Let's say I have a ratio of polynomials as follows


The polynomials are finite. Is there a procedure to convert it into a polynomial

$P(x) = A_0 + A_1 f(x) + A_2g(x) + ...$

where $f(x)$ and $g(x)$ are some power functions of x. Maybe there is a way to expand it in series if it is not divisible...


2 Answers 2


Let's re-index the polynomials, writing $$ a(x) = \sum_{j=0}^{n} a_{j} x^{j},\qquad b(x) = \sum_{j=0}^{m} b_{j} x^{j},\qquad P(x) = \frac{a(x)}{b(x)}. $$ (That is, let the coefficient index match the power of $x$; start with the constant terms.)

  1. If $b_{0} = b(0) \neq 0$, the rational function $P$ is analytic in a neighborhood of $0$. Put $c_{j} = b_{j}/b_{0}$, and write $$ b(x) = b_{0}\Bigl(1 + \sum_{j=1}^{m} c_{j} x^{j}\Bigr) = b_{0}\bigl(1 + c(x)\bigr). $$ In an open neighborhood of $0$ where $|c(x)| < 1$, the reciprocal may be expanded as a geometric series: $$ \frac{1}{b(x)} = \frac{1}{b_{0}\bigl(1 + c(x)\bigr)} = \frac{1}{b_{0}} \sum_{k=0}^{\infty} \bigl(-c(x)\bigr)^{k}, $$ or $$ P(x) = \frac{a(x)}{b_{0}} \sum_{k=0}^{\infty} \bigl(-c(x)\bigr)^{k}. $$ Since $x$ divides $c(x)$, the series on the right may be expanded explicitly to any desired finite order $\ell$ by adding up the first $\ell$ terms, a purely algebraic process (no infinite series).

  2. If $b(0) = 0$ but $b$ is not identically zero, $P(x)$ has a pole at $0$. Factor out the largest power of $x$ dividing $b(x)$, $$ b(x) = \sum_{k=\ell}^{m} b_{k} x^{k} = x^{\ell} \sum_{k=0}^{m-\ell} b_{k+\ell} x^{k} = x^{\ell} \bar{b}(x),\quad b_{\ell} = \bar{b}(0) \neq 0, $$ then proceed as in 1. (with the factor $x^{\ell}$ in the denominator): $$ P(x) = \frac{1}{x^{\ell}} \frac{a(x)}{\bar{b}(x)}. $$ This isn't a power series, of course, but a Laurent series.


This is not possible. Think of $P(x)=1/x$. This does not have a Taylor expansion, so even your second questions, the series expansion is not possible.


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