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Let $P(z)=a_nz^n+\cdots+a_0$ 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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I think this is related to the Schur-Cohn criterion –  Cocopuffs Aug 28 '12 at 19:28
It's known as the Eneström–Kakeya theorem. See this question: math.stackexchange.com/questions/185818/enestrom-kakeya-theorem –  Hans Lundmark Aug 28 '12 at 19:40
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1 Answer

up vote 11 down vote accepted

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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The proof looks short enough that you could present it as an answer. External links might be broken some time in the future... –  Fabian Aug 28 '12 at 19:37
How did you get the idea to construct $Q(z)$? –  MJD Aug 28 '12 at 20:17
+1 Very nice proof. –  DonAntonio Aug 29 '12 at 2:52
yea, thanks, the construction of $Q(z)$, makes the condition $a_i$ are monotonic into practice. –  van abel Sep 6 '12 at 8:59
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