In this post, we show that the ubiquitous constant is transcendental: there does not exist a polynomial such that . 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 is transcendental.
Proof. Let be of degree , and define
By the chain rule, for any differentiable function , , so that
since for all . We apply the intermediate value theorem to on the interval to obtain that
for some . Multiplying both sides , we obtain
Suppose is algebraic, so that there exist integers such that
We will derive a contradiction by finding a prime number and a polynomial such that
- for .
- so that , since and therefore .
Assertions (2) and (3) are contradictory since is an integer less than and therefore , but .
Define a polynomial by
Note that the first derivativess of vanish at the points . Multiplying out the expresssion defining , we can write
If is a polynomial with integer coefficients, then for any , the derivative of is a polynomial with integer coefficients divisible by . The first derivatives of vanish at and the first derivatives of vanish at . and since , . Hence, the function evaluated at is an integer divisible by .
Lastly, we need an estimate for in order to know that we can make a sufficiently large choice for . Substituting in the definition of , we have
Since and , we see that
by Stirling’s formula.