Mathematical and Physical Journal
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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
|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 =

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


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

Problems in Mathematics of KöMaL, September 2008