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Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0<a_0<a_1<\cdots<a_n.$$

I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).

Any ideas?

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  • $\begingroup$ I think this is related to the Schur-Cohn criterion $\endgroup$ – Cocopuffs Aug 28 '12 at 19:28
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    $\begingroup$ It's known as the Eneström–Kakeya theorem. See this question: math.stackexchange.com/questions/185818/enestrom-kakeya-theorem $\endgroup$ – Hans Lundmark Aug 28 '12 at 19:40
  • $\begingroup$ @HansLundmark: That (marginally) older question is now (curiously) closed as a duplicate of this one. This comment is just to dissuade people who like me want to mark this question as a duplicate of that one. $\endgroup$ – Marc van Leeuwen Jan 16 '15 at 10:04
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The thing to do is to look instead at the polynomial $$Q(z) = (1-z)P(z) = (1-z)\left(\sum_{i=0}^n a_iz^i \right) = a_0 -a_n z^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})z^i$$ Now, let $|z|>1$ be a root of $P(z)$, and hence a root of $Q(z)$. Therefore, we have $a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i = a_n z^{n+1}$ Then, we have \begin{aligned} |a_n z^{n+1}| &= |a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|z^i| \\ & < a_0|z^n| + \sum_{i=1}^n (a_i-a_{i-1})|z^n| \\ & = |a_n z^n|\end{aligned} a contradiction.

For a nice article on integer polynomials, see here. (Your problem is Proposition 10)

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  • $\begingroup$ How did you get the idea to construct $Q(z)$? $\endgroup$ – MJD Aug 28 '12 at 20:17
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    $\begingroup$ yea, thanks, the construction of $Q(z)$, makes the condition $a_i$ are monotonic into practice. $\endgroup$ – van abel Sep 6 '12 at 8:59
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    $\begingroup$ This shows that the zeros lie in the closed unit disk, but I think the question is to show that the zeros lie in the open unit disk. $\endgroup$ – Alex Ortiz Apr 26 '18 at 17:18
  • $\begingroup$ But how do you prove that $|z|<1$, and not just $|z|\leq1$? $\endgroup$ – Maja Blumenstein Oct 19 '19 at 22:27
  • $\begingroup$ @MajaBlumenstein: Compare this recent question: math.stackexchange.com/q/3536695/42969. $\endgroup$ – Martin R Feb 6 at 15:19

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