Recurrence relation for Chebyshev polynomials of the first kind

Question

I tried to derive recurrence relation for Chebyshev polynomials from their generating function $$\frac{1-xt}{1-2xt+t^2}=\sum_{n=0}^{\infty}T_{n}(x)t^n.$$ I've differentiated both sides with respect to t to obtain: $$\frac{x-t}{(1-2tx+t^2)^{3/2}}-\frac{x(1-t^2)}{(1-2tx+t^2)^2}=\sum_{n=1}^{\infty}T_{n}(x)nt^{n-1}.$$I don't know how to proceed.

Answer 1

\begin{eqnarray*} T_n( \cos( \theta))= \cos(n \theta). \end{eqnarray*} ... \begin{eqnarray*} \sum_{n=0}^{\infty} T_n( \cos( \theta)) t^n &=& \sum_{n=0}^{\infty} \cos(n \theta) t^n \\ &=& \Re \left( \sum_{n=0}^{\infty} ( e^{i \theta} t) ^n ) \right) \\ &=& \Re \left( \frac{1}{(1- e^{i \theta} t) } \right) \\ &=& \Re \left( \frac{1-e^{-i \theta}t}{(1- e^{i \theta} t)(1-e^{-i \theta}t) } \right) \\ &=& \Re \left( \frac{1-e^{-i \theta}t}{1- 2\cos(\theta)t+t^2 } \right)\\ &=& \frac{1-\cos(\theta) t}{1- 2\cos(\theta)t+t^2 }. \\ \end{eqnarray*}