Is there a name for the remainder of long polynomial division in sigma notation?

I was dividing polynomials and forgot to stop when the remainder's degree was lower than that of the dividend's. I had never done this before, so I continued and a very predictable sequence emerged. I thought about this and realized that since repeating decimals can be written in sigma notation, then it should also work with polynomials. And indeed it does. It seems to me that this could be used to study particularly interesting sequences of numbers (like the harmonic sequence). I tried searching for this, but since I do not know the name for something like this, I found no answers to my burning questions. What is the proper name for this? I would also appreciate if anyone could provide any references to websites or books where I could learn more about this. Thank you.

• How were you dividing, synthetic division? Could you give an illustrative example? – J. M. is a poor mathematician May 1 '11 at 22:09
• How did you continue? Did you allow rational functions as the quotient? – Arturo Magidin May 1 '11 at 22:10
• @Arturo: I suspect he let some loop run too long... – J. M. is a poor mathematician May 1 '11 at 22:11
• @J.M. Like you, I'd like to see an example to figure out what the OP is refering to. – Arturo Magidin May 1 '11 at 22:16
• @Arturo: I did allow rational functions as the quotient. – Hautdesert May 2 '11 at 1:49

I think what you may be talking about is a Laurent expansion: if $p(x)$ and $q(x)$ are polynomials of degrees $m$ and $n$ respectively, $m \ge 1$, you can write $p(x)/q(x) = \sum_{k=-\infty}^{m-n} a_k x^k$ where $a_j$ are constants, converging for $|x|$ greater than the absolute values of all zeros of $q(x)$. This is essentially the same as the Maclaurin series of the quotient of the reverse polynomials: if $P(z) = z^m p(1/z)$ and $Q(z) = z^n q(1/z)$, then $P(z)/Q(z) = z^{m-n} p(1/z)/q(1/z) = \sum_{j=0}^\infty a_{m-n-j} z^j$.
For example, $\frac{1}{x+2} = \frac{1}{x} - \frac{2}{x^2} + \frac{4}{x^3} - \ldots$ corresponding to $\frac{1}{2 z + 1} = 1 - 2 z + 4 z^2 - \ldots$.