Find the Fourier series representation of $f(t)=\sin(3\pi t)$

Question

Find the Fourier series representation of

$$f(x)=\sin(3\pi t)\qquad \text{for }-1\leq t\leq1$$

When I calculate the coefficients, I always get $0$. Why is that? Is the series indeed zero?

Answer 1

For the general Fourier series \begin{align} f(x) = A_{0} + \sum_{n=1}^{\infty} \left( A_{n} \, \cos\left( \frac{n \pi x}{L} \right) + B_{n} \, \sin\left( \frac{n \pi x}{L} \right) \right) \end{align} where the coefficients are \begin{align} A_{0} &= \frac{1}{2L} \int_{-L}^{L} f(x) \, dx \\ A_{n} &= \frac{1}{L} \int_{-L}^{L} f(x) \, \cos\left( \frac{n \pi x}{L} \right) \, dx \\ B_{n} &= \frac{1}{L} \int_{-L}^{L} f(x) \, \sin\left( \frac{n \pi x}{L} \right) \, dx \end{align} leads to, for $x \in [-1,1]$ and $f(x) = \sin(3 \pi x)$, the following. \begin{align} A_{0} &= \frac{1}{2} \int_{-1}^{1} \sin(3 \pi x) \, dx = \frac{1}{2} \left[ - \frac{\cos(3 \pi x)}{3 \pi} \right]_{-1}^{1} = 0 \\ A_{n} &= \int_{-1}^{1} \sin(3 \pi x) \, \cos(n \pi x) \, dx = \frac{1}{2\pi} \, \left[\frac{\cos(\pi (n-3) x))}{n-3} - \frac{cos(\pi (n+3) x))}{n+3} \right]_{-1}^{1} \end{align} If $n \neq 3$ then the value of this is zero. For the case of $n=3$ it is seen that \begin{align} A_{3} &= \int_{-1}^{1} \sin(3\pi x) \, \cos(3\pi x) \, dx = - \frac{1}{12 \pi} \left[ \cos(6 \pi x) \right]_{-1}^{1} = 0. \end{align} The remaining coefficients are \begin{align} B_{n \neq 3} &= \int_{-1}^{1} \sin(3 \pi x) \sin(n \pi x) \, dx = - \frac{6 \, \sin(n \pi)}{\pi (n^{2} -9)} \end{align} which is zero for $n$ being an integer. The last term is \begin{align} B_{3} = \int_{-1}^{1} \sin^{2}(3\pi x) \, dx = 1. \end{align} The Fourier series is then given by \begin{align} \sin(3 \pi x) = \sin(3 \pi x) \end{align}