Vtyjkyuk

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

$$|| sin(nx) ||^2 = int_0^pi sin^2(nx) dx \ = frac12 int_0^pi (1 - cos(2nx)) dx \ = frac12 left[ x - fracsin(2nx)2n right]^pi_0 \ = fracpi2$$

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.

The $mathscrL^2$ norm on $[0,pi]$ is
$$ Vert f Vert_mathscrL^2 = left(int_0^pi |f(x)|^2 ,dxright)^1/2. $$
It is induced by the $mathscrL^2$ inner product
$$ langle f,g rangle_mathscrL^2 := int_0^pi f(x)overlineg(x) ,dx. $$
One usually works in $mathscrL^2$ when dealing with Fourier series because $mathscrL^2$ 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
$$ langle sin(nx),sin(mx) rangle_mathscrL^2 = int_0^pi sin(nx)sin(mx) = 0 $$
whenever $n not= m$.

Now you can see directly from the above that
$$ Vert sin(nx) Vert_mathscrL^2^2 = int_0^pi sin^2(nx) ,dx = fracpi2. $$
Note that this shows that the set $leftsqrtfrac2pisin(nx)right_n=1^infty$ is orthonormal.