text:
\begin{document}

\title{Infinite products related to generalized Thue-Morse sequences}

\author{Yao-Qiang Li}
\address{Institut de Math\'ematiques de Jussieu-Paris Rive Gauche \\
         Sorbonne Universit\'e \\
         Paris, 75005 \\
         France}
\email{yaoqiang.li@etu.upmc.fr\quad yaoqiang.li@imj-prg.fr}
\address{School of Mathematics \\
         South China University of Technology \\
         Guangzhou, 510641 \\
         P.R. China}
\email{scutyaoqiangli@qq.com\quad scutyaoqiangli@gmail.com}

\subjclass[2010]{Primary 11B83; Secondary 11B85, 68R15.}
\keywords{Thue-Morse sequence, morphic sequence, closed formulas for infinite products, Woods and Robbins product}
\date{}

\begin{abstract}
Given an integer $q\ge2$ and $\theta_1,\cdots,\theta_{q-1}\in\{0,1\}$, let $(\theta_n)_{n\ge0}$ be the generalized Thue-Morse sequence, defined to be the unique fixed point of the morphism
$$0\mapsto0\theta_1\cdots\theta_{q-1}$$
$$1\mapsto1\overline{\theta}_1\cdots\overline{\theta}_{q-1}$$
beginning with $\theta_0:=0$, where $\overline{0}:=1$ and $\overline{1}:=0$. For rational functions $R$, we study infinite products of the forms
$$\prod_{n=1}^\infty\Big(R(n)\Big)^{(-1)^{\theta_n}}\quad\text{and}\quad\prod_{n=1}^\infty\Big(R(n)\Big)^{\theta_n}.$$
This generalizes relevant results given by Allouche, Riasat and Shallit in 2019 on infinite products related to the famous Thue-Morse sequence $(t_n)_{n\ge0}$ of the forms
$$\prod_{n=1}^\infty\Big(R(n)\Big)^{(-1)^{t_n}}\quad\text{and}\quad\prod_{n=1}^\infty\Big(R(n)\Big)^{t_n}.$$
\end{abstract}

\maketitle

\section{Introduction}

For any integer $q\ge2$ and $\theta_1,\cdots,\theta_{q-1}\in\{0,1\}$, define the \textit{$(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence} $(\theta_n)_{n\ge0}$ to be the unique fixed point of the morphism
$$0\mapsto0\theta_1\cdots\theta_{q-1}$$
$$1\mapsto1\overline{\theta}_1\cdots\overline{\theta}_{q-1}$$
beginning with $\theta_0:=0$, where $\overline{0}:=1$ and $\overline{1}:=0$. Note that the classical Thue-Morse sequence $(t_n)_{n\ge0}$ (see \cite{AS99,AS03}) is exactly the $(0,1)$-Thue-Morse sequence in our terms. Recently Allouche, Riasat and Shallit \cite{ARS19} studied infinite products of the general form $\prod_{n=1}^\infty(R(n))^{(-1)^{t_n}}$ for rational functions $R$, and obtained a class of equalities involving variables in \cite[Theorem 2.2 and Corollary 2.3]{ARS19} together with many concrete equalities in \cite[Corollary2.4]{ARS19}. Besides, they began to study infinite products of the form $\prod_{n=1}^\infty(R(n))^{t_n}$ and obtained three concrete equalities in \cite[Theorem 4.2]{ARS19}. In this paper, we generalize these results by studying infinite products of the forms
$$\prod_{n=1}^\infty\Big(R(n)\Big)^{(-1)^{\theta_n}}\quad\text{and}\quad\prod_{n=1}^\infty\Big(R(n)\Big)^{\theta_n}.$$

Let $\N$, $\N_0$ and $\C$ be the sets of positive integers $1,2,3,\cdots$, non-negative integers $0,1,2,\cdots$ and complex numbers respectively. Moreover, for simplification we denote $\delta_n:=(-1)^{\theta_n}\in\{+1,-1\}$ for all $n\in\N_0$ throughout this paper.

First we have the following convergence theorem, which is a generalization of \cite[Lemma 2.1 and 4.1]{ARS19} (see also \cite[Lemma 1]{R18}) and guarantees the convergence of all the infinite products given in the results in this paper.

\begin{theorem}\label{main-con} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$, $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence and $R\in\C(X)$ be a rational function such that the values $R(n)$ are defined and non-zero for all $n\in\N$. Then:
\begin{itemize}
\item[\emph{(1)}] the infinite product $\prod_{n=1}^\infty (R(n))^{\delta_n}$ converges if and only if the numerator and the denominator of $R$ have the same degree and the same leading coefficient;
\item[\emph{(2)}] the infinite product $\prod_{n=1}^\infty (R(n))^{\theta_n}$ converges if and only if the numerator and the denominator of $R$ have the same degree, the same leading coefficient and the same sum of roots (in $\C$).
\end{itemize}
\end{theorem}

Although Theorem \ref{main-con} is a natural generalization of \cite[Lemma 2.1 and 4.1]{ARS19}, the proof is more intricate and relies on Proposition \ref{partial-sum} as we will see.

In the following Subsection 1.1 and 1.2, we introduce our results on the forms $\prod (R(n))^{\delta_n}$ and $\prod (R(n))^{\theta_n}$ respectively. Then we give some preliminaries in Section 2 and prove all the results in Section 3.

\subsection{Results on the form $\prod (R(n))^{\delta_n}$}

In order to study the infinite product $\prod_{n=1}^\infty(R(n))^{\delta_n}$, by Theorem \ref{main-con} (1), it suffices to study products of the form
$$f(a,b):=\prod_{n=1}^\infty\Big(\frac{n+a}{n+b}\Big)^{\delta_n},$$
where $a,b\in\C\setminus\{-1,-2,-3,\cdots\}$. For the $(0,1)$-Thue-Morse sequence $(t_n)_{n\ge0}$, the special $f(\frac{x}{2},\frac{x+1}{2})=\prod_{n=1}^\infty(\frac{2n+x}{2n+x+1})^{(-1)^{t_n}}$ is used to define new functions in \cite[Theorem 2.2]{ARS19} and \cite[Definition 1]{R18} and further studied (see also \cite[Remark 6.5]{CS13}). For infinite products involving the first $2^m$ terms of $(t_n)_{n\ge0}$, see the equalities (23) and (24) in \cite[Section 6]{CS13}.

As the first main result in this paper, the following theorem generalizes \cite[Theorem 2.2 and Corollary 2.3 (i)]{ARS19} (see also \cite[Lemma 2]{R18} and the equalities (6) and (7) in \cite[Section 4]{R18}).

\begin{theorem}\label{main-1st} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence. Then for all $a,b\in\C\setminus\{-1,-2,-3,\cdots\}$, we have
\begin{small}
$$f(a,b)=\Big(\frac{a+1}{b+1}\Big)^{\delta_1}\cdots\Big(\frac{a+q-1}{b+q-1}\Big)^{\delta_{q-1}}f(\frac{a}{q},\frac{b}{q})\Big(f(\frac{a+1}{q},\frac{b+1}{q})\Big)^{\delta_1}\cdots\Big(f(\frac{a+q-1}{q},\frac{b+q-1}{q})\Big)^{\delta_{q-1}},$$
\end{small}
which is equivalent to
$$\prod_{n=1}^\infty\Big(\frac{n+a}{n+b}\cdot\frac{qn+b}{qn+a}\big(\frac{qn+b+1}{qn+a+1}\big)^{\delta_1}\cdots\big(\frac{qn+b+q-1}{qn+a+q-1}\big)^{\delta_{q-1}}\Big)^{\delta_n}=\Big(\frac{a+1}{b+1}\Big)^{\delta_1}\cdots\Big(\frac{a+q-1}{b+q-1}\Big)^{\delta_{q-1}}.$$
\end{theorem}

This theorem implies many neat equalities.

\begin{corollary}\label{cor1-1} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence.
\begin{itemize}
\item[\emph{(1)}] For all $a,b\in\C\setminus\{0,-1,-2,\cdots\}$, we have
$$\prod_{n=0}^\infty\Big(\frac{n+a}{n+b}\cdot\frac{qn+b}{qn+a}\big(\frac{qn+b+1}{qn+a+1}\big)^{\delta_1}\cdots\big(\frac{qn+b+q-1}{qn+a+q-1}\big)^{\delta_{q-1}}\Big)^{\delta_n}=1.$$
\item[\emph{(2)}] For all $a\in\C\setminus\{0,-1,-2,\cdots\}$, we have
$$\prod_{n=0}^\infty\Big(\frac{n+a}{n+a+1}\cdot\frac{qn+a+1}{qn+a}\big(\frac{qn+a+2}{qn+a+1}\big)^{\delta_1}\big(\frac{qn+a+3}{qn+a+2}\big)^{\delta_2}\cdots\big(\frac{qn+a+q}{qn+a+q-1}\big)^{\delta_{q-1}}\Big)^{\delta_n}=1$$
and
$$\prod_{n=0}^\infty\Big(\frac{qn+qa}{qn+a}\big(\frac{qn+1}{qn+a+1}\big)^{\delta_1}\big(\frac{qn+2}{qn+a+2}\big)^{\delta_2}\cdots\big(\frac{qn+q-1}{qn+a+q-1}\big)^{\delta_{q-1}}\Big)^{\delta_n}=q.$$
\item[\emph{(3)}] We have
$$\prod_{n=0}^\infty\Big(\frac{qn+q}{qn+1}\big(\frac{qn+1}{qn+2}\big)^{\delta_1}\big(\frac{qn+2}{qn+3}\big)^{\delta_2}\cdots\big(\frac{qn+q-1}{qn+q}\big)^{\delta_{q-1}}\Big)^{\delta_n}=q.$$
\end{itemize}
\end{corollary}

\begin{remark} It seems that Corollary \ref{cor1-1} is weaker than Theorem \ref{main-1st} by the domains of $a$ and $b$. In fact they are equivalent, since Corollary \ref{cor1-1} (1) is the case that $a\neq0$ and $b\neq0$ in Theorem \ref{main-1st}, the second equality in (2) of Corollary \ref{cor1-1} is the case that $a\neq0$ and $b=0$ (the same as $a=0$ and $b\neq0$) in Theorem \ref{main-1st}, and obviously the case that $a=b=0$ in Theorem \ref{main-1st} is trivial.
\end{remark}

Let $q\ge2$ be an integer. For $k=1,2,\cdots,q-1$, define $N_{k,q}(n)$ to be the number of occurrences of the digit $k$ in the base $q$ expansion of the non-negative integer $n$, and let
$$s_q(n):=\sum_{k=1}^{q-1}kN_{k,q}(n)$$
be the sum of digits. It is obtained in \cite[Example 11 and Corollary 5]{AS08} (see also \cite{S06,SG08}) respectively that
\begin{equation}\label{g1}
\prod_{n=0}^\infty\Big(\frac{qn+k}{qn+k+1}\Big)^{(-1)^{N_{k,q}(n)}}=\frac{1}{\sqrt{q}}
\end{equation}
for $k=1,2,\cdots,q-1$, and
\begin{equation}\label{g2}
\prod_{n=0}^\infty\prod_{\substack{0<k<q\\k\text{ odd}}}\Big(\frac{qn+k}{qn+k+1}\Big)^{(-1)^{s_q(n)}}=\frac{1}{\sqrt{q}}.
\end{equation}
For more infinite products related to $(s_q(n))_{n\ge0}$, see for example \cite[Proposition 6 and 7]{L19}. Equalities (\ref{g1}) and (\ref{g2}) are two ways to represent $\frac{1}{\sqrt{q}}$ in the form of infinite products and generalize the well known Woods-Robbins product \cite{R79,W78}
\begin{equation}\label{W-R}
\prod_{n=0}^\infty\Big(\frac{2n+1}{2n+2}\Big)^{(-1)^{t_n}}=\frac{1}{\sqrt{2}}
\end{equation}
where $(t_n)_{n\ge0}$ is the $(0,1)$-Thue-Morse sequence. We give one more such way in the first equality in the following corollary.

\begin{corollary}\label{cor1-2} Let $q\ge2$ be an integer, $k\in\{1,2,\cdots,q-1\}$, $\theta_0=\theta_1=\cdots=\theta_{k-1}=0$, $\theta_k=\theta_{k+1}=\cdots=\theta_{q-1}=1$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence. Then
$$\prod_{n=0}^\infty\Big(\frac{qn+k}{qn+q}\Big)^{\delta_n}=\frac{1}{\sqrt{q}}$$
and
$$\prod_{n=0}^\infty\Big(\frac{(n+a)(qn+a+k)^2}{(n+a+1)(qn+a)(qn+a+q)}\Big)^{\delta_n}=1$$
for all $a\in\C\setminus\{0,-1,-2,\cdots\}$.
\end{corollary}

For more generalizations of the Woods-Robbins product (\ref{W-R}), we refer the reader to \cite{AC85,AS89,S85}.

