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\begin{center}
 { \bf  SOME THETA FUNCTIONS IDENTITIES ASSOCIATED 
WITH THE MODULAR EQUATIONS OF DEGREE ${\bf 5}$}
\vskip 20pt
{\bf Zhi-Guo Liu}\\ 
{\smallit  Xinxiang Education College, Xinxiang, Henan 453000, 
People's Republic of China}\\
{\tt xxlzg@public.xxptt.ha.cn}\\
\end{center}
\vskip 30pt
\centerline{\smallit Received:  2/24/00, Revised: 2/21/01 Accepted: 
3/10/01, Published:  3/19/01}
\vskip 30pt


\centerline{\bf Abstract}

\noindent
In this paper the author proves a general theta functions
identity by using  the theory of elliptic functions. 
 This identity allows the author to derive four interesting theta 
functions identities. These identities lead to new proofs of   
some well-known identities of Ramanujan  associated with 
the modular equations of degree $5$. Some new identities are also
discussed.


\pagestyle{myheadings}
\markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF
COMBINATORIAL NUMBER THEORY \smalltt 1(2001), \#A03\hfill}

\thispagestyle{empty}
\baselineskip=15pt
\vskip 30pt

\section*{\normalsize 1. Introduction}
\addtocounter{section}{+1}
\setcounter{equation}{0}
We suppose throughout that  $q:=e^{2\pi i\tau}$,  Im $(\tau)>0$;
this condition ensures that all the sums and products that
appear here converge.
 We will use the standard $q$-notations:
\begin{eqnarray*}
(a; q)_\infty&=&\prod_{n=1}^\infty (1-aq^{n-1}),
\\
(a_1, a_2\dots a_m; q)_\infty
&=&\prod_{k=1}^m (a_k; q)_\infty.
\end{eqnarray*}

\noindent
 The well-known Jacobi's triple product identity 
\cite [pp. 21-22]{Andrews},
\cite [p. 35] {Ber}, \cite {Hir, Lewis} is
\begin{equation}
(q, z, q/z; q)_\infty=\sum_{n=-\infty}^\infty
(-1)^n q^{\fr{n(n-1)}{2}}z^n.
\label{0-eq1}
\end{equation}

\noindent
Using the above identity,  M. D. Hirschhorn \cite{Hir} derived the following
 septagonal  numbers identity 
\begin {eqnarray}
&&(a^3-a-a^{-1}+a^{-3}) (a^2q, a^{-2}q, 
a^4q, a^{-4}q; q)_\infty (q; q)^2_\infty
\nonumber\\
&&\qquad=(q^2, q^3, q^5;  q^5)_\infty
\sum_{n=-\infty}^\infty(-1)^n (a^{10n-3}+a^{-10n+3})q^{\fr{n(5n-3)}{2}}
\nonumber\\
&&\qquad-(q, q^4,  q^5; q^5)_\infty
\sum_{n=-\infty}^\infty (-1)^n (a^{10n-1}+a^{-10n+1})q^{\fr{n(5n-1)}{2}}.
\label{0-eq2}
\end{eqnarray}

Using this identity, Hirschhorn provided a simpler proof of 
the following identity of Ramanujan \cite[p. 139]{Raml},
\cite{Chan, Hir, Rav}:
\begin{equation}
1-5 \sum_{n=1}^\infty \lt(\fr{n}{5}\rt) 
\fr{nq^n}{1-q^n}
=\fr{(q; q)^5_\infty}{(q^5; q^5)_\infty},
\label{0-eq3}
\end{equation}
where $\lt(  \fr{n}{5}\rt)$ denote the Legendre symbol.

In \cite{F+K},  H. M. Farkas and I. Kra  rediscovered (\ref{0-eq2}) using the 
theory of theta functions with rational characteristics.  
F. G. Garvan \cite {Gar} provided an interesting generalization of
(\ref{0-eq2}) and  his proof depends only on the triple product identity.

Using the Jacobi theta function $\T_1(z|q)$ ((\ref{1-eq1}) below), 
(\ref{0-eq2}) can be reformulated as
\begin{eqnarray}
iq^{-\fr{1}{2}} \T_1(x | q) \T_1(2x | q) =
&\hskip -20pt \T_1(2\pi\tau |
q^5)\lt\{e^{2ix} \T_1(5x+ \pi\tau | q^5) -e^{-2ix}\T_1(5x-\pi\tau |
q^5)\rt\}
\nonumber\\
&\hskip -5pt - \T_1(\pi\tau | q^5)\lt\{e^{4ix} \T_1(5x+ 2\pi\tau | q^5)
-e^{-4ix}\T_1(5x-2\pi\tau | q^5)\rt\}.
\label{0-eq4}
\end{eqnarray}
 
In Section $3$ of this paper we will  prove a very general identity 
(identity (\ref{2-eq2}) below)  involving theta functions.
 
In Section $4$, we will derive  (\ref{0-eq4}) from identity (\ref{2-eq2}),
and then prove (\ref{0-eq3}) using (\ref{0-eq4}). We will also prove
the following important result of Ramanujan:
\begin{equation}
\fr{1}{R(q^5)}-1-R(q^5)=\fr{(q; q)_\infty}{q(q^{25}; q^{25})_\infty},
\label{0-eq5}
\end{equation}
   where
\begin{equation}
q^{-\fr{1}{5}}R(q)=\fr{(q,  q^4; q^5)_\infty}{(q^2,  q^3; q^5)_\infty}.
\label{0-eq6}
\end{equation}

  The following identity will be established 
in Section $5$ using (\ref{2-eq2}):
\begin{eqnarray}
5\T_1(x | q^5) \T_1(2x | q^5)
&=&\T_1(\fr{2\pi}{5} | q)\lt\{\T_1(x+ \fr{\pi} {5}| q)
-\T_1(x-\fr{\pi}{5} | q)\rt\}
\nonumber\\
&-&\T_1(\fr{\pi}{5}| q)\lt\{ \T_1(x+\fr {2\pi}{5}| q)
-\T_1(x-\fr{2\pi} {5} | q)\rt\}.
\label{0-eq7}
\end{eqnarray}



\noindent
>From this identity, we can obtain the following identity of 
Ramanujan \cite[p. 139]{Raml}, \cite{Chan, Rav}:
\begin{equation} 
\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{q^n}{(1-q^n)^2}
=q\fr{(q^5; q^5)^5_\infty}{(q; q)_\infty}.
\label{0-eq8}
\end{equation}

