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Division of Power Series

  1. Mar 6, 2005 #1
    I'm trying to show that the quotient of two power series Sum(n=o, infinity)[an*z^n] and Sum(n=0, infinity)[bn*z^n] is the power series Sum(n=0, infinity)[cn*z^n] where c0=a0/b0 and b0cn= (an-Sum(k=0, infinity)[bk*c(n-k)]).
    Is there a way of showing this by (Sum[bn*z^n])(Sum[cn*z^n])=Sum[an*z^n] rigorously?
  2. jcsd
  3. Mar 6, 2005 #2
    By the usual distributive law: [tex] (\sum_{n=0}^{\infty}b_n z^n) (\sum_{n=0}^{\infty}c_n z^n) = \sum_{n=0}^{\infty}a_n z^n [/tex]
    where [tex]a_n=\sum_{k=0}^{n} b_k c_{n-k}[/tex].

    So [tex]a_0=b_0 c_0[/tex] and for n>0: [tex]b_0 c_n = a_n - \sum_{k=1}^{n}b_k c_{n-k}[/tex]
  4. Mar 6, 2005 #3
    I still don't see how that proves
    it seems like a restatement of the hypothesis.
  5. Mar 7, 2005 #4
    It just follows from the way we multiply polynomials:

    [tex] (a_0 + a_1 z + a_2 z^2 + ...)(b_0 + b_1 z + b_2 z^2 + ...) = a_0 b_0 + a_1 b_0 z + a_0 b_1 z + a_0 b_2 z^2 + a_1 b_1 z^2 + a_2 b_0 z^2 + ... = ( a_0 b_0 ) + (a_0 b_1 + a_1 b_0) z + (a_0 b_2 + a_1 b_1 + a_2 b_0) z^2 + .... [/tex]

    (Ofcourse you don't know if this power series will converge, but that's a different question)
  6. Nov 11, 2007 #5
    That sounds good

    cetin hakimoglu
  7. Nov 11, 2007 #6


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    if you know how to substitute one power series into another, since you know the reciprocal of 1-x is 1+x + x^2 + x^3+....

    then the reciprocal of any power series with leading term 1, say 1 - u, is 1+u + u^2 + .... where u is a power series with leading term 0.

    a nice treatment of power series is in the first few chapters of henri cartan's book on complex analysis.
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