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\begin{document}
\begin{CJK*}{GBK}{song}
\begin{center}
 {\huge\textbf{Some Rapidly Converging Series for $\zeta(2n+1)$ from Abstract Operators}}
 \end{center}
\vspace{0.3cm}

\begin{center}
\rm Guang-Qing Bi
\end{center}

\begin{abstract}
The author derives new family of series representations for the values of the Riemann Zeta function $\zeta(s)$ at positive odd integers. For $n\in\mathbb{N}$, each of these series representing $\zeta(2n+1)$ converges remarkably rapidly with its general term having the order estimate:
$$O(m^{-2k}\cdot k^{-2n+1})\qquad(k\rightarrow\infty;\quad m=3,4,6).$$
The method is based on the mapping relationships between analytic functions and periodic functions using the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$, including the mapping relationships between power series and trigonometric series, if each coefficient of a power series is respectively equal to that of a trigonometric series. Thus we obtain a general method to find the sum of the Dirichlet series of integer variables. By defining the Zeta function in an abstract operators form, we have further generalized these results on the whole complex plane.
\end{abstract}

\footnotetext{\hspace*{-.45cm} \noindent{\bf 2010 \emph{Mathematics Subject Classification}.}\ \ Primary 11M06, 35S05; Secondary 42A24, 33E20, 40A30.\\
\noindent {\bf \emph{Key words and phrases}.}\ \ Riemann Zeta functions, Rapidly convergent series, Pseudo-differential operators, Abstract operators, Fourier series.}

\section{Introduction}
\noindent
\renewcommand\theequation{1.\arabic{equation}}

The trigonometric series, especially the Fourier series, is of great importance in both mathematics and physics. Expanding periodic functions into Fourier series has become a very mature theory. During early years of the last century, researchers had recognized the significance of the inverse problem, which is how to find the sum function of a certain Fourier series.

A class of pseudo-differential operators extended from $\mathscr{S}(\mathbb{R}^n)$ to $C^\infty(\Omega)$ is called the abstract operators such as $\cos(h\partial_x)$ and $\sin(h\partial_x)$. The concept of abstract operators is based on the analytic continuity fundamental theorem contained in a 1997 paper entitled \emph{Applications of abstract operators to partial differential equations} (See \cite[p. 7]{bi97}). In other words, if two abstract operators $A$ and $B$ acts on $e^{\xi{x}},\,x\in\mathbb{R}^n,\,\xi\in\mathbb{R}^n$ such that $Ae^{\xi{x}}=Be^{\xi{x}}$, then $Ag(x)=Bg(x),\,\forall{g(x)}\in{C^\infty(\Omega)},\,\Omega\subset\mathbb{R}^n$. Therefore, the abstract operators $f(\partial_x),\,\partial_x:=(\partial_{x_1},\partial_{x_2},\cdots,\partial_{x_n})$ is defined by (See \cite[p. 8]{bi97})
\begin{equation}\label{abs}
 f(\partial_x)e^{\xi{x}}:=f(\xi)e^{\xi{x}},\quad\forall f(\xi)\in C^\infty(\mathbb{R}^n),
\end{equation}
where $f(\xi),\xi\in\mathbb{R}^n$ is called the symbols of abstract operators $f(\partial_x)$.  From this, we can derive the operation rules of abstract operators, such as
\begin{equation}\label{e0}
f(\partial_x)(e^{\xi_0x}g(x))=g(\partial_\xi)(e^{\xi{x}}f(\xi))|_{\xi=\xi_0},
\end{equation}
\begin{equation}\label{e1}
f(\partial_x)(e^{\xi{x}}g(x))=e^{\xi{x}}f(\xi+\partial_x)g(x),
\end{equation}
\begin{equation}\label{e1'}
 e^{h\partial_x}f(x)=f(x+h),\quad h\partial_x:=h_1\partial_{x_1}+h_2\partial_{x_2}+\cdots+h_n\partial_{x_n}
\end{equation}
and so on. The concept of abstract operators and the theory of partial differential equations expressed in terms of abstract operators have been established in \cite{bi18}-\cite{bi11}. We believe that our findings here can be useful for investing the number theoretical properties for $\zeta(2n+1)$. In this paper we describe the operation rules of the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$, and derive a series of operation formulas for these two operators by their operation rules. In particular, the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$ can express the mapping relations between periodic functions and analytic functions more explicitly. According to these mapping relations, if we know the sum function of the power series, we can obtain that of the trigonometric series, and the non-analytical points of which are also determined at the same time. Thus we can use the sum of power series to obtain that of corresponding trigonometric series. A summation method of trigonometric series expressed by abstract operators is proposed.

In fact, the trigonometric series has been applied to the Riemann Zeta function by many mathematicians such as Choi \cite{Choi}, Katsurada \cite{Kats} and Tsumura \cite{Tsu} etc. By using the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$,  we can see that our summation method of trigonometric series is quite suitable to find the sum of the Dirichlet series of integer variables including the Zeta function. The Riemann Zeta function $\zeta(s)$ is defined by
\begin{equation}\label{00}
  \zeta(s):=\sum^\infty_{n=1}\frac{1}{n^s}\qquad(\Re(s)>1),
\end{equation}
which can be continued analytically to the whole $s$-plane except for a simple pole at $s=1$ with its residue $1$.

Recall $\zeta(0)=-1/2$. In 1749, Euler proved
\begin{equation}\label{0}
  \zeta(2n)=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}\qquad(n\in\mathbb{N}),
\end{equation}
where $B_m$ are the Bernoulli numbers. Since $B_m$ are rational numbers, the expression (\ref{0}) shows that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$. Yet, there is no analogue for $\zeta(2n+1)$. In 1978, Ap\'{e}ry proved that $\zeta(3)$ is irrational. Since then, it is known that infinitely many $\zeta(2n+1)$ are irrational. Instead, many mathematicians have tried to present rapidly converging series for $\zeta(2n+1)$. In this paper, we aim to derive certain new rapidly converging series for $\zeta(2n+1)$, by using the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$. We will see that the concept of abstract operators can be applied in defining Riemann Zeta function on the whole complex plane without the experience of analytic continuation process. So we can extend the related results of $\zeta(2n+1)$ to the whole complex plane.

\section{The summation method of Fourier series expressed by abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$}
\noindent\setcounter{equation}{0}
\renewcommand\theequation{2.\arabic{equation}}

Let $h\partial_x=\langle{h,\partial_x}\rangle=h_1\partial_{x_1}+h_2\partial_{x_2}+\cdots+h_n\partial_{x_n}$. Then $\cos(h\partial_x)$ and $\sin(h\partial_x)$ are the abstract operators taking $\cos(hb)$ and $\sin(hb)$ as the symbols respectively, namely (See \cite[p. 8]{bi97})
\begin{equation}\label{1}
    \cos(h\partial_x)e^{bx}:=\cos(bh)e^{bx},\quad\sin(h\partial_x)e^{bx}:=\sin(bh)e^{bx}.
\end{equation}
Here $bx=b_1x_1+b_2x_2+\cdots+b_nx_n,\;bh=b_1h_1+b_2h_2+\cdots+b_nh_n$. Further, their operation rules can be expressed as the following three groups of operator relationships by Guang-Qing Bi \cite[pp. 7-9]{bi97}:

\textbf{Theorem 2.1.} (See \cite[p. 9, Theorem 3, 4, and 6]{bi97}) Let $x\in\mathbb{R}^n,\;h\in\mathbb{R}^n$, $h\partial_x=\langle{h,\partial_x}\rangle=h_1\partial_{x_1}+\cdots+h_n\partial_{x_n}$. For the abstract operators $\cos(h\partial_x)\,\mbox{and}\,\sin(h\partial_x)$, we have
\begin{equation}\label{y0}
  \cos(h\partial_x)f(x)=\Re[f(x+ih)],\quad\sin(h\partial_x)f(x)=\Im[f(x+ih)],
\end{equation}
$\forall{f(z)}\in{C}^\infty(\Omega),\,z=x+iy\in\Omega\subset\mathbb{C}^n$;

\parbox{11cm}{\begin{eqnarray*}\label{y1}
                \sin(h\partial_x)(uv) &=& \cos(h\partial_x)v\cdot\sin(h\partial_x)u+\sin(h\partial_x)v\cdot\cos(h\partial_x)u,\\
                \cos(h\partial_x)(uv) &=& \cos(h\partial_x)v\cdot\cos(h\partial_x)u-\sin(h\partial_x)v\cdot\sin(h\partial_x)u;
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\parbox{11cm}{\begin{eqnarray*}\label{y2}
                \sin(h\partial_x)\frac{u}{v} &=& \frac{\cos(h\partial_x)v\cdot\sin(h\partial_x)u-\sin(h\partial_x)v\cdot\cos(h\partial_x)u}
                {(\cos(h\partial_x)v)^2+(\sin(h\partial_x)v)^2}, \\
                \cos(h\partial_x)\frac{u}{v} &=& \frac{\cos(h\partial_x)v\cdot\cos(h\partial_x)u+\sin(h\partial_x)v\cdot\sin(h\partial_x)u}
                {(\cos(h\partial_x)v)^2+(\sin(h\partial_x)v)^2}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\textbf{Theorem 2.2.} (See \cite[p. 9, Theorem 5]{bi97}) Let $h_0\in\mathbb{R},\;x(t)\in\mathbb{R}^n,\;t\in\mathbb{R}^1,\;X\in\mathbb{R}^n,\,Y\in\mathbb{R}^n$,
$Y\partial_X=\langle{Y,\partial_X}\rangle=Y_1\partial_{X_1}+\cdots+Y_n\partial_{X_n}$. Then we have

\parbox{11cm}{\begin{eqnarray*}\label{y3}
 \sin(h_0\partial_t)f(x(t)) &=& \sin(Y\partial_X)f(X),\\
 \cos(h_0\partial_t)f(x(t)) &=& \cos(Y\partial_X)f(X),
\end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}
where $X_j=\cos(h_0\partial_t)x_j(t),\;Y_j=\sin(h_0\partial_t)x_j(t),\;j=1,\cdots,n$.

In the special case when $n=1$, (\ref{y3}) can easily be restated as

\parbox{11cm}{\begin{eqnarray*}\label{yb1}
\sin\left(h_0\frac{d}{dt}\right)f(x(t)) &=& \sin\left(Y\frac{\partial}{\partial{X}}\right)f(X),\\
\cos\left(h_0\frac{d}{dt}\right)f(x(t)) &=& \cos\left(Y\frac{\partial}{\partial{X}}\right)f(X),
\end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}
where
$Y=\sin(h_0\partial_t)x(t),\;X=\cos(h_0\partial_t)x(t),\;t\in\mathbb{R}^1,\;h_0\in\mathbb{R}$.

\textbf{Theorem 2.3.} (See \cite[p. 9, Theorem 7]{bi97}) Let $u=g(y)$ be a monotonic function on its domain. If $y=f(bx)$ is the inverse function of $bx=g(y)$ such that $g(f(bx))=bx$, where $bx=b_1x_1+b_2x_2+\cdots+b_nx_n,\;bh=b_1h_1+b_2h_2+\cdots+b_nh_n$, then $\sin(h\partial_x)f(bx)$ (denoted by $Y$) and
$\cos(h\partial_x)f(bx)$ (denoted by $X$) can be determined by the following set of equations:
\begin{equation}\label{y4}
    \left\{\begin{array}{l@{\qquad}l}\displaystyle
    \cos\left(Y\frac{\partial}{\partial{X}}\right)g(X)=bx,&x\in\mathbb{R}^n,\;b\in\mathbb{R}^n,\\\displaystyle
    \sin\left(Y\frac{\partial}{\partial{X}}\right)g(X)=bh, &h\in\mathbb{R}^n.
    \end{array}\right.
\end{equation}

When acting on elementary functions, the abstract operators $\cos(h\partial_x)$ and $\sin(h\partial_x)$ have complete operation formulas as follows:

By using (\ref{y0})  we have

\parbox{10cm}{\begin{eqnarray*}\label{2a}
                \cos(h\partial_x)\sin{bx} &=& \cosh(bh)\sin{bx}, \\
                \sin(h\partial_x)\sin{bx} &=& \sinh(bh)\cos{bx}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\parbox{10cm}{\begin{eqnarray*}\label{2b}
                \cos(h\partial_x)\cos{bx} &=& \cosh(bh)\cos{bx}, \\
                \sin(h\partial_x)\cos{bx} &=& -\sinh(bh)\sin{bx}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

Based on (\ref{2a}) and (\ref{2b}), by using (\ref{y2}) we obtain

\parbox{10cm}{\begin{eqnarray*}\label{2c}
                \cos(h\partial_x)\tan{bx} &=& \frac{\sin(2bx)}{\cosh(2bh)+\cos(2bx)}, \\
                \sin(h\partial_x)\tan{bx} &=& \frac{\sinh(2bh)}{\cosh(2bh)+\cos(2bx)}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\parbox{10cm}{\begin{eqnarray*}\label{4'}
                \cos(h\partial_x)\cot{bx} &=& \frac{\sin(2bx)}{\cosh(2bh)-\cos(2bx)},\\
                \sin(h\partial_x)\cot{bx} &=& \frac{\sinh(2bh)}{\cos(2bx)-\cosh(2bh)}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

By using (\ref{1}) and (\ref{y4}), we have
\begin{eqnarray*}
    \left\{\begin{array}{l@{\qquad}l}\displaystyle
    e^X\cos{Y} = bx, & X=\cos(h\partial_x)\ln(bx),\\\displaystyle
    e^X\sin{Y} = bh, & Y=\sin(h\partial_x)\ln(bx).
    \end{array}\right.
\end{eqnarray*}
By solving this set of equations we obtain

\parbox{10cm}{\begin{eqnarray*}\label{7}
                \cos(h\partial_x)\ln(bx) &=& \frac{1}{2}\ln\left((bx)^2+(bh)^2\right), \\
                \sin(h\partial_x)\ln(bx) &=&\textrm{arccot}\frac{bx}{bh}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

