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\begin{large}
\begin{center}
{\bf Corrections to Solution Set 5, Problem 6}
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As some of you might have noticed, there were some sign errors in the 
solution to problem 6 of problem set 5.  Here is the corrected solution.
I would like to thank Guy Fisher for pointing out the mistakes and 
supplying the corrections.

\bigskip

{\bf Problem 6}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Problem 6

Given that $x<1$ and $y<1$, express the following in closed form

\begin{Large}
\begin{equation*}
\prod_{p=1}^{\infty} \prod_{q=p}^{\infty} e^{ -{pq \over 2}x^{p/2}y^q}
\end{equation*}
\end{Large}

\bigskip


\begin{eqnarray*}
\prod_{p=1}^{\infty} \prod_{q=p}^{\infty} e^{ -{pq \over 2}x^{p/2}y^q} =\\
e^{ - \sum_{p=1}^{\infty} \sum_{q=p}^{\infty}{pq \over 2}x^{p/2}y^q}\\ 
\end{eqnarray*}

Now we compute  
$\sum_{p=1}^{\infty} \sum_{q=p}^{\infty} {pq \over 2}x^{p/2}y^q$.

\begin{eqnarray*}
\sum_{q=p}^{\infty}{pq \over 2}x^{p/2}y^q & = &
{p \over 2}x^{p/2} \sum_{q=p}^{\infty}q y^q\\
& = & {p \over 2}x^{p/2} \left( \sum_{q=1}^{\infty}q y^q
- \sum_{q=1}^{p-1}q y^q \right)\\
& = & {p \over 2}x^{p/2} y 
\left( {py^{p-1} \over 1-y} + {y^p \over (1-y)^2} \right)\\
\end{eqnarray*}


We know from tutorial that

$$
\sum_{k=1}^{\infty} k^2 a^k = {2a^2 \over (1-a)^3} + {a \over (1-a)^2}
$$

We will use this to compute the sum over $p$.

\begin{eqnarray*}
\sum_{p=1}^{\infty} \sum_{q=p}^{\infty} {pq \over 2}x^{p/2}y^q & = &
\sum_{p=1}^{\infty} {p \over 2}x^{p/2} y
\left( {py^{p-1} \over 1-y} + {y^p \over (1-y)^2} \right)\\
& = & {1 \over 2(1-y)} \sum_{p=1}^{\infty} p^2 (yx^{1/2})^p
+ {y \over 2(1-y)^2} \sum_{p=1}^{\infty} p(yx^{1/2})^p\\ 
& = & {1 \over 2(1-y)} \left( {2xy^2 \over (1-yx^{1/2})^2} 
+ {yx^{1/2} \over (1-yx^{1/2})^2} \right) 
+ {y \over 2(1-y)^2} {yx^{1/2} \over (1-yx^{1/2})^2}
\end{eqnarray*}

Therefore the final result is

\[
\prod_{p=1}^{\infty} \prod_{q=p}^{\infty} e^{ -{pq \over 2}x^{p/2}y^q} =
exp{-\left(  {1 \over 2(1-y)} \left( {2xy^2 \over (1-yx^{1/2})^2} 
+ {yx^{1/2} \over (1-yx^{1/2})^2} \right) 
+ {y \over 2(1-y)^2} {yx^{1/2} \over (1-yx^{1/2})^2} \right) }
\]


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