So I have a Chebyshev polynomial
where I am trying to prove that
$T_n(t)=\cos{(n \space \arccos{t})}, \space n=0,1,2...$ to form a system of orthogonal polynomials under the weighted inner product $\lt f,g\gt=\int_{-1}^1 \frac{f(t)g(t)dt}{\sqrt{1-t^2}}$ . and then finally show $||T_n||$ and write out the formulae for $T_{0}(t),...,T_{6}(t)$
My attempt is this:
I know that in order to prove orthogonality we have to use the change of variables $t=\cos \theta$ in the inner product integral such that $dt=-\sin ( \theta \space d \theta) $ such that $d \theta=\frac{dt}{\sqrt{1-t^2}}$ Where I can derive the inner product and write out the formulae for
$\lt T_m,T_n \gt = \int_{-1}^1 \frac{\cos{ (m \space \arccos \space t)} \cos{(n \space \arccos \space t})}{\sqrt{1-t^2}}=\int _{0}^{\pi} \cos (\space m \theta \space) \cos{(n \theta )}\space d \theta$
I know that orthogonal space has to equal zero so my results of the norm should involve $m=n=0, m=n>0,$ and $n\neq m $ but having trouble writing this as proof.
Ive seen similar posts that did the algebra or proving a genalized form but I am more interested in explaining why this works in a generalized case.
Thoughts and feedback are greatly appreciated !
You are almost there, now consider for various $n$ and $m$ the value of your integral.
in the case where $n$ and $m$ both equal zero
$\int_0^\pi cos(n\theta)cos(m\theta) d\theta = \int_0^\pi d\theta = \pi$
but in the case where $n,m \neq 0$ , remember
$cos(n\theta)cos(m\theta) = \frac{1}{2}(cos((n+m)\theta) + cos((n-m)\theta))$
so that
$\int_0^\pi cos(n\theta)cos(m\theta) d\theta = \frac{1}{2}\int_0^\pi cos((n+m)\theta) + cos((n-m)\theta) d\theta$
there are two cases to consider here. When $n=m$ the integral evaluates to $\pi/2$, and when $n\neq m$ the integral evaluates to zero.
Then you have orthogonality
$<T_n,T_m> = \begin{cases} 0, n \neq m \\ \pi, n = m = 0 \\ \pi/2, n=m \neq 0 \end{cases}$