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How to show that function $\left(\operatorname{sinc}{x}\right)^{n}=\left\{\dfrac{\sin\left(x\right)}{x}\right\}^{n}$ is infinitely differentiable at $0$?

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You only need to show it for $\frac{sin x}{x}$ itself; its powers are then $C^\infty$ because the product of two $C^\infty$ functions is itself $C^\infty$. – Henning Makholm Oct 5 '11 at 20:36
If your class has covered Taylor series, then you will see that $\sin x/x$ is represented by an everywhere converging Taylor series, and therefore ... The rest then follows from Henning's comment. – Jyrki Lahtonen Oct 5 '11 at 20:52
This… maybe also helpful – David Jun 7 '12 at 3:16
up vote 5 down vote accepted

Use the Taylor series expansion as mentioned in the comments to get: $$ \text{sinc}(x)=\frac{\sin x}{x}=\frac{-\sum_{k=0}^\infty { (-x)^{2k+1} \over {(2k+1)!}} }{x}=-\sum_{k=0}^\infty {(-1)^{2k+1}(x)^{2k} \over {(2k+1)!}}=1-{x^2 \over 3!}+{x^4 \over 5!}-\cdots, $$ which is infinitely often differentiable at $0$. You'll get $$ \frac{d^m}{dx^m} \text{sinc}(x) \Biggr|_{x=0}= \cases{ {(-1)^{m}\over(m+1)} & \text{if $m$ is even}\\ \phantom{case}&\\ \;\;\;\;\,\,0 & \text{if $m$ odd.} \\ } $$ Because the product of two $C^∞$ functions is itself $C^∞$, we are done.

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