Higher-order derivatives of $(e^x + e^{-x})^{-1}$

MathOverflow Asked by tobias on November 28, 2020

I am currently trying to build the derivatives of $$f(x) = frac{1}{e^x+e^{-x}}.$$
It is fairly straightforward to obtain
$$ frac{d^n f}{dx^n} = frac{P_n(e^x)}{e^{(n-1)cdot x} (e^x+e^{-x})^{n+1}}, $$
where $P_n(x)$ is given by the recursive relationship $P_0(x) = 1$ and
$$P_{n+1}(x) = P_n'(x) cdot x cdot (x^2+1) – P_n(x)((2cdot n +1)cdot x^2-1). $$
If we represent $P_n(x)$ by $sum_{i = 0}^n a^{(n)}_{2i} x^{2i}$, then we have $a_0^{(n)} = 1$ for all $n$, $a_{2k}^{(n)} = 0$ for all $k > n$, and $$a_{2i}^{(n+1)} = (2i+1) cdot a_{2i}^{(n)} – (2(n-i)+3) cdot a_{2 cdot (i-1)}^{(n)}$$ for all $n geq 0$ and $0 < i leq n+1$.
So we can obtain with this relationship, that $a_{2n}^{(n)} = (-1)^{n}$ and that $a_2^{(n)} = -3^n +k +1$.
However, I am wondering whether we can say something about the maximum of $|P_n(e^x)/e^{2n}|$ for each $n$ or the maximum of $max_{0 leq i leq n} |a_{2i}^{(n)}|$ over each $n$.

Edit: It seems that $sum_{i=0}^n |a_{2i}^{(n)}| = n! cdot 2^n$.

One Answer

Using the tried-and-true method of calculating small examples and plugging them into the OEIS, one finds that the $P_n(x)$ are, up to sign, known as MacMahon polynomials, and their coefficients are given by Eulerian numbers of type B. The OEIS also has a separate entry for the maximal coefficients that you are asking about, although it doesn't list a formula for the asymptotic growth. But there is a long list of references that will hopefully be helpful.

Correct answer by Timothy Chow on November 28, 2020

Add your own answers!

Ask a Question

Get help from others!

© 2024 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP