Differential equation and Appell polynomials

[pawlo392]

Hello!
Let $n$ be a natural number, $P_n(x)$ be a polynomial with rational coefficients, and $\deg P_n(x) = n$. Let $P_0(x)$ be a constant polynomial that is not equal to zero. We define the sequence ${P_n(x)}_{n \geq 0}$ as an Appell sequence if the following relation holds:
\begin{equation}
P_n'(x) = nP_{n-1}(x) \quad \text{for all } n \in \mathbb{N}.
\end{equation}

In the literature, we can also find the following definition:
\begin{df}\textit{
A sequence $\{P_n(x) \}_{n \ge 0}$ is called an Appell sequence (and $P_n(x)$ is called the $n$-th Appell polynomial) if
\begin{equation}
A(t)e^{xt}=\sum_{n=0}^\infty \frac{P_n(x)}{n!}t^n, \ A(0) \neq 0,
\end{equation}
where $A(t)=\sum_{n=0}^\infty \frac{\mathcal{P}_n}{n!}t^n$ is the generating function with $\mathcal{P}_n=P_n(0)$. }I would like to show implication 1 definition to 2 definition.
Defining the generating function $$h(x) = \sum_{n=0}^{\infty} \frac{P_n(x)}{n!}t^n$$ differentiating with respect to $x$ and using the first definition, we obtain a linear first-order differential equation $$h'(x)=\sum_{n=0}^\infty \frac{P_{n}'(x)}{n!}t^n = \sum_{n=1}^\infty \frac{nP_{n-1}(x)}{n!}t^n = \sum_{n=1}^\infty \frac{P_{n-1}(x)}{(n-1)!}t^n. $$

I don't know how to show that $A(t)e^{xt}$ is a solution.

 

[pasmith]

You have [tex]
h(x,t) = \sum_{n=0}^\infty \frac{P_n(x)}{n!}t^n[/tex] and you wnat to show show that [itex]h(x,t) = A(t)e^{xt}[/itex] for a specific [itex]A(t)[/itex].

You found after applying the recurrence relation that [tex]
\frac{\partial h}{\partial x} = \sum_{n=1}^\infty \frac{P_{n-1}(x)}{(n-1)!}t^n.[/tex] But the right hand side is equal to [itex]th(x,t)[/itex] (set [itex]m = n-1[/itex]) so that [tex]
\frac{\partial h}{\partial x} = th[/tex] is the PDE you need to solve for [itex]h[/itex].