# orthogonal chebyshev polyhomials

This Theorem says

statement: $$\int_{-1}^{1} \frac{T_n(x)T_m(x)}{ \sqrt{1+x^2} } dx = 0 ;$$ when $$n\ne m$$

proof: "substitute $$x= cos \theta$$ " and that's it.

$$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+x^2}} dx$$ or with this

$$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+cos^2 \theta}} dx$$ I order to verify the proof.

note: T(x) is Chebyshev polynomial.

When you do this, you apply the substitution rule for integrals, fully. Everything with $$x$$ in it, including the denominator, the $$dx$$, and the limits, transform.
On the other hand, the statement that you're trying to prove is incorrect. The correct form has $$\sqrt{1-x^2}$$ in the denominator instead.
• $\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+cos^2 \theta}} dx = \int_{0}^{\pi} cos(\theta m). cos(\theta n) d\theta$ I wonder how this come about when we change $x$ to $cos(\theta)$ where does the denominator go. – tt z Mar 10 '19 at 4:42
• It goes into the $d\theta$. Well, it does if you use the correct version with a minus sign in that square root. – jmerry Mar 10 '19 at 4:44