How does the orthogonality of sine and cosine figure in the Fourier series?

Question

Okay so this is my first time asking a question so if I've made a mistake pls inform

context: I'm in high school and I'm going to use the Fourier Series for easily doable applications, that isn't the point of the question. But i've been asked to explain it, and I don't really get orthogonality or how it works with the Fourier transform. This question might sound really silly but I've never dealt with either of these two before so bear with me. Thanks!

Answer 1

first let see in $\mathbb{R}^2$ what it mean orthogonality. two vectors $u,v$ are saying to be orthogonal if $\langle u,v \rangle=0$ and orthonormal if both $\langle u,u \rangle$ & $\langle v,v \rangle$ are equal to one. for example $u=(1,0)$ and $v=(0,1)$ form an orthonormal basis in $\mathbb{R}^2$ because every vector $w$ in $\mathbb{R^2}$ can be deduced by it projection on $u$ and $v$, we write : $$ w=\langle w,u \rangle u+\langle w,v \rangle v $$

pythagorean theorem is

$$ \langle w,w\rangle = (\langle w,v \rangle)^2 +(\langle w,u \rangle)^2 $$

if we passes to $\mathbb{R^3}$ you can see that your two vector $u$ and $v$ admit a representation as $u=(1,0,0)$ and $v=(0,1,0)$ and that they form a proper plan in your space but if we add the vector $z=(0,0,1)$ the system verify : $$ \langle u,v \rangle=\langle u,z \rangle=\langle v,z \rangle=0\\ \langle u,u \rangle=\langle z,z \rangle=\langle v,v \rangle=1 $$ and they became an orthonormal system. and every vector $w\in\mathbb{R^3}$ can be expressed as $$ w=\langle w,u \rangle u+\langle w,v \rangle v+\langle w,z \rangle z $$ as you can deduce the number of vector depend of dimension of your space (for $\mathbb{R^2}$ it was 2, but in $\mathbb{R}^3$ it's 3).

pythagorean theorem is

$$ \langle w,w\rangle = (\langle w,v \rangle)^2 +(\langle w,u \rangle)^2+(\langle w,z \rangle)^2 $$

So for fourier case we can take the space of differentiable function on $[-\pi,\pi]$, as above we need to define a inner product so the product is given by : $$ \langle f, g \rangle_{fourier}=\frac{1}{\pi}\int_{-\pi}^\pi f(t)g(t) dt $$ so you can ( using the propriety of integral )verify that

1.$\langle f, f \rangle_{fourier}=\frac{1}{\pi}\int_{-\pi}^\pi f(t)^2 dt \geq 0$

2. If $\langle f, f \rangle_{fourier}=\frac{1}{\pi}\int_{-\pi}^\pi f(t)^2 dt= 0 \implies f=0$

3. $\langle f, g \rangle_{fourier}=\langle f, g \rangle_{fourier}$

4. $\langle (f+\alpha h), g \rangle_{fourier}=\langle f, g \rangle_{fourier}+\alpha \langle h, g \rangle_{fourier}$

(you see the same propriety of scalar product)

and we need this result :

$\int_{-\pi}^\pi \cos(nt)dt=0$ for all $n\geq 1$

(see that if $n\neq 0$ the primitive of $\cos(nt)$ is $\frac{\sin(nt)}{n}$)

Now we can prove the orthogonality in fourier analysis :

1.$\langle \sin(nt),\cos(mt)\rangle_{fourier} =0$ for all $n,m\geq 0$

2.$\langle \cos(nt),\cos(mt)\rangle_{fourier} =0$ for all $n\neq m$

3.$\langle \sin(nt),\sin(mt)\rangle_{fourier} =0$ for all $n\neq m$

4. $\langle \sin(nt),\sin(nt)\rangle_{fourier} =\langle \cos(mt),\cos(mt)\rangle_{fourier}=1$ for all $m,n\geq 1$

5. $\langle \cos(0),\cos(0)\rangle_{fourier}=2$

For the proof of 1. see that the function $ \sin(nt)\cos(mt)$ is odd then the integral vanish.

For 2 and 3 and 4 we use

$$ 2\cos(x)\cos(y)=\cos(x+y)+\cos(x-y) \\ 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) $$

So as in $\mathbb{R^3}$ we want to write our function $f$ in function of its projection we can write think's to Dirichlet theorem: $$ f(x)=\sum_{n\geq 1} \langle f(t), \sin(nt)\rangle_{fourier} \sin(nx)+\sum_{n\geq 0} \langle f(t), \cos(nt)\rangle_{fourier} \cos(nx) $$ if we put $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)dt$ $a_k=\langle f(t), \cos(kt)\rangle_{fourier}=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\cos(kt)dt$ and $b_k=\langle f(t), \sin(kt)\rangle_{fourier}=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\sin(kt)dt$ we get :

$$ f(x)=a_0+ \sum_{k\geq 1}(a_k \cos(kt)+b_k\sin(kx)) $$ the equivalent of pythagorean theorem in this case is called Parseval theorem : $$ \frac{1}{\pi}\int_{-\pi}^\pi f(t)^2dt=\langle f(t), f(t)\rangle_{fourier}= \sum_{n\geq 1} (\langle f(t), \sin(nt)\rangle_{fourier})^2 +\sum_{n\geq 0} (\langle f(t), \cos(nt)\rangle_{fourier})^2=a_0^2+ \frac{1}{2}\sum_{k\geq 1}(a_k^2 +b_k^2) $$