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Thread: Eigenfunction Expansion

  1. #1
    Super Member Aryth's Avatar
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    Eigenfunction Expansion

    Use the orthogonality relation to find expressions for the coefficients $\displaystyle c_n$ in the eigenfunction expansion of a function f(x):

    $\displaystyle f(x) \approx \sum_{n=1}^{\infty} c_n\phi_n$
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    MHF Contributor Bruno J.'s Avatar
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    Hint : what is $\displaystyle <f, \phi_n>$?
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    Super Member Aryth's Avatar
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    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.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Aryth View Post
    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?
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    Super Member Aryth's Avatar
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    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?
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    MHF Contributor Bruno J.'s Avatar
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    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 View Post
    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 $\displaystyle <\phi_n, \phi_n>$.
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    Super Member Aryth's Avatar
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    Ah, I see. I appreciate the help. Thanks a lot.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Aryth View Post
    Ah, I see. I appreciate the help. Thanks a lot.
    Most welcome! Good luck.
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