Can You Solve the 1999 Putnam Competition Problem A-5?

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SUMMARY

The discussion centers on the 1999 Putnam Competition Problem A-5, which requires proving the existence of a constant \( C \) such that for any polynomial \( p(x) \) of degree 2017, the inequality \( |p(0)| \leq C \int_{-1}^1 |p(x)|\,dx \) holds. Opalg provided a correct solution to this problem, earning congratulations from the forum, while Kiwi received an honorable mention for their contribution. This problem is part of the Problem of the Week series on the Math Help Boards.

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  • Familiarity with integral calculus, specifically definite integrals
  • Knowledge of the Cauchy-Schwarz inequality in the context of function spaces
  • Experience with mathematical competition problems and proof techniques
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  • Study the Cauchy-Schwarz inequality and its applications in polynomial analysis
  • Explore techniques for bounding polynomial values using integrals
  • Review previous Putnam Competition problems for similar proof strategies
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Mathematics students, competitive problem solvers, and educators interested in advanced polynomial analysis and proof techniques will benefit from this discussion.

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Here is this week's POTW:

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Prove that there is a constant $C$ such that, if $p(x)$ is a polynomial of degree $2017,$ then
\[|p(0)|\leq C \int_{-1}^1 |p(x)|\,dx.\]

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 253 - Feb 16, 2017

This was essentially Problem A-5 in the 1999 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows. Also, honorable mention to Kiwi.

Let $\mathcal P$ be the vector space of all polynomials of degree at most 2017. Then $\mathcal P$ is finite-dimensional (in fact, its dimension is 2018).
For $p(x) \in \mathcal P$, define $$\|p(x)\|_1 = \int_{-1}^1|p(x)|\,dx$$ and $$\|p(x)\|_\infty = \max_{-1\leqslant x \leqslant1}|p(x)|$$. Then $\|\,.\,\|_1$ and $\|\,.\,\|_\infty$ are both norms on $\mathcal P$. But there is a theorem that any two norms on a finite-dimensional vector space are equivalent. In other words, there exists a constant $C$ such that $\|p(x)\|_\infty \leqslant C\|p(x)\|_1$ for all $p(x) \in \mathcal P$. It follows that $|p(0)| \leqslant \|p(x)\|_\infty \leqslant C\|p(x)\|_1$.
 

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