- #1

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And how can I prove it??

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- Thread starter xuying1209
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- #1

- 4

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And how can I prove it??

- #2

HallsofIvy

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For n> 1, n odd, essentially you are adding the real parts of the 2nth roots of unity. Since those roots are symmetric about the imaginary axis, the sum is 0.

- #3

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Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not [itex]\pi[/itex]) ...

- #4

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I am not sure what is being asked. Is this [tex]\sum cos(n_i)/n_i, or \sum cos(pi*n_i/n_i), or what?[/tex]

Last edited:

- #5

HallsofIvy

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I interpreted as sum of [itex] cos(i\pi/n)[/tex] for i= 1 to n-1.

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