text: # 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 Commented Aug 28, 2012 at 19:28
  • It's known as the Eneström–Kakeya theorem. See this question: math.stackexchange.com/questions/185818/enestrom-kakeya-theorem Commented Aug 28, 2012 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. Commented Jan 16, 2015 at 10:04
  
  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
  Commented Aug 28, 2012 at 20:17
  • yea, thanks, the construction of $Q(z)$, makes the condition $a_i$ are monotonic into practice. Commented Sep 6, 2012 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. Commented Apr 26, 2018 at 17:18
  • But how do you prove that $|z|<1$, and not just $|z|\leq1$? Commented Oct 19, 2019 at 22:27
  • @MajaBlumenstein: Compare this recent question: math.stackexchange.com/q/3536695/42969. Commented Feb 6, 2020 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$$.
  
  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 $$ax^2+bx+c,a>b>c>0$$
explanation: The document is a discussion about a mathematical problem, specifically about showing that the roots of a polynomial with descending positive coefficients lie within the unit disc. The main content is a question posed by the user, followed by several comments and a detailed solution provided by another user. The solution involves constructing a new polynomial \( Q(z) \) and using inequalities to show that the roots of the original polynomial \( P(z) \) must lie within the unit disc.
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distilabel_metadata.role: system
distilabel_metadata.content: # 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 Commented Aug 28, 2012 at 19:28
  • It's known as the Eneström–Kakeya theorem. See this question: math.stackexchange.com/questions/185818/enestrom-kakeya-theorem Commented Aug 28, 2012 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. Commented Jan 16, 2015 at 10:04
  
  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
  Commented Aug 28, 2012 at 20:17
  • yea, thanks, the construction of $Q(z)$, makes the condition $a_i$ are monotonic into practice. Commented Sep 6, 2012 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. Commented Apr 26, 2018 at 17:18
  • But how do you prove that $|z|<1$, and not just $|z|\leq1$? Commented Oct 19, 2019 at 22:27
  • @MajaBlumenstein: Compare this recent question: math.stackexchange.com/q/3536695/42969. Commented Feb 6, 2020 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$$.
  
  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 $$ax^2+bx+c,a>b>c>0$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion about a mathematical problem, specifically about showing that the roots of a polynomial with descending positive coefficients lie within the unit disc. The main content is a question posed by the user, followed by several comments and a detailed solution provided by another user. The solution involves constructing a new polynomial \( Q(z) \) and using inequalities to show that the roots of the original polynomial \( P(z) \) must lie within the unit disc.
  
  Label: question-and-answer