How can Fourier series formulas be derived without just memorizing them?

[lukaszh]

Hello,
everywhere I can see this
$$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \cos(nt)\, dt$$
$$b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \sin(nt)\, dt$$
etc... I can't find, how to derive this formulas. I'm really tired and a bit confused of this formulas, because I can't find possible way to derive them. I don't like only formula application, but I want to know, what is that formula about.
Thank you...

 

[HallsofIvy]

If you have an "inner product space", that is, an vector space with an inner product defined on it, together with an orthonormal basis, $v_1, v_2, ...$, that is such that $<v_i, v_j>= 0$ if $i\ne j$ and $<v_i, v_i>= 1$ for all i, and want to write v as a linear combination, $v= a_1v_1+ a_2v_2+ ...+ a_nv_n$, then \(\displaystyle a_i= <v, v_i[/itex]. What you have is a vector space with basis $cos(nx)$, $sin(nx)$ with inner product $<f, g>= \frac{1}{\pi}\int_{-\pi}^\pi f(t)g(t)dt$ which leads to the given formulas.\)

 

[jtbell]

It works because the functions sin(nt) for different values of n are orthogonal to each other, that is,

$$\int^{\pi}_{-\pi} {\sin(nt) \sin (mt) dt} = 0$$

for $n \ne m$, and

$$\int^{\pi}_{-\pi} {\sin^2(nt) dt} = \pi$$

Likewise for cosines. Try a few examples if you like. Therefore if you have a function

$$f(t) = b_1 \sin (t) + b_2 \sin (2t) + b_3 \sin (3t) + ...$$

then, for example, letting n = 2:

$$\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \int^{\pi}_{-\pi} {\sin (t) \sin (2t) dt} + b_2 \int^{\pi}_{-\pi} {\sin^2 (2t) dt} + b_3 \int^{\pi}_{-\pi} {\sin(3t) \sin (2t) dt} + ...$$

$$\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \cdot 0 + b_2 \cdot \pi + b_3 \cdot 0 + ...$$

 

[lukaszh]

Thank you. Now I understand. Thanks Thanks

 

[Count Iblis]

It is easier to work with the basis functions

e_n(x) = exp(i n x)

and define the inner product as

<f,g> = 1/(2 pi) Integral from minus pi to pi of f(x)g*(x) dx