Taylor expansion of $1/(1+x^2)^r$ around $x_0$

Question

Let $r>0$, $k\geq 0$. We can write $$\left(\frac{1}{(1+x^2)^r}\right)^{(k)} = \frac{P_k(x)}{(1+x^2)^{r+k}},$$ where $P_k\in \mathbb{Z}\lbrack x\rbrack$. It is clear that $P_k$ satisfies the recurrence formula $$P_{k+1}(x) = P_k'(x) (1+ x^2) + 2 (r+k)\cdot x P_k(x).$$

Are the polynomials $P_k$ of some very well-known kind that I simply don't recognize? (They certainly feel that way!)

(Obviously $1/(1+x^2) = \arctan'(x)$.)

It would be nice, for instance, to have an estimate for where the zeroes of $P_{k+1}(x)$ lie, and so for what the maximum of $P_k(x)/(1+x^2)^{r+k}$ is.

Answer 1

Let $r>0$ and $k\ge0$. Then \begin{align*} \biggl[\frac{1}{(1+x^2)^r}\biggr]^{(k)}&=\sum_{j=0}^{k}\frac{\textrm{d}^j}{\textrm{d} u^j}\biggl(\frac{1}{u^r}\biggr) B_{k,j}(2x,2,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\frac{\langle-r\rangle_j}{u^{r+j}} 2^jB_{k,j}(x,1,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\frac{\langle-r\rangle_j}{(1+x^2)^{r+j}} 2^j \frac{1}{2^{k-j}}\frac{k!}{j!}\binom{j}{k-j}x^{2j-k}\\ &=\frac{k!}{2^kx^k(1+x^2)^r} \sum_{j=0}^{k}\langle-r\rangle_j\frac{2^{2j}}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1+x^2)^{j}}, \end{align*} where $u=u(x)=1+x^2$, the notation $B_{k,j}$ denotes the Bell polynomials of the second, and $\langle-r\rangle_j$ stands for the falling factorial. This means that \begin{equation*} P_k(x)=\frac{k!}{2^k}\frac{(1+x^2)^k}{x^k} \sum_{j=0}^{k}\langle-r\rangle_j\frac{2^{2j}}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1+x^2)^{j}} \end{equation*} for $k\ge0$. I believe that this result would be useful for answering this question.

For knowledge of those notions, terminology, and concepts employed above, please refer to the following reference.

Reference

1. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.