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\def\Z{{\mathbb Z}}
\def\lm{{\lambda/\mu}}


\begin{document}


\begin{slide}{}
\begin{center}
{\large The  $q$-Zeilberger  Algorithm and \\[8pt]
the $q$-WZ Method for Infinite Series}

\vskip 20pt {\yellow William Y.C. Chen and Ernest X.W. Xia}

\vskip 10pt  (presented by Doron  Zeilberger)

\end{center}

\vskip 25pt

{\scriptsize Center for Combinatorics \\
Nankai University \\
Tianjin, 300071\\
email: chen@nankai.edu.cn}
\end{slide}


%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The Main Result}
 The main result of this paper is to extend the  $q$-WZ method to
 prove nonterminating basic hypergeometric series. The essential
 ingredient  is the $q$-Gosper algorithm. The motivation is the firm
 believe that the $q$-WZ method should be applicable to infinite
 series. The discovery is based on an observation that the
 Andrews-Warnnar identities have telescoping proofs.
\end{slide}}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Some Notation}
We always assume $|q|<1$. The $q$-shifted factorials $(a;q)_n$
 and $(a;q)_\infty$ are defined by
 {\small
 \begin{eqnarray*}
(a;q)_n & = &\left\{
     \begin{array}{ll}
         1, \qquad & \mbox{if $n=0$}, \\ [4pt]
        (1-a) \cdots (1-aq^{n-1}), \quad
              & \mbox{if $n\geq 1$},
     \end{array} \right. \\ [5pt]
     (a;q)_{-n}&=&\frac{1}{(aq^{-n};q)_n}, \qquad \mbox{if $n\geq
     1$},\\[4pt]
(a;q)_\infty & = & (1-a) (1-aq)  (1-aq^2) \cdots,
\end{eqnarray*}
and
 $(a_1,\ldots,a_k;q)_n=(a_1;q)_n\cdots (a_k;q)_n.$}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Some Notation}
 An $_r\phi_{s}$ {\it basic hypergeometric  series} is defined  by
 {\small
\begin{align}\label{phi}
&_r\phi_{s}\left[{a_1,a_2,\ldots,a_r \atop
b_1,b_2,\ldots,b_{s}};q,z\right] \nonumber\\[8pt]
&:= \sum_{n=0}^\infty\frac{(a_1,a_2,\ldots,a_r;q)_n} {(q,b_1,\ldots,
b_{s};q)_n}\left[(-1)^nq^{n\choose 2}\right]^{1+s-r}z^n,
\end{align}}
where $q\neq0$ when $r>s+1$.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Some Notation}
An $_r\psi_s$ {\it bilateral basic hypergeometric  series} is
defined by  {\small
\begin{align}\label{psi}
{_r\phi_s}&\left[ \begin{array}{c}a_1,\ldots,a_r
\\b_1,\ldots,b_s
\end{array};q,x \right]  \nonumber \\
&: = \sum_{k=0}^\infty {(a_1;q)_k \cdots (a_r;q)_k \over (b_1;q)_k
\cdots (b_s;q)_k} {x^k \over (q;q)_k} \left( (-1)^k q^{k \choose 2}
\right)^{s-r+1} .
\end{align}}
 It is assumed that $q$, $z$ and the parameters are such that each
 term of the series is well-defined.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Hypergeometric Term}
We say that $t_k$ is a hypergeometric term if $\frac{t_{k+1}}{t_k}$
is a rational  function in $k$, i.e.,
\[
\frac{t_{k+1}}{t_k}=\frac{P(k)}{Q(k)},
\]
where $P(k)$ and $Q(k)$ are polynomials in $k$.

 For example,
 \[
  t(k)=k!, \   a^k,\ \frac{(3k+1)!}{(5k+4)!}.
 \]

\end{slide}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{$q$-hypergeometric Term}
Let $x=q^k$. We say that $t_k$ is a $q$-hypergeometric term if
\[
\frac{t_{k+1}}{t_k}=r(x),
\]
where $r(x)$ is a rational function in $x$.

For example,
  \[
    t_k=\frac{(a;q)_k}{(q;q)_k}z^k, \qquad k>0.
  \]
  where
  $(b;q)_k=(1-b)(1-bq)\cdots (1-bq^{k-1})$.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The Gosper's Algorithm}
{\small
\begin{itemize}
\item  The  Gosper's algorithm is a milestone for proving
hypergeometric identities. Zeilberger built up a powerful machinery
based on the Gosper's algorithm.

 \vspace{0.2cm}
 R.W. Gosper, Decision procedure for indefinite
 hypergeometric summation, Proc. Natl. Acad. Sci. USA
       {\bf75(1)} (1978) 40--42.

