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$$\sum_{k=-N}^{N} e^{2 \pi i k t}=\frac{\sin [(2 N+1) \pi t]}{\sin (\pi t)}$$

I am trying to solve the above question. But I have literally no idea to where to start. How can a logarithmic expression be equal to an sinusoidal expression? Can you give me an idea? Thank you from now :)

Daniel Fischer
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Jake Evergreen
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1 Answers1

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For $e^{2\pi it}\ne1,$

$$\sum_{k=-N}^N(e^{2\pi it})^k=e^{-2\pi Nit}\cdot\dfrac{1-(e^{2\pi it})^{2N+1}}{1-e^{2\pi it}}=\dfrac{e^{2\pi(N+1) it}-e^{-2\pi N it}}{e^{2\pi it}-1}=\dfrac{e^{\frac{2\pi it(2N+1)}2}}{e^{2\pi it/2}}\cdot\dfrac{e^{\pi(2N+1)it}-e^{-\pi(2N+1)it}}{e^{i\pi t}-e^{-i\pi t}}$$

Now use $e^{ix}-e^{-ix}=2i\sin x$ and

How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

lab bhattacharjee
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  • I coulnd't understand how you tranlate \frac{e^{2 \pi(N+1) i t}-e^{-2 \pi N i t}}{e^{2 \pi i t}-1} into =\frac{e^{\frac{2 \pi i t(2 N+1)}{2}}}{e^{2 \pi i t / 2}} \cdot \frac{e^{\pi(2 N+1) i t}-e^{-\pi(2 N+1) i t}}{e^{i \pi t}-e^{-i \pi t}} – Jake Evergreen Jun 07 '20 at 16:33
  • I coulnd't understand how you tranlate \frac{e^{2 \pi(N+1) i t}-e^{-2 \pi N i t}}{e^{2 \pi i t}-1} into =\frac{e^{\frac{2 \pi i t(2 N+1)}{2}}}{e^{2 \pi i t / 2}} \cdot \frac{e^{\pi(2 N+1) i t}-e^{-\pi(2 N+1) i t}}{e^{i \pi t}-e^{-i \pi t}}

    I don't know how to convert latex so sorry

     – Jake Evergreen Jun 07 '20 at 16:34