Note that for any integer $q\ge2$, the $(\overbrace{0,\cdots,0}^q)$-Thue-Morse sequence is the trivial $0^\infty$. For $q=2$, the only nontrivial case, related to the $(0,1)$-Thue-Morse sequence, is already studied in \cite{R18} and \cite[Section 2]{ARS19}. In the following three examples, we study nontrivial cases for $q=3$ in detail, related to the $(0,0,1)$, $(0,1,1)$ and $(0,1,0)$-Thue-Morse sequences.

\begin{example}\label{ex1} Let $(\theta_n)_{n\ge0}$ be the $(0,0,1)$-Thue-Morse sequence.
\newline(1) For all $a,b\in\C\setminus\{0,-1,-2,\cdots\}$ we have
$$\prod_{n=0}^\infty\Big(\frac{(n+a)(3n+b)(3n+b+1)(3n+a+2)}{(n+b)(3n+a)(3n+a+1)(3n+b+2)}\Big)^{\delta_n}=1.$$
(2) For all $a\in\C\setminus\{0,-1,-2,\cdots\}$ we have
\begin{equation*}
\begin{aligned}
\textcircled{\tiny{1}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+a)(3n+a+2)^2}{(n+a+1)(3n+a)(3n+a+3)}\Big)^{\delta_n}=1,\\
\textcircled{\tiny{2}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(3n+3a)(3n+a+2)}{(3n+2)(3n+a)(3n+a+1)}\Big)^{\delta_n}=3,\\
\textcircled{\tiny{3}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(3n+3a)(3n+a+2)}{(3n+3)(3n+a)(3n+a+1)}\Big)^{\delta_n}=\sqrt{3},\\
\textcircled{\tiny{4}}\quad&\prod_{n=0}^\infty\Big(\frac{(6n+1)(3n+3a)(3n+a+2)}{(6n+5)(3n+a)(3n+a+1)}\Big)^{\delta_n}=1.
\end{aligned}
\end{equation*}
(3) The following concrete equalities hold.
\begin{small}
$$\begin{aligned}
\text{\textcircled{\tiny{1}}}\quad&\prod_{n=0}^\infty\Big(\frac{3n+2}{3n+3}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{2}}}\quad&\prod_{n=0}^\infty\Big(\frac{(6n-3)(6n+3)}{(6n-1)(6n+5)}\Big)^{\delta_n}=1,\\
\text{\textcircled{\tiny{3}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(6n+5)}{(3n+2)(6n+1)}\Big)^{\delta_n}=3,\quad&\text{\textcircled{\tiny{4}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(6n+5)}{(3n+3)(6n+1)}\Big)^{\delta_n}=\sqrt{3},\\
\text{\textcircled{\tiny{5}}}\quad&\prod_{n=0}^\infty\Big(\frac{(6n+7)^2}{(6n+3)(6n+15)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{6}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+3)(9n+8)}{(9n+2)(9n+5)}\Big)^{\delta_n}=3,\\
\text{\textcircled{\tiny{7}}}\quad&\prod_{n=0}^\infty\Big(\frac{(18n+3)(18n+17)}{(18n+5)(18n+11)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{8}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+3)(3n+1)(6n+7)}{(2n+1)(3n+2)(6n+5)}\Big)^{\delta_n}=3,\\
\text{\textcircled{\tiny{9}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(3n+3)^2}{(n+2)(3n+1)(3n+4)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{10}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(3n+2)(3n+3)}{(n+2)(3n+1)(3n+4)}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\\
\text{\textcircled{\tiny{11}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(3n+2)^2}{(n+2)(3n+1)(3n+4)}\Big)^{\delta_n}=\frac{1}{3},\quad&\text{\textcircled{\tiny{12}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+2)^3}{(3n+1)(3n+4)(3n+6)}\Big)^{\delta_n}=\frac{1}{3\sqrt{3}},\\
\text{\textcircled{\tiny{13}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(3n+4)^2}{(n+3)(3n+2)(3n+5)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{14}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(3n+4)^2}{(n+3)(3n+3)(3n+5)}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\\
\text{\textcircled{\tiny{15}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(9n+4)(9n+7)}{(n+1)(9n+6)(9n+10)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{16}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(6n+3)(6n-3)}{(3n+2)(6n+1)(6n-1)}\Big)^{\delta_n}=3.
\end{aligned}$$
\end{small}
\end{example}

\begin{example}\label{ex2} Let $(\theta_n)_{n\ge0}$ be the $(0,1,1)$-Thue-Morse sequence.
\newline(1) For all $a,b\in\C\setminus\{0,-1,-2,\cdots\}$ we have
$$\prod_{n=0}^\infty\Big(\frac{(n+a)(3n+b)(3n+a+1)(3n+a+2)}{(n+b)(3n+a)(3n+b+1)(3n+b+2)}\Big)^{\delta_n}=1.$$
(2) For all $a\in\C\setminus\{0,-1,-2,\cdots\}$ we have
\begin{equation*}
\begin{aligned}
\text{\textcircled{\tiny{1}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+a)(3n+a+1)^2}{(n+a+1)(3n+a)(3n+a+3)}\Big)^{\delta_n}=1,\\
\text{\textcircled{\tiny{2}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+a+1)(3n+a+2)(3n+3a)}{(3n+1)(3n+2)(3n+a)}\Big)^{\delta_n}=3,\\
\text{\textcircled{\tiny{3}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+a+1)(3n+a+2)(3n+3a)}{(3n+2)(3n+3)(3n+a)}\Big)^{\delta_n}=\sqrt{3}.
\end{aligned}
\end{equation*}
(3) The following concrete equalities hold.
\begin{small}
$$\begin{aligned}
\text{\textcircled{\tiny{1}}}\quad&\prod_{n=0}^\infty\Big(\frac{3n+1}{3n+3}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{2}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+4)(3n+6)}{(3n+2)^2}\Big)^{\delta_n}=\sqrt{3},\\
\text{\textcircled{\tiny{3}}}\quad&\prod_{n=0}^\infty\Big(\frac{(6n+5)^2}{(6n+3)(6n+15)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{4}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+4)(9n+7)}{(9n+1)(9n+6)}\Big)^{\delta_n}=3,\\
\text{\textcircled{\tiny{5}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+5)(9n+8)}{(9n+2)(9n+3)}\Big)^{\delta_n}=3,\quad&\text{\textcircled{\tiny{6}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+5)(9n+8)}{(9n+2)(9n+9)}\Big)^{\delta_n}=\sqrt{3},\\
\text{\textcircled{\tiny{7}}}\quad&\prod_{n=0}^\infty\Big(\frac{(24n+7)(24n+13)}{(24n+5)(24n+23)}\Big)^{\delta_n}=1,\quad&\text{\textcircled{\tiny{8}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(3n+3)^2}{(n+3)(3n+2)(3n+5)}\Big)^{\delta_n}=1,\\
\text{\textcircled{\tiny{9}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(3n+1)(3n+3)}{(n+3)(3n+2)(3n+5)}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{10}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)(3n+1)^2}{(n+3)(3n+2)(3n+5)}\Big)^{\delta_n}=\frac{1}{3},\\
\text{\textcircled{\tiny{11}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+3)(3n+4)(3n+5)}{(n+1)(3n+1)(3n+2)}\Big)^{\delta_n}=3,\quad&\text{\textcircled{\tiny{12}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+3)(3n+4)(3n+5)}{(n+1)(3n+2)(3n+3)}\Big)^{\delta_n}=\sqrt{3},\\
\text{\textcircled{\tiny{13}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+4)(3n+5)(3n+9)}{(3n+1)^2(3n+2)}\Big)^{\delta_n}=3\sqrt{3},\quad&\text{\textcircled{\tiny{14}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n-1)(6n+1)^2}{(2n+1)(6n-1)(6n+5)}\Big)^{\delta_n}=1,\\
\text{\textcircled{\tiny{15}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+2)(6n+1)(6n+4)}{(2n+1)(6n+3)(6n+5)}\Big)^{\delta_n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{16}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+3)(6n+5)(6n+7)}{(2n+1)(6n+2)(6n+4)}\Big)^{\delta_n}=3.
\end{aligned}$$
\end{small}
\end{example}

Note that the $(0,1,0)$-Thue-Morse sequence is exactly $01010101\cdots$, which implies $\delta_n:=(-1)^{\theta_n}=(-1)^n$ for all $n\in\N_0$. The next example is deduced from Corollary \ref{cor1-1}, and can also be deduced from Theorem \ref{Gamma prod} and Proposition \ref{Gamma properties}, which are classical results on the Gamma function.

\begin{example}\label{ex3} (1) For all odd $q\ge3$, we have
$$\prod_{n=0}^\infty\Big(\frac{(qn+1)(qn+3)\cdots(qn+q-2)}{(qn+2)(qn+4)\cdots(qn+q-1)}\Big)^{(-1)^n}=\frac{1}{\sqrt{q}}.$$
(2) For all odd $q\ge3$ and all $a\in\C\setminus\{0,-1,-2,\cdots\}$ we have
$$\prod_{n=0}^\infty\Big(\frac{(qn+a)(qn+a+2)(qn+a+4)\cdots(qn+a+q-1)}{(qn+qa)(qn+a+1)(qn+a+3)\cdots(qn+a+q-2)}\Big)^{(-1)^n}=\frac{1}{\sqrt{q}}.$$
(3) The following concrete equalities hold.
\begin{small}
$$\begin{aligned}
\text{\textcircled{\tiny{1}}}\quad&\prod_{n=0}^\infty\Big(\frac{3n+1}{3n+2}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{2}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(3n+5)}{(n+3)(3n+4)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\\
\text{\textcircled{\tiny{3}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+2)(3n+4)}{(3n+3)(3n+6)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{4}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(3n+5)}{(3n+6)(3n+9)}\Big)^{(-1)^n}=\frac{1}{3\sqrt{3}},\\
\text{\textcircled{\tiny{5}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+2)(9n+8)}{(9n+5)(9n+6)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{6}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+2)(9n+8)}{(9n+3)(9n+5)}\Big)^{(-1)^n}=1,\\
\text{\textcircled{\tiny{7}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+1)(9n+7)}{(9n+3)(9n+4)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{8}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+1)(9n+7)}{(9n+4)(9n+6)}\Big)^{(-1)^n}=\frac{1}{3},\\
\text{\textcircled{\tiny{9}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+2)(9n+11)}{(9n+3)(9n+15)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{10}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+1)(9n+10)}{(9n+6)(9n+12)}\Big)^{(-1)^n}=\frac{1}{3\sqrt{3}},\\
\text{\textcircled{\tiny{11}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n+5)(9n+11)}{(9n+8)(9n+15)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\quad&\text{\textcircled{\tiny{12}}}\quad&\prod_{n=0}^\infty\Big(\frac{(9n-1)(9n+8)}{(9n-3)(9n+3)}\Big)^{(-1)^n}=\frac{1}{\sqrt{3}},\\
\text{\textcircled{\tiny{13}}}\quad&\prod_{n=0}^\infty\Big(\frac{(5n+1)(5n+3)}{(5n+2)(5n+4)}\Big)^{(-1)^n}=\frac{1}{\sqrt{5}},\quad&\text{\textcircled{\tiny{14}}}\quad&\prod_{n=0}^\infty\Big(\frac{(10n+1)(10n+9)}{(10n+3)(10n+7)}\Big)^{(-1)^n}=\frac{1}{\sqrt{5}},\\
\text{\textcircled{\tiny{15}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(5n+3)(5n+7)}{(n+3)(5n+4)(5n+6)}\Big)^{(-1)^n}=\frac{1}{\sqrt{5}},\quad&\text{\textcircled{\tiny{16}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(5n+2)(5n+7)}{(n+3)(5n+1)(5n+6)}\Big)^{(-1)^n}=1.
\end{aligned}$$
\end{small}
\end{example}

In \cite{H16} Hu studied infinite sums of the form
$$\sum_{n\ge0}\Big((-1)^{a_{w,B}(n)}\sum_{(l,c_l)\in L_{w,B}}c_lf(l(n))\Big)$$
where $a_{w,B}(n)$ denote the number of occurrences of the word $w$ in the base $B$ expansion of the non-negative integer $n$, $f$ is any function that verifies certain convergence conditions, and $L_{w,B}$ is a computable finite set of pairs $(l,c_l)$ where $l$ is a polynomial with integer coefficients of degree $1$ and $c_l$ is an integer. If $f$ is taken to be an appropriate composition of a logarithmic function and a rational function, after exponentiating, some infinite products of the form $\prod_n(R(n))^{(-1)^{a_{w,B}(n)}}$ can be obtained, where $R$ is a rational function depending on the sequence $(a_{w,B}(n))_{n\ge0}$. For instance the above Example \ref{ex3} (3) \textcircled{\tiny{1}} is also obtained in \cite[Section 5]{H16} (see also \cite[Section 4.4]{AS08}).

\subsection{Results on the form $\prod (R(n))^{\theta_n}$}

In order to study the infinite product $\prod_{n=1}^\infty(R(n))^{\theta_n}$, by Theorem \ref{main-con} (2), it suffices to study products of the form
$$\mathfrak{f}(a_1,\cdots,a_d;b_1,\cdots,b_d):=\prod_{n=1}^\infty\Big(\frac{(n+a_1)\cdots(n+a_d)}{(n+b_1)\cdots(n+b_d)}\Big)^{\theta_n}$$
where $d\in\N$ and $a_1,\cdots,a_d,b_1,\cdots,b_d\in\C\setminus\{-1,-2,-3,\cdots\}$ satisfy $a_1+\cdots+a_d=b_1+\cdots+b_d$. As the second main result in this paper, the following theorem (which implies Corollary \ref{TM}) generalizes \cite[Theorem 4.2]{ARS19}.

\begin{theorem}\label{main-2nd} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence. Then for all $d\in\N$ and $a_1,\cdots,a_d,b_1,\cdots,b_d\in\C\setminus\{-1,-2,-3,\cdots\}$ satisfying $a_1+\cdots+a_d=b_1+\cdots+b_d$, we have
\begin{small}
$$\mathfrak{f}(a_1,\cdots,a_d;b_1,\cdots,b_d)=\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{\Gamma(\frac{b_i+k}{q})}{\Gamma(\frac{a_i+k}{q})}\Big)^{\theta_k}\cdot\prod_{k=0}^{q-1}\Big(\mathfrak{f}\big(\frac{a_1+k}{q},\cdots,\frac{a_d+k}{q};\frac{b_1+k}{q},\cdots,\frac{b_d+k}{q}\big)\Big)^{(-1)^{\theta_k}},$$
\end{small}
which is equivalent to
$$\prod_{n=1}^\infty\Big(\prod_{i=1}^d\Big(\frac{n+a_i}{n+b_i}\cdot\prod_{k=0}^{q-1}\big(\frac{qn+b_i+k}{qn+a_i+k}\big)^{(-1)^{\theta_k}}\Big)\Big)^{\theta_n}=\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{\Gamma(\frac{b_i+k}{q})}{\Gamma(\frac{a_i+k}{q})}\Big)^{\theta_k}.$$
\end{theorem}

This theorem implies a large number of equalities for products of the form $\prod (R(n))^{\theta_n}$ as we will see in the following corollaries, which can also be viewed as special examples.

\begin{corollary}\label{cor2-1} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence.
\newline\emph{(1)} For all $a,b,c\in\C$ such that $a,b,a+c,b+c\notin\{-1,-2,-3,\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+a)(n+b+c)}{(n+b)(n+a+c)}\cdot\prod_{k=0}^{q-1}\Big(\frac{(qn+b+k)(qn+a+c+k)}{(qn+a+k)(qn+b+c+k)}\Big)^{(-1)^{\theta_k}}\Big)^{\theta_n}=\prod_{k=1}^{q-1}\Big(\frac{\Gamma(\frac{b+k}{q})\Gamma(\frac{a+c+k}{q})}{\Gamma(\frac{a+k}{q})\Gamma(\frac{b+c+k}{q})}\Big)^{\theta_k}.$$
\newline\emph{(2)} For all $d\in\N$ and $a_1,\cdots,a_d\in\C\setminus\{-1,-2,-3,\cdots\}$ such that $a_1+\cdots+a_d=0$ we have
$$\prod_{n=1}^\infty\Big(\prod_{i=1}^d\Big(\frac{qn+qa_i}{qn+a_i}\cdot\prod_{k=1}^{q-1}\Big(\frac{qn+k}{qn+a_i+k}\Big)^{(-1)^{\theta_k}}\Big)\Big)^{\theta_n}=\prod_{k=1}^{q-1}\Big(\frac{(\Gamma(\frac{k}{q}))^d}{\Gamma(\frac{a_1+k}{q})\cdots\Gamma(\frac{a_d+k}{q})}\Big)^{\theta_k}.$$
\newline\emph{(3)} For all $a\in\C\setminus\Z$ we have
$$\prod_{n=1}^\infty\Big(\frac{(qn+qa)(qn-qa)}{(qn+a)(qn-a)}\cdot\prod_{k=1}^{q-1}\Big(\frac{(qn+k)^2}{(qn+a+k)(qn-a+k)}\Big)^{(-1)^{\theta_k}}\Big)^{\theta_n}=\prod_{k=1}^{q-1}\Big(\frac{(\Gamma(\frac{k}{q}))^2}{\Gamma(\frac{k+a}{q})\Gamma(\frac{k-a}{q})}\Big)^{\theta_k}.$$
\end{corollary}