 \vspace{0.2cm}
 \item  The $q$-Gosper algorithm  is $q$-analogue  of the
  Gosper's algorithm, which was introduced by {\red Koornwinder}.
 \vspace{0.1cm}

 T.H. Koornwinder, On Zeilberger's algorithm and its q-analogue, J.
     Comput. Appl. Math. {\bf48} (1993) 91--111.

\end{itemize}}
\end{slide}
}


%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{ {\normalsize The Zeilberger's
 Algorithm and the WZ-Method}}


\begin{itemize}
\item {\red H.S.~Wilf} and {\red D.~Zeilberger} developed
 the Zeilberger's algorithm and the WZ-method for proving
  identities on hypergeometric series. \\[15pt]

They were awarded the {\yellow Steele Prize} for {\yellow Seminal
Contribution} (1998).
\\[15pt]

\item {\red B\"{o}ing-Koepf} and {\red T.H.~Koornwinder} gave
detailed discussions about the $q$-analogues.\\[15pt]

\end{itemize}


\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The  Package}
There are many packages to implement
 the Gosper's algorithm, the Zeilberger's algorithm, the WZ-method and their
 $q$-analogues.
 \vspace{0.3cm}
 We use the package {\it \red qsum6.mpl} to
implement the
 $q$-Gosper algorithm,  which is maintained at the following site
 \vspace{0.3cm}
\begin{center}
{\it
http://www.mathematik.uni-kassel.de/~koepf/Publikationen/index.html{\#}down}
\end{center}
\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Zeilberger's Algorithm}
The Zeilberger's algorithm uses the {\yellow Gosper's algorithm} to
find polynomials $a_i(n)$ which are {\yellow free of $k$} and a
hypergeometric term $G(n,k)$ such that {\small
\begin{multline*}
a_0(n)F(n,k) + a_1(n)F(n+1,k) + \cdots + a_d(n)F(n+d,k) \\
= G(n,k+1) - G(n,k).
\end{multline*}}
Summing over $k$, we obtain a recursion:
\[
a_0(n)f(n) + \cdots + a_d(n)f(n+d) = 0
\]
where $f(n) = \sum\limits_k F(n,k)$.
\end{slide}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Applications of Zeilberger's algorithm}
The identities are verified by finding recurrences and comparing
initial values. {\small
\begin{itemize}
\item Chinese identity ({\red Li, Shan-Lan} 1811--1882):
\[
\sum_{k=0}^n {n \choose k}^2 {m+2n-k \choose 2n} = {m+n \choose
n}^2,
\]
\item Its $q$-analogue ({\red Shi})
\[
\sum_{k=0}^n {n \brack k}^2 {m+2n-k \brack 2m} q^{k^2} = {m+n \brack
n}^2,
\]
\end{itemize}}
It is a special case of the $q$-Saalsch\"{u}tz identity.
\end{slide}


%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Andrews-Warnaar Identities}
We utilize the {\yellow telescoping method} to prove  the following
two identities
 \begin{align}
&\bigg(\sum_{n=0}^\infty(-1)^na^nq^{n\choose2
}\bigg)\bigg(\sum_{n=0}^\infty(-1)^nb^nq^{n\choose2 }\bigg)
\nonumber\\[4pt]
&\qquad\qquad
=(q,a,b;q)_\infty\sum_{n=0}^\infty\frac{(abq^{n-1};q)_n}
{(q,a,b;q)_n}q^n, \label{Andrews-Warnaar-1}
\end{align}
and
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Andrews-Warnaar Identities}
 \begin{align}
 1+&\sum_{n=1}^\infty(-1)^nq^{n\choose
2}(a^n+b^n)\nonumber\\[4pt]
&\qquad =(a,b,q;q)_\infty \sum_{n=0}^\infty\frac{(ab/q;q)_{2n}}
{(q,a,b,ab;q)_n}q^n, \label{Andrews-Warnaar-2}
\end{align}
 which were proved by {\red  Andrews} and {\red Warnaar}.
 \vspace{0.3cm}

G.E. Andrews and S. Ole Warnaar, The product of
   partial theta functions.  Adv. Applied Math.
   {\bf 39} (2007) 116--120.



\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}}
{\it Proof.}
 Let $g(a)$ and $f(a)$ denote the left side  and right side
 of \eqref{Andrews-Warnaar-1}, respectively. Note that the left hand
 side contains two factors, one is concerned with $a$, and the
 other is involved with $b$.
 Comparing
the coefficients of $a^i$, $i=0,1,2,\ldots$, yields the iteration
relation:
 \begin{align}
 g(a)=(1-a)g(aq)+aqg(aq^2).
\end{align}
However, it is not clear how to derive the same iteration relation
for the right hand side, because it involves both $a$ and $b$.
\end{slide}
}

