Fourier series - derivation

1. May 21, 2009

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...

2. May 21, 2009

HallsofIvy

Staff Emeritus
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 [math]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.

3. May 22, 2009

Staff: Mentor

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:

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

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

4. May 23, 2009

lukaszh

Thank you. Now I understand. Thanks Thanks

5. May 23, 2009

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