# Eigenfunction Expansion

• May 4th 2010, 11:17 AM
Aryth
Eigenfunction Expansion
Use the orthogonality relation to find expressions for the coefficients $c_n$ in the eigenfunction expansion of a function f(x):

$f(x) \approx \sum_{n=1}^{\infty} c_n\phi_n$
• May 4th 2010, 11:19 AM
Bruno J.
Hint : what is $$?
• May 4th 2010, 11:40 AM
Aryth
Isn't it: $\int_a^b w(x)f(x)\phi_n(x)~dx$?

Sorry if I don't understand how it helps... I do know that $c_n$ is a quotient of inner products... I'm afraid I don't know how to show it.
• May 4th 2010, 11:48 AM
Bruno J.
Quote:

Originally Posted by Aryth
Isn't it: $\int_a^b w(x)f(x)\phi_n(x)~dx$?

Sorry if I don't understand how it helps... I do know that $c_n$ is a quotient of inner products... I'm afraid I don't know how to show it.

Well that's the left side of the equation, indeed. But what if you replace $f$ by its expansion in the expression $$, and then use the linearity of the inner product?
• May 4th 2010, 11:56 AM
Aryth
Ok... So... We get:

$ = \sum_{n=1}^{\infty}c_n \int_a^b w(x)\phi_n(x)^2~dx = \sum_{n=1}^{\infty}c_n <\phi_n,\phi_n>$

Which means that:

$\frac{}{<\phi_n,\phi_n>} = \sum_{n=1}^{\infty} c_n$

Is that along the right lines?
• May 4th 2010, 12:04 PM
Bruno J.
Be careful! You're using $n$ both as a fixed integer and as a summation index! Here's what your corrected post looks like :

Quote:

Originally Posted by Aryth
Ok... So... We get:

$ = \sum_{j=1}^{\infty}c_j <\phi_j,\phi_n> = c_n <\phi_n, \phi_n>$

because all the terms cancel (by orthogonality), except $<\phi_n, \phi_n>$.
• May 4th 2010, 12:05 PM
Aryth
Ah, I see. I appreciate the help. Thanks a lot.
• May 4th 2010, 12:06 PM
Bruno J.
Quote:

Originally Posted by Aryth
Ah, I see. I appreciate the help. Thanks a lot.

Most welcome! Good luck. (Wink)