Show that the function in the Fourier series of a periodic function of period $2L$ are periodic of period $2L$ and orthogonals.
My question is about orthogonality.
$$f(x) = \frac{a_{0}}{2} + \sum_{1}^{\infty}\left(\underbrace{a_{n}\cos\left(\frac{n\pi x}{L}\right)}_{g(x)} + \underbrace{b_{n}\sin\left(\frac{n\pi x}{L}\right)}_{h(x)}\right).$$
So, we need to show that
$$\int_{-L}^{L}g(x)\cdot h(x)\mathrm{d}x = 0.$$
But $h(x)$ and $g(x)$ depends implicitly of $f(x)$. How to prove without know $f(x)$? Probably, I think wrong. Maybe I must to prove that
$$\int_{-L}^{L}\cos\left(\frac{n\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)\mathrm{d}x = 0,$$
and this is obvious. Can someone help me?
Best Answer
First, we have
$$ \int_{-L}^{L} \cos(\frac{n \pi x}{L}) \sin(\frac{n \pi x}{L}) dx = 0 \tag{1}$$
now we use the following trigonometric identity
$$ \sin(\alpha)\cos(\beta) = \frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2} \tag{2} $$
which turns the integrand into
$$ \cos(\frac{n \pi x}{L}) \sin(\frac{n \pi x}{L}) = \frac{\sin(\frac{2 \pi n x}{L}) +\sin(0 )}{2} = \frac{\sin(\frac{2 \pi n x}{L}) }{2} \tag{3} $$
so our integral becomes
$$ \frac{1}{2}\int_{-L}^{L} \sin(\frac{2 \pi n x}{L}) dx = \frac{-L\cos(\frac{2\pi nx}{L})}{4 \pi n}\Big|_{-L}^{L} \tag{4}$$
so we get
$$ \frac{-L}{4\pi n } \bigg( \cos(\frac{2\pi nL}{L}) - \cos(\frac{-2\pi nL}{L}) \bigg) \tag{5}$$
$$ \frac{-L}{4\pi n } \bigg( \cos(2n\pi) - \cos(-2n \pi) \bigg) = 0 \tag{6}$$
You only need $f(x)$ to get the coefficients.