# Differential equation and Appell polynomials

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• pawlo392
In summary, an Appell sequence is a sequence of polynomials that follows the relation $P_n'(x) = nP_{n-1}(x)$ and can also be defined using a generating function. To show the equivalence of these definitions, one can differentiate the generating function and use the first definition to obtain a linear first-order differential equation. To prove that the generating function is a solution, one can solve the obtained partial differential equation $\frac{\partial h}{\partial x} = th$.
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:

P_n'(x) = nP_{n-1}(x) \quad \text{for all } n \in \mathbb{N}.

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

A(t)e^{xt}=\sum_{n=0}^\infty \frac{P_n(x)}{n!}t^n, \ A(0) \neq 0,

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.

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

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

Last edited:
pawlo392

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