orthogonality of Chebyshev polynomials from recursion

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


with initial conditionsMathworldPlanetmath T0(x)=1 and T1(x)=x. The relation we seek to demonstrate is


when mn.

We start with the observation that Tn is an even function when n is even and an odd function when n is odd. That this is true for T0 and T1 follows immediately from their definitions. When n>1, we may induce this from the recursion. Suppose that Tm(-x)=(-1)mTm(x) when m<n. Then we have

Tn+1(-x) =2(-x)Tn(-x)-Tn-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 Tm(-x)Tn(-x)=-Tm(x)Tn(x), so we have


because the integrand is an odd function of x.

To cover the remaining cases, we shall proceed by inductionMathworldPlanetmath. Assume that Tk is orthogonalPlanetmathPlanetmath to Tm whenever mn and kn and mk. By the conclusionsMathworldPlanetmath of last paragraph, we know that Tn+1 is orthogonal to Tn. Assume then that mn-1. Using the recursion, we have

-1+1𝑑xTm(x)Tn+1(x)1-x2 =2-1+1𝑑xxTm(x)Tn(x)1-x2--1+1𝑑xTm(x)Tn-1(x)1-x2

By our assumptionPlanetmathPlanetmath, each of the three integrals is zero, hence Tn+1 is orthogonal to Tm, so we conclude that Tk is orthogonal to Tm when mn+1 and kn+1 and mk.

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