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.

If $f$ is a $C^1$ function of period $2\pi$ and $f(0)=0$, and $g(x) = f(x)/(e^{ix} -1)$,

then let $C_n$ be the complex fourier coefficients of $f(x)$ and $D_n$ the coefficients of $g(x)$. How can we show that $D_n \to 0$?

ii) And how can we show that $C_n = D_{n-1} - D_n$, so that the series $\sum C_n$ is telescoping.

So, for the first one, I'm thinking of uniform continuity and how to show that $g(x)$ is continuous in it.

share|improve this question
    
$D_n \to 0$ looks like the Riemann-Lebesgue lemma... –  Siminore Oct 28 '12 at 9:50

1 Answer 1

For ii), write $g(x)=f(x)/(\def\eix{\mathrm e^{\mathrm ix}}\eix-1)$ as $f(x)=g(x)(\eix-1)$. Now if you compute the Fourier coefficients on both sides, the factor $\eix$ shifts the coefficients of $g$ by $1$, yielding $C_n=D_{n-1}-D_n$.

For the first question, you can conclude from $C_n=D_{n-1}-D_n$ that $\sum_{k=j}^nC_k=D_j-D_n$ by induction, then let $j\to-\infty$ and $n\to\infty$ to get $\lim_{n\to\infty}(D_{-n}-D_n)=\sum_{k=-\infty}^\infty C_k=f(0)=0$. Contrary to what I wrote in the first version of this answer, that only establishes that the imaginary part of $D_n$, which is odd, goes to $0$. To conclude that all of $D_n$ goes to $0$, either note that $g$ is continuous at $0$ (and thus everywhere), and conclude that $\sum_{k=-\infty}^\infty D_k=g(0)$, so $D_n$ must go to zero for the series to converge; or invoke the Riemann-Lebesgue lemma, as Siminore suggested in a comment under the question.

share|improve this answer
    
@mary: Didn't or doesn't? Does it make sense now? –  joriki Oct 28 '12 at 19:57
    
@mary: Could you please a) use $\TeX$ in the comments, too, to make them readable, and b) explain how your comment relates to my answer? Neither of those, as far as I can decypher them, seems to be related to what I wrote. –  joriki Oct 28 '12 at 21:44
    
@mary: I've updated the answer. Apologies for the error in the first version. –  joriki Oct 29 '12 at 0:17
    
It looks good, can you make the answer more complete by helping out with computing the fourier coefficients and induction? I'm sure I will make a mistake on this part –  mary Oct 29 '12 at 0:39
    
@mary: To be honest, it's beginning to feel like I'm doing your work for you. How about you do it and then if you're not sure about some part you can ask again? –  joriki Oct 29 '12 at 0:51

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.