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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
    Commented Aug 28, 2012 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 $\endgroup$ Commented Aug 28, 2012 at 19:40
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    $\begingroup$ @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. $\endgroup$ Commented Jan 16, 2015 at 10:04

3 Answers 3

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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}| &= \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)

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  • $\begingroup$ How did you get the idea to construct $Q(z)$? $\endgroup$
    – MJD
    Commented Aug 28, 2012 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
    Commented Sep 6, 2012 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
    Commented Apr 26, 2018 at 17:18
  • $\begingroup$ But how do you prove that $|z|<1$, and not just $|z|\leq1$? $\endgroup$ Commented Oct 19, 2019 at 22:27
  • $\begingroup$ @MajaBlumenstein: Compare this recent question: math.stackexchange.com/q/3536695/42969. $\endgroup$
    – Martin R
    Commented Feb 6, 2020 at 15:19
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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 $0<a_0\le Ra_1 \le R^2a_2\le ... \le R^na_n$. 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 $|z|\le R<1$.

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A very classical proof

Similar idea to other but slightly different point of view (symetric polynomial):

I let the proof as an exercise :

  1. Considering the geometric series $$Q(x)=\sum_{i=0}^1x^i/a^i,a\in(0,1]$$ show the roots are in the unit circle .

  2. Show that every polynomial :$P(x)$ (degree $n$ with all real roots) with decreasing coefficient and positive can be written as the product of polynomials $Q_i(x)=a_ix+1,a_i>1$ with $C>0$ a positive constant or :

$$P(x)=C\prod_{i=1}^{n}Q_i(x)$$

Hint use fundamental theorem of algebra.

  1. As the propotionality doesn't affect the origin of the roots conclude .

  2. Generalize with $ax^2+bx+c,a>b>c>0$

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