Proof of \(\displaystyle{\sum_{{{n}=-{k}}}^{{{k}}}}{e}^{{{2}\pi{i}{n}{x}}}={\frac{{{\sin{{\left({2}{k}+{1}\right)}}}\pi{x}}}{{{\sin{\pi}}{x}}}}\)

Nancy Richmond

Nancy Richmond

Answered question

2022-03-30

Proof of n=kke2πinx=sin(2k+1)πxsinπx

Answer & Explanation

SofZookywookeoybd

SofZookywookeoybd

Beginner2022-03-31Added 15 answers

n=kke2πx=e2kπix+e(2(1k)πix)+e(2(2k)πix)
n=kke2πx=e2kπix+e(2kπix)+2πix+e(2kπix)+4πix
This is a geometric series with common ratio e2πix
To prove the identity, you just need to find the sum of this series using the formula for a geometric series.
U1(rn1)r1
You know that U1=e2kπix, so this should lead you to your answer.

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