We introduce the Jacobi Triple Product Formula [1] here. Actually it can be easily obtained from the Ramanujan ${}_1\psi_1$ Summation Formula [2] with some proper coefficients. Here we denote $f$ as the Ramanujan Theta Function which is defined in [3] as
$$
f(a,b) = \sum_{k=-\infty}^\infty a^{k(k+1)/2} b^{k(k-1)/2}
$$
for $|ab|<1$. Then we have the following.
Theorem. (Jacobi Triple Product Formula) $$ f(a,b) = (-a; ab)_\infty (-b; ab)_\infty (ab; ab)_\infty $$ |
Proof. First, we recall the Ramanujan ${}_1\psi_1$ Summation Formula:
\begin{align}
&1+\sum_{k=1}^\infty \frac{(\frac{1}{\alpha}; q^2)_k (-\alpha q)^k}{(\beta q^2;q^2)_k} z^k
+
\sum_{k=1}^\infty \frac{(\frac{1}{\beta}; q^2)_k (-\beta q)^k}{(\alpha q^2;q^2)_k} z^{-k}\\
&=
\left( \frac{(-qz;q^2)_\infty (-q/z;q^2)_\infty}{(-\alpha qz; q^2)_\infty (-\beta q/z; q^2)_\infty} \right)
\left( \frac{(q^2;q^2)_\infty (\alpha \beta q^2;q^2)_\infty}{(\alpha q^2; q^2)_\infty (\beta q^2; q^2)_\infty} \right).
\end{align}
Setting $a = qz$, $b = q/z$ and $\alpha, \beta \to 0$ so that $q^2 = ab$ in (1), we deduce that
\begin{eqnarray*}
(\text{lhs})
&=& \sum_{k=-\infty}^\infty q^{k^2} z^k \\
&=& \sum_{k=-\infty}^\infty (qz)^{(k(k+1)/2} (qz^{-1})^{k(k-1)/2}\\
&=& \sum_{k=-\infty}^\infty a^{k(k+1)/2} b^{k(k-1)/2} \\
&=& f(a,b)
\end{eqnarray*}
and
\begin{eqnarray*}
(\text{rhs})
&=&
(-qz; q^2)_\infty (-q/z; q^2)_\infty (q^2; q^2)_\infty\\
&=&
(-a; ab)_\infty (-b; ab)_\infty (ab; ab)_\infty,
\end{eqnarray*}
which proves Theorem.$\square$
References.
[1] http://en.wikipedia.org/wiki/Jacobi_triple_product
[2] Leun Kim, [Calculation 16] Ramanujan ‘s 1ψ1 (1-psi-1) Summation Formula.
[3] Leun Kim, [Calculation 15] Introduction to the Ramanujan Theta Functions.
[4] Burce C. Berndt, Ramanujan’s Notebooks Part III, Springer-Verlag, p. 35.
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