Suppose that

  • $f$ and $g$ are polynomials with nonnegative coefficients,
  • the degree of $g$ is greater than the degree of $f$,
  • $g + f$ have no zeros on the right half plane $\mathbb{C}_+ = \{z \in \mathbb{C}\colon \Re z \ge 0 \}$.

Then $$ h = \frac{f}{g + f} $$ is a proper rational function that is analytic on $\mathbb{C}_+$. Hence, by the partial fraction decomposition, there exists $C > 0$ such that $$ |h^{(n)}(x)| \le \frac{Cn!}{x^{n+1}}, \qquad n \ge 0,\ x > 0.\tag{1} $$

Can we bound $C$ in terms of the coefficients of $f$ and $g$?

Or, in a more general setup:

Can we bound the coefficients of a partial fraction decomposition by a function of coefficients of a rational function?

My goal is to prove $(1)$ for fixed $C$, when coefficients of $f$ change. In particular:

Is it true that there exists $C > 0$ such that for every positive integer $k$ it follows that $$ \left|\left( \frac{kf}{g + kf} \right)^{(n)}(x)\right| \le \frac{Cn!}{x^{n+1}}, \qquad n \ge 0,\ x > 0? $$

This question is in a way a follow-up of Upper bound for the $n$-th derivative of a rational function $\frac{f}{f+g}$.

  • 3
    $\begingroup$ For the last question, the answer is no for the very simple example of $g(z) = z, f(z) = 1$, it is not hard to see that the best possible constant is $C = k$ (and in fact I think I can show that this is a more or less general phenomenon). $\endgroup$ Mar 4 at 17:58
  • $\begingroup$ Aleksei Kulikov: what is $k$ in your comment? $\endgroup$ Mar 5 at 12:46
  • $\begingroup$ @AlexandreEremenko the $k$ from the OP in "... such that for every positive integer $k$ it follows that..." $\endgroup$ Mar 5 at 15:16

1 Answer 1


I will address the second, more general question (the answer implies the answer to the first one). Let us write the rational function as $p/q$, and assume for simplicity that zeros of $q$ are simple, so that $$\frac{p}{q}=\sum_{k=1}^n\frac{c_k}{z-z_k}.$$ We want to estimate the residues $c_k$ in terms of coefficients of $p$ and $q$. We have $$c_k=p(z_k)/q'(z_k).$$ Since $p(z_k)$ is easy to estimate from above in terms of coefficients of $p$, we need an estimate of $q'(z_k)$ from below. To do this, we use the Euclid algorithm on the mutually prime polynomials $q$ and $q'$ and find polynomials $r,s$ such that $$rq+sq'=1.$$ Plugging $z_k$ we obtain $1/q'(z_k)=s(z_k)$, and then we obtain a required estimate.

So the algorithms is as follows. 1. Localize the zeros of $q$. That is find $C$ such that $|z_k|\leq C$ for all $k$. This is easy to do in terms of coefficients. 2. Compute the polynomial $s$ by applying Euclid's algorithm for $q,q'$. 3. Find the upper estimate for $q(z_k)$ which is easy if you know an estimate on coefficients and the estimate $C$.

The assumption of simplicity of roots of $q$ is essential. If you have a denominator with multiple zero, then in any neighborhood of $p/q$ there is a function with arbitrarily large residue.

  • $\begingroup$ Thanks. I hoped finding some useful bound without knowing the poles would be possible. Unfortunately, I am dealing with high-degree polynomials with coefficients depending on a parameter. $\endgroup$
    – xen
    Mar 6 at 13:44
  • 1
    $\begingroup$ When two zeros of denominator come close together, the estimate deteriorates. So there is no estimate of the sort you ask, even when your denominator is quadratic. $\endgroup$ Mar 6 at 13:46
  • $\begingroup$ Or maybe I should try the other way (I know this is a different question): are you aware of a sufficient condition for a function to satisfy such an estimate? Widder's theorem says that $f$ satisfies $|f^{(n)}(x)|\le Mn!/x^{n+1}$ if and only if the inverse Laplace transform of $f$ is bounded, but this is rather difficult to check. Are there "checkable" sufficient conditions? And, does it help if we want $f$ to satisfy $|f^{(n)}(x)|\le Mn!/(x-x_0)^{n+1}$ for some $x_0$? For rational functions this is clear. $\endgroup$
    – xen
    Mar 6 at 14:45

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