<s>[INST] Chebyshev polynomials of first kind I know the chebyshev polynomials of the first kind can be approximated using the cosine function, where $T_n(\cos \theta)=\cos(n \theta)$ and I know that chebyshev polynomials are a family of orthogonal polynomials. How would I prove that $$\int_{-1}^1 \frac{T_h(x)T_i(x)}{\sqrt{1-x^2}}dx=0, h \neq i$$
I can use $x=\cos(\theta)$, but I don't understand how to prove, and what this really means. Isn't it trivial because they're orthogonal?
Can anyone clarify?
Thank you. 
 [/INST] Using Tschebyscheff differential equation in it's autoadjoint form we have, for $T_n$ and $T_m$
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right\} + \frac{n^2}{\sqrt{1-x^2}} T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\} + \frac{m^2}{\sqrt{1-x^2}} T_m &= 0
\end{align}
Multiplying the first equation by $T_m$ and the second by $T_n$ and substracting them
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right \}T_m - \frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\}T_n + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} (T_n'T_m - T_m'T_n)\right\} + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0
\end{align}
Now, the wronskian $W(T_n,T_m)$ is
$$
W(T_n,T_m) = T_n T_m' - T_n' T_m = \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{\sqrt{1-x^2}}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \int_{-1}^1 \frac{d}{d x} \left\{\frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2}\right\}dx
$$
If $n \neq m$, it's easy to see that the integral is zero. If $n \to m$, we have that
$$
\lim_{n \to m} \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2} = \begin{cases}-\frac{\theta}{2}, & m \neq 0, \\ \\ - \theta, & m = 0 \end{cases}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \begin{cases} 0, & m \neq n, \\ \\
\frac{\pi}{2} & m = n \neq 0, \\ \\ \pi & m = n = 0. \end{cases}
$$
</s>