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Firstly, I consider the Fourier series of the $2\pi$-periodic function $f=\cos{\frac{3}{2}\pi}$ if $-\pi\leq x\leq \pi.$ And I get $$\cos\frac{3}{2}t \sim -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt.$$ Since I find that $f\in C^2(T),$ i.e. $f,f’$ and $f’'$ are continuous on $T,$ then $$\cos\frac{3}{2}t = -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt \text{ (converge uniformly)}.$$ Let $x=\frac{\pi}{2},$ then \begin{align} \cos\frac{3}{2}(\frac{\pi}{2})&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{\pi}{2}\notag\\ &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag \end{align} Let $x=\frac{3}{2}\pi,$ then \begin{align} \cos\frac{3}{2}(\frac{3}{2}\pi)&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{3}{2}\pi\notag\\ &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag \end{align} I find that these two same series $$-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi} $$ converge to 2 different values, which are $$\cos\frac{3}{2}(\frac{\pi}{2}) \text{ and } \cos\frac{3}{2}(\frac{3}{2}\pi).$$ Did I make any mistake? Thanks~

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No, but your function is not $cos(\frac{3x}{2})$ as such, it is a periodic extension of that defined in $-\pi$ to $\pi$. If we call this function $f(x)$ then you can see in the graph below that $f(\frac{\pi}{2})$ and $f(\frac{3 \pi}{2})$ are the same.

enter image description here

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  • $\begingroup$ Thanks, I get it. $\endgroup$
    – Neo
    Mar 10, 2021 at 15:58

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