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I computed the term $(T_{3})$ in the Chebyshev polynomials on Wolfram Alpha:

http://www.wolframalpha.com/input/?i=integrate%28%284x%5E3-3x%29%2F%281-x%5E2%29%5E%281%2F2%29%2Cx%2C-1%2C1%29

After viewing this integral, I am wondering if each polynomial in the Chebyshev polynomial sequence, $T_{n}$, is orthogonal with respect to the weight function $(1-x^{2})^{-1/2}$. If so, is that the reason why each product of two different Tchebychev polynomials are orthogonal to eachother with respect to this weight function?

CuriousGeorge119
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1 Answers1

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Indeed, Chebyshev polynomials are orthogonal with respect to the $\sqrt{1-x^2}^{-1}.$ The "reason" behind it is that the sequence $\cos{nx},$ $n\ge 1$ is orthogonal on $[0,2\pi].$ More precisely, recall that $T_n(\cos x)=\cos{nx}$ and thus for $m\ne n,$ making change of variables in the integral below leads to $$\int_{-1}^1T_n(x)T_m(x)\frac{1}{\sqrt{1-x^2}}dx=\int_{0}^{\pi}\cos{ny}\cdot\cos{my}dy=0.$$

For more information see wikipedia page devoted to the Chebyshev polynomials: see here

leshik
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  • Thanks, leshik. I see you have used two Chebyshev polynomials and the weight function. In the integral I linked, there is only one Chebyshev polynomial and the weight function: Would this integral bear any significance? – CuriousGeorge119 Oct 03 '13 at 14:09
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    absolutely, the same holds for one integral. Just let $n=0$ in the above and recall that $T_0(x)=1.$ – leshik Oct 03 '13 at 14:11