By using (\ref{2c}) and (\ref{y4}), we have
\begin{eqnarray*}
    \left\{\begin{array}{l@{\qquad}l}\displaystyle
    \frac{\sin2X}{\cosh2Y+\cos2X}=bx, & X=\cos(h\partial_x)\arctan{bx},\\\displaystyle
    \frac{\sinh2Y}{\cosh2Y+\cos2X}=bh, & Y=\sin(h\partial_x)\arctan{bx}.
    \end{array}\right.
\end{eqnarray*}
By solving this set of equations we have
\[1+(bx)^2+(bh)^2=1+\frac{\sin^22X+\sinh^22Y}{(\cosh2Y+\cos2X)^2}=\frac{2\cosh2Y}{\cosh2Y+\cos2X}=\frac{2bh\cosh2Y}{\sinh2Y}.\]
and
\[1-(bx)^2-(bh)^2=1-\frac{\sin^22X+\sinh^22Y}{(\cosh2Y+\cos2X)^2}=\frac{2\cos2X}{\cosh2Y+\cos2X}=\frac{2bx\cos2X}{\sin2X}.\]
So we have
\[\tanh2Y=\frac{2bh}{1+(bx)^2+(bh)^2}\quad\mbox{and}\quad\tan2X=\frac{2bx}{1-(bx)^2-(bh)^2}.\]
Thus we obtain

\parbox{10cm}{\begin{eqnarray*}\label{8}
                \sin(h\partial_x)\arctan{bx} &=& \frac{1}{2}\textrm{tanh}^{-1}\frac{2bh}{1+(bx)^2+(bh)^2}, \\
                \cos(h\partial_x)\arctan{bx} &=& \frac{1}{2}\arctan\frac{2bx}{1-(bx)^2-(bh)^2}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

Similarly,

\parbox{10cm}{\begin{eqnarray*}\label{9}
                \sin(h\partial_x)\textrm{arccot}\,bx &=& -\frac{1}{2}\textrm{coth}^{-1}\frac{1+(bx)^2+(bh)^2}{2bh}, \\
                \cos(h\partial_x)\textrm{arccot}\,bx &=& \frac{1}{2}\textrm{arccot}\frac{(bx)^2+(bh)^2-1}{2bx}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\parbox{10cm}{\begin{eqnarray*}\label{11}
                \sin(h\partial_x)\sqrt{bx} &=& \sqrt{\frac{\sqrt{(bx)^2+(bh)^2}-bx}{2}}, \\
                \cos(h\partial_x)\sqrt{bx} &=&\sqrt{\frac{\sqrt{(bx)^2+(bh)^2}+bx}{2}}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

For hyperbolic and inverse hyperbolic functions, we can also derive the corresponding basic formulas. For instance, correspondingly to
(\ref{2a}) and (\ref{2b}), we have

\parbox{10cm}{\begin{eqnarray*}\label{10a}
                \cos(h\partial_x)\sinh{bx} &=& \cos(bh)\sinh{bx}, \\
               \sin(h\partial_x)\sinh{bx} &=& \sin(bh)\cosh{bx}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\parbox{10cm}{\begin{eqnarray*}\label{10b}
                \cos(h\partial_x)\cosh{bx} &=& \cos(bh)\cosh{bx}, \\
                \sin(h\partial_x)\cosh{bx} &=& \sin(bh)\sinh{bx}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

\textbf{Theorem 2.4.}  Let $f(x)\in{L^2}[-c,c]$ be the sum function of the Fourier cosine series, and
$g(x)\in{L^2}[-c,c]$ be that of the corresponding Fourier  sine series, namely
\[ f(x)=\sum^\infty_{n=0}a_n\cos\frac{n\pi{x}}{c}\quad\mbox{and}\quad g(x)=\sum^\infty_{n=0}a_n\sin\frac{n\pi{x}}{c},\]
where $c>0,\;x\in\Omega\subset\mathbb{R}^1$. If $S(t)$ is the sum function of the corresponding power series $\sum^\infty_{n=0}a_nt^n$, namely
\[S(t)=\sum^\infty_{n=0}a_nt^n,\quad(t\in\mathbb{R}^1,\;|t|<r,\;0<r<+\infty),\]
then for $a<x<b$, we have the following mapping relationships:

\parbox{11.5cm}{\begin{eqnarray*}\label{12}
                f(x) &=& \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^z)\right|_{z=0}, \\
                g(x) &=& \left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^z)\right|_{z=0}.
              \end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}
Here interval $(a,b)\subset\Omega$. The endpoints $a$ and $b$ of the interval $a<x<b$ are non-analytical points (singularities) of Fourier series, which can
be uniquely determined by the detailed computation of the right-hand side of (\ref{12}).

\textbf{Proof.} By substituting $S(e^z)=\sum^\infty_{n=0}a_ne^{nz}$ into (\ref{12}), we can prove Theorem 2.4 easily.

\textbf{Note 2.1.} It is easily seen from (\ref{12}) that
\begin{equation}\label{12ct1}
 \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^z)\right|_{z=0}=\sum^\infty_{n=0}a_n\cos\frac{n\pi{x}}{c};
\end{equation}
\begin{equation}\label{12ct2}
 \left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^z)\right|_{z=0}=\sum^\infty_{n=0}a_n\sin\frac{n\pi{x}}{c}.
\end{equation}
Here $\Omega\in\mathbb{R}^1$. The $\Omega$ can be uniquely determined by the detailed computation of the left-hand side of (\ref{12ct1}) and (\ref{12ct2}) respectively.
For example, according to the proof of Lemma 3.2 in this paper, if $S(e^z)=-\ln(1-e^z)$, then we have $\Omega:=\{x\in\mathbb{R}^1|\sin(\pi{x}/(2c))>0\}$ for (\ref{12ct1}) and $\Omega:=\{x\in\mathbb{R}^1|\sin(\pi{x}/(2c))\neq0\}$ for (\ref{12ct2}) respectively.

In the application of Theorem 2.4, the following theorem can be particularly useful:

\textbf{Theorem 2.5.}  Let $S(t)$ be an arbitrary analytic function integrable in the interval $[0,1]$. Then we have
\begin{equation}\label{13}
  \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\int^{e^z}_0\!\!\!S(e^z)\,de^z\right|_{z=0}
 = \int^1_0\!\!S(\xi)\,d\xi-\frac{\pi}{c}\int^x_0\!\!\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)
 [S(e^z)\,e^z]\right|_{z=0}dx;
\end{equation}
\begin{equation}\label{14}
 \left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\int^{e^z}_0\!\!S(e^z)\,de^z\right|_{z=0}=
\frac{\pi}{c}\int^x_0\!\!\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)[S(e^z)\,e^z]\right|_{z=0}dx.
\end{equation}

\textbf{Proof.} According to the analytic continuous fundamental theorem, we only need to prove this set of formulas when $S(t)=t^n$, $n\in\mathbb{N}_0:=\{0,1,2,\cdots\}$, this is obvious.

\textbf{Note 2.2.} The analytic continuous fundamental theorem is contained in a 1997 paper entitled \emph{Applications of abstract operators to partial differential equations} by Guang-Qing Bi \cite[p. 7]{bi97}.

\textbf{Theorem 2.6.}   Let $f(x)\in{L^2}[-c,c]$ be the sum function of the Fourier cosine series, and $g(x)\in{L^2}[-c,c]$ be that of the corresponding Fourier sine series, namely
\[f(x)=\sum^\infty_{n=1}a_n\cos\frac{n\pi{x}}{c}\quad\mbox{and}\quad g(x)=\sum^\infty_{n=1}a_n\sin\frac{n\pi{x}}{c},\]
where $c>0,\;x\in\Omega\subset\mathbb{R}^1$. If $S(t)$ is an analytic function integrable in the interval $[0,1]$, and
$\int^t_0S(t)\,dt$ is the sum function of the corresponding power series $\sum^\infty_{n=1}a_nt^n$, namely
\[\int^t_0\!\!S(t)\,dt=\sum^\infty_{n=1}a_nt^n\qquad(t\in\mathbb{R}^1,\;|t|<r,\;0<r<+\infty),\]
then for $a<x<b$ we have the following mapping relationships:

\parbox{11.5cm}{\begin{eqnarray*}\label{15}
f(x) &=&
\int^1_0\!\!S(\xi)\,d\xi-\frac{\pi}{c}\int^x_0\!\!\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)
[S(e^z)\,e^z]\right|_{z=0}dx, \\
g(x)&=&\frac{\pi}{c}\int^x_0\!\!\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)[S(e^z)\,e^z]\right|_{z=0}dx.
\end{eqnarray*}}\hfill\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}
Here interval $(a,b)\subset\Omega$. The endpoints $a$ and $b$ of the interval $a<x<b$ are non-analytical points (singularities) of Fourier series, which can
be uniquely determined by the detailed computation of the right-hand side of (\ref{15}).

\textbf{Proof.} Combining Theorem 2.4 with Theorem 2.5 will lead us to the proof.

\textbf{Example.}   Let $\omega$ be a positive real number, $f(\omega t)\in{L^2}[-\pi,\pi]$ defined by
$$f(\omega t):=\sum^\infty_{n=1}\frac{(-1)^{n-1}}{(3n-1)(3n+1)}\cos(3n\omega{t})\quad\omega{t}\in\Omega\subset\mathbb{R}^1.$$
Then $f(\omega t)=(\pi/(3\sqrt{3}))\cos\omega{t}-1/2$ for $\omega{t}\in(-\pi/3,\pi/3)\subset\Omega$, namely
\begin{equation}\label{22}
    \cos\omega{t}=\frac{3\sqrt{3}}{\pi}\left[\frac{1}{2}+\sum^\infty_{n=1}\frac{(-1)^{n-1}}{(3n-1)(3n+1)}\cos(3n\omega{t})\right]\qquad
    -\frac{\pi}{3}<\omega{t}<\frac{\pi}{3}.
\end{equation}
Here $\Omega:=\{\omega{t}\in\mathbb{R}^1|2\cos\omega{t}-1\neq0\;\mbox{and}\;\cos(\omega{t}/2)\neq0\}$.

\textbf{Proof.}  By using Theorem 2.4, let
\begin{eqnarray*}
  S(x) &=& \sum^\infty_{n=1}(-1)^{n-1}\frac{x^{3n}}{(3n-1)(3n+1)}\qquad{x}\in\mathbb{R}^1,\;|x|\leq1,\\
  f(\omega t) &=& \sum^\infty_{n=1}\frac{(-1)^{n-1}}{(3n-1)(3n+1)}\cos(3n\omega{t})\qquad\omega{t}\in\Omega\subset\mathbb{R}^1.
\end{eqnarray*}
As it is difficult to obtain the sum function $S(x)$ directly, we can use the operators $\frac{d}{dx}(x\cdot)$ and
$\frac{d}{dx}(\frac{1}{x}\cdot)$ to transform the power series into the series familiar to us, namely
\[\frac{d}{dx}\left(\frac{1}{x}\frac{d}{dx}(xS(x))\right)=\sum^\infty_{n=1}(-1)^{n-1}x^{3n-2}=\frac{x}{1+x^3}.\]
Then we can obtain $S(x)$, and then $f(\omega t)$. To describe concisely, the following results will be given directly:
\begin{eqnarray*}
  S(x) &=& \frac{1}{x}\int^x_0\!\!x\,dx\int^x_0\!\!\frac{x}{1+x^3}\,dx=\left(\frac{x}{12}-\frac{1}{12x}\right)\ln\left(x^2-x+1\right)\\
       & & -\left(\frac{x}{6}-\frac{1}{6x}\right)\ln(1+x)+
       \left(\frac{x}{4}+\frac{1}{4x}\right)\frac{2}{\sqrt{3}}\left(\arctan\frac{2x-1}{\sqrt{3}}+\frac{\pi}{6}\right)-\frac{1}{2}.
\end{eqnarray*}

Applying (\ref{1}), (\ref{y1}), (\ref{yb1}), (\ref{7}), (\ref{8}), (\ref{10a}) and (\ref{10b}), we have
\begin{eqnarray*}

f(\omega t) &=& \left.\cos\left(\omega{t}\frac{\partial}{\partial{z}}\right)S(e^z)\right|_{z=0} \\
&=& \frac{1}{6}\left.\cos\left(\omega{t}\frac{\partial}{\partial{z}}\right)\left[\,\sinh{z}\ln\left(e^{2z}-e^z+1\right)\right]\right|_{z=0}\\
& &-\frac{1}{3}\left.\cos\left(\omega{t}\frac{\partial}{\partial{z}}\right)[\,\sinh{z}\ln\left(1+e^z\right)]\right|_{z=0} \\
& & +\frac{1}{\sqrt{3}}\left.\cos\left(\omega{t}\frac{\partial}{\partial{z}}\right)\left[\,\cosh{z}\left(\arctan\frac{2e^z-1}{\sqrt{3}}
  +\frac{\pi}{6}\right)\right]\right|_{z=0}-\frac{1}{2}\\
&=& -\frac{1}{6}\sin\omega{t}\,\textrm{arccot}\frac{(2\cos\omega{t}-1)\cos\omega{t}}{(2\cos\omega{t}-1)\sin\omega{t}}
+\frac{1}{3}\sin\omega{t}\,\textrm{arccot}\frac{\cos^2(\omega{t}/2)}{\sin(\omega{t}/2)\cos(\omega{t}/2)} \\
& & +\frac{1}{2\sqrt{3}}\cos\omega{t}\arctan\frac{\sqrt{3}(2\cos\omega{t}-1)}{2\cos\omega{t}-1}+\frac{\pi}{6\sqrt{3}}\cos\omega{t}-\frac{1}{2}
\qquad\omega{t}\in\Omega.
\end{eqnarray*}

When $2\cos\omega{t}-1\neq0$ and $\cos(\omega{t}/2)\neq0$, we obtain
\begin{eqnarray*}

  f(\omega t) &=& -\frac{1}{6}\sin\omega{t}\,\textrm{arccot}\cot\omega{t}+\frac{1}{3}\sin\omega{t}\,\textrm{arccot}\cot\frac{\omega{t}}{2} \\
       & & +\frac{1}{2\sqrt{3}}\cos\omega{t}\arctan\sqrt{3}+\frac{\pi}{6\sqrt{3}}\cos\omega{t}-\frac{1}{2}
       =\frac{\pi}{3\sqrt{3}}\cos\omega{t}-\frac{1}{2}.
\end{eqnarray*}
Since $2\cos\omega{t}-1\neq0$ and $\cos(\omega{t}/2)\neq0$, there are four non-analytical points: $\omega{t}=-\pi,-\pi/3,\pi/3,\pi$ in the interval $[-\pi,\pi]$. Thus we obtain (\ref{22}) for $-\pi/3<\omega{t}<\pi/3$.

