Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a polynomial with real coefficients, that satisfies $\forall x \in \left[ -1,1 \right]: \ |p(x)|\leqslant 1$, I have to show, that its leading coefficient, $a_m$ satisfies $a_m \leqslant 2^{m-1}$.

I have obtained some easy partial results: for example under the given conditions the sum of all even coefficients has to be in $\left[ -1,1 \right]$ (and the sum of the odd in $\left[ -2,2 \right]$), but that didn't help me much...

share|improve this question

2 Answers 2

up vote 2 down vote accepted

This is directly related to Chebyshev polynomials.

share|improve this answer
1  
But isn't there also a more "low-level" proof ? The argument using the Chebyshev polynomials isn't really giving an explanation, why this holds - I mean, it is giving an explanation, but just using some other "high-level" argument. It would interest me to also find a proof that uses just elementary methods. –  resu May 18 '11 at 8:33

This is a property of Chebyshev polynomials. See the section "Minimal $\infty$-norm" there, where there's also a proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.