Series expansion of $\left(\frac{\sin x}{x}\right)^a$ at $x=0$ I need a series expansion at $x=0$ (with nth term please) for $$f(x)= \left(\frac{\sin x}{x}\right)^a$$ where $a>0$ is a real number.
We know that $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$
So we have $$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-...$$
Now if we take $y= -\frac{x^2}{3!}+\frac{x^4}{5!}-...$ then $$\left(\frac{\sin x}{x}\right)^a=(1+y)^a$$
Then by Binomial theorem we have $$\left(\frac{\sin x}{x}\right)^a=1+a y+\frac{a(a-1)}{2!}y^2+ +\frac{a(a-1)(a-2)}{3!}y^3+...   $$
Now I am stuck to calculate $y^2,y^3,...$
Another approach
$$f(x)= \left(\frac{\sin x}{x}\right)^a$$
So on differentiation $$f'(x)=a  \left(\frac{\sin x}{x}\right)^{a-1} \left(\frac{x\cos x-\sin x}{x^2}\right) $$
So we have taking the limit of above function as $x\to 0$ $$f'(0)=0$$
This is again where I am stuck. Any help would be surely appreciated. Thank you.
 A: Start with
$$
\operatorname{sinc}x=\frac{\sin x}{x}=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k}
$$
Now I'm not sure about $a>0$ but for $a\in\Bbb N$ we may use this formula to write
$$
\operatorname{sinc}^ax=\sum_{m=0}^\infty c_m x^{2m},
$$
where $c_0=1$ and
$$
c_m=\frac{1}{m}\sum_{k=1}^m(ka-m+k)\frac{(-1)^k}{(2k+1)!}c_{m-k}.
$$
Here is a Mathematica implementation of this series expansion:
f[x_, a_] := Sinc[x]^a;
c[0, a_] := 1;
c[m_, a_] := 1/m Sum[(k a - m + k) (-1)^k/(2 k + 1)! c[m - k, a], {k, 1, m}];
g[x_, a_, M_] := Sum[c[m, a] x^(2 m), {m, 0, M}];

Interestingly enough, when substituting non-integer values of $a$ into both Mathematica functions I get the same answer, e.g.
f[0.3, 2.5] = 0.943296
g[0.3, 2.5, 8] = 0.943296

Here is a plot comparing $f(x,a)=\operatorname{sinc}^ax$ to the series expansion $g(x,a,M)$ for $a=2.5$ and $M=4,6$.

Finally, here is a table of the first six coefficients $c_m$:
$$
\left(
\begin{array}{cc}
 m & c_m\\
 0 & 1 \\
 1 & -\frac{a}{6} \\
 2 & \frac{1}{360} a (5 a-2) \\
 3 & -\frac{a (7 a (5 a-6)+16)}{45360} \\
 4 & \frac{a (5 a-4) (7 a (5 a-8)+36)}{5443200} \\
 5 & -\frac{a (11 (a-2) a (35 (a-2) a+104)+768)}{359251200} \\
\end{array}
\right)
$$
