= Solution
{c}
For period two, write $f\sim a_0/2+\sum_{n\geq1}[a_n\cos(n\pi x)+b_n\sin(n\pi x)]$. Direct integration gives $a_0=1/2$. For $0<x<1$, $f(x)+f(-x)=2-3x$ and $f(x)-f(-x)=x$, so
$$
a_n=\int_0^1(2-3x)\cos(n\pi x)\,dx=\frac{3(1-(-1)^n)}{n^2\pi^2},\qquad
b_n=\int_0^1x\sin(n\pi x)\,dx=\frac{(-1)^{n+1}}{n\pi}.
$$
Thus the <Fourier series> is
$$
\boxed{\frac14+\sum_{n=1}^\infty\left[
\frac{3(1-(-1)^n)}{n^2\pi^2}\cos(n\pi x)+\frac{(-1)^{n+1}}{n\pi}\sin(n\pi x)\right].}
$$
It converges to the values specified in part <4/a>. Changing the assigned values at the jumps would not change these <integral> coefficients.
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