Suppose $\hat{p}/\hat{q}$ and $p/q$ are two rational functions where $p,q,\hat{p},\hat{q}$ are of degree $n$. Suppose they satisfy that $|p(z)/q(z) - \hat{p}(z)/\hat{q}(z)| < \epsilon$ for any $z$ on unit circle.

Moreover, we assume that $p$ has all its roots inside unit circle and $q$ has all its roots outside unit circle, and that $C \ge |q(z)| \ge c > 0$, $C \ge |\hat{q}(z)| \ge c > 0$for any $z$ on unit circle.

The question is whether we can prove the following statement: suppose $\hat{p},\hat{q},p,q$ satisfy conditions above, then for any $z$ on unit circle, $|q(z)- \hat{q}(z)| < \epsilon' = \epsilon^{c_1}c^{c_2}C^{c_3}n^{c_4}$ for postive constant $c_1$ and constant $c_2,c_3,c_4$.

(In general I am wondering what conditions on $p,q$ can make the statement true. As you may notice, all the conditions assumed above attempt to rule out the possibility that $p$ cancel some factor with $q$ in some way. The lower bound and upper bound on $q,\hat{q}$ may help us remove the denominator. Other similar assumptions on $p$ and $q$ would be also acceptable. )

(PS: for my problem, it suffices to prove that $\Re(q(z)/\hat{q}(z)) > 0$ for all $z$ on unit circle. I am not sure whether this is really easier).


  • 1
    $\begingroup$ How do these conditions rule out, e.g., the possibility that $\hat{p} = 2 p$ and $\hat{q} = 2 q$? $\endgroup$ – Robert Israel Feb 19 '16 at 21:46
  • $\begingroup$ Ah.. yeah, just assume that $p$ and $\hat{p}$ has leading coefficient 1. This should be rather a minor thing though. $\endgroup$ – Alex Wenxin Xu Feb 21 '16 at 1:36
  • $\begingroup$ Hmmm, this sort of has the flavor of a converse to Runge's theorem: If a sequence of rational functions $\widehat{r_i}$ (in your example these are choices of $\widehat{p}/\widehat{q}$) converges to a meromrophic function $r$ (in your example $p/q$) on a compact set $K$ (in your example the unit circle), and the degree of all of the rational functions $r_i$ are bounded by some number $n$, then the zeros of $r_i$ must converge and the poles of $r_i$ must converge. I will think about it. Have you gotten a solution to the problem you posted yourself yet? $\endgroup$ – Trevor J Richards Mar 22 '17 at 22:00
  • $\begingroup$ Though now I see that you want some control on the speed of convergence, forcing $\epsilon'$ to have a special form. This is a more complicated question than I realized. By the way, I do not see how the condition $Re(q/\widehat{q})>0$ is related to $|q-\widehat{q}|<\epsilon'$. Could you explain that? $\endgroup$ – Trevor J Richards Mar 22 '17 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.