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Frobenius Method

  1. Aug 14, 2008 #1
    I'm a little confused with ODEs. After two weeks of trying to figure out Frobenius I have finally realized that there seems to be two different power set used by all of my three books for the y substitution but I am unsure when to use either one. Here are the two sets that i'm talking about:

    Set 1 from section 6.1

    [tex] y = \sum_{n=0}^\infty C_n x^n [/tex]

    [tex] y\prime = \sum_{n=1}^\infty n C_n x^{n-1}[/tex]

    [tex] y\prime\prime = \sum_{n=2}^\infty n(n-1) C_n x^{n-2}[/tex]

    set 2 from section 6.2

    [tex] y = \sum_{n=0}^\infty n C_n x^{n-1} [/tex]

    [tex] y\prime = \sum_{n=0}^\infty n C_n x^{n-1} [/tex]

    [tex] y\prime\prime = \sum_{n=0}^\infty n(n-1) C_n x^{n-2} [/tex]

    I think I have come to understand that I should use set 1 if and only if all of the singular points are irregular and set 2 when I have at least one regular singular point. Is this correct? If not, when is it appropriate to use set 1 rather than set 2? Or is set 2 only designed to work with Frobenius' method while set 1 only works lacking a Taylor power series expansion?
  2. jcsd
  3. Aug 14, 2008 #2
    In your 2nd set, why are you starting the series from 0? for y' you would get 0 for n = 0 and for y'' you would get 0 for n = 0 and n = 1. It just seems like they are shifting indices around.
  4. Aug 14, 2008 #3
    That's what I thought as well when I first realized that little difference within the DE books I am basing the sets on. Considering that they we're using this method to get something backed by primary source papers I stopped trying to pass it off as a typo within my books. Apparently with the solutions manual I have available to me, both sets are being used, when and why this is the case, I really have no idea. Hopefully somebody does.
  5. Aug 15, 2008 #4


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    Neither of those is correct for Frobenius' method.

    If you have a regular singular point then Frobenius' method uses
    [tex]\sum_{n=0}^\infty C_n x^{n+ c}[/tex]
    where c is determined from the indicial equation, essentially requiring that C0 not be 0. c in not necessarily positive or even an integer.

    If you have an irregular singular point, there may not be a series solution at all.
  6. Aug 16, 2008 #5
    I agree with HallsofIvy.

    If x=0 is an ordinary point then either set 1 or set 2 is a valid expansion. However when using set 2 we must assume that Cn=0 if n is a negative integer (in case we want to replace the dummy index n).
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