In particular for the well known $(0,1)$-Thue-Morse sequence, we have the following corollary, in which (5) \textcircled{\tiny{2}}, \textcircled{\tiny{3}} and \textcircled{\tiny{4}} recover \cite[Theorem 4.2]{ARS19}.

\begin{corollary}\label{TM} Let $(t_n)_{n\ge0}$ be the $(0,1)$-Thue-Morse sequence.
\newline\emph{(1)} For all $d\in\N$ and $a_1,\cdots,a_d,b_1,\cdots,b_d\in\C\setminus\{-1,-2,-3,\cdots\}$ such that $a_1+\cdots+a_d=b_1+\cdots+b_d$ we have
$$\prod_{n=1}^\infty\Big(\prod_{i=1}^d\frac{(n+a_i)(2n+b_i)(2n+a_i+1)}{(n+b_i)(2n+a_i)(2n+b_i+1)}\Big)^{t_n}=\prod_{i=1}^d\frac{\Gamma(\frac{b_i+1}{2})}{\Gamma(\frac{a_i+1}{2})}.$$
\emph{(2)} For all $a,b,c\in\C$ such that $a,b,a+c,b+c\notin\{-1,-2,-3,\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+a)(n+b+c)(2n+b)(2n+a+1)(2n+a+c)(2n+b+c+1)}{(n+b)(n+a+c)(2n+a)(2n+b+1)(2n+b+c)(2n+a+c+1)}\Big)^{t_n}=\frac{\Gamma(\frac{b+1}{2})\Gamma(\frac{a+c+1}{2})}{\Gamma(\frac{a+1}{2})\Gamma(\frac{b+c+1}{2})}.$$
\begin{itemize}
\item[\emph{(3)} \textcircled{\tiny{1}}] For all $a,b\in\C$ such that $a,b,a+b\notin\{-1,-2,-3,\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{2(n+a)(n+b)(2n+a+1)(2n+b+1)(2n+a+b)}{(2n+1)(n+a+b)(2n+a)(2n+b)(2n+a+b+1)}\Big)^{t_n}=\frac{\sqrt{\pi}\text{ }\Gamma(\frac{a+b+1}{2})}{\Gamma(\frac{a+1}{2})\Gamma(\frac{b+1}{2})}.$$
\item[\textcircled{\tiny{2}}] For all $a,b\in\C$ such that $a,b,2a+1,a+b\notin\{-1,-2,-3,\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+a+b)(2n+a+2)(2n+2a+1)(2n+b)(2n+a+b+1)}{(n+2a+1)(2n+a+1)(2n+b+1)(2n+2b)(2n+a+b)}\Big)^{t_n}=\frac{2^a\Gamma(\frac{a+1}{2})\Gamma(\frac{b+1}{2})}{\sqrt{\pi}\text{ }\Gamma(\frac{a+b+1}{2})}.$$
\end{itemize}
\begin{itemize}
\item[\emph{(4)} \textcircled{\tiny{1}}] For all $a\in\C\setminus\{-1,-\frac{3}{2},-2,-\frac{5}{2},\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+a)(2n+a+2)(2n+2a+1)}{(n+2a+1)(2n+1)(2n+a)}\Big)^{t_n}=2^a.$$
\item[\textcircled{\tiny{2}}] For all $a\in\C\setminus\{-1,-\frac{3}{2},-2,-\frac{5}{2},\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+1)(n+a+2)(2n+a+3)(2n+2a+1)}{(n+2)(n+2a+1)(2n+3)(2n+a+1)}\Big)^{t_n}=\frac{2^a}{a+1}.$$
\item[\textcircled{\tiny{3}}] For all $a\in\C\setminus\Z$ we have
$$\prod_{n=1}^\infty\Big(\frac{(2n+a+1)(2n-a+1)(2n+2a)(2n-2a)}{(2n+1)^2(2n+a)(2n-a)}\Big)^{t_n}=\cos\frac{\pi a}{2}.$$
\item[\textcircled{\tiny{4}}] For all $a\in\C\setminus(\Z\cup\{\frac{3}{2},\frac{5}{2},\frac{7}{2},\cdots\})$ we have
$$\prod_{n=1}^\infty\Big(\frac{(2n+a+1)(2n-a+1)(2n+2a)(2n-4a+2)}{(2n+1)(2n+a)(2n-a+2)(2n-2a+1)}\Big)^{t_n}=2^a\cos\frac{\pi a}{2}.$$
\item[\textcircled{\tiny{5}}] For all $a\in\C\setminus\{\pm3,\pm5,\pm7,\cdots\}$ we have
$$\prod_{n=1}^\infty\Big(\frac{(2n+a+1)(2n-a+1)(4n+a+3)(4n-a+3)}{(2n+2)^2(4n+a+1)(4n-a+1)}\Big)^{t_n}=\frac{\sqrt{\pi}}{\Gamma(\frac{3+a}{4})\Gamma(\frac{3-a}{4})}.$$
\item[\textcircled{\tiny{6}}] For all $d\in\N$ we have
$$\prod_{n=1}^\infty\Big(\frac{(n+1)(2n+d)(2n+2)^{2d-1}}{(n+d)(2n+d+1)(2n+1)^{2d-1}}\Big)^{t_n}=\pi^{\frac{d-1}{2}}\Gamma(\frac{d+1}{2}).$$
\end{itemize}
\emph{(5)} The following concrete equalities hold.
\begin{footnotesize}
$$\begin{aligned}
\text{\textcircled{\tiny{1}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+1)(4n-1)}{(2n-1)(4n+3)}\Big)^{t_n}=\sqrt{2},\quad&\text{\textcircled{\tiny{2}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+1)(4n+3)}{(2n+2)(4n+1)}\Big)^{t_n}=\frac{\Gamma(\frac{1}{4})}{\sqrt{2}\pi^{\frac{3}{4}}},\\
\text{\textcircled{\tiny{3}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(4n+5)}{(n+2)(4n+1)}\Big)^{t_n}=\sqrt{2},\quad&\text{\textcircled{\tiny{4}}}\quad&\prod_{n=0}^\infty\Big(\frac{(8n+1)(8n+7)}{(8n+3)(8n+5)}\Big)^{t_n}=\sqrt{2\sqrt{2}-2},\\
\text{\textcircled{\tiny{5}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(2n+3)^2}{(n+3)(2n+1)^2}\Big)^{t_n}=2,\quad&\text{\textcircled{\tiny{6}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+2)^2(6n+5)}{(3n+3)^2(6n+1)}\Big)^{t_n}=\frac{\sqrt{3}\text{ }\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{4\pi^{\frac{3}{2}}},\\
\text{\textcircled{\tiny{7}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+2)^2(2n+5)}{(n+1)(n+5)(2n+1)}\Big)^{t_n}=4,\quad&\text{\textcircled{\tiny{8}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+1)^2(4n-1)}{(2n-1)(2n+2)(4n+1)}\Big)^{t_n}=\frac{\Gamma(\frac{1}{4})}{\pi^{\frac{3}{4}}},\\
\text{\textcircled{\tiny{9}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n+3)^2(4n-1)}{(2n-1)(2n+6)(4n+1)}\Big)^{t_n}=\frac{2\Gamma(\frac{1}{4})}{\pi^{\frac{3}{4}}},\quad&\text{\textcircled{\tiny{10}}}\quad&\prod_{n=0}^\infty\Big(\frac{(2n-1)(4n+3)^2}{(2n+2)(4n+1)(4n-1)}\Big)^{t_n}=\frac{\Gamma(\frac{1}{4})}{2\pi^{\frac{3}{4}}},\\
\text{\textcircled{\tiny{11}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n-1)^2(6n+3)}{(3n+2)(3n-2)(6n-1)}\Big)^{t_n}=2^{\frac{2}{3}},\quad&\text{\textcircled{\tiny{12}}}\quad&\prod_{n=0}^\infty\Big(\frac{(4n+2)^2(8n-1)}{(4n-1)(4n+1)(8n+7)}\Big)^{t_n}=2^{\frac{1}{4}},\\
\text{\textcircled{\tiny{13}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(2n+7)(4n+9)}{(n+4)(2n+3)(4n+5)}\Big)^{t_n}=\frac{4\sqrt{2}}{5},\quad&\text{\textcircled{\tiny{14}}}\quad&\prod_{n=0}^\infty\Big(\frac{(n+1)(3n+7)(6n+5)}{(n+2)(3n+2)(6n+9)}\Big)^{t_n}=3\cdot2^{-\frac{5}{3}},\\
\text{\textcircled{\tiny{15}}}\quad&\prod_{n=0}^\infty\Big(\frac{(3n+1)(6n-1)(6n+3)}{(3n-1)(6n+1)(6n+5)}\Big)^{t_n}=2^{\frac{1}{3}},\quad&\text{\textcircled{\tiny{16}}}\quad&\prod_{n=0}^\infty\Big(\frac{(5n+4)(10n+1)(10n+5)}{(5n+2)(10n+3)(10n+7)}\Big)^{t_n}=\frac{\sqrt{5}-1}{2^{\frac{2}{5}}}.\\
\end{aligned}$$
\end{footnotesize}
\end{corollary}

\section{Preliminaries}

First we need the following concept.

\begin{definition}[\cite{AS08,U99}]
Let $q\ge2$ be an integer. A sequence $u=(u_n)_{n\ge0}\in\C^{\N_0}$ is called \textit{strongly $q$-multiplicative} if $u_0=1$ and
$$u_{nq+k}=u_nu_k$$
for all $k\in\{0,1,\cdots,q-1\}$ and $n\in\N_0$.
\end{definition}

The following theorem is a classical result on the Gamma function $\Gamma$ (see for examples \cite[Theorem 1.1]{CS13} and \cite[Section 12.13]{WW96}).

\begin{theorem}\label{Gamma prod} Let $d\in\N$ and $a_1,a_2,\cdots,a_d,b_1,b_2,\cdots,b_d\in\C\setminus\{0,-1,-2,\cdots\}$. If $a_1+a_2+\cdots+a_d=b_1+b_2+\cdots+b_d$, then
$$\prod_{n=0}^\infty\frac{(n+a_1)(n+a_2)\cdots(n+a_d)}{(n+b_1)(n+b_2)\cdots(n+b_d)}=\frac{\Gamma(b_1)\Gamma(b_2)\cdots\Gamma(b_d)}{\Gamma(a_1)\Gamma(a_2)\cdots\Gamma(a_d)}.$$
\end{theorem}

Besides, we need the properties on the Gamma function gathered in the following proposition.

\begin{proposition}[\cite{A64,B17,W}]\label{Gamma properties} \emph{(1)} For all $n\in\N$ and $z\in\C\setminus\{0,-\frac{1}{n},-\frac{2}{n},-\frac{3}{n},-\frac{4}{n},\cdots\}$ we have
$$\Gamma(z)\Gamma(z+\frac{1}{n})\Gamma(z+\frac{2}{n})\cdots\Gamma(z+\frac{n-1}{n})=(2\pi)^{\frac{n-1}{2}}n^{\frac{1}{2}-nz}\Gamma(nz).$$
\emph{(2)} For all $z\in\C\setminus\{0,-1,-2,\cdots\}$ we have
$$\Gamma(z+1)=z\Gamma(z)$$
and
$$\Gamma(\frac{z}{2})\Gamma(\frac{z+1}{2})=2^{1-z}\sqrt{\pi}\text{ }\Gamma(z).$$
\emph{(3)} For all $z\in\C\setminus\Z$ we have
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}.$$
\emph{(4)} We have
$$\Gamma(1)=\Gamma(2)=1,\quad\Gamma(\frac{1}{2})=\sqrt{\pi}\quad\text{and}\quad\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}.$$
\end{proposition}

\section{Proofs of the results}

Let $q\ge2$ be an integer, $\theta_0=0$, $\theta_1,\cdots,\theta_{q-1}\in\{0,1\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence. Recall that $(\delta_n)_{n\ge0}$ is defined by $\delta_n=(-1)^{\theta_n}$ for all $n\in\N_0$. At the same time $(\delta_n)_{n\ge0}$ can be view as the unique fixed point of the morphism
\begin{equation}\label{+-morphism}
\begin{aligned}
+1\mapsto(+1)(+\delta_1)\cdots(+\delta_{q-1})\\
-1\mapsto(-1)(-\delta_1)\cdots(-\delta_{q-1})
\end{aligned}
\end{equation}
beginning with $\delta_0=+1$. Define the sequence of partial sums of $(\delta_n)_{n\ge0}$ by
$$\Delta_0:=0\quad\text{and}\quad\Delta_n:=\delta_0+\delta_1+\cdots+\delta_{n-1}\quad\text{for all }n\ge1.$$
Note that $(\Delta_n)_{n\ge0}$ depends on the choice of $(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}$. Before proving Theorem \ref{main-con}, we need the following proposition on $(\Delta_n)_{n\ge0}$, which is itself valuable.

