# Bound on difference of roots between two “close” polynomials

Let $$p(z) = z^n + a_{n-1}z^{n-1}+\dots+a_0$$ and $$q(z) = z^n + b_{n-1}z^{n-1}+\dots+b_0$$ be two complex monic polynomials with $$|a_i - b_i| < \epsilon < 1$$. Furthermore, assume that all the coefficients of both polynomials are bounded in modulus by some fixed number $$K$$. Then I believe that the Hausdorff distance between the roots of $$p$$ and $$q$$ considered as sets in the complex plane is atmost $$C(K,n)\times \epsilon^{1/n}$$ where $$C(K,n)$$ is a constant that depends on $$K$$ and $$n$$. Is this true?

• You might look at Wilkinson's polynomial where a tiny change in one coefficient makes a large change in the roots. It is not quite a counterexample because the coefficients are large, but it is enlightening. – Ross Millikan Nov 12 '18 at 6:23
• @RossMillikan Thanks! This is interesting. From what I understand, it is still possible that my claim above is still correct in spite of this example? – Jaikrishnan Nov 12 '18 at 6:34
• Yes, if $a$ is an $m$-fold root of $p$, $p(z)=(z-a)^mp_1(z)$, then a perturbation $p+ϵr$ has in first order roots close to $a$ at $0=(z-a)^mp_1(a)+ϵr(a)$, $z=a+q_m^k\sqrt[m]{ϵr(a)/p_1(a)}$ with $q_m$ the $m$-th unit root and $p_1(a)=p^{(m)}(a)/m!$. The number $C(K,n)$ has then to be a bound $|m!r(a)/p^{(m)}(a)|^{1/m}$ for all possible $m$, plus a little more to account for the higher order terms. – LutzL Nov 12 '18 at 11:29