Problem A. 460. (September 2008)

A. 460. If p(x) is a polynomial with degree less than 2n, show that

$|p(n)| \le 2\sqrt{n} \cdot \max\big( |p(0)|, |p(1)|, \ldots, |p(n-1)|, |p(n+1)|, |p(n+2)|, \ldots, |p(2n)| \big).$

(5 pont)

Deadline expired on November 17, 2008.

Solution (outline). As is well-known

$\sum_{k=0}^m (-1)^k \binom{m}{k} p(k) = 0$

for all polynomials p with degree less than m. We will apply this for m=2n.

Moreover, it can be proved easily that $\binom{2n}{n}\ge\frac{2^{2n-1}}{\sqrt{n}}$ .

Let $M=\max\big(|p(0)|, |p(1)|, \ldots, |p(n-1)|, |p(n+1)|, |p(n+2)|, \ldots, |p(2n)| \big)$ . Then

$\binom{2n}{n}|p(n)| = \left| \sum_{0\le k\le2n,~k\ne n} (-1)^k \binom{2n}{k} p(k) \right| \le \sum_{0\le k\le2n,~k\ne n} \binom{2n}{k} |p(k)| \le$

$\le \left(\sum_{0\le k\le2n,~k\ne n} \binom{2n}{k}\right) M = \left(2^{2n}-\binom{2n}{n}\right)M,$

$|p(n)| \le \frac{2^{2n}-\binom{2n}{n}}{\binom{2n}{n}} M= \left(\frac{2^{2n}}{\binom{2n}{n}}-1\right) M \le \left(2\sqrt{n}-1\right)M.$

7 students sent a solution.

5 points: Backhausz Tibor, Bodor Bertalan, Nagy 235 János, Nagy 314 Dániel, Tomon István.

0 point: 2 students.

7 students sent a solution.
5 points:	Backhausz Tibor, Bodor Bertalan, Nagy 235 János, Nagy 314 Dániel, Tomon István.
0 point:	2 students.

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