Does anyone have a reference for the exponential generating function for the Chebyshev polynomials of the second kind?
$$
\sum_{n=0}^{\infty}\frac{t^n U_n(x)}{n!}= etc
$$
I know what it is from the Wikipedia page, https://en.wikipedia.org/wiki/Chebyshev_polynomials , I just cannot find a proper reference for it.
I could not find it in the books by Rivlin and Mason & Handscomb, and my googling skills are not up to the task.
Best Answer
We can compute the generating function directly from the definition $$U_n(\cos \theta) = \frac{\sin (n+1)\theta}{\sin \theta}.$$
Write $z = e^{i \theta}$. Then the RHS is $\frac{z^{n+1} - z^{-n-1}}{z - z^{-1}}$, and so the exponential generating function is
$$\begin{align*} \sum_{n \ge 0} U_n(\cos \theta) \frac{t^n}{n!} &= \sum_{n \ge 0} \frac{z^{n+1} - z^{-n-1}}{z - z^{-1}} \frac{t^n}{n!} \\ &= \frac{ze^{tz} - z^{-1} e^{tz^{-1}}}{z - z^{-1}} \\ &= \frac{(\cos \theta + i \sin \theta) e^{t(\cos \theta + i \sin \theta)} - (\cos \theta - i \sin \theta) e^{t(\cos \theta - i \sin \theta)}}{2i \sin \theta} \\ &= e^{t \cos \theta} \frac{(\cos \theta + i \sin \theta) e^{it \sin \theta} - (\cos \theta - i \sin \theta) e^{-it \sin \theta}}{2i \sin \theta} \\ &= e^{t \cos \theta} \frac{2i \cos \theta \sin (t \sin \theta) + 2i \sin \theta \cos (t \sin \theta)}{2i \sin \theta} \\ &= e^{t \cos \theta} \left( \cos (t \sin \theta) + \cot \theta \sin (t \sin \theta) \right) \\ &= e^{tx} \left( \cos (t \sqrt{1 - x^2}) + \frac{x}{\sqrt{1 - x^2}} \sin (t \sqrt{1 - x^2}) \right) \end{align*}$$
where $x = \cos \theta$. To convert to the hyperbolic sine and cosine form given on Wikipedia write $\sqrt{1 - x^2} = i \sqrt{x^2 - 1}$ and use that $\cos (it) = \cosh t$ and $\sin it = i \sinh t$; I don't know why Wikipedia writes it that way.
For some corroboration, in Clemente Cesarano's Identities and generating functions for Chebyshev polynomials, you can find an identity for $\sum U_{n-1}(x) \frac{t^n}{n!}$ (Proposition 2) which appears to be equivalent to this one although one needs to differentiate it in $t$.