{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}}
  We will first verify
that $f(a)$ satisfies the same iteration relation as $g(a)$ by
constructing a telescoping relation, which gives a hint of the
possibility that one can use the $q$-Gosper algorithm to achieve the
same goal. This incidental observation was in fact the starting
point of this paper.
 Let
 {\small
 \[f(a)=\sum_{n=0}^\infty D_n(a),\ \ \ \
 D_n(a)=(q,a,b;q)_\infty
\frac{(abq^{n-1};q)_nq^n} {(q,a,b;q)_n}.\]}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1} and
\eqref{Andrews-Warnaar-2}} {\small
 If we can find $u_n$ satisfying
 \begin{align*}
 D_n(a)-(1-a)D_n(aq)-aqD_n(aq^2)=u_{n+1}-u_n
\end{align*}
and
\[
\lim_{n\rightarrow +\infty}u_n=u_0,
\]
  then  sum  $n$ over the non-negative integers and show that
  $f(a)$
 satisfies  the same recurrence  relation as $g(a)$.  $u_n$
 can be
  solved  by applying the  {\yellow $q$-Gasper algorithm} to
\[
D_n(a)-(1-a)D_n(aq)-aqD_n(aq^2).
\]}
\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}}
 In fact, we obtain
\[
u_n=-\frac{(1-q^n)(1-bq^{n-1})(abq^n;q)_n(q,a,b;q)_\infty
aq^n}{(1-aq^n)(1-abq^{2n-1})(q,a,b;q)_n}.
\]
Let $G(a)=g(a)-f(a)$, then $G(a)$ satisfies the recurrence relation
\[
  G(a)=(1-q)G(aq)+aqG(aq^2).
\]
Iterating the above identity, we find
\[
 G(a)=A_nG(aq^{n+1})+B_nG(aq^{n+2}),
\]
\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}}
where {\footnotesize
\begin{align*}
A_0&=(1-q), \qquad\qquad\qquad A_1=(1-a)(1-aq)+aq, \\[4pt]
B_0&=aq,\qquad\qquad\qquad\qquad  B_1=(1-a)aq^2, \\[4pt]
 A_{n+1}&=(1-aq^{n+1})A_n+aq^{n+1}A_{n-1},\qquad
 B_{n+1}=aq^{n+2}A_n \qquad n \geq 1.
\end{align*}}
We can rewrite the recurrence relation of $A_n$ as
\[
A_{n+1}-A_{n}=-aq^{n+1}(A_n-A_{n-1}).
\]

\end{slide}
} {
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}} Iterating the above identity, we have
\begin{align*}
|A_{n+1}&-A_{n}|=|-aq^{n+1}(A_n-A_{n-1})| =\cdots
\\[8pt]
&=|(-1)^na^nq^{(n+1)!}(A_1-A_0)| \\[8pt]
&\leq
|(-1)^na^nq^{(n+1)!}|(|A_1|+|A_0|).
\end{align*}
So, for fixed $a$, $q$, the limit $\lim\limits_{n\rightarrow
+\infty}A_n$ exists. Since $B_{n+1}=aq^{n+2}A_n$, then  the limit
$\lim\limits_{n\rightarrow +\infty}B_n$ exists.
\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Telescoping for \eqref{Andrews-Warnaar-1}
and \eqref{Andrews-Warnaar-2}} It is easy to verify that
$G(0)=g(0)-f(0)=0$. So we can deduce that
 {
\[
  G(a)=G(0)\left(\lim _{n\rightarrow +\infty} A_n+
  \lim _{n\rightarrow +\infty} B_n\right)=0. \quad {\yellow \qed}
\]}

\vspace{0.3cm}
 Using a similar method, we
can prove the identity \eqref{Andrews-Warnaar-2}.



\end{slide}
}

{
\begin{slide}{
The Key Idea} For the identities \eqref{Andrews-Warnaar-1} and
\eqref{Andrews-Warnaar-2},  the recurrence relations of their left
hand sides are easy to establish.  Although their right hand sides
contain infinite $q$-shifted factorials, we can still use the
$q$-Gosper algorithm to solve $u_n$.

\vspace{0.3cm}
 The key idea is that
  we can deal with
infinite $q$-shifted factorials. If we go through the procedure of
the $q$-Gosper algorithm, we will see that the $q$-shifted
factorials will result in rational functions after divisions.
\end{slide}
}

{
\begin{slide}{
The Key Idea} For example, for the identity
\eqref{Andrews-Warnaar-1},  $D_n(q)$ contains infinite $q$-shifted
factorials $(q,a,b;q)_\infty$, but the ratio
\[
\frac{D_{n+1}(a)-(1-a)D_{n+1}(aq)-aqD_{n+1}(aq^2)}
{D_n(a)-(1-a)D_n(aq)-aqD_n(aq^2)}
\]
is a rational function in $q^n$, it follows that $D_n(q)$ is a
$q$-hypergeometric term. Therefore,
   we can directly use
 the $q$-Gosper to solve $u_n$.
\end{slide}
}



%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The Key Idea} {\small As will be seen later, even if
 the recurrence relation of any side of an identity
  is not known, Chen, Hou and
 Mu provided  a method to  get the recurrence relations of both
 sides of the identity.

