= Solution
The <function> is even, so every sine coefficient is zero. Its mean coefficient is
$$
a_0=\frac2\pi\int_0^\pi\sin x\,\mathrm dx=\frac4\pi.
$$
For $n\geq1$, compute $a_n=(2/\pi)\int_0^\pi\sin x\cos(nx)\,\mathrm dx$. The product-to-sum identity makes this integral zero for odd $n$ and gives $-4/(\pi(n^2-1))$ for even $n$. Thus the missing <Fourier coefficients> are \b[$a_n=0$ for odd $n$ and $b_n=0$ for every $n$], with $a_0=4/\pi$. Equivalently,
$$
|\sin x|=\frac2\pi-\frac4\pi\sum_{m=1}^\infty\frac{\cos(2mx)}{4m^2-1}.
$$
The <Fourier series> converges to this continuous periodic <function>; there is no jump value to average.
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