数学代写|组合学代写Combinatorics代考|The Fubini numbers count chains of subsets

Let us fix a set $S$ of $n$ elements. We say that a sequence
$$U_1 \subset U_2 \subset \cdots \subset U_k$$
of subsets of $S$ is a chain if $U_i \subset U_{i+1}$ and $U_i \neq U_{i+1}$ for all possible $i$ s. The length of the chain (6.4) is defined to be $k$.

Let a chain be called full if $U_1=\emptyset$ and $U_k=S$. We are going to see that the number of all the full chains on an $n$ element set $S$ is the $n$-th Fubini number.

Indeed, $U_1=\emptyset$, and we would like to choose an $U_2$, as the next element of the chain. If the chain is of length two, we choose $U_2$ to be $S$. If it is longer, then we have to choose a non-empty proper subset of $S$. If the chain is of length three, we must choose $U_3=S$. Otherwise, we have to fix the proper subset $U_3$ of $S$ such that it is disjoint from $U_2$. We continue the process in this manner until eventually we reach $U_k=S$. Obviously, $U_2 \backslash U_1\left(=U_1\right), U_3 \backslash$ $U_2, \ldots, U_k \backslash U_{k-1}$ form a partition of $S$. In the case of chains, the order in which we have chosen our chain elements (blocks) matters, since, for example, the two chains
$$\emptyset \subset{1,2} \subset{1,2,3,4}=S$$
and
$$\emptyset \subset{3,4} \subset{1,2,3,4}=S$$
are different (even if the corresponding partitions, 1,2|3,4 and $3,4 \mid 1,2$, are the same). For this reason, the number of chains of length $k$ for a set of $n$ elements is $k !\left{\begin{array}{l}n \ k\end{array}\right}$. Summing over $k$, we get the number of all the possible full chains for an $n$ element set, and this is the number $F_n$.

数学代写|组合学代写Combinatorics代考|The generating function of the Fubini numbers

In order to find the exponential generating function of the Fubini numbers, we use (6.1). Summing over $n$ after multiplying with $\frac{x^n}{n !}$ we get that
$$\sum_{n=0}^{\infty} F_n \frac{x^n}{n !}=\sum_{n=0}^{\infty}\left(\sum_{k=1}^n k !\left{\begin{array}{l} n \ k \end{array}\right}\right) \frac{x^n}{n !}$$

Here $k$ can run from 0 to infinity, so that we can interchange the sums:
$$\begin{gathered} \sum_{n=0}^{\infty}\left(\sum_{k=1}^n k !\left{\begin{array}{l} n \ k \end{array}\right}\right) \frac{x^n}{n !}=\sum_{k=0}^{\infty} k !\left(\sum_{n=0}^{\infty}\left{\begin{array}{l} n \ k \end{array}\right} \frac{x^n}{n !}\right)=\sum_{k=0}^{\infty} k ! \frac{\left(e^x-1\right)^k}{k !}= \ \sum_{k=0}^{\infty}\left(e^x-1\right)^k=\frac{1}{1-\left(e^x-1\right)} . \end{gathered}$$
The last sum is a simple geometric series (see (2.2)). The exponential generating function is therefore
$$\sum_{n=0}^{\infty} F_n \frac{x^n}{n !}=\frac{1}{2-e^x}$$
This results in a nice infinite sum representation for the Fubini numbers, which can be considered as the analogue of the Dobiński formula (2.17). Rewriting the generating function:
$$\begin{gathered} \frac{1}{2-e^x}=\frac{1}{2} \frac{1}{1-\frac{e^x}{2}}=\frac{1}{2} \sum_{k=0}^{\infty}\left(\frac{e^x}{2}\right)^k=\frac{1}{2} \sum_{k=0}^{\infty} \frac{e^{k x}}{2^k}= \ \sum_{k=0}^{\infty} \frac{1}{2^{k+1}} \sum_{n=0}^{\infty} \frac{k^n x^n}{n !}=\sum_{n=0}^{\infty} \frac{x^n}{n !} \sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}} . \end{gathered}$$
Comparing the coefficients,
$$F_n=\sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}} .$$
As the Fubini numbers are also called ordered Bell numbers, we may call this nice result an ordered Dobiniski formula.

组合学代写

数学代写|组合学代写Combinatorics代考|The Fubini numbers count chains of subsets

$$U_1 \subset U_2 \subset \cdots \subset U_k$$

$$\emptyset \subset 1,2 \subset 1,2,3,4=S$$

$$\emptyset \subset 3,4 \subset 1,2,3,4=S$$

数学代写|组合学代写Combinatorics代考|The generating function of the Fubini numbers

$\lfloor\text { sum_{n }=0}^{\wedge}{$ infty $} _F_{-} n \backslash$ frac $\left{x^{\wedge} n\right}{n !}=\backslash$ sum_{n=0 $}^{\wedge}{$ infty $} \backslash$ left $\left(\right.$ sum_${k=1}^{\wedge} n k ! \backslash$ left: $\backslash$ begin ${$ array $\left.} \mid\right} n$

$$\sum_{n=0}^{\infty} F_n \frac{x^n}{n !}=\frac{1}{2-e^x}$$

$$\frac{1}{2-e^x}=\frac{1}{2} \frac{1}{1-\frac{e^x}{2}}=\frac{1}{2} \sum_{k=0}^{\infty}\left(\frac{e^x}{2}\right)^k=\frac{1}{2} \sum_{k=0}^{\infty} \frac{e^{k x}}{2^k}=\sum_{k=0}^{\infty} \frac{1}{2^{k+1}} \sum_{n=0}^{\infty} \frac{k^n x^n}{n !}=\sum_{n=0}^{\infty} \frac{x^n}{n !}$$

$$F_n=\sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}$$

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