The series is
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I checked it for some values in the interval and the identity does hold.
I'm asked to show that for , but if that were true then the function should be independent of on the interval... This seems odd. And going to Wolfram|Alpha, it tells me that, at .5, this equality doesn't hold... Am I misunderstand something here? This seems bizarre.
When one develops for and for (the value at zero doesn't really matter) as a Fourier series over one sees that only the coefficients of the sine will appear since is an odd function, so one gets:
where so the Fourier series for f becomes .
There is a constant there that doesn't appear in your sum...
But
for any constants c and k since is an odd function
EDIT: Nevermind, i missed that f(x) is negative on
Now that I look at it, isn't it the case that
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And then the sum becomes
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Notice that the last sum starts at k=0 and that's the proper range.
You need to sum from n = 0 not n = 1. I think that's the problem in your calculation on wolfram alpha.
EDIT: link sum of sin((2n+1)x)/(2n+1 ) from n = 0 to infinity - Wolfram|Alpha
You can see that the sum is constant from to . That is precisely
This is so frustrating... I tried just doing the Fourier series of the constant function and I don't get it. Since I can just use , right? And , , no? If we integrate from 0 to , then don't we get , so that no matter what the sum is, if it's not 0 then we get a contradiction--and if it is, we get no information?
Okay, I've done all the computations and I seem to get basically the right thing, but in the step , I see why we only take the odd values in the top. The exponent is odd iff the numerator is 0 and the exponent is odd iff k is even.
\strikethrough{However, why does this carry over to the denominator, since in the LHS the denominator is always even. So even if we're skipping some terms because the numerator is 0, the denominator should still be even on the other terms too.}
I see. When the k is odd, the numerator has 2, cancelling the 2 on bottom. Nevermind.