In this entry, we shall demonstrate the orthogonality relation of the Chebyshev polynomials from their recursion relation. Recall that this relation reads as $$ T_{n+1} (x) - 2 x T_n (x) + T_{n-1} = 0 $$ with initial conditions $T_0 (x) = 1$ and $T_1 (x) = x$ . The relation we seek to demonstrate is $$ \int_{-1}^{+1} dx \, {T_m (x) T_n (x) \over \sqrt {1 - x^2}} = 0 $$ when $m \neq n$ .

We start with the observation that $T_n$ is an even function when $n$ is even and an odd function when $n$ is odd. That this is true for $T_0$ and $T_1$ follows immediately from their definitions. When $n > 1$ , we may induce this from the recursion. Suppose that $T_m (-x) = (-1)^m T_m (x)$ when $m < n$ . Then we have

To cover the remaining cases, we shall proceed by induction. Assume that $T_k$ is orthogonal to $T_m$ whenever $m \le n$ and $k \le n$ and $m \neq k$ . By the conclusions of last paragraph, we know that $T_{n+1}$ is orthogonal to $T_n$ . Assume then that $m \le n-1$ . Using the recursion, we have