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

This entry was posted in math.CA, math.NT and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s