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Frobenius series

  1. Apr 25, 2009 #1
    Why do we assume that the first term c0 in a frobenius series render.cgi?x%5E%7Br%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20c_%7Bn%7Dx%5E%7Bn%7D%5Cnocache.png cannot equal 0?

  2. jcsd
  3. Apr 25, 2009 #2

    Ben Niehoff

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    Gold Member

    Latex seems to be misbehaving, so I'll write in plain text:

    In the Frobenius substitution, the x dependence of the first term is already factored out:

    y(x) = x^r Sum (a_k x^k)

    So, the first term in the series is actually

    a_0 x^r

    and when we plug the series into the differential equation, the question we are asking is "What is the smallest r for which a_0 does not vanish?" The answer is given by the indicial equation.

    After solving the indicial equation for r, we are then equipped to ask the next question: "Given that a_0 does not vanish, can I find some sequence a_k such that my formal sum converges and solves the differential equation?"
  4. Apr 25, 2009 #3
    Thanks for the response. Why is it necessary though that a0 not vanish?
  5. May 5, 2009 #4
    Let say the first term that we obtained on substituting the Frobenius series into the DE as

    Aa0xr + .... is identically zero.

    This implies Aa0=0.
    We may assume a0 to be zero or nonzero. But if it is zero then A can be any number. Not an interesting result.
    Last edited: May 5, 2009
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