\begin{proposition}\label{partial-sum} Let $q\ge2$ be an integer.
\begin{itemize}
\item[\emph{(1)}] For all $k,s\in\N_0$ and $t\in\{0,1,\cdots,q^k-1\}$ we have
$$\delta_{sq^k+t}=\delta_s\delta_t\quad\text{and}\quad\Delta_{sq^k+t}=\Delta_s\Delta_{q^k}+\delta_s\Delta_t.$$
\item[\emph{(2)}] With the convention $0^0:=1$, for all $k\in\N_0$ we have
$$\Delta_{q^k}=\Delta_q^k,$$
$$\max\Big\{|\Delta_{q^k}|:(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}\setminus\{(+1)^{q-1}\}\Big\}=(q-2)^k,$$
$$\max\Big\{|\Delta_n|:0\le n\le q^k,(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}\setminus\{(+1)^{q-1}\}\Big\}=1+(q-2)+\cdots+(q-2)^k.$$
\item[\emph{(3)}] If $(\delta_1,\cdots,\delta_{q-1})\neq(+1)^{q-1}$, then for all $n$ large enough we have
$$|\Delta_n|\le n^{\log_q(q-1)}.$$
\end{itemize}
\end{proposition}
\begin{proof} (1) \textcircled{\tiny{1}} Prove $\delta_{sq^k+t}=\delta_s\delta_t$ for all $k,s\in\N_0$ and $t\in\{0,1,\cdots,q^k-1\}$.
\begin{itemize}
\item[i)] Prove that $(\delta_n)_{n\ge0}$ is strongly $q$-multiplicative, i.e.,
$$\delta_{sq+t}=\delta_s\delta_t\quad\text{for all }s\in\N_0\text{ and }t\in\{0,1,\cdots,q-1\}.$$
Denote the morphism (\ref{+-morphism}) by $\psi$. Then by $\psi((\delta_0,\delta_1,\delta_2,\cdots))=(\delta_0,\delta_1,\delta_2,\cdots)$ we get $\psi(\delta_s)=(\delta_{sq},\delta_{sq+1},\cdots,\delta_{(s+1)q-1})$ for all $s\in\N_0$. It follows from $\psi(+1)=(+1,+\delta_1,\cdots,+\delta_{q-1})$ and $\psi(-1)=(-1,-\delta_1,\cdots,-\delta_{q-1})$ that $\delta_{sq+t}=\delta_s\delta_t$ for all $t\in\{0,\cdots,q-1\}$.
\item[ii)] Let $k\in\N$, $s\in\N_0$ and $t\in\{0,\cdots,q^k-1\}$. Then there exist $l\in\N_0$ and $s_l,\cdots,s_1,s_0$, $t_{k-1},\cdots,t_1,t_0$ $\in\{0,1,\cdots,q-1\}$ such that
$$s=s_lq^l+\cdots+s_1q+s_0\quad\text{and}\quad t=t_{k-1}q^{k-1}+\cdots+t_1q+t_0.$$
By i) and \cite[Proposition 1]{AS08} we get
$$\delta_{sq^k+t}=\delta_{s_l}\cdots\delta_{s_1}\delta_{s_0}\delta_{t_{k-1}}\cdots\delta_{t_1}\delta_{t_0},$$
$$\delta_s=\delta_{s_l}\cdots\delta_{s_1}\delta_{s_0}\quad\text{and}\quad\delta_t=\delta_{t_{k-1}}\cdots\delta_{t_1}\delta_{t_0}.$$
Thus $\delta_{sq^k+t}=\delta_s\delta_t$.
\end{itemize}
\textcircled{\tiny{2}} Prove $\Delta_{sq^k+t}=\Delta_s\Delta_{q^k}+\delta_s\Delta_t$ for all $k,s\in\N_0$ and $t\in\{0,1,\cdots,q^k-1\}$. In fact, we have
\begin{eqnarray*}
\Delta_{sq^k+t}&=&(\delta_0+\delta_1+\cdots+\delta_{q^k-1})+(\delta_{q^k}+\delta_{q^k+1}+\cdots+\delta_{q^k+(q^k-1)})\\
& &+\cdots+(\delta_{(s-1)q^k}+\delta_{(s-1)q^k+1}+\cdots+\delta_{(s-1)q^k+(q^k-1)})\\
& &+(\delta_{sq^k}+\delta_{sq^k+1}+\cdots+\delta_{sq^k+t-1})\\
&=&\delta_0(\delta_0+\delta_1+\cdots+\delta_{q^k-1})+\delta_1(\delta_0+\delta_1+\cdots+\delta_{q^k-1})\\
& &+\cdots+\delta_{s-1}(\delta_0+\delta_1+\cdots+\delta_{q^k-1})\\
& &+\delta_s(\delta_0+\delta_1+\cdots+\delta_{t-1})\\
&=&\Delta_s\Delta_{q^k}+\delta_s\Delta_t
\end{eqnarray*}
where the second equality follows from \textcircled{\tiny{1}}.
\newline(2) \textcircled{\tiny{1}} We have $\Delta_{q^k}=\Delta_q^k$ for all $k\in\N_0$ since (1) \textcircled{\tiny{2}} implies $\Delta_{q\cdot q^l}=\Delta_q\Delta_{q^l}$ for all $l\in\N_0$.
\newline\textcircled{\tiny{2}} For all $k\in\N_0$, the fact
$$\max\Big\{|\Delta_{q^k}|:(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}\setminus\{(+1)^{q-1}\}\Big\}=(q-2)^k$$
follows from \textcircled{\tiny{1}} and
$$\max\Big\{|\Delta_q|:(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}\setminus\{(+1)^{q-1}\}\Big\}=q-2.$$
\textcircled{\tiny{3}} In order to prove the last equality in statement (2), since the case $k=0$ is trivial and \textcircled{\tiny{2}} implies $|\Delta_{q^k}|\le1+(q-2)+(q-2)^2+\cdots+(q-2)^k$, it suffices to verify that for all $k\in\N$, we have
\newline\leftline{$\max\Big\{|\Delta_n|:0\le n\le q^k-1,(\delta_1,\cdots,\delta_{q-1})\in\{+1,-1\}^{q-1}\setminus\{(+1)^{q-1}\}\Big\}$}
\rightline{$=1+(q-2)+(q-2)^2+\cdots+(q-2)^k.$}
\begin{itemize}
\item[\boxed{\le}] (By induction on $k$) For $k=1$, obviously we have $|\Delta_0|,|\Delta_1|,\cdots,|\Delta_{q-1}|\le q-1$. Suppose that for some $k\in\N$ and all $l\in\{0,1,\cdots,k\}$, we have already had
$$|\Delta_0|,|\Delta_1|,\cdots,|\Delta_{q^l-1}|\le1+(q-2)+(q-2)^2+\cdots+(q-2)^l.$$
Let $n\in\{0,1,\cdots,q^{k+1}-1\}$. It suffices to prove
\begin{equation}\label{to prove}
|\Delta_n|\le1+(q-2)+(q-2)^2+\cdots+(q-2)^{k+1}.
\end{equation}
If $n\le q^k-1$, this follows immediately from the inductive hypothesis. We only need to consider $q^k\le n\le q^{k+1}-1$ in the following. Let $s\in\{1,\cdots,q-1\}$ and $t\in\{0,1,\cdots,q^k-1\}$ such that $n=sq^k+t$. By (1) \textcircled{\tiny{2}} we get
$$\Delta_n=\Delta_s\Delta_{q^k}+\delta_s\Delta_t.$$
If $s\le q-2$, then
\begin{eqnarray*}
|\Delta_n|&\le&|\Delta_s|\cdot|\Delta_{q^k}|+|\Delta_t|\\
&\le&s(q-2)^k+(1+(q-2)+(q-2)^2+\cdots+(q-2)^k)\\
&\le&1+(q-2)+(q-2)^2+\cdots+(q-2)^{k+1}
\end{eqnarray*}
where the second inequality follows from \textcircled{\tiny{2}} and the inductive hypothesis. In the following we only need to consider $s=q-1$. It means that
$$\Delta_n=\Delta_{q-1}\Delta_{q^k}+\delta_{q-1}\Delta_t.$$
If there exists $p\in\{0,1,\cdots,q-2\}$ such that $\delta_p=-1$, then $|\Delta_{q-1}|\le q-3$ and
\begin{eqnarray*}
|\Delta_n|&\le&(q-3)|\Delta_{q^k}|+|\Delta_t|\\
&\le&(q-3)(q-2)^k+(1+(q-2)+(q-2)^2+\cdots+(q-2)^k)\\
&\le&1+(q-2)+(q-2)^2+\cdots+(q-2)^{k+1}
\end{eqnarray*}
where the second inequality follows from \textcircled{\tiny{2}} and the inductive hypothesis. Thus it suffices to consider $\delta_0=\delta_1=\cdots=\delta_{q-2}=+1$ in the following. By $(\delta_1,\cdots,\delta_{q-1})\neq(+1)^{q-1}$ we get $\delta_{q-1}=-1$. It follows from $\Delta_{q-1}=q-1$ and $\Delta_{q^k}=\Delta_q^k=(q-2)^k$ that
$$\Delta_n=(q-1)(q-2)^k-\Delta_t.$$
Thus proving (\ref{to prove}) is equivalent to verifying
$$-1-(q-2)-\cdots-(q-2)^{k-1}\le\Delta_t\le1+(q-2)+\cdots+(q-2)^{k+1}+(q-1)(q-2)^k.$$
Since the second inequality follows immediately from the inductive hypothesis, we only need to prove the first inequality. Let $u\in\{0,1,\cdots,q-1\}$ and $v\in\{0,1,\cdots,q^{k-1}-1\}$ such that $t=uq^{k-1}+v$. By (1) \textcircled{\tiny{2}} we get
$$\Delta_t=\Delta_u\Delta_{q^{k-1}}+\delta_u\Delta_v.$$
Since $\delta_0=\delta_1=\cdots=\delta_{q-2}=+1$, $\delta_{q-1}=-1$ and $0\le u\le q-1$ imply $\Delta_u=u\ge0$, $\Delta_q=q-2$ and $\Delta_{q^{k-1}}=\Delta_q^{k-1}=(q-2)^{k-1}\ge0$, by $\delta_u\in\{+1,-1\}$ we get
$$\Delta_t\ge-|\Delta_v|\ge-1-(q-2)-\cdots-(q-2)^{k-1}$$
where the last inequality follows from the inductive hypothesis.
\item[\boxed{\ge}] Let $\delta_1=\delta_2=\cdots=\delta_{q-2}=+1$ and $\delta_{q-1}=-1$. It suffices to prove that for all $k\in\N$ we have
\begin{equation}\label{to prove example}
\Delta_{q^k-q^{k-1}-\cdots-q-1}=(q-2)^k+\cdots+(q-2)^2+(q-2)+1.
\end{equation}
(By induction) For $k=1$ we have $\Delta_{q-1}=q-1$. Suppose that (\ref{to prove example}) is true for some $k\in\N$. Then for $k+1$, we have
\begin{eqnarray*}
\Delta_{q^{k+1}-q^k-q^{k-1}-\cdots-q-1}&=&\Delta_{(q-2)q^k+(q^k-q^{k-1}-\cdots-q-1)}\\
&=&\Delta_{q-2}\Delta_{q^k}+\delta_{q-2}\Delta_{q^k-q^{k-1}-\cdots-q-1}\\
&=&(q-2)\Delta_q^k+(q-2)^k+\cdots+(q-2)^2+(q-2)+1\\
&=&(q-2)^{k+1}+(q-2)^k+\cdots+(q-2)^2+(q-2)+1
\end{eqnarray*}
where the second equality follows from (1) \textcircled{\tiny{2}} and the third equality follows from \textcircled{\tiny{1}} and the inductive hypothesis.
\end{itemize}
(3) For $n\in\N$ large enough, there exists $k\in\N$ large enough such that $q^k+1\le n\le q^{k+1}$. By (2) \textcircled{\tiny{3}} we get
$$|\Delta_n|\le1+(q-2)+\cdots+(q-2)^{k+1}\le(q-1)^k=(q^k)^{\log_q(q-1)}\le n^{\log_q(q-1)}$$
where the second inequality can be verified straightforwardly for $k$ large enough.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-con}] Since (2) follows in the same way as in the proof of \cite[Lemma 4.1]{ARS19} by applying (1), we only need to prove (1) in the following.
\newline\boxed{\Rightarrow} Suppose that $\prod_{n=1}^\infty (R(n))^{\delta_n}$ converges. Then $(R(n))^{\delta_n}\to1$ as $n\to\infty$. Since $\delta_n\in\{+1,-1\}$ for all $n\in\N$, we get $R(n)\to1$ as $n\to\infty$. Thus the numerator and the denominator of $R$ have the same degree and the same leading coefficient.
\newline\boxed{\Leftarrow} Suppose that the numerator and the denominator of $R$ have the same leading coefficient and the same degree. Decompose them into factors of degree $1$. To prove that $\prod_{n=1}^\infty (R(n))^{\delta_n}$ converges, it suffices to show that $\prod_{n=1}^\infty(\frac{n+a}{n+b})^{\delta_n}$ converges for all $a,b\in\C$ satisfying $n+a\neq0$ and $n+b\neq0$ for all $n\in\N$ (that is, $a,b\in\C\setminus\{-1,-2,-3,\cdots\}$). Since $(\frac{n+a}{n+b})^{\delta_n}\to1$ as $n\to\infty$, we only need to prove that
$$\prod_{n=1}^\infty\Big(\big(\frac{qn+a}{qn+b}\big)^{\delta_{qn}}\big(\frac{qn+1+a}{qn+1+b}\big)^{\delta_{qn+1}}\cdots\big(\frac{qn+q-1+a}{qn+q-1+b}\big)^{\delta_{qn+q-1}}\Big)$$
converges. Since Proposition \ref{partial-sum} (1) implies $\delta_{qn}=\delta_n\delta_0$, $\delta_{qn+1}=\delta_n\delta_1$, $\cdots$, $\delta_{qn+q-1}=\delta_n\delta_{q-1}$, it suffices to show that
$$\prod_{n=1}^\infty\big(r(n)\big)^{\delta_n}$$
converges, where
$$r(n):=\big(\frac{qn+a}{qn+b}\big)^{\delta_0}\big(\frac{qn+1+a}{qn+1+b}\big)^{\delta_1}\cdots\big(\frac{qn+q-1+a}{qn+q-1+b}\big)^{\delta_{q-1}}.$$
This is equivalent to show that
\begin{equation}\label{sum}
\sum_{n=1}^\infty\delta_n\ln r(n)
\end{equation}
converges. Since there exist $c_0,c_1,\cdots,c_{q-1},d_0,d_1,\cdots,d_{q-1}\in\C$ such that
\begin{small}
$$r(n)=\frac{q^qn^q+c_{q-1}n^{q-1}+\cdots+c_1n+c_0}{q^qn^q+d_{q-1}n^{q-1}+\cdots+d_1n+d_0}=1+\frac{(c_{q-1}-d_{q-1})n^{q-1}+\cdots+(c_1-d_1)n+(c_0-d_0)}{q^qn^q+d_{q-1}n^{q-1}+\cdots+d_1n+d_0},$$
\end{small}
we get
$$\ln r(n)-\frac{c_{q-1}-d_{q-1}}{q^qn}=\cO(\frac{1}{n^2}),$$
which implies that
$$\sum_{n=1}^\infty\delta_n\big(\ln r(n)-\frac{c_{q-1}-d_{q-1}}{q^qn}\big)$$
converges absolutely. In order to prove that (\ref{sum}) converges, we only need to show that
$$\sum_{n=1}^\infty\frac{\delta_n}{n}$$
converges. Enlightened by partial summation (see for example the equality (6.5) in \cite{BMS20} related to the Thue-Morse sequence), we consider the following \textcircled{\tiny{1}} and \textcircled{\tiny{2}}, which complete the proof.
\begin{itemize}
\item[\textcircled{\tiny{1}}] Prove that
$$\sum_{n=1}^\infty\frac{\delta_1+\cdots+\delta_n}{n(n+1)}$$
converges. In fact, since Proposition \ref{partial-sum} (3) implies
$$\frac{|\Delta_n|}{n^2}\le\frac{1}{n^{2-\log_q(q-1)}}\quad\text{for all }n\text{ large enough},$$
where $2-\log_q(q-1)>1$, it follows that $\sum_{n=1}^\infty\frac{\Delta_n}{n^2}$ converges absolutely. So does $\sum_{n=1}^\infty\frac{\Delta_n}{n(n+1)}$. Thus we only need to check that $\sum_{n=1}^\infty(\frac{\delta_1+\cdots+\delta_n}{n(n+1)}-\frac{\Delta_n}{n(n+1)})$ converges. This follows immediately from $|\delta_1+\cdots+\delta_n-\Delta_n|=|\delta_n-\delta_0|\le2$.
\item[\textcircled{\tiny{2}}] Prove that
$$\sum_{n=1}^\infty(\frac{\delta_n}{n}-\frac{\delta_1+\cdots+\delta_n}{n(n+1)})$$
converges to $0$. In fact, for all $N\in\N$ we have
\begin{eqnarray*}
\sum_{n=1}^N\frac{\delta_1+\cdots+\delta_n}{n(n+1)}&=&\sum_{n=1}^N(\delta_1+\cdots+\delta_n)(\frac{1}{n}-\frac{1}{n+1})\\
&=&\delta_1\sum_{n=1}^N(\frac{1}{n}-\frac{1}{n+1})+\delta_2\sum_{n=2}^N(\frac{1}{n}-\frac{1}{n+1})+\cdots+\delta_N\sum_{n=N}^N(\frac{1}{n}-\frac{1}{n+1})\\
&=&\delta_1(1-\frac{1}{N+1})+\delta_2(\frac{1}{2}-\frac{1}{N+1})+\cdots+\delta_N(\frac{1}{N}-\frac{1}{N+1})\\
&=&\sum_{n=1}^N\frac{\delta_n}{n}-\frac{\delta_1+\delta_2+\cdots+\delta_N}{N+1},
\end{eqnarray*}
which implies
$$\sum_{n=1}^N(\frac{\delta_n}{n}-\frac{\delta_1+\cdots+\delta_n}{n(n+1)})=\frac{\delta_1+\delta_2+\cdots+\delta_N}{N+1}=\frac{\Delta_{N+1}-1}{N+1}.$$
Since Proposition \ref{partial-sum} (3) implies
$$\frac{|\Delta_{N+1}|}{N+1}\le\frac{1}{(N+1)^{1-\log_q(q-1)}}\quad\text{for all }N\text{ large enough},$$
where $1-\log_q(q-1)>0$, as $N\to\infty$ we get $\frac{\Delta_{N+1}}{N+1}\to0$ and then $\sum_{n=1}^N(\frac{\delta_n}{n}-\frac{\delta_1+\cdots+\delta_n}{n(n+1)})\to0$.
\end{itemize}
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-1st}] Since Proposition \ref{partial-sum} (1) implies $\delta_{qn}=\delta_n\delta_0,\delta_{qn+1}=\delta_n\delta_1,\cdots,\delta_{qn+q-1}=\delta_n\delta_{q-1}$ for all $n\in\N_0$, we get $f(a,b)$
\begin{small}
$$\begin{aligned}
&=\prod_{n=1}^\infty\Big(\frac{qn+a}{qn+b}\Big)^{\delta_{qn}}\prod_{n=0}^\infty\Big(\frac{qn+1+a}{qn+1+b}\Big)^{\delta_{qn+1}}\cdots\prod_{n=0}^\infty\Big(\frac{qn+q-1+a}{qn+q-1+b}\Big)^{\delta_{qn+q-1}}\\
&=\prod_{n=1}^\infty\Big(\frac{qn+a}{qn+b}\Big)^{\delta_n\delta_0}\prod_{n=0}^\infty\Big(\frac{qn+a+1}{qn+b+1}\Big)^{\delta_n\delta_1}\cdots\prod_{n=0}^\infty\Big(\frac{qn+a+q-1}{qn+b+q-1}\Big)^{\delta_n\delta_{q-1}}\\
&=\Big(\frac{a+1}{b+1}\Big)^{\delta_0\delta_1}\cdots\Big(\frac{a+q-1}{b+q-1}\Big)^{\delta_0\delta_{q-1}}\prod_{n=1}^\infty\Big(\frac{n+\frac{a}{q}}{n+\frac{b}{q}}\Big)^{\delta_n\delta_0}\prod_{n=1}^\infty\Big(\frac{n+\frac{a+1}{q}}{n+\frac{b+1}{q}}\Big)^{\delta_n\delta_1}\cdots\prod_{n=1}^\infty\Big(\frac{n+\frac{a+q-1}{q}}{n+\frac{b+q-1}{q}}\Big)^{\delta_n\delta_{q-1}}\\
&=\Big(\frac{a+1}{b+1}\Big)^{\delta_1}\cdots\Big(\frac{a+q-1}{b+q-1}\Big)^{\delta_{q-1}}f(\frac{a}{q},\frac{b}{q})\Big(f(\frac{a+1}{q},\frac{b+1}{q})\Big)^{\delta_1}\cdots\Big(f(\frac{a+q-1}{q},\frac{b+q-1}{q})\Big)^{\delta_{q-1}}.
\end{aligned}$$
\end{small}
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor1-1}] (1) follows from Theorem \ref{main-1st} after multiplying by the factor corresponding to $n=0$. The first equality in (2) follows from taking $b=a+1$ in (1). The second equality in (2) follows from taking $b=0$ in Theorem \ref{main-1st} and then multiplying the factor corresponding to $n=0$. We should note that it does not follow from taking $b=0$ in (1). Finally (3) follows immediately from taking $a=1$ in the second equality in (2).
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor1-2}] These two equalities follow from Corollary \ref{cor1-1} (3) and the first equality in (2) of Corollary \ref{cor1-1} respectively.
\end{proof}

