## e Is Transcendental

In this post, we show that the ubiquitous constant $e:= \sum_{n=0}^{\infty} \frac{1}{n!} = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^{n}$ is transcendental: there does not exist a polynomial $q \in \mathbb{Z}[x]$ such that $q(e) = 0$. The proof we will give was first brought to my attention in an analysis course taught by Professor Shlomo Sternberg–many thanks to him. You can find a PDF of the contents of this post here.

Theorem 1. The constant $e$ is transcendental.

Proof. Let $f \in \mathbb{R}[x]$ be of degree $r$, and define

$\displaystyle F:= f+f^{(1)}+\cdots+f^{(r)}$

By the chain rule, for any differentiable function $g$, $\frac{\partial}{\partial x}(e^{-x}g(x)) = e^{-x}(g' - g)$, so that

$\displaystyle \frac{\partial }{\partial x}(e^{-x}F(x)) = e^{-x}\left[(f^{(1)} + \cdots + f^{(r)}) - (f + \cdots + f^{(r)})\right] = -e^{-x}f(x)$

since $f^{(k)} = 0$ for all $k > r$. We apply the intermediate value theorem to $e^{-x}F(x)$ on the interval $[0,k]$ to obtain that

$\displaystyle e^{-k}F(k)-F(0)=-ke^{-\theta_{k}k}f(k\theta_{k})$

for some $\theta_{k} \in (0,1)$. Multiplying both sides $e^{k}$, we obtain

$\displaystyle F(k) - e^{k}F(0) = -ke^{k(1-\theta_{k})}f(k\theta_{k}) =: \epsilon_{k}$

Suppose $e$ is algebraic, so that there exist integers $0 \neq c_{0}, c_{1},\cdots,c_{n}$ such that

$\displaystyle c_{n}e^{n} + c_{n-1}e^{n-1} + \cdots + c_{0} = 0$

Observe that

$\displaystyle c_{1}\epsilon_{1}+\cdots+c_{n}\epsilon_{n} = \sum_{k=1}^{n}c_{k}[F(k) - e^{k}F(0)] = \sum_{k=1}^{n}c_{k}F(k) - \underbrace{(c_{1}e + \cdots + c_{n}e^{n})}_{-c_{0}}F(0)$

We will derive a contradiction by finding a prime number $p > \max\left\{n,c_{0}\right\}$ and a polynomial $f = f_{p}$ such that

1. $p \mid F(j)$ for $j = 1,\cdots,n$.
2. $p \nmid F(0)$ so that $p \nmid \sum_{k=0}^{n}c_{k}F(k)$, since $p > c_{0}$ and therefore $p \nmid c_{0}$.
3. $\left|c_{1}\epsilon_{1} + \cdots + c_{n}\epsilon_{n}\right| < 1$.

Assertions (2) and (3) are contradictory since $\left|c_{0}\epsilon_{1} + \cdots + c_{n}\epsilon_{n}\right|$ is an integer less than $1$ and therefore $0$, but $p \mid 0$.

Define a polynomial $f_{p}$ by

$\displaystyle f_{p}(x) := \frac{1}{(p-1)!}x^{p-1}(1-x)^{p}(2-x)^{p}\cdots (n-x)^{p}$

Note that the first $(p-1)^{th}$ derivativess of $f_{p}$ vanish at the points $x = 1,2,\cdots,n$. Multiplying out the expresssion defining $f$, we can write

$\displaystyle f_{p}(x) = \frac{1}{(p-1)!}\left[(n!)x^{p-1} + a_{0}x^{p} + a_{1}x^{p+1} + \cdots + a_{pn-1}x^{p(n+1) - 1}\right]$

for $a_{0},\cdots,a_{pn-1} \in \mathbb{Z}$.

If $g$ is a polynomial with integer coefficients, then for any $j \geq p$, the $j^{th}$ derivative of $\frac{1}{(p-1)!}g$ is a polynomial with integer coefficients divisible by $p$. The first $p-2$ derivatives of $f_{p}$ vanish at $x = 0,1,\cdots,n$ and the first $p-1$ derivatives of $f_{p}$ vanish at $x = 1,\cdots,n$. $f_{p}^{(p-1)}(0) = (n!)^{p}$ and since $p > n$, $p \nmid f_{p}^{(p-1)}(0)$. Hence, the function $F$ evaluated at $x = 1,2,\cdots,n$ is an integer divisible by $p$.

Lastly, we need an estimate for $\epsilon_{k}$ in order to know that we can make a sufficiently large choice for $p$. Substituting in the definition of $\epsilon_{k}$, we have

$\displaystyle \epsilon_{k} = -ke^{(1-\theta_{k})k}f(k\theta_{k}) = -\frac{ke^{(1-\theta_{k})k}(k\theta_{k})^{p-1}(1-k\theta_{k})^{p}\cdots (n-k\theta_{k})^{p}}{(p-1)!}$

Since $\theta_{k} \in (0,1)$ and $k \leq n$, we see that

$\displaystyle \left|\epsilon_{k}\right| \leq \frac{e^{n}n^{2p}(n!)^{p}}{(p-1)!} \rightarrow 0, p \rightarrow \infty$

by Stirling’s formula. $\Box$