Discussion
Let
\[ f(x) = \begin{cases} \pi + x, & -\pi \leq x \lt 0, \\ 0, & 0 \leq x \lt \pi, \end{cases} \]
with \(f(x + 2\pi) = f(x)\). If \(F(x)\) represents the Fourier series of \(f(x)\), then the value of \(F\left(-\frac{\pi}{2}\right) + F(0)\) is
Answer: C
The value is \(\pi\).
By Dirichlet's theorem the Fourier series equals \(f(x)\) where f is continuous and the average of the left and right limits at a jump.
At \(x = -\pi/2\), f is continuous and equal to \(\pi + x\), so \(F(-\pi/2) = \pi - \pi/2 = \pi/2\). At \(x = 0\) there is a jump from \(f(0^-) = \pi\) to \(f(0^+) = 0\), so
\[ \begin{aligned} F(0) &= \frac{\pi + 0}{2} = \frac{\pi}{2} \\ F(-\tfrac{\pi}{2}) + F(0) &= \frac{\pi}{2} + \frac{\pi}{2} = \pi \end{aligned} \]
Taking \(F(0) = f(0) = 0\) gives \(\pi/2\), the trap of ignoring the jump, and taking \(F(0) = \pi\) gives \(3\pi/2\).
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