\begin{proof}[Proof of Example \ref{ex1}] (1) follows from Corollary \ref{cor1-1} (1).
\begin{itemize}
\item[(2)]\textcircled{\tiny{1}} and \textcircled{\tiny{2}} follow from Corollary \ref{cor1-1} (2).
\item[]\textcircled{\tiny{3}} follows from \textcircled{\tiny{2}} and the fact that the first equality in Corollary \ref{cor1-2} implies
\begin{equation}\label{above}
\prod_{n=0}^\infty\Big(\frac{3n+2}{3n+3}\Big)^{\delta_n}=\frac{1}{\sqrt{3}}.
\end{equation}
\item[]\textcircled{\tiny{4}} follows from taking $b=\frac{1}{2}$ in (1).
\item[(3)] \textcircled{\tiny{1}} is the above equality (\ref{above}).
\item[]\textcircled{\tiny{2}}, \textcircled{\tiny{5}}, \textcircled{\tiny{9}} and \textcircled{\tiny{13}} follow from taking $a=-\frac{1}{2},\frac{3}{2},1$ and $2$ respectively in (2) \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{3}}, \textcircled{\tiny{6}}, \textcircled{\tiny{8}} and \textcircled{\tiny{16}} follow from taking $a=\frac{1}{2},\frac{2}{3},\frac{3}{2}$ and $-\frac{1}{2}$ respectively in (2) \textcircled{\tiny{2}}.
\item[]\textcircled{\tiny{4}} follows from multiplying \textcircled{\tiny{3}} and \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{7}} follows from taking $a=\frac{5}{6}$ in (2) \textcircled{\tiny{4}}.
\item[]\textcircled{\tiny{10}}, \textcircled{\tiny{11}}, \textcircled{\tiny{12}} and \textcircled{\tiny{14}} follow respectively from \textcircled{\tiny{9}}, \textcircled{\tiny{10}}, \textcircled{\tiny{11}} and \textcircled{\tiny{13}} by applying \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{15}} follows from taking $a=2$ and $b=\frac{4}{3}$ in (1).
\end{itemize}
\end{proof}

\begin{proof}[Proof of Example \ref{ex2}] (1) follows from Corollary \ref{cor1-1} (1).
\begin{itemize}
\item[(2)]\textcircled{\tiny{1}} and \textcircled{\tiny{2}} follow from Corollary \ref{cor1-1} (2).
\item[]\textcircled{\tiny{3}} follows from \textcircled{\tiny{2}} and the fact that the first equality in Corollary \ref{cor1-2} implies
$$\prod_{n=0}^\infty\Big(\frac{3n+1}{3n+3}\Big)^{\delta_n}=\frac{1}{\sqrt{3}}.$$
\item[(3)]\textcircled{\tiny{1}} is the above equality.
\item[]\textcircled{\tiny{2}} follows from taking $a=2$ in (2) \textcircled{\tiny{3}}.
\item[]\textcircled{\tiny{3}} and \textcircled{\tiny{8}} follow from taking $a=\frac{3}{2}$ and $2$ respectively in (2) \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{4}}, \textcircled{\tiny{5}}, \textcircled{\tiny{11}} and \textcircled{\tiny{16}} follow from taking $a=\frac{1}{3},\frac{2}{3},3$ and $\frac{3}{2}$ respectively in (2) \textcircled{\tiny{2}}.
\item[]\textcircled{\tiny{6}}, \textcircled{\tiny{9}} and \textcircled{\tiny{10}} follow respectively from \textcircled{\tiny{5}}, \textcircled{\tiny{8}} and \textcircled{\tiny{9}} by applying \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{7}} follows from taking $a=\frac{5}{8}$ and $b=\frac{7}{8}$ in (1).
\item[]\textcircled{\tiny{12}} and \textcircled{\tiny{13}} follow respectively from multiplying and dividing \textcircled{\tiny{11}} by \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{14}} follows from combining the results of taking $a=\frac{1}{2}$ and $-\frac{1}{2}$ in (2) \textcircled{\tiny{2}}.
\item[]\textcircled{\tiny{15}} follows from taking $a=1,b=\frac{1}{2}$ in (1) and then multiplying by \textcircled{\tiny{1}}.
\end{itemize}
\end{proof}

\begin{proof}[Proof of Example \ref{ex3}] For odd $q\ge3$, let $\theta_1=\theta_3=\cdots=\theta_{q-2}=1$ and $\theta_2=\theta_4=\cdots=\theta_{q-1}=0$. Then the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence $(\theta_n)_{n\ge0}$ is exactly $(01)^\infty$. It follows that $\delta_n:=(-1)^{\theta_n}=(-1)^n$ for all $n\ge0$.
\newline(1) By the second equality in Corollary \ref{cor1-1} (2) we get
\begin{footnotesize}
\begin{equation}\label{q}
\prod_{n=0}^\infty\Big(\frac{(qn+qa)(qn+a+1)(qn+2)(qn+a+3)(qn+4)\cdots(qn+a+q-2)(qn+q-1)}{(qn+a)(qn+1)(qn+a+2)(qn+3)(qn+a+4)\cdots(qn+q-2)(qn+a+q-1)}\Big)^{(-1)^n}=q
\end{equation}
\end{footnotesize}
for all $a\in\C\setminus\{0,-1,-2,\cdots\}$. Then we conclude (1) by taking $a=1$ in (\ref{q}).
\newline(2) follows from (\ref{q}) and (1).
\newline(3) Note that for all $q\in\N$ and $a\in\C\setminus\{0,-1,-2,\cdots\}$ we have
\begin{equation}\label{qa}
\prod_{n=0}^\infty\Big(\frac{(qn+a)(qn+a+q)}{(qn+qa)(qn+qa+q)}\Big)^{(-1)^n}=\frac{1}{q}
\end{equation}
since the left hand side is
$$\lim_{k\to\infty}\frac{a}{qa}\cdot\frac{a+q}{qa+q}\cdot\Big(\frac{a+q}{qa+q}\cdot\frac{a+2q}{qa+2q}\Big)^{-1}\cdot\frac{a+2q}{qa+2q}\cdot\frac{a+3q}{qa+3q}\cdots\Big(\frac{a+kq}{qa+kq}\cdot\frac{a+(k+1)q}{qa+(k+1)q}\Big)^{(-1)^k}$$
$$=\lim_{k\to\infty}\frac{a}{qa}\cdot\Big(\frac{a+(k+1)q}{qa+(k+1)q}\Big)^{(-1)^k}=\frac{1}{q}.$$
We prove the concrete equalities in the following.
\newline\textcircled{\tiny{1}} and \textcircled{\tiny{13}} follow from taking $q=3$ and $5$ respectively in (1).
\newline\textcircled{\tiny{2}}, \textcircled{\tiny{3}}, \textcircled{\tiny{5}} and \textcircled{\tiny{7}} follow from taking $q=3$, and then $a=3,2,\frac{2}{3}$ and $\frac{1}{3}$ respectively in (2).
\newline\textcircled{\tiny{4}}, \textcircled{\tiny{9}}, \textcircled{\tiny{10}} and \textcircled{\tiny{12}} are deduced by applying \textcircled{\tiny{1}} noting that (\ref{qa}) with $q=3$ and then $a=2,\frac{2}{3},\frac{1}{3}$ and $-\frac{1}{3}$ give respectively
    $$\prod_{n=0}^\infty\Big(\frac{(3n+2)(3n+5)}{(3n+6)(3n+9)}\Big)^{(-1)^n}=\frac{1}{3},\quad\prod_{n=0}^\infty\Big(\frac{(9n+2)(9n+11)}{(9n+6)(9n+15)}\Big)^{(-1)^n}=\frac{1}{3},$$
    $$\prod_{n=0}^\infty\Big(\frac{(9n+1)(9n+10)}{(9n+3)(9n+12)}\Big)^{(-1)^n}=\frac{1}{3}\quad\text{and}\quad\prod_{n=0}^\infty\Big(\frac{(9n-1)(9n+8)}{(9n-3)(9n+6)}\Big)^{(-1)^n}=\frac{1}{3}.$$
\newline\textcircled{\tiny{6}}, \textcircled{\tiny{8}}, \textcircled{\tiny{11}} and \textcircled{\tiny{16}} follow respectively from dividing \textcircled{\tiny{5}} by \textcircled{\tiny{1}}, multiplying \textcircled{\tiny{7}} by \textcircled{\tiny{1}}, dividing \textcircled{\tiny{9}} by \textcircled{\tiny{6}} and dividing \textcircled{\tiny{15}} by \textcircled{\tiny{13}}.
\newline\textcircled{\tiny{14}} and \textcircled{\tiny{15}} follow from taking $q=5$, and then $a=\frac{1}{2}$ and $3$ respectively in (2).
\end{proof}

Before proving Theorem \ref{main-2nd}, we need the following proposition.

