Using Mathematica, I need an expansion with $n$th term of $$ f(x)=\left(\frac{\sin (x)}{x}\right)^a $$ about $x=0$ where $a\geq 0$ or if $$f(x)=\sum_{n=0}^{\infty} b_{2n }x^{2n}$$ then I need a formula for $b_{2n}$ .

By wolfram alpha (see:https://www.wolframalpha.com/input?i=series+%28%28sin+x%29%2Fx%29%5Ea+at+x%3D0) we have $$f(x)=1-\frac{a}{6}x^2+\frac{a (5 a-2)}{360} x^4-\frac{a \left(35 a^2-42 a+16\right) }{45360}x^6+O(x^8)$$

Any help would be appreciated. Thank you.

  • $\begingroup$ The question is??? You don't believe the result? $\endgroup$
    – Ulrich Neumann
    Mar 22 at 12:01
  • $\begingroup$ No, I believe the result. Question is the nth term of the expansion. $\endgroup$
    – Max
    Mar 22 at 12:02
  • $\begingroup$ Series[(Sin[x]/x)^a, {x, 0, 6}] // Normal // Map[Factor] ? $\endgroup$
    – Syed
    Mar 22 at 12:09
  • $\begingroup$ @Max Coefficient for variable n ? $\endgroup$
    – Ulrich Neumann
    Mar 22 at 12:12
  • 1
    $\begingroup$ Maple's best is $$\sum _{k=0}^{\infty } \left( {\frac {27\,{{\rm e}^{k\ln \left( 3 \right) }}\cos \left( 1/2\,k\pi \right) }{4\, \left( k+3 \right) !}}- 3/4\,{\frac {\cos \left( 1/2\,k\pi \right) }{ \left( k+3 \right) !}} \right) {x}^{k} $$ for $a=3$ by convert((sin(x)/x)^3, FPS). Maple fails with $a=4$. $\endgroup$
    – user64494
    Mar 22 at 13:44

1 Answer 1


Too long for a comment.

Answering your question on Mathematics Stack Excahane, @Aaron Hendrickson provided what I would consider as the only viable solution

$$c_m(a)=\frac{1}{m}\sum_{k=1}^m(ka-m+k)\,\frac{(-1)^k}{(2k+1)!}\,c_{m-k}(a)$$ with $c_0(a)=1$.

May be, for a given value of $a$, something could be done but for general $a$ this could just be a dream.

  • 1
    $\begingroup$ @user64494. Being blind, it is extremely difficult to write in comments. $\endgroup$ Mar 22 at 15:27
  • $\begingroup$ I don't think so. You may type your comment in any editor, making use of glasses, and then copy&paste. $\endgroup$
    – user64494
    Mar 22 at 15:38

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