Exponential generating function of the falling factorial

Question

Let $\alpha$ be a real number. Define the sequence $(a_n)_n$ by $a_0=1$ and $a_n=\alpha(\alpha-1)\cdots(\alpha - (n-1))$ for $n\geq 1$. Find the exponential generating function of this sequence.

We have that $a_n=(\alpha-(n-1))a_{n-1}$ for $n\geq1$, so \begin{align*} A(x)&=\sum_{n\geq 0}a_n\frac{x^n}{n!}=a_0+\sum_{n\geq 1}a_n\frac{x^n}{n!}\\ &=1+\sum_{n\geq 1}(\alpha+n-1)a_{n-1}\frac{x^n}{n!}\\ &=1+\sum_{n\geq 0}(n+\alpha)a_{n}\frac{x^{n+1}}{(n+1)!}=1+\alpha\int_0^xA(t)dt+\sum_{n\geq 0}n\frac{x^{n+1}}{(n+1)!}\end{align*} I'm stuck here. I tried to write the last sum as an integral and then solve a differential equation for $A(x)$, but it didn't work.

Should I search for another recurrence relation that $a_n$ satisfies?

Answer 1

We have \begin{eqnarray*} A_{\alpha}(x)= 1 +\alpha x +\alpha (\alpha-1) \frac{x^2}{2!} +\alpha (\alpha-1) (\alpha-2)\frac{x^3}{3!} +\cdots. \end{eqnarray*} Differentiate this wrt to $x$ \begin{eqnarray*} \frac{d }{dx} A_{\alpha}(x) = \alpha \left(1+ (\alpha-1) x +(\alpha-1) (\alpha-2) \frac{x^2}{2!} +\cdots \right). \end{eqnarray*} So \begin{eqnarray*} \frac{d }{dx} A_{\alpha}(x) = \alpha A_{\alpha-1}(x). \end{eqnarray*} Solving this differential equation inductively will rapidly give \begin{eqnarray*} A_{\alpha}(x) = (1+x)^{\alpha } . \end{eqnarray*}

Answer 2

This is just the binomial series $$\sum_{n\ge0}\alpha^{\underline n}\frac{x^n}{n!}= \sum_{n\ge0}\frac{\alpha^{\underline n}}{n!}x^n= \sum_{n\ge0}\binom{\alpha}nx^n=(1+x)^\alpha. $$

Answer 3

Recall two useful facts about exponential generating functions: First, if $$f(x) = \sum_{k \geq 0} \frac{a_k}{k!} x^k$$ is the exponential generating function for $a_n$, then the exponential generating function for $P(n) a_n$, where $P$ is any polynomial in $n$, is $P(xD) f(x)$, where $D$ is the differentiation operator. For example, $$\sum_{k \geq 0} \frac{ka_k}{k!} x^k = \sum_{k \geq 0} xD \frac{a_k}{k!} x^k = xD f(x) = x f'(x).$$

Second, if $f$ is the egf of $a_n$, then $f'$ is the egf of $a_{n + 1}$.

In your situation, you have $a_{n + 1} = (\alpha - n)a_n$ for $n \geq 0$. (Note the useful shift in boundary conditions.) Taking the egf of both sides yields $$f' = (\alpha - xD)f,$$ or $$f' = \alpha f - xf'.$$ This is a linear differential equation. Can you solve it?