Let $p$ be a polynomial of minimal degree to which the following is true:

$p(0) > \sum\limits_{i=1}^n \lvert p(i) \rvert + \sum\limits_{i=1}^m \lvert p(-i) \rvert$

Give upper and lower bounds for $deg(p)$ (for sufficently large $n$).

I constructed bounds for $p(0) > \sum\limits_{i=1}^n \lvert p(i) \rvert$ using Chebyshev-polynomials, which is $c_1\sqrt{n} < deg(p) < c_2\sqrt{n}$, where $c_1$ and $c_2$ are constants. However, I'm having trouble with both sums.

  • $\begingroup$ Is there anything missing from the question? It is possible to get polynomials of arbitrarily high degree with $p(0)=1$ and $p(\pm i)=0$ for $|i|\leq\max\{m,n\}$. So there cannot be an upper bound. $\endgroup$ – Martin Argerami Mar 20 '14 at 1:44
  • $\begingroup$ @MartinArgerami Sorry, you're right. Fixed. $\endgroup$ – mathaway__ Mar 20 '14 at 4:30

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