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.
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.
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Reference