Biomedical Imaging

As the base clause, we can use

$U_0\left(s\right)=1,\left(4.85\right)$

$U_1\left(s\right)=\frac{\sin 2\theta }{\sin \theta }=\frac{2\sin \theta \cos \theta }{\sin \theta }=2\cos \theta =2s.\left(4.86\right)$

Since these two expressions are polynomials in s, this is also true for all other Un(s). The orthogonality of Chebyshev polynomials is shown with equation (4.72)

${\displaystyle \underset{-1}{\overset{1}{\int }}\text{ds}}\sqrt{1-s^2}{U}_n\left(s\right)U_m\left(s\right)\left(4.87\right)$

$\stackrel{s\to \cos \theta }{=}{\displaystyle \underset{0}{\overset{\pi }{\int }}\text{d}\theta }\sin \theta \underset{\sqrt{1-s^2}}{\underbrace{\sin \theta }}{U}_n\left(\cos \theta \right)U_m\left(\cos \theta \right)\left(4.88\right)$

$\underset{=}{\left(4.76\right)}{\displaystyle \underset{0}{\overset{\pi }{\int }}\text{d}\theta }\sin \left(\left(n+1\right)\theta \right)\sin \left(\left(m+1\right)\theta \right)\left(4.89\right)$

$={\displaystyle \underset{0}{\overset{\pi }{\int }}\text{d}\theta \frac{1}{2i}\left(e^{i\left(n+1\right)\theta }-e^{-i\left(n+1\right)\theta }\right)}\frac{1}{2i}\left(e^{i\left(m+1\right)\theta }-e^{-i\left(m+1\right)\theta }\right)\left(4.90\right)$

$=-\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi }{\int }}\text{d}\theta \left[e^{i\left(n+m+2\right)\theta }+e^{-i\left(n+m+2\right)\theta }-e^{i\left(n-m\right)\theta }-e^{-i\left(n-m\right)\theta }\right]}\left(4.91\right)$

$=$