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Problem A. 574. (November 2012)

A. 574. Let n\ge2 and let p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0 be a polynomial with real coefficients. Prove that if for some positive integer k the polynomial (x-1)k+1 divides p(x) then \sum_{\ell=0}^{n-1} |a_\ell| >
1+\frac{2k^2}{n}.

CIIM 2012, Guanajuato, Mexico

(5 pont)

Deadline expired on 10 December 2012.


Solution. For convenience, define the leading coefficient an=1 also.

Lemma. For every polynomial q(y) with degree at most k, we have \sum\limits_{\ell=0}^n a_\ell \, q(\ell) = 0.

Proof. Let \varphi0(y)=1 and let \varphi_\nu(y)=y(y-1)\ldots(y-\nu+1) for \nu=1,2,\ldots. By (x-1)k|p(x), for 0\le\nu\lek we have


    \sum_{\ell=0}^n a_\ell \, \varphi_\nu(\ell) = f^{(\nu)}(1) = 0.

Every polynomial with degree at most k can be repesented as a linear combination of the polynomials \varphi_0(y),\ldots,\varphi_k(y), so q(y)=\sum\limits_{\nu=0}^k c_\nu \varphi_\nu(y) with some real numbers c_0,\ldots,c_k. Then


    \sum_{\ell=0}^n a_\ell \, q(\ell)
    = \sum_{\ell=0}^n a_\ell \left(
      \sum_{\nu=0}^k c_\nu \, \varphi_\nu(\ell) \right)
    = \sum_{\nu=0}^k c_\nu \left(
      \sum_{\ell=0}^n a_\ell \, \varphi_\nu(\ell) \right)
    = 0.

To prove the problem statement, let Tk be the kth Chebyshev-polynomial, and choose


  q(y) = T_k\left(\frac2{n-1}y-1\right).

Then q(0),\ldots,q(n-1)\in T_k\big([-1,1]\big)=[-1,1], and


    q(n) = T_k\left(\frac{n+1}{n-1}\right) 
    = \cosh \left( k\cdot \cosh^{-1} \frac{n+1}{n-1} \right)
    = \cosh \left( k\cdot \ln \left(
        \frac{n+1}{n-1} + 
        \sqrt{\left(\frac{n+1}{n-1}\right)^2-1}
      \right) \right)


    = \cosh \left( k\cdot \ln \frac{(\sqrt{n}+1)^2}{n-1} \right)
    = \cosh \left( k\cdot
      \ln \frac{1+\frac1{\sqrt{n}}}{1-\frac1{\sqrt{n}}} \right)
    > \cosh \frac{2k}{\sqrt{n}}.

(In the last step we applied the inequality \ln\frac{1+x}{1-x}>2x.)

By applying the lemma,


  \sum_{\ell=0}^{n-1} |a_\ell|
  \ge \sum_{\ell=0}^{n-1} a_\ell\big(-q(\ell)\big)
  = q(n) > \cosh \frac{2k}{\sqrt{n}} > 1+\frac{2k^2}{n}.


Statistics:

>
3 students sent a solution.
5 points:Ioan Laurentiu Ploscaru.
1 point:2 students.

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