Problem 796 (difficulty: 10/10)

Let \(\displaystyle p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\) be a polynomial with real coefficients and \(\displaystyle n\ge2\), and suppose that the polynomial \(\displaystyle (x-1)^{k+1}\) divides \(\displaystyle p(x)\) with some positive integer \(\displaystyle k\). Prove that

\(\displaystyle \sum_{\ell=0}^{n-1} |a_\ell| > 1+\frac{2k^2}{n}. \)

CIIM 4, Guanajuato, Mexico, 2012

Solution:

For convenience, define the leading coefficient \(\displaystyle a_n=1\) also.

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

Proof: Let \(\displaystyle \varphi_0(y)=1\) and let \(\displaystyle \varphi_\nu(y)=y(y-1)\ldots(y-\nu+1)\) for \(\displaystyle \nu=1,2,\ldots\). By \(\displaystyle (x-1)^k\big|p(x)\), for \(\displaystyle 0\le\nu\le k\) we have

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

The polynomials \(\displaystyle \varphi_0(y),\ldots,\varphi_k(y)\) form a basis of the vector space of polynomials with degree at most \(\displaystyle k\), so \(\displaystyle q(y)=\sum\limits_{\nu=0}^k c_\nu \varphi_\nu(y)\) with some real numbers \(\displaystyle c_0,\ldots,c_k\). Then

To prove the problem statement, let \(\displaystyle T_k\) be the \(\displaystyle k\)th Chebyshev-polynomial, and choose

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

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

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

\(\displaystyle = \cosh \left( k\cdot \ln \frac{(\sqrt{n}+1)^2}{n-1} \right) = \cosh \left( k\cdot \ln \frac{1+\tfrac1{\sqrt{n}}}{1-\tfrac1{\sqrt{n}}} \right) > \cosh \frac{2k}{\sqrt{n}}. \)

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

By applying the lemma,

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

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