Magyar Information Contest Journal Articles

# 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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