# Can anyone explain this regarding to fourier series and bessel series expansion?

1. May 18, 2010

### yungman

For finding series expansion solution of problems like

f(x) = h(x) for 0<x<1
f(x) = 0 for 1<x<2

0<x<2

Where the fourier series expansion only integrate from x=0 to x=1 only and totally ignor the portion of x=1 to x=2.

This is also true for fourier bessel series expansion also.

I never see the prove, this only show up in the work problems. Can anyone show me the prove of this.

Last edited: May 18, 2010
2. May 18, 2010

### jasonRF

$$a_n = \int_0^2 f(x) q_n(x) dx$$
where $$q_n$$ is a sin, cos, bessel, etc.

If f is zero over [1,2], how can you simplify this integral?

jason

3. May 18, 2010

### yungman

Thanks for the reply, I know exactly how to solve the problem and I know the answer is 0 when integrate from x=1 to x=2.

I just can not "see" it in English!!

4. May 19, 2010

### jasonRF

I am probably missing something, but to me it looks like you just wrote it in English. So I'm not sure what more you are hoping to "see".

jason

5. May 19, 2010

### LCKurtz

I think I see what is bothering you. In general in eigenfunction expansions, whatever their form, you have a set of eigenfunctions φn satisfying an orthogonality property with respect to some weight function w on an interval (a,b), and you want to express some function f in an eigenfunction expansion.

$$f(x) = \sum_{k=1}^\infty c_k\phi_k(x)$$

You multiply by φn and integrate termwise with respect to the weight function:

$$\int_a^b f(x)\phi_n(x)w(x)dx=\sum_{k=1}^\infty\int_a^b c_k\phi_k(x)\phi_n(x)w(x)dx= c_n\int_a^b \phi_n^2(x)w(x)dx$$

using the orthogonality of the φn's. This gives you the formula for cn:

$$c_n=\frac{\int_a^b f(x)\phi_n(x)w(x)dx}{\int_a^b \phi_n^2(x)w(x)dx}$$

The denominator is frequently constant as, for example, in the classical sine-cosine series it is $2\pi$, and we tend to forget where it came from.

Now to get to what I think is bothering you. You are thinking that if f(x) is zero on part of the (a,b) interval so you only integrate over part of the interval, so what the φn's are on the rest of the interval don't matter. But notice, in the full formula for the cn, you still go from a to b in the denominator. Again, that may be a constant so you don't notice it. It's like in the ordinary sine cosine series, you don't change the $1/2\pi$ out front when you integrate over just part of the interval because f(x) is partly zero.

Last edited: May 20, 2010
6. May 21, 2010

### yungman

Thanks for the reply. I have been busy on another problem, have not been able to get to this until now.

So you mean the denominator is the full range( say $2\pi$) even the f(x) only non zero from o<x<$\pi$ This will cover it according to the formula of fourier/bessel series expansion.

Thanks