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Thus we have proved Runge's theorem.

\subsubsection*{Consequences}

We prove lots of examples like the following:

There exists polynomials $P_n$ such that $P_n(z)
\to 1$ for $\Im z \ge 0$, but $P_n(z) \to 0$ for
$\Im z < 0$. (Contrast: uniform limit of analytic
functions is analytic).

Define:
\begin{align*}
  K_n
  &= \left\{ z - \frac{i}{n} : |z| \le n,
  \Im z \ge 0 \right\} \\
  K_n'
  &= \left\{ z - \frac{4i}{n} : |z| \le n,
  \Im z \le 0 \right\} \\
  \Omega_n
  &= \left\{ z - \frac{2i}{n} : |z| < n + 1, \Im z
  > 0\right\} \\
  \Omega_n'
  &= \left\{ z - \frac{3i}{n} : |z| < n + 1, \Im z
  < 0\right\}
\end{align*}
Observe that:
\begin{itemize}
  \item $K_n \cup K_n'$ is compact.
  \item $\Omega_n, \Omega_n'$ are open, and
    disjoint (and hence their union is open).
  \item $\Omega_n \cup \Omega_n' \supseteq K_n
    \cup K_n'$.
\end{itemize}
Furthermore $(K_n \cup K_n')^c$ is path-connected.
Define
\[ f_n(z) = \begin{cases}
  1 & z \in \Omega_n \\
  0 & z \in \Omega_n'
\end{cases} \]
$f_n$ is analytic on $\Omega_n \cup \Omega_n'$
(since locally constant). So we can find a
polynomial $P_n$ with
\[ |P_n(z) - f_n(z)| \le 2^{-n} \]
for all $z \in K_n \cup K_n'$. If $z^*$ is fixed
and $\Im z^* \ge 0$ then $z^* \in \Omega_n$ for
$n$ sufficiently large, so since $f_n(z^*) =1$,
$P_n(z^*) \to 1$ as $n \to \infty$. If $z^*$ is
fixed and $\Im z^* < 0$ then $z^* \in \Omega_n'$
for $n$ sufficiently large $n$, so since $f_n(z^*)
= 0$ for $n$ sufficiently large, $P_n(z^*) \to 0$
as $n \to \infty$.

\newpage
\section{Irrational and Transcendental Numbers}

Proof that
\[ e = \sum_{n = 0}^\infty \frac{1}{n!} \]
is irrational.

\begin{remark*}
  We shall use the fact that $1$ is the smallest
  strictly positive integer.
\end{remark*}

\vspace{-1em}
Suppose that $e$ is rational. Then (since $e >
0$), $e = \frac{p}{q}$ for some $p, q$ coprime,
with $p, q \ge 1$. Calculate:
\begin{align*}
  e
  &= \sum_{r = 0}^\infty \frac{1}{r!} \\
  \left( e - \sum_{r = 0}^q \frac{1}{r!} \right)
  &= \sum_{r = q + 1}^\infty \frac{1}{r!} \\
  q! \left( e - \sum_{r = 0}^q \frac{1}{r!} \right)
  &= q! \sum_{r = p + 1}^\infty \frac{1}{r!} \\
  q! \left( e - \sum_{n = 0}^q \frac{1}{r!}
  \right)
  &= q!e - \sum_{r = 0}^q \frac{q!}{r!} \in \NN
\end{align*}
but also
\[ q! \left( e - \sum_{r = 0}^q \frac{1}{r!}
\right) > 0 .\]
But
\[ \sum_{r = q + 1}^\infty < \sum_{s = 1}^\infty
\frac{1}{(p + 1)^r} = \frac{1}{p + 1} \frac{1}{1 -
\frac{p}{p + 1}} = 1 .\]
So we have found an integer between $0$ and $1$,
which is nonsense!

A similar idea gives us $\pi$ is irrational.
\begin{flashcard}[pi-irrational]
\prompt{Prove that $\pi$ is irrational.}

\begin{proof}
  \cloze{Consider $S_n = \int_0^\pi x_n(\pi - x)^n \sin x
  \dd x$. We will show that $S_n$ is a polynomial
  in $\pi$ with coefficient of the form $A_n n!$
  with $A_n \in \ZZ$. First step is to
  ``evaluate'' the integral and we shall use the
  fact that if
  $f_n(x) = x^n (\pi - x)^n$ then:
  \begin{itemize}
    \item $f_n^{(r)}(0) = 0$ if $0 \le r \le n -
      1$
    \item $f_n^{(r)}(0) = B_n n! \pi^{2\pi - r}$
      with $B_n$ ($n \le r \le 2n$.
    \item $f_n^{(r)}(0)=  0$ if $r > 2n$.
  \end{itemize}
  \[ f_n(x) = \sum_k {n \choose k} x^{n + k}
  \pi^{n - k} (-1)^{n - k} .\]
  $f_n^{(r)}(0) = 0$ unless $n \le r \le 2n$.
  \begin{align*}
    f_n^{(k)}(0)
    &= {n \choose k} (n + k)! \pi^{n -
    k} (-1)^{n - k} \\
    &= B_k n! \pi^{n - k}
  \end{align*}
  By symmetry about $\frac{\pi}{2}$, we get
  similar results for derivatives of $f$ when
  evaluated at $\pi$.

  Now calculate:
  \begin{align*}
    \int_0^\pi f_n^{(r)}(x) \sin x \dd x
    &= [f_n^{(r + 1)} (x) \sin x]_0^\pi +
    \int_0^\pi f_n^{(r + 1)} (x) \cos x \dd x \\
    &= \int_0^\pi f_n^{(r + 1)} (x) \cos x \dd x \\
    \int_0^\pi f_n^{(r)} (x) \cos x \dd x
    &= [f_n^{(r + 1)}(x) \cos x]_0^\pi -
    \int_0^\pi f_n^{(r + 1)} (x) \sin x \dd x \\
    &= -f_n^{(r + 1)}(\pi) - f_n^{(r + 1)}(0) -
    \int_0^\pi f_n^{(r + 1)}(x) \cos (x) \dd x
  \end{align*}
  so by repeated integration by parts
  \[ \int_0^\pi x^n(\pi - x)^n \dd x = \sum_{r =
  0}^n C_r \pi^r n! \]
  with $C_r \in \ZZ$.
  \[ \frac{1}{n!} \int_0^\pi x^n (\pi - x)^n \dd x
  = \sum_{r = 0}^n C_r \pi^r .\]
  Now
  \[ x^n(\pi - x)^n \le \left( \frac{\pi}{2}
  \right)^{-n} \]
  (since $x(\pi - x) \le \frac{\pi^2}{4}$ by
  AM-GM), so if $\pi = \frac{p}{q}$ with $p, q >
  0$, $p, q \in \ZZ$ then we have $q^n \sum_r C_r
  \pi^r \in \ZZ$ and
  \[ q^n \sum_r C_r \pi^r \le \frac{q^n}{n!}
  \int_0^\pi x^n (\pi - x)^n \dd x \le
  \frac{q^n}{n!} \pi \left( \frac{\pi}{2}
  \right)^{-n} = u_n \]
  Then $\frac{u_{n + 1}}{u_n} = \frac{\pi^2 q}{(n
  + 1)} \to 0$ as $n \to \infty$, so if $n$ is
  large then $u_n < 1$ so
  \[ 0 < q^n \sum_r C_r \pi^r < 1 \]
  which contradicts the fact that it is an
  integer.}
\end{proof}
\end{flashcard}
