Question. If $f(z)\in\mathbb{C}[z]$ is a monic polynomial of degree $n$, is it true that $$\max\{\,\vert f(x)\vert: \, -1\leq x\leq 1\}\geq 2^{1-n} \,\, ?$$
Context. This came up while working on some convexity problems.
$\begingroup$
$\endgroup$
asked Feb 15, 2017 at 12:18
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
Yes. Proved by Chebyshev. The extreme case ($\max = 2^{1-n}$) is given by $2^{n-1}\cdot f(x)$ being a Chebyshev polynomial of first kind. https://en.wikipedia.org/wiki/Chebyshev_polynomials
answered Feb 15, 2017 at 12:54
-
-
$\begingroup$ Check the lead section of Wikipedia. $\endgroup$ Feb 15, 2017 at 12:58
-
1$\begingroup$ Cool. Maybe one adds a line of argument to complex polynomials. But, that is okay. $\endgroup$ Feb 15, 2017 at 13:12
-
$\begingroup$ Interpolate your polynomial in the extrema points of $T_n$ and look at the leading coefficient. $\endgroup$ Feb 15, 2017 at 17:58
$\begingroup$
$\endgroup$
You are essentially asking to approximate $x^n$ on $[-1,1]$ by a polynomial of degree less than $n$. The Remez algorithm will find the best possible aproximation. In Maple you can enter
numapprox[minimax](x^n,x=-1..1,[n-1,0]);
If you take $n=10$ or so and work to 100 digits you will quickly convince yourself that the maximum error is precisely $2^{1-n}$. It also looks as though the optimal approximation $f(x)$ has $2^{1-n}f(x)\in\mathbb{Z}[x]$. One could probably find a formula for the coefficients experimentally with the help of OEIS and then work back to a proper proof. But sadly I need to go and do less interesting things now.
answered Feb 15, 2017 at 12:49