 \vspace{0.3cm}
Examining the telescoping proofs of \eqref{Andrews-Warnaar-1} and
\eqref{Andrews-Warnaar-2}, we find that the {\yellow $q$-Gosper
algorithm} does apply to $q$-hypergeometric terms which contain
infinite $q$-shifted factorials. {\red  Consequently, the
$q$-Zeilberger algorithm should not be afraid of infinite
$q$-shifted factorials}. This is indeed the idea for the {\yellow
$q$-WZ method} for infinite $q$-series.}
\end{slide}
}



\begin{slide}{Nonterminating Series}
{ Note that the  $q$-Zeilberger algorithm  can not be directly used
to prove nonterminating basic hypergeometric identities. The reason
is that
 the summand is
 not a bivariate $q$-hypergeometric  term in the strict sense.

For nonterminating hypergeometric identities:
\begin{itemize}
\item the Gauss' summation formula {\red(I.M.~Gessel)}

\item Saalsch\"{u}tz summation formula {\red(T.H.~Koornwinder)}
\end{itemize}
}
\end{slide}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Chen-Hou-Mu's Method}
   Recently, Chen, Hou
and Mu  provided  a systematic method for proving nonterminating
basic hypergeometric identities by means of
 the $q$-Zeilberger algorithm.
 \vspace{0.5cm}

  W.Y.C. Chen, Q.H. Hou and Y.P. Mu,  Nonterminating
 basic hypergeometric series
  and the q-Zeilberger algorithm,
   Proc. Edinburgh Math. Soc.  to appear.

\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Chen-Hou-Mu's Method}
 Chen, Hou and Mu's method  can be
 stated as follows. Let
{\small
 \[
   f(x_1,x_2,\ldots, x_l)=\sum_{k=0}^\infty t_k(x_1,x_2,\ldots,
   x_l),
 \]}
where $t_k(x_1,x_2,\ldots,
   x_l)$ is a $q$-hypergeometric term.
   They first {\red set one or more
   parameters $x_1,x_2,\ldots, x_l$ to $x_1q^n,x_2q^n,\dots,
   x_lq^n$}. This is the key step, it makes the summand
   $t_k(x_1q^n,x_2q^n,\ldots,
   x_lq^n)$ become a bivariate $q$-hypergeometric term.


\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Chen-Hou-Mu's Method}
Then
   the {\yellow $q$-Zeilberger algorithm} is utilized to obtain the
    recurrence
   relation. Once they obtain the recurrence  relations
   of both sides,
    the following theorem is used to prove the identity.

\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Chen-Hou-Mu's Method }{\small
\begin{theo}[Chen, Hou and Mu, 2007] \label{CHMTheorem}
Suppose that {\footnotesize
\[
\label{eq-rec} f(z)= a_1(z)f(zq) + a_2(z) f(zq^2) + \cdots +
a_d(z)f(zq^d),
\]}
and there exist $w_1,\ldots,w_d \in {\Bbb C}$ and $M>0$ such that
{\footnotesize
\[
|a_i(z) - w_i| \le M |z|, \quad 1 \le i \le d,
\]}
and {\footnotesize\[ |w_d| + |w_{d-1}+w_d| + \cdots +
|w_2+\cdots+w_d| < 1. \]}
 Then $f(z)$ is uniquely determined by $f(0)$ and the
functions $a_i(z)$.
\end{theo}}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{{\large Chen-Hou-Mu's Method for
\eqref{Andrews-Warnaar-1}
 and \eqref{Andrews-Warnaar-2}}}
 The Andrews-Warnnar identities \eqref{Andrews-Warnaar-1}
  and \eqref{Andrews-Warnaar-2} can be proved by Chen, Hou and Mu's
  method.

\vspace{0.3cm}
  For \eqref{Andrews-Warnaar-1}, set $a$ to $aq^n$, we can
   get that fact that both sides of \eqref{Andrews-Warnaar-1}
   satisfy
    the same recurrence relation. Utilizing the above theorem, we
    can verifty this identity.

\vspace{0.3cm}
    For \eqref{Andrews-Warnaar-2},
      we first
     assume that $|a|<1$, using the
    same method, we can prove that \eqref{Andrews-Warnaar-2}
    holds
    for $|a|<1$.  By {\yellow analytic continuation}, we may drop the
    assumption
that $|a|<1$.

\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
Now we have a version of the $q$-WZ
method for infinite series, keeping in mind that the
$q$-WZ  method
has the advantage of generating certificates to
verify identities.

