Generating Functions
Employed is only the recurrence relation which is valid for all of the
solutions of the Chebyshev differential equation:
$$
S_{n+1}(x) + S_{n-1}(x) = 2 x S_n(x)
$$
Let $F(z,x)$ be the Generating Function, then by definition:
$$
F(z,x) = S_0(x) + S_1(x).z + S_2(x).z^2 + S_3(x).z^3 + \cdots
$$
Hence:
$$
z^2 F(z,x) = S_0(x).z^2 + S_1(x).z^3 + S_2(x).z^4 + S_3(x).z^5 + \cdots
$$
Add these two equations together:
$$
F(z,x) + z^2.F(z,x) = S_0(x) + S_1(x).z \; + \\
\left[S_2(x) + S_0(x)\right] z^2 +
\left[S_3(x) + S_1(x)\right] z^3 + \cdots
$$
And use the recurrence relation:
$$
F(z,x) + z^2.F(z,x) = S_0(x) + S_1(x).z - 2 x z . S_0(x) \, + \\
2 x S_0(x) z + 2 x S_1(x) z^2 + 2 x S_2(x) z^3 \, + \cdots \\
= (1-2xz)S_0(x) + z.S_1(x) + 2xz\left[\;S_0(x)+S_1(x).z+S_2(x).z^2\, + \cdots\;\right] \\
= (1-2xz)S_0(x) + z.S_1(x) + 2xz.F(z,x)
$$
Solve for $F(z,x)$ :
$$
F(z,x) (1 - 2 x z + z^2) = (1 - 2 x z)S_0(x) + z.S_1(x) \quad \Longrightarrow \\
F(z,x) = \frac{ (1 - 2 x z)S_0(x) + z.S_1(x) }{ 1 - 2 x z + z^2 }
$$
We have found the following initializations for Chebyshev polynomials:
$$
S_0(x) = 1 \quad \wedge \quad S_1(x) = \left\{ \begin{array}{ll}
x & \quad(\mbox{first kind})\\
2 x & \quad(\mbox{second kind})\\
2 x - 1 & \quad(\mbox{half integer})\\
2 x + 1 & \quad(\mbox{half integer}) \end{array} \right.
$$
Respectively resulting in the following set of Generating Functions:
$$ \Large
F(z,x) = \left\{ \begin{array}{l}
\frac{1 - x z}{1 - 2 x z + z^2} \\
\frac{1}{1 - 2 x z + z^2} \\
\frac{1 - z}{1 - 2 x z + z^2} \\
\frac{1 + z}{1 - 2 x z + z^2} \end{array} \right.
$$
We can do some more with the general formula
if we factorize the denominator:
$$
z^2 - 2 x z + 1 = (z-\alpha)(z-\beta)
\quad \Longleftrightarrow \quad \alpha+\beta = 2 x \quad \wedge \quad \alpha\beta = 1
$$
Where we find:
$$
\alpha = x + \sqrt{x^2-1} \quad \wedge \quad \beta = x - \sqrt{x^2-1}
$$
Split into partial fractions:
$$
F(z,x) = \frac{A}{z-\alpha} + \frac{B}{z-\beta}
= \frac{A(z-\beta) + B(z-\alpha)}{(z-\alpha)(z-\beta)} \\
= \frac{(A+B)z - (A\beta+B\alpha)}{(z-\alpha)(z-\beta)}
$$
On the other hand:
$$
F(z,x) = \frac{\left[S_1(x)-2 x S_0(x)\right]z + S_0(x)}{z^2 - 2 x z + 1} \\
\begin{cases}
A + B &=& S_1(x)-2 x S_0(x) \\
\beta A + \alpha B &=& - S_0(x)
\end{cases}
$$
Two equations with two unknowns. The solution is:
$$
A = + \frac{\alpha\left[S_1(x)-2 x S_0(x)\right] + S_0(x)}{\alpha-\beta}
= + \frac{\alpha S_1(x) + \left[1-2\alpha x\right]S_0(x)}{\alpha-\beta}\\
B = - \frac{\beta\left[S_1(x)-2 x S_0(x)\right] + S_0(x)}{\alpha-\beta}
= - \frac{\beta S_1(x) + \left[1-2\beta x\right]S_0(x)}{\alpha-\beta}
$$
This can be simplified even further:
$$
1-2x\alpha = \alpha\beta-(\alpha+\beta)\alpha = -\alpha^2 \quad \Longrightarrow \quad
A = \alpha \frac{S_1(x)-\alpha S_0(x)}{\alpha-\beta} \\
1-2x\beta = \alpha\beta-(\alpha+\beta)\beta = -\beta^2 \quad \Longrightarrow \quad
B = \beta \frac{-S_1(x)+\beta S_0(x)}{\alpha-\beta}
$$
We proceed as follows:
$$
F(z,x) = \frac{A}{z-\alpha} + \frac{B}{z-\beta}
= - \frac{A}{\alpha}\frac{1}{1-z/\alpha}
- \frac{B}{\beta}\frac{1}{1-z/\beta} \\
= - \frac{A}{\alpha}\left[\; 1 + \frac{1}{\alpha} z + \frac{1}{\alpha^2} z^2
+ \frac{1}{\alpha^3} z^3 + \frac{1}{\alpha^4} z^4 + \cdots\; \right] \\
- \frac{B}{\beta}\left[\; 1 + \frac{1}{\beta} z + \frac{1}{\beta^2} z^2
+ \frac{1}{\beta^3} z^3 + \frac{1}{\beta^4} z^4 + \cdots\; \right] \\
= - \left[ \; \frac{A}{\alpha} + \frac{B}{\beta} \right]
- \left[ \frac{A}{\alpha^2} + \frac{B}{\beta^2} \right] z
- \left[ \frac{A}{\alpha^3} + \frac{B}{\beta^3} \right] z^2
- \left[ \frac{A}{\alpha^4} + \frac{B}{\beta^4} \right] z^3 + \cdots
$$
On the other hand we have:
$$
F(z,x) = S_0(x) + S_1(x).z + S_2(x).z^2 + S_3(x).z^3 + \cdots
$$
Conclusion:
$$
S_n(x) = - \left[ \frac{A}{\alpha}\frac{1}{\alpha^n}
+ \frac{B}{\beta}\frac{1}{\beta^n} \right]
= \frac{-S_1(x)+\alpha S_0(x)}{\alpha-\beta}\frac{1}{\alpha^n}
+ \frac{ S_1(x)-\beta S_0(x)}{\alpha-\beta}\frac{1}{\beta^n}
$$
In all cases known to us we have that $S_0(x) = 1$ . Hence:
$$
S_n(x) = \frac{-S_1(x)+x+\sqrt{x^2-1}}{2\sqrt{x^2-1}}(x-\sqrt{x^2-1})^n \\
+ \frac{ S_1(x)-x+\sqrt{x^2-1}}{2\sqrt{x^2-1}}(x+\sqrt{x^2-1})^n
$$
The most well-known result is the one for Chebyshev polynomials of the first
kind. In that case $S_1(x) = x$ and $S_n(x) = T_n(x)$ . Hence:
$$
T_n(x) = \frac{(x-\sqrt{x^2-1})^n + (x+\sqrt{x^2-1})^n}{2}
$$
Which appears in many other forms in the litterature and on the Internet, for example in:
Chebyshev_polynomials.
The only new thing here may be that such formulas can also be derived for the
polynomials in the half integer solutions of the Chebyshev differential equation.