Chebyshev polynomial. Show : $ T_n(x) = \frac{1}{2} ((x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n)$ Consider $T_n(x) = \cos ( n \cdot \arccos(x)) $ on $ I = [-1,1]$.
Show:
a: $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) $
b : The $T_n$ are orthogonal with $(f,g) = \int_{-1}^{1} f(x)g(x)\frac{1}{\sqrt{1-x^2}} dx $.
c: $T_n(x) = \frac{1}{2} ((x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n)$
So I solved a and b, but I can't solve c. I tried to rewrite $(x+\sqrt{x^2-1})^n$ with the binomial theorem, but it didn't work out so well. Maybe we could use some identities of $\cos$ or $\arccos$? Thank you for your help.
 A: Use $\pm\sqrt{x^2-1}= \pm i \sqrt{1-x^2}$. So if $x = \cos t$ then 
$$(x \pm \sqrt{x^2-1})= \cos t \pm i \sin t$$
A: You could consider the function
$$
G(t;x)=\sum_{k=0}^\infty T_n(x)t^n \\
$$
by $\bf{a}$ you know that
$$
T_n(x) = 2xT_{n-1}(x) - T_{n-2}(x)
$$
so
$$
G(t;x)=\sum_{k=0}^\infty (2xT_{n-1}(x) - T_{n-2}(x))t^n \\
G(t;x)=2x\sum_{k=0}^\infty T_{n-1}(x)t^n - \sum_{k=0}^\infty T_{n-2}(x)t^n \\
G(t;x)=1 - tx + 2x\sum_{k=1}^\infty T_{n-1}(x)t^n - \sum_{k=2}^\infty T_{n-2}(x)t^n \\
G(t;x)=1-tx + 2txG(t;x) - t^2G(t;x) \\
G(t;x) = \frac{1-tx}{1-2tx+t^2} 
$$
now consider the coefficients of $t^n$ of that generating function.
A: Let
$f_n(x) 
= \frac{1}{2} ((x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n)
$,
and let
$x = \cosh(y)$.
Since
$\sqrt{x^2-1}
=\sqrt{\cosh^2(x)-1}
=\sinh(x)
$,
$\begin{array}\\
f_n(x)
&=f_n(\cosh(y))\\
&=\frac12((\cosh(y)+\sinh(y))^n+(\cosh(y)-\sinh(y))^n)\\
&=\frac12((e^y)^n+(e^{-y})^n)\\
&=\frac12(e^{ny}+e^{-ny})\\
&=\cosh(ny)\\
&=\cosh(n\cosh^{-1}(x))\\
\end{array}
$
Since
$|x| < 1$,
$y$ is imaginary,
so
$y = i\cos^{-1}(x)
$
and
$f_n(x)
=\cosh(ni\cos^{-1}(x))
=\cos(n\cos^{-1}(x))
$
which is
one definition
of the Chebychev
polynomials.