 \vspace{0.2cm}
Our method can be described as follows. We aim to prove
\begin{align}\label{equation1}
\sum_{k=N_0}^\infty F_k(a_1,a_2,\ldots, a_l) =r(a_1,a_2,\ldots,a_l),
\end{align}

\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
where $l$ is an positive integer, $N_0=0$ or $N_0=-\infty$,
$\sum\limits_{k=N_0}^\infty F_k(a_1,a_2,\ldots, a_l)$ is the form of
the right hand side of \eqref{phi} or \eqref{psi} and {\small
\begin{align*}
 r(a_1,a_2,\ldots,a_l)=
 \frac{\prod\limits_{i=1}^\gamma
  \left(c_i(a_1,a_2,\ldots,a_l);q\right)_\infty}{\prod\limits_{j=1}^
  \lambda\left(d_j(a_1,a_2,\ldots,a_l);q\right)_\infty},
\end{align*}}
$c_i(a_1,a_2,\ldots,a_l)$ and $d_j(a_1,a_2,\ldots,a_l)$ are
functions decided by $a_1,a_2,\ldots,a_l$.


\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
First, we set some parameters $a_1,\ldots,a_p$,  $1\leq p\leq l$
(without loss of generality) to $a_1q^n,\ldots, a_pq^n$, i.e.,
{\small
\begin{align*}
\sum_{k=N_0}^\infty &F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l) \nonumber\\[4pt]
&=r(a_1q^n,\ldots, a_pq^n,a_{p+1},\ldots,a_l).
\end{align*}}
 We remark that the choice of the parameters $a_1, a_2,\ldots, a_p$
is made by human considerations. Nevertheless, there are not many
choices for a few parameters.




\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
 If $r(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)\neq 0$, then let  \[
F(n,k)=\frac{F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)}{r(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)}.
\]
Otherwise, for $r(a_1q^n,\ldots, a_pq^n,a_{p+1},\ldots,a_l)=0$, put
\[
F(n,k)={F_k(a_1q^n,\ldots, a_pq^n,a_{p+1},\ldots,a_l)}.
\]

\end{slide}
}



%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
Once having established that
\begin{align}\label{Constant}
 \sum_{k=N_0}^\infty F(n,k)= { constant}, \qquad
n=0,1,2,\ldots.
\end{align}
we can let $n=0$ and select special cases of $a_1,a_2,\ldots, a_l$
to determine  the constant. This completes the proof of identity
\eqref{equation1}.
\end{slide}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
To use the $q$-WZ method, let $f(n)$ denote the left hand side of
\eqref{Constant}, i.e., $f(n)=\sum\limits_{k=N_0}^\infty F(n,k)$.
Then, we try to prove that $f(n+1)-f(n)=0$ for every nonnegative
integer $n$. One way to achieve this goal is find a function
$G(n,k)$ such that
\begin{align}
\label{WZPair}
F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k).
\end{align}
Then, sum both sides of \eqref{WZPair}   from $k=N_0$ to $+\infty$,
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
{\small under suitable hypotheses, we can show $f(n+1)-f(n)=0$ for
every nonnegative integer $n$.
 A pair of functions $(F(n,k),G(n,k))$ that satisfy \eqref{WZPair} is
called a {\it \red  WZ pair}.
 \vspace{0.3cm}

The question is how  to find  $G(n,k)$?  In fact, it can be solved
by applying the {\yellow $q$-Gasper algorithm} to $F(n+1,k)-F(n,k)$.

 \vspace{0.3cm}
 To do so,  we
shall show that
  $F(n+1,k)-F(n,k)$ is a $q$-hypergeometric term.
   If $r(a_1,\ldots,a_l)= 0$, obviously,
   $F(n+1,k)-F(n,k)$ is a $q$-hypergeometric term.}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
If $r(a_1,\ldots,a_l)\neq0$, then let
 \begin{align*}
 M_1&=\frac{r(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)}{r(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)},\\[4pt]
M_2 &=\frac{F_{k+1}(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)},\\[4pt]
M_3&=\frac{F_{k+1}(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)},\\[4pt]
M_4&=\frac{F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_l)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)}.
 \end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
Since $M_1$ is a rational function in $q^n$ and is independent on
 $k$,
 $M_2, M_3, M_4$ are rational functions in $q^k$,  then
\begin{align*}
 \frac{F(n+1,k+1)-F(n,k+1)}{F(n+1,k)-F(n,k)}=
\frac{M_2-M_1M_3}{1-M_1M_4}
\end{align*}
is a rational function in $q^k$,  i.e., {\red  $F(n+1,k)-F(n,k)$ is
a $q$-hypergeometric term}. So we can employ the $q$-Gosper
algorithm to decide whether such a $G(n,k)$ exists or not.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
 The following
 theorem is used to prove identities and discover
 new identities, which  was provided by
 {\red H.S. Wilf} and {\red D. Zeilberger}.

