= Solution
Divide the differential equation by $\sqrt{1-x^2}$. Its <Sturm-Liouville form> is
$$
\boxed{\frac d{dx}\left(\sqrt{1-x^2}\,y'\right)
+\frac{n^2}{\sqrt{1-x^2}}y=0.}
$$
Here $p(x)=\sqrt{1-x^2}$ and weight $w(x)=(1-x^2)^{-1/2}$. Multiply the equations for $T_n,T_m$ by the other polynomial and subtract. <Integration by parts> gives a boundary term $p(T_mT_n'-T_nT_m')$, which vanishes at both endpoints because polynomial derivatives are finite and $p\to0$. Hence
$$
(n^2-m^2)\int_{-1}^1\frac{T_nT_m}{\sqrt{1-x^2}}dx=0.
$$
With the standard nonnegative degree indices $n,m$, this proves \b[orthogonality for $n\ne m$]. If signed integer indices are used literally, the condition is $|n|\ne|m|$: $T_{-n}=T_n$, so merely $n\ne m$ would not suffice.
Set $x=\cos\theta$ and $Y(\theta)=y(\cos\theta)$. The chain rule gives $(1-x^2)y''-xy'=Y''$, so $Y''+n^2Y=0$. For $n>0$, $Y=A\cos(n\theta)+B\sin(n\theta)$. A polynomial $y$ has $Y'(0)=0$, forcing $B=0$. At $n=0$, $Y=A+B\theta$ and the same endpoint condition gives $B=0$. Therefore every polynomial solution is proportional to
$$
\boxed{T_n(x)=\cos(n\arccos x).}
$$
The cosine recurrence yields $T_{n+1}=2xT_n-T_{n-1}$, $T_0=1,T_1=x$, confirming that these functions are polynomials. In this normalization the <Chebyshev polynomial> weight identity is
$$
\int_{-1}^1\frac{T_nT_m}{\sqrt{1-x^2}}dx=
\begin{cases}0,&n\ne m,\\\pi,&n=m=0,\\\pi/2,&n=m>0.\end{cases}
$$
It is simply cosine orthogonality after $x=\cos\theta$.
Back to article page
