\infty \frac{\cos 2 \pi ma \cos 2 \pi na}{m^s n^s} da
= \frac{\zeta (2s)}{2}
$$
from $P(s,a) = P (s,1-a)$. Hence, for any $s \in {\mathbb{C}}$ with $\sigma > 1$, there is $0 < a < 1/2$ such that $P(s,a) \ne 0$. By using the functional equation of $P(1-s,a)$ in Lemma \ref{lem:fe1} and fact proved above that for any $0, 1 \ne s \in {\mathbb{C}}$ with $\sigma > -1$, there is $0< a < 1/2$ such that $Z(s,a)$ does not vanish, we can see that for any $s \in {\mathbb{C}}$ with $\sigma < 2$ and $-s \not \in \{-1,0\} \cup 2{\mathbb{N}}$, there exists $0 < a < 1/2$ such that $P(s,a) \ne 0$. From \cite[(4.12)]{NPCZ}, one has the following equations
$$
P(1,a) = -2 \log (2\sin \pi a ), \qquad P(0,a) =-1.
$$
Thus, we have $P(s,a) \equiv 0$ for all $0 < a < 1/2$ if and only if $s$ is a negative even integer.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:d1} for $Y(s,a)$ and $O(s,a)$]
When $\sigma >0$, we can show $|Y(s,a)| \to \infty$ as $a \to +0$ by modifying the proof of (\ref{asy:Z}). For $\sigma >1$, we have
\begin{equation*}
\begin{split}
&\int_0^1 Y(1-s,a)^2 da = \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2 \int_0^1 O(s,a)^2 da \\
= & \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2
\int_0^1 \sum_{m,n=1}^\infty \frac{\sin 2 \pi ma \sin 2 \pi na}{m^s n^s} da =
 \biggl( \frac{2\Gamma (s)}{(2\pi )^s} \sin \Bigl( \frac{\pi s}{2} \Bigr) \biggr)^2 \frac{\zeta (2s)}{2}
\end{split}
\end{equation*}
by the functional equation of $Y(1-s,a)$ in Lemma \ref{lem:fe1} and the equation $2 \sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta)$. It should be mentioned that the integral $\int_0^1 Y(1-s,a)^2 da$ converges when $\sigma > 1/2$ by the Hermite formula and
$$
\int_0^1 Y(1-s,a)^2 da \ll_s \int_0^1 a^{2s-2} da = O_s (1).
$$
Furthermore, it holds that
$$
0 \ne \int_0^1 Y(1-s,a)^2 da = 2 \int_0^{1/2} Y(1-s,a)^2 da
$$
by $Y(s,a) = Y (s,1-a)$. Hence we can prove that for any $-s \not \in 2{\mathbb{N}}-1$, there is $0<a<1/2$ such that $Y(s,a) \ne 0$.

When $\sigma >1$, it holds that
$$
2 \int_0^{1/2} O(s,a)^2 da = \int_0^1 O(s,a)^2 da = \int_0^1 \sum_{m,n=1}^\infty \frac{\sin 2 \pi ma \sin 2 \pi na}{m^s n^s} da
= \frac{\zeta (2s)}{2}.
$$
Thus, for any $s \in {\mathbb{C}}$ with $\sigma > 1$, there is $0 < a < 1/2$ such that $O(s,a) \ne 0$. From the functional equation of $O(1-s,a)$ in Lemma \ref{lem:fe1} and the fact proved above that for any $s \in {\mathbb{C}}$ with $\sigma > -1$, there is $0< a < 1/2$ such that $Y(s,a)$ does not vanish, we can see that for any $s \in {\mathbb{C}}$ with $\sigma < 2$ and $-s \not \in 2{\mathbb{N}}-1$, there exists $0 < a < 1/2$ such that $O(s,a) \ne 0$. Therefore, we have $O(s,a) \equiv 0$ for all $0 < a < 1/2$ if and only if $s$ is a negative odd integer.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:d1} for $Q(s,a)$ and $X(s,a)$]
According to the functional equation of $P(s,a)$ in Lemma \ref{lem:fe1}, one has
\begin{equation}\label{eq:Q0ZP1}
\begin{split}
2Q(s,a) &= Z(s,a) + P(s,a) = Z(s,a) + \frac{(2\pi )^s}{2\Gamma (s) \cos (\pi s/2)} Z(1-s,a) \\
&= Z(s,a) +  2 (2\pi )^{s-1} \Gamma (1-s) \sin \Bigl( \frac{\pi s}{2} \Bigr) Z(1-s,a) .
\end{split}
\end{equation}
From (\ref{eq:1+a}) and the Hermite formula, we have
$$
Z(s,a) = a^{-s} + O_s(1), \qquad Z(1-s,a) = a^{s -1} + O_s(1)
$$
when $a \to +0$, $s \ne 1$ and $\sigma > 1/2$. Hence, for any $1 \ne s \in {\mathbb{C}}$ with $\sigma > 1/2$, there exists $0 < a< 1/2$ such that $Q(s,a) \ne 0$ by (\ref{eq:Q0ZP1}). From (\ref{eq:Q0ZP1}) and the equation $Z(1/2- it,a) = Z(\overline{1/2+ it},a) = \overline{Z(1/2+ it,a)}$, one has
\begin{equation*}
\begin{split}
&\quad \,\, 2Q(1/2+it,a)\\
&= Z(1/2+it,a) +  2 (2\pi )^{-1/2-it} \Gamma (1/2-it) \sin \Bigl( \frac{\pi (1+2it)}{4} \Bigr) \overline{Z(1/2+ it,a)} \\
&= a^{-1/2-it} + 2 (2\pi )^{-1/2-it} \Gamma (1/2-it) \sin \Bigl( \frac{\pi (1+2it)}{4} \Bigr) a^{-1/2+it} + O_t(1).
\end{split}
\end{equation*}
Therefore, for any $t \in {\mathbb{R}}$, there exist $0 < a < 1/2$ such that $Q(1/2+it,a) \ne 0$. Hence, for any $1 \ne s \in {\mathbb{C}}$ with $\sigma \ge 1/2$, there exists $0 < a< 1/2$ such that $Q(s,a)$ does not vanish. According to \cite[(2.4)]{NPRCZ}, that we have
$$
Q(0,a) = -1/2 = \zeta (0) \ne 0.
$$
Thus, by using the functional equation of $Q(s,a)$ in Lemma \ref{lem:fe1}, we have that for any $1 \ne s \in {\mathbb{C}}$ with $-\sigma \not \in 2{\mathbb{N}}$, there exists $0 < a< 1/2$ such that $Q(s,a) \ne 0$. We can similarly show that for any $1 \ne s \in {\mathbb{C}}$ with $-\sigma \not \in 2{\mathbb{N}}-1$, there exists $0 < a< 1/2$ such that $X(s,a)$ does not vanish.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:d2}]
Let $1 \ne s \in {\mathbb{C}}$ with $\sigma >0$. Then, there exists $0 < a < 1/2$ such that $\zeta(s,a) \ne 0$ since we have $|\zeta (s,a)| \to \infty$ as $a \to +0$ by modifying the proof of (\ref{asy:Z}). When $\sigma >1$, one has
\begin{equation*}
\begin{split}
\int_0^1 \zeta (1-s,a)^2 da = \frac{2\Gamma (s)^2}{(2\pi)^{2s}} \zeta (2s)
\end{split}
\end{equation*}
according to the functional equation of $\zeta(1-s,a)$ in Lemma \ref{lem:fe1}. The integral converges absolutely when $\sigma > 1/2$ from (\ref{eq:1+a}) and the Hermite formula. Hence for any $s \in {\mathbb{C}}$ with $\sigma < 1/2$, there is $0 < a < 1$ such that $\zeta(s,a) \ne 0$. In addition, we have
$$
\int_0^{1/2} \zeta (1-s,a) da + \int_{1/2}^1 \zeta (1-s,a) da = \int_0^1 \zeta (1-s,a) da =0
$$
by the functional equation of $\zeta(1-s,a)$ and $\int_0^1 {\rm{Li}}_s (e^{2\pi ia}) da =0$ for $\sigma >1$. The integral $\int_0^1 \zeta (1-s,a) da$ converges absolutely when $\sigma > 0$ from (\ref{eq:1+a}) and the Hermite formula. Thus, for any $s \in {\mathbb{C}}$ with $\sigma < 1/2$, there is $0 < a < 1/2$ such that $\zeta(s,a)$ does not vanish.

Suppose $\sigma >1$. Then we have
\begin{equation*}
\begin{split}
&2 \int_0^{1/2} {\rm{Li}}_s (e^{2\pi ia}) {\rm{Li}}_s (e^{2\pi i(1-a)}) da \\
= & \int_0^{1/2} {\rm{Li}}_s (e^{2\pi ia}) {\rm{Li}}_s (e^{2\pi i(1-a)}) da
+ \int_{1/2}^1 {\rm{Li}}_s (e^{2\pi ia}) {\rm{Li}}_s (e^{2\pi i(1-a)}) da \\
=& \int_0^1 {\rm{Li}}_s (e^{2\pi ia}) {\rm{Li}}_s (e^{2\pi i(1-a)}) da = \zeta (2s).
\end{split}
\end{equation*}
Hence, for any $s \in {\mathbb{C}}$ with $\sigma > 1$, there is $0 < a < 1/2$ such that ${\rm{Li}}_s (e^{2\pi ia}) \ne 0$. By using (\ref{eq:1+a}), the Hermit formula and functional equation of ${\rm{Li}}_{1-s} (e^{2\pi ia})$ in Lemma \ref{lem:fe1}, we have
$$
{\rm{Li}}_{1-s} (e^{2\pi ia}) = \frac{\Gamma (s)}{(2\pi)^s} \Bigl( e^{\pi i s/2} a^{-s} + O_s (1) \Bigr), \qquad a \to +0
$$
when $\sigma >0$. Hence, for any $s \in {\mathbb{C}}$ with $\sigma < 1$, there is $0 < a < 1/2$ such that ${\rm{Li}}_s (e^{2\pi ia})$ does not vanish. Furthermore, it holds that
$$
\frac{\partial}{\partial a} {\rm{Li}}_s (e^{2\pi ia}) = 2\pi i {\rm{Li}}_{s-1} (e^{2\pi ia}), \qquad 0 < a <1
$$
which implies
$$
{\rm{Li}}_{1+it} (e^{2\pi ia}) = \frac{1}{2 \pi i} \frac{\partial}{\partial a} {\rm{Li}}_{2+it} (e^{2\pi ia}) , \qquad 0 < a <1.
$$
On the other hand, one has ${\rm{Li}}_{2+it} (1) = \zeta (2+it) \ne 0$ and
$$
{\rm{Li}}_{2+it} (e^{\pi i}) = \bigl( 2^{-1-it} -1 \bigr) \zeta (2+it) \ne {\rm{Li}}_{2+it} (1)
$$
which is proved by
$$
{\rm{Li}}_s (e^{\pi i}) = \sum_{n=1}^\infty \frac{(-1)^n}{n^s} = \frac{-1}{1^s} + \frac{1}{2^s} + \frac{-1}{3^s} + \frac{-1}{4^s} + \cdots
= -\zeta (s) + 2 \cdot 2^{-s} \zeta (s), \qquad \sigma >1 .
$$
Hence, there is $0 < a < 1/2$ such that $(\partial / \partial){\rm{Li}}_{1+it} (e^{2\pi ia}) \ne 0$ by ${\rm{Li}}_{2+it} (1) \ne {\rm{Li}}_{2+it} (e^{\pi i})$. Therefore, for any  $t \in {\mathbb{R}}$, there exists $0 < a < 1/2$ such that ${\rm{Li}}_{1+it} (e^{2\pi ia}) \ne 0$.
\end{proof}

\begin{remark}
The condition $a \in (0,1/2)$ in Theorems \ref{th:d1} and \ref{th:d2} can be replaced by $a \in I$, where $I \subset (0,1/2)$ is an open interval by the identity theorem and fact that the functions $(s-1)\zeta (s,a)$ and ${\rm{Li}}_s (e^{2\pi ia})$ are real analytic with respect to $a \in (0,1/2)$.
\end{remark}

\section{$Z(s,a)$ and stationary self-similar distribution}

We first define one-dimensional stationary self-similar distributions (see \cite[Section 1]{Sinai}). Let $X$ be the space of realizations of a one-dimensional random field $x:= \{ x_l : l \in {\mathbb{Z}} \}$. Note that each random variable $x$ takes on real values, and the space $X$ is a vector space. There is a group $\{ T_l : l \in {\mathbb{Z}} \}$ of translations acting naturally on the space $X$. The symbols ${\mathfrak{M}}$ and ${\mathfrak{M}}^{\rm{st}}$ denote the space of all probability distributions on $X$ and all stationary distributions on $X$ (namely, distributions invariant with respect to the group $\{ T_l^* : l \in {\mathbb{Z}} \}$ of translations, where $\{ T_l^* : l \in {\mathbb{Z}} \}$ is the group adjoint to $\{ T_l : l \in {\mathbb{Z}} \}$ which acts on ${\mathfrak{M}}$), respectively.

For each $1 < \lambda <2$, we introduce the multiplicative semigroup $A_k (\lambda)= A_k$, where $k \in {\mathbb{N}}$, of linear endmomorphisms of $X$ whose action is given by the formula
$$
\tilde{x_l} = (A_k x)_l := \frac{1}{k^{\lambda/2}} \sum_{lk \le r < (l+1)k} x_r, \qquad l \in {\mathbb{Z}} .
$$
Let $\{ A_k : k \in {\mathbb{N}} \}$ denote the adjoint semigroup acting on the space ${\mathfrak{M}}$, namely,
$$
(A_k^* P) (C) = P (A_k^{-1} C), \qquad C \subset X, \quad P \in {\mathfrak{M}}.
$$

\begin{dfn}
A probability distribution $P \in {\mathfrak{M}}$ is called a self-similar distribution (s.d.) if one has
$$
A_k^* P = P \quad \mbox{for all} \quad k \in {\mathbb{N}}.
$$
\end{dfn}
In other words, an s.d.~is a fixed point of the semigroup $\{ A_k^* : k \in {\mathbb{N}} \}$ acting on the space ${\mathfrak{M}}$. On the other hand, It follows from the definition of $A_k$ that $A_k T_{lk} = T_l A_k$. Hence, if  $P \in {\mathfrak{M}}^{\rm{st}}$, then $A_k^* P \in {\mathfrak{M}}^{\rm{st}}$ for any $k \in {\mathbb{N}}$.
\begin{dfn}
An s.d.~distribution $P \in {\mathfrak{M}}$ is called a stationary self-similar distribution (s.s.d.) if $P \in {\mathfrak{M}}^{\rm{st}}$.
\end{dfn}

Now let $P$ be a one-dimensional stationary Gaussian distribution on $X$ with ${\mathbb{E}} x_l =0$, where ${\mathbb{E}} x_l$ is the expected value of $x_l$. Then we have the following.
\begin{thm}[{\cite[Theorem 2.1]{Sinai}}]
The distribution $P$ is an s.s.d, if and only if its spectral density $\rho_\lambda(\alpha)$ has the form
$$
\rho_\lambda(\alpha):= C\bigl| e^{2\pi i \alpha} -1 \bigr|^2 \sum_{n \in {\mathbb{Z}}} \frac{1}{|n+\alpha|^{\lambda+1}},
\qquad -1/2 \le \alpha \le 1/2,
$$
where $C>0$ is a constant.
\end{thm}