 \vspace{0.5cm}
H.S. Wilf and D. Zeilberger, Rational functions certify
combinatorial identities, J. Amer. Math. Soc. {\bf3(1)} (1990)
147--158.

In fact, we extension looks the same as the original $q$-WZ method.
We just need to pretend to treat the infinite $q$-shifted factorials
by the finite counterparts after some parameters $a_i$ to $a_iq^n$.


\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
 The following
conditions are used in the theorem:
 \vspace{0.3cm}

(C1) For each integer $n\geq 0$, $\lim\limits_{k\rightarrow\pm
\infty}G(n,k)=0$.

(C2) For each integer $k$, the limit
\begin{align}\label{limit}
f_k=\lim_{n\rightarrow \infty}F(n,k)
\end{align}
\ \ \ \ \ \ \ \ \ \  exists and is finite.

(C3) We have $\lim\limits_{L\rightarrow \infty}\sum\limits_{n\geq0}
G(n,-L)=0$.

\end{slide}
}



%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $q$-WZ method for Infinite Series}
{\small
\begin{theo}[H.S. Wilf and D. Zeilberger, 1990]
Let $(F(n,k),G(n,k))$ satisfy \eqref{WZPair}.
 If  (C1) holds then we have
 the identity
\begin{align}\label{Con1}
\sum_{k}F(n,k)= constant,\ \ \ \ \ \ \ n=0,1,2,\ldots.
\end{align}
If (C2) and (C3) hold then we have the identity (companion identity)
\begin{align}\label{Identity}
\sum_{n=0}^\infty G(n,k)=\sum_{j\leq k-1}(f_j-F(0,j)),
\end{align}
 where $f_j$ is defined by \eqref{limit}.
\end{theo}}
\end{slide}
}

%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{$q$-Gauss sum}
We give some examples.


 \vspace{0.5cm}
\noindent{\bf {Example 1.}}
 The $q$-Gauss sum  is
\begin{align*}
\sum_{k=0}^\infty \frac{(a,b;q)_k}{(q,c;q)_k}
\left(\frac{c}{ab}\right)^k =\frac{(c/a,c/b;q)_\infty}
{(c,c/ab;q)_\infty},
\end{align*}
where $|c/ab|<1$.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{$q$-Gauss sum}
Consider the pair functions
\begin{align*}
F(n,k)&=\frac{(aq^n,b;q)_k(cq^n,c/ab;q)_\infty}
{(q,cq^n;q)_k(c/a,cq^n/b;q)_\infty}
\left(\frac{c}{ab}\right)^k,\\[8pt]
G(n,k)&=-\frac{(a-aq^k)(aq^n,b;q)_k(cq^n,c/ab;q)_\infty}
{(q,cq^n;q)_k(c/a,cq^n/b;q)_\infty(1-aq^n)}
\left(\frac{c}{ab}\right)^k q^n.
\end{align*}
Since $|c/ab|<1$, it is easy to verify that the two functions
$(F(n,k),G(n,k))$ satisfy the relation \eqref{WZPair} and conditions
(C1), (C2) and $C(3)$.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{$q$-Gauss sum}
By \eqref{Con1}, we have
\[
\sum_{k=0}^\infty F(n,k)=constant, \qquad n=0,1,2,\ldots.
\]
In order to determine the constant, let $c=0$ and $n=0$, then the
constant is 1, so we have
\[
\sum_{k=0}^\infty F(0,k)=1,
\]
which is $q$-Gauss sum.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{{\large The Companion Identity of the $q$-Gauss sum}}
By \eqref{Identity}, we obtain the companion identity of the
$q$-Gauss sum
 is
 \begin{align*}
  \sum_{j=0}^k&\frac{(a,b;q)_j}{(q,c;q)_j}\left(\frac{c}{ab}\right)^j
  =\frac{(c/b;q)_\infty}{(c;q)_\infty}\sum_{j=0}^k
  \frac{(b;q)_j}{(q;q)_j}\left(\frac{c}{ab}\right)^j\\[8pt]
  & \quad +\frac{(a,b;q)_{k+1}c^{k+1}}{(q;q)_k(c;q)_{k+1}a^kb^{k+1}}
  \sum_{n=0}^\infty
  \frac{(aq^{k+1},c/b;q)_n}{(a;q)_{n+1}(cq^{k+1};q)_n}q^n.
 \end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The $_6\phi_5$ Sum}
\noindent{\bf {Example 2.}}

