orthogonality of Chebyshev polynomials from recursion

In this entry, we shall demonstrate the orthogonality relation of the Chebyshev polynomials from their recursion relation. Recall that this relation reads as

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

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 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 . 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 half of orthogonality. Suppose that $m$ and $n$ are nonnegative integers 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 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

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

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

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

Title	orthogonality of Chebyshev polynomials from recursion
Canonical name	OrthogonalityOfChebyshevPolynomialsFromRecursion
Date of creation	2013-03-22 18:54:46
Last modified on	2013-03-22 18:54:46
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 33C45
Classification	msc 33D45
Classification	msc 42C05