How to use long division to compute the reciprocal $1/Q(s)$ of a generic polynomial?

Question

In these notes (pag 3, pdf alert) the author observes that, given a complex polynomial $Q(s)=q_0 + q_1 s+ ... + q_m s^m$ with $q_0\neq0$, "using the long-division algorithm", we have

$$\frac{1}{Q(s)}=\left(\frac{1}{q_0}-\frac{q_1 s}{q_0^2}\right) + s^2 \frac{R(s)}{Q(s)},\tag1$$ for some polynomial $R$.

I'm aware of what polynomial long division is, but I don't see how to apply it here. Normally, if I were to compute $p/q$, I would find a polynomial that multiplied by $q$ gives the same leading order term as $p$ and so on. But this assumes that the order of $p$ is larger than that of $q$, which is not the case for $1/Q(s)$ if I am to think about it as "$1$ divided by $Q$".

This actually looks closer to a partial fraction decomposition, but then again that would usually involve polynomials in the denominator of both factors in the RHS, which we do not have here.

So where does (1) come from?

Answer 1

We apply long division starting from the constant term, and then the $s$ term:

$$ \require{enclose} \begin{array}{r} \dfrac1{q_0}-\dfrac{q_1}{q_0^2}s+\dots \\[-1pt] q_0+q_1s+\dots \enclose{longdiv}{\hspace{7pt}1+\hspace{7pt}0s+\dots} \\[-1pt] \underline{1+\dfrac{q_1}{q_0}s+\dots} \\[1pt] -\dfrac{q_1}{q_0}s+\dots \\[1pt] \underline{-\dfrac{q_1}{q_0}s+\dots} \\[1pt] \dots & = s^2R(s) \end{array} $$

where "$\dots$" are the $\mathcal O(s^2)$ remainders.

Answer 2

The equation is equivalent to $$Q(s)\left(\frac{1}{q_0} -\frac{q_1}{q_0^2}s\right) = 1 - s^2 R(s).$$

To see this, use the expansion $Q(s)=\sum_{k=0}^n q_k s^k$: $$ Q(s)\left(\frac{1}{q_0} -\frac{q_1}{q_0^2}s\right) = \\ 1 + \sum_{k=1}^n \left( \frac{q_k}{q_0} - \frac{q_1 q_k}{q_0^2} s \right) s^k = 1 + s^2 \underbrace{\sum_{k=2}^n \left[\frac{(q_0 - q_1) q_k}{q_0^2}\right] s^{k-2}}_{\equiv -R(s)}. $$ As pointed out in the comments of question, this process can be understood from a more general perspective as a "division" with respect to a different order of the monomials.

We can also generalise this idea: we can always find a polynomial $P$ of degree $1\le \ell\le\deg Q$ such that $$Q(s)P(s) = 1 + s^\ell R(s)$$ where $R$ is a polynomial $R(0)\neq0$. We can understand this procedure as finding approximate polynomial inverses of $Q$.

A concrete example of this type of division is $$\frac{1}{3+2x+x^2}=\left(\frac13 - \frac29 x\right) + \frac{x^2}{9(3+2x+x^2)},$$ or equivalently, $$(3+2x+x^2)\left(\frac13 - \frac29 x\right) = 1 - \frac{1}{9}x^2.$$ Using a polynomial $P$ of degree $\ell=3$ we would instead get $$(3+2x+x^2)\frac19\left(3-2x + \frac13 x^2\right) = 1 + x^3 R(x).$$

This post is also somewhat related.