\noindent 
Using (\ref{0-eq7}),  we can also derive  the identity
 \begin{equation}
\fr{\T_1(\fr{2\pi}{5}|q)}{\T_1(\fr{\pi}{5}|q)}
-\fr{\T_1(\fr{\pi}{5}|q)}{\T_1(\fr{2\pi}{5}|q)}
=1+5q\fr{(q^{25}; q^{25})_\infty}{(q; q)_\infty}.
\label{0-eq9}
\end{equation}

In Section $6$, we will prove
\begin{eqnarray}
&C(q)\T_1(x | q) \T_1(2x | q)=&\T^5_1(\fr{2\pi}{5} | q)\lt\{\T^5_1(x+
\fr{\pi} {5}| q) -\T^5_1(x-\fr{\pi}{5} | q)\rt\}
\nonumber\\
&&-\T^5_1(\fr{\pi}{5}| q)\lt\{ \T^5_1(x+\fr {2\pi}{5}| q)
-\T^5_1(x-\fr{2\pi} {5} | q)\rt\},
\label{0-eq10}
\end{eqnarray}
where 
\begin{equation}
C(q)=250 q (q; q)^4_\infty (q^5; q^5)^4_\infty
+3125 q^2 \fr{(q^5; q^5)^{10}_\infty}{(q; q)^2_\infty},
\label{0-eq11} 
\end{equation}

\noindent
Using (\ref{0-eq10}), we can obtain
\begin{equation}
\fr{\T^5_1(\fr{2\pi}{5}|q)}{\T^5_1(\fr{\pi}{5}|q)}
-\fr{\T^5_1(\fr{\pi}{5}|q)}{\T^5_1(\fr{2\pi}{5}|q)}
=11+125q\fr{(q^5; q^5)^6_\infty}{(q; q)^6_\infty}.
\label{0-eq12}
\end{equation} 

 Section $7$  is devoted  to the proof of the following identity:
\begin{eqnarray}
\hskip -20pt C(q)\T_1(x | q^5) \T_1(2x | q^5)=&\hskip -25pt
\T^5_1(2\pi\tau | q^5)\lt\{e^{2ix}
\T^5_1(x+ \pi\tau | q^5) -e^{-2ix}\T^5_1(x-\pi\tau | q^5)\rt\}
\\
&\hskip -5pt -\T^5_1(\pi\tau | q^5)\lt\{e^{4ix} \T^5_1(x+ 2\pi\tau | q^5)
-e^{-4ix}\T^5_1(x-2\pi\tau | q^5)\rt\},
\label{0-eq13}
\end{eqnarray}
where
\begin{equation}
q^{\fr{5}{2}}C(q)=10q (q; q)^4_\infty(q^5; q^5)^4_\infty
+\fr{(q; q)^{10}_\infty}{(q^5; q^5)^2_\infty}.
\label{0-eq14}
\end{equation}

\noindent
Using this identity, we can rederive the following identity of Ramanujan:
\begin{equation}
\fr{1}{R^5(q)}-11-R^5(q)=\fr{(q; q)^6_\infty}{q(q^5; q^5)^6_\infty}.
\label{0-eq15}
\end{equation}

It should be pointed out that we also need the method of 
L. -C. Shen \cite{Shen} 
in deriving  (\ref{0-eq4}), (\ref{0-eq7}), (\ref{0-eq10}) and (\ref{0-eq13}).

Identities (\ref{0-eq5}) and (\ref{0-eq15}) can be found in Ramanujan's second 
notebook \cite[pp. 265-267]{Rams} and were first proved by G. N. Watson \cite{Wat}
for the purpose of establishing some of Ramanujan's claims about $R(q)$ made
in his first two letters to Hardy \cite[pp. xxvii, xxviii]{Rama}. They were used
by B. C.   Berndt,  H. H. Chan, and L. -C.  Zhang \cite{B+C+Z} in deriving 
general formulas for the explicit evaluation of $R(q).$  
They were also used by the author and 
R. P. Lewis \cite{L+L} to provide simpler proofs of two Lambert series
 identities of Ramanujan.

In \cite {Chan},  Chan  utilized the Hecke correspondence
 between Dirichlet series and Fourier expansions of modular forms to show that
(\ref{0-eq3}) and (\ref{0-eq4}) are equivalent.
\vskip 30pt


\section* {\normalsize 2.   Some Basic Facts  About  $\T_1(z | q)$}
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For $q=e^{2\pi i \tau}$,  with Im$(\tau)>0$, 
 the Jacobi theta function $\T_1(z | q)$ is  
defined by \cite[p. 463]{W-W}
\begin{eqnarray}
\T_1(z|q):&=&-iq^{\fr{1}{8}} \sum_{n=-\infty}^\infty (-1)^n
q^{\fr{1}{2}n(n+1)} e^{(2n+1)iz}\nonumber\\
&=&2q^{\fr{1}{8}}\sum_{n=0}^\infty (-1)^n
q^{\fr{1}{2}n(n+1)}  \sin (2n+1)z.
\label{1-eq1}
\end{eqnarray}
In terms of infinite products \cite[p. 469]{W-W},
\begin{eqnarray}
\T_1(z|q)&=&2q^{\fr{1}{8}}(\sin z) (q; q)_\infty(q e^{2iz};
q)_\infty
(q e^{-2iz}; q)_\infty \nonumber\\
&=&i q^{\fr{1}{8}}e^{-iz}
(q; q)_\infty ( e^{2iz}; q)_\infty (q e^{-2iz}; q)_\infty.
\label{1-eq2}
\end{eqnarray}
One deduces easily from (\ref{1-eq1}) that (see,  for example, \cite
{Shen})
\begin{equation}
\fr {{\partial}^2 {\T_1}}{\partial z^2}=-8q
\fr{\partial \T_1}{\partial q}.
\label{1-eq3}
\end{equation}

