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## Thursday, December 30, 2010

### Stirling numbers in Discrete Calculus

#### Definition

Stirling numbers of the second kind represent the number of k-partitions of an n-set and are recursively defined as $\left\{ 0,0 \right\} = 1$, and $\left\{ n,k \right\} = \left\{ n-1,k-1 \right\} + k \cdot \left\{ n-1,k \right\}$. They are used in the Discrete Calculus to convert powers to factorial powers, i.e. $n^2 = n^{\underline{1}} + n^{\underline{2}}$.

The matrix below shows the Stirling numbers for $n=0$, to $n=5$.
$\left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 1 & 0 & 0 \\ 0 & 1 & 7 & 6 & 1 & 0 \\ 0 & 1 & 15 & 25 & 10 & 1 \end{array} \right)$

#### Example

Calculate $\sum_{k=1}^{n} k^5$.

$\Sigma \Delta k^5$
$=\Sigma ((n+1)^{\underline{1}} + 15(n+1)^{\underline{2}} + 25(n+1)^{\underline{3}} + 10(n+1)^{\underline{4}} + (n+1)^{\underline{5}} )$
$=\frac{1}{2}(n+1)^{\underline{2}} + 5(n+1)^{\underline{3}} + \frac{25}{4}(n+1)^{\underline{4}} + 2(n+1)^{\underline{5}} + \frac{1}{6}(n+1)^{\underline{6}}$
$=-\frac{n^2}{12}+\frac{5 n^4}{12}+\frac{n^5}{2}+\frac{n^6}{6}$
$=\frac{1}{12} n^2 (1+n)^2 \left(-1+2 n+2 n^2\right)$

\begin{array}{lll}
\underline{n} & \underline{n^5} & \underline{\frac{1}{12} n^2 (1+n)^2 \left(-1+2 n+2 n^2\right)} \\
1 & 1 & 1 \\
2 & 32 & 33 \\
3 & 243 & 276 \\
4 & 1024 & 1300 \\
5 & 3125 & 4425
\end{array}

#### 1 comment:

1. I am here to share some simple information about precalculus that is,It is a part of mathematics which prepares a student to learn and clear the basics of calculus and by learning this its simple for students to deal with calculus,It is just like a first step towards calculus.

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