Find the generating function of Chebyshev polynomials

Question

The Chebyshev polynomials of the first kind are: $$T_n(x)=\cos(n\theta)$$ where $x=\cos(\theta)$.

Prove that the generating function of Chebyshev polynomials is: $$\sum_{n=0}^{\infty}T_n(x)t^n=\frac{1-xt}{1-2xt+t^2}$$

I tried to prove using De Moivre's formula but I don't get something that brings the relation that we have to prove.

Does anyone have any idea or proof of this?

Answer 1

The series is the real part of

$$\sum_{n=0}^\infty \exp(in\theta)t^n=\sum_{n=0}^\infty \exp(n(i\theta+\ln t))=\frac{1}{1-te^{i\theta}}=\frac{1-te^{-i\theta}}{1-2t\cos(\theta)+t^2},$$

where we multipled by $1-te^{-\theta}$ for the last step. Can you finish from here?

Answer 2

It follows directly from $T_{n}(x)=2xT_{n-1}(x)-T_{n-2}(x)$ for $n\ge2$: $$ f(t) = T_0(x)+T_1(x)t + 2xt(f(t)-T_0(x))-t^2f(t) $$ Now solve for $f(t)$ using $T_0(x)=1$ and $T_1(x)=x$.