   In the following, $\T'_1$ and $\T''_1$  denote the first  and second
partial derivatives of $\T_1$ with respect to $z$.
We can derive the following two important identities 
\cite{Shen} from (\ref{1-eq2}):
\begin{eqnarray}
\fr{\T'_1}{\T_1}(z | q)&=&
\cot z +4\sum_{n=1}^\infty \fr{q^n}{1-q^n}\sin 2nz,
\label{1-eq4} \\
\fr{\T''_1}{\T_1}(z | q)&=&
-1+8\sum_{n=1}^\infty\fr{nq^n e^{2iz}}{1-q^n e^{2iz}}
+ 8\sum_{n=1}^\infty\fr{nq^n e^{-2iz}}{1-q^n e^{-2iz}}
+8\sum_{n=1}^\infty \fr{nq^n }{1-q^n }
\nonumber\\
 &=&-1+16\sum_{n=1}^\infty \fr{q^n}{(1-q^n)^2}\cos 2nz
+8\sum_{n=1}^\infty \fr{n q^n}{1-q^n}.
\label{1-eq5} 
\end{eqnarray}

 Differentiating (\ref{1-eq4}) with respect to $z$, we obtain
\begin {equation}
  \lt(\fr{\T'_1}{\T_1}\rt)'(z | q)=
-\csc^2 z +8\sum_{n=1}^\infty \fr{nq^n}{1-q^n}\cos 2nz.
\label{1-eq6}
\end{equation}
Using  these identities and the simple differential identity
\begin{equation}
\lt(\T'_1/ \T_1 \rt)^2=\T''_1/ \T_1-\lt(\T'_1/ \T_1\rt)',
\label{1-eq7}
\end{equation}
an interesting proof of the following trigonometric series 
identity of Ramanujan  \cite {Ram}, \cite[pp. 134-135]
{Hardy} has recently been given by  Shen \cite{Shen}:
\begin{eqnarray}
&&\lt\{\fr{1}{4}\cot z +\sum_{n=1}^\infty \fr{q^n\sin 2nz}{1-q^n} \rt\}^2
\nonumber\\
&=&\lt\{\fr{1}{4}\cot z\rt\}^2
+\sum_{n=1}^\infty \fr{q^n \cos 2nz}{ (1-q^n)^2}
+\fr{1}{2}\sum_{n=1}^\infty \fr{nq^n (1-\cos 2nz)}{1-q^n}.
\label{1-eq8}
\end{eqnarray}  

By some elementary calculations we find the following  trigonometric
 functions identities:
\begin{eqnarray}
\csc^2 \fr{\pi}{5}-\csc^2 \fr{2\pi}{5}&=&\fr{4}{\sqrt{5}},
\label{1-eq9}\\
\cos \fr{2n\pi}{5} -\cos \fr{4n\pi}{5}
&=&\fr{\sqrt{5}}{2}\lt(\fr{n}{5}\rt).
\label{1-eq10} 
\end{eqnarray}
Replacing $z$ by $\fr{\pi}{5}$ and $\fr{2\pi}{5}$ in (\ref{1-eq4})
 respectively and then subtracting the two resulting equations
 and finally using  (\ref{1-eq9}) we find that
\begin{equation}
\fr{\T''_1}{\T_1}\lt(\fr {\pi}{5}\Bigl | q \rt)
  -\fr{\T''_1}{\T_1} \lt(\fr{2 \pi}{5} \Bigl | q \rt)
=8\sqrt{5}\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{q^n}{(1-q^n)^2}.
\label{1-eq11}
\end{equation}

  Replacing $z$ by $\fr{\pi}{5}$ and $\fr{2\pi}{5}$ in (\ref{1-eq5}) respectively
and then subtracting the two resulting equations and finally using
(\ref{1-eq9}), (\ref{1-eq10}) and (\ref{0-eq3}) we find that
\begin{eqnarray}
  \lt(\fr{\T'_1}{\T_1}\rt)'\lt(\fr {\pi}{5}\Bigl | q \rt)-
  \lt(\fr{\T'_1}{\T_1}\rt)' \lt(\fr{2 \pi}{5} \Bigl | q \rt)
&=&-\fr{4}{\sqrt{5}}
 \lt\{1-5 \sum_{n=1}^\infty \lt(\fr{n}{5}\rt) 
\fr{nq^n}{1-q^n}\rt\}
\nonumber\\
&=&-\fr{4}{\sqrt{5}}\fr{(q; q)^5_\infty}{(q^5; q^5)_\infty}.
\label{1-eq12}
\end{eqnarray}

 Applying  logarithmic differentiation
 to (\ref{1-eq2}),  we find that
\begin{equation}
\fr{\T'_1}{\T_1}(z|q)=
-i-2i\sum_{n=0}^\infty\fr{q^n e^{2iz}}{1-q^n e^{2iz}}
+2i\sum_{n=1}^\infty\fr{q^n e^{-2iz}}{1-q^n e^{-2iz}}.
\label{1-eq13}
\end{equation}
Differentiating (\ref{1-eq13}) with respect to $z$, we find that
\begin{equation}
\lt(\fr{\T'_1}{\T_1}\rt)'(z | q)=
4\sum_{n=0}^\infty\fr{q^n e^{2iz}}{(1-q^n e^{2iz})^2}
+4\sum_{n=1}^\infty\fr{q^n e^{-2iz}}{(1-q^n e^{-2iz})^2}.
\label{1-eq14}
\end{equation}
Replacing $q$ by $q^5$, setting
 $z=\pi\tau$ and  $z=2\pi\tau$ respectively,  we obtain
\begin{eqnarray*}
\lt(\fr{\T'_1}{\T_1}\rt)'(\pi\tau | q^5)&=&
4\sum_{n=0}^\infty\fr{q^{5n+1} }{(1-q^{5n+1 })^2}
+4\sum_{n=0}^\infty\fr{q^{5n+4} }{(1-q^{5n+4} )^2},
\\
\lt(\fr{\T'_1}{\T_1}\rt)'(2\pi\tau | q^5)&=&
4\sum_{n=0}^\infty\fr{q^{5n+2} }{(1-q^{5n+2 })^2}
+4\sum_{n=0}^\infty\fr{q^{5n+3} }{(1-q^{5n+3} )^2}.
\end{eqnarray*}
Therefore, we have
\begin{equation}
\lt(\fr{\T'_1}{\T_1}\rt)'(\pi\tau | q^5)
-\lt(\fr{\T'_1}{\T_1}\rt)'(2\pi\tau | q^5)
=4\sum_{n=1}^\infty\lt(\fr{n}{5}  \rt) \fr{q^n}{(1-q^n)^2}.
\label{1-eq15}
\end{equation}

