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 $$sum_{i=0}^n |a_{2i}^{(n)}| = n! cdot 2^n$$.

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

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