By the next proposition, we can easily see that the spectral density $\rho_\lambda(\alpha)$ above is written by $Z(\lambda,|\alpha|)$.
\begin{proposition}\label{pro:Si}
When $\sigma >1$ and $\alpha \ne 0$, one has
$$
\sum_{n \in {\mathbb{Z}}} \frac{1}{|n+\alpha|^s} = Z(s,|\alpha|),
\qquad -1/2 \le \alpha \le 1/2.
$$
\end{proposition}
\begin{proof}
For $0 <  \alpha \le 1/2$, we have
\begin{equation*}
\begin{split}
\sum_{n \in {\mathbb{Z}}} \frac{1}{|n+\alpha|^s} &=
\sum_{n=0}^\infty \frac{1}{|n+\alpha|^s} + \sum_{n=-1}^{-\infty} \frac{1}{|n+\alpha|^s}
= \sum_{n=0}^\infty \frac{1}{(n+\alpha)^s} + \sum_{n=0}^{\infty} \frac{1}{(n+1-\alpha)^s} \\
&= \zeta (s,\alpha) + \zeta (s,1-\alpha) = Z(s,\alpha) = Z(s,|\alpha|).
\end{split}
\end{equation*}
When $-1/2 \le  \alpha < 0$, it holds that
\begin{equation*}
\begin{split}
\sum_{n \in {\mathbb{Z}}} \frac{1}{|n+\alpha|^s} &=
\sum_{n=1}^\infty \frac{1}{|n+\alpha|^s} + \sum_{n=0}^{-\infty} \frac{1}{|n+\alpha|^s}
= \sum_{n=0}^\infty \frac{1}{(n+1-|\alpha|)^s} + \sum_{n=0}^{\infty} \frac{1}{(n+|\alpha|)^s} \\
&= \zeta (s,1-|\alpha|) + \zeta (s,|\alpha|) = Z(s,|\alpha|).
\end{split}
\end{equation*}
The equations above imply Proposition \ref{pro:Si}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{th:apSi1}]
This is easily proved by Theorem E and Proposition \ref{pro:Si}.
\end{proof}

\begin{remark}
Fukasawa and Takabatake \cite[p.~1877]{FT} considered a sequence of $n$-dimensional centered Gaussian random vectors which covariance functions are characterized by the following spectral density:
$$
\rho^2 \delta_n^{2H} \frac{\Gamma(2H+1) \sin (\pi H)}{(2\pi)^{2+2H+2\psi}} \bigl( 2 - 2 \cos (2\pi \alpha) \bigr)^{\psi +1}
\sum_{n \in {\mathbb{Z}}} \frac{1}{|n+\alpha|^{1+2H+2\psi}},
$$
where $\rho, \psi >0$, $0 < H \le 1$, $n$ is the sample size and $\delta_n$ is the length of sampling intervals. Note that the infinite series above coincides with $Z(1+2H+2\psi,|\alpha|)$ by Proposition \ref{pro:Si}. Moreover, the function $Z(1+2H+2\psi,|\alpha|)$ can be expressed as a rational function with rational coefficients of $\exp (2 \pi i |\alpha|)$ from Proposition \ref{Pro:Z1} (see also Section 3.1).
\end{remark}

\subsection*{Acknowledgments}

The author was partially supported by JSPS grant 16K05077. I would like to express my gratitude to Professor Fukasawa for his comments on Sinai's paper and related topics.\\


TITLE: $p_1+p_2+p_3 = 1234$ and $p_1p_2p_3 = 87654321$
QUESTION [2 upvotes]: Is there any $3$ primes such that their sum is $1234$ and product is $87654321$.

I can factorize $87654321$ by wolfram-alpha - $3^2 \times 1997 \times 4877$. So there is no solution.
But I dont know how I can do this in hand. Is there any simpler way to prove this?

REPLY [0 votes]: $87654321$ is divisible by $9$, so two of the three primes would have to be $3$. But then the third one would have to be $1234-3-3=1228$ which is not a prime, thus the problem has no solutions.


text:
Heuristically, the goal of approximating \eqref{eq:SGD} via \eqref{eq:GD} is to reduce the difficulty of the direct analysis of the former by leveraging the strong convergence properties of the latter.
of the latter, which is relatively straightforward to analyze, to the former, which is much more difficult.
However, the notion of an \ac{APT} does not suffice in this regard:
in the supplement, we provide an example where a discrete-time \ac{APT} has a completely different behavior relative to the underlying flow.
As such, a considerable part of our analysis below focuses on tightening the guarantees provided by the \ac{APT} approximation scheme.

\subsection{Boundedness and stability of the approximation}
\label{sec:stability}

The basic point of failure in the \acl{SA} approach is that \acp{APT} may escape to infinity, rendering the whole scheme useless, \cf \citep{Ben99,Bor08} and references therein.
It is for this reason that a large part of the literature on \ac{SGD} explicitly assumes that the trajectories of the process are bounded (precompact), \ie
\begin{equation}
\label{eq:stability}
\txs
\sup_{\ctime\geq\cstart} \norm{\apt{\ctime}}
	< \infty
	\quad
	\as.
\end{equation}
However, this is a prohibitively strong assumption for \eqref{eq:opt}:
unless certified ahead of time, any theoretical result relying on this assumption would be of limited practical value.

Our first result below provides exactly this certification by establishing that \eqref{eq:stability} is solely an implication of our underlying \crefrange{asm:reg}{asm:noise}.
It is a non-trivial outcome which provides the key to unlocking the potential of \acl{SA} techniques in the sequel.

\begin{restatable}{theorem}{stability}
\label{thm:stability}
Suppose that \crefrange{asm:reg}{asm:noise} hold
and
\eqref{eq:SGD} is run with a variable step-size sequence
of the form $\curr[\step]\propto 1/\run^{\pexp}$ for some $\pexp \in (2/(\qexp + 2),1]$.

Then, with probability $1$, every \ac{APT} $\apt{\ctime}$ of \eqref{eq:GD} that is induced by \eqref{eq:SGD} has $\sup_{\ctime\geq\cstart} \norm{\apt{\ctime}} < \infty$.
\end{restatable}

Because of the generality of our assumptions, the proof of \cref{thm:stability} involves a delicate combination of  non-standard techniques;
for completeness, we provide a short sketch below and refer the reader to the supplement for the details.

\begin{proof}[Sketch of proof of \cref{thm:stability}]
The main reasoning evolves along the following lines:

\begin{enumerate}[Step 1.,leftmargin=*]
\item
We first show that, under the stated assumptions, there exists a \textpar{possibly random} subsequence $\state_{\run_{\runalt}}$ of $\curr$ that converges to $\sols$;
formally, $\liminf_{\run\to\infty} \dist(\curr,\sols) = 0$ \as.
As a result, $\apt{\ctime}$ eventually reaches a sublevel set $\flvl_{\eps}$ whose elements are arbitrarily close to $\sols$, \ie there exists some $\ctime_{\eps} > 0$ such that $\apt{\ctime_{\eps}} \in \flvl_{\eps}$.

\item
By a technical argument relying on the regularity assumptions for $\obj$ (\cf \cref{asm:reg}), it can be shown that there exists some \emph{uniform} time window $\tau$ such that $\apt{\ctime}$ remains within uniformly bounded distance to $\flvl_{\eps}$ for all $\ctime \in [\ctime_{\eps},\ctime_{\eps} + \tau]$.
Thus, once $\apt{\ctime}$ gets close to $\flvl_{\eps}$, it will not escape too far within a fixed length of time.

\item
An additional technical argument reveals that, under the stated assumptions for $\obj$, the trajectories of \eqref{eq:GD} either descend the objective by a uniform amount, or they have reached a neighborhood of the critical set $\sols$ where further descent is impossible (or irrelevant).

