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Recursion formula for power series solution

  1. Jan 30, 2005 #1
    Hi, I'm trying to solve a differential equation and I'm supposed to obtain a recursion formula for the coefficients of the power series solution of the following equation:

    w'' + (1/(1+z^2)) w = 0.

    The term 1/(1+z^2) I recognize as a geometric series and can be expressed as sum of 0 to infinity of: (-z^2)^n.

    But I'm having trouble multiplying it with w, which is also a power series. And also what is the radius of convergence for general initial conditions?

    Thanks.
     
  2. jcsd
  3. Jan 30, 2005 #2

    dextercioby

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    For the second part,well i'm not really keen on mathematical subtleties...The series definitely have to converge and then u'll find the condition from the general recurrence relation.

    As for the first,i cannot really understand your difficulties.U have to multiply 2 sums (never mind the sigma symbol and the infinite no.of terms).Can't u do it??

    And then the recurrence relation is fairly easy to find...

    Daniel.
     
  4. Jan 30, 2005 #3
    I'm just a little rusty on multiplication of series, and especially in this case, you have different powers of z. For the first series: z to the power 2n, and the second series, z to the power n. I know the formula when the powers of z are the same. Would you mind showing me how it's done for this case?
     
  5. Jan 30, 2005 #4

    dextercioby

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    What is the following product:
    [tex] (Ax^{n})(Bx^{m}) [/tex]

    If u know that,it's more than enough...

    Daniel.
     
  6. Jan 30, 2005 #5
    Ok. ABx^(m+n). Thanks.
     
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