# Bound for the $n$-th derivative of a proper rational function with no poles on the right half-plane

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 $$\frac{f}{f+g}$$.

• 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). Mar 4 at 17:58
• Aleksei Kulikov: what is $k$ in your comment? Mar 5 at 12:46
• @AlexandreEremenko the $k$ from the OP in "... such that for every positive integer $k$ it follows that..." Mar 5 at 15:16

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.

• 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.
– xen
Mar 6 at 13:44
• 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. Mar 6 at 13:46
• 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.
– xen
Mar 6 at 14:45