>From (\ref{1-eq1}),  we find that
\begin{eqnarray}
e^{2iz} \T_1(5z+\pi \tau | q^5)
&=&iq^{\fr{1}{8}}\sum_{n=-\infty}^\infty(-1)^n q^{\fr{n(5n-3)}{2}} e^{(10n-3)iz},
\label{a}\\
e^{4iz} \T_1(5z+2\pi \tau | q^5)
&=&iq^{-{\fr{3}{8}}}\sum_{n=-\infty}^\infty(-1)^n q^{\fr{n(5n-1)}{2}} e^{(10n-1)iz}.
\label{b}
\end {eqnarray}

Differentiating both sides of  (\ref{a}) twice with respect to $z$, and
then setting $z=0$ in the resulting equation, we obtain
\begin{eqnarray}
&&-4\T_1(\pi\tau|q^5)+20i\T'_1(\pi \tau|q^5)
+25\T''_1(\pi\tau|q^5)
\nonumber\\
&=&-iq^{\fr{1}{8}}\sum_{n=-\infty}^\infty (-1)^n (10n-3)^2q^{\fr{n(5n-3)}{2}}
\nonumber\\
&=&-iq^{\fr{1}{8}} \lt( 9+40q\fr{d}{dq} \rt)
\sum_{n=-\infty}^\infty (-1)^n q^{\fr{n(5n-3)}{2}}
\nonumber\\
&=&-iq^{\fr{1}{8}}\lt( 9+40q\fr{d}{dq} \rt) (q, q^4, q^5; q^5)_\infty.
\label{c}
\end{eqnarray}
By logarithmic differentiation, we have
\begin{eqnarray}
&&(q, q^4, q^5; q^5)^{-1}_\infty \lt( 9+40q\fr{d}{dq} \rt)
(q, q^4, q^5; q^5)_\infty
\nonumber\\
 &=&  9-40\sum_{n=1}^\infty \lt(\fr{5nq^{5n}}{1-q^{5n}}
+\fr{(5n-4)q^{5n-4}}{1-q^{5n-4}}+\fr{(5n-1)q^{5n-1}}{1-q^{5n-1}}\rt).
\label{d}
\end{eqnarray}
Substituting (\ref{d}) into (\ref{c}) and using (\ref{eq18}) below  we
obtain
\begin{eqnarray}
&&4-20i\fr{\T'_1}{\T_1}(\pi\tau|q^5)-25\fr{\T''_1}{\T_1}(\pi\tau|q^5)
\nonumber\\
&=& 9-40\sum_{n=1}^\infty \lt(\fr{5nq^{5n}}{1-q^{5n}}
+\fr{(5n-4)q^{5n-4}}{1-q^{5n-4}}+\fr{(5n-1)q^{5n-1}}{1-q^{5n-1}}\rt).
\label{e}
\end{eqnarray}
Similarly, from (\ref{b}) we obtain
 \begin{eqnarray}
&&16-40i\fr{\T'_1}{\T_1}(2\pi\tau|q^5)-25\fr{\T''_1}{\T_1}(2\pi\tau|q^5)
\nonumber\\
&=& 1-40\sum_{n=1}^\infty \lt(\fr{5nq^{5n}}{1-q^{5n}}
+\fr{(5n-3)q^{5n-3}}{1-q^{5n-3}}+\fr{(5n-2)q^{5n-2}}{1-q^{5n-2}}\rt).
\label{f}
\end{eqnarray}
Substracting (\ref{e}) from (\ref{f}) we obtain
\begin{eqnarray}
12+20i \lt\{ \fr{\T'_1}{\T_1}(\pi\tau|q^5)-2\fr{\T'_1}{\T_1}(2\pi\tau|q^5) \rt\}
+25\lt\{\fr{\T''_1}{\T_1}(\pi\tau|q^5)-\fr{\T''_1}{\T_1}(2\pi\tau|q^5)\rt\}
\nonumber\\
=-8\lt\{1-5\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{nq^n}{1-q^n}\rt\}.
\label{1-eq16}
\end{eqnarray}

Differentiating (\ref{1-eq2}) with respect to $z$ and
 then putting $z=0$ we obtain the identity
\begin{equation}
\T'_1(q):=\T_1'(0|q)=2q^{\fr{1}{8}}(q; q)_\infty^3.
\label{1-eq17}
\end{equation}

Replacing $q$ by $q^5$ in (\ref{1-eq2}) and then taking
$z=\pi \tau$, and $2\pi\tau$ in the resulting equation   
respectively we find that
\begin{eqnarray}
\T_1(\pi\tau|q^5)&=&iq^{\fr{1}{8}}(q; q^5)_\infty
(q^4; q^5)_\infty(q^5; q^5)_\infty,
\label{eq18}\\
\T_1(2\pi\tau|q^5)&=&iq^{-{\fr{3}{8}}}(q^2; q^5)_\infty
(q^3; q^5)_\infty(q^5; q^5)_\infty.
\label{1-eq18}
\end{eqnarray}
Multiplying the above two equations, 
   after  some  manipulation  we find that
\begin{equation}
\T_1(\pi\tau|q^5)\T_1(2\pi\tau|q^5)
=-q^{-{\fr{1}{4}}}(q; q)_\infty (q^5; q^5)_\infty.
\label{1-eq19}
\end{equation}

>From (\ref{1-eq2}) we  have the following special values
of $\T_1(z|q)$
\begin{eqnarray}
\T_1(\fr{2\pi}{5}|q)&=&2q^{\fr{1}{8}}(\sin \fr{\pi}{5})
(q; q)_\infty (qe^{\fr{2\pi i}{5}}; q)_\infty
(qe^{\fr{-2\pi i}{5}}; q)_\infty,
\label{eq20}\\
\T_1(\fr{2\pi}{5}|q)&=&
2q^{\fr{1}{8}}(\sin \fr{\pi}{5})
(q; q)_\infty (qe^{\fr{4\pi i}{5}}; q)_\infty
(qe^{\fr{-4\pi i}{5}}; q)_\infty.
\label{1-eq20}
\end{eqnarray}
Multiplying the above equations and then using the
elementary facts
 \[
\sin \fr{\pi}{5}\sin \fr{2\pi}{5}=\fr{\sqrt{5}}{4}
\]
and
\[
(1-x)(1-xe^{\fr{2\pi i}{5}})
(1-xe^{\fr{2\pi i}{5}})(1-xe^{\fr{2\pi i}{5}})
(1-xe^{\fr{2\pi i}{5}})=1-x^5,
\]
in the resulting equation,  we find that
\begin{equation}
\T_1(\fr{\pi}{5}|q)\T_1(\fr{2\pi}{5}|q)
=\sqrt{5}q^{\fr{1}{4}}(q; q)_\infty (q^5; q^5)_\infty.
\label{1-eq21}
\end{equation}