\textbf{Note 2.3.}  For $f(\omega t)\in{L^2}[-\pi,\pi]$, since $2\cos\omega{t}-1\neq0$ and $\cos(\omega{t}/2)\neq0$, there are countable infinite non-analytical points on $-\infty<\omega{t}<\infty$, namely $\Omega:=\{\omega{t}\in\mathbb{R}^1|2\cos\omega{t}-1\neq0\;\mbox{and}\;\cos(\omega{t}/2)\neq0\}$. The open interval $(-\pi/3,\pi/3)\subset\Omega$.

\section{Preliminary applications}
\noindent\setcounter{equation}{0}
\renewcommand\theequation{3.\arabic{equation}}

Let $S_0(t)$ be a function analytic in the neighborhood of $t=0$ and
\[S_0(t)=\sum^\infty_{n=1}a_nt^n\qquad(t\in\mathbb{R}^1,\;|t|<r,\;0<r<+\infty),\]
where $a_n$ are rational numbers. Then the sum function $S_m(t)$ is defined as
\begin{equation}\label{23}
S_m(t):=\underbrace{\int^t_0\frac{dt}{t}\cdots}_m\int^t_0S_0(t)\,\frac{dt}{t}=\sum^\infty_{n=1}a_n\frac{t^n}{n^m}\qquad m\in\mathbb{N}_0:=\{0,1,2,\cdots\}.
\end{equation}
Apparently $S_m(1)$ is the sum function of the Dirichlet series taking $m$ as the variable.

\textbf{Lemma 3.1.} $S_m(t)$ satisfies the following recurrence relation:
\begin{equation}\label{24}
\int^t_0S_{m-1}(t)\,\frac{dt}{t}=S_m(t)\qquad m\in\mathbb{N}:=\{1,2,\cdots\}.
\end{equation}

\textbf{Lemma 3.2.} Let $x\in\mathbb{R}^1$ with $0<x<2c$. Then
\begin{equation}\label{31a}
\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}=
 \frac{\pi}{2}-\frac{\pi{x}}{2c};
\end{equation}
\begin{equation}\label{31b}
 \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}=
 -\ln\left(2\sin\frac{\pi{x}}{2c}\right).
\end{equation}

\textbf{Proof.} By using (\ref{1}), (\ref{yb1}) and (\ref{7}), we have
\begin{eqnarray*}
   & & \left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}\,=\,
   -\left.\sin\left(Y\frac{\partial}{\partial{X}}\right)\ln{X}\right|_{z=0}\\
   &=& -\left.\textrm{arccot}\frac{X}{Y}\right|_{z=0}=-\left.\arctan\frac{Y}{X}\right|_{z=0}=-\arctan\frac{-\sin(\pi{x}/c)}{1-\cos(\pi{x}/c)}\\
   &=& \arctan\frac{\sin(\pi x/(2c))\cos(\pi x/(2c))}{\sin^2(\pi x/(2c))}.
\end{eqnarray*}
When $\sin(\pi{x}/(2c))\neq0$ or $x\neq0,\;x\neq2c$, the above expression can be written in the form:
\[\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}=
\arctan\cot\frac{\pi{x}}{2c}=\frac{\pi}{2}-\frac{\pi{x}}{2c}\qquad(0<x<2c).\]
Similarly, for $\sin(\pi{x}/(2c))>0$ (or $0<x<2c$), we have
\begin{eqnarray*}
   & & \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}\,=\,
   -\left.\cos\left(Y\frac{\partial}{\partial{X}}\right)\ln{X}\right|_{z=0}\\
   &=& -\left.\frac{1}{2}\ln(X^2+Y^2)\right|_{z=0}= -\frac{1}{2}\ln\left(\left(1-\cos\frac{\pi{x}}{c}\right)^2+\sin^2\frac{\pi{x}}{c}\right)=-\ln\left(2\sin\frac{\pi{x}}{2c}\right).
\end{eqnarray*}

\textbf{Lemma 3.3.} Let $m\in\mathbb{N}$. We have the following integral formula for $\ln{x}$:
\begin{equation}\label{37}
\underbrace{\int^x_0dx\cdots}_m\int^x_0\ln{x}\,dx=\frac{x^m}{m!}\left(\ln{x}-H_m\right),
\end{equation}
where $H_m$ denotes the familiar harmonic numbers defined by
\begin{equation}\label{har}
  H_m:=\sum^m_{j=1}\frac{1}{j}\qquad(m\in\mathbb{N}).
\end{equation}

\textbf{Proof.} Let $x\in\Omega\subset\mathbb{R}^1,\;n\in\mathbb{N}$. We can use the mathematical induction to prove the following formula
\begin{equation}\label{1*}
v\frac{d^nu}{dx^n}=\sum^n_{j=0}(-1)^j{n\choose{j}}\frac{d^{n-j}}{dx^{n-j}}\left(u\frac{d^jv}{dx^j}\right)\quad\forall{v,u}\in{C^n}(\Omega).
\end{equation}

Letting $u=x^n/n!$ in (\ref{1*}), we have
\begin{equation}\label{1*+}
v=\sum^n_{j=0}(-1)^j{n\choose{j}}\frac{d^{n-j}}{dx^{n-j}}\left(\frac{x^n}{n!}\frac{d^jv}{dx^j}\right)\quad\forall{v}\in{C^n}(\Omega).
\end{equation}
Taking $(\int_0^xdx)^n$ of both sides gives
\begin{equation}\label{1*++}
\underbrace{\int^x_0dx\cdots}_n\int^x_0vdx=\sum^n_{j=0}(-1)^j{n\choose{j}}\underbrace{\int^x_0dx\cdots}_j\int^x_0\frac{x^n}{n!}\frac{d^jv}{dx^j}dx
\quad\forall{v}\in{C^n}(\Omega).
\end{equation}
Letting $v=\ln{x}$ in (\ref{1*++}), then it is proved.

\textbf{Lemma 3.4.} Let $x\in\mathbb{R}^1$ with $0<x<\pi$. Then
\begin{equation}\label{37j}
\ln\frac{\sin x}{x}=-\sum^\infty_{k=1}\frac{1}{k}\frac{\zeta(2k)}{\pi^{2k}}x^{2k}.
\end{equation}

\textbf{Theorem 3.1.}  Let $m\in\mathbb{N},\,c>0$. The sum function $S_m(t)$ has the following recurrence property:
\begin{equation}\label{25}
\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_m(e^z)\right|_{z=0}=S_m(1)
-\frac{\pi}{c}\int^x_0\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-1}(e^z)\right|_{z=0}dx,
\end{equation}
\begin{equation}\label{26}
\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_m(e^z)\right|_{z=0}=
\frac{\pi}{c}\int^x_0\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-1}(e^z)\right|_{z=0}dx.
\end{equation}

\textbf{Proof.}  Taking $S(x)=S_{m-1}(x)/x$ in Theorem 2.5, then it is proved by using Lemma 3.1.

\textbf{Theorem 3.2.} For $m,r\in\mathbb{N}$, the sum function $S_m(t)$ has the following recurrence property:
\begin{eqnarray}\label{28}
                & & \left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-1}(e^z)\right|_{z=0}\nonumber\\
                &=& \sum^{r-1}_{k=1}(-1)^{k-1}\frac{1}{(2k-1)!}\left(\frac{\pi{x}}{c}\right)^{2k-1}S_{m-2k}(1)\nonumber\\
                & & +\,(-1)^{r-1}\left(\frac{\pi}{c}\right)^{2r-1}
\underbrace{\int^x_0dx\cdots}_{2r-1}\int^x_0\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-2r}(e^z)\right|_{z=0}dx.
\end{eqnarray}
Here (and elsewhere in this work) an empty sum is to be interpreted (as usual) to be zero.

\textbf{Proof.}  We can use the mathematical induction to prove it. According to (\ref{26}) of Theorem 3.1, it is obviously tenable when $r=1$ in (\ref{28}). Now we inductively hypothesize that it is tenable when $r=K$. Using Theorem 3.1, then it is tenable when $r=K+1$, thus Theorem 3.2 is proved.

Similarly,

\textbf{Theorem 3.3.} For $m,r\in\mathbb{N}$, the sum function $S_m(t)$ has the following recurrence property:
\begin{eqnarray}\label{30}
                & & \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-1}(e^z)\right|_{z=0}\nonumber\\
                &=& \sum^{r-1}_{k=0}(-1)^k\frac{1}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}S_{m-2k-1}(1)\nonumber\\
                & & +\,(-1)^r\left(\frac{\pi}{c}\right)^{2r}
\underbrace{\int^x_0dx\cdots}_{2r}\int^x_0\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S_{m-2r-1}(e^z)\right|_{z=0}dx.
\end{eqnarray}

\textbf{Theorem 3.4.}  Let $r\in\mathbb{N}$, $x\in\mathbb{R}^1$ with $0<x<2c$. Then we have the following Fourier series expressions:
\begin{eqnarray}\label{36}
 \sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}&=&
 \sum^{r-1}_{k=0}(-1)^{k}\frac{1}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^{r-1}\left(\frac{\pi}{c}\right)^{2r}\underbrace{\int^x_0dx\cdots}_{2r}\int^x_0\ln\left(2\sin\frac{\pi{x}}{2c}\right)\,dx
\end{eqnarray}
and
\begin{eqnarray}\label{36j}
 \sum^\infty_{n=1}\frac{1}{n^{2r}}\sin\frac{n\pi{x}}{c}&=&
 \sum^{r-1}_{k=1}(-1)^{k-1}\frac{1}{(2k-1)!}\left(\frac{\pi{x}}{c}\right)^{2k-1}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^r\left(\frac{\pi}{c}\right)^{2r-1}\underbrace{\int^x_0dx\cdots}_{2r-1}\int^x_0\ln\left(2\sin\frac{\pi{x}}{2c}\right)\,dx.
\end{eqnarray}

\textbf{Proof.} By using the relationship (\ref{23}), if $S_0(t)=-\ln(1-t)$, then $S_{2r-2k}(1)=\zeta(2r+1-2k)$. Letting $m=2r+1$ in Theorem 3.3, and by using Theorem 2.4, we have
\begin{eqnarray*}
 & & \sum^{\infty}_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}\\
 &=& \sum^{r-1}_{k=0}(-1)^k\frac{1}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\\
 & & +\,(-1)^r\left(\frac{\pi}{c}\right)^{2r}
\underbrace{\int^x_0dx\cdots}_{2r}\int^x_0\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^z))\right|_{z=0}dx.
\end{eqnarray*}
Then (\ref{36}) is proved using Lemma 3.2. Letting $m=2r,\;S_0(t)=-\ln(1-t)$ in Theorem 3.2, similarly we obtain (\ref{36j}).

In Theorem 3.4, since  $r\in\mathbb{N}$, the Fourier series expressions are also tenable at the endpoints of the interval $0<x<2c$.

\textbf{Theorem 3.5.} Let $r\in\mathbb{N}$, $x\in\mathbb{R}^1$ with $0\leq{x}\leq2c$. Then
\begin{eqnarray}\label{36'}
 \sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}
 &=& \sum^{r-1}_{k=0}(-1)^k\frac{1}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^r\frac{1}{(2r)!}\left(\frac{\pi{x}}{c}\right)^{2r}\left(H_{2r}-\ln\frac{\pi{x}}{c}\right)\nonumber\\
 & & +\,(-1)^r2\sum^\infty_{k=1}\frac{(2k-1)!}{(2r+2k)!}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi{x}}{c}\right)^{2r+2k}
\end{eqnarray}
and
\begin{eqnarray}\label{yl7}
 \sum^\infty_{n=1}\frac{1}{n^{2r}}\sin\frac{n\pi{x}}{c}&=&
 \sum^{r-1}_{k=1}(-1)^{k-1}\frac{1}{(2k-1)!}\left(\frac{\pi{x}}{c}\right)^{2k-1}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^{r-1}\frac{1}{(2r-1)!}\left(\frac{\pi{x}}{c}\right)^{2r-1}\left(H_{2r-1}-\ln\frac{\pi{x}}{c}\right)\nonumber\\
 & & +\,(-1)^{r-1}2\sum^\infty_{k=1}\frac{(2k-1)!}{(2r+2k-1)!}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi x}{c}\right)^{2r+2k-1}.
\end{eqnarray}
Here the formula (\ref{yl7}) is a slightly modified version of a result proven in a significantly different way by Katsurada \cite[p. 81, Theorem 2]{Kats}.

