$n$th derivative of $\tanh$

Question

Successive derivative of $\tanh u$ can be expressed as polynomial functions of $\tanh u$: \begin{align} \frac{d}{du}\tanh u&=1-\tanh^2u\\ \frac{d^2}{du^2}\tanh u&=-2\tanh u\left(1-\tanh^2u\right)\\ \frac{d^3}{du^3}\tanh u&=2\left(1-\tanh^2u\right)\left(3\tanh^2u-1\right)\\ \frac{d^4}{du^4}\tanh u&=-8\tanh u\left(1-\tanh^2u\right)\left(3\tanh^2u-2\right)\\ \ldots \end{align} I am specially interested in the derivatives of even order. Their expressions can be found in this article by Boyadzhiev: Denoting $z=\tanh u$ and $C_{2n}\left( \tanh u \right)=\tfrac{d^{2n}}{du^{2n}}\left( \tanh u \right) $, \begin{equation} C_{2n}\left(z \right)=2^{2n}\left( 1+ z\right)\sum_{k=0}^{2n}\frac{k!}{2^k}S\left( 2n,k \right)\left( z-1 \right)^k \end{equation} where $S(n,k)$ are the Stirling numbers of the second kind. We have also $C_{2n}(0)=C_{2n}(\pm1)=0$. As $\tanh u$ is an odd function of $u$, they can also be expressed as \begin{equation} C_{2n}\left(z \right)=z\left( 1-z^2 \right)Q_{2n-2}(z^2) \tag{A}\label{eqA} \end{equation} for $n=1,2,\ldots$. Then, $Q_0(z)=-2$, $Q_2(z)=-8\left( 3z^2-2 \right)$... One may establish the recurrence relation (with $Z=z^2 $): \begin{equation} Q_{2n}=4Z\left( 1-Z \right)^2Q''_{2n-2}-6\left( 1-Z \right)\left( 3Z-1 \right)Q'_{2n-2}+4\left( 3Z-2 \right)Q_{2n-2} \end{equation} What else can be found about the family $Q_{2n}$? Are they related to a well-known family of polynomials?

Context: Answering a MO question, an integral representation for $\zeta(3)$ appeared: \begin{equation} \int_{0}^\infty \frac{u-\tanh u}{u^3}\,du=\frac{7}{\pi^2}\zeta(3) \end{equation} It can easily be checked using the residue theorem, taking advantage of the parity of the integrand. This result can be generalized for $n=1,2,3,\ldots$ as: \begin{equation} \int_{0}^\infty \frac{T_{2n-1}(u)-\tanh u}{u^{2n+1}}\,du=(-1)^{n+1}\frac{2^{2n+1}-1}{\pi^{2n}}\zeta(2n+1) \end{equation} where $T_{2n-1}(u)$ is the odd polynomial identical to the Maclaurin expansion for $\tanh u$: \begin{equation} \tanh u=u-\frac{1}{3}u^3+\frac{2}{15}u^5-\frac{17}{315}u^7+\ldots \end{equation} up to the $u^{2n-1}$ term. This identity can also be checked using the residue theorem.

Now, integrating $2n$ times by parts, one obtains the following representation \begin{equation} \int_0^\infty C_{2n}\left(\tanh u \right)\frac{du}{u}=(-1)^{n}\frac{2^{2n+1}-1}{\pi^{2n}}\left( 2n \right)!\zeta(2n+1) \end{equation} With the representation \eqref{eqA} and the substitution $u=e^{-2v}$, one obtains \begin{equation} \int_0^1 \frac{v-1}{\left( v+ 1\right)^3}Q_{2n-2}\left( \left( \frac{1-v}{1+v} \right)^2 \right)\frac{dv}{\ln v}=(-1)^{n}\frac{2^{2n+1}-1}{4\pi^{2n}}\left( 2n \right)!\zeta(2n+1) \end{equation} which provides another integral representation for $\zeta(2n+1)$.

Answer 1

Derivative polynomial of the hyperbolic tangent function

It is known that $$ \tan z=\operatorname{i}\tanh(\operatorname{i}z). $$ So, from the derivative polynomial of the tangent function $\tan z$, we can derive the derivative polynomial of the hyperbolic tangent function $\tanh z$.

It is not difficult to see that if $f$ is a function whose derivative is a polynomial in $f$, that is, $f'(x)=P_1(f(x))$ for some polynomial $P_1$, then all the higher order derivatives of $f$ are also polynomials in $f$, so we have a sequence of polynomials $P_n$ defined by $f^{(n)}(x)=P_n(f(x))$ for $n\ge0$. As usual, we call $P_n(u)$ the derivative polynomials of $f$. In fact, the polynomials $P_n$ are determined by \begin{equation*} P_0(u)=u, \quad P_{n+1}(u)=P'_n(u)P_1(u), \quad n\in\mathbb{N}. \end{equation*}

Theorem. For $n\ge0$, the derivative polynomials $P_n(u)$ of the tangent function $u=\tan x$ can be explicitly computed by \begin{equation}\label{Tan-Deriv-Polyn-Eq} P_n(u)=\sum_{k=0}^{\frac12\bigl[n+\frac{1-(-1)^n}{2}\bigr]}a_{n,n+1-2k}u^{n+1-2k} \end{equation} with \begin{equation}\label{Tan-No-Exp-Form} a_{2m-1,0}=(-1)^{m} \sum_{\ell=1}^{2m} (-1)^{\ell} 2^{2m-\ell} (\ell-1)! S(2m,\ell) \end{equation} for $m\ge1$ and \begin{equation*} a_{n,n+1-2k}=(-1)^{k-1}\sum_{\ell=n+1-2k}^{n+1} (-1)^{n-\ell} 2^{n+1-\ell} (\ell-1)! \binom{\ell}{n+1-2k} S(n+1,\ell) \end{equation*} for $0\le k\le\frac{1}{2}\bigl[n-\frac{1-(-1)^n}{2}\bigr]$, where $S(n,k)$ for $n\ge k\ge1$ stand for the Stirling numbers of the second kind which can be analytically generated by \begin{equation*} \frac{(\operatorname{e}^x-1)^k}{k!}=\sum_{n=k}^\infty S(n,k)\frac{x^n}{n!}, \quad k\in\mathbb{N}. \end{equation*}

References

1. K. N. Boyadzhiev, Derivative polynomials for tanh, tan, sech and sec in explicit form, Fibonacci Quart. 45 (2007), no. 4, 291--303 (2008); available online at https://arxiv.org/abs/0903.0117.
2. Feng Qi and Bai-Ni Guo, An explicit formula for derivative polynomials of the tangent function, Acta Universitatis Sapientiae Mathematica 9 (2017), no. 2, 348--359; available online at https://doi.org/10.1515/ausm-2017-0026.