[ $\left[\begin{array}{l}\text { Generating functions } \\ {\left[\begin{array}{l}\text { Chebyshev Polynomials }\end{array}\right]} \\ \text { ] }\end{array}\right]$ [ Special polynomials (other).]

Find the generating functions of the sequences of Chebyshev polynomials of the first and second kind:

$$
F_{T}(x, z)=\sum_{n=0}^{\infty} T_{n}(x) z^{n} ; \quad F_{U}(x, z)=\sum_{n=0}^{\infty} U_{n}(x) z^{n}
$$

Definitions of Chebyshev polynomials can be found in the reference manual.
According to the problem $\underline{61471} T_{n}(x)=1 / 2(-i)^{n} L_{n}(2 i x), U_{n}(x)=(-i) n F_{n+1}(2 i x)$, where $L_{n}(x)$ and $F_{n}(x)$ are the Lucas and Fibonacci polynomials. Therefore,

$$
\begin{aligned}
& F_{T}(x, z)=\frac{1}{2} L(2 i x,-i z)=\frac{1}{2} \cdot \frac{2-2 i x(-i z)}{1-2 i x(-i z)-(-i z)^{2}}=\frac{1-x z}{1-2 x z+z^{2}} \\
& F_{U}(x, z)=i z^{-1} F(2 i x,-i z)=\frac{i}{z} \cdot \frac{-i z}{1-2 x z+z^{2}}=\frac{1}{1-2 x z+z^{2}}
\end{aligned}
$$

(The generating functions $L(x, z)$ and $F(x, z)$ of the Lucas and Fibonacci polynomials are computed in problem $\underline{61506}$.)

## Answer

$F_{T}(x, z)=(1-x z)\left(1-2 x z+z^{2}\right)^{-1}, \quad F_{U}(x, z)=\left(1-2 x z+z^{2}\right)^{-1}$.