\begin{proposition}\label{theta qn+k} Let $q\ge2$ be an integer, $\theta_0=0$, $(\theta_1,\cdots,\theta_{q-1})\in\{0,1\}^{q-1}\setminus\{0^{q-1}\}$ and $(\theta_n)_{n\ge0}$ be the $(0,\theta_1,\cdots,\theta_{q-1})$-Thue-Morse sequence. Then for all $n\in\N_0$ and $k\in\{0,1,\cdots,q-1\}$ we have
$$\theta_{nq+k}=\theta_n(-1)^{\theta_k}+\theta_k.$$
\end{proposition}
\begin{proof} Let $h$ denote the morphism
$$0\mapsto0\theta_1\cdots\theta_{q-1}$$
$$1\mapsto1\overline{\theta}_1\cdots\overline{\theta}_{q-1}$$
where $\overline{0}:=1$ and $\overline{1}:=0$. By $h(\theta_0\theta_1\theta_2\cdots)=\theta_0\theta_1\theta_2\cdots$ we get
$$h(\theta_n)=\theta_{nq}\theta_{nq+1}\cdots\theta_{nq+q-1}$$
for all $n\in\N_0$. It follows from $h(0)=\theta_0\theta_1\cdots\theta_{q-1}$ and $h(1)=\overline{\theta}_0\overline{\theta}_1\cdots\overline{\theta}_{q-1}$ that
$$\theta_{nq+k}=\left\{\begin{array}{ll}
\theta_k & \text{if } \theta_n=0\\
\overline{\theta}_k & \text{if } \theta_n=1
\end{array}\right.\quad=\theta_n(-1)^{\theta_k}+\theta_k\quad\text{for all }k\in\{0,1,\cdots,q-1\}.$$
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-2nd}] We have $\mathfrak{f}(a_1,\cdots,a_d;b_1,\cdots,b_d)$
\begin{eqnarray*}
&=&\prod_{n=1}^\infty\Big(\prod_{i=1}^d\frac{n+a_i}{n+b_i}\Big)^{\theta_n}\\
&=&\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{k+a_i}{k+b_i}\Big)^{\theta_k}\cdot\prod_{n=1}^\infty\prod_{k=0}^{q-1}\Big(\prod_{i=1}^d\frac{nq+k+a_i}{nq+k+b_i}\Big)^{\theta_{nq+k}}\\
&\overset{(\star)}{=}&\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{a_i+k}{b_i+k}\Big)^{\theta_k}\cdot\prod_{n=1}^\infty\prod_{k=0}^{q-1}\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_n(-1)^{\theta_k}+\theta_k}\\
&=&\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{a_i+k}{b_i+k}\Big)^{\theta_k}\cdot\prod_{n=1}^\infty\prod_{k=0}^{q-1}\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_k}\cdot\prod_{n=1}^\infty\prod_{k=0}^{q-1}\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_n(-1)^{\theta_k}}\\
&\overset{(\star\star)}{=}&\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{a_i+k}{b_i+k}\Big)^{\theta_k}\cdot\prod_{n=1}^\infty\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_k}\cdot\prod_{k=0}^{q-1}\prod_{n=1}^\infty\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_n(-1)^{\theta_k}}\\
&=&\prod_{n=0}^\infty\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_k}\cdot\prod_{k=0}^{q-1}\Big(\prod_{n=1}^\infty\Big(\prod_{i=1}^d\frac{qn+a_i+k}{qn+b_i+k}\Big)^{\theta_n}\Big)^{(-1)^{\theta_k}}\\
&=&\prod_{k=1}^{q-1}\Big(\prod_{n=0}^\infty\prod_{i=1}^d\frac{n+\frac{a_i+k}{q}}{n+\frac{b_i+k}{q}}\Big)^{\theta_k}\cdot\prod_{k=0}^{q-1}\Big(\prod_{n=1}^\infty\Big(\prod_{i=1}^d\frac{n+\frac{a_i+k}{q}}{n+\frac{b_i+k}{q}}\Big)^{\theta_n}\Big)^{(-1)^{\theta_k}}\\
&\overset{(\star\star\star)}{=}&\prod_{k=1}^{q-1}\Big(\prod_{i=1}^d\frac{\Gamma(\frac{b_i+k}{q})}{\Gamma(\frac{a_i+k}{q})}\Big)^{\theta_k}\cdot\prod_{k=0}^{q-1}\Big(\mathfrak{f}\big(\frac{a_1+k}{q},\cdots,\frac{a_d+k}{q};\frac{b_1+k}{q},\cdots,\frac{b_d+k}{q}\big)\Big)^{(-1)^{\theta_k}},
\end{eqnarray*}
where ($\star$), ($\star\star$) and ($\star\star\star$) follow from Proposition \ref{theta qn+k}, $\theta_0=0$ and Theorem \ref{Gamma prod} respectively.
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor2-1}] (1) follows from taking $d=2$, $a_1=a$, $a_2=b+c$, $b_1=b$ and $b_2=a+c$ in Theorem \ref{main-2nd}.
\newline(2) follows from taking $b_1=\cdots=b_d=0$ in Theorem \ref{main-2nd}.
\newline(3) follows from taking $d=2$, $a_1=a$ and $a_2=-a$ in (2).
\end{proof}

\begin{proof}[Proof of Corollary \ref{TM}] In the following proof, for calculations related to the Gamma function, we use Proposition \ref{Gamma properties} frequently without citations. (1) and (2) follow from Theorem \ref{main-2nd} and Corollary \ref{cor2-1} (1) respectively.
\begin{itemize}
\item[(3)]\textcircled{\tiny{1}} follows from taking $b=0$ in (2) and then replacing all $c$ by $b$.
\item[]\textcircled{\tiny{2}} follows from taking $c=a-1$ in (2) and then replacing all $a$ by $a+1$.
\item[(4)]\textcircled{\tiny{1}} follows from multiplying (3) \textcircled{\tiny{1}} and \textcircled{\tiny{2}}.
\item[]\textcircled{\tiny{2}} follows from taking $b=2$ in (3) \textcircled{\tiny{2}}.
\item[]\textcircled{\tiny{3}} and \textcircled{\tiny{4}} follow from taking $b=-a$ and $1-2a$ respectively in (3) \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{5}} follows from taking $d=2,a_1=\frac{1+a}{2},a_2=\frac{1-a}{2},b_1=0$ and $b_2=1$ in (1).
\item[]\textcircled{\tiny{6}} follows from taking $a_1=\cdots=a_d=1,b_1=d$ and $b_2=\cdots=b_d=0$ in (1).
\item[(5)]\textcircled{\tiny{1}} follows from taking $a=\frac{1}{2}$ in (4) \textcircled{\tiny{3}}.
\item[]\textcircled{\tiny{2}} and \textcircled{\tiny{6}} follow from taking $a=0$ and $\frac{1}{3}$ respectively in (4) \textcircled{\tiny{5}}.
\item[]\textcircled{\tiny{3}}, \textcircled{\tiny{5}}, \textcircled{\tiny{7}}, \textcircled{\tiny{11}} and \textcircled{\tiny{12}} follow from taking $a=\frac{1}{2}$, $1$, $2$, $-\frac{2}{3}$ and $-\frac{1}{4}$ respectively in (4) \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{4}}, \textcircled{\tiny{15}} and \textcircled{\tiny{16}} follow from taking $a=\frac{1}{4}$, $\frac{2}{3}$ and $\frac{2}{5}$ respectively in (4) \textcircled{\tiny{4}}.
\item[]\textcircled{\tiny{8}}, \textcircled{\tiny{9}} and \textcircled{\tiny{10}} follow respectively from multiplying \textcircled{\tiny{1}} by \textcircled{\tiny{2}}, multiplying \textcircled{\tiny{5}} by \textcircled{\tiny{8}} and dividing \textcircled{\tiny{2}} by \textcircled{\tiny{1}}.
\item[]\textcircled{\tiny{13}} and \textcircled{\tiny{14}} follow from taking $a=\frac{3}{2}$ and $\frac{1}{3}$ respectively in (4) \textcircled{\tiny{2}}.
\end{itemize}
\end{proof}

\begin{ack}
The author thanks Prof. Jean-Paul Allouche for his advices, and thanks the Oversea Study Program of Guangzhou Elite Project (GEP) for financial support (JY201815). While the author was preparing this paper, he learned that Dr. Shuo Li was working on infinite products related to $\phi$-Thue-Morse sequence, which is another generalization of the classical Thue-Morse sequence, and in particular for the classical one, some new equalities are obtained.
\end{ack}

config: arxiv


TITLE: Is the given function injective, surjective?
QUESTION [0 upvotes]: The following function is obviously injective, surjective and
  bijective, but I'm not really sure how to prove this. N = {0,1,2,...}

$f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, $f(x,y)=x+y$

REPLY [4 votes]: In this case, the function is clearly surjective, yes. Note that for any $n \in \mathbb{N}$ you have $(0,n) \in \mathbb{N} \times \mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)\neq(0,3)$, thus not injective (and in turn, not bijective).


TITLE: How many employees are there in the office
QUESTION [1 upvotes]: In an office, $37$ people like $X$ drink, $32$ people like $Y$ drink and $41$ people like $Z$ drink. Also, $5$ employees like all the types of drinks and $15$ people like at least two types of drinks. If each of the employees like at least one type of drink , then how many employees are there in the office?

Question says that $15$ people like at least two types of drinks. Mathematically does it mean
$n(X \cap Y)+n(Y \cap Z)+n(Z \cap X)+n(X \cap Y \cap Z)=15$?

REPLY [0 votes]: No. Your expression:
$$n(X \cap Y)+n(Y \cap Z)+n(Z \cap X)+n(X \cap Y \cap Z)=15$$
counts the three-way intersection a total of $4$ times. Instead you need to remove it twice from the total of the three two-way intersections:
$$n(X \cap Y)+n(Y \cap Z)+n(Z \cap X)-2\times n(X \cap Y \cap Z)=15$$


TITLE: Linear Programming Problem/Integer Programming Problem Bounds
QUESTION [0 upvotes]: Consider the Integer Programming Problem (IPP) $$\text{minimize}\, \langle c,x\rangle \\ \text{subject to}\, Ax \geq b,\\ x \geq 0, \quad x-\text{integer} $$
in which the matrix $A$ has integer entries. I need to prove that its optimal value is not smaller than the optimal value of the Linear Programming Problem (LPP) $$\text{maximize}\, \langle\lceil b \rceil , \lambda\rangle \\ \text{subject to}\, A^{T}\lambda  \leq c\\ \lambda \geq 0$$
where $\lceil b \rceil$ denotes the smallest integer vector greater than or equal to $b$, and $A^T$ denotes the transpose of the matrix $A$.
In attempting to prove this myself, I have approached it from myriad ways - one of which was to find the dual of the IPP, which is the following:
$$ \text{maximize} \langle b, \lambda \rangle \\ \text{subject to}\,A^{T}\leq c, \\ \lambda \geq 0 $$
but has only $b$ and not $\lceil b \rceil$. So clearly, I cannot say that a duality relation exists between the IPP and the LPP.
From here, I had thought of maybe trying to relax the integer condition of the IPP so that it becomes a standard LPP and then trying to find the dual, but 1) I don't know how to do that, and 2) I'm not sure what good it would do me. So, I am kind of stuck.
I also thought to myself, if I can show that the LPP here is the dual to a primal problem almost identical to the IPP given here, except with $A$ not necessarily having integral entries, and with $\lceil b \rceil$ instead of just $b$, I would by weak duality have that $d^{*} \leq p^{*}$ (where $d^{*}$ is the optimal solution to the LPP, and p^{*} is the solution to the problem whose dual it is). Then, could I somehow relate $p^{*}$ to the optimal solution of the given IPP?
Could somebody please help me figure out how to proceed? And please be as direct and obvious as possible. I thank you ahead of time for your patience.

REPLY [1 votes]: You can use weak-duality, as you've done, to prove the desired result after noting that the feasible set of (IPP) remains the same when the constraint $Ax \geq b$ is replaced by $Ax \geq \lceil b \rceil$ (why?).


TITLE: Question on Cauchy Estimates.
QUESTION [1 upvotes]: The functions $f_{k}(x) = \sin(kx)$ are infinitely differentiable on $[-1,1]$ and bounded by 1. However their derivatives at 0 are unbounded.
If we have functions $f_{k}(z) = \sin(kz)$ on the unit disc, the Cauchy Estimates provide bounds for $(\frac{\partial}{\partial z }) f_{k}(0)$. Why are these examples not contradictory?
Well I am not sure, but what I think is that the z's are complex numbers and they have an imaginary component. However, on the real numbers, we are looking at functions from two dimensions; so there is only one direction. I am not sure. Try not to work out this problem completely. Thanks guys!!!

REPLY [1 votes]: The Cauchy estimates provide bounds for $f_k'(0)$ for each $k\in\mathbb N$. There is no contradiction between that and the assertion that the set $\{f_k'(0)\mid k\in\mathbb N\}$ is unbounded.
Besides, you wrote that $f_k$ is bounded by $1$. If that means that $(\forall z\in\mathbb C):\bigl\lvert\sin(kz)\bigr\rvert\leqslant1$, then you are wrong. In fact, no $f_k$ is a bounded function.


TITLE: Solving Optics problem with and without differentiation result in different results
QUESTION [1 upvotes]: In a recent class of physics-calculus, I and my friends were discussing about a certain problem about optical lens. The problem was: "I a certain laboratory setup, there are a lens of focus $f$, an object of distance $s$ from the lens, and a resulting image of distance $s'$ from the lens. The object is moving at a speed of $v$ from the lens. Find the speed of the resulting image, $v'$.

Our class went on and solve the problem. However, we didn't come to the same conclusion. Here is where we do agree:
$$
\frac{1}{f}=\frac{1}{s}+\frac{1}{s'} \\
\frac{1}{s'}=\frac{1}{f}-\frac{1}{s} \\
\frac{1}{s'}=\frac{s-f}{sf} \\
s' = \frac{sf}{s-f}
$$
This is where the solutions diverge:
Solution 1
$$
\frac{d}{dt}s' = \frac{d}{dt}\frac{sf}{s-f} \\
\frac{d}{dt}s' = f\frac{(s-f)v-sv}{(s-f)^2} \\
v' = \frac{-f^2v}{(s-f)^2} \\
$$
Solution 2
From: $ v=s/t$
$$
v' = \frac{s'}{t}=\frac{sf}{t(s-f)} \\
v' = \frac{vf}{s-f}
$$
Both solutions are consistent with dimensional analysis, surprisingly.
So, how does one explain this inconsistency?

REPLY [3 votes]: Solution 2 has a problem right at the beginning: the correct definition of velocity is $v=\frac{ds}{dt}$, not $v=\frac st$. The two expressions are, in general, only equivalent for objects moving at a constant velocity (and located at $s=0$ at time $t=0$). But in your system, $s'$ is not a linear function of $s$, so the object and the image cannot both be moving at a constant velocity.
Solution 1 looks reasonable at a glance, but it's a bit hard to follow, so let me just work out the solution with more detail and see what comes out.  We'll start with:
$$v' = \frac{d}{dt} s' = \frac{d}{dt} \frac{sf}{s-f}$$
Clearly, since $f$ is constant, $\frac{d}{dt} sf = vf$ and $\frac{d}{dt} (s-f) = \frac{d}{dt} s = v$.  Thus, applying the quotient rule, we have:
$$v' = \frac{vf \cdot (s-f) - sf \cdot v}{(s-f)^2} = \frac{vfs - vf^2 - sfv}{(s-f)^2} = \frac{-vf^2}{(s-f)^2} = -v \left(\frac{f}{s-f}\right)^2$$
just as in your solution 1.


TITLE: What does the symbol $\forall$ mean?
QUESTION [3 upvotes]: What does the symbol $\forall$ mean?  I am sure this is an easy question, but it's not really conducive for googling because I don't know the name of the symbol.  If you can just give me the name or a link where I can read about it that will probably be more than sufficient.
Context:
In the coursera course Introduction to Mathematical Thinking
by Dr. Keith Devlin I came across this sentence in the set supplement reading:

If we have to prove that the sets $A$ and $B$ are equal, we usually split the proof into two parts: $(a)$ Show that every member of $A$ is a member of $B$. $(b)$ Show that every member of $B$ is a member of $A$.
  Taken together, $(a)$ and $(b)$ clearly imply $A = B$. (The proof of both $(a)$ and $(b)$ is usually of the ‘take an arbitrary element’ variety. To prove $(a)$, for instance, we must prove $(\forall x \in A)(x \in B)$; so we take an arbitrary element $x$ of $A$ and show that $x$ must be an element of $B$.)

