Do approximately the same polynomials have approximately the same roots? [closed]

"If $U$ is an open subset of the complex plane, then matrices $X\in\textrm{M}(n,\mathbb C)$ all of whose eigenvalues belong to $U$ make up an open subset of $\textrm{M}(n,\mathbb C)$." Trying to prove this by using the argument principle, I was led to the following question: do approximately the same polynomials have approximately the same roots? More precisely, fix a complex polynomial $p(z)=a_0+a_1 z+\ldots+a_n z^n$ with $a_n\neq 0$. Let us say that a polynomial $q(z)=b_0+b_1 z+\ldots+b_n z^n$ is in the $\varepsilon$-neighborhood of $p(z)$ if $|b_i-a_i|<\varepsilon$ for all $0\leq i\leq n$. Now suppose all roots of $p(z)$ lie in, say, the open unit disk. Does $p(z)$ have an $\varepsilon$-neighborhood consisting of polynomials with the same property?

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closed as too localized by quid, Dmitri Pavlov, Chris Godsil, Qiaochu Yuan, Felipe VolochAug 4 '12 at 18:53

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Yes, the roots of a polynomial depend continously on the coefficients; see en.wikipedia.org/wiki/Properties_of_polynomial_roots . This is better asked on math.stackexchange.com IMO. – user9072 Aug 3 '12 at 15:39

I think so, yes. Suppose all the roots lie inside the unit disc. By compactness $|p(z)|$ takes a positive minimum $\delta$ on the contour (the unit circle). For $q(z)$ in the epsilon-neighbourhood, $q(z)-p(z)$ is a polynomial with epsilon coefficients, so we can take epsilon small enough so that on the contour $|q(z)-p(z)|<\delta<|p(z)|$. Then appeal to Rouché's theorem to learn that $p(z)$ and $q(z)$ have the same number of zeros in the region.

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How is this related to the question? – Qfwfq Aug 3 '12 at 20:47
It simply solves the more-precisely-formulated question above, which generalizes at once to give a proof of the proposition in quotation marks. – Murat Güngör Aug 4 '12 at 11:18

The question about matrices is analyzed to death in Kato's "perturbation theory of linear operators", chapters 1 and 2. The question in the title than is answered (in the affirmative by associating to each polynomial its companion matrix. Note that the roots depend analytically on the coefficients when they are all distinct, but only $C^0$ when there are multiple roots.

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For the headline question, no. See, for example, Wilkinson's polynomial: http://en.wikipedia.org/wiki/Wilkinson's_polynomial.

Edit: From the response to this answer, it seems that some expansion is needed. The OP asks two questions. In the the title, and repeated in the text, the question is: Do approximately the same polynomials have approximately the same roots? The answer to this question is no, as Wilkinson shows with his monic degree-20 polynomial, which has well separated roots of moderate size, namely the integers 1 to 20. A decrement of $2^{-23}$ in the coefficient of the degree-19 term (which is $-210$)---a proportional change of less than 1 in 1.7 billion---induces substantial changes in the roots: for example, the root pair $16.5\pm 0.5$ becomes the root pair $16.73024\pm 2.81262\mathrm i$ (to 5 d.p.). The second question asked by the OP is whether the coefficients-to-roots map is continuous, which was already answered (affirmatively) by others.

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This is a beautiful example, but perturbing the coefficients still perturbs the roots. It's just that the derivative can be far larger than one would naively expect. – Henry Cohn Aug 3 '12 at 16:30
The OP does not ask two questions, the same question is phrased formally and informally. – user9072 Aug 5 '12 at 12:17