In class we found the exponential generating function for the Bell numbers $B_n$ which are defined by the recurrence $B(0) = 1$, $B(1) = 1$ and $B(n+1) =\sum_{i=1}^n\dbinom{n}{i}B(n-i)$ for all$ n\geq 1$. We found that $B(x)=\sum_{n\geq 0}B_n\frac{x^n}{n!}=e^{e^x−1}$. Also the Stirling numbers of the second kind $S_{n,k}$ are defined as the number of set partitions into $k$ parts. They are defined recursively as $S_{0,0}= 1$,$S_{n,1}=S_{n,n}= 1$ for all $n\geq 1$, and $S_{n,k}= 0$ if $k > n$. Moreover $S_{n+1,k}=kS_{n,k}+S_{n,k−1}$ for $n\geq 0$ and $1 \leq k \leq n$.
Now I am trying to refine the computation that gives the formula for $B(x)=\sum_{n\geq 0}B_n\frac{x^n}{n!}= e^{e^x−1}$
(that is we show that $B(x)$ satisfies a differential equation and we know that $B(0)=B_0$ and $B′(0) =B_1$ to show that
\begin{equation*}
S(x,q) =\sum_{n\geq 0}\sum_{k\geq 0}S_{n,k}q^k\frac{x^n}{n!}=e^{q(e^x−1)}
\end{equation*}.
Now this is a homework question but i am completely stuck. How to Show that an equivalency relation occurs between a set of integers (Stirling Numbers) and a function with euler values? I will try induction but then for the product of summations equivalence, i wouldn't induct.
Best Answer
Let \begin{eqnarray*} S_k(x)=\sum_{n=k}^{\infty} S_{n,k} \frac{x^n}{n!}. \end{eqnarray*} The Stirling number satisfy the recurrence relation $S_{n+1,k}=kS_{n,k} +S_{n,k-1}$, multiply this equation by $x^n$ and sum $n$ from $k$ to infinity & we get \begin{eqnarray*} \frac{d}{dx} S_k(x)=kS_k(x)+S_{k-1}(x). \end{eqnarray*} $S_0(x)=1$ and $S_1(x)= e^x-1$, it is easy to show by induction \begin{eqnarray*} S_k(x)=\frac{(e^x-1)^k}{k!}. \end{eqnarray*} So \begin{eqnarray*} S(q,x) =\sum_{n=0}^{\infty} \sum_{k=0}^{n} S_{n,k} q^k \frac{x^n}{n!} = \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} S_{n,k} q^k \frac{x^n}{n!} = \sum_{k=0}^{\infty}q^k \frac{(e^x-1)^k}{k!}= \exp(q(e^x-1)). \end{eqnarray*}