>From the definition of $\T_1(z|q)$ we  readily  find
the following functional equations
\begin{eqnarray}
\T_1(z+n\pi| q)&=&(-1)^n\T_1(z|q),
\label{1-eq22}\\
 \T_1(z+n\pi\tau | q)&=&(-1)^n q^{-{\fr{n^2}{2}}}
e^{-2n\pi iz}\T_1(z|q),
\label{1-eq23}
\end{eqnarray}
where $n$ is any integer. Differentiating the above equations 
with respect to $z$ and then setting $z=0$ in the resulting equations,
 we   find that
\begin{equation}
\T'_1(n\pi | q)=(-1)^n\T'_1(q)
\quad
 \mbox{and}
\quad
\T'_1(n\pi\tau | q)=(-1)^n q^{-{\fr{n^2}{2}}}\T'_1(q).
\label{1-eq24}
\end{equation}

\vskip 30pt


\section*{\normalsize 3. A General Identity  For $\T_1(z|q)$ }
\addtocounter{section}{+1}
\setcounter{equation}{0}

In this section we will prove the following theta functions identity.
\begin{thm} \label {t0} If  $f_1(z)$ and $f_2(z)$ are two different
entire functions satisfying the functional equations 
\begin {equation}
f(z+\pi)=-f(z) \qquad {\mathrm{and}} \qquad
f(z+\pi\tau)=-q^{-\fr{5}{2}}e^{-10iz}f(z),
\label{2-eq1}
\end{equation}
with $f_1(0)\ne {0}$,  $f_2(0)\ne {0}$,
then there is a constant $C(q)$ such that
\begin{eqnarray}
C(q)\T_1(x | q)\T_1(2x | q)
=f_2(0) \lt(f_1(x)+f_1(-x)\rt)
-f_1(0)\lt(f_2(x)+f_2(-x)\rt).
\label{2-eq2}
\end{eqnarray}
\end{thm}

To prove the above identity  we require the following 
fundamental theorem of elliptic functions
\cite[p. 22, Theorem 2]{Chand}:

\begin{thm}\label{t1}
The sum of  all the residues of an elliptic
function vanishes in the period parallelogram.
\end{thm}

\noindent
 The idea is  to construct an elliptic function whose poles are
known and then compute the residues of the elliptic function at these
poles. Set the sum of the residues to zero to obtain 
the desired identity for theta functions. 
 We will use L'H\^{o}pital's rule to compute the residues.

\noindent
 {\it  Proof.} Let $0<x,  y<\pi$ be two distinct parameters 
 different from the zeros of $f_1(z)$.
By (\ref{1-eq23}) and (\ref{1-eq24}) we readily verify that
\[
\T_1(z | q) \T_1(z-x | q)\T_1(z+x | q)\T_1(z-y | q)\T_1(z+y | q)
\]
satisfies  (\ref{2-eq1}). Therefore,
\[
E(z)=\fr{f_1(z)}
{\T_1(z|q)\T_1(z-x|q)\T_1(z+x|q)\T_1(z-y|q)\T_1(z+y|q)},
\]
is an elliptic function with periods $\pi$ and $\pi\tau$.
 The poles of $E(z)$ are $0, x,  \pi-x,  y,$ and  $\pi-y$, 
all of which are simple poles. Let res$(E; \alpha)$ denote the 
residue of $E(z)$ at $\alpha$. From Theorem \ref{t1}, we have 
\begin{equation}
\res(E; 0)+\res(E; x)+\res(E; \pi-x)+\res(E; y)+\res(E; \pi-y)=0.
\label{2-eq3}
\end{equation}
Now 
\begin{eqnarray}
\res(E;0)&=&\lim_{z\to 0}zE(z)
\nonumber\\
  &=&\lim_{z\to 0} \fr{f_1(z)}
          {\T_1(z-x|q)\T_1(z+x|q)\T_1(z-y|q)\T_1(z+y|q)}
\times\lim_{z\to 0} \fr{z}{\T_1(z|q)}
\nonumber\\
&=&\fr{f_1(0)}{\T_1'(q)\T_1^2(x|q)\T_1^2(y|q)}.
\label{2-eq4}
\end{eqnarray}
\begin{eqnarray}
\res(E;x)&=&\lim_{z\to x}(z-x)E(z)
\nonumber\\
  &=&\lim_{z\to x} \fr{f_1(z|q)}
          {\T_1(z|q)\T_1(z+x|q)\T_1(z-y|q)\T_1(z+y|q)}
\times\lim_{z\to x} \fr{z-x}{\T_1(z-x|q)}
\nonumber\\
&=&\fr{f_1(x)}{\T'_1(q)\T_1(x | q)\T_1(2x|q)\T_1(x-y|q)\T_1(x+y|q)}.
\label{2-eq5}
\end{eqnarray}
\begin{eqnarray}
\res(E;\pi-x)&=&\lim_{z\to{\pi- x}}(z-\pi+x)E(z)
\nonumber\\
  &=&\lim_{z\to{\pi- x}} \fr{f_1(z)}
          {\T_1(z|q)\T_1(z-x|q)\T_1(z-y|q)\T_1(z+y|q)}
\times\lim_{z\to {\pi-x}} \fr{z-\pi+x}{\T_1(z+x|q)}
\nonumber\\
&=&-\fr{f_1(\pi-x)}
{\T_1' (\pi|q)\T_1(\pi-x | q)\T_1(\pi-2x|q)\T_1(\pi-x-y|q)\T_1(\pi-x+y|q) }
\nonumber\\
&=&\fr{f_1(-x)}{\T'_1(q)\T_1(x | q)\T_1(2x|q)\T_1(x-y|q)\T_1(x+y|q)}.
\label{2-eq6}
\end{eqnarray}
Similarly, we have
\begin{eqnarray}
\res(E; y)&=&
-\fr{f_1(y)}{\T_1'(q)\T_1(y|q)\T_1(2y | q)\T_1(x-y|q)\T_1(x+y|q)},
\label{2-eq7}
\\
\res(E; \pi-y)&=&
-\fr{f_1(-y)}{\T_1'(q)\T_1(y|q)\T_1(2y | q)\T_1(x-y|q)\T_1(x+y|q)}.
\label{2-eq8}
\end{eqnarray}
Substituting (\ref{2-eq5}), (\ref{2-eq6}), (\ref{2-eq7}) 
and  (\ref{2-eq8}) into (\ref{2-eq3}) we obtain
\begin {eqnarray}
\fr{f_1(y)+f_1(-y)}{f_1(0)\T_1(y|q)\T_1(2y|q)}
&-&\fr{f_1(x)+f_1(-x)}{f_1(0)\T_1(x|q)\T_1(2x|q)}
=\fr{\T_1(x+y|q)\T_1(x-y|q)}{\T_1^2(x|q)\T_1^2(y|q)}.
\label{2-eq9}
\end{eqnarray}

