How do I find the fourier series for the function f(t) = sin t?

Question

How do I find a fourier series for the function of period $2\pi$ satisfying $$f(t)= \begin{cases}\sin t &0 \le t<\pi\\0 &\pi\le t<2\pi\end{cases} $$

Do I find $b_n$ as usual (because it's an odd function) and then give the Dirichlet conditions? I'm a bit thrown by the zero.

You can save yourself a bit of calculation by noting that$$f(x) = \max(\sin x, 0) = \tfrac{1}{2}(\sin x + |\sin x|),$$the right-hand side being the decomposition into odd and even parts. — Andrew D. Hwang, Jun 12 '17 at 11:31

score 3 · Answer 1 · answered Jun 12 '17 at 10:49

3

The function is not odd. Since its period is given as $2\pi$, the function is also $0$ from $-\pi$ to $0$. So you'll have to grind out the integrals

$$a_n =\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos \frac{n\pi x}{\pi} \; dx =\frac{1}{\pi} \int_{0}^{\pi} \sin(x) \cos nx\; dx$$

and so on.

answered Jun 12 '17 at 10:49

B. Goddard

Can't the question have defined $f$ to be odd, implying that it's $\sin x$ for $-\pi < x < 0$, $0$ for $-2 \pi < x \leq -\pi$ etc?
https://www.desmos.com/calculator/zkxyg1sukn – Twenty-six colours Jun 12 '17 at 10:58
1

@Twenty-sixcolours That would make the period $4\pi.$ – B. Goddard Jun 12 '17 at 11:11
Ah my bad, thanks! – Twenty-six colours Jun 12 '17 at 11:19
I hoping to avoid having to find both $a_n$ and $b_n$, what do I do with the zero? That's my main concern – Ellise Jun 12 '17 at 11:19
The difference between the first and second integrals in my answer is the interval of integration. The zero eliminates the part from $-\pi$ to $\pi.$ – B. Goddard Jun 12 '17 at 12:46

1 Answers1