\textbf{Proof.}  According to the expression (\ref{36}) of Theorem 3.4, using Lemma 3.3 and Lemma 3.4, if $r\in\mathbb{N}$, then in the closed interval $[0,2c]$ we have
\begin{eqnarray*}

   & & \sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c} \\
   &=& \sum^{r-1}_{k=0}(-1)^{k}\frac{1}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)
   +\frac{(-1)^{r-1}}{(2r)!}\left(\frac{\pi{x}}{c}\right)^{2r}\ln2 \\
   & & +\,(-1)^{r-1}\left(\frac{\pi}{c}\right)^{2r}\underbrace{\int^x_0dx\cdots}_{2r}\int^x_0\left(\ln\frac{\pi{x}}{2c}
   +\ln\frac{\sin\frac{\pi{x}}{2c}}{\frac{\pi{x}}{2c}}\right)\,dx \\
   &=&\sum^{r-1}_{k=0}\frac{(-1)^{k}}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)
   +\frac{(-1)^{r}}{(2r)!}\left(\frac{\pi{x}}{c}\right)^{2r}\left(H_{2r}-\ln\frac{\pi{x}}{c}\right)\\
   & &+\,(-1)^{r-1}\left(\frac{\pi}{c}\right)^{2r}\underbrace{\int^x_0dx\cdots}_{2r}\int^x_0
   -\sum^\infty_{k=1}\frac{1}{k}\frac{\zeta(2k)}{\pi^{2k}}\left(\frac{\pi{x}}{2c}\right)^{2k}\,dx\\
   &=& \sum^{r-1}_{k=0}\frac{(-1)^{k}}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)
   +\frac{(-1)^{r}}{(2r)!}\left(\frac{\pi{x}}{c}\right)^{2r}\left(H_{2r}-\ln\frac{\pi{x}}{c}\right)\\
   & &+\,(-1)^r\sum^\infty_{k=1}\frac{1}{k}\frac{(2k)!\,\zeta(2k)}{(2\pi)^{2k}(2r+2k)!}\left(\frac{\pi x}{c}\right)^{2r+2k}.
\end{eqnarray*}
Thus (\ref{36'}) is proved. According to expression (\ref{36j}), similarly we have (\ref{yl7}) by using Lemma 3.3 and Lemma 3.4.

\section{Main results}
\noindent\setcounter{equation}{0}
\renewcommand\theequation{4.\arabic{equation}}

\textbf{Lemma 4.1.} (See Srivastava and Tsumura \cite[p. 329, Lemma 4]{sri}) For $\Re(s)>1$
\begin{equation}\label{yc2}
\sum^\infty_{n=1}\frac{1}{n^s}\cos\frac{2n\pi}{3}=\frac{1}{2}(3^{1-s}-1)\zeta(s);
\end{equation}
\begin{equation}\label{yc4}
\sum^\infty_{n=1}\frac{1}{n^s}\cos\frac{n\pi}{2}=2^{-s}(2^{1-s}-1)\zeta(s);
\end{equation}
\begin{equation}\label{yc1}
\sum^\infty_{n=1}\frac{1}{n^s}\cos\frac{n\pi}{3}=\frac{1}{2}(6^{1-s}-3^{1-s}-2^{1-s}+1)\zeta(s).
\end{equation}

\textbf{Lemma 4.2.} Let $r\in\mathbb{N}$, $x\in\mathbb{R}^1$ with $-2c\leq{x}\leq2c$ ($c>0$ be a given real number). Then we have the following Fourier series relationships:
\begin{eqnarray}\label{yl8}
 & & r\sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}+\frac{\pi x}{2c}\sum^\infty_{n=1}\frac{1}{n^{2r}}\sin\frac{n\pi x}{c}\nonumber\\
 &=& \sum^{r-1}_{k=0}(-1)^k\,\frac{r-k}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^{r-1}\sum^\infty_{k=0}\frac{(2k)!}{(2r+2k)!}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi{x}}{c}\right)^{2r+2k},
\end{eqnarray}
which is a slightly modified version of a result proven in a significantly different way by Katsurada \cite[p. 81, Theorem 1]{Kats}.

\textbf{Proof.} By $\frac{\pi x}{c}\times(\ref{yl7})+2r\times(\ref{36'})$, since $\zeta(0)=-1/2$ and
\[\frac{(2k)!}{(2r+2k)!}=\frac{(2k-1)!}{(2r+2k-1)!}-2r\frac{(2k-1)!}{(2r+2k)!}\qquad(r,k\in\mathbb{N}),\]
then we can prove (\ref{yl8}).

By applying Lemma 4.2 with $x=c$, we have
\begin{eqnarray}\label{yl11}
 \zeta(2r+1)&=&\frac{2^{2r}}{r(2^{2r+1}-1)}\sum^{r-1}_{k=1}(-1)^{k-1}\frac{(r-k)\pi^{2k}}{(2k)!}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^r\frac{(2\pi)^{2r}}{r(2^{2r+1}-1)}\sum^\infty_{k=0}\frac{(2k)!}{(2r+2k)!}\frac{\zeta(2k)}{2^{2k}}\quad(r\in\mathbb{N}),
\end{eqnarray}
which was given earlier by Cvijovi\'{c} and Klinowski \cite[p. 1265, Theorem A]{Cvij}. In particular, when $r=1$, (\ref{yl11}) immediately yields Ewell's formula:
\[\zeta(3)=-\frac{4\pi^2}{7}\sum^\infty_{k=0}\frac{\zeta(2k)}{(2k+1)(2k+2)2^{2k}},\]
which was found by Euler in 1772 (See, e.g., \cite[p. 1080, Section 7]{Ay}), was rediscovered by Ramaswami \cite{Ra} and by Ewell \cite{Ew}.

Since
\[\zeta(2k)=(-1)^{k-1}\frac{(2\pi)^{2k}}{2(2k)!}B_{2k}\quad\mbox{and}\quad\zeta(-2k+1)=-\frac{1}{2k}B_{2k}\quad(k\in\mathbb{N}),\]
the last term on the right-hand side of (\ref{yl8}) can be written in the form:
\begin{eqnarray*}

  & & \sum^\infty_{k=r}(-1)^k\,\frac{r-k}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k) \\
  &=& \lim_{k\rightarrow0}a_k+\sum^\infty_{k=1}(-1)^{r+k}\,\frac{-k}{(2r+2k)!}\left(\frac{\pi{x}}{c}\right)^{2r+2k}\zeta(-2k+1) \\
  &=& \sum^\infty_{k=0}(-1)^{r+k}\,\frac{B_{2k}}{2(2r+2k)!}\left(\frac{\pi{x}}{c}\right)^{2r+2k} \\
  &=& (-1)^{r-1}\sum^\infty_{k=0}\frac{(2k)!}{(2r+2k)!}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi{x}}{c}\right)^{2r+2k},
\end{eqnarray*}
where
\[\lim_{k\rightarrow0}a_k=\lim_{k\rightarrow0}\left[(-1)^{r+k}\,\frac{-k}{(2r+2k)!}\left(\frac{\pi{x}}{c}\right)^{2r+2k}\zeta(-2k+1)\right]
=(-1)^r\,\frac{B_0}{2(2r)!}\left(\frac{\pi{x}}{c}\right)^{2r}.\]
Thus the Katsurada's formula (\ref{yl8}) and  Cvijovi\'{c}-Klinowski's formula (\ref{yl11}) can be written in the form:
\begin{eqnarray}\label{yl8g}
 & & r\sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}+\frac{\pi x}{2c}\sum^\infty_{n=1}\frac{1}{n^{2r}}\sin\frac{n\pi x}{c}\nonumber\\
 &=& \sum^\infty_{k=0}(-1)^k\,\frac{r-k}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\quad(r\in\mathbb{N},\,|x|\leq2c);
\end{eqnarray}
\begin{equation}\label{yl11g}
 \zeta(2r+1)=\frac{2^{2r}}{r(2^{2r+1}-1)}\sum^\infty_{k=1}(-1)^{k-1}\frac{(r-k)\pi^{2k}}{(2k)!}\zeta(2r+1-2k)\quad(r\in\mathbb{N}).
\end{equation}

Since each series in (\ref{yl8}) is uniformly convergent with respect to $x$ on the interval $[-2c,2c]$ for $r\in\mathbb{N}$, multiplying the both side of (\ref{yl8}) by $(\pi x/c)^{2r-1}$, and executing termwise differentiation in them with respect to $x$, namely
$\partial_x[(\pi x/c)^{2r-1}\times(\ref{yl8})]$, then the sine series term is counteracted. Thus we can obtain the following theorem:

\textbf{Theorem 4.1.} Let $r\in\mathbb{N}$, $x\in\mathbb{R}^1$ with $-2c\leq{x}<2c$ (or $-2c\leq x\leq2c$ for $r\in\mathbb{N}\setminus\{1\}$). Then we have the following Fourier series relationship:
\begin{eqnarray}\label{yl15}
 & & r(2r-1)\sum^\infty_{n=1}\frac{1}{n^{2r+1}}\cos\frac{n\pi{x}}{c}+\frac{1}{2}\left(\frac{\pi x}{c}\right)^2\sum^\infty_{n=1}\frac{1}{n^{2r-1}}\cos\frac{n\pi x}{c}\nonumber\\
 &=& \sum^{r-1}_{k=0}(-1)^k\frac{(r-k)(2r+2k-1)}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(2r+1-2k)\nonumber\\
 & & +\,(-1)^{r-1}\sum^\infty_{k=0}\frac{(2k)!(4r+2k-1)}{(2r+2k)!}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi{x}}{c}\right)^{2r+2k}.
\end{eqnarray}

In the special cases of (\ref{yl15}) when $r=1$, since
\begin{equation}\label{yl16}
  \sum^\infty_{n=1}\frac{1}{n}\cos\frac{n\pi{x}}{c}=-\ln\left(2\sin\frac{\pi x}{2c}\right)\qquad(0<x<2c),
\end{equation}
the Theorem 4.1 readily yield

\textbf{Theorem 4.2.} Let $x\in\mathbb{R}^1$ with $0<x<2c$. Then
\begin{eqnarray}\label{yl17}
 & & \sum^\infty_{n=1}\frac{1}{n^3}\cos\frac{n\pi{x}}{c}-\frac{1}{2}\left(\frac{\pi x}{c}\right)^2\ln\left(2\sin\frac{\pi x}{2c}\right)\nonumber\\
 &=& \zeta(3)+\sum^\infty_{k=0}\frac{2k+3}{(2k+1)(2k+2)}\frac{\zeta(2k)}{(2\pi)^{2k}}\left(\frac{\pi{x}}{c}\right)^{2k+2}.
\end{eqnarray}

\textbf{Theorem 4.3.} We have the following series representations for $\zeta(3)$
\begin{equation}\label{yl18}
  \zeta(3)=-\frac{\pi^2}{13}\ln3-\frac{4\pi^2}{13}\sum^\infty_{k=0}\frac{2k+3}{(2k+1)(2k+2)}\frac{\zeta(2k)}{3^{2k}};
\end{equation}
\begin{equation}\label{yl19}
  \zeta(3)=-\frac{2\pi^2}{35}\ln2-\frac{8\pi^2}{35}\sum^\infty_{k=0}\frac{2k+3}{(2k+1)(2k+2)}\frac{\zeta(2k)}{4^{2k}};
\end{equation}
\begin{equation}\label{yl20}
  \zeta(3)=-\frac{\pi^2}{6}\sum^\infty_{k=0}\frac{2k+3}{(2k+1)(2k+2)}\frac{\zeta(2k)}{6^{2k}}.
\end{equation}

\textbf{Proof.} Letting $x=2c/3$ in (\ref{yl17}), and by using (\ref{yc2}) we have (\ref{yl18}). Similarly, we can prove (\ref{yl19}) and (\ref{yl20}).

\textbf{Theorem 4.4.} For $r\in\mathbb{N}\setminus\{1\}$
\begin{eqnarray}\label{yl21}
 \zeta(2r+1)&=& -\frac{2\pi^2(3^{2r-2}-1)}{r(2r-1)(3^{2r+1}-1)}\,\zeta(2r-1)+\frac{2\times3^{2r}}{r(2r-1)(3^{2r+1}-1)}\nonumber\\
 & & \times\left[\,\sum^{r-1}_{k=1}(-1)^{k-1}\frac{(r-k)(2r+2k-1)}{(2k)!}\left(\frac{2\pi}{3}\right)^{2k}\zeta(2r+1-2k)\right.\nonumber\\
 & & \left.+\,(-1)^r\left(\frac{2\pi}{3}\right)^{2r}\sum^\infty_{k=0}\frac{(2k)!(4r+2k-1)}{(2r+2k)!}\frac{\zeta(2k)}{3^{2k}}\right].
\end{eqnarray}

\textbf{Proof.} Letting $x=2c/3$ in (\ref{yl15}), and by using (\ref{yc2}) we have (\ref{yl21}).

When $r=2$ and $3$ we obtain
\begin{equation}\label{yl24}
  \zeta(5)=\frac{41\pi^2}{363}\,\zeta(3)+\frac{8\pi^4}{363}\sum^\infty_{k=0}\frac{(2k+7)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+4)3^{2k}}.
\end{equation}
\begin{equation}\label{yl25}
\zeta(7)=\frac{2188\pi^2}{16395}\,\zeta(5)-\frac{18\pi^4}{5465}\,\zeta(3)
-\frac{64\pi^6}{16395}\sum^\infty_{k=0}\frac{(2k+11)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+6)3^{2k}}.
\end{equation}

\textbf{Theorem 4.5.} For $r\in\mathbb{N}\setminus\{1\}$
\begin{eqnarray}\label{yl22}
 \zeta(2r+1)&=& -\frac{(2^{2r-1}-2)\pi^2}{r(2r-1)(2^{4r+1}+2^{2r}-1)}\,\zeta(2r-1)+\frac{2^{4r+1}}{r(2r-1)(2^{4r+1}+2^{2r}-1)}\nonumber\\
 & & \times\left[\,\sum^{r-1}_{k=1}(-1)^{k-1}\frac{(r-k)(2r+2k-1)}{(2k)!}\left(\frac{\pi}{2}\right)^{2k}\zeta(2r+1-2k)\right.\nonumber\\
 & & \left.+\,(-1)^r\left(\frac{\pi}{2}\right)^{2r}\sum^\infty_{k=0}\frac{(2k)!(4r+2k-1)}{(2r+2k)!}\frac{\zeta(2k)}{4^{2k}}\right].
\end{eqnarray}

\textbf{Proof.} Letting $x=c/2$ in (\ref{yl15}), and by using (\ref{yc4}) we have (\ref{yl22}).

