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Can someone help direct me towards information on calculating the polynomial remainder?

I know how to do the basics, but what if I don't have access to the polynomial in expanded form?

For example, $\frac{1}{1-x}$ represents the polynomial $\sum_{i=0}^\infty{x^i}$. I want to be able to find the remainder using an expression like the fraction.

Please help explain where I can find more resources on this. I'm not a math expert, so something very readable would be great. However, I'm willing to learn the math if I need to.

Could someone help me find resources with the equivalent and generating functions?

Just to show that this operation is well defined, Dr. Herwig Hauser has written a paper concerning the operation here

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$\frac{1}{1-x}$ is a power series not a polynomial. You can still perform division with these but there is no remainder, to derive the 'formula' for division just invert multiplication (look at the diagonals of the infinite product). –  anon Oct 19 '10 at 17:22
    
I was able to get a remainder in Mathematica using their PolynoialRemainder method with two generating functions as the polynomials. I'm interested in this result. –  Matt Groff Oct 19 '10 at 17:32
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This is not the right site to ask questions about Mathematica. –  anon Oct 19 '10 at 17:45
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Please pardon my comment. I'm simply trying to state that I believe the operation is well defined. I'm adding a link to a paper detailing this above. –  Matt Groff Oct 19 '10 at 18:18

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Perhaps what you're after is on page 31 in Wilf's generatingfunctionology:

Proposition. A formal power series $f=\sum_{n \geq 0} a_n x^n$ has a reciprocal if and only if $a_0 \neq 0$. In that case the reciprocal is unique.

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That may help. You have me wondering how to calculate the reciprocal. Unfortunately, I'm working with a "power series" where the powers range from $-n$ to $n$; it may be easier to think of it as a "special" Laurent series. I'm exploring methods to find the constant term. I've heard that there's an easy way to extract the residue for the Laurent series. Unfortunately, I remember that attempting to modify that method failed for some reason. –  Matt Groff Nov 19 '10 at 16:48

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