Question

(Chebyshev Polynomials)(Orthogonality) I'm having a bit of trouble or difficulty thinking I'm taking the right steps in proving the orthogonality between something like a Chebyshev polynomial, more info below.

Answer 1

Original prompt:

\[\text{(a) Is the set $F_n(x)=\sin[(n+1)\arccos(x)]$ on $(-1,1), n \in \mathbf{N_0}$, orthogonal}\]\[\text{with respect to the weight function $(1-x^2)^{1/2}?$}\]

Answer 2

Workings so far: \[\int\limits_{-1}^{1}\sin[(n+1)\arccos(x)]\sqrt{1-x^2}dx, \ \ \ \text{Let $x=\cos(u), \ dx=-\sin(u)du$};\]

Answer 3

\[=\int\limits_{-1}^{1}\sin[(n+1)\arccos(\cos(u))]\sqrt{1-\cos^2(u)}(-\sin(u)du)\]\[=\int\limits_{-1}^{1}\sin[(n+1)u]\sin(u)(-\sin(u))du=-\int\limits_{-1}^{1}=\sin[(n+1)u]\sin^2(u)du\]

Answer 4

Is this so far the right direction? @dan815

Answer 5

Looks ok so far

Answer 6

Alright, in that case, I'm a little confused about how to continue from here; do I have to use integration by parts, or do I have other options?

Answer 7

@JFraser

Answer 8

It looks like you have to explore trig identities ? If I get an idea, I'll post it.

Answer 9

I have no idea, sorry

Answer 10

\[\int_{-1}^1\sin\bigg((n+1)\arccos x\bigg)\sqrt{1-x^2}\,dx\]
Setting \(x=\cos u\) gives \(dx=-\sin u\,du\):
\[-\int_{\arccos(-1)}^{\arccos1}\sin\bigg((n+1)u\bigg)\sqrt{1-\cos^2u}\sin u\,du\\
\int_0^\pi\sin\bigg((n+1)u\bigg)|\sin u|\sin u\,du\]
Over \([0,\pi]\), we have \(\sin u>0\), so \(|\sin u|=\sin u\) and you have
\[\int_0^\pi\sin\bigg((n+1)u\bigg)\sin^2 u\,du\]
Recall that
\[\sin^2u=\frac{1}{2}(1-\cos2u)\]
so the integral is equivalent to
\[\frac{1}{2}\left(\int_0^\pi\sin\bigg((n+1)u\bigg)\,du-\int_0^\pi\sin\bigg((n+1)u\bigg)\cos2 u\,du\right)\]
To deal with the second integral, recall another set of identities:
\[\left.\begin{cases}
\sin(x+y)=\sin x\cos y+\sin y\cos x\\
\sin(x-y)=\sin x\cos y-\sin y\cos x
\end{cases}\right\}\stackrel{\text{adding}}{\implies}\sin x\cos y=\frac{\sin(x+y)+\sin(x-y)}{2}\]
which means
\[\sin\bigg((n+1)u\bigg)\cos2u=\frac{\sin\bigg((n+3)u\bigg)+\sin\bigg((n-1)u\bigg)}{2}\]
so you're left to evaluate
\[\frac{1}{2}\left[\int_0^\pi\sin\bigg((n+1)u\bigg)\,du-\frac{1}{2}\int_0^\pi\left(\sin\bigg((n+3)u\bigg)+\sin\bigg((n-1)u\bigg)\right)\,du\right]\]