# Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

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?

• I think this is related to the Schur-Cohn criterion 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 Aug 28 '12 at 19:40
• @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. Jan 16 '15 at 10:04

## 2 Answers

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}| &= \left|a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i\right| \\ & \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)

• How did you get the idea to construct $Q(z)$?
– MJD
Aug 28 '12 at 20:17
• yea, thanks, the construction of $Q(z)$, makes the condition $a_i$ are monotonic into practice. Sep 6 '12 at 8:59
• 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. Apr 26 '18 at 17:18
• But how do you prove that $|z|<1$, and not just $|z|\leq1$? Oct 19 '19 at 22:27
• @MajaBlumenstein: Compare this recent question: math.stackexchange.com/q/3536695/42969. Feb 6 '20 at 15:19

The previous answer establishes that roots lie inside or on the unit circle. By a slight modification to $$Q(z)=(1-z)P(Rz)$$ this argument remains true as long as $$0. Let $$R$$ be minimal with this property, then $$R<1$$. Now if $$z$$ is any root of $$P$$, then $$z/R$$ is a root of $$Q$$ and thus in the unit disk. Thus $$|z|\le R<1$$.