\item
By combining the two previous steps, we conclude that $\apt{\ctime_{\eps} + \tau} \in \flvl_{\eps}$ at the end of said window.
This argument may then be iterated ad infinitum to show inductively that $\apt{\ctime} \in \flvl_{\eps}$ for all intervals of the form $[\ctime_{\eps} + \runalt\tau, \ctime_{\eps} + (\runalt+1)\tau]$.
\end{enumerate}

Since $\flvl_{\eps}$ is bounded (by \cref{asm:sublvl}), we conclude that $\apt{\ctime}$ remains in a compact set for all $\ctime\geq0$, \ie $\apt{\ctime}$ is precompact.
The conclusion of \Cref{thm:stability} then follows by \cref{cor:APT-step}.
\end{proof}

\subsection{Almost sure convergence}
\label{sec:convergence}

By virtue of \cref{thm:stability}, we are now in a position to state our almost sure convergence result:

\begin{theorem}
\label{thm:convergence}
Suppose that \crefrange{asm:reg}{asm:noise} hold
and
\eqref{eq:SGD} is run with a variable step-size sequence
of the form $\curr[\step] = \Theta(1/\run^{\pexp})$ for some $\pexp \in (2/(\qexp + 2),1]$.

Then, with probability $1$, $\curr$ converges to a \textpar{possibly random} connected component $\sols_{\infty}$ of $\sols$ over which $\obj$ is constant.
\end{theorem}

\begin{corollary}
\label{cor:convergence}
With assumptions as in \cref{thm:convergence}, we have the following:
\begin{enumerate}
\item
$\obj(\curr)$ converges \as to some critical value $\obj_{\infty}$.
\item
Any limit point of $\curr$ is \as a critical point of $\obj$.
\end{enumerate}
\end{corollary}

\Cref{thm:convergence} extends a range of existing treatments of \eqref{eq:SGD} under explicit boundedness assumptions of the form \eqref{eq:stability}, \cf \citep{Lju77,Ben99,Bor08} and references therein.
It also improves on a similar result by \citet{BT00} who use a completely different analysis to dispense of boundedness requirements via the use of more restrictive step-size policies.
Specifically, \citet{BT00} require the \acl{RM} summability conditions $\sum_{\run} \curr[\step] = \infty$ and $\sum_{\run} \curr[\step]^{2} < \infty$ under a bounded variance assumption.
In this regard, our analysis extends to more general step-size policies, while that of \citet{BT00} cannot because of its reliance on the Robbins-Siegmund theorem for almost-supermartingales \citep{RS71}.
Among other benefits, this added degree of flexibility is a key advantage of the \ac{APT} approach.

The heavy lifting in the proof of \cref{thm:convergence} is provided by \cref{thm:stability}.
Thanks to this boundedness certificate,

the total chain of implications is relatively short, so we provide it in full below.

\begin{proof}[Proof of \cref{thm:convergence}]
Under the stated assumptions, $\obj$ is a strict Lyapunov function for \eqref{eq:GD} in the sense of \citet[Chap.~6.2]{Ben99}. Specifically, this means that $\obj(\flow{\ctime}{\point})$ is strictly decreasing in $\ctime$ unless $\point$ is a stationary point of \eqref{eq:GD}.
Furthermore, by Sard's theorem \citep[Chap.~2]{Mil65}, the set $\obj(\sols)$ of critical values of $\obj$ has Lebesgue measure zero \textendash\ and hence, empty topological interior.
Therefore, applying Theorem 5.7 and Proposition 6.4 of \citet{Ben99} in tandem, we conclude that any precompact \acl{APT} of \eqref{eq:GD} converges to a connected component $\sols_{\infty}$ of $\sols$ over which $\obj$ is constant.
Since \cref{thm:stability} guarantees that the \acp{APT} of \eqref{eq:GD} induced by \eqref{eq:SGD} are bounded with probability $1$, our claim follows.
\end{proof}

\subsection{Avoidance analysis}
\label{sec:avoidance}

\Cref{thm:convergence} represents a strong convergence guarantee but, at the same time, it does not characterize the component of $\sols$ to which $\curr$ converges.
The rest of this section is devoted to showing that $\curr$ does not converge to a component of $\sols$ that only consists of saddle points (a \emph{saddle-point manifold}).

Specifically, we will make precise the following informal statement:
\begin{center}
\emph{\eqref{eq:SGD} avoids strict saddles \textendash\ and sets thereof \textendash\ with probability $1$}.
\end{center}
To set the stage for the analysis to come, we begin by reviewing some classical and recent results on the avoidance of saddle points.
We then present our general results towards the end of the section.

To begin, a crucial role will be played in the sequel by the \emph{Hessian matrix} of $\obj$, viz.
\begin{equation}
\label{eq:Hessian}
\hmat(\point)
	\equiv \nabla^{2}\obj(\point)
	\equiv (\pd_{i} \pd_{j} \obj(\point))_{i,j=1,\dotsc,\vdim}.
\end{equation}
Since $\hmat(\point)$ is symmetric, all of its eigenvalues are real.
If $\saddle$ is a critical point of $\obj$ and

$\lambda_{\min}(\hmat(\saddle)) < 0$, we say that $\saddle$ is a \emph{strict saddle point} \citep{LSJR16,LPPS+19}.

By standard results in center manifold theory \citep{Tes12}, the space around strict saddle points admits a decomposition into a \emph{stable}, \emph{center} and \emph{unstable} manifold (each of the former two possibly of dimension zero; the latter of dimension at least $1$ given that $\lambda_{\min}(\hmat(\saddle)) < 0$).
Heuristically, under the continuous-time dynamics \eqref{eq:GD}, directions along the stable manifold of $\saddle$ are attracted to $\saddle$ at a linear rate, while those along the unstable manifold are repelled (again at a linear rate);
the dynamics along the center manifold could be considerably more complicated, but, in the presence of unstable directions, they only emerge from a measure zero of initial conditions.

As a result, if $\saddle$ is a strict saddle point of $\obj$, it stands to reason that \eqref{eq:SGD} should ``probably'' avoid it as well.

In the case of \emph{deterministic} gradient descent with step-size $\step < 1/\smooth$, this intuition was made precise by \citet{LSJR16,LPPS+19} who proved that all but a measure zero of initializations of gradient descent avoid strict saddles.
As we discussed in the introduction, this result was then extended to various deterministic settings, with different assumptions for the gradient oracle, the method's step-size, or the structure of the saddle-point manifold, see \eg \cite{JGNK+17,LPPS+19,DJLJ+17,FVP19b,PP17,PPW19} and the references therein.

In the stochastic regime, the situation is considerably more involved.
\citet{Pem90} and \citet{BD96} were the first to establish the avoidance of hyperbolic unstable equilibria in general \acl{SA} schemes.
However, a key requirement in the analysis of these works is that of \emph{hyperbolicity}, which in our setting amounts to asking that $\hmat(\saddle)$ is \emph{invertible}.
In particular, this means the saddle point in question cannot be isolated, nor can it have a center manifold:
both hypotheses are too stringent for applications of \ac{SGD} to contemporary machine learning models, such as deep net training, so their results do not apply in many cases of practical interest.

More relevant for our purposes is the recent result of \citet{GHJY15}, who provided the following guarantee.
Suppose that $\obj$ is \emph{$(\alpha,\beta,\eps,\delta)$-strict saddle}, \ie for all $\point\in\points$, one of the following holds:
\begin{enumerate*}
[(\itshape i\hspace*{.5pt}\upshape)]
\item
$\norm{\nabla\obj(\point)} \geq \eps$;
\item
$\lambda_{\min}(\hmat(\point)) \leq - \beta$;
or
\item
$\point$ is $\delta$-close to a local minimum $\point_{c}$ around which $\obj$ is $\alpha$-strongly convex.
\end{enumerate*}
Suppose further that $\obj$ is bounded, $\smooth$-Lipschitz smooth, and $\hmat(\point)$ is $\rho$-Lipschitz continuous;
finally, assume that the noise in the gradient oracle \eqref{eq:SFO} is finite \as and contains a component uniformly sampled from the unit sphere.
Then, given a confidence level $\zeta>0$, and assuming that \eqref{eq:SGD} is run with \emph{constant} step-size $\step = \bigoh(1/\log(1/\zeta))$, the algorithm produces after a given number of iterations a point which is $\bigoh(\sqrt{\gamma \log(1/(\gamma\zeta))})$-close to $\point_{c}$, and hence away from any strict saddle of $\obj$, with probability at least $1-\zeta$.

