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 have arrived at the following expression:

$$\sum_{n=0}^{\infty} \sqrt{n+1}D_{n}\left(x\right)$$

Where $D_n\left(x\right) = \frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{\sin\left(x/2\right)}=1+2\sum_{k=1}^{n} \cos\left(kx\right)$ is the Derichlet kernel. I expected this expression to essentially be a Dirac delta function, but the connection seems a bit unclear to me. Does a sum of Derichlet kernels have any significance or direct relation to Dirac delta functions?

The only progress I've made is by using the second definition of the Derichlet kernel and re-arranging the terms in the double sum:

$$\sum_{n=0}^{\infty} \sqrt{n+1}D_{n}\left(x\right)=\sum_{n=0}^{\infty}\sqrt{n+1}\left[1+2\sum_{k=1}^{n} \cos\left(kx\right)\right]$$

Then if we truncate the series at $N$:

$$=\left(\sqrt{1}+\sqrt{3}+\dots\sqrt{2N+1}\right)1+\dots$$ $$\dots\left(\sqrt{1}+\sqrt{3}+\dots\sqrt{2\left(N-1\right)+1}\right)\cos\left(x\right)+\dots$$ $$\dots\left(\sqrt{1}+\sqrt{3}+\dots\sqrt{2\left(N-2\right)+1}\right)\cos\left(2x\right)$$

Which looks like:

$$\sum_{n=0}^{N} \sqrt{n+1}D_{n}\left(x\right)=\sum_{k=0}^{N} a_k\cos\left(kx\right)$$

with $a_k=2\sum_{m=0}^{N-k}\sqrt{2m+1}$

Which, being a cosine series seems sort of like it would describe a delta function, but its still a bit hand wavy.

share|improve this question
    
Perhaps you've already done this, but I would start by suggesting you take a look at en.wikipedia.org/wiki/Dirichlet_kernel Please comment if you have further questions. –  Christopher A. Wong Oct 4 '12 at 7:31
    
I have looked at that article, but it is still unclear. Obviously $D_{\infty}\left(x\right)=2\pi\delta\left(x\right)$, but since $D_n\left(x\right)$ are the partial sums, I'm confused what a sum of partial sums could mean? –  okj Oct 4 '12 at 15:32
    
Incidentally, the sum does appear to have a closed form solution, in terms of Lerch transcendent functions, which, when evaluated looks a lot like a delta function except that the spread is finite: $$\frac{\sqrt{2}}{e^{ix}-1}\left[e^{ix}\Phi\left(e^{ix},-\frac{1}{2},\frac{1}{2}‌​\right)-\Phi\left(e^{-ix},-\frac{1}{2},\frac{1}{2}\right)\right]$$ –  okj Oct 4 '12 at 15:37
1  
@okj It may be of help, did you see Fejer Kernel? –  Rajesh D Oct 5 '12 at 11:15
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.