In the same way we can obtain the identity for $f_2(z)$:

\begin {eqnarray}
\fr{f_2(y)+f_2(-y)}{f_2(0)\T_1(y|q)\T_1(2y|q)}
&-&\fr{f_2(x)+f_2(-x)}{f_2(0)\T_1(x|q)\T_1(2x|q)}
=\fr{\T_1(x+y|q)\T_1(x-y|q)}{\T_1^2(x|q)\T_1^2(y|q)}.
\label{2-eq10}
\end{eqnarray}

By comparing (\ref{2-eq9}) and (\ref{2-eq10}), we obtain
\begin {eqnarray}
&&\fr{f_1(x)+f_1(-x)}{f_1(0)\T_1(x|q)\T_1(2x|q)}
-\fr{f_2(x)+f_2(-x)}{f_2(0)\T_1(x|q)\T_1(2x|q)}
\nonumber\\
&=& \fr{f_1(y)+f_1(-y)}{f_1(0)\T_1(y|q)\T_1(2y|q)}
-\fr{f_2(y)+f_2(-y)}{f_2(0)\T_1(y|q)\T_1(2y|q)}.
\label{2-eq11}
\end{eqnarray}
>From this identity we readily find that
\[
\fr{f_1(x)+f_1(-x)}{f_1(0)\T_1(x|q)\T_1(2x|q)}
-\fr{f_2(x)+f_2(-x)}{f_2(0)\T_1(x|q)\T_1(2x|q)}
\]
is independent of $x$, so it must be a constant,
say $\fr{C(q)}{f_1(q)f_2(q)}$. This completes
the proof of Theorem \ref{t0}.
 By  analytic continuation, 
we know (\ref{2-eq2}) holds for all $x$.

  Identity (\ref{2-eq2}) is a very general identity  involving theta
functions.
 In the next sections we will choose some special functions $f_1(z)$ and $f_2(z)$
to obtain some interesting identities for theta functions.
\vskip 30pt                             


\section* {\normalsize 4.  The Proofs of  (\ref{0-eq3}), (\ref{0-eq4}),
and (\ref{0-eq5}) }
\addtocounter{section}{+1}
\setcounter {equation} {0}

Using (\ref{1-eq23}), it is easy to verify that 
 $f_1(z)=e^{2iz}\T_1(5z+\pi\tau|q^5)$ 
and $f_2(z)=e^{4iz}\T_1(5z+2\pi\tau|q^5)$
satisfy all the conditions of Theorem \ref{t0}.
  Taking $f_1(z)=e^{2iz}\T_1(5z+\pi\tau|q^5)$ 
and $f_2(z)=e^{2iz}\T_1(5z+2\pi\tau|q^5)$ in Theorem
\ref{t0}, we find that
\begin{eqnarray}
C(q) \T_1(x | q) \T_1(2x | q)=& \hskip -25pt \T_1(2\pi\tau |
q^5)\lt\{e^{2ix}
\T_1(5x+
\pi\tau | q^5) -e^{-2ix}\T_1(5x-\pi\tau | q^5)\rt\}
\nonumber\\
&\hskip -5pt -\T_1(\pi\tau | q^5)\lt\{e^{4ix} \T_1(5x+ 2\pi\tau | q^5)
-e^{-4ix}\T_1(5x-2\pi\tau | q^5)\rt\}.
\label{3-eq1}
\end{eqnarray} 
To determine $C(q)$, we take $x=\fr{\pi}{5}$ in (\ref{3-eq1}).
After some simple calculations we find that
\begin{equation}
C(q)\T_1(\fr{\pi}{5}|q)\T_1(\fr{2\pi}{5}|q)
=2i (\cos \fr{4\pi}{5}-\cos \fr{2\pi}{5})
\T_1(\pi\tau|q^5)\T_1(2\pi\tau|q^5).
\label{3-eq2}
\end{equation}
Substituting (\ref{1-eq19}) and (\ref{1-eq21}) into the  above identity we
find that
\begin{equation}
C(q)=\fr{2}{\sqrt{5}}i (\cos \fr{2\pi}{5}-\cos \fr{4\pi}{5})q^{-\fr{1}{2}}
=iq^{-\fr{1}{2}}.
\label{3-eq3}
\end{equation}
Using (\ref{3-eq3}) in (\ref{3-eq1}) we obtain (\ref{0-eq4}).

Replacing $q$ by $q^5$, setting $x=\pi\tau$, 
and  using (\ref{1-eq18}) we obtain (\ref{0-eq5}).

Differentiating both sides of (\ref{0-eq4}) twice with respect to $x$ and
setting $x=0$, we find that
\begin{eqnarray}
12+20i \lt\{ \fr{\T'_1}{\T_1}(\pi\tau|q^5)-2\fr{\T'_1}{\T_1}(2\pi\tau|q^5) \rt\}
+25\lt\{\fr{\T''_1}{\T_1}(\pi\tau|q^5)-\fr{\T''_1}{\T_1}(2\pi\tau|q^5)\rt\}
\nonumber\\
=\fr{4q^{-\fr{1}{4}} \T'_1(q)^2}
{\T_1(\pi\tau|q^5) \T_1(2\pi\tau|q^5)}.
\label{3-eq4}
\end{eqnarray}
Substituting (\ref{1-eq16}), (\ref{1-eq17}), and (\ref{1-eq19}) into
the above identity we obtain (\ref{0-eq3}).
\vskip 30pt                             


\section* {\normalsize 5. The Proofs of  (\ref{0-eq7}),
 (\ref{0-eq8}), and
(\ref{0-eq9}) }
\addtocounter{section}{+1}
\setcounter {equation} {0} 

