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I have the Fourier series

$$\sum_{n=1}^{\infty}\frac{\sin nx\sin^2n\alpha}{n}$$

which represents some function $f$ to be found, and I aim to determine the points at which the Fourier series is discontinuous. The answer is that it is some constant for $0<x<2\alpha$ and zero for $2\alpha<x<\pi.$

To do this I used the method of Stokes. I divided the interval $(0,\pi)$ at $n$ points $k_1,k_2,\dots,k_n,$ and if we break up the integrals for $a_n$ (the cosine coefficients) and $b_n$ (the sine coefficients) then we get

$$na_n=A_n-b'_n,~nb_n=B_n+a'_n$$

where $a'_n$ and $b'_n$ represent the Fourier coefficients of the derivative, and $A_n,$ $B_n$ given by

$$\pi A_n=\sum_{r=1}^{n}\{f(k_r^-)-f(k_r^+)\}\sin nk_r,~\pi B_n=-\sum_{r=1}^{n}\{f(k_r^-)-f(k_r^+)\}\cos nk_r$$

will vanish when $f$ is continuous at all points.

I thought that since $f$ has a sine series it should be an odd function of $x$ and therefore its derivative should also be an odd function of $x.$ So that would mean $a'_n=0$ for all $n$ and therefore $nb_n=B_n=\sin^2n\alpha=\tfrac12(1-\cos2n\alpha),$ which seems to suggest that the points of discontinuity are only $k_1=2\alpha.$ Then if I take

$$\pi A_n=\{f(2\alpha^-)-f(2\alpha^+)\}\sin2n\alpha$$

then if $0<2\alpha<\pi,$ the sine term cannot vanish, and I have just claimed that $f(2\alpha^-)\neq f(2\alpha^+),$ so this means $A_n\neq0.$ But then that demands $b'_n\neq0,$ and therefore a nonconstant $f.$

On the other hand, I can't see how to show that $b'_n=0$ for this Fourier series.

Would someone be able to let me know how to see when a Fourier series is discontinuous when we know only the series, but not the function? I would appreciate either correcting any mistakes I have made (I am not a professional or a maths student and I'm aware this method might be outdated or cumbersome) or showing me a better way to tackle these sorts of problems.

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  • $\begingroup$ The derivative of an odd function is an even function. $\endgroup$ – Julián Aguirre May 30 '18 at 17:44

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