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?
Best Answer
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*}