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!
Best Answer
It is straightforward to to see that the leading coefficient of $T_1$ is $2^0$, and the degree of $T_1$ is $1$.
So suppose the degree of $T_n$ is $n$, and the leading coefficient of $T_n$ is $2^{n-1}$. Then since $T_{n+1}(x) = 2 x T_n(x) - T_{n-1}(x)$, it is clear that the degree of $T_{n+1}$ is $n+1$ and the leading coefficient is $2 . 2^{n-1} = 2^{n}$. Hence it is true for all $n$.
Clarification: To see why $T_{n+1}$ has degree $n+1$, suppose for $k < n+1$ that $T_k$ has degree $k$, and leading coefficient $2^{k-1}$. Then $T_k$ will have the form $T_k(x) = 2^{k-1} x^k + \cdots$, where the $\cdots$ means a polynomial of degree $< k$. Then $T_{n+1}(x) = 2 x T_n(x) - T_{n-1}(x) = 2 x (2^{n-1} x^n + \cdots) - (2^{n-1} x^{n-1} + \cdots) = 2^{n} x^{n+1} + \cdots$. Hence $T_{n+1}$ has degree $n+1$ and leading coefficient $2^n$.