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\centerline
{\bf A Maple Rendition of George Polya's Elementary Proof\footnote{$^\alpha$}
{\eightrm G. Polya, Sitz. der Phys.-Math. Klasse, Berlin (1927), 158-161. See also 
R. Bellman, ``A Brief Introduction to  Theta Functions'' (1961), pp. 40-41.}
of the Modularity of the Theta Function
}
\bigskip
\centerline
{
\it Dr. Z.'s Experimental Math Spring 2018 class
\footnote{$^\beta$}{\eightrm Yonah Biers-Ariel, Matthew P. Charnley, Amit Harel, George D. Hauser, Edna L. Jones, Ahsan Z. Khan, Brooke Logan, Yukun Yao,
Anthony Zaleski. Instructor: Doron Zeilberger.}
}

{\bf Trivial Fact 1\footnote{$^\gamma$}{\eightrm Binomial Theorem.}: 
$$
(z^{\frac{1}{2}}+z^{-\frac{1}{2}})^{2m} = \sum_{v=-m}^{m} {{2m} \choose {m+v}}\, z^v \quad .
$$

{\bf Trivial Fact 2\footnote{$^\delta$}
{\eightrm Sum of a geometric series $\sum_{-l/2 < v < l/2}  z^v=\frac{z^{l/2}-z^{-l/2}}{z^{1/2}-z^{-1/2}}$ \quad .}} \quad }:
For any (Laurent) polynomial $P(z)={{\sum} \atop {N_1<v<N_2}} \, a_v \, z^v$ and any positive odd integer $l$,  if $\omega=e^{2\pi i/l}$,
$$
\sum_{-\frac{l}{2} < v < \frac{l}{2} } P(\omega^v z) \, =\, l \, \sum_{\frac{N_1}{l} < v < \frac{N_2}{l}}  a_{lv} z^{lv} \quad .
$$
Combining (and dividing by $2^{2m}$):
$$
\sum_{-\frac{l}{2} < v <\frac{l}{2}}  \frac{1}{2^{2m}} \,((\omega^v \, z)^{1/2} +  (\omega^v \, z)^{-1/2} )^{2\,m} \, = \,
l \, \sum_{ - \frac{m}{l} <v < \frac{m}{l}} \frac{1}{2^{2m}}\, {{2m} \choose {m+l\,v}} z^{lv} \quad .
$$
Now let $z=e^{s/l}$ and $l=\lfloor \sqrt{ m \, t} \rfloor$ , and take the limits as $m \rightarrow \infty$.
Go to  {\tt maple}, and {\it copy-and-paste}:

{\tt l:=sqrt(m*t): z:=exp(s/l): omega:=exp(2*Pi*I/l):
L:= ( ( ((omega**v*z)**(1/2) +  (omega**v*z)**(-1/2)))/2)**(2*m): R:= l*binomial(2*m,m+l*v)*z**(l*v)/2**(2*m): \hfill\break
factor(limit(L, m=infinity)); simplify(limit(R,m=infinity));
}

and get, respectively,
$$
{{\rm e}^{{\frac { -\left( 2\,\pi \,v-is \right) ^{2}}{4t}}}} \quad , \quad 
{\frac {\sqrt {t}\,{{\rm e}^{v \left( -tv+s \right) }}}{\sqrt {\pi }}} \quad ,
$$
yielding:
$$
\sum_{v=-\infty}^{\infty} e^{(s+2\pi i v)^2/(4t)} \, = \, (\frac{t}{\pi})^{1/2} \sum_{v=-\infty}^{\infty} e^{-tv^2+sv} \quad .  \quad \halmos 
$$

{\eightrm April 24, 2018.}

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