 The sum of a very-well-poised $_6\phi_5$
series is
\begin{align}\label{65}
\sum_{k=0}^\infty &
\frac{(1-aq^{2k})(a,b,c,d;q)_k}{(1-a)(q,aq/b,aq/c,aq/d)_k}
\left(\frac{aq}{bcd}\right)^k \nonumber\\[8pt]
&=\frac{(aq,aq/bc,aq/bd,aq/cd;q)_\infty}
{(aq/b,aq/c,aq/d,aq/bcd;q)_\infty},\ \ \ \ \ \ \ |aq/bcd|<1.
\end{align}
\end{slide}
}


%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The  $_6\phi_5$ Sum}
 It is easy to  check that when $|aq/bcd|<1$ the two functions
\begin{align*}
F(n,k)=&\frac{(1-aq^{n+2k})(aq^n,bq^n,c,d;q)_k}
{(1-aq^n)(q,aq/b,aq^{n+1}/c,aq^{n+1}/d;q)_k}\\[4pt]
& \times\frac{(aq/b,aq^{n+1}/c,aq^{n+1}/d,aq/bcd;q)_\infty}
{(aq^{n+1},aq/bc,aq/bd,aq^{n+1}/cd;q)_\infty}
\left(\frac{aq}{bcd}\right)^k
\end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The  $_6\phi_5$ Sum}
and
\begin{align*}
G(n,k)=&\frac{(1-q^k)(c,d;q)_k(a/b,a/bcd;q)_\infty}
{(aq^n/c,aq^n/d;q)_k(aq^n,aq^n/cd;q)_\infty}\\[4pt]
&\times
\frac{(aq^n,bq^n;q)_k(aq^n/c,aq^n/d;q)_\infty}{(a/bd,a/bc;q)_\infty
(aq^{n+k}-c)(aq^{n+k}-d)}\\[4pt]
&\times \frac{(a-bc)(a-bd)(aq^n-cd)}{(a/b,q;q)_k(a-bcd)(bq^n-1)}
\left(\frac{aq}{bcd}\right)^kq^n
\end{align*}
are a WZ-pair and satisfy the conditions (C1), (C2) and (C3).
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The  $_6\phi_5$ Sum}
 Then,
by \eqref{Con1}, $\sum\limits_{k=0}^\infty F(n,k)$ is a constant.
Let $n=0$ and $a=0$, so the constant is 1, then  we have
\[ \sum_{k=0}^\infty F(0,k)={ constant}=1,\]
which completes the proof. By \eqref{Identity}, we obtain the
companion identity of \eqref{65}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{The Companion Identity of  $_6\phi_5$}
\begin{align*}
\sum_{j=0}^k&\frac{(1-aq^{2j})(a,b,c,d;q)_j }
{(1-a)(q,aq/b,aq/c,aq/d;q)_j}
\left(\frac{aq}{bcd}\right)^k\\
=&\frac{(aq,aq/cd;q)_\infty}{(aq/c,aq/d;q)_\infty}
\sum_{j=0}^k\frac{(c,d;q)_j}
{(q,aq/b;q)_j}\left(\frac{aq}{bcd}\right)^j\\
&+\frac{b(aq;q)_k(b,c,d;q)_{k+1}}{(q,aq/b;q)_k(aq/c,aq/d;q)_{k+1}}
\left(\frac{aq}{bcd}\right)^{k+1}\\
&\times\sum_{n=0}^\infty
\frac{(aq^{k+1},bq^{k+1};q)_n(aq/cd;q)_n}{(b;q)_{n+1}
(aq^{k+2}/c,aq^{k+2}/d;q)_{n}}q^n.
\end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Ramanujan's  $_1\psi_1 $ Sum}
\noindent{\bf {Example 3.}} The Ramanujan's $_1\psi_1$ sum  is
\begin{align*}
\sum_{k=-\infty}^\infty \frac{(a;q)_k}{(b;q)_k}
z^k=\frac{(q,b/a,az,q/az;q)_\infty}{(b,q/a,z,b/a;q)_\infty},
\end{align*}
where $|b/a|<|z|<1$.
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Ramanujan's $_1\psi_1 $ Sum}
 The  functions
\begin{align*}
F(n,k)&=\frac{(aq^n;q)_k(bq^n,q^{1-n}/a,z,b/az;q)_\infty}
{(bq^n;q)_k(q,b/a,azq^n,q^{1-n/az};q)_\infty}z^k,\\[8pt]
G(n,k)&=\frac{(aq^n;q)_k(bq^n,q^{-n}/a,z,b/az;q)_\infty(1-azq^n)}
{(bq^n;q)_k(q,b/a,azq^n,q^{-n/az};q)_\infty(z-azq^n)}z^k
\end{align*}
 are a WZ pair. Since $|b/a|<|z|<1$, then we can verify that
 $G(n,k)$ satisfies the condition (C1), so by \eqref{Con1},
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Ramanujan's  $_1\psi_1 $ Sum}
 \[
   \sum_{k=-\infty}^\infty F(n,k)= constant, \qquad n=0,1,2,\ldots.
 \]
  In order to determine the constant, let $n=0$ and $b=q$ and utilize
the $q$-binomial theorem, we get the constant is 1. Inserting the
constant 1 and $n=0$ into the above  identity, we obtain
  the Ramanujan's $_1\psi_1$ sum.
\end{slide}
}



