= Fourier sine series of a symmetric tent
{c}
{title2=$\min(x,L-x)=\frac{4L}{\pi^2}\sum_{n\geq1}\frac{\sin(n\pi/2)}{n^2}\sin(n\pi x/L)$}
On $0\leq x\leq L$, with $L>0$, the tent $f(x)=\min(x,L-x)$ has <Fourier sine series> coefficients $b_n=4L\sin(n\pi/2)/(\pi^2n^2)$. Split the coefficient integral at $L/2$ and integrate its two linear pieces by parts; their cosine endpoint terms cancel and their sine terms add. Reflection about the midpoint makes every even-index coefficient zero, and the remaining odd-index coefficients alternate in sign. The $n^{-2}$ bound gives <absolute convergence> and <uniform convergence>; the series equals the continuous tent, including its zero endpoints.
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