# Exponential generating function of the falling factorial

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?

• I suspect you are supposed to recall that $a_n = n! \dbinom{\alpha}{n}$. But your integral equation is a good step in a different path to the answer, provided that you solve it (hint: simplify the sum on the right hand side, and transform the integral equation into a differential equation). – darij grinberg Nov 25 '19 at 0:18
• @darijgrinberg I'll try to simplify the sum, thanks – alex118 Nov 25 '19 at 0:21
• I've edited your post to (I think) correct your recurrence. You were a little off after the "$\dots$", and wrote the rising factorial expression. – Robert D-B Nov 25 '19 at 0:43

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 $$x$$ $$\begin{eqnarray*} \frac{d }{dx} A_{\alpha}(x) = \alpha \left( (\alpha-1) x +\alpha (\alpha-1) \frac{x^2}{2!} +\cdots \right). \end{eqnarray*}$$ So $$\begin{eqnarray*} \frac{d }{dx} A_{\alpha}(x) = \alpha A_{\alpha-1}(x). \end{eqnarray*}$$ Solving this differential equation inductively will rapidly give $$\begin{eqnarray*} A_{\alpha}(x) = (1+x)^{\alpha } . \end{eqnarray*}$$
This is just the binomial series $$\sum_{n\ge0}\alpha^{\underline n}\frac{x^n}{n!}= \sum_{n\ge0}\frac{\alpha^{\underline n}}{n!}x^n= \sum_{n\ge0}\binom{\alpha}nx^n=(1+x)^\alpha.$$
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?
• Ah, got it. Using separation of variables we have that $\ln{f}$=$\ln{((1+x)^{\alpha})}$ then $f=(1+x)^{\alpha}$ – alex118 Nov 25 '19 at 0:51