When $r=2$ and $3$ we obtain
\begin{equation}\label{yl26}
  \zeta(5)=\frac{157\pi^2}{1581}\,\zeta(3)+\frac{16\pi^4}{1581}\sum^\infty_{k=0}\frac{(2k+7)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+4)4^{2k}}.
\end{equation}
\begin{equation}\label{yl27}
\zeta(7)=\frac{14306\pi^2}{123825}\,\zeta(5)-\frac{64\pi^4}{41275}\,\zeta(3)
-\frac{128\pi^6}{123825}\sum^\infty_{k=0}\frac{(2k+11)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+6)4^{2k}}.
\end{equation}

\textbf{Theorem 4.6.} For $r\in\mathbb{N}\setminus\{1\}$
\begin{eqnarray}\label{yl23}
 \zeta(2r+1)&=& \frac{2\pi^2(6^{2r-2}-3^{2r-2}-2^{2r-2}+1)}{r(2r-1)(3^{2r}(2^{2r}+1)+2^{2r}-1)}\,\zeta(2r-1)\nonumber\\
 & & +\,\frac{2\times6^{2r}}{r(2r-1)(3^{2r}(2^{2r}+1)+2^{2r}-1)}\nonumber\\
 & & \times\left[\,\sum^{r-1}_{k=1}(-1)^{k-1}\frac{(r-k)(2r+2k-1)}{(2k)!}\left(\frac{\pi}{3}\right)^{2k}\zeta(2r+1-2k)\right.\nonumber\\
 & & \left.+\,(-1)^r\left(\frac{\pi}{3}\right)^{2r}\sum^\infty_{k=0}\frac{(2k)!(4r+2k-1)}{(2r+2k)!}\frac{\zeta(2k)}{6^{2k}}\right].
\end{eqnarray}

\textbf{Proof.} Letting $x=c/3$ in (\ref{yl15}), and by using (\ref{yc1}) we have (\ref{yl23}).

When $r=2$ and $3$ we obtain
\begin{equation}\label{yl28}
  \zeta(5)=\frac{8\pi^2}{87}\,\zeta(3)+\frac{\pi^4}{261}\sum^\infty_{k=0}\frac{(2k+7)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+4)6^{2k}}.
\end{equation}
\begin{equation}\label{yl29}
\zeta(7)=\frac{3124\pi^2}{29655}\,\zeta(5)-\frac{2\pi^4}{3295}\,\zeta(3)
-\frac{16\pi^6}{88965}\sum^\infty_{k=0}\frac{(2k+11)\zeta(2k)}{(2k+1)(2k+2)\cdots(2k+6)6^{2k}}.
\end{equation}

Since $\zeta(2k)\rightarrow1$ as $k\rightarrow\infty$, the general term in our series representations has the order estimate:
\[O(m^{-2k}\cdot k^{-2r+1})\qquad(k\rightarrow\infty;\quad m=3,4,6;\quad r\in\mathbb{N}),\]
whereas the general term in each of these earlier similar series representations has the order estimate:
\[O(2^{-2k}\cdot k^{-2r})\qquad(k\rightarrow\infty;\quad r\in\mathbb{N}).\]
Thus, even in the special case when $m=3$, the series representing $\zeta(2r+1)$ converges faster in (\ref{yl21}) than in (\ref{yl11}).

\textbf{Definition 4.1.} The Riemann Zeta function $\zeta(s)$ can be defined in the complete form (This definition is valid for all complex $s$):
\begin{equation}\label{rize}
 \zeta(s):=\frac{1}{s-1}\left(\frac{\partial}{\partial z}\right)^{1-s}\!\!\left.\frac{ze^z}{e^z-1}\right|_{z=0}\quad(s\in\Omega:=\{s\in\mathbb{C}|s\neq1\}).
\end{equation}
Here $\partial_z^{1-s},\,z\in\mathbb{R}^1$ is an abstract operators taking $\xi^{1-s}$ as the symbols, namely
\begin{equation}\label{cxsz}
 \left(\frac{\partial}{\partial z}\right)^{1-s}e^{\xi z}:=\xi^{1-s}e^{\xi z}\quad(\xi\in\mathbb{R}^1,\,s\in\mathbb{C}).
\end{equation}

In the special case when $\Re(s)>1$, by applying (\ref{e0}), (\ref{e1}) and (\ref{e1'}) to (\ref{rize}), we have
\begin{eqnarray*}

 \zeta(s)&=&\frac{1}{s-1}\left(\frac{\partial}{\partial z}\right)^{1-s}\!\!\left.\frac{ze^z}{e^z-1}\right|_{z=0}
 =\sum^\infty_{k=0}\left(\frac{\partial}{\partial z}\right)^{1-s}\!\!\left.\frac{ze^{(1+k)z}}{1-s}\right|_{z=0}\\
 &=&\sum^\infty_{k=0}\frac{\partial}{\partial\xi}e^{(1+k)\frac{\partial}{\partial\xi}}\left.\frac{\xi^{1-s}e^{\xi{z}}}{1-s}\right|_{\xi=0,z=0}\\
 &=&\sum^\infty_{k=0}\left(z+\frac{\partial}{\partial\xi}\right)e^{(1+k)\left(z+\frac{\partial}{\partial\xi}\right)}\left.\frac{\xi^{1-s}}{1-s}\right|_{\xi=0,z=0}\\
 &=&\sum^\infty_{k=0}\left(z+\frac{\partial}{\partial\xi}\right)\left.\frac{e^{(1+k)z}(\xi+1+k)^{1-s}}{1-s}\right|_{\xi=0,z=0}\\
 &=&\sum^\infty_{k=0}\left.\left(z\frac{e^{(1+k)z}(1+k)^{1-s}}{1-s}+e^{(1+k)z}(1+k)^{-s}\right)\right|_{z=0}\\
 &=&\sum^\infty_{k=0}\frac{1}{(1+k)^s}=\sum^\infty_{k=1}\frac{1}{k^s}\quad(\Re(s)>1).
\end{eqnarray*}
The result of this calculation shows that our definition is reasonable.

When $s=-n,\,n\in\mathbb{N}_0$ in (\ref{rize}), we obtain
\begin{equation}\label{hurn}
 \zeta(-n)=-\frac{1}{n+1}\frac{\partial^{n+1}}{\partial z^{n+1}}\!\left.\frac{ze^z}{e^z-1}\right|_{z=0}=-\frac{1}{n+1}B_{n+1}(1),
\end{equation}
where $B_n(x)$ are the Bernoulli polynomials defined by the generating functions:
\begin{equation}\label{bnld}
  \frac{ze^{xz}}{e^z-1}=\sum^\infty_{n=0}B_n(x)\frac{z^n}{n!}\quad(|z|<2\pi).
\end{equation}
It is easily seen from the definition (\ref{rize}) that
\begin{equation}\label{djd3}
 \lim_{s\rightarrow1}[(s-1)\zeta(s)]=\lim_{s\rightarrow1}\left(\frac{\partial}{\partial z}\right)^{1-s}\!\!\left.\frac{ze^z}{e^z-1}\right|_{z=0}
 =\left.\frac{ze^z}{e^z-1}\right|_{z=0}=B_0(1)=1.
\end{equation}
In other words, $\zeta(s)$ has a simple pole at $s=1$, and $B_0(1)=1$ is the residue of $\zeta(s)$ at the simple pole $s=1$.

\textbf{Definition 4.2.} Let $\zeta_z(s)$ be an analytic function defined by
\begin{equation}\label{rize'}
 \zeta_z(s):=\frac{1}{s-1}\left(\frac{\partial}{\partial z}\right)^{1-s}\!\!\frac{ze^z}{e^z-1}\quad(s\in\Omega:=\{s\in\mathbb{C}|s\neq1\},\,z\in\mathbb{R}^1).
\end{equation}

In the special case when $\Re(s)>2$ and $z\leq0$, we have
\begin{equation}\label{rize"}
 \zeta_z(s)=\frac{1}{1-s}\sum^\infty_{n=1}\frac{ze^{nz}}{n^{s-1}}+\sum^\infty_{n=1}\frac{e^{nz}}{n^s}\quad(\Re(s)>2,\,z\leq0).
\end{equation}

By Definition 4.1, we have $\left.\zeta_z(s)\right|_{z=0}=\zeta(s)$ and
\begin{equation}\label{wfze}
 \left.\frac{\partial^k}{\partial z^k}\zeta_z(s)\right|_{z=0}=\frac{s-1-k}{s-1}\zeta(s-k)\quad(k\in\mathbb{N},\,s\neq1).
\end{equation}

From the differential relation (\ref{wfze}), we have for $k\in\mathbb{N},\,s\neq1$
\[ \left.\frac{\partial^{2k-1}}{\partial x^{2k-1}}\cos\left(\frac{\pi x}{c}\frac{\partial}{\partial z}\right)\zeta_z(s)\right|_{z=0,x=0}=0;\]
\[ \left.\frac{\partial^{2k}}{\partial x^{2k}}\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\zeta_z(s)\right|_{z=0,x=0}
=(-1)^k\left(\frac{\pi}{c}\right)^{2k}\frac{s-1-2k}{s-1}\zeta(s-2k).\]
Here $c>0$ is a given real number. Therefore, we have the following Taylor expansion in the neighborhood of $x=0$:
\begin{equation}\label{taze}
 \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\zeta_z(s)\right|_{z=0}=\sum^\infty_{k=0}\frac{(-1)^k}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}
 \frac{s-1-2k}{s-1}\zeta(s-2k)\quad(s\neq1).
\end{equation}
On the other hand, by using (\ref{rize"}) and Theorem 2.4, since
\begin{eqnarray*}

 & &\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(ze^{nz})\right|_{z=0}\\
 &=&\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)z\cdot\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)e^{nz}\right|_{z=0}
-\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)z\cdot\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)e^{nz}\right|_{z=0}\\
 &=&-\frac{\pi{x}}{c}\sin\frac{n\pi x}{c}\quad(n\in\mathbb{N}),
\end{eqnarray*}
we obtain the following summation formula for $\Re(s)>2,\,x\in\Omega\subset\mathbb{R}^1$:
\begin{equation}\label{sjqh}
 \left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\zeta_z(s)\right|_{z=0}=\sum^\infty_{n=1}\frac{1}{n^s}\cos\frac{n\pi x}{c}
 +\frac{\pi{x}/c}{s-1}\sum^\infty_{n=1}\frac{1}{n^{s-1}}\sin\frac{n\pi x}{c}.
\end{equation}

By making use of (\ref{taze}), (\ref{sjqh}) and (\ref{yl8g}), we obtain the following theorem:

\textbf{Theorem 4.7.} Let $\Re(s)>1$. For $x\in\mathbb{R}^1$ with $|x|\leq2c$, we have
\begin{eqnarray}\label{zegh}
 & & s\sum^\infty_{n=1}\frac{1}{n^{s+1}}\cos\frac{n\pi x}{c}+\frac{\pi{x}}{c}\sum^\infty_{n=1}\frac{1}{n^s}\sin\frac{n\pi x}{c}\nonumber\\
 &=& \sum^\infty_{k=0}(-1)^k\frac{s-2k}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(s+1-2k)\quad(\Re(s)>1),
\end{eqnarray}
which generalize the Katsurada's formula (\ref{yl8}) to the half complex plane $\Re(s)>1$; When $x=c$ in (\ref{zegh}), we obtain
\begin{equation}\label{tgck}
 \zeta(s+1)=\frac{2^s}{s(2^{s+1}-1)}\sum^\infty_{k=1}(-1)^{k-1}\frac{(s-2k)\pi^{2k}}{(2k)!}\zeta(s+1-2k)\quad(s\neq0),
\end{equation}
which generalize the Cvijovi\'{c}-Klinowski's formula (\ref{yl11}) to the whole complex plane.

Since each series in (\ref{zegh}) is uniformly convergent with respect to $x$ on the closed interval $[-2c,2c]$ for $\Re(s)>2$, multiplying the both side of (\ref{zegh}) by $(\pi x/c)^{s-1}$, and executing termwise differentiation in them with respect to $x$, namely $\partial_x[(\pi x/c)^{s-1}\times(\ref{zegh})]$, then the sine series term is counteracted. Thus we can generalize Theorem 4.1 to the half complex plane $\Re(s)>2$:

\textbf{Theorem 4.8.} Let $\Re(s)>2$. For $x\in\mathbb{R}^1$ with $|x|\leq2c$, we have
\begin{eqnarray}\label{zegh'}
 & & s(s-1)\sum^\infty_{n=1}\frac{1}{n^{s+1}}\cos\frac{n\pi{x}}{c}+\left(\frac{\pi{x}}{c}\right)^2\sum^\infty_{n=1}\frac{1}{n^{s-1}}\cos\frac{n\pi{x}}{c}\nonumber\\
 &=&\sum^\infty_{k=0}(-1)^k\frac{(s-2k)(s+2k-1)}{(2k)!}\left(\frac{\pi{x}}{c}\right)^{2k}\zeta(s+1-2k)\quad(\Re(s)>2).
\end{eqnarray}

When $x=2c/3,c/2,c/3$ in (\ref{zegh'}), by applying Lemma 4.1 to Theorem 4.8, we can generalize the Theorem 4.4, Theorem 4.5 and Theorem 4.6 to the whole complex plane:

\textbf{Theorem 4.9.} For $s\in\mathbb{C}$
\begin{eqnarray}\label{zegh1}
 & & s(s-1)\frac{3^{-s}-1}{2}\zeta(s+1)+\left(\frac{2\pi}{3}\right)^2\frac{3^{2-s}-1}{2}\zeta(s-1)\nonumber\\
 &=& \sum^\infty_{k=0}(-1)^k\frac{(s-2k)(s+2k-1)}{(2k)!}\left(\frac{2\pi}{3}\right)^{2k}\zeta(s+1-2k);
\end{eqnarray}
\begin{eqnarray}\label{zegh2}
 & & s(s-1)\frac{2^{-s}-1}{2^{s+1}}\zeta(s+1)+\left(\frac{\pi}{2}\right)^2\frac{2^{2-s}-1}{2^{s-1}}\zeta(s-1)\nonumber\\
 &=& \sum^\infty_{k=0}(-1)^k\frac{(s-2k)(s+2k-1)}{(2k)!}\left(\frac{\pi}{2}\right)^{2k}\zeta(s+1-2k);
\end{eqnarray}
\begin{eqnarray}\label{zegh3}
 & & s(s-1)\frac{6^{-s}-3^{-s}-2^{-s}+1}{2}\zeta(s+1)+\left(\frac{\pi}{3}\right)^2\frac{6^{2-s}-3^{2-s}-2^{2-s}+1}{2}\zeta(s-1)\nonumber\\
 &=& \sum^\infty_{k=0}(-1)^k\frac{(s-2k)(s+2k-1)}{(2k)!}\left(\frac{\pi}{3}\right)^{2k}\zeta(s+1-2k).
\end{eqnarray}

In other words, in the special case of Theorem 4.9 when $s=2r,\,r\in\mathbb{N}$, we can obtain the Theorem 4.4, Theorem 4.5 and Theorem 4.6.


