# Chebyshev Polynomials

Given $$-\frac{1}2<a<\alpha<0<\beta<b<+\frac{1}2$$ $$+\frac{1}2<c<\gamma<1<\delta<d<+\frac{3}2$$ I want to find a polynomial $f(x)\in\Bbb R[x]$ such that $f([a,b])\subseteq[\alpha,\beta]$, $f([c,d])\subseteq[\gamma,\delta]$. What is minimum degree polynomial that is needed and maximum degree that will suffice?

I believe Chebyshev polynomials play a role here. What is degree as function of $a,b,c,d,\alpha,\beta,\gamma,\delta$ that is necessary and sufficient using Chebyshev polynomials?

Is there an explicit formula (not a computational solution) that gives a good enough answer within constants?

• Is that second $-1/2$ supposed to be $+1/2$? And the $0$ supposed to be $1$? – Robert Israel Jun 3 '15 at 21:39
• Still not quite right: $1/2 < \ldots < 0$? – Robert Israel Jun 3 '15 at 21:43
• corrected both $+$, $1$ values. – 1.. Jun 3 '15 at 21:44

We obtained asymptotic estimates (when the degree is large, $\beta-\alpha$ and $\delta-\gamma$ small. But it is clear from these papers how to obtain minimal degree for given $a,b,\alpha,\beta,c,d,\gamma,\delta$ in the cases when this degree is not large, using a computer. We have pictures of some extremal polynomials, and some representation of them.
• Thank you could you summarize the degree dependency on $a,\dots,\delta$ for completeness? – 1.. Jun 4 '15 at 9:19