# Eigenfunction Expansion

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

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

Sorry if I don't understand how it helps... I do know that $\displaystyle c_n$ 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: $\displaystyle \int_a^b w(x)f(x)\phi_n(x)~dx$?

Sorry if I don't understand how it helps... I do know that $\displaystyle c_n$ 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 $\displaystyle f$ by its expansion in the expression $\displaystyle <f, \phi_n>$, and then use the linearity of the inner product?
• May 4th 2010, 11:56 AM
Aryth
Ok... So... We get:

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

Which means that:

$\displaystyle \frac{<f,\phi_n(x)>}{<\phi_n,\phi_n>} = \sum_{n=1}^{\infty} c_n$

Is that along the right lines?
• May 4th 2010, 12:04 PM
Bruno J.
Be careful! You're using $\displaystyle n$ 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:

$\displaystyle <f,\phi_n(x)> = \sum_{j=1}^{\infty}c_j <\phi_j,\phi_n> = c_n <\phi_n, \phi_n>$

because all the terms cancel (by orthogonality), except $\displaystyle <\phi_n, \phi_n>$.
• 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)