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 &lt;v_i, v_j&gt;= 0 if i\ne j and &lt;v_i, v_i&gt;= 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 &lt;f, g&gt;= \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}<br /> + b_2 \int^{\pi}_{-\pi} {\sin^2 (2t) dt} <br /> + 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