orthogonality of Chebyshev polynomials from recursion

In this entry, we shall demonstrate the orthogonalityrelation of the Chebyshev polynomials from theirrecursion relation. Recall that this relation reads as

$$T_{n+1}(x)-2x{T}_n(x)+T_{n-1}=0$$

with initial conditions $T_0(x)=1$ and $T_1(x)=x$.The relation we seek to demonstrate is

$${\int }_{-1}^{+1}𝑑x\frac{T_m(x)T_n(x)}{\sqrt{1-x^2}}=0$$

when $m\ne n$.

We start with the observation that $T_n$ is an even functionwhen $n$ is even and an odd function when $n$ is odd. Thatthis is true for $T_0$ and $T_1$ follows immediately from theirdefinitions. When $n>1$, we may induce this from therecursion. Suppose that $T_m(-x)={(-1)}^m{T}_m(x)$ when. Then we have

$T_{n+1}(-x)$$=2(-x)T_n(-x)-T_{n-1}(-x)$

$=-{(-1)}^{n}2x{T}_n(x)-{(-1)}^{n-1}{T}_{n-1}(x)$

$={(-1)}^{n+1}(2x{T}_n(x)-T_{n-1}(x))$

$={(-1)}^{n+1}{T}_{n+1}(x).$

From this observation, we may immediately conclude halfof orthogonality. Suppose that $m$ and $n$ are nonnegativeintegers whose difference is odd. Then $T_m(-x)T_n(-x)=-T_m(x)T_n(x)$, so we have

$${\int }_{-1}^{+1}𝑑x\frac{T_m(x)T_n(x)}{\sqrt{1-x^2}}=0$$

because the integrand is an odd function of $x$.

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 $m\ne k$. By the conclusions of lastparagraph, we know that $T_{n+1}$ is orthogonal to $T_n$.Assume then that $m\le n-1$. Using the recursion, we have

${\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{T_m(x)T_{n+1}(x)}{\sqrt{1-x^2}}}$$=2{\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{x{T}_m(x)T_n(x)}{\sqrt{1-x^2}}}-{\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{T_m(x)T_{n-1}(x)}{\sqrt{1-x^2}}}$

$={\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{T_{m+1}(x)T_n(x)}{\sqrt{1-x^2}}}+{\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{T_{m-1}(x)T_n(x)}{\sqrt{1-x^2}}}-{\displaystyle {\int }_{-1}^{+1}}𝑑x{\displaystyle \frac{T_m(x)T_{n-1}(x)}{\sqrt{1-x^2}}}$

By our assumption, each of the three integrals is zero,hence $T_{n+1}$ is orthogonal to $T_m$, so we concludethat $T_k$ is orthogonal to $T_m$ when $m\le n+1$ and$k\le n+1$ and $m\ne k$.