TITLE: sum of a positive series
QUESTION [4 upvotes]: For sufficiently large $m$, I want to calculate the following series.

$\displaystyle\sum_{i=0}^{3m}\sum_{j=0}^m \sum_{k=0}^m\sum_{l=0}^m \min(i,j+k+l)$

Except considering all case by case, is there any clever trick?

REPLY [2 votes]: Hint:
\begin{eqnarray*}
\sum_{i=0}^{3m}\sum_{j=0}^m \sum_{k=0}^m\sum_{l=0}^m \min(i,j+k+l)
&=& \sum_{j,k,l=0}^m \left[\sum_{i=0}^{j+k+l} i
+ \sum_{i=j+k+l+1}^{3m}(j+k+l) \right]\\
&=& \sum_{j,k,l=0}^m \bigg[ \frac{1}{2}(j+k+l)(j+k+l+1) \\
&&+ (j+k+l)(3m-j-k-l)\bigg]
\end{eqnarray*}
The sum can now be found using only
\begin{eqnarray*}
\sum_{k=0}^n 1 &=& n \\
\sum_{k=0}^n k &=& \frac{n(n+1)}{2} \\
\sum_{k=0}^n k^2 &=& \frac{n(n+1)(2n+1)}{6}.
\end{eqnarray*}


text:
\begin{document}
\input{zcom}
\setlength{\arraycolsep}{0.05cm}
\newcommand{\rest}{{T_\text{rest}}}
\newcommand{\zetahat}{\hat{\zeta}}

\newcommand{\tDelta}{{\tilde{\Delta}}}
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\newtheorem{sigmodel}{Signal Model}

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\newcommand{\delbound}{{b}}
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\newcommand{\dett}{{\text{det}}}
\newcommand{\CSres}{{\text{CSres}}}
\newcommand{\diff}{{\text{diff}}}

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\newcommand{\Subsection}[1]{  \vspace{-0.12in} \subsection{#1}  \vspace{-0.08in} }
\newcommand{\Subsubsection}[1]{   \subsubsection{#1} }

\date{}
\maketitle

\newcommand{\Aset}{{\cal A}}
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\newcommand{\sm}{e^2}
\newcommand{\smd}{\sm}

\begin{abstract}
In this work, we study the ``stability" of our recently proposed algorithm, LS-CS (compressive sensing on least squares residual), designed for recursive reconstruction of sparse signal sequences from noisy measurements. By ``stability" we mean that the number of extra elements and the number of misses from the current support estimate remains bounded by a small value compared to the current signal support size. This, in turn, implies that the LS-CS reconstruction error remains bounded by a small value compared to the total signal power. We show that ``stability" holds under mild assumptions (bounded noise, high enough SNR and enough measurements at every time) for a simple deterministic signal model with fixed signal power and support size, support changes (additions and removals) at every time, and gradual coefficient magnitude increase/decrease.

\end{abstract}

{\bf Index Terms: } compressed sensing, sparse reconstruction

\Section{Introduction}
Consider the problem of recursively and causally reconstructing time sequences of spatially  sparse signals (with unknown and time-varying sparsity patterns) from a limited number of linear incoherent measurements with additive noise. The signals are sparse in some transform domain referred to as the ``sparsity basis" \cite{dantzig}. Important applications include real-time dynamic MRI, video compression, single-pixel video imaging and spatial field sensing over time. Due to strong temporal dependency in the signal sequence, it is usually valid to assume that its {\em sparsity pattern (support of the sparsity transform vector) changes slowly over time}. This was empirically verified in \cite{icipmodcs} for dynamic MRI.
The solution to the static version of the above problem is provided by compressed sensing (CS). CS for noisy observations, e.g. Dantzig selector  \cite{dantzig} or Basis Pursuit Denoising (BPDN) \cite{bpdn}, has been shown to have small error as long as incoherence assumptions hold.
{\em Note that, in this paper, whenever we say ``CS", we refer to the Dantzig selector \cite{dantzig}.}

In \cite{kfcsicip,kfcspap}, we proposed the first recursive solution to the dynamic problem, CS on Least Squares residual (LS-CS) and CS on Kalman filtering residual (KF-CS). The key idea of LS-CS is to replace CS on the observation (simple CS) by CS on the least squares (LS) residual computed using the previous support estimate. Doing this significantly improves the reconstruction error of LS-CS when the number of measurements available at any time is small.

Other related work includes \cite{giannakis} (mostly focusses on the case when support does not change) \cite{rozell,romberg} (use past reconstructions to speed up current optimization but not to improve reconstruction error), and \cite{jung_etal} (causal but batch solution for dynamic MRI).

In \cite{kfcspap,lscspaper} we bounded LS-CS error and compared it with the CS bound. In \cite{kfcspap} we also studied LS-CS and KF-CS stability  but under strong assumptions and for a random walk model with increasing support size. By ``stability" we mean that the extras and misses in the support estimate are bounded by a small value compared to the support size and the reconstruction error remains small compared to the signal power. In \cite{lscspaper}, we proved LS-CS stability under mild  assumptions and for a bounded SNR and bounded support size model. But we still assumed that the delay between support addition/removal times is more than one. In many real applications, particularly if we define the support of a compressible signal using a smaller threshold, {\em the delay is actually one, i.e. the support changes occur at every time \cite{kfcsmri}. This important case is the focus of the current work.}

The paper is organized as follows.
 We give the LS-CS algorithm and the error bound from \cite{lscspaper} in Sec. \ref{lscs}. Our stability result and its implications and the outline of its proof is given in Sec. \ref{stability}. Simulation results are discussed in Sec. \ref{sims} and conclusions in Sec. \ref{conclusions}.

\Subsection{Problem Definition and Notation}
The set operations $\cup$, $\cap$, $\setminus$ have the usual meanings. $\phi$ denotes the empty set. $T^c$ denotes the complement of $T$ w.r.t. $[1,m]:=[1,2,\dots m]$, i.e. $T^c := [1,m] \setminus T$. $|T|$ denotes the cardinality of $T$.

For a vector, $v$, and a set, $T$, $v_T$ denotes the $|T|$ length sub-vector containing the elements of $v$ corresponding to the indices in the set $T$. $\|v\|_k$ denotes the $\ell_k$ norm of a vector $v$. If just $\|v\|$ is used, it refers to $\|v\|_2$.

For a matrix $M$, $\|M\|_k$ denotes its induced $k$-norm, while just $\|M\|$ refers to $\|M\|_2$. $M'$ denotes the transpose of $M$. For a tall matrix, $M$, $M^\dag : = (M'M)^{-1}M'$.

For a fat matrix $A$, $A_T$ denotes the sub-matrix obtained by extracting the columns of $A$ corresponding to the indices in $T$.
The $S$-restricted isometry constant \cite{dantzig}, $\delta_S$, for an $n \times m$ matrix (with $n<m$), $A$, and the $S,S'$ restricted orthogonality constant \cite{dantzig}, $\theta_{S,S'}$, are as defined in \cite[eq 1.3]{dantzig} and \cite[eq 1.5]{dantzig} respectively.

Let $(z_t)_{m \times 1}$ denote the spatial signal at time $t$ and $(y_t)_{n \times 1}$, with $n<m$, denote its noise-corrupted observation vector at $t$, i.e. $y_t = H z_t + w_t$. The signal, $z_t$, is sparse in a given sparsity basis (e.g. wavelet) with orthonormal basis matrix, $\Phi_{m \times m}$, i.e. $x_t \defn \Phi' z_t$ is a sparse vector. We denote its support by $N_t$ and we use $S_t:= |N_t|$ to denote its size.
Thus the observation model is
\bea
y_t = A x_t + w_t, \ A \defn H \Phi, \ \  \E[w_t]=0, \  \E[w_t w_t']= \sigma^2 I \ \ \ \

\label{obsmod}
\eea

We assume that $A$ has unit norm columns. The observation noise, $w_t$, is independent identically distributed (i.i.d.) over $t$ and is independent of $x_t$. Our goal is to recursively obtain the best estimates of $x_t$ (or equivalently of the signal, $z_t = \Phi x_t$) using $y_1, \dots y_t$. By {\em recursively}, we mean, use only $y_t$ and the estimate from $t-1$, $\xhat_{t-1}$, to compute the estimate at $t$.

To quantify the incoherence (approximate orthogonality) of $A$, define the following quantities.
\bd[$\sone$, $\sinf$]
\vspace{-0.08in}
\label{delta_bnd}
For $A:=H \Phi$,
\ben
\label{deltas}
\item let $\sone$ denote the largest $S$ for which $\delta_S < 1/2$,
\label{delta3s}
\item let $\sinf$ denote the largest $S$ for which $\delta_{2S}+ \theta_{S,2S} < 1$.
\een

\vspace{-0.1in}
\ed
We assume that the support changes slowly over time, i.e. the additions, $|N_t \setminus N_{t-1}| \le S_a  \ll  S_t$ and the removals, $|N_{t-1} \setminus N_t| \le S_a  \ll  S_t$. This was verified in \cite{kfcsmri}. Also, $A$ is incoherent enough to at least satisfy $S_t + S_a < \sone$ and $S_a < \sinf$.

{\em Further Notation. }
We use $\xhat_t$ to denote the estimate of $x_t$ given by our algorithm at time $t$ and $\Nhat_t$ to denote its support estimate.
We will use the following sets often.
\bd[$T$, $\Delta$, $\Delta_e$]
\vspace{-0.05in}
We use $T  : = \Nhat_{t-1}$ to denote the support estimate from the previous time.
We use $\Delta := N_t \setminus T$ to denote the unknown part of the support at the current time and $\Delta_e := T \setminus N_t$ to denote the ``erroneous" part of $T$. We attach the subscript $t$ to the set, i.e. we use $T_t$, $\Delta_t$ or $\Delta_{e,t}$ where necessary, e.g. in Sec. \ref{proofs}.

\ed
\bd[$\tT$, $\tDelta$, $\tDelta_e$]
\vspace{-0.05in}
We use $\tT := \Nhat_t$ to denote the final estimate of the current support; $\tDelta : = N_t \setminus \tT$ to denote the ``misses" in the final estimate and $\tDelta_e := \tT \setminus N_t$ to denote the ``extras".
\vspace{-0.08in}
\ed

\Section{CS on Least Squares Residual (LS-CS)}
\label{lscs}
Given observation, $y$, CS (Dantzig selector) \cite{dantzig} solves
\bea
\min_{\zeta} \|\zeta\|_1 \ \text{s.t.} \ \|A'(y - A \zeta)\|_\infty < \lambda
\label{simplecs}
\eea
The main idea of LS-CS is to replace CS on $y_t$ (simple CS) by CS on the observation residual computed using the previous support estimate, $T:=\Nhat_{t-1}$. The residual is computed as follows:
\bea
(\xhat_{t,\text{init}})_{T} \se {A_{T}}^\dag y_t, \ \ (\xhat_{t,\text{init}})_{T^c} = 0 \nn \\
\tty \se  y_t - A \xhat_{t,\text{init}}
\label{deftty0}
\eea
Notice that the residual can be rewritten as $\tty = A\beta_t + w_t$ where $\beta_t$ is a $|T \cup \Delta|$-sparse vector with $(\beta_t)_{(T \cup \Delta)^c} = 0$,
\bea
(\beta_t)_T \se (x_t)_T - {A_T}^\dag y_t  = -  {A_{T}}^\dag (w_t + A_\Delta (x_t)_\Delta), \nn \\
(\beta_t)_\Delta \se (x_t)_\Delta - 0 = (x_t)_\Delta
\label{defbeta}
\eea
If $|T| \le \sone$,
$\|(\beta_t)_T\|^2 \le  2 \frac{{\theta_{|T|,|\Delta|}}^2}{(1-\delta_{|T|})^2} \|x_\Delta\|^2 + \frac{2}{(1-\delta_{|T|})} \|w\|^2 \le 8{\theta_{|T|,|\Delta|}}^2 \|x_\Delta\|^2 + 4 \|w\|^2$. Thus, if $|\Delta|$ is small enough (i.e. $S_a$ is small enough and $T \approx N_{t-1}$) and if the noise is small enough, then $\|(\beta_t)_T\|$ will be small, i.e. $\beta_t$ will be only $|\Delta|$-approximately-sparse. Thus, doing CS on $\tty$ incurs much less error than simple CS, which needs to reconstruct a $|N_t|$-sparse signal, $x_t$ [see Fig. \ref{fig1}].

