Different domains and ranges of the same Chebyshev polynomial?

Question

Suppose I have a Chebyshev polynomial $T_2=\cos{(2\arccos(x))}$. Its range is $[-1,1]$ and its domain is $[-1,1]$.

At the same time: $T_2=\cos{(2\arccos(x))}=2\cos^2{(\arccos(x))}-1=2x^2-1$ which has range of $(-\infty,+\infty)$ and domain of $[-1, +\infty)$.

How is this possible? Why can the same function have different domains and ranges? Probably missing something basic here.

Answer 1

The Chebyshev polynomials are defined for any $x \in \Bbb R$ (or even $x \in \Bbb C$), e.g. via the recurrence relation $$ T_0(x) = 1 \\ T_1(x) = x \\ T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) $$

They satisfy the equation $$ T_n(\cos \theta) = \cos (n \theta) $$ for all $\theta \in \Bbb R$ (and this characterizes the Chebyshev polynomials uniquely).

For $x \in [-1, 1]$ you can set $\theta = \arccos x$ and get $$ T_n(x) = \cos (n \arccos x) \quad \text{ for } {-1} \le x \le 1 $$ which shows that the restriction of $T_n$ to $ [-1, 1]$ maps this interval onto $ [-1, 1]$.