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For the fourier series of $\sin(x)$ for the given domain, I got

$$-\,\frac{8}{\pi}\sum^{\infty}_{n = 1}\left(-1\right)^{n} \,{n \over 4n^{2} - 1}\,\sin\left(2nx\right)$$ Now after using this result to calculate the first four plots ( approximations for part $\left(\mathrm{b}\right)$ ), as I increase the n value in the fourier approximation, I'm getting results that look less and less like $\sin\left(x\right)$. Shouldn't the fourier series of $\sin\left(x\right)$ ultimately converge to $\sin\left(x\right)$ ?. Does this mean I did something wrong ?.

Felix Marin
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Damon Williams
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  • I think the function you want to converge to will have jump discontinuities at every odd multiple of $\frac{\pi}{2}$ – WW1 Sep 12 '16 at 03:20
  • Your series approximates $\sin$ in the first place in the interval $-{\pi\over2}<x<{\pi\over2}$, then in some more selected intervals of length $\pi$. Note that you forced the series to have period $\pi$, whereas $\sin$ has period $2\pi$. – Christian Blatter Sep 12 '16 at 11:52

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