# Thread: An upper bound for roots of certain polynomials

1. ## An upper bound for roots of certain polynomials

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

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

Source: Alex Lupas

Note: The problem has a simple generalization.

2. Originally Posted by NonCommAlg
Happy new year to everybody. Ok, see if you like this problem as much as I do:

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

Source: Alex Lupas

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

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

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

3. Originally Posted by DeMath
If we make a substitution $\displaystyle t = \frac{1}{x}$, multiply by $\displaystyle t^n$, and the estimate for the module, then the problem equivalent of this problem $\displaystyle \sum\limits_{k = 1}^n {\frac{{{k^2} - \frac{4}{3}k + 1}}{{{3^k}}}} < 1$.
yes, proving the inequality will solve the problem!

4. 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.

5. in general:

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

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

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

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

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