Proof for k=0n1cos(2k+12n+1π)=12\displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12

This note is to provide a proof for

k=0n1cos(2k+12n+1π)=12\sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12

Proof:

S=k=0n1cos(2k+12n+1π)={k=0n1e2k+12n+1πi}={eπi2n+1k=0n1e2kπi2n+1}={eπi2n+1(1e2nπi2n+11e2πi2n+1)}={eπi2n+1eπi1e2πi2n+1}={eπi2n+1+1(1+eπi2n+1)(1eπi2n+1)}={11eπi2n+1}={11cosπ2n+1isinπ2n+1}={1cosπ2n+1+isinπ2n+1(1cosπ2n+1)2+sin2π2n+1}={1cosπ2n+1+isinπ2n+122cosπ2n+1}=12 \begin{aligned} S & = \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) \\ & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k\pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left(\frac {1-e^{\frac {2n\pi i}{2n+1}}}{1-e^{\frac {2\pi i}{2n+1}}}\right) \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}}-e^{\pi i}}{1-e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}} + 1}{\left(1+e^{\frac {\pi i}{2n+1}} \right) \left(1-e^{\frac {\pi i}{2n+1}} \right)} \right \} \\ & = \Re \left \{ \frac 1{1-e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac 1{1-\cos \frac {\pi}{2n+1} - i\sin \frac {\pi}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi}{2n+1} + i\sin \frac {\pi}{2n+1}}{\left(1-\cos \frac {\pi}{2n+1}\right)^2 + \sin^2 \frac {\pi}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi}{2n+1} + i\sin \frac {\pi}{2n+1}}{2-2\cos \frac {\pi}{2n+1}} \right \} \\ & = \frac 12 \ \square \end{aligned}

Since cos(πx)=cosx\cos (\pi - x) = - \cos x, we have: k=0n1cos(π2k+12n+1π)=k=0n1cos(2n2k2n+1π)=k=1ncos(2k2n+1π)=12\displaystyle \sum_{k=0}^{n-1} \cos \left(\pi - \frac {2k+1}{2n+1}\pi \right) = \sum_{k=0}^{n-1} \cos \left(\frac {2n-2k}{2n+1}\pi \right) = \sum_{k=1}^n \cos \left(\frac {2k}{2n+1}\pi \right) = - \frac 12.

Note by Chew-Seong Cheong
2 years, 2 months ago

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good. A minor typo in the next-to-last three lines: delete the ii from the sine and cosine args that resulted from a LATEX copy-and-paste error.

Wesley Zumino - 2 years, 2 months ago

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Thanks a lot.

Chew-Seong Cheong - 2 years, 2 months ago

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Given what you've written up, can you provide a one-line proof to demonstrate that k=0n1cos(2k2n+1π)=12?\sum_{k=0}^{n-1} \cos \left( \frac {2k}{2n+1}\pi \right) = -\frac 12?

Pi Han Goh - 2 years, 2 months ago

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Thanks, now I see it.

Chew-Seong Cheong - 2 years, 2 months ago

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I think you meant Σk=1n\Sigma_{k=1}^{n} because otherwise it would return 3/2

Mehdi K. - 2 months, 1 week ago

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OK, for the last one, you mean.

Chew-Seong Cheong - 2 months, 1 week ago

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@Chew-Seong Cheong yeah I was replying to Pi Han, I went through the proof multiple times..just find it a cool property :)

Mehdi K. - 2 months, 1 week ago

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big fan, sir

Mehdi K. - 2 months, 1 week ago

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Thanks. I will keep up my work.

Chew-Seong Cheong - 2 months, 1 week ago

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