Fourier Sine series uses the orthogonal set $\{\sin(nx)\}^\infty_{n = 1}$ on $0 \le x \le \pi$

Question

I came across the following explanation:

The Fourier Sine series uses the orthogonal set $\{\sin(nx)\}^\infty_{n = 1}$ on $0 \le x \le \pi$.

Here, the norm squared is

$$|| \sin(nx) ||^2 = \int_0^\pi \sin^2(nx) \ dx \\ = \frac{1}{2} \int_0^\pi (1 - \cos(2nx)) \ dx \\ = \frac{1}{2} \left[ x - \frac{\sin(2nx)}{2n} \right]^\pi_0 \\ = \frac{\pi}{2}$$

What does it mean to say that the Fourier Sine series uses the orthogonal set $\{\sin(nx)\}^\infty_{n = 1}$ on $0 \le x \le \pi$? And how did they get and why is the norm squared $|| \sin(nx) ||^2 = \int_0^\pi \sin^2(nx) \ dx$?

I would appreciate it if people could please take the time to clarify this.

Answer 1

The $\mathscr{L}^2$ norm on $[0,\pi]$ is $$ \Vert f \Vert_{\mathscr{L}^2} = \left(\int_0^\pi |f(x)|^2 \,dx\right)^{1/2}. $$ It is induced by the $\mathscr{L}^2$ inner product $$ \langle f,g \rangle_{\mathscr{L}^2} := \int_0^\pi f(x)\overline{g(x)} \,dx. $$ One usually works in $\mathscr{L}^2$ when dealing with Fourier series because $\mathscr{L}^2$ is a Hilbert space, so it makes sense to speak about orthogonality and similar things.

The set $\{\sin(nx)\}_{n=1}^\infty$ is orthogonal because $$ \langle \sin(nx),\sin(mx) \rangle_{\mathscr{L}^2} = \int_0^\pi \sin(nx)\sin(mx) = 0 $$ whenever $n \not= m$.

Now you can see directly from the above that $$ \Vert \sin(nx) \Vert_{\mathscr{L}^2}^2 = \int_0^\pi \sin^2(nx) \,dx = \frac{\pi}{2}. $$ Note that this shows that the set $\left\{\sqrt{\frac{2}{\pi}}\sin(nx)\right\}_{n=1}^\infty$ is orthonormal.