At each time $t$, LS-CS computes $\tty$ using (\ref{deftty0}) and solves (\ref{simplecs}) with $y = \tty$. It uses the output to compute additions to $T$ to get a new support estimate, $\tT_\dett$, followed by LS estimation using $\tT_\dett$ to get $\xhat_{t,\dett}$. Finally it uses $\xhat_{t,\dett}$ to compute deletions from $\tT_\dett$ to get the final support estimate, $\Nhat_t$, followed by LS estimation to get $\xhat_t$.

\Subsection{LS-CS algorithm}
\label{lscsalgo}
We list the LS-CS algorithm \cite{lscspaper,kfcspap} below.
(In LS-CS, at $t=0$ we run simple CS (or equivalently CS-residual with $T=\phi$). This needs  a large enough number of measurements, $n_0 > n$ so that the simple CS estimate is accurate enough. Thus at $t=0$ we use $A = H_0 \Phi$ where $H_0$ is an $n_0 \times m$ measurement matrix.)

\noindent Set $\Nhat_{-1}=\phi$.

For $t \ge 0$ do,

\ben
\item If $t=0$, set $A := H_0 \Phi$, else set $A := H \Phi$.
\label{step1}

\item {\em Initial LS. } Use $T:=\Nhat_{t-1}$ to compute the initial LS estimate, $\xhat_{t,\text{init}}$, and the LS residual, $\tty$ using (\ref{deftty0}).
\label{initls}

\item {\em CS-residual. } Do CS (Dantzig selector) on the LS residual, i.e.  solve (\ref{simplecs}) with $y= \tty$ and denote its output by $\betahat_t$. Compute $\xhat_{t,\CSres}$ using
$
\xhat_{t,\CSres} : = \betahat_t + \xhat_{t,\text{init}}.
\label{xhatcsres}
$

\item {\em Detection and LS. } Compute additions to the support using
\label{addls}
\bea
\Nhat_{t,\dett} = T \cup \{ i \in [1,m] : |(\xhat_{t,\CSres})_i| > \alpha \}
\eea
Compute the LS estimate, $\xhat_{t,\dett}$, using $\tT_{\dett}:=\Nhat_{t,\dett}$ as
\bea
(\xhat_{t,\dett})_{\tT_{\dett}} \se {A_{\tT_{\dett}}}^\dag  y_t, \ \ (\xhat_{t,\dett})_{(\tT_{\dett})^c} = 0
\label{secls}
\eea

\item {\em Deletion and LS. } Compute deletions from the support using
\label{delete}
\bea
\Nhat_t \se  \tT_{\dett} \setminus \{ i \in  \tT_{\dett} : |(\xhat_{t,\dett})_i| \le \alpha_{del} \}
\label{delstep}
\eea
Compute the LS estimate, $\xhat_t$, using $\tT:=\Nhat_t$ as
\bea
(\xhat_{t})_{\tT} \se {A_{\tT}}^\dag  y_t, \ \ (\xhat_{t})_{\tT^c} = 0
\label{finalls}
\eea

\item Output $\xhat_t$ and $\hat{z}_t = \Phi \xhat_t$.  Feedback $\Nhat_t$.
\een
Increment $t$ and go to step \ref{step1}.

We define the following sets which will be used in Sec. \ref{stability}.
\bd[$\tT_{\dett}$, $\tDelta_{\dett}$, $\tDelta_{e,\dett}$]
\vspace{-0.07in}
We define $\tT_{\dett} := \Nhat_{t,\dett}$, $\tDelta_{\dett}:= N_t \setminus \tT_{\dett}$ and $\tDelta_{e,\dett}:= \tT_{\dett} \setminus N_t$.
\vspace{-0.05in}
\ed

\Subsection{Reconstruction Error Bound for CS-residual step}
\label{csresbnd}
In \cite[Theorem 1, Corollary 1]{lscspaper}, we bounded CS-residual error and showed that (a) our bounds hold under weaker assumptions than those of \cite{dantzig} and (b) under these weaker assumptions, the bound on CS error is much looser.
We restate \cite[Corollary 1]{lscspaper} below.
\begin{lemma}
Assume that  $\|w\|_\infty \le {\lambda}/{\|A\|_1}$, $|\Delta| \le \sinf$, $|T| \le \sone$.

The CS-residual error satisfies
\label{cor1}
\bea
\|x - \xhat_{\CSres}\|^2 \sle C' + C''{\theta_{|T|,|\Delta|}}^2 \|x_\Delta\|^2, \ \text{where} \nn \\
 C' \equiv C'(|T|,|\Delta|) \sdefn C_2(|\Delta|) |\Delta| \lambda^2 +  C_3(|\Delta|) |T| \frac{4n\lambda^2}{\|A\|_1^2}, \nn \\
\label{defcprime}
 C'' \equiv C''(|T|,|\Delta|) \sdefn C_3(|\Delta|) |T|8,  \\
C_2(S) \sdefn \frac{48}{(1-\delta_{2S} - \theta_{S,2S})^2}, \nn \\
\label{defc2c3}
C_3(S) \sdefn 8 + \frac{24 \theta_{S,2S}^2}{(1-\delta_{2S} - \theta_{S,2S})^2}
\eea

\end{lemma}

\begin{remark}
Notice that $C'(|T|,|\Delta|)$ and $C''(|T|,|\Delta|)$ are {\em non-decreasing functions of $|T|$, $|\Delta|$}.

\label{cor1_rem}
\end{remark}

\Section{LS-CS Stability}
\label{stability}
In Sec. \ref{csresbnd} we bounded the CS-residual error as a function of $|T|,|\Delta|$. The bound is small as long as $|\Delta_e|$ and $|\Delta|$ are small. Similarly a bound on the final LS-CS error as a function of $|\tT|,|\tDelta|$ is easy to obtain.
The next question is: under what conditions on the measurement model, {\em will $|\tDelta_e|$, $|\tDelta|$ (and hence $|\Delta_e|$, $|\Delta|$) remain stable, i.e. remain bounded by a small value compared to the signal support size, at all times? This, in turn, will imply that the LS-CS error is small compared to the signal power at all times.}

\Subsection{LS-CS Stability Result}
For studying stability, we need to assume a signal model. We assume a simple deterministic model that (a) allows equal and nonzero number of additions/removals from the support at every time, (b) allows a new coefficient magnitude to gradually increase from zero for sometime and finally reach a constant value, $M$, (c) allows coefficients to gradually decrease and become zero (get removed from support) at the same rate, and (d) has constant signal power and support size.

\begin{sigmodel} Assume the following.
\ben

\item At $t=0$, the support size is $S_0$. The support contains $2S_a$ elements each with magnitude $M/d,2M/d, \dots (d-1)M/d$, and $S_0-(2d-2)S_a$ elements with magnitude $M$.

\item At each $t>0$, $S_a$ coefficients get added to the support at magnitude $M/d$.

\item At each $t>0$, $S_a$ coefficients which had magnitude $M/d$ at $t-1$ get removed from the support (magnitude becomes zero).

\item At each $t>0$, the magnitude of $S_a$ coefficients which had magnitude $j M/d$ at $t-1$ increases to $(j+1)M/d$. This occurs for all $1 \le j \le d-1$.

\item At each $t>0$, the magnitude of $S_a$ coefficients which had magnitude $j M/d$ at $t-1$ decreases to $(j-1)M/d$. This occurs for all $2 \le j \le d$.

\een
\label{sigmod2}
\end{sigmodel}

From the above model, the composition of the support at any time $t$ is the same as that at $t=0$. It consists of $(d-1)S_a$ increasing and $(d-1)S_a$ decreasing coefficients, and $S_0- (2d-2)S_a$ constant coefficients (with magnitude $M$). Also $|N_t|=S_0$ for all $t$.

To understand the model better, let us define the following sets.
\bd Define the following, for any $d_0 \le d$
\vspace{-0.04in}
\ben
\item Newly added set, $\Aset_t:= \{i: |x_{t,i}|=M/d,  \  |x_{t-1,i}|=0 \}$.
\item Newly removed set, $\Rset_t:= \{i: |x_{t,i}|=0,  \  |x_{t-1,i}|=M/d \}$.

\item Small coefficients set, $\Sset_t(d_0):= \{i:  0 < |x_{t,i}| < d_0M/d \}$.
\item $\Dset_{t}(d_0): = \{i: |x_{t,i}| = (d_0-1)M/d, \ |x_{t-1,i}| = d_0M/d \}$.
\item $\Iset_{t}(d_0): = \{i: |x_{t,i}| = d_0M/d, \ |x_{t-1,i}| = (d_0-1)M/d \}$.
\een
\vspace{-0.04in}
\ed

From our model, at any $t$, $S_a$ elements enter the set $\Sset_t(d_0)$ from the bottom (set $\Aset_t$) and $S_a$ enter from the top (set $\Dset_{t}(d_0)$); $S_a$ elements leave  $\Sset_t(d_0)$ from the bottom  (set $\Rset_t$) and $S_a$ from the top (set $\Iset_{t}(d_0)$), i.e.
\bea
\Sset_t(d_0) = \Sset_{t-1}(d_0) \ \cup \ (\Aset_t \cup \Dset_{t}(d_0)) \ \setminus  \ (\Rset_t \cup \Iset_{t}(d_0))
\label{sseteq}
\eea

Our approach to show stability is as follows. We use Lemma \ref{cor1} to obtain conditions for an element of $\Delta$ to definitely get detected at $t$. Next, we use a bound on the error in the LS step after the detection step to obtain conditions for a nonzero element of $\tT_{\dett}$ to not get falsely deleted and for all elements of $\tDelta_{e,\dett}$ to get deleted.

Using the above conditions, we obtain sufficient conditions to ensure that at the end of LS-CS at time $t$, $\tDelta \subseteq \Sset_t(d_0)$ (i.e. $|\tDelta| \le (2d_0-2)S_a$) and $|\tDelta_{e}|=0$ for some $d_0 \le d$. In other words, all elements with current magnitude greater than or equal to $d_0M/d$ are correctly detected and not falsely deleted, while all zero elements (removals or false additions) are deleted. This leads to the following result.

\begin{theorem}
For any $d_0$, let \\
$k^a:=\max(1,(2d_0-2))S_a$, ~ $k^d:=\max(0,(2d_0-3))S_a$,
\label{stabres2}
\bea
&& \sm:= \frac{S_a (\sum_{j=1}^{d_0-1} j^2 + \sum_{j=1}^{d_0-2} j^2) M^2}{d^2}, \nn \\
&& \theta^a:=\theta_{S_0,k^a}, \ \ \theta^d:= \theta_{S_0+S_a+f,k^d}, \ \nn \\
&& C':=C'(S_0,k^a), \ \ C'':=C''(S_0,k^a),
\eea
Under Signal Model \ref{sigmod2}, if the following hold for some $d_0 \le d$,
\ben
\item the addition threshold, $\alpha$, is set large enough so that there are at most $f$ false additions per unit time,
\label{addthresh}

\item  the deletion threshold, $\alpha_{del}^2  = (4n\lambda^2/\|A\|_1^2) + 8 (\theta^d)^2 \sm $,
\label{delthresh}

\item $\|w\|_\infty \le \lambda/\|A\|_1$, $k^a S_a \le \sinf$, $S_0+S_a + f \le \sone$,
\label{measmodel}

\item $2S_a (\theta^a)^2 C'' < 1$ and $(d_0M/d)^2 \ge \max(G1,G2)$, where
\label{add_del}
\bea
G1 \se \frac{2 \alpha^2 + 2C' + 2(\theta^a)^2  \sm C''}{1 - 2S_a (\theta^a)^2 C''} \nn \\
G2 \se 2\alpha_{del}^2 + \frac{8n\lambda^2}{\|A\|_1^2} + 16 (\theta^d)^2 \smd
\eea

\item and, at $t=0$, we use enough measurements, $n_0$, so that $\tDelta  \subseteq \Sset_0(d_0)$, $|\tDelta| \le (2d_0-2)S_a$, $|\tDelta_e| =0$, and $|\tT| \le S_0$,
\label{initass}

\een

then,
\ben
\item at all $t \ge 0$, $\tDelta \subseteq \Sset_t(d_0)$, i.e. no element with magnitude $d_0M/d$ or larger is part of the unknown support.
\item at all $t \ge 0$, $|\tDelta| \le (2d_0-2)S_a$,  $|\tDelta_e| =0$, and $|\tT| \le S_0$,
\een
\end{theorem}

\begin{corollary}
Under the assumptions of Theorem \ref{stabres2},
\label{stabres2_cor}
\ben
\item at all $t > 0$, $|\Delta| \le k^a$,
$|\Delta_e| \le S_a$, $|T| \le S_0$ and  $|\tDelta_\dett| \le k^d$,
$|\tDelta_{e,\dett}| \le S_a+f$, $|\tT_\dett| \le S_0+S_a+f$,

\item at all $t \ge 0$, the LS-CS error, $\|x-\xhat\|^2 \le \frac{4n\lambda^2}{\|A\|_1^2} + (8(\theta_{S_0,(2d_0-2)S_a})^2+1) (\smd+S_a (d_0-1)^2M^2/d^2)$.

\item at all $t \ge 0$, the CS-residual error, $\|x - \xhat_\CSres\|^2 \le C' + (\theta^a)^2 (\sm + S_a d_0^2 M^2/d^2) C''$.
\een
\end{corollary}

\begin{remark}
Note that the initialization condition (condition \ref{initass}) is not restrictive. Denote the bound given by Theorem 1.1 of \cite{dantzig} for $S=S_0$ by $B_1$. It can be shown that condition \ref{initass} will hold if assumptions of Theorem 1.1 of \cite{dantzig} hold ($n_0$ is large enough); $\alpha_0$ is large enough to ensure at most $f$ false additions; $\alpha_{del,0} = ({4n\lambda^2}/{\|A\|_1^2})+8 (\theta_{S_0+f,k^a})^2 \sm$; and $M^2 > \min((\alpha_0+B_1)^2,4\alpha_{del,0}^2)$. Here $\alpha_0$ ($\alpha_{del,0}$) are the addition (deletion) thresholds used at $t=0$.
\label{initremark}
\end{remark}

\Subsection{Implications}
Under Signal Model \ref{sigmod2}, at all $t$, $|N_t|= S_0$; there are $S_a$ additions and removals; and the signal power is $\frac{S_a\sum_{j=1}^{d-1} j^2 M^2}{d^2} + (S_0-(2d-2)S_a)M^2$.