%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Bailey's $_6\psi_6 $ Sum}
\noindent{\bf {Example 4.}} The Bailey $_6\psi_6$ sum  is
\begin{align*}
\sum_{k=-\infty}^\infty&\frac{(1-aq^{2k})(b,c,d,e;q)_k}
{(1-a)(aq/b,aq/c,aq/d,aq/e;q)_k}
\left(\frac{a^2q}{bcde}\right)^k \nonumber\\[8pt]
=&\frac{(aq,aq/bc,aq/bd,aq/be,aq/cd;q)_\infty}
{(aq/b,aq/c,aq/d,aq/e,q/b;q)_\infty}\nonumber\\[8pt]
&\times\frac{aq/ce,aq/de,q,q/a;q)_\infty}
{(q/c,q/d,q/e,a^2q/bcde;q)_\infty}.
\end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Bailey's $_6\psi_6 $ Sum}
The WZ-pair that works here is
\begin{align*}
F(n,k)&=\frac{(1-aq^{n+2k}) (aq/b,aq/c,aq^{n+1}/d;q)_\infty}
{(1-aq^n)
(aq^{n+1},aq^{1-n}/bc,aq/be,aq/ce;q)_\infty}\\[8pt]
& \quad \times \frac{(bq^n,cq^n,d,e;q)_k( aq^{n+1}/e,q^{1-n}/b;q
)_\infty}{(aq/b,aq/c,aq^{n+1}/d,aq^{n+1}/e;q)_k(
aq/bd;q)_\infty}\\[8pt]
&\quad \times\frac{(q^{1-n}/c,q/d,q/e,a^2q/bcde;q)_\infty}{(
aq/cd,aq^{n+1}/de,q,q^{1-n}/a;q)_\infty} \left(\frac{a^2q}
{bcde}\right)^k
\end{align*}
and
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Bailey's $_6\psi_6 $ Sum}
\begin{align*}
G(n,k)&=\frac{(bq^n,cq^n,d,e;q)_k(a/b,a/c;q)_\infty}{(1-bq^n)
(1-cq^n)(a-ad)(1-e)(a^2-bcde)}\\[4pt]
&\times \frac{(aq^n/d,aq^n/e,q^{-n}/b ,q^{-n}/c,a^2/bcde;q)_\infty}
{(a/b,a/c;q)_k(a/bd,a/be,aq^n,aq^{-n}/bc;q)_\infty}\\[4pt]
&\times \frac{(-1+aq^n)(a-bd)(1/d,1/e;q)_\infty}
{(aq^{n+k}-d)(aq^n/d;q)_k(aq^{n+k}-e)(aq^n/e;q)_k
}\\[4pt]
&\times\frac{(a-be)(a-cd)(a-ce)(aq^n-de)q^n}
{(a/cd,a/ce,aq^n/de,q,q^{-n}/a;q)_\infty}\left(\frac{a^2q}
{bcde}\right)^k.
\end{align*}
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Bailey's $_6\psi_6 $ Sum}
Since $|a^2q/bcde|<1$,  we can verify that  $G(n,k)$ satisfies the
condition (C1), so by \eqref{Con1},
\begin{align} \label{Ra2}
\sum_{k=-\infty}^\infty F(n,k)=  constant, \ \ \ \ \ n=0,1,2,\ldots.
\end{align}
In order to determine  the constant, let $n=0$ and $b=a$,   by the
 the sum of a very-well-poised $_6\phi_5$ series  \eqref{65}, we have
\end{slide}
}
%------------------------------------------------------------------------------
%---------------------------------------------------------------------- SLIDE -
{
\begin{slide}{Bailey's $_6\psi_6 $ Sum}
{\small
\begin{align*}
&\sum_{k=0}^\infty\frac{(1-aq^{2k})(a,c,d,e;q)_k}
{(1-a)(aq/c,aq/d,aq/e;q)_k
}\\[2pt]
&\quad \times \frac{(aq,aq/cd,aq/ce,aq/de;q)_\infty}
{(aq/c,aq/d,aq/e,aq/cde;q)_\infty}\left(\frac{aq}{cde}\right)^k=1.
\end{align*}}
Then, let $n=0$, we have {\small
\[
\sum_{k=-\infty}^\infty F(0,k)=constant=1,
\]}
which completes the proof.
\end{slide}
}




\begin{slide}{}
\vspace{2cm}
\begin{center}
{\LARGE Thank You}
\end{center}
\end{slide}

\end{document}