Taking $f_1(z)=\T_1(z+\fr{\pi}{5}|q^{\fr{1}{5}})$ 
and $f_2(z)=\T_1(z+\fr{2\pi}{5}|q^\fr{1}{5})$ in Theorem
\ref{t0}, we find that

\begin{eqnarray}
&C(q) \T_1(x | q) \T_1(2x | q)
=&\T_1(\fr{2\pi}{5} | q^{\fr{1}{5}})
\lt\{ \T_1(x+ \fr{\pi}{5} | q^{\fr{1}{5}})
-\T_1(x-\fr{\pi}{5} | q^{\fr{1}{5}})\rt\}
\nonumber\\
&&  -\T_1(\fr{\pi}{5}| q^{\fr{1}{5}})
\lt\{ \T_1(x+ \fr{2\pi}{5} | q^{\fr{1}{5}})
-\T_1(x-\fr{2\pi}{5} | q^{\fr{1}{5}})\rt\}.
\label{4-eq1}
\end{eqnarray} 
Replacing $q$ by $q^5$ we obtain
\begin{eqnarray}
&C(q^5) \T_1(x | q^5) \T_1(2x | q^5)
=&\T_1(\fr{2\pi}{5} | q)
\lt\{ \T_1(x+ \fr{\pi}{5} | q)
-\T_1(x-\fr{\pi}{5} | q)\rt\}
\nonumber\\
&&-\T_1(\fr{\pi}{5}| q)
\lt\{ \T_1(x+ \fr{2\pi}{5} | q)
-\T_1(x-\fr{2\pi}{5} | q)\rt\}.
\label{4-eq2}
\end{eqnarray}

To determine $C(q^5)$, we take $x=\pi\tau$ in (\ref{4-eq2}).
Using (\ref{1-eq23}), after some simple calculations, we find that
\begin{equation}
C(q^5)\T_1(\pi\tau|q^5)\T_1(2\pi\tau|q^5)
=2q^{-\fr{1}{2}} (\cos \fr{4\pi}{5}-\cos \fr{2\pi}{5})
\T_1(\fr{\pi}{5}|q)\T_1(\fr{2\pi}{5}|q).
\label{4-eq3}
\end{equation}
Substituting (\ref{1-eq19}) and (\ref{1-eq21}) into the  above identity we
find that
\begin{equation}
C(q^5)=2{\sqrt{5}} (\cos \fr{2\pi}{5}-\cos \fr{4\pi}{5})q^{-\fr{1}{2}}
=5.
\label{4-eq4}
\end{equation}
Using (\ref{4-eq4}) in (\ref{4-eq2}) we obtain (\ref{0-eq7}).

Setting $x=\fr{\pi}{5}$ in (\ref{0-eq7})
and using (\ref{1-eq21}) we obtain (\ref{0-eq8}).

Differentiating both sides of (\ref{0-eq7}) twice with respect to $x$ and
setting $x=0$, we find that
\begin{equation}
\fr{\T''_1}{\T_1}\lt(\fr {\pi}{5}\Bigl | q \rt)
  -\fr{\T''_1}{\T_1} \lt(\fr{2 \pi}{5} \Bigl | q \rt)
=\fr{10 \T'_1(q^5)^2}
{\T_1(\fr{\pi}{5}|q) \T_1(\fr{2\pi}{5}|q)}.
\label{4-eq5}
\end{equation}
Substituting (\ref{1-eq11}), (\ref{1-eq17}), and (\ref{1-eq21}) in
the above identity we obtain (\ref{0-eq9})
\vskip 30pt                             


\section* {\normalsize 6.  The Proofs of  (\ref{0-eq10})
and (\ref{0-eq12})}
\addtocounter{section}{+1}
\setcounter {equation} {0}
 
Taking $f_1(z)=\T^{5}_1(z+\fr{\pi}{5}|q)$ 
and $f_2(z)=\T^{5}_1(z+\fr{2\pi}{5}|q)$ in Theorem
\ref{t0}, we find that
\begin{eqnarray}
&C(q) \T_1(x | q) \T_1(2x | q)
=&\T^5_1(\fr{2\pi}{5} | q)
\lt\{ \T^5_1(x+ \fr{\pi}{5} | q)
-\T^5_1(x-\fr{\pi}{5} | q)\rt\}
\nonumber\\
&&-\T^5_1(\fr{\pi}{5}| q)
\lt\{ \T^5_1(x+ \fr{2\pi}{5} | q)
-\T^5_1(x-\fr{2\pi}{5} | q)\rt\}.
\label{5-eq1}
\end{eqnarray} 

 Differentiating both sides of (\ref{5-eq1}) 
twice with respect to $x$ and
setting $x=0$, we find that
\begin{eqnarray}
\fr{\T''_1}{\T_1}\lt(\fr {\pi}{5}\Bigl | q \rt)
  -\fr{\T''_1}{\T_1} \lt(\fr{2 \pi}{5} \Bigl | q \rt)
+4\lt(\fr{\T'_1}{\T_1}\rt)^2\lt(\fr{\pi}{5} \Bigl|q\rt)
-4\lt(\fr{\T'_1}{\T_1}\rt)^2\lt(\fr{2\pi}{5}\Bigl|q\rt)
=\fr{2C(q) \T'_1(q)^2}
{5\T^5_1(\fr{\pi}{5}|q) \T^5_1(\fr{2\pi}{5}|q)}.
\label{5-eq2}
\end{eqnarray}
Using (\ref{1-eq7}) we have
 \begin{eqnarray}
5\lt\{\fr{\T''_1}{\T_1}\lt(\fr {\pi}{5}\Bigl | q \rt)
\!\!  - \!\fr{\T''_1}{\T_1} \lt(\fr{2 \pi}{5} \Bigl | q \rt)
\rt\} \!
-4\lt\{\lt(\fr{\T'_1}{\T_1}\rt)' \!\! \lt(\fr{\pi}{5}\Bigl|q\rt) \!\!
- \!\!\lt(\fr{\T'_1}{\T_1}\rt)' \!\! \lt(\fr{2\pi}{5}\Bigl|q\rt)\rt\}
=\fr{2C(q) \T'_1(q)^2}
{5\T^5_1(\fr{\pi}{5}|q) \T^5_1(\fr{2\pi}{5}|q)}.
\label{5-eq3}
\end{eqnarray}
Substituting (\ref{1-eq11}), (\ref{1-eq12}), (\ref{1-eq17})
and (\ref{1-eq21}) into the above equation we obtain
\begin{eqnarray}
40\sqrt{5}\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{q^n}{(1-q^n)^2}
-\fr{16}{\sqrt{5}}
\lt\{1-5\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{n q^n}{1-q^n}\rt\}
=\fr{8q^{-1}C(q)(q; q)_\infty}{125\sqrt{5}(q^5; q^5)^5_\infty}.
\label{5-eq4}
\end{eqnarray}
Substituting (\ref{0-eq3}) and (\ref{0-eq8})
into the above equation we obtain
\begin{equation}
C(q)=250 q (q; q)^4_\infty (q^5; q^5)^4_\infty
+3125 q^2 \fr{(q^5; q^5)^{10}_\infty}{(q; q)^2_\infty},
\label{5-eq5} 
\end{equation}
thereby completing the proof of (\ref{0-eq10}). 
Taking $x=\fr{\pi}{5}$ in (\ref{0-eq10})
and using (\ref{1-eq21}) we obtain (\ref{0-eq12}).
\vskip 30pt                              


