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For a question like finding the fourier series of:
$$f(x) = \begin{cases} x & 0 \leq x < \pi \\ 0 & -\pi \leq x\leq 0 \\ \frac{\pi}{2} & x = -\pi \end{cases} $$ with period $2\pi$, the fourier series is:

$$Sf(x) = \frac{\pi}{4} - \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} \cos((2k-1)x) + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sin(kx)$$ ,
I graphed the first 9 terms of each sum, and this is what it looks like:
http://www.wolframalpha.com/input/?i=sketch+y+%3D+pi%2F4+-+2%2Fpi+(cosx+%2B+1%2F9+cos3x+%2B+1%2F25+cos5x+%2B+1%2F49+cos7x+%2B+1%2F64+cos8x+%2B+1%2F81+cos9x)+%2B+(-sinx+%2B+sin2x+%2F2+-+sin3x+%2F3+%2B+sin4x+%2F4+-+sin5x+%2F5+%2B+sin6x+%2F+6)
(sorry about the link!)

Upon sketching the actual function $f$, it really seems that it won't converge towards my series. Namely, the interval $(0,\pi)$.
My function is a straight line from $(0,0)$ to $(\pi,\pi)$. But the fourier series looks like it'd converge towards the line $y=0$ for that interval $(0,\pi)$, whereas this should actually occur for say an interval $(-\pi,0)$.
Can someone clarify why the fourier series doesn't look like it'll converge towards $f$?

OneGapLater
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  • Yes, because upon calculating it, the even $k$ values will cause the $k$ terms in the cosine series to be $0$.
    In particular, $a_k = \frac{(-1)^k - 1}{k^2 \pi}$.     – OneGapLater Jun 01 '17 at 02:53
  • Hm, that's strange. I did it by hand and this video also confirmed it:
    https://www.youtube.com/watch?v=KeT6CB6Qi10&index=2&list=PL1B727B06A221E026
    (at 27:18).
    What should the $a_k$ coefficients be?     – OneGapLater Jun 01 '17 at 02:56
  • Your graph looks like $f(-x).$ Which means that if you exchange $-x$ for $x$ in the Fourier series you have calculated, you will reflect is across the y axis. Since $\cos x$ is an even function, nothing happens there. And $\sin x$ is an odd function so $\sin -x = - \sin x$ and your signs flip. – Doug M Jun 01 '17 at 02:57
  • Actually, the equations you show above are correct. The signs are flipped in the wolfram alpha entry. – Doug M Jun 01 '17 at 02:58
  • @DougM I see. Such a stupid mistake, I expanded the summation out wrong... Thanks! – OneGapLater Jun 01 '17 at 02:59
  • http://www.wolframalpha.com/input/?i=sketch+y+%3D+pi%2F4+-+2%2Fpi+(cosx+%2B+1%2F9+cos3x+%2B+1%2F25+cos5x+%2B+1%2F49+cos7x+%2B+1%2F64+cos8x+%2B+1%2F81+cos9x)+-+(-sinx+%2B+sin2x+%2F2+-+sin3x+%2F3+%2B+sin4x+%2F4+-+sin5x+%2F5+%2B+sin6x+%2F+6) – Doug M Jun 01 '17 at 03:01
  • Thanks! Also, a bit related subquestion: What if each "splitting" point wasn't assigned a value? Say, we didn't have the last line for the definition of $f$? So we have the values $f(\pi)$ and $f(-\pi)$ values unassigned, and upon graphing it, we have the same graph except just open circles at those points (and each multiple of $2\pi$ after it as well). WIll there be a fourier series for this, and if not is there a way to articulate it (or a resource online)? – OneGapLater Jun 01 '17 at 03:01
  • Might've rambled a bit but that question is basically asking:
    If $f(-\pi + k 2\pi)$ wasn't assigned a value for every integer $k$, would we have a fourier?     – OneGapLater Jun 01 '17 at 03:03
  • If your function has a jump discontinuity, perhaps happening at the the endpoints of the period, the Fourier series will go through the midpoint of the jump. – Doug M Jun 01 '17 at 03:05
  • I see but I think my case I'm describing isn't a jump discontinuity. As in, the function $f$ has $no$ value assigned at $f(-\pi + k 2\pi)$. So since there's no value assigned at all, we can't really take the midpoint of these values (because there aren't any values assigned at all) right? (Sorry if I sound sarcastic, but I don't mean to) – OneGapLater Jun 01 '17 at 03:08
  • You have a left hand limit and a right hand limit and the Fourier series is going to split the difference. Your function can be continuous almost everywhere, while the series is indeed continuous everywhere. – Doug M Jun 01 '17 at 03:09

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