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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.

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    $\begingroup$ Do you intend $\sqrt{1-x^2}$ instead? Anyway, plug in $x=\cos(t)$ and see what you can do. $\endgroup$ – Ian May 29 '18 at 7:56
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    $\begingroup$ You can also prove $c$ by induction using $a$. $\endgroup$ – C. Dubussy May 29 '18 at 8:20
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    $\begingroup$ Perhaps try induction ... show that (c) satisfies the recurrence (a). Or, alternatively: to find (c) from scratch, solve the recurrence (a), which has constant (i.e. not depending on $n$) coefficients. $\endgroup$ – GEdgar May 29 '18 at 13:31
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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.

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Use $\pm\sqrt{x^2-1}= \pm i \sqrt{1-x^2}$. So if $x = \cos t$ then $$(x \pm \sqrt{x^2-1})= \cos t \pm i \sin t$$

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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.

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