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Series expansion

  1. Apr 11, 2006 #1
    Let,s suppose we have a function f(x) wich is not on [tex] L^{2} [/tex] space but that we choose a basis of orthononormal functions so the coefficients:

    [tex] c_{n}=\int_{0}^{\infty}dxf(x)\phi_{n}(x) [/tex] are finite.

    would be valid to expand the series into this basis in the form:

    [tex] f(x)=\sum_{n=0}^{\infty}\phi_{n}(x) [/tex] of course the sum:

    [tex] \sum_{n=0}^{\infty}|c_{n}|^{2} [/tex] would diverge
  2. jcsd
  3. Apr 11, 2006 #2


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    I presume you meant to include the coefficients (cn) in the infinite series for f(x). Depending on the properties of f(x), the series may or may not converge for any specific x.
  4. Apr 11, 2006 #3

    matt grime

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    1. Pick a basis of what?

    2. No that function is not equal to that sum for any reason at all, now wouldit be if you even put the c_n in as you meant to

    3. It is not even true that an L^2 function is equal to its Fourier series
  5. Apr 14, 2006 #4
    -I said a basis of orthonormal function (they are on L^{2} but f(x) isn,t)

    -If the integral [tex] c_{n}=\int_{0}^{\infty}dxf(x)\phi_{n}(x) [/tex] is finite then every c coefficient exist.

    -then when it would the equality hold?....[tex] f(x)\sim\sum_{n=0}^{\infty}c_{n}\phi_{n}(x) [/tex]

    perhapsh we could consider it to be an "asymptotic" representation of the function by means of eigenfunctions in the sense that you take only a few finite coefficients to approximate the function.
  6. Apr 15, 2006 #5

    matt grime

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    1. You didn't, and still haven't said what they are a basis of. L^2 of what?

    2. Why say equal and then write ~? Define your equivalence relation.
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