Bounding coefficients of the polynomial $(1-x^2)^n (1-x)^{-m}$.

Question

I have been trying to get some upper bound on the coefficient of $x^k$ in the polynomial $$(1-x^2)^n (1-x)^{-m}, \text{ $m \le n$}.$$

A straightforward calculation shows that for even $k$, the coefficient can be expressed as $$\sum_{i=0}^{k/2} (-1)^i \binom{n}{i} (-1)^{k-2i} \binom{-m}{k-2i} = \sum_{i=0}^{k/2} (-1)^i\binom{n}{i} \binom{m+k-2i-1}{k-2i}$$ and therefore simply using $\binom{n}{k} \le n^{k}$, one gets a bound of $$(k/2+1) (n+(m+k)^2)^{\frac{k}{2}} .$$

I'm wondering if one could get a better bound, ideally with a better dependence on $k$?

Answer 1

$(1-X^2)^n(1-X)^{-m}=(1+X)^n(1-X)^{n-m}=\sum_{k=0}^{2n-m}a_kX^k$ where $$ a_k=\sum_{i=0}^k (-1)^i\binom{n-m}{i}\binom{n}{k-i} $$ Using $\binom{n}{k}\leqslant\frac{n^k}{k!}$, we have $$ |a_k|\leqslant\sum_{i=0}^k\frac{(n-m)^i}{i!}\frac{n^{k-i}}{(k-i)!}\leqslant n^k\sum_{i=0}^k\frac{1}{i!(k-i)!}=\frac{(2n)^k}{k!} $$

Answer 2

One can get a better bound of $$\frac{2^k}{(k/2)!} \bigl(n + (m+k)^2\bigr)^{k/2} .$$

Using $\binom{n}{k} \le n^k/k!\,$, we have \begin{align} \sum_{i=0}^{k/2} (-1)^i\binom{n}{i} \binom{m+k-2i-1}{k-2i} &\le \sum_{i=0}^{k/2} \frac{n^i}{i!} \frac{(m+k-2i-1)^{k-2i}}{(k-2i)!} \\ &\le (n + (m+k)^2)^{k} \sum_{i=0}^{k/2} \frac{1}{i! (k-2i)!} . \end{align} Following @Tuvasbien's argument, the summation can be bounded by \begin{align} \sum_{i=0}^{k/2} \frac{1}{i! (k-2i)!} &=\sum_{i=0}^{k/2} \frac{(2i)!}{i!k!} \frac{k!}{(2i)! (k-2i)!} \\ &\le \frac{1}{(k/2)!}\sum_{i=0}^{k/2} \frac{k!}{(2i)! (k-2i)!} \\ &\le \frac{2^k}{(k/2)!} . \end{align}