In a more recent paper, \citet{VS19} examined the convergence of \eqref{eq:SGD} to second-order stationary points.
More precisely, they showed that \eqref{eq:SGD} guarantees \emph{expected} descent for strict saddle points in a finite number of iterations, and with high probability, \eqref{eq:SGD} iterates reach a set of \emph{approximate} second-order stationary points in finite time.

The theory of \citet{Pem90} and the result of \citet{GHJY15} paint a complementary picture to the above:
\citet{Pem90} shows that saddle points are avoided with probability $1$, provided they are hyperbolic (\ie $\det \nabla^{2}\obj(\saddle) \neq 0$);
on the other hand, \citet{GHJY15} require much less structure on the saddle point, but they only provide a result with high probability (and $\zeta$ cannot be taken to zero because the range of allowable step-sizes would also vanish).
\footnote{\citet{Pem90} employs a vanishing step-size, which is more relevant for us:
\eqref{eq:SGD} with persistent noise and a constant step-size is an irreducible ergodic Markov chain whose trajectories do not converge \emph{anywhere} \citep{Ben99}.}
Our objective in the sequel is to provide a result that combines the ``best of both worlds'', \ie almost sure avoidance of strict saddle points (and sets thereof) with probability $1$.

To that end, we make the following assumption for the noise:
\begin{assumption}
\label{asm:exciting}
The error term $\noise \equiv \noise(\point;\sample)$ of \eqref{eq:SFO} is \emph{uniformly exciting}, \ie there exists some $\const>0$ such that
\begin{equation}
\label{eq:exciting}
\exof{\braket{\noise(\point;\sample)}{\unitvec}^{+}}
	\geq \const
\end{equation}
for all $\point\in\points$ and all unit vectors $\unitvec \in \sphere^{\vdim-1}$.
\end{assumption}

This assumption simply means that the average projection of the noise along every ray in $\points$ is uniformly positive;
in other words, $\noise$ ``excites'' all directions uniformly \textendash\ though not necessarily \emph{isotropically}.
As such, \cref{asm:exciting} is automatically satisfied by noisy \acl{GD} (\eg as in \citet{GHJY15}), generic finite sum objectives with at least $\vdim$ summands, etc.

\begin{figure}[tbp]
\centering
\includegraphics[width=.5\textwidth]{Figures/Ridge.pdf}
\caption{A strict saddle manifold (a ridge), typical of ResNet loss landscapes \citep{LXTS+18}.}
\label{fig:saddle}
\end{figure}

With all this in hand, we say that $\saddles$ is a \emph{strict saddle manifold} of $\obj$ if it is a smooth connected component of $\sols$ such that:
\begin{enumerate}
\item
Every $\saddle\in\saddles$ is a strict saddle point of $\obj$ \textpar{\ie $\lambda_{\min}(\hmat(\saddle)) < 0$}.
\item
There exist $\const_{-},\const_{+}>0$ such that, for all $\saddle\in\saddles$, all negative eigenvalues of $\hmat(\saddle)$ are bounded from above by $-\const_{-} < 0$, and any positive eigenvalues (\emph{if} they exist) are bounded from below by $\const_{+}$.
\end{enumerate}

Somewhat informally, the definition of a strict saddle manifold implies that the eigenspaces of $\hmat(\saddle)$ corresponding to zero, positive, and negative eigenvalues decompose smoothly along $\saddles$

and $\saddles$ can be seen as an ``integral manifold'' of the nullspace of the Hessian of $\obj$.

With all this in hand, we are finally in a position to state our main avoidance result.

\begin{restatable}{theorem}{avoidance}
\label{thm:avoidance}
Suppose that \eqref{eq:SGD} is run with a variable step-size sequence of the form $\curr[\step]\propto 1/\run^{\pexp}$ for some $\pexp \in (0,1]$.
If
\crefrange{asm:reg}{asm:exciting} hold \textpar{with $\qexp=\infty$ for \cref{asm:noise}},
and
$\saddles$ is a strict saddle manifold of $\obj$,
we have

\(
\probof{\curr \to \saddles \; \text{as $\run\to\infty$}}
	= 0.
\)

\end{restatable}
\Cref{thm:avoidance} is the formal version of the avoidance principle that we stated in the beginning of this section.
Importantly, it makes \emph{no} assumptions regarding the initialization of \eqref{eq:SGD} and holds for \emph{any} initial condition.

The proof of \cref{thm:avoidance} relies on two basic components.

The first is a probabilistic estimate, originally due to \citet{Pem90}, that shows that a certain class of stochastic processes avoid zero with probability $1$.

The second is a differential-geometric argument, building on \citet{BH95} and \citet{Ben99}, and relying on center manifold theory to isolate the center/stable and unstable manifolds of $\saddles$.

Combining these two components, it is possible to show that even ambulatory random walks along the stable manifold of $\saddles$ will eventually be expelled from a neighborhood of $\saddles$.
We provide the details of this argument in the paper's supplement.

\subsection{Rate of convergence}
\label{sec:rate}

We conclude our analysis of \eqref{eq:SGD} by establishing the algorithm's rate of convergence, as stated in \cref{thm:rate} below.
Since $\obj$ is non-convex, any convergence rate analysis of this type must be a fortiori local;
in view of this, we will examine the algorithm's convergence to local minimizers $\sol\in\sols$ that are regular in the sense of Hurwicz, \ie $\hmat(\sol) \mg 0$.

Because we are primarily interested in the convergence of the algorithm's trajectories, we focus here on the distance $\curr[\breg] = \norm{\curr - \sol}^{2}/2$ between the iterates of \eqref{eq:SGD} and a local minimizer of $\obj$.
In this light, our rate guarantee (which we state below), differs substantially from other results in the literature, in both scope and type, as it does not concern the ergodic average $\bar\state_{\run} = \run^{-1} \sum_{\runalt=\start}^{\run} \iter$ or the ``best iterate'' $\curr^{\text{best}} = \argmin_{\runalt=\start,\dotsc,\run} \norm{\nabla\obj(\iter)}$ of \eqref{eq:SGD}:
the former has very weak convergence in convex settings (if at all), while the latter cannot be calculated with access to perfect gradient information for the entire run of the process (in which case, stochastic \acl{GD} would become \emph{ordinary} \acl{GD}).

\begin{restatable}{theorem}{rate}
\label{thm:rate}
Fix some tolerance level $\delta>0$,
let $\sol$ be a regular minimizer of $\obj$,
and
suppose that \cref{asm:noise} holds.
Assume further that \eqref{eq:SGD} is run with a step-size schedule of the form $\curr[\step] = \step/(\run + \offset)^{\pexp}$ for some $\pexp \in (2/(\qexp + 2),1]$ and large enough $\offset,\step > 0$.
Then:
\begin{enumerate}
\setlength{\itemsep}{\smallskipamount}
\item
There exist neighborhoods $\nhd$ and $\nhd_{1}$ of $\sol$ such that, if $\init\in\nhd_{1}$, the event
\begin{equation}
\label{eq:stay}
\event_{\nhd}
	= \{\curr\in\nhd\;\text{for all $\run=\running$}\}
\end{equation}
occurs with probability at least $1-\delta$.
\item
Conditioned on $\event_{\nhd}$, we have
\begin{equation}
\label{eq:rate}
\exof{\norm{\curr - \sol}^{2} \given \event_{\nhd}}
	 = \bigoh(1/\run^{\pexp}).
\end{equation}
\end{enumerate}
\end{restatable}

\begin{remark*}
Note that \cref{thm:rate} does not presuppose \cref{asm:reg,asm:sublvl,asm:gradlvl};
since the rate analysis is local, the differentiability of $\obj$ suffices.
\end{remark*}

