# Division of Power Series

1. Mar 6, 2005

### spacediablo

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. Mar 6, 2005

### Timbuqtu

By the usual distributive law: $$(\sum_{n=0}^{\infty}b_n z^n) (\sum_{n=0}^{\infty}c_n z^n) = \sum_{n=0}^{\infty}a_n z^n$$
where $$a_n=\sum_{k=0}^{n} b_k c_{n-k}$$.

So $$a_0=b_0 c_0$$ and for n>0: $$b_0 c_n = a_n - \sum_{k=1}^{n}b_k c_{n-k}$$

3. Mar 6, 2005

### spacediablo

I still don't see how that proves
it seems like a restatement of the hypothesis.

4. Mar 7, 2005

### Timbuqtu

It just follows from the way we multiply polynomials:

$$(a_0 + a_1 z + a_2 z^2 + ...)(b_0 + b_1 z + b_2 z^2 + ...) = a_0 b_0 + a_1 b_0 z + a_0 b_1 z + a_0 b_2 z^2 + a_1 b_1 z^2 + a_2 b_0 z^2 + ... = ( a_0 b_0 ) + (a_0 b_1 + a_1 b_0) z + (a_0 b_2 + a_1 b_1 + a_2 b_0) z^2 + ....$$

(Ofcourse you don't know if this power series will converge, but that's a different question)

5. Nov 11, 2007

### DavidSmith

That sounds good

cetin hakimoglu

6. Nov 11, 2007

### mathwonk

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.