I have this quantity:
$ T_n(x) = \exp\big[i\cdot n\arccos(x)\big] $
How can i represent the polynomials of first and second kind?
I tried substituting $n = 0$ and $n = 1$ but i don't know when to stop and plot the generative function
EDIT using MathLaTeX
$ T_0(x) = e^{i0\arccos(x)} = e ^ 0 = 1 $
$ T_1(x) = e^{i1\arccos(x)} = 1 $
$ T_2(x) = 2x(1)-1 = 2x-1$
$ T_3(x) = 2x(T_2(x))-T_1(x) = 2x-1(2x-1)-1=4x^2-2x-1$
$ T_4(x) = 2x(T_3(x))-T_2(x) = 2x(4x^2-2x-1)-(2x-1) = 8x^3-4x^2-4x+1$
Using $$\exp(i\theta)=e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ the definition turns to $$T_n(x)= e^{in\cos^{-1}(x)}=\cos(n\cos^{-1}(x))+i\sin(n\cos^{-1}(x))$$ Thus, for example: $$T_1(x)=\cos(\cos^{-1}(x))+i\sin(\cos^{-1}(x))$$
Edit: Now we have $$T_{n+1}(x)=e^{i(n+1)\cos^{-1}(x)}=e^{in\cos^{-1}(x)+{i\cos^{-1}(x)}}\\ =e^{in\cos^{-1}(x)}e^{i\cos^{-1}(x)}\\=T_n(x)T_1(x)$$ can you continue?
Also we have $$T_{n+1}(x)-T_{n-1}(x)=e^{i(n+1)\cos^{-1}(x)}-e^{i(n-1)\cos^{-1}(x)}\\= e^{in\cos^{-1}(x)}(e^{i\cos^{-1}(x)}-e^{-i\cos^{-1}(x)})\\= 2iT_n(x) \sin(\cos^{-1}(x))$$ Now you can find the generating function easily.