Thanks!

REPLY [4 votes]: It means "for all". Look up "universal quantifier" on Wikipedia.

REPLY [3 votes]: It means "for all"/"for each"/"for every." For example, $$(\forall x\in A)(x\in B)$$ translates to "For all $x\in A,$ $x\in B,$" meaning every element of $A$ is an element of $B$.

REPLY [2 votes]: $\forall$ = "For all",  $\exists$ = "There exists", Any other questions?


TITLE: Consistency of definition of weak derivative with classical derivative
QUESTION [3 upvotes]: I know the definitions of both weak and classical derivative. But I am trying to see the classical derivative as a weak derivative. When we have
$\int f' \varphi = -\int f\varphi'$ for all $\varphi\in C^\infty$ with compact support. Is this definition consistent with the one for classical derivative, I mean, if $f\in C^1$ how can we obtain $f'$(classical) from the def. above?

REPLY [2 votes]: I am guessing Mark's question is that how we know that the classical derivative is the weak derivative as functions. Or, more precisely, if there's a function $g$ (say in $L^p$) such that $\int g \phi = \int f' \phi = - \int f \phi'$ for all test functions $\phi$, is it necessary that $f' = g$ (almost everywhere)?
The question boils down to whether two functions that coincide when viewed as distributions are in fact the same as functions. Being the same as distributions is a priori a weaker notion than pointwise equality as functions. But one might resort to an approximation argument by taking the test function sequence $\{ \phi_n \}$ approaching the Dirac function in some sense.
Given a test function $\phi$, we can construct a convolution approximate unit $\{ \phi_t\}$ by defining $\phi_t(x) = \frac{1}{t} \phi(\frac{x}{t})$.
Fact: If $|\phi(x)| \leq \frac{C}{(1+|x|)^{1+\epsilon}}$ for some $C, \epsilon >0$, $\int \phi = 1$, and $f \in L^p$ ($1 \leq p \leq \infty$), then $f * \phi_t \rightarrow f$ pointwise on the Lebesgue set of $f$. In particular, we have pointwise convergence almost everywhere and where $f$ is continuous.
Such a $\phi$ with suitable decay can always be found. So if $f' \in L^p$ for some $p$ then any $g \in L^p$ that acts as the weak derivative of $f$ must be equal to $f'$ almost everywhere.
As for "computation", you can compute $f'$ in the usual way and quote uniqueness above.
Hope this helps.


TITLE: Finding the norm of a complex trigonometric function?
QUESTION [0 upvotes]: Given that the complex norm $|z| = 1$, how would I go about proving that $|cos(z)| \leq e$?  Just a hint would be helpful.

REPLY [1 votes]: $z=e^{i\theta}=\cos\theta+i\sin\theta$.
$\cos z=\frac12(e^{ie^{i\theta}}+e^{-ie^{i\theta}})=\ldots$

REPLY [0 votes]: Write out $$\cos{z} = \frac{e^{iz} + e^{-iz}}{2} $$ and apply the triangle inequality.
 There will be some minor details to work out, but they should be fairly straightforward :)


TITLE: Write ODE in Polar Coordinates
QUESTION [2 upvotes]: I want to write this ODE system in polar coordinates (r,$\theta$).
$$\dot x =x-y-x^3 $$
$$\dot y = x+y-y^3$$

REPLY [5 votes]: We use:
$$x = r \cos \theta, y = r \sin \theta, r^2 = x^2 + y^2, \theta = \tan^{-1}\left(\frac{y}{x}\right)$$
From:
$$x^2 + y^2 = r^2 \implies x x' + y y' = r r'$$
We have:
$$\begin{align} x x' + y y' &= x(x-y-x^3) + y(x+y-y^3) \\ &= r \cos \theta(r \cos \theta-r \sin \theta - (r \cos \theta)^3) + r \sin \theta(r \cos \theta + r \sin \theta -(r \sin \theta)^3) \\&= r r'\end{align}$$
We also have:
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) \implies \theta' = \dfrac{xy'-yx'}{r^2}$$
Make the appropriate substitutions, do some trig simplifications and cleanup to derive $r'$ and $\theta'$.
Update
You should end up with:
$$r' = -\frac{1}{4} r \left(r^2 \cos (4 \theta)+3 r^2-4\right)$$
$$\theta' = -\left(\left(r^2-1\right) \cos (2 \theta)+\sin (2 \theta)\right)$$


TITLE: Convergence of series with trigonometric function and factorial
QUESTION [0 upvotes]: I'm stuck on this.. Given $\sum_{n=1}^{\infty}\frac{n!}{n^n}\cdot sin(n^2)$ I have to determine if it's convergent or not.
I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $\frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test
Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test
Any hints?

REPLY [1 votes]: Hint
$$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot...\cdot\frac{n}{n}\leq \frac{1}{n}\cdot\frac{2}{n}\cdot 1 \cdot 1 \cdot... \cdot 1=\frac{2}{n^2}$$


TITLE: What is the coefficient of this term in multinomial expansion?
QUESTION [2 upvotes]: I am trying to find the coefficient of the term $(a+d) ^{11} b^4  c^2$
in the expansion of $(a - 2b + 3c + d)^{17}$
I understand the multinomial theorem can be used and tried to solve it that way but got confused because of the $(a + d) ^{11}$ within the term. How would I go about this?

REPLY [1 votes]: Tips:
$$(a-2b+3c+d)^{17}=[(a+d)+(-2)b+3c]^{17}$$
Then use multinomial theorem

REPLY [1 votes]: Write $a-2b+3c+d=a+d-2b+3c$.
Then $(a-2b+3c+d)^{17}=(a+d-2b+3c)^{17}$, and hence the coefficient of $(a+d)^{11}b^4c^2$ is $\frac{17!}{11!4!2!}(-2)^43^2$.

Hope this helps.


TITLE: Stabilisers of maximal ideals over an integral domain
QUESTION [5 upvotes]: Suppose that $R = \mathbb{C}[x_1, \dots, x_n]/I$ is a finitely generated commutative $\mathbb{C}$-algebra which is an integral domain, and suppose that $G \leqslant \text{GL}(n, \mathbb{C})$ is a finite group acting faithfully on $R$. Is it possible for every maximal ideal of $R$ to have a nontrivial stabiliser?

Obviously, by the Nullstellensatz, every maximal ideal has the form $( x_1- \alpha_1, \dots, x_n - \alpha_n)/I$, where $I \subseteq ( x_1- \alpha_1, \dots, x_n - \alpha_n)$. I feel like there's some easy density argument that I'm missing, along the lines of the $\alpha_i$ having to satisfy a polynomial equation.
I originally thought that $S_2$ acting naturally on $\mathbb{C}[x,y]/(x-y)$ gave a counterexample, but here $S_2$ is actually acting trivially! The integral domain hypothesis might not even be necessary, but I thought that I had a counterexample when $R$ wasn't a domain.

REPLY [2 votes]: Let $G$ be a finite group acting faithfully by regular maps on an irreducible complex algebraic affine variety $X$. Then some closed point in $X$ has trivial stabilizer.
Proof: For $g \in G$ other than the identity, the subvariety $\{x \in X \mid g.x=x\}$ is a proper subvariety (a priori, not necessarily irreducible) in $X$, so it has strictly smaller dimension than $X$. An irreducible complex variety cannot be a finite union of smaller dimension subvarieties. So some point in $X$ is outside the union of these fixed point subvarieties.


TITLE: If $\{a_{n}\} \to \pm \infty$, then show that $\liminf a_{n} =\limsup a_{n} = \pm \infty$
QUESTION [0 upvotes]: If we define a cluster point of a sequence $\{a_{n}\}$ to be an extended real number $c$ (possibly $\pm \infty$ ) such that a subsequence of $a_{n}$, say $a_{n_{k}}$ converges to it, then proving that a sequence in $\mathbb{R}$ converges to an extended real $a \in \mathbb{R}$ iff set of cluster points is a singleton reduces to showing that a sequence converges iff all its subsequences converge to the same value.
Right now, I'm having difficulty with the "all the $a_{n_{k}}$ converge to the same value implies $a_{n}$ converges part", specifically in the case where we're dealing with convergence (technically divergence) to $\pm \infty$.
I'll share what I have done so far, but I think it's utter crap:

(Just showing the case for $\infty$, but I had the same exact difficulty [mentioned below the quoted text] for $-\infty$): In order to show that if all subsequences $\{ a_{n_{k}} \}$ of $\{a_{n}\}$ converge to $\infty$ implies that $\{a_{n}  \} \to \infty$, we will assume that $\{a_{n} \} \nrightarrow \infty$ and show that $\exists$ subsequence $\{a_{n_{k}} \}$ that also $\nrightarrow \infty$.

Now, $\{a_{n} \} \nrightarrow \infty$ means that $\exists M \in \mathbb{R}$, $\forall N \in \mathbb{N}$ such that $\exists n \geq N$ giving us $a_{n} \leq M$, so there are infinitely many such $N \leq n$ such that this is the case.
Running through them, we have for $N = 1$, $n_{1} \geq 1$ and $a_{n_{1}}\leq M$; for $N = 2$, $n_{2} \geq 2$, and $a_{n_{2}}\leq M$.  Continuing on in this fashion, we obtain a subsequence $\{ a_{n_{k}}\} \leq M$, $\forall k \in \mathbb{N}$. Then, $\limsup \{a_{n_{k}}\}\neq \infty$, since $\limsup \{a_{n_{k}}\}= \lim_{n\to \infty}[\sup \{ a_{l_{k}}| l \geq n\}]=M$.
Then, for $\liminf\{a_{n_{k}} \} = \lim_{n \to \infty} \{\inf\{a_{l_{k}}|l\geq n\}]$, we have two cases:

$\liminf \{a_{n_{k}}\} = M$ as well, then $\lim_{n \to \infty}\{a_{n_{k}}\} = M \neq \infty$.
If $\liminf \{a_{n_{k}}\}\neq M$, then $\lim_{n \to \infty} \{a_{n_{k}}\}$ still does not converge to $\infty$, since we need $\liminf = \limsup$.

Now, here I had a real problem. I was able to prove that $\{a_{n}\} \to a$ implies $\liminf = \limsup$ for [nonextended] real $a$, but proving it for $a = \pm \infty$ is giving me a lot of problems. So, my question is, could somebody please show me (in detail - pretend like you're talking to a clueless person, which you are) how explicitly to prove that $\{a_{n}\} \to \infty$ implies $\liminf = \limsup$ and $\{a_{n}\} \to -\infty$ implies $\liminf = \limsup$ (as separate cases - please don't just do it for either $\infty$ or $-\infty$ and say that the other case follows; remember, you're talking to a clueless person)
If you can do this for me, I will be eternally, eternally grateful!

REPLY [1 votes]: If $a_n \to \infty$, then clearly $\lim \sup a_n = \infty$. To show $\lim \inf a_n = \infty$, note that for any finite real $C$ you can find $N$ so that $a_n \geq C$ for all $n \geq N$. Thus for any $C$, you have $\lim \inf a_n \geq C$. This shows $\lim \inf a_n = \infty$.
If $a_n \to - \infty$, then clearly $\lim \inf a_n = - \infty$. To show $\lim \sup a_n = - \infty$, note that for any finite real $C$ you can find $N$ so that $a_n \leq C$ for all $n \geq N$. Thus for any $C$, you have $\lim \sup a_n \leq C$. This shows $\lim \sup a_n = - \infty$.


TITLE: Geometric interpretation of Hilbert's Nullstellensatz via gluing of Galois conjugates
QUESTION [2 upvotes]: In Vakil's The Rising Sea, he gives the following version of Hilbert's Nullstellensatz:

Theorem 3.2.6: If $k$ is any field, every maximal ideal of $k[x_1, \dots, x_n]$ has residue field a finite extension of $k$.

He later says the following:

If the field is not algebraically closed, the traditional [closed] points are glued together into clumps by Galois conjugation ... This is a geometric interpretation of Hilbert's Nullstellensatz 3.2.6.

I am trying to understand how this is a geometric interpretation, and I think I am stuck because he hasn't yet explicitly laid out the geometric interpretation of residue fields. My understanding from a few examples is that the closed points of $k[x_1, \dots, x_n]$ are the maximal ideals, and from the Nullstellensatz we get that the quotient $k[x_1, \dots, x_n] / \mathfrak{m}$ is a finite (hence algebraic) extension of $k[x_1, \dots, x_n]$. Somehow this should then tell us that the closed point $\mathfrak{m}$ is generated by irreducible polynomials and therefore corresponds to gluing together Galois conjugates, but I'm not really able to fill in the details.
Could anyone share a clear explanation of how the Nullstellensatz gets interpreted geometrically in terms of gluing Galois conjugates?

REPLY [1 votes]: Let $L/K$ be a finite Galois extension. We have a morphism of $K$-schemes $ \Bbb A^n_{L}\to\Bbb A^n_K$ which induces a map:
$$L^n=\Bbb A^n_{L}(L) \to \Bbb A^n_K(L)$$
This map is surjective: any $K$-algebra homomorphism $K[x_1, \dots, x_n] \to L$ can be factored $K[x_1, \dots, x_n] \to L[x_1, \dots, x_n] \to L$. We have an action of $G:=\mathrm{Gal}(L/K)$ on $\Bbb A^n_{K}(L)$. It's not hard to see that elements in the same orbit get mapped to the same element in $\Bbb A^n_K(L)$
The ring extension $K[x_1, \dots, x_n] \hookrightarrow L[x_1, \dots, x_n]$ is integral (even finite) becaues $K \hookrightarrow L$ is algebraic (finite). Now let $a,b \in L$ such that their associated maximal ideals $\mathfrak{m}_a$ and $\mathfrak{m}_b$ satisfy $k[x_1, \dots, x_n] \cap \mathfrak{m}_a= k[x_1, \dots, x_n] \cap \mathfrak{m}_b$. Take any $f \in \mathfrak{m}_a$. Then by Galois theory, we have
$$\prod_{\sigma \in G} \sigma(f) \in K[x_1, \dots, x_n] \cap \mathfrak{m}_a= K[x_1, \dots, x_n] \cap \mathfrak{m}_b \subset\mathfrak{m}_b$$
Because $\mathfrak{m}_b$ is a prime ideal, we get $\sigma(f) \in \mathfrak{m}_b$ for some $\sigma$, or equivalently $f \in \sigma^{-1}(\mathfrak{m}_b)$. Because this holds for any $f$, we get
$$\mathfrak{m}_a \subset \bigcup_{\sigma \in G} \sigma(\mathfrak{m}_b)$$
The prime avoidance lemma implies that $\mathfrak{m}_a \subset \sigma(\mathfrak{m}_b)$ for some $\sigma$ and by incomparability of prime ideals in integral extensions, we get $\mathfrak{m}_a = \sigma(\mathfrak{m}_b)$. So we have shown that elements in the same fiber are in the same Galois orbit. We can sum this up in a commutative diagram:
$$\require{AMScd} \begin{CD}
L^n @>{\cong}>> \Bbb{A}^n_L(L)\\ @VVV @VVV\\
L^n/G @>>{\cong}> \Bbb{A}^n_K(L)
\end{CD}$$
Now if $\overline{K}/K$ is a fixed algebraic closure (and we assume for simplicity that $K$ is perfect), then we want to take a colimit of this diagram in a suitable sense. This works because of the Hilbert Nullstellensatz you mention: any element of $\Bbb A^n_K(\overline{K})$ is a closed point and hence has residue field a finite extension of $K$ and hence lies in $\Bbb A^n_K(L)$ for some finite Galois extension $L/K$. We obtain $\Bbb A^n_K(\overline{K})=\displaystyle \bigcup \Bbb A^n_K(L)$. We also have that $\overline{K}^n = \displaystyle \bigcup L^n$.
Thus to define a map $\overline{K}^n/\mathrm{Gal}(\overline{K}/K) \to \Bbb A^n_K(\overline{K})$, we can proceed as follows: for an orbit $[x]$ take any representative $x$, then this lies in some $L^n$, then using the diagram at finite level we can map this $L^n \to \Bbb A^n_L(L) \to \Bbb A^n_K(L) \subset \Bbb A^n_K(\overline{K})$. Using the fact that we have a bijection $L^n/G \to \Bbb A^n_K(L)$ for all $L$ and that the commutative diagrams are compatible for varying $L$, one obtains a commutative diagram:
$$\require{AMScd} \begin{CD}
\overline{K}^n @>{\cong}>> \Bbb{A}^n_\overline{K}(\overline{K})\\ @VVV @VVV\\
\overline{K}^n/\mathrm{Gal}(\overline{K}/K) @>>{\cong}> \Bbb{A}^n_K(\overline{K})
\end{CD}$$
Again by the version of the Nullstellensatz in the OP, we see that every closed point of $\Bbb A^n_K$ is in fact $\overline{K}$-rational (and conversely every $\overline{K}$-rational point is closed).


