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
<br /> a_n = \int_0^2 f(x) q_n(x) dx<br />
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
<br /> a_n = \int_0^2 f(x) q_n(x) dx<br />
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