Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I see the comment by Henning Makholm may be relevant to my question on Derivative of sinc function. But I don't quite get the background. Could anybody show me to prove all the derivatives of $\mathrm{sinc}(x)$ are continuous at $x=0$ without omission of details? Thanks!

share|improve this question
2  
I think Jyrki Lahtonen's comment in that thread would be more useful to you. –  Neal Jul 6 '12 at 2:24
add comment

1 Answer

Expanding on @JyrkiLahtonen's comment:

Since $\sin x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$, you have $\mathbb{sinc}\, x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}$. It is straightforward to show that the series converges uniformly on any bounded set, hence it is $C^{\infty}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.