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Math Help - An upper bound for roots of certain polynomials

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    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 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.
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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.
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    Quote Originally Posted by DeMath View Post
    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!
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  4. #4
    Senior Member DeMath's Avatar
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    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.
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    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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