Complex analysis using Kronecker delta..

Asked Sep 10 '19 at 10:34

Active Sep 10 '19 at 10:39

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I have some trouble with proving the formula below;

$$\sum_{k=1}^n \sin\Big(\frac{kr \pi}{n+1}\Big)\sin\Big(\frac{ks \pi}{n+1}\Big) = \frac{n+1}{2}\delta_{rs}$$

by making use of complex analysis for any $r, s=1, 2, 3, \ldots, n$.

To solve this, I wanted to use the Kronecker delta's property, $$\delta_{rs} = \frac{1}{n}\sum_{k=1}^ n{e}^{2\pi i \frac{k}{n} (r-s)}$$

So I changed the first formula using addition formulas for sinusoidal functions, $$\sum_{k=1}^n \frac{1}{2}\left[\cos\Big(\frac{k\pi}{n+1}(r-s)\Big)-\cos\Big(\frac{k\pi}{n+1}(r+s)\Big)\right]$$

I'm stuck in this situation. How can I get started..?

edited Sep 10 '19 at 10:39

Bernard

179,256

asked Sep 10 '19 at 10:34

Hyo

Did you try the $r=s$ case first? Then the Kronecker-$\delta$ is just zero and this case should be easier to prove. – quarague Sep 10 '19 at 10:40
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I thought I did the case when r=s, but I think my solution is incomplete.. Oh, and if r=s, isn't the kronecker delta be 1? – Hyo Sep 10 '19 at 10:58
It's 1, you're right. It is zero if $r \neq s$. – quarague Sep 10 '19 at 11:25
How can I get started..? I can't get any idea of it..,, quite difficult to me.. – Hyo Sep 10 '19 at 11:34
1

This issue is called "discrete orthogonality". Take a look at the answer I gave to this question – Jean Marie Sep 10 '19 at 11:48

Complex analysis using Kronecker delta..

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