Also notice that if $2dS_a \ll S_0$ (valid to assume that only a small number of coefficients are increasing or decreasing), then Theorem \ref{stabres2} implies that $|\tDelta| \le (2d_0-2)S_a < 2dS_a \ll S_0$, i.e. $|\tDelta|$ is small compared to the support size, $S_0$. Clearly, $|\tDelta_{e}|=0 \ll S_0$. Also, if the SNR is large enough, clearly the LS-CS error bound is much smaller than the signal power. In other words, LS-CS is ``stable".

From Theorem \ref{stabres2}, under Signal Model \ref{sigmod2}, ``stability" is ensured under the following mild assumptions: (a) the addition/deletion thresholds are appropriately set (conditions \ref{addthresh} and \ref{delthresh});  (b) the noise is bounded and (c) the SNR is high enough and there are enough measurements (for a given support change rate, $S_a$, and magnitude change rate, $M/d$) so that $\sone, \sinf$ are large enough for condition \ref{measmodel} to hold and $\theta^2$, $C',C''$ are small enough for condition \ref{add_del} to hold.

Notice that if $M/d$ is large enough and if $f \ll S_0$ and $(2d_0-2)S_a \ll S_0$, then the requirements of Theorem \ref{stabres2} on the measurements (condition \ref{measmodel}) are {\em significantly weaker} than those for the corresponding CS results \cite[Theorems 1.1,1.2]{dantzig} which require $S_0 \le \sinf$.

\Subsection{Key Lemmas for Theorem \ref{stabres2}}
The following facts follow easily from the LS-CS algorithm.
\ben
\item An $i \in \Delta$ will definitely get detected if $x_i^2 > 2\alpha^2 + 2\|x - \xhat_\CSres\|^2$.
\label{det1}

\item An $i \in (\tT_\dett \setminus \tDelta_{e,\dett})$ will definitely not get falsely deleted if $x_i^2 > 2 \alpha_{del}^2 + 2\|(x - \xhat_\dett)_{\tT_\dett}\|^2$.
\label{nofalsedel1}

\item All $i \in \tDelta_{e,\dett}$ will get deleted if $\alpha_{del}^2 \ge \|(x - \xhat_\dett)_{\tT_\dett}\|^2$.
\label{truedel1}

\item If $\|w\|_\infty \le \lambda/\|A\|_1$ and $|\tT_\dett| \le \sone$, then $\|(x - \xhat_\dett)_{\tT_\dett}\|^2 \le (4n\lambda^2/\|A\|_1^2) + 8 {\theta_{|\tT_\dett|,|\tDelta_\dett|}}^2 \|x_{\tDelta_\dett}\|^2$. This follows using an approach similar to the bound on $\|\beta_T\|$ (given below (\ref{defbeta})).
\label{errls1}

\item The bound in fact \ref{errls1} is non-decreasing in $|\tT_\dett|$ and $|\tDelta_\dett|$.
\label{nondec}
\een

Using the above facts, we can prove the following lemmas.
\vspace{-0.05in}
\begin{lemma}[Detection condition]
Assume that $\|w\|_\infty \le \lambda/\|A\|_1$, $|T| \le S_T \le  \sone$ and $|\Delta| \le S_\Delta \le \sinf$. Let $b:=\|x_\Delta\|_\infty$. For a $\gamma \le 1$, let $\Delta_1:=\{i \in \Delta: \gamma b^2 \le x_i^2 \le b^2 \}$ and $\Delta_2:= \Delta \setminus \Delta_1$. All $i \in \Delta_1$ will definitely get detected at the current time if
\label{detectcond}
\bea
{2|\Delta_1|C''(S_T,S_\Delta)} {\theta_{S_T,S_\Delta}}^2 & < & {\gamma} \ \ \text{and} \nn \\
 \frac{2 \alpha^2 + 2C'(S_T,S_\Delta) + 2{\theta_{S_T,S_\Delta}}^2C''(S_T,S_\Delta) \|x_{\Delta_2}\|^2}{\gamma - 2{\theta_{S_T,S_\Delta}}^2 |\Delta_1| C''(S_T,S_\Delta)} & < & b^2
\label{detcond}
\eea

\end{lemma}

{\em Proof: }
Notice that $\|x_\Delta\|^2 \le |\Delta_1| b^2 + \|x_{\Delta_2}\|^2$. Using Lemma \ref{cor1}, if
$\|w\|_\infty \le \lambda/\|A\|_1$, $|T| \le \sone$ and $|\Delta| \le \sinf$, $\|x - \xhat_\CSres\|^2 \le C' + \theta^2 C''\|x_\Delta\|^2 \le C' + \theta^2 C''(|\Delta_1| b^2 + \|x_{\Delta_2}\|^2)$. Using fact \ref{det1} and the fact that for all $i \in \Delta_1$, $x_i^2 \ge \gamma b^2$, we can conclude that all $i \in \Delta_1$ will get detected if ${2|\Delta_1|C''(|T|,|\Delta|)} {\theta_{|T|,|\Delta|}}^2 < {\gamma}$ and $\frac{2 \alpha^2 + 2C'(|T|,|\Delta|) + 2{\theta_{|T|,|\Delta|}}^2C''(|T|,|\Delta|) \|x_{\Delta_2}\|^2}{\gamma - 2{\theta_{|T|,|\Delta|}}^2 |\Delta_1| C''(|T|,|\Delta|)}  < b^2$. Using Remark \ref{cor1_rem}, the left hand sides of both inequalities are non-decreasing functions of $|T|,|\Delta|$. Thus if $|T| \le S_T$ and $|\Delta| \le S_\Delta$, and if (\ref{detcond}) holds, then all $i \in \Delta_1$ will get detected, i.e. Lemma \ref{detectcond} follows.

\begin{lemma}[No false deletion condition]
Assume that $\|w\|_\infty \le \lambda/\|A\|_1$, $|\tT_\dett| \le S_T \le \sone$ and $|\tDelta_\dett| \le S_\Delta$. For a given $b_1$, let $T_1:=\{i \in \tT_\dett:x_i^2 \ge b_1\}$. All $i \in T_1$ will not get (falsely) deleted at the current time if
\label{nofalsedelscond}
$
b_1^2 >2 \alpha_{del}^2 + \frac{8n\lambda^2}{\|A\|_1^2} + 16 {\theta_{S_T,S_\Delta}}^2 \|x_{\tDelta_\dett}\|^2.
$
\end{lemma}
{\em Proof: } The lemma follows directly from facts \ref{nofalsedel1}, \ref{errls1} and \ref{nondec}.

\begin{lemma}[Deletion condition]
Assume that $\|w\|_\infty \le \lambda/\|A\|_1$, $|\tT_\dett| \le S_T \le \sone$ and $|\tDelta_\dett| \le S_\Delta$. All elements of $\tDelta_{e,\dett}$ will get deleted if
\label{truedelscond}
$
\alpha_{del}^2 \ge \frac{4n\lambda^2}{\|A\|_1^2} + 8 {\theta_{S_T,S_\Delta}}^2 \|x_{\tDelta_\dett}\|^2.
$
\end{lemma}
{\em Proof: } The lemma follows directly from facts \ref{truedel1}, \ref{errls1} and \ref{nondec}.

\begin{figure}[t!]
\centerline{
\epsfig{file = C:/Code/KFCS/LSCS_KFCS_code/icassp10_CS_large_delthresh.eps, width=5cm, height=3cm}

}
\vspace{-0.2in}
\caption{\small{Stability of LS-CS, CS-residual error \& comparison with CS.}}
\vspace{-0.15in}
\label{fig1}
\end{figure}

\Subsection{Outline of Proof of Theorem \ref{stabres2} and Corollary \ref{stabres2_cor}}
\label{proofs}

The proof of Theorem \ref{stabres2} follows by induction. Using condition \ref{initass}, the result holds for $t=0$. This proves the base case.
For the induction step, assume that $\tDelta_{t-1} \subseteq \Sset_{t-1}$, $|\tDelta_{e,t-1}|=0$ and $|\tT_{t-1}| \le S_0$. We use the signal model to show that $\Delta_{t} \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus \Rset_t$ and $\Delta_{e,t} \subseteq \Rset_t$. Thus, $|\Delta_{t}| \le
k^a$ and $|\Delta_{e,t}| \le S_a$. Also, $|T_t|=|\tT_{t-1}| \le S_0$.
Next, we use this and conditions \ref{measmodel} and \ref{add_del} in Lemma \ref{detectcond} (applied with $\gamma=1$) to show that $\tDelta_{\dett,t} \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus (\Rset_t \cup \Iset_{t}(d_0)) = \Sset_{t}(d_0) \setminus \Dset_t(d_0)$, i.e. the elements that just increased to $d_0M/d$ definitely get detected (the last equality follows from (\ref{sseteq})). Thus $|\tDelta_{\dett,t}| \le k^d$.
 Because of condition \ref{addthresh}, $|\tDelta_{e,\dett,t}| \le S_a+f$ and hence $|\tT_{\dett,t}| \le |N_t|+ |\tDelta_{e,\dett,t}| \le S_0+S_a+f$.
We use this and conditions \ref{measmodel} and \ref{add_del} in Lemma \ref{nofalsedelscond} (applied with $b_1=d_0M/d$) to show that only the elements that have magnitude $(d_0-1)M/d$ or lesser may get falsely deleted (and no others), i.e. $\tDelta_t \subseteq \Sset_{t-1}(d_0) \cup (\Aset_t \cup \Dset_{t}(d_0)) \setminus (\Rset_t \cup \Iset_{t}(d_0)) = \Sset_{t}(d_0)$. Thus, $|\tDelta_t| \le (2d_0-2)S_a$. Finally, we use condition \ref{delthresh} in Lemma \ref{truedelscond} to show that $|\tDelta_{e,t}|=0$ and thus $|\tT_t| \le S_0$. This completes the proof of the induction step.

For the first part of Corollary \ref{stabres2_cor}, use Theorem \ref{stabres2} to bound $|\tDelta_{t-1}|,|\tT_{t-1}|,|\tDelta_{e,t-1}|$. The discussion above then gives the desired bounds on $|\Delta_t|,|T_t|,|\Delta_{e,t}|$ and $|\tDelta_{\dett,t}|,|\tT_{\dett,t}|,|\tDelta_{e,\dett,t}|$.
The second and third part follow using $\tDelta \subseteq \Sset_t(d_0)$ and $\Delta \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus \Rset_t$.

The complete proof is given in the Appendix of the long version of this paper (available at \url{http://www.ece.iastate.edu/~namrata/LSCS_stability.pdf}).

\Section{Simulation Results}
\label{sims}
We ran 50 simulations of Signal Model \ref{sigmod2} with $m=200$, $S_0=20$, $S_a=2$, $d=1$, $M=3$, and the observation model of (\ref{obsmod}) with $n=59$, a random Gaussian $A$ and $w_t \sim unif(-c,c)$ with $c=0.1266$. LS-CS was run with $\alpha=0.0633$, $\alpha_{del}=14\alpha=0.8862$ and $\lambda=c \|A\|_1$. As can be seen from Fig. \ref{fig1}, the LS-CS and CS-residual error are stable at 0.1\% and 8\%. The CS error (both Dantzig and Gauss-Dantzig selector with $\lambda= c \|A\|_1/10$, which is the best choice of $\lambda$) is between 40-50\%.

Also,  $|\tDelta_{e}|=0$ and $|\tDelta|=0$ in all simulations, at all $t$, i.e. Theorem \ref{stabres2} holds with $d_0=d=1$.

\Section{Conclusions}
\label{conclusions}
We showed the stability of  LS-CS under fairly mild assumptions (bounded noise, high enough SNR and enough measurements at every time) for a simple deterministic signal model with fixed signal power; fixed support size, $S_0$; support changes (additions and removals) at every time at rate $S_a \ll S_0$; and gradual coefficient magnitude increase/decrease at rate $M/d$. By ``stability" we mean that the extras, $|\tDelta_e|$, and the misses, $|\tDelta|$, in the support estimate, remain bounded by a small value compared to $S_0$. This also implies that the LS-CS error remains small compared to the signal power. Future work includes studying KF-CS stability for a similar signal model.

\bibliographystyle{IEEEbib}
\bibliography{tipnewpfmt_kfcsfullpap}

\clearpage \input{LSCS_stab_appendix}

\end{document}

config: arxiv


TITLE: Finding the number of roots of the equation
QUESTION [0 upvotes]: The equation $x^{13}-e^{-x}+x-\sin{x}=0$ has

No real root
More than two real roots.
Exactly two real roots.
Exactly one real root.

I tried doing with the odd derivative  and check whether the derivative change sign or remain same(positive or negative) so i find $f^{(15)}(x)=e^{-x}+cos(x)$ now i am not able to determine how this will change the sign. Is there any other way to do the same problem thanks

REPLY [3 votes]: Let $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^{13}-e^{-x}+x-\sin(x)$.
Then,
$$\lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}x^{13}-e^{-x}+x-\sin(x)\ge\lim_{x\to +\infty}x^{13}-e^{-x}+x-1=\infty$$
In the same way, $\lim_{x\to -\infty}f(x)=-\infty$.
More over $f^\prime(x)=13x^{12}+e^{-x}+1-\cos(x)>0$. Thus, $f$ is strictly increasing. By the Intermediate Value Theorem, $f$ has only one root.