TITLE: Function with divergence, curl and normal trace on boundary equals zero is zero
QUESTION [5 upvotes]: Let $u\in H^1(\Omega)$ with $\nabla\times u=0$ in $\Omega\subset\mathbb{R}^3$ (open bounded domain), $u\times n=0$ on $\partial\Omega$ (where $n$ is a a normal vector to $\partial\Omega$), $\operatorname{div}(u)=0$ in $\Omega$ and $u\cdot n=0$ on $\partial\Omega$.
Prove that $u=0$.
Thanks.

REPLY [1 votes]: I could not comment in the above answer by tuhin, but strangely it assumes $u=0$ to prove $u=0$!!!!
This result is actually far from trivial. It is not necessarily true if you only assume $\Omega$ to be just open and bounded. However, if you assume $\partial\Omega$ to be $C^{2},$ the result is true.
See Theorem 5.21 in the book "The Pullback Equation for Differential Forms" by Csato, Dacorogna, Kneuss,  Progress in Nonlinear Differential Equations and their Applications, 83. Birkhäuser/Springer, New York, 2012.
The estimate, simplified to the case of the question is
$$ \lVert u \rVert_{H^1(\Omega)} \leq C \left( \lVert \operatorname*{curl}u \rVert_{L^2(\Omega)} + \lVert \operatorname*{div}u \rVert_{L^2(\Omega)} + \lVert n \times u \rVert_{H^{1/2}(\partial\Omega)}
+ \lVert n \cdot u \rVert_{L^1(\partial\Omega)} \right).$$
The estimate itself is a corollary of Theorem 2 in Bolik's paper--- Bolik, Jürgen, "H. Weyls Boundary Value Problems for Differential Forms",  Differential Integral Equations 14 (2001), no. 8, 937–952.


TITLE: dimension of a vector space of harmonic functions among polynomials in two variables
QUESTION [1 upvotes]: I'm searching for harmonic functions inside a set of HOMOGENIOUS polynomials in two variables. Let's say that $$P_n = \{\sum_{i+j = n} a_{ij} x^i y^j \quad|\quad a_{ij} \in \mathbb{R}\}$$
Let's write $H_n$ for a set of all harmonic functions from $P_n$.
So, obviously $H_n \subset P_n$.
Show that $dim(H_n) = 2$ for all $n$.
I tried it using induction on n but I couldn't prove it. Then I tried it by separating polynomials like this:
$$ p(x, y) \in P_n \\
p(x, y) = x a(x, y) + y b(x, y)$$
But again couldn't come far, because now $ a(x, y), b(x, y)$ aren't necessarily homogenous.

REPLY [1 votes]: Write $x+iy=z$ so $2x=z+\bar z, 2iy=z-\bar z$ and express the polynomial as $P(z,\bar z)=\sum_{j+k=n} c_{jk}z^k \bar z^j, c_{jk} \in \mathbb C$
Since a polynomial in $z, \bar z$ is harmonic iff all the mixed terms are zero, we get $P(x,y)=P(z, \bar z)=cz^n+d\bar z^n=c(x+iy)^n+d(x-iy)^n, c, d \in \mathbb C$
But the original coefficients are real so $P$ is real hence $P(z, \bar z)=\bar P(z, \bar z)$ or $d=\bar c$ so the complex dimension is $1$ hence the real dimension is $2$ since clearly the polynomials for $c=1,i$ or if you want $\Re z^n, \Im z^n$ are harmonic, homogeneous of degree $n$ and independent over the reals. Done!


TITLE: Why is it okay to decompose forces?
QUESTION [0 upvotes]: I guess the title says it all, it doesn't seem intuitive that we can consider something as abstract as forces to be decomposable vectors. An example would be a mass sliding off an inclined plane, we consider the tangential component of the weight to the plane. What guarantees us in this case that our results will be correct?

REPLY [1 votes]: It is okay to do so because we have experimentally verified that force is indeed a vector (in the sense that it satisfies the triangle rule of addition). For instance (try this), pushing an object with a force of $1 $ N east and a force of $1 $N north simultaneously has the same effect (in terms of the acceleration) as pushing it with a force of $\sqrt 2 \approx 1.4$N north of east.
Another way to think of this: if you pulled (with a force $F$) on a toy train (mass $m$) along a straight railway (at an angle of $\theta$) and the train was somehow restricted to move along the railway and you measured the acceleration of the train, you would get $\frac{F \cos \theta}{m}$. If the railway is at right angles to the original direction, the acceleration is $\frac{F \sin \theta}{m}$. Now, if the railway is in the direction that you're pulling it in, then the acceleration is $\frac{F}{m}$.


TITLE: How would you approximate $y$ using implicit functions
QUESTION [0 upvotes]: The variables $x$ and $y$ satisfy $f(x,y) = x^5 + xy^4 + y + x^2$. In a neighbourhood of (0,0), $y$ is approximately
(i) $1+x+x^2$
(ii) $x+2x^2$
(iii) $-2x^2$
(iv) $-x^2$

I don't know how to proceed using implicit function theorem. I found that $f(y) = 1$, not 0 hence there must exist a neighborhood such that $y=g(x)$ and I can rewrite $f(x,y)=0$ as $f(x,g(x))=0$ and I know the formula for computing derivatives: $f'(x,y)= \frac{-f'_x(x,y)}{f'_y(x,y)}$ but I don't get how I am supposed to use that to solve the problem.

REPLY [1 votes]: If I were you I would find first $m:=\displaystyle\left.\frac{dy}{dx}\right|_{(0,0)}$ and then $a:=\displaystyle\left.\frac{d^2y}{dx^2}\right|_{(0,0)}$.
Then
$$y(x)\approx \frac{a}{2}x^2+mx+0.$$
This is the second order Taylor Series approximation.


TITLE: Probability that some that m points are probable given the probability of subsets.
QUESTION [0 upvotes]: Working in a problem in analysis I came across this combinatorial/discrete probability question. I would appreciate if someone knows how to approach this problem or knows if this problem is related with something else.
Let $X=\left\{  x_{1},x_{2},...,x_{N}\right\}  $ and $m<M.$ We denote the
subsets of size $m$ as $S_{1},...,S_{\binom{N}{m}}.$ Assume $P$ is a real
function such that $\sum_{i=1}^{_{\binom{N}{m}}}P(S_{i})=1.$ Now we define
$p(x_{j})=\sum_{S_{i}\ni x_{j}}P(s_{i}).$
The question is does there exist $m$ points  $x\in X$ such that $p(x)\geq
m/N?$ (if it is not true what if we write $2^{-N/m}$ instead of $m/N$)

REPLY [1 votes]: $$\sum_{j=1}^{N}{p(x_j)} = \sum_{i=1}^{\binom{N}{m}}{m P(S_i)} = m.$$
The first equality is because each $P(S_i)$ appears in exactly $m$ of the $p(x_j)$ terms (since each $S_i$ has $m$ members).
Dividing both ends of the equation by $N$, we see the average of $p(x_j)$ is $\dfrac{m}{N}$. Therefore, for some $j,\; p(x_j) \geq \dfrac{m}{N}$.
Edit #1: A simple counterexample: $N=3, m=2, X=\{1,2,3\}$.
\begin{eqnarray*}
S_1 = \{1,2\} && P(S_1) = 1/2 \\
S_2 = \{1,3\} && P(S_2) = 1/2 \\
S_3 = \{2,3\} && P(S_3) = 0. \\
&& \\
\therefore  \frac{m}{N} = \frac{2}{3} &&\text{ but } p(1) = 1,\; p(2) = \frac{1}{2},\; p(3) = \frac{1}{2}.
\end{eqnarray*}
Edit #2: A more complete answer to the question of does there exist a given number of points $x$ such that $p(x) \geq $ some value.
Firstly, we know that the average of all the $p(x)$ is $\frac{m}{N}$ and also that numbers are in the interval $[0,1]$. If we arrange the $N$ $P(x)$ values in ascending order and they had the following values:
$$\underbrace{a,a,\ldots,a}_{N-k \text{ times}},\underbrace{1,1,\ldots,1}_{k \text{ times}}.$$
then we could say that there exists $k+1$ points $x$ with $p(x) \geq a$ for any arrangement where the numbers had that average and those upper and lower bounds.
Since the average is $\dfrac{m}{N}$, we must have $a = \dfrac{m-k}{N-k}$. This is as long as $k \leq m$, otherwise we get $a \lt 0$.
Note that such an arrangement is achievable in terms of the $P(S_i)$ values as follows. Assume $k \leq m$ and nominate a given $k$ elements of $X$. There are $\binom{N-k}{m-k}$ sets $S_i$ containing all $k$ nominated elements since we need to choose the remaining $m-k$ elements of $S_i$ from $N-k$ choices. Set $P(S_i) = \binom{N-k}{m-k}^{-1}$ for each of these. For all other $S_i$, set $P(S_i) = 0$. Then, if $x$ is one of the $k$ nominated elements, we have $p(x) = 1$. If $y$ is not one of the $k$ nominated elements, it appears in exactly $\binom{N-k-1}{m-k-1}$ of the $S_i$ with positive probability since such an $S_i$ contains the $k$ nominated elements and also $y$ and there are $\binom{N-k-1}{m-k-1}$ ways to choose the remaining $m-k-1$ elements. Therefore $p(y) = \binom{N-k-1}{m-k-1} \bigg/ \binom{N-k}{m-k} = \dfrac{m-k}{N-k}$. This completes the above arrangement.
$\\$
To address your particular question, setting $k=m-1$, we get $a = p(y) = \dfrac{m-(m-1)}{N-(m-1)} =\dfrac{1}{N-m+1}$. Therefore, we can say that, for any assignment of the $P(S_i)$ values, there exists $m$ points $x$ such that $p(x) \geq \dfrac{1}{N-m+1}$.


TITLE: An analogue to Euler's sum of powers conjecture
QUESTION [4 upvotes]: What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \\
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \\
\vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \\
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \\
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \\
\vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.

REPLY [2 votes]: Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, z\in \mathbb{Z}$ with $z\neq \pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $m\geq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1\leq t \leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + \cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $y\neq \pm 1$, which will give a parametrization
$$
(y^{m})^{k} + \cdots + (y^{m})^{k} + (y^{m+1})^{k} + \cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + \cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.


TITLE: Ricci Tensor, Curvature and Scalar Curvature computation from definition
QUESTION [4 upvotes]: I am studying a little of Riemannian geometry and I am having some problem in making the connection between two expressions of Ricci tensor, curvature and scalar curvature.
Well, in the book that I am studying I find that:
Let $L_{X,Y} : T_xM \to T_xM$, we define the Ricci tensor like:
$$Q'(X,Y) = \operatorname{trace} L_{X,Y}.$$
Where $L_{X,Y} = R(\cdot,X)Y$, where $R(Z,X)Y$  is the Riemann tensor.
Thats way, we define the Ricci curvature as $\operatorname{Ricci}(X) = Q'(X,X)$.
So, if we define $$Q(X) = \sum_i R(X,\partial_i)\partial_i$$ where $\{\partial_i\}$ is one orthonormal basis, it is easy to verify that $$Q'(X,Y) = \langle Q(X), Y \rangle,$$ for any $X,Y \in T_xM$.
In particular, if we use $X = \partial_j$ and compute $$Q'(X,X) = \operatorname{Ricci}(X) = \sum_i K(\partial_j, \partial_i),$$ that is, the sum of all secctional curvatures,
it is easy to see that if we denote $S$ the scalar curvature as $S = \operatorname{trace} Q(X)$, we find that $$S = \sum_i \operatorname{Ricci}(\partial_i).$$
My question is, how do I obtain the expression using Christoffel symbols from these definitions? What is the connection between these expressions and the expression in coordinates?
Thanks a lot for the help.
Cordially,
Leonardo

REPLY [4 votes]: Using the definition of the Riemann curvature tensor
$$R_{XY}Z=\nabla_{[X,Y]}Z-[\nabla_X,\nabla_Y]Z,$$
we can compute that
$$R^a_{\;bcd}=\partial_d\Gamma^a_{cb}-\partial_c\Gamma^a_{db}+\Gamma^a_{de}\Gamma^e_{bc}-\Gamma^a_{ce}\Gamma^e_{bd},$$
(note that my convention is $R_{\partial_c\partial_d}\partial_b=R^a_{\;bcd}\partial_a$).  Now using our definitions of the Ricci curvature
$$R_{ab}=R^c_{\;acb}$$
and of the scalar curvature
$$R=R^a_{\;a},$$
you would just substitute and calculate.
I'm not sure I understand your second question, since anything expressed in terms of Christoffel symbols is also an expression in coordinates (recall that $\nabla_{\partial_a}\partial_b=\Gamma^c_{ab}\partial_c$).


TITLE: Pressure derivative of bulk modulus
QUESTION [0 upvotes]: Hi all what is the definition of pressure derivative of bulk modulus if it is a pressure derivative of bulk modulus at zero pressure. if the pressure is zero how it is derivative by pressure?

REPLY [1 votes]: I guess the definition of derivative is still applicable: you take the limit of "modulus at pressure = $\triangle p$ minus modulus at pressure = 0" divided by $\triangle p$ for $\triangle p\rightarrow 0$.