= Periodic sine with an endpoint-correcting linear term
{title2=$b_n=2(-1)^n\sin L/\{n\pi[1-(n\pi/L)^2]\}$}
For $L>0$, extend $f_L(x)=\sin x-x\sin L/L$ from $[-L,L]$ periodically. The linear term makes both endpoint values zero and the endpoint first <derivatives> match. The function is <odd>, so its <Fourier series> has only sine terms, with the displayed <Fourier coefficients>. Indeed, for $k_n=n\pi/L\ne1$,
$$
\int_0^L\sin x\sin(k_nx)dx=\frac{(-1)^nk_n\sin L}{1-k_n^2},\qquad \int_0^Lx\sin(k_nx)dx=-\frac{L(-1)^n}{k_n};
$$
subtracting and multiplying by $2/L$ proves the formula. The coefficients decay as $O(n^{-3})$. At $L=m\pi$, the single coefficient $b_m$ has removable <limit> $1$, with all others zero. Multiplying $b_n$ by $e^{-\kappa k_n^2t}$ solves the corresponding <heat equation> on the <circle>.
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