Problem A. 654. (November 2015)
A. 654. Let \(\displaystyle p(x)\) be a polynomial of degree at most \(\displaystyle n\) such that \(\displaystyle \bigp(x)\big\le\frac{1}{\sqrt{x}}\) for \(\displaystyle 0<x\le 1\). Prove that \(\displaystyle \bigp(0)\big\le 2n+1\).
(5 pont)
Deadline expired on 10 December 2015.
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