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I have an example from a paper (listed below) that I cannot figure out. I can divide normal polynomials, but the alternative ways to divide Laurent polynomials is beyond me at the moment. The paper lists two polynomials, and three possible quotient/remainder pairs. The polynomials are,

\begin{equation} a(z) = z^{-1} + 6 + z, \quad b(z) = 4 + 4z, \end{equation}

and the three sets of quotients and remainders are,

\begin{equation} q(z) = \frac{1}{4}(z^{-1} + 5) , \quad r(z) = -4z, \end{equation}

\begin{equation} q(z) = \frac{1}{4}(z^{-1} + 1), \quad r(z) = 4, \end{equation}

\begin{equation} q(z) = \frac{1}{4}(5z^{-1} + 1), \quad r(z) = -4z^{-1}. \end{equation}

I understand how to obtain the last quotient and remainder, but I do not understand how the first two were obtained. A worked example of one or the other would be fantastic, thanks. This is from the Daubechies and Sweldens paper on factoring wavelet transforms.

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up vote 2 down vote accepted

The answer depends on the order in which you try to eliminate the terms. If you start with the $z^{-1}$ term, you can get rid of it by $\frac{1}{4}z^{-1}b(z)$ and you are left with

$$ a(z)-\frac{1}{4}z^{-1}b(z) = 5+z $$

For this polynomial you can again chose which one you want to eliminate with a fitting multiple of $b(z)$.

Eliminating the $5$ $$ a(z)-\frac{1}{4}z^{-1}b(z)-\frac{5}{4}b(z) = -4z = r(z) $$

eliminating the $z$ $$ a(z)-\frac{1}{4}z^{-1}b(z)-\frac{1}{4}b(z) = 4 = r(z) $$

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Voila! That does it! Thanks so much! – cjohnson318 Mar 28 '12 at 19:33

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