Consider the Chebyshev polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by $T_0(x) = 1; \quad T_1(x) = x; \quad T_n(x) = 2x\cdot T_{n−1}(x) − T_{n−2}(x)$ for $n = 2, 3, \ldots$. I need to show that
$T_n(x) = 2^{n−1}\cdot (x − x_0)(x − x_1) . . . (x − x_{n−1})$
I think I need to do this by induction, but am not entirely sure. Does anyone have some insight into this that would help me work through it?
I know I start with the $n = 1$ case, in which case, $T_n(x) = 2^0\cdot(x-x_0)$ (I think)
Then I am guessing I need to assume it works for $n$ cases, and prove for $n+1$?
Thanks!