2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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*}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.