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\begin{document}

\section{Diversion on $\pi$}

It's $\pi$ day.  Someone asked me to prove something about $\pi$.
Here we go:

\begin{theorem} $\pi^2$ is irrational
\end{theorem}

To prove this theorem I will use a lemma about a special function.
\begin{lemma}
Define 
\[
f(x) = \frac{x^n(1-x)^n}{n!}.
\]
Then
\begin{enumerate}
\item \label{i:poly} $f(x)$ is a polynomial
  $\frac{1}{n!}\sum_{i=n}^{2n}c_ix^i$.
\item \label{i:small} For $-<x<1$ we have $0<f(x)<1/n!$.
\item \label{i:int} For any $k$, we the $k^{th}$ derivatives
  $f^{(k)}(0)$ and $f^{(k)}(1)$ are integers
\end{enumerate}
\end{lemma}
\begin{proof}
  For (\ref{i:poly}), ignore the $n!$ term.  The coefficients of
  $x^n(1-x)^n$ are integers by induction on $n$.  Also $x^n$ divides
  every term, so the smallest term is $x^n$, and the max degree is
  $2n$.

  For (\ref{i:small}), note that $x(1-x)<1$ so clearly any integer
  power is less than 1 (induction again).

  For (\ref{i:int}), note that by (\ref{i:poly}) the $k^{th}$
  derivative is 0 unless $n \le k \le 2n$.  In this range the
  derivative $f^{(k)}(0) = \frac{k!}{n!}c_k$ is an integer since $k>n$
  and $c_k$ is an integer.  Also since $f(x)=f(1-x)$ we have
  $f^{(k)}(x) = (-1)^kf^{(k)}(1-x)$ so $f^{(k)}(1)=(-1)^{k}f^{(k)}(0)$
  is an integer too.
\end{proof}

With this lemma we can prove the theorem.  Let's assume $\pi^2=a/b$
for integer $a$ and $b$.  Consider the polynomial
\[
F(x) = b^n\sum_{k=0^n}(-1)^k\pi^{2(n-k)}f^{(k)}(x).
\]
Note that $b^n\pi^{2(n-k)}$ is an integer for every $k \le n$, so this
is actually an integer sum of $f$ terms.
And from (\ref{i:int}) in the lemma each $f^{(k)}(x)$ is an integer at 0
and 1.  Thus $F(0)$ and $F(1)$ are integers.

Elementary differentiation says
\[
\frac{d}{dx}[F'(x)\sin(\pi x)-\pi F(x)\cos(\pi x)] =
b^n\pi^{2n+2}f(x)\sin(\pi x)=\pi^2a^n\sin(\pi x).
\]
It follows that
\begin{align*}
N &= \pi \int_0^1 a^n f(x)\sin(\pi x)dx\\
&= \left[\frac1\pi F'(x)\sin(\pi x)-F(x)\cos(\pi x)\right]_0^1\\
&= F(0)+F(1)\\
\end{align*}
is an integer.
Also, $N$ is positive since it is the integral of a function that is
positive (except at 0 and 1 where it is 0).  We are going to show that
also $N<1$, this contradicts $N$ being an integer.

To see that, recall that $f(x) < 1/n!$.  So the integral is at most
\begin{align*}
N &= \pi \int_0^1 a^n f(x)\sin(\pi x)dx\\
&<\pi a^n/n!\\
&<\pi a^n(e/n)^n/\sqrt{2\pi}e^{1/12n}
&<(ae/n)^n.
\end{align*}
Note, however, that by taking $n>ae$. this is at most 1.

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