An upper bound for roots of certain polynomials

**NonCommAlg** · December 31st 2008, 03:21 PM

Happy new year to everybody. Ok, see if you like this problem as much as I do:

Let $p(x)=x^n + a_1x^{n-1} + \cdots +a_{n-1}x + a_n \in \mathbb{C}[x],$ and suppose that $\forall k \leq n: \ \ |a_k| \leq k^2 - \frac{4}{3}k + 1.$ Prove that if $p(\alpha)=0,$ then $|\alpha| < 3.$

Source: Alex Lupas

Note: The problem has a simple generalization.

**DeMath** · February 4th 2009, 07:02 AM

Originally Posted by NonCommAlg

Happy new year to everybody. Ok, see if you like this problem as much as I do:

Let $p(x)=x^n + a_1x^{n-1} + \cdots +a_{n-1}x + a_n \in \mathbb{C}[x],$ and suppose that $\forall k \leq n: \ \ |a_k| \leq k^2 - \frac{4}{3}k + 1.$ Prove that if $p(\alpha)=0,$ then $|\alpha| < 3.$

Source: Alex Lupas

Note: The problem has a simple generalization.

If we make a substitution $t = \frac{1}{x}$ , multiply by $t^n$ , and the estimate for the module, then the problem equivalent of this problem $\sum\limits_{k = 1}^n {\frac{{{k^2} - \frac{4}{3}k + 1}}{{{3^k}}}} < 1$ .

This amount is considered to be easy, and with $n \to \infty$ and it's equal to 1.

I apologize if I don't quite understood the condition of your problem.

**NonCommAlg** · February 7th 2009, 12:51 PM

Originally Posted by DeMath

If we make a substitution $t = \frac{1}{x}$ , multiply by $t^n$ , and the estimate for the module, then the problem equivalent of this problem $\sum\limits_{k = 1}^n {\frac{{{k^2} - \frac{4}{3}k + 1}}{{{3^k}}}} < 1$ .

yes, proving the inequality will solve the problem!

**DeMath** · February 14th 2009, 10:57 AM

Also we can not difficult to prove your problem by methods of complex analysis, using Rouche's theorem.

P.S. I would be interested to see your proof.

**NonCommAlg** · February 14th 2009, 03:11 PM

in general:

let $r > 0$ and $a, b, c \in \mathbb{R}$ such that: $a(r + 1)(r + 2) + br(r + 1) + cr^2 \leq r^3.$ suppose also that $p(x)=x^n + a_1x^{n-1} + \cdots + a_n \in \mathbb{C}[x]$ and: $|a_k| \leq ak^2 + bk + c, \ \forall k.$

then all roots of $p(x)$ lie in the disc $\{z: \ |z| < r + 1 \}.$ to prove this, suppose that $p(\alpha)=0$ and $|\alpha| \geq r+1.$ first note that if $|z| < 1,$ then:

$(*) \ \ \ \sum_{k \geq 1}z^k= \frac{z}{1-z}, \ \ \sum_{k \geq 1}kz^k = \frac{z}{(1-z)^2}, \ \ \sum_{k \geq 1} k^2z^k = \frac{z(z+1)}{(1 - z)^3}.$

now we have: $|\alpha|^n=|a_1\alpha^{n-1} + \cdots + a_n| \leq |a_1||\alpha|^{n-1} + \cdots + |a_n|.$ therefore: $1 \leq \sum_{k=1}^n \frac{|a_k|}{|\alpha|^k}<br /> \leq \sum_{k=1}^n \frac{ak^2 + b k +c}{(r+1)^k} < \sum_{k=1}^{\infty} \frac{ak^2 + b k +c}{(r+1)^k}.$ but by $(*)$ and our assumption:

$\sum_{k=1}^{\infty} \frac{ak^2 + b k +c}{(r+1)^k}=\frac{a(r+1)(r+2) + br(r+1)+cr^2}{r^3} \leq 1.$ contradiction!

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