The proof of \cref{thm:rate} relies on showing that

\begin{enumerate*}
[\itshape a\upshape)]
\item
$\sol$ is \emph{stochastically stable}, \ie with high probability, any initialization that is close enough to $\sol$ remains close enough;
and
\item
conditioned on this event, the distance $\curr[\breg] = (1/2) \norm{\curr - \sol}^{2}$ to a regular local minimizers behaves as an ``almost'' supermartingale.
\end{enumerate*}
A major complication that arises here is that this conditioning changes the statistics of the noise, so the martingale property ceases to hold.
Overcoming this difficulty requires an intricate probablistic argument that we present in the supplement (where we also provide explicit expressions of the constants in the estimate of \cref{thm:rate}).

config: arxiv


TITLE: Solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$
QUESTION [1 upvotes]: How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,
where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.
I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular case.
As a side comment: the values of $N$ I am working with are $N\approx 10^6$.

REPLY [3 votes]: Let
$f(x)
= e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}-1
=\sum_{i=1}^N e^{k_i/x}-1
$,
and let
$K = \sum_{i=1}^N k_i
$.
The restrictions that
$x > 0$
and
$k_i < 0$
are important in what follows.
$f'(x)
=\sum -\frac{k_i}{x^2}e^{k_i/x}
=-\frac1{x^2}\sum k_ie^{k_i/x}
=\frac1{x^2}\sum |k_i|e^{k_i/x}
$,
so
$f'(x) > 0$.
This means that
your function has
at most one root.
Since $f(x) < 0$
for small $x$
and
$f(x) > 0$
for large $x$,
$f$ has exactly one positive root.
To get simple bounds
on the root of
$f(x) = 0$,
let
$k_{min} = \min(k_i)$
and
$k_{max} = \max(k_i)$.
Note that,
since the $k_i < 0$,
$|k_{min}|
>|k_{max}|
$.
Then
$e^{k_{min}/x}
\le e^{k_i/x}
\le e^{k_{max}/x}
$
so
$e^{k_{min}/x}
\le \dfrac{\sum_{i=1}^N e^{k_i/x}}{N}
\le e^{k_{max}/x}
$.
If
$f(x) = 0
$, then
$e^{k_{min}/x}
\le \dfrac{1}{N}
\le e^{k_{max}/x}
$
or
$k_{min}/x
\le -\ln(N)
\le k_{max}/x
$
or,
since $k_i < 0$,
$|k_{min}|/x
\ge \ln(N)
\ge |k_{max}|/x
$
or
$\frac{|k_{max}|}{\ln(N)}
\le x
\le \frac{|k_{min}|}{\ln(N)}
$.
Another bound can be gotten using
$e^z \ge 1+z$.
$f(x)
\ge \sum (1+k_i/x) - 1
= N+\frac1{x}\sum k_i - 1
= N+\frac{K}{x} - 1
$.
If
$f(x) = 0$,
then
$1-N
\ge \frac{K}{x}
= -\frac{|K|}{x}
$
or
$N-1
\le \frac{|K|}{x}
$
or
$x \le \frac{|K|}{N-1}
$.
This can be improved
by using the inequality
between the arithmetic and geometric means.
$\begin{array}\\
\dfrac{\sum_{i=1}^N e^{k_i/x}}{N}
&\ge \left(\prod_{i=1}^N e^{k_i/x}\right)^{1/N}\\
&= e^{\sum_{i=1}^N k_i/(Nx)}\\
&= e^{K/(Nx)}\\
\end{array}
$
Therefore,
if $f(x) = 0$,
$\dfrac1{N}
\ge e^{K/(Nx)}
$
or
$-\ln(N)
\ge K/(Nx)
$
or
$\ln(N) \le |K|/(Nx)
$
or
$x \le \dfrac{|K|}{N\ln(N)}
$.
So an initial $x$
could be
$x_0
=\frac{|K|}{N\ln(N)}
$.
We can then apply
Newton's iteration,
$x_{n+1}
=x_n - \frac{f(x_n)}{f'(x_n)}
$
a few times
and see what happens.
It should be OK.


TITLE: Modular form on $\Gamma_0(N)$
QUESTION [7 upvotes]: I recently asked this question on Math.StackExchange with no answer so far. So I thought maybe I can find an answer here.
Let $M(k,\Gamma_0(N))$ be a space of modular forms of weight $k$ on $\Gamma_0(N)$.
Each $f \in M(k,\Gamma_0(N))$ has a Fourier expansion of the form
$$ f(\tau)=\sum_{n\in \mathbb{N}}a(n)\,q^n \quad \text{where}\quad q=\mathrm{e}^{2\pi \mathrm{i}\tau} . $$
Now, let $g(\tau)$ be a function obtained from $f(\tau)$ by omitting all $a(n)$ such that $\gcd(n,N)\neq 1$, i.e.
$$ g(\tau)=\sum_{\substack{n\in \mathbb{N}\\(n,N)=1}}a(n)\,q^n . $$
Question: Is $g(\tau)$ a modular form? What is its level?
Many thanks.

REPLY [5 votes]: $$ g(\tau)=f(\tau)\otimes \left(\tfrac{N^2}{\cdot}\right)=\sum_{n\in \mathbb{N}}\left(\tfrac{N^2}{n}\right)a(n)\,q^n \quad \text{where}\left(\tfrac{N^2}{\cdot}\right) \text{ is the Kronecker symbol}.$$
This means that $g(\tau)$ is just the twist of $f(\tau)$ by a principle character. Indeed we have $$g(\tau)\in M(k,\Gamma_0(N^3)).$$


TITLE: Show that $X = L^p \cap L^q$ with norm $||f|| = ||f||_p + ||f||_q$ is a Banach space.
QUESTION [0 upvotes]: I have difficulties understanding the solution of the following problem: ($L^p$ denotes $L^p(\mathbb{R})$)
Let $1 \le p \lt q\lt \infty$.
Show that $X = L^p \cap L^q$ with norm $||f|| = ||f||_p + ||f||_q$ is a Banach space.
That it is a normed space is clear to already. It is the completeness part that I'm not sure about.
Let $f_k$ be Cauchy in $X$. That implies that $f_k$ is Cauchy in both $L^p$ and $L^q$. Since they are complete, $\exists h \in L^p$ and $g\in L^q$ such that $||h - f_k||_p \to 0$ and $||g - f_k||_q \to 0$.
The goal is to show that $h = g$, hence $X$ is Banach.
Since convergence in $L^p$ implies convergence in measure, and convergence in measure implies that the existence of a subsequence that converges a.e., we get that $\exists f_{n_k}$ and $f_{l_k}$ such that
$$f_{n_k} \to g \ \text{  a.e.}$$
and
$$f_{l_k} \to h \ \text{ a.e.}$$
Now comes the part I'm not sure about. In the solution I have, they claim that:
1: $g=h$ a.e.  (why?)
2: Hence, $g$ and $h$ belong to the same equivalence class. (why?)

REPLY [1 votes]: Following your notation, $h$ denotes the limit in the $L^p$ norm and $g$ denotes the one in the $L^q$ norm. You have established that there is a subsequence $(f_{n_k})_k$ such that $f_{n_k}$ converges a.e. to $g$. To conclude you just need to use Fatou's lemma:
$$\int_{\mathbb{R}}|g-h|^p d\mu\leq\liminf_{k\to+\infty}\int_{\mathbb{R}}|h-f_{n_k}|^pd\mu = 0,$$
which means $g=h$ a.e.


