How to derive the Taylor expansion form of a polynomial expression? I want to change this polynomial into the form $\sum_{k=0}^m a_k x^k$
$$q_m(x)=\sum_{k=0}^m(-1)^k\binom{2m+1}{2k+1}x^k(1-x)^{m-k}$$
I see no way to do this as I fear one might need intricate binomial identities if I try to expand $(1-x)^{m-k}$ and collect the powers of $x$. The way to use calculus directly to get the Taylor expansion seems also not viable to me.
 A: Let we set:
$$q_m(x)=(1-x)^m\sum_{k=0}^{m}\binom{2m+1}{2k+1}\left(\frac{x}{x-1}\right)^k .$$
Since:
$$\sum_{k=0}^{m}\binom{2m+1}{2k+1}w^{2k+1} = \frac{1}{2}\left(\left(1+w\right)^{2m+1}-\left(1-w\right)^{2m+1}\right)$$
we have:
$$ q_m(x)=(1-x)^m \frac{\left(1+\sqrt{\frac{x}{x-1}}\right)^{2m+1}-\left(1-\sqrt{\frac{x}{x-1}}\right)^{2m+1}}{2\sqrt{\frac{x}{x-1}}}$$
or:
$$ q_m(x) = (-1)^m \frac{\left(\sqrt{x-1}+\sqrt{x}\right)^{2m+1}-\left(\sqrt{x-1}-\sqrt{x}\right)^{2m+1}}{2\sqrt{x}} $$
from which:

$$ q_m(x) = U_{2m}(\sqrt{1-x}) \tag{1}$$

where $U_{2m}$ is a Chebyshev polynomial of the second kind. That implies:
$$\begin{eqnarray*} q_m(x)&=&\sum_{k=0}^{m}\binom{2m-k}{k}(-1)^k 4^{m-k}(1-x)^{m-k}\\&=&\sum_{k=0}^{m}\binom{m+k}{m-k}(-1)^{m-k} 4^k (1-x)^k \tag{2}\end{eqnarray*} $$
so we have the Taylor series of $q_m(x)$ in a neighbourhood of $x=1$:

$$ q_m(x)=(-1)^m \sum_{k=0}^{m}\binom{m+k}{m-k}4^k (x-1)^k. \tag{3}$$

We may also notice that:
$$ q_m(\sin^2\theta) = \frac{\sin((2m+1)\theta)}{\sin\theta},\qquad \begin{eqnarray*}x\cdot q_m(x^2)&=&\sin\left((2m+1)\arcsin x\right)\\&=&\text{Im}\left(\left(\sqrt{1-x^2}+i x\right)^{2m+1}\right).\end{eqnarray*}$$
A: Notice that $a_0=q(0)$, $a_1=q'(0)$, $a_2={1\over2}q''(0)$ and so on. It should be easy to evaluate the $a_k$ this way.
