Let $p(x)$ be a polynomial with complex coefficients and $p(\tilde x)=0$. Choose $\delta>0$ small enough, such that $\tilde x$ is the only root of $p$ in $B_\delta(\tilde x)$. I want to show that there exists a constant $c$ such that for all $x\in\partial B_{\frac{\delta}{\lambda}}(\tilde x)$ with $\lambda\geq 1$ there holds

$$\vert p(x)\vert\geq c\lambda^{-1}$$.

I'm thankful for every kind of help :)

  • $\begingroup$ Presumably you mean $|p(x)| \ge c \lambda^{-1}$. $\endgroup$ – Robert Israel Apr 22 '16 at 18:44
  • $\begingroup$ yes,! sorry for that $\endgroup$ – lbf_1994 Apr 22 '16 at 18:45

It's not true. Try $p(x) = x^2$. Or do you mean that $\overline{x}$ is a simple root?

EDIT: With the assumption that $\overline{x}$ is a simple root, $$0 \ne p'(\overline{x}) = \lim_{x \to \overline{x}} \dfrac{p(x)}{x-\overline{x}}$$ Take $c = |p'(\overline{x})|/2$. There is some $\eta > 0$ such that for $|x - \overline{x}| \le \eta$ we have $$\left| \dfrac{p(x)}{x-\overline{x}} - p'(\overline{x}) \right| < c$$ and thus $|p(x)| \ge c |x - \overline{x}|$. You want to take $1/\lambda \le \eta$.
(I presume you meant "... for some $\lambda$ ...". It certainly can't work for all $\lambda \ge 1$: e.g. you might have another zero at distance $1$ from $\overline{x}$.

  • $\begingroup$ I guess 'the only root' indicates it should be simple... $\endgroup$ – Thomas Apr 22 '16 at 18:47
  • $\begingroup$ yes it should be a simple root. $\endgroup$ – lbf_1994 Apr 22 '16 at 18:50

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.