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

Observe 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

for .

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.

### Like this:

Like Loading...

*Related*