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\noindent 
\hspace{0.3in}\textsf{\textbf{{\Large Mathematics 403A}}}
\hspace{0.3in}\textsf{\textbf{{\Large Winter 2005}}}

\begin{theorem}
$e$ is transcendental.
\end{theorem}

I will need the following formulation of the Mean Value Theorem.

\begin{theorem}[Mean Value Theorem]\label{mvt}
If $g(x)$ is a continuously differential function on the interval
$[x_{1}, x_{2}]$, then for some $\theta \in (0,1)$, 
\[
\frac{g(x_{1})-g(x_{2})}{x_{1}-x_{2}} = g'(x_{1} + \theta
(x_{2}-x_{1})).
\]
\end{theorem}

\begin{proof}[Proof that $e$ is transcendental]
[This is from \emph{Topics in Algebra} by Herstein.]

Given a polynomial $f(x) \in \R[x]$ of degree $r$, let 
\[
F(x) = f(x) + f'(x) + f''(x) + \dots + f^{(r)}(x).
\]
Then one can check that $\frac{d}{dx}(e^{-x} F(x)) = -e^{-x}f(x)$.

Apply the Mean Value Theorem to $e^{-x}F(x)$, on the interval $[0,k]$,
for each positive integer $k$: there are numbers $\theta_{k} \in
(0,1)$ such that 
\[
e^{-k}F(k)-F(0) = -ke^{(1-\theta_{k})k} f(\theta_{k} k).
\]
Also, define $\varepsilon_{k}$ to be $\varepsilon_{k} =
F(k)-e^{k}F(0)$.  That is, 
\begin{gather*}
\varepsilon_{1} = F(1)- eF(0)=-e^{(1-\theta_{1})} f(\theta_{1}), \\
\varepsilon_{2} = F(2)- e^{2}F(0)=-2e^{2(1-\theta_{2})} f(2\theta_{2}), \\
\vdots \\ 
\varepsilon_{n} = F(n)- e^{n}F(0)=-ne^{n(1-\theta_{n})} f(n\theta_{n}).
\end{gather*}
Now suppose that $e$ is algebraic: suppose there are integers $c_{i}$
such that 
\[
c_{n} e^{n} + c_{n-1}e^{n-1} + \dots + c_{1} e + c_{0} = 0.
\]
We will (eventually) derive a contradiction.

Multiply the equation defining $\varepsilon_{i}$ by $c_{i}$ and add up
the resulting equations:
\[
c_{1} F(1) + c_{2} F(2) + \dots + c_{n}F(n) - F(0) (c_{1} e + c_{2}
e^{2} + \dots + c_{n}e^{n}) = c_{1} \varepsilon_{1} + c_{2}
\varepsilon_{2} + \dots + c_{n} \varepsilon_{n}.
\]
The term in parentheses equals $-c_{0}$, so I can rewrite this as 
\[
c_{0} F(0) + c_{1} F(1) + c_{2} F(2) + \dots + c_{n}F(n) = c_{1}
\varepsilon_{1} + c_{2} \varepsilon_{2} + \dots + c_{n}
\varepsilon_{n}. 
\]
This holds for any $f(x)$; now we pick a particular one.  Let $p$ be a
prime number such that $p>n$ and $p>c_{0}$.  (Later on I'll assume
that $p$ is even larger, but this is good enough for now.)  Let
\[
f(x) = \frac{1}{(p-1)!} x^{p-1} (1-x)^{p} (2-x)^{p} \dots (n-x)^{p}.
\]
Then 
\[
f(x) = \frac{(n!)^{p}}{(p-1)!} x^{p-1} + \frac{a_{0}}{(p-1)!} x^{p} +
\frac{a_{1}}{(p-1)!} x^{p+1} + \dots 
\]
for some integers $a_{0}$, $a_{1}$, \dots.

\begin{claim}
For each $i \geq p$, $f^{(i)}(x)$ is a polynomial with integer
coefficients, each of which is divisible by $p$.
\end{claim}

\begin{proof}[Proof of claim]
Exercise.
\end{proof}

For $j=1, 2, \dots, n$, $f(j)=0$ with multiplicity $p$, so
$f^{(i)}(j)=0$ when $i \leq p-1$.  Thus 
\[
F(j) = \underbrace{f(j) + f'(j) + \dots + f^{(p-1)}(j)} + 
\underbrace{f^{(p)}(j) + \dots + f^{(r)}(j)} 
\]
is an integer which is divisible by $p$: the first bunch of terms are
zero, and the second bunch are all divisible by $p$, by the claim.

On the other hand, consider 
\[
F(0) = \underbrace{f(0) + f'(0) + \dots + f^{(p-2)}(0)} + 
f^{(p-1)}(0) + \underbrace{f^{(p)}(0) + \dots }.
\]
The first bunch of terms are all zero, and the last bunch are all
divisible by $p$, but $f^{(p-1)}(0) = (n!)^{p}$.  Since $p$ was chosen
to be larger than $n$, this is not divisible by $p$.

Also, $p$ was chosen to be larger than $c_{0}$, and so doesn't divide
$c_{0}$.  As a result, $p$ doesn't divide the integer $c_{0}F(0) +
c_{1}F(1) + \dots + c_{n}F(n)$.

Recall from before that $c_{0} F(0) + c_{1} F(1) + c_{2} F(2) + \dots
+ c_{n}F(n) = c_{1} \varepsilon_{1} + c_{2} \varepsilon_{2} + \dots +
c_{n} \varepsilon_{n}$.  The left side of this equation is a
nonnegative integer (since it is not divisible by $p$).  I will show
that the right side has absolute value less than 1, and this will be a
contradiction.

Fix $i$ with $1 \leq i \leq n$.  From the definition of
$\varepsilon_{i}$ and the formula for $f(x)$, we see that
\[
\varepsilon_{i} = -ie^{i(1-\theta_{i})} f(i\theta_{i}) = 
\frac{-ie^{i(1-\theta_{i})} (i\theta_{i})^{p-1} (1-i\theta_{1})^{p}
\dots (n-i\theta_{i})^{p}}{(p-1)!}.
\]
Since $0 < \theta_{i} < 1$, we have
\[
|\varepsilon_{i}| \leq \frac{e^{n} n^{p} (n!)^{p}}{(p-1)!}.
\]
As $p$ goes to $\infty$, the right hand side goes to zero, so for $p$
a sufficiently large prime, 
\[
|c_{1} \varepsilon_{1} + \dots + c_{n} \varepsilon_{n}| < 1.
\]
But $c_{1} \varepsilon_{1} + \dots + c_{n} \varepsilon_{n}$ must be a
nonzero integer.  This is the contradiction we were looking for.
\end{proof}

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