orthogonality of Chebyshev polynomials

By expanding the of de Moivre identity

$$\cos n\phi ={(\cos \phi +i\sin \phi )}^n$$

to sum, one obtains as real part certain terms containing power products of $\cos \phi$ and $\sin \phi$, the latter ones only with even exponents.  When these are expressed with cosines (${\sin }^2\phi =1-{\cos }^2\phi$), the real part becomes a polynomial $T_n$ of degree $n$ in the argument (http://planetmath.org/Argument2) $\cos \phi$:

$\cos n\phi =T_n(\cos \phi )$(1)

This can be written equivalently (http://planetmath.org/Equivalent3)

$T_n(x)=\cos (n\arccos x).$(2)

It’s a question of Chebyshev polynomial of first kind and of $n$ (cf. special cases of hypergeometric function).

For showing the orthogonality of $T_m$ and $T_n$ we start from the integral${\displaystyle {\int }_0^{\pi }}\cos m\phi \cos n\phi d\phi$, which via the substitution

$$\cos \phi :=x,dx=-\sin \phi d\phi =-\sqrt{1-x^2}d\phi$$

changes to

${\displaystyle {\int }_0^{\pi }}\cos m\phi \cos n\phi d\phi =-{\displaystyle {\int }_1^{-1}}{T}_m(x)T_n(x){\displaystyle \frac{dx}{\sqrt{1-x^2}}}.$(3)

The left of this equation is evaluated by using the product formula in the entry trigonometric identities:

${\displaystyle {\int }_0^{\pi }}\cos m\phi \cos n\phi d\phi ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi }}(\cos (m-n)\phi +\cos (m+n)\phi )𝑑\phi =\{\begin{array}{cc}0\text{ for }m\ne n,\hfill & \\ \frac{\pi }{2}\text{ for }m=n\ne 0.\hfill & \end{array}$

By (3), we thus have

${\displaystyle {\int }_{-1}^1}{T}_m(x)T_n(x){\displaystyle \frac{dx}{\sqrt{1-x^2}}}=\{\begin{array}{cc}0\text{ for }m\ne n,\hfill & \\ \frac{\pi }{2}\text{ for }m=n\ne 0,\hfill & \end{array}$

which means the orthogonality of the polynomials $T_m(x)$ and $T_n(x)$ weighted by $\frac{1}{\sqrt{1-x^2}}$.

Any Riemann integrable real function $f$, defined on  $[-1,\mathrm{\hspace{0.17em}1}]$,  may be expanded to the series

$$f(x)=\frac{a_0}{2}{T}_0(x)+\sum _{j=1}^{\mathrm{\infty }}{a}_j{T}_j(x),$$

where

$$a_j=\frac{2}{\pi }{\int }_{-1}^{1}f(x)T_j(x)\frac{dx}{\sqrt{1-x^2}}\mathit{\hspace{1em}\hspace{1em}}(j=0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots })$$

This concerns especially the polynomials  $f(x):=x^n$,  for which we obtain

$x^n$$\mathrm{\hspace{0.33em}}={\cos }^n\phi ={\cosh }^{n}i\phi ={\mathrm{\hspace{0.33em}2}}^{-n}(e^{i\phi }+e^{-i\phi })$

$\mathrm{\hspace{0.33em}}={\mathrm{\hspace{0.33em}2}}^{-n}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)(e^{ni\phi }+e^{-ni\phi })+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{1}}\right)(e^{(n-2)i\phi }+e^{-(n-2)i\phi })+\mathrm{\dots }\right]$

$\mathrm{\hspace{0.33em}}=2^{1-n}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)T_n(x)+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{1}}\right)T_{n-2}(x)+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)T_{n-4}(x)+\mathrm{\dots }\right].$

(If $n$ is even, the last term contains $T_0(x)$ but its coefficient is only a half of the middle number of the Pascal’s triangle row in question.)  Explicitly:

$1=T_0$
$x=T_1$
$x^2={\mathrm{\hspace{0.33em}2}}^{-1}(T_2+T_0)$
$x^3={\mathrm{\hspace{0.33em}2}}^{-2}(T_3+3T_1)$
$x^4={\mathrm{\hspace{0.33em}2}}^{-3}(T_4+4T_2+3T_0)$
$x^5={\mathrm{\hspace{0.33em}2}}^{-4}(T_5+5T_3+10T_1)$
$x^6={\mathrm{\hspace{0.33em}2}}^{-5}(T_6+6T_4+15T_2+10T_0)$
$x^7={\mathrm{\hspace{0.33em}2}}^{-6}(T_7+7T_5+21T_3+35T_1)$
$x^8={\mathrm{\hspace{0.33em}2}}^{-7}(T_8+8T_6+28T_4+56T_2+36T_0)$
$x^9={\mathrm{\hspace{0.33em}2}}^{-8}(T_9+9T_7+36T_5+84T_3+126T_1)$
$\mathrm{\cdots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\cdots }$