Euler polynomials

The previous post looked at the logistic function and showed how it illustrates Runge’s phenomenon: adding interpolation points makes the interpolation error worse.

Here’s another interesting thing about the logistic function: it’s power series coefficients involve Euler polynomials.

$\frac{e^x}{1 + e^x} = \sum_{n=0}^\infty \frac{x^n E_n(1)}{2 \, n!}$

So, what are Euler polynomials? I don’t have a good answer to that, but I can say they’re closely related to several things I’ve written about lately.

There are multiple ways of defining the Euler polynomials, the most succinct being in terms of their generating function. That tells you what they are, in a sense, but it doesn’t answer the question of why Euler was interested in them, or why anyone else would be interested in them.

One thing I can say is they seem to pop up occasionally in combinatorics problems. They come up, for example, in the calculus of finite differences. Another is that they’re closely related to Bernoulli polynomials, and Bernoulli polynomials pop up even more often in combinatorics.

Three weeks ago I wrote about the sawtooth function and how it satisfies the multiplicative identity

$f(nx) = \sum_{i=0}^{n-1} f\left(x + \frac{i}{n} \right )$

Then a few days later I wrote about more general multiplication theorems. I mentioned the Bernoulli polynomials and how they satisfy the same identity, except for a factor that is independent of x.

$B_m(nx) = a(m, n) \sum_{i=0}^{n-1} B_m\left(x + \frac{i}{n} \right)$

The Euler polynomials satisfy an equation of the same form, but with an alternating term inside the sum.

$E_m(nx) = a(m, n) \sum_{i=0}^{n-1} (-1)^i E_m\left(x + \frac{i}{n} \right)$

For Bernoulli polynomials a(m, n) = n^m-1.

For Euler polynomials a(m, n) = n^mfor odd n and -2/(m + 1) times that for even n.