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

[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.

 

[jasonRF]

Your nth coefficient in your expansion is something like
$$a_n = \int_0^2 f(x) q_n(x) dx$$
where $$q_n$$ is a sin, cos, bessel, etc.

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

jason

 

[yungman]

Your nth coefficient in your expansion is something like
$$a_n = \int_0^2 f(x) q_n(x) dx$$
where $$q_n$$ is a sin, cos, bessel, etc.

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

jason

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!

 

[jasonRF]

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!

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

 

[LCKurtz]

I think I see what is bothering you. In general in eigenfunction expansions, whatever their form, you have a set of eigenfunctions φ_n 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 φ_n and integrate termwise with respect to the weight function:

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

using the orthogonality of the φ_n's. This gives you the formula for c_n:

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

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 φ_n's are on the rest of the interval don't matter. But notice, in the full formula for the c_n, 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.

 

[yungman]

I think I see what is bothering you. In general in eigenfunction expansions, whatever their form, you have a set of eigenfunctions φ_n 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 φ_n and integrate termwise with respect to the weight function:

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

using the orthogonality of the φ_n's. This gives you the formula for c_n:

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

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 φ_n's are on the rest of the interval don't matter. But notice, in the full formula for the c_n, 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.

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