TITLE: Elements in the group von Neumann algebra which are not summable
QUESTION [3 upvotes]: Let $G$ be a discrete group. I am wondering if there is a recipe which can be applied to find elements in the group von Neumann algebra which are not absolutely summable, i.e. $T \in VN(G)$ while $T \notin \ell^1(G)$?
PS Once I heard that for $G=\Bbb{Z}$, $(a_n)_{n \in \Bbb{Z}}$, where $a_n=1/n$ for $n \neq 0$ and $a_0=0$, is such an element. Is this a particular example of a general result on abelian discrete groups? Or it is proved only for $\Bbb{Z}$?

REPLY [6 votes]: The example in your PS is probably related to the Fourier series of the "sawtooth function" $f : [-\pi, \pi) \to {\bf R}$, $f(t)=t/\pi$, when we regard $f$ as an element of $L^\infty({\bf T})\cong {\rm VN}({\bf Z})$. I don't recall the precise formula for the Fourier series, but it certainly gives something which does not belong to $\ell^1({\bf Z})$.
More generally: an integrable function $h$ on ${\bf T}$ whose Fourier series belongs to $\ell^1({\bf Z})$ must be equal a.e. to a continuous function, and so any element of $L^\infty({\bf T})\setminus C({\bf T})$ would also suffice. It is worth noting that there exist functions in the disc algebra (i.e. continuous on the closed unit disc and analytic in the interior) whose Taylor series do not converge absolutely on the unit circle, and these give elements of $C({\bf T})\setminus {\ell^1}({\bf Z})$ (where by abuse of notation I am identifying $\ell^1({\bf Z})$ with a space of functions on the circle).
Anyway, regarding your original question: if $G$ contains an element of infinite order then this gives an inclusion of groups ${\bf Z} \to G$ and hence (since we are dealing with discrete groups) an inclusion of von Neumann algebras ${\rm VN}({\bf Z}) \to {\rm VN}(G)$, so you can embed the counterexample for ${\rm VN}({\bf Z})$ to get a counterexample for ${\rm VN}(G)$.


TITLE: References for Experimental Validation Of Lorentz Force
QUESTION [3 upvotes]: I am writing the paper that involves the history of Lorentz force and I would very much like to add some specific references that involve experimental validation of Lorentz force. The only one I was able to find was J.J. Thomson 1897 paper "Cathode Rays" published in "Philosophical Magazine". However, in that paper the Lorentz force is not mentioned although it is evident that the deflection of the electron beam must be due to the Lorentz force. Can someone please point me to historic papers that describe the experiments that validate the Lorentz force?

REPLY [1 votes]: So I am personally interested in electromagnetic forces applied to condensed matter, and while these systems perhaps do not exhibit the simplest manifestations of Lorentz forces, I can recommend two historical papers:
Edwin Hall (1879): "On a New Action of the Magnet on Electric Currents"
This paper described the discovery of the Hall Effect, which of course is a manifestation of the Lorentz force. Its a great read just to see how they wrote scientific papers back then (rather informally), but it's also historically interesting to see the motivation for his experiment: It was controversial whether the "magnetic force" acts upon the electric current or conductor which carries it. Maxwell thought it was on the conductor, and Hall seemed to show it was on the current.
Nichols and Hull (1901): "A Preliminary Communication on the Pressure of Heat and Light Radiation"
This is one of the earliest papers on quantitative measurement of radiation pressure, which of course can be understood in terms of Lorentz forces (the magnetic force term is what applies the radiation pressure for normally-incident light on matter). See also Rothman and Boughn, American Journal of Physics 77, 122 (2009) (a more pedagogical take) for more on this.


TITLE: Integral $\int \sin(x^3) {\rm d}x$
QUESTION [1 upvotes]: I'm new here :)
I'd like to know why do apps like maple and wolframalpha freak out at this integral please(see pic). It's supposed to be a basic one, is it not?

Thanks,
Andre

REPLY [1 votes]: Another line of input to add to the answers here: There was a good post on MathOverflow called "Why is differentiating mechanics and integrating art?":
https://mathoverflow.net/questions/66377/why-is-differentiating-mechanics-and-integration-art
This is a multi-faceted question with a number of different parts. The most upvoted answer basically says that differentiation has enough rules that let you break down pretty much any expression, but integration does not. To me though that's not entirely satisfactory since it doesn't really detail what the "deep" reason is why integration doesn't have those rules but differentiation does. Other answers provide other pieces of the answer: differentiation is a local operation, but integration is global. Integration involves taking into account the behavior at a distant point from the point the integral is evaluated at, but differentiation is only about what is around that particular point. Another one, and perhaps the easiest, is just that some sets are closed under one operation but not its inverse: consider the natural numbers: the number 2 is easily squareable, but its square root is an irrational number. The elementary functions are the same way viz. differentiation and integration.
Regarding your specific integral, you should think about what it would look like if it "could" be integrated. In particular, were the integrand $3x^2 \sin(x^3)$, then integration is simple, by substitution. You can use the "reverse chain rule" of integration by substitution to get back to an antiderivative. But if I leave off the $3x^2$, that is no longer possible: we just have $\sin(x^3)$, and we cannot:
a) use linearity, as it's not a sum (involutary inverse of differentiation linear laws)
b) use integration by parts, as it's not a product (the only possible choice is to take $\sin(x^3)$ as $u$, then $dv$ must be $dx$, but that doesn't help the situation any if you try it) (inverse of differentiation product rule)
c) use partial fractions, as it's not a rational function,
d) use trigonometric integrals as it has a power inside the trig function.
In other words, none of the elementary integrating methods, which are essentially reverses of differentiation rules, will work. (FWIW trig integrals are essentially just further application of parts and substitution so are not really anything "new".)
To understand Wolfram's answer you need two things. One is the function it used: $\Gamma(a, x)$ which is called the "incomplete gamma function". The gamma function is a generalization of the factorial $n!$ to real numbers $x!$. Bernoulli first posed that problem, then Euler gave the solution, back in the 1700s. The gamma function is given by
$$\Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1} dt$$
then $x! = \Gamma(x+1)$, while the incomplete gamma function is a further generalization, derived later by someone else whose name this author doesn't know (if it is known at all), given by
$$\Gamma(s, x) = \int_{x}^{\infty} e^{-t} t^{s-1} dt$$
This incomplete gamma function provides, essentially, a new differentiation rule:
$$\frac{d}{dx} \Gamma(s, x) = e^{-x} x^{s-1}$$
which turns out to be sufficient to solve the problem. The other component you will need is something known as Euler's formula, which is a key result of complex variables. Yes, to deal with this problem you have to go outside the real numbers into the complex numbers. Euler's formula is
$$e^{ix} = \cos(x) + i \sin(x)$$
and also note that from this, we have
$$e^{-ix} = \cos(x) - i \sin(x)$$
Now you can derive some formulas for $\cos$ and $\sin$ in terms of $e^{ix}$ and $e^{-ix}$. That is, we can convert sines and cosines to exponentials, which are what appear in the incomplete gamma function. I will leave it for the thread starter, to put these pieces together and find Wolfram's solution by hand.


TITLE: understanding a proof on ricci soliton
QUESTION [2 upvotes]: Recently I am reading the book The Ricci Flow: techniques and applications: Part I: geometric aspects by Bennett Chow et al.. Here I have encountered a result (Proposition 1.13) due to Hamilton which states that "Any expanding or steady Ricci soliton $(g,X)$ on a closed manifold $M^n$ is Einstein. Any shrinking Ricci solution on a closed $n$-manifold has positive scalar curvature."
In the proof, the general Ricci soliton equation is given by $$\Delta R + 2|\mathrm{Rc}|^2=\mathcal{L}_X(R)-\epsilon R,$$ which can be written as $$\Delta R + 2|\mathrm{Rc}|^2-\langle\nabla R,X\rangle+\epsilon R=0.$$ The proof says that the last equation can be rewritten as $$\Delta\left(R+\frac {n\epsilon} {2}\right)+2\,\left|\mathrm{Rc}+\frac {\epsilon} {2}g\right|^2-\left\langle\nabla\left(R+\frac {n\epsilon} {2}\right),X\right\rangle-\epsilon\left(R+\frac {n\epsilon} {2}\right)=0.$$ I can not understand the last equation. Please some one help me to understand this.