\section* {\normalsize 7. The Proofs of  (\ref{0-eq13})
and (\ref{0-eq15}) }
\addtocounter{section}{+1}
\setcounter {equation} {0} 

Taking $f_1(z)=e^{2iz}\T^{5}_1(z+\fr{\pi\tau}{5}|q)$ 
and $f_2(z)=e^{4iz}\T^{5}_1(z+\fr{2\pi\tau}{5}|q)$ in Theorem
\ref{t0}, we find that
\begin{eqnarray}
&\hskip -10pt C(q) \T_1(x | q) \T_1(2x | q)
&\!\!\!=\T^5_1(\fr{2\pi\tau}{5} | q)
\lt\{ e^{2ix}\T^5_1(x+ \fr{\pi\tau}{5} | q)
-e^{-2ix}\T^5_1(x-\fr{\pi\tau}{5} | q)\rt\}
\nonumber\\
&&\hskip 8pt -\T^5_1(\fr{\pi\tau}{5}| q)
\lt\{ e^{4ix}\T^5_1(x+ \fr{2\pi\tau}{5} | q)
-e^{-4ix}\T^5_1(x-\fr{2\pi\tau}{5} | q)\rt\} \!\!.
\label{6-eq1}
\end{eqnarray}
Replacing $q$ by $q^5$, the above identity becomes
 \begin{eqnarray}
&\hskip -25pt C(q^5) \T_1(x | q^5) \T_1(2x | q^5)
&\!\!\!=\T^5_1(2\pi\tau | q^5)
\lt\{ e^{2ix}\T^5_1(x+ \pi\tau | q^5)
-e^{-2ix}\T^5_1(x-\pi\tau | q^5)\rt\}
\nonumber\\
&&\hskip 3pt -\T^5_1(\pi\tau| q^5)
\lt\{ e^{4ix}\T^5_1(x+ 2\pi\tau | q^5)
-e^{-4ix}\T^5_1(x-2\pi\tau | q^5)\rt\} \!\!.
\label{6-eq2}
\end{eqnarray}
 Differentiating both sides of (\ref{6-eq2}), twice, with respect to $x$,
setting $x=0$, we find that
\begin{eqnarray}
12+20i\lt\{\fr{\T'_1}{\T_1}(\pi\tau|q^5)
-2 \fr{\T'_1}{\T_1}(2\pi\tau|q^5)  \rt\}
+5\lt\{\fr{\T''_1}{\T_1}(\pi\tau | q^5 )
  -\fr{\T''_1}{\T_1} (2 \pi\tau | q^5 )\rt\}
\nonumber\\
+20\lt\{\lt(\fr{\T'_1}{\T_1}\rt)^2(\pi\tau|q^5)
-\lt(\fr{\T'_1}{\T_1}\rt)^2(2\pi\tau|q^5)\rt\}
=\fr{2C(q^5) \T'_1(q^5)^2}
{\T^5_1(\pi\tau|q^5) \T^5_1(2\pi\tau|q^5)}.
\label{6-eq3}
\end{eqnarray}
Using (\ref{1-eq7}) we have
 \begin{eqnarray}
12+20i\lt\{\fr{\T'_1}{\T_1}(\pi\tau|q^5)
-2 \fr{\T'_1}{\T_1}(2\pi\tau|q^5)  \rt\}
+25\lt\{\fr{\T''_1}{\T_1}(\pi\tau | q^5 )
  -\fr{\T''_1}{\T_1} (2 \pi\tau | q^5 )\rt\}
\nonumber\\
-20\lt\{\lt(\fr{\T'_1}{\T_1}\rt)'(\pi\tau|q^5)
-\lt(\fr{\T'_1}{\T_1}\rt)'(2\pi\tau|q^5)\rt\}
=\fr{2C(q^5) \T'_1(q^5)^2}
{\T^5_1(\pi\tau|q^5) \T^5_1(2\pi\tau|q^5)}.
\label{6-eq4}
\end{eqnarray}
Substituting (\ref{1-eq15}), (\ref{1-eq16}), (\ref{1-eq17})
and (\ref{1-eq19}) into the above equation we obtain
\begin{eqnarray}
-8\lt\{1-5\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{n q^n}{1-q^n}\rt\}
-80\sum_{n=1}^\infty \lt(\fr{n}{5}\rt)
\fr{q^n}{(1-q^n)^2}
=-\fr{8q^{\fr{5}{2}}C(q^5)(q^5; q^5)_\infty}{(q; q)^5_\infty}.
\label{6-eq5}
\end{eqnarray}
Using (\ref{0-eq3}) and (\ref{0-eq8})
in the above equation we obtain
\begin{equation}
q^{\fr{5}{2}}C(q^5)=10 q (q; q)^4_\infty (q^5; q^5)^4_\infty
+ q \fr{(q; q)^5_\infty}{(q^5; q^5)_\infty},
\label{6-eq6} 
\end{equation}
thereby completing the proof of (\ref{0-eq13}).
 Taking $x=\pi\tau$ in (\ref{0-eq13})
and using (\ref{1-eq18}) we obtain (\ref{0-eq15}).
\vskip 30pt 

\section*{\normalsize Acknowledgements}
I  would like to thank the referee and 
the managing editor for their valuable comments.
\vskip 30pt 


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