- #1

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Is there a way of showing this by (Sum[bn*z^n])(Sum[cn*z^n])=Sum[an*z^n] rigorously?

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- Thread starter spacediablo
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- #1

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Is there a way of showing this by (Sum[bn*z^n])(Sum[cn*z^n])=Sum[an*z^n] rigorously?

- #2

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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]

- #3

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it seems like a restatement of the hypothesis.spacediablo said: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)]).

- #4

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[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)

- #5

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That sounds good

cetin hakimoglu

cetin hakimoglu

- #6

mathwonk

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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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