Puzzle... $|a_{2018}|+|a_{2017}|\le 2^{2017}$

Here is the problem :

$f(x)=a_{2018}x^{2018}+a_{2017}x^{2017}+\cdots +a_{1}x+a_{0}$ is a polynomial whose coefficients are all real numbers( $a_{2018}\neq 0$ ).For all $x\in [-1,1]$ we have $|f(x)|\le 1$ .Show that $|a_{2018}|+|a_{2017}|\le 2^{2017}$ .

I've tried Lagrange Interpolation but failed.

I would be very grateful if you could give me some inspiration !

#Algebra

Easy Math Editor

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Comments

I notice that you get equality with the Chebychev polynomials of the first kind, if that helps.

http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html

Jon Haussmann - 2 years, 10 months ago

Thank you so much.

Haosen Chen - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Puzzle... ∣a2018∣+∣a2017∣≤22017|a_{2018}|+|a_{2017}|\le 2^{2017}∣a2018​∣+∣a2017​∣≤22017

Comments

Puzzle... $|a_{2018}|+|a_{2017}|\le 2^{2017}$