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Problem A. 656. (December 2015)

A. 656. Let \(\displaystyle p(x)=a_0+a_1x+\dots+a_nx^n\) be a polynomial with real coefficients such that \(\displaystyle p(x)\ge0\) for \(\displaystyle x\ge0\). Prove that for every pair of positive numbers \(\displaystyle c\) and \(\displaystyle d\), \(\displaystyle a_0 + a_1(c+d) + a_2(c+d)(c+2d) + \dots + a_n(c+d)(c+2d)\dots(c+nd) \ge0\).

(5 pont)

Deadline expired on 11 January 2016.


Solution. For \(\displaystyle A,B>0\) we have

\(\displaystyle \int_0^\infty x^A e^{-Bx} \,\mathrm{d}x = \frac1{B^{A+1}} \int_0^\infty t^A e^{-t} \,\mathrm{d}t = \frac{\Gamma(A+1)}{B^{A+1}}, \)

so

\(\displaystyle 0 \le \int_0^\infty p(x) \cdot x^A e^{-Bx} \,\mathrm{d}x = \)

\(\displaystyle = \int_0^\infty \left( \sum_{k=0}^n a_kx^k \right) x^A e^{-Bx} \,\mathrm{d}x = \sum_{k=0}^n a_k \int_0^\infty x^{A+k} e^{-Bx} \,\mathrm{d}x = \)

\(\displaystyle = \sum_{k=0}^na_k \frac{\Gamma(A+k+1)}{B^{A+k+1}} = \frac{\Gamma(A+1)}{B^{A+1}} \sum_{k=0}^na_k \frac{(A+1)(A+2)\cdots(A+k)}{B^k} = \)

\(\displaystyle = \frac{\Gamma(A+1)}{B^{A+1}} \sum_{k=0}^na_k \left(\frac{A}{B}+\frac1{B}\right) \left(\frac{A}{B}+\frac2{B}\right) \cdots \left(\frac{A}{B}+\frac{k}{B}\right). \)

By choosing \(\displaystyle A=\frac{c}{d}\) and \(\displaystyle B=\frac1{d}\) we obtain the problem statement.


Statistics:

2 students sent a solution.
5 points:Bukva Balázs.
1 point:1 student.

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