REPLY [1 votes]: It is a direct simple computation We use the followings :
$$\Delta C=\nabla C=0$$ for constant $C$
$$|Rc +Cg|^2=|Rc|^2+ 2C(Rc,g)+C^2 |g|^2=|Rc|^2+2CR +nC^2$$


TITLE: From the viewpoint of modern geometry, is there a "best" definition of the term "triangle"?
QUESTION [6 upvotes]: I can think of at least six different possible definitions of the term "triangle" in Euclidean geometry.

a subset of $\mathbb{R}^n$ that can be expressed as the convex hull of three or fewer points.
a subset of $\mathbb{R}^n$ that can be expressed as the convex hull of three points, but no fewer.
A function $\{0,1,2\} \rightarrow \mathbb{R}^n$.
A function $\{0,1,2\} \rightarrow \mathbb{R}^n$ whose multiset image is an affinely independent subset of $\mathbb{R}^n$.
A point $p \in \mathbb{R}^n$ together with two vectors in $T_p,$ the tangent space at $p$.
A point $p \in \mathbb{R}^n$ together with two linearly independent vectors in $T_p$.

From the viewpoint of modern geometry, is there a "best" definition of the term "triangle" that generalizes most easily to e.g. Riemannian manifolds and beyond? And what framework (e.g. Riemannian manifolds) does this definition belong to?

REPLY [3 votes]: The most common notion of a "triangle" used in the modern metric and Riemannian geometry is "none of the above". The relevant notion is the one of a "geodesic triangle".
Definition 1. Suppose that $(X,d)$ is a metric space. A geodesic triangle in $X$ is a concatenation of three geodesic segments $\gamma_1, \gamma_2, \gamma_3$ where $\gamma_i$ connects $x_i, x_{i+1}$, $i$ is taken modulo 3.
In this definition, geodesic means an isometric embedding of an interval in the real line into $X$.
Definition 2. Suppose that $(M,g)$ is a Riemannian manifold. A geodesic triangle in $X$ is a concatenation of three geodesic segments $\gamma_1, \gamma_2, \gamma_3$ where $\gamma_i$ connects $x_i, x_{i+1}$, $i$ is taken modulo 3.
In this definition, geodesic means a smooth path $\gamma$ of zero acceleration: $\gamma''=0$, where the second derivative is taken via the Levi-Civita connection on $(M,g)$.
These notions of triangles provide a key tool to:
(1) Defining upper and lower curvature bounds in metric spaces which are not Riemannian manifolds.
(2) Connecting upper and lower curvature bounds for Riemannian metrics to the global geometric and topological properties of the manifolds. ("Comparison Geometry".)
(3) Introducing geometric ideas into the combinatorial group theory.
Read more in:

Cheeger and Ebin, "Comparison Theorems in Riemannian Geometry".
Burago, Burago, Ivanov, "A Course in Metric Geometry".
Bridson and Haefliger, "Metric Spaces of Non-Positive Curvature".


TITLE: reference request: $K$ is relatively weakly compact in $C(X)$ iff $K$ is relatively pointwise compact in $C(X)$
QUESTION [2 upvotes]: Let $X$ be a compact Hausdorff topological space and $K$ be a norm-bounded subset of $C(X)$(The space of all bounded continuous real valued function on $X$). Then $K$ is relatively compact in $C(X)$ for the weak topology if and only if $K$ is relatively compact in $C(X)$ for the topology of pointwise convergence on $X$.
Proof: $\implies $: since the topology of pointwise convergence is weaker than the weak topology  so If $K$ is relatively weakly compact then these two topologies coincide, therefor $K$ is relatively pointwise compact.
the inverse is not clear for me.
Also I would be grateful if you could introduce a reference about this.

REPLY [1 votes]: Let $K$ be relatively compact for the topology of pointwise convergence. By Eberlian Smulian Theorem it suffices to prove that $K$ is relatively weakly sequentially compact. Let $f_n$ be a sequence in $K$. since $K$ is relatively pointwise compact, $f_n$ has a pointwise convergent subsequence $f_{n_k}$ to a function $f\in C(X)$. Let $x^*\in C(X)^*$ we show that $x^*(f_{n_k})\to x^*(f)$. By Riesz Representation Theorem there exists a regular Borel measure $\mu$ on $X$ such that
$$x^*(g)=\int_X gd\mu$$
then by Lebesgue Dominated Convergence Theorem:
$$\lim_k x^*(f_{n_k})=\lim \int _X f_{n_k}d\mu=\int_Xfd\mu=x^*(f)$$


TITLE: Reference request: Deligne's conjecture (cycles)
QUESTION [1 upvotes]: In "The work of Tate" Milne says:

"The relation between the two conjectures has been greatly clarified by the work
  of Deligne. He defines the notion of an absolute Hodge class on a (complete smooth)
  variety over a field of characteristic zero, and conjectures that every Hodge class on
  a variety over C is absolutely Hodge. The Tate conjecture for a variety implies that
  all absolute Hodge classes on the variety are algebraic. Therefore, in the presence of
  Deligne’s conjecture, the Tate conjecture implies the Hodge conjecture. As Deligne
  has proved his conjecture for abelian varieties, this gives another proof of Piatetski-
  Shapiro’s theorem."

In which work (articles) Deligne shows this conjecture?
Is there some book (or expository article) about this conjecture and its connections with Tate and Hodge conjectures?

REPLY [6 votes]: Deligne, Hodge Cycles on Abelian Varieties, in Hodge Cycles, Motives, and Shimura Varieties,
Lecture Notes in Math. 900, 1981, pp. 9-100. Maybe you could use Google to find these references by yourself?


TITLE: Minimum energy required to roll a cuboid through an angle of 90°
QUESTION [1 upvotes]: I'm stuck on this textbook question about the minimum energy required to roll a cuboid through 90° from an upright position to a horizontal position (see image). I believe the correct answer is A, 200 J.

I really don't know which approach to use here. Even my physics teacher is stumped. I don't think I can use difference in potential energy here since I'm not lifting the box, just rolling it, so the weight cannot be 4000N in any case. I'm wondering why the value of the answer is so small. Is the principle of moments applicable here? Or is it something to do with the change in centre of gravity? Can someone give me some hints to get to the correct answer?

REPLY [1 votes]: When you rotate the block, the center of mass (which lies midway on the diagonal) will move until the diagonal is perpendicular.  Therefore, the center of mass must rise from the midpoint of the vertical edge of the block to the midpoint of the (now vertical) diagonal.
If you push horizontally on the top edge of the block, you've got a lever equal to the diagonal of the block pivoting on the fulcrum of the edge on which the block rolls.  But this will not affect the energy required to roll the block.


TITLE: Analytic at a point x implies analytic at every point
QUESTION [2 upvotes]: I know the definition of analyticity of a function at a point. But if a function f is analytic at a point x then does it imply that f is analytic at every point in some open neighborhood of the point x. In other words is it possible or not for a function to be analytic at an isolated point.
Any help please. Thanks in advance.

REPLY [0 votes]: Assuming that you say that a function is analytic at some point $x_0$ if $f^{(n)}$ exists for every $n\in\Bbb Z_+$ and if$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\tag1$$in a neighborhood $V$ of $x_0$, is $f(x)$, then $f$ will be analytic too in every non-isolated point of $V$, precisely because $(1)$ holds in $V$.