Help with Baby Rudin Chapter 8 Question 12

Question

The question:

Suppose $0< \delta < \pi$, $f(x) = 1$ if $|x| \leq \delta$, $f(x) = 0$ if $\delta < |x| \leq \pi$, and $f(x + 2 \pi) = f(x)$ for all $x$.

(a) Compute the Fourier Coefficients for $f$.

(b) Conclude that $$\sum_{n=1}^\infty \frac{\sin(n \delta)}{n} = \frac{\pi - \delta}{2}$$

I've found the coefficients as $c_n = \frac{1}{2 \pi} \int^\pi_\pi f(x) e^{-inx} = \frac{\sin (n \delta)}{i \pi n}$. But I can't seem to see what I should be looking for to prove (b). Parseval's Theorem gives it for $|c_n|^2$ and even then it doesn't give the answer I'm looking for. A hint would be appreciated.

Answer 1

For part (a), it might be a bit more helpful to notice that $f$ is even and real-valued so that you may use the real version of the Fourier series. Also because $f$ is even, $b_n=0$ for all $n$. You should then be able to calculate $a_n$ for $n \geq 0$ (the answer, no work, below)

$a_0=\frac{2\delta}{\pi}$ and $a_n=\frac{2\sin n\delta}{\pi n}$

For (b), think of what extra conditions $f$ satisfies and look at Theorem 8.14 with $x=0$.

Answer 2

There is a small mistake in the calculation of $\hat {f} (n)$. Also, your calculation does not hold for $n=0$. Calculate $\hat {f} (0)$ separately and use the fact that $\sum \hat {f} (n) =f(0)$. You will also have change $n$ to $-n$ when you sum over negative values of $n$.