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Let $S \subseteq \mathbb{R}$ denote a cofinite subset of $\mathbb{R}$, and suppose $r : S \rightarrow \mathbb{R}$ is a rational function. Suppose $a$ is an element of $S^c$ (i.e. suppose $a$ is a pole of $r$.) Does there necessarily exist a polynomial function $p : \mathbb{R} \rightarrow \mathbb{R}$ such that the following limit exists and equals zero? $$\lim_{x \rightarrow a}\left|r(x)-p\left(\frac{1}{x-a}\right)\right|=0$$

The idea is that $p\left(\frac{1}{x-a}\right)$ ought to work kind of like a vertical asymptote to $r(x)$ at $x=a,$ while providing more information than a mere vertical line about the behaviour of $r$ near this point.

goblin GONE
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2 Answers2

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To complement Robert Israel’s answer, your polynomial $p$ can be computed rather simply. Write $y=\frac{1}{x-a}$, and $r(x)=\frac{U(x)}{V(x)}$ where $U$ and $V$ are two coprime polynomials. Expanding $U(x)=U(a+\frac{1}{y})$ and putting to the same denominator, we find a polynomial $A$ such that $A(0)\neq 0$ and $U(x)=\frac{A(y)}{y^u}$ with $u\in{\mathbb Z}$. Similarly we find a polynomial $B$ such that $A(0)\neq 0$ and $V(x)=\frac{B(y)}{y^v}$ with $v\in{\mathbb Z}$. In the end, $r(x)=y^e\frac{A(y)}{B(y)}$ with $e\in{\mathbb Z},A(0)\neq 0,B(0)\neq 0$. Then :

If $e\leq 0$, perform the Euclidean division of $A(y)$ by $B(y)$ : this yields polynomials $Q,R$ with $A(y)=Q(y)B(y)+R(y)$ and ${\sf deg}(R)<{\sf deg}(B)$. Then $r(x)=\frac{Q(y)}{y^{|e|}}+\frac{R(y)}{y^{|e|}B(y)}$. Write $Q(y)=\sum_{i=0}^d q_i y^i$. Then you may take $p(y)=\sum_{i=|e|}^{d} q_iy^{i-|e|}$ (note that in several cases this sum will be empty so that $p$ will be $0$).

If $e > 0$, perform the Euclidean division of $y^eA(y)$ by $B(y)$ : this yields polynomials $Q,R$ with $y^eA(y)=Q(y)B(y)+R(y)$ and ${\sf deg}(R)<{\sf deg}(B)$. Then $r(x)=Q(y)+\frac{R(y)}{B(y)}$, so that you may take $p=Q$.

Ewan Delanoy
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Yes, Phrases to look up: Laurent series, principal part.

Robert Israel
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