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## Proof: (related to "Recursive Formula for the Stirling Numbers of the Second Kind")

• Let $n\ge 1$ be a natural number
• By definition of the Stirling numbers of the second kind \begin{align}x^{n}&=\sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\} x^\underline{r}\tag{1},\\x^{n+1}&=\sum_{r=1}^{n+1}\left\{\begin{array}{c}n+1\\r\end{array}\right\} x^\underline{r}\tag{2}.\end{align}
• From the exponentiation law $x^nx=x^{n+1}$ we can conclude
\begin{align}\sum_{r=1}^{n+1}\left\{\begin{array}{c}n+1\\r\end{array}\right\} x^\underline{r}&=x\sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\} x^\underline{r}\tag{3}.\end{align}
• Moreover, by definition of falling factorial powers \begin{align}x^\underline{q}(x-q)^\underline{p}&=x(x-1)(x-2)\cdots(x-q+1)\cdot\nonumber\\&\quad\cdot (x-q)(x-q-1)\cdots(x-q-(p-q)+1)\nonumber\\&=x(x-1)(x-2)\cdots(x-p+1)\nonumber\\&=x^\underline{p}\tag{4}\end{align} for any integers $p,q\in\mathbb Z.$
• For $p=r+1$ and $q=r,$ we get \begin{align}x^\underline{r+1}&=x^\underline{r}(x-r)^\underline{1}=xx^\underline{r}-rx^\underline{r}\tag{5}.\end{align}
• Result $(5)$ allows us to write $(3)$ as \begin{align}\underbrace{\sum_{r=1}^{n+1}\left\{\begin{array}{c}n+1\\r\end{array}\right\} x^r}_{=:A}&=\underbrace{\sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\} x^{r+1}}_{=:B}+\underbrace{r\sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\} x^r}_{=:C}.\tag{6} \end{align}
• Now, \begin{align}B&=\sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\} x^{r+1}=\sum_{k=2}^{n+1}\left\{\begin{array}{c}n\\k-1\end{array}\right\} x^{k}=\sum_{k=2}^n\left\{\begin{array}{c}n\\k-1\end{array}\right\} x^{k} + \left\{\begin{array}{c}n\\n\end{array}\right\} x^{n+1},\tag{7}\\A&=\left\{\begin{array}{c}n+1\\1\end{array}\right\}x+\sum_{k=2}^n\left\{\begin{array}{c}n+1\\k\end{array}\right\} x^{k}+\left\{\begin{array}{c}n+1\\n+1\end{array}\right\} x^{n+1}\tag{8},\\C&=\left\{\begin{array}{c}n\\1\end{array}\right\}x+\sum_{k=2}^nr\left\{\begin{array}{c}n\\k\end{array}\right\} x^{k}.\tag{9}\end{align}
• From $(6)$ it follows $0=A-C-B,$ and thus \begin{align}0&=\left(\left\{\begin{array}{c}n+1\\1\end{array}\right\}-\left\{\begin{array}{c}n\\1\end{array}\right\}\right)x\nonumber\\&\quad+\sum_{k=2}^n\left(\left\{\begin{array}{c}n+1\\k\end{array}\right\}-k\left\{\begin{array}{c}n\\k\end{array}\right\}-\left\{\begin{array}{c}n\\k-1\end{array}\right\}\right)x^k\nonumber\\&\quad+\left(\left\{\begin{array}{c}n+1\\k+1\end{array}\right\}-\left\{\begin{array}{c}n\\n\end{array}\right\}\right)x^{n+1}.\tag{10}\end{align}
• Sufficient for $(10)$ are the conditions \begin{align}0&=\left\{\begin{array}{c}n+1\\1\end{array}\right\}-\left\{\begin{array}{c}n\\1\end{array}\right\},\tag{11}\\0&=\left\{\begin{array}{c}n+1\\k\end{array}\right\}-k\left\{\begin{array}{c}n\\k\end{array}\right\}-\left\{\begin{array}{c}n\\k-1\end{array}\right\},\quad k=2,3,\ldots,n\tag{12}\\0&=\left\{\begin{array}{c}n+1\\k+1\end{array}\right\}-\left\{\begin{array}{c}n\\n\end{array}\right\}\tag{13}.\end{align}
• Given the conventions $\left\{\begin{array}{c}n\\0\end{array}\right\}:=0$ and $\left\{\begin{array}{c}n\\k\end{array}\right\}:=0$ for $k > n,$ we can re-write those as \begin{align}\left\{\begin{array}{c}n+1\\1\end{array}\right\}&=\left\{\begin{array}{c}n\\0\end{array}\right\}+1\left\{\begin{array}{c}n\\1\end{array}\right\},\tag{14}\\ \left\{\begin{array}{c}n+1\\k\end{array}\right\}&=\left\{\begin{array}{c}n\\k-1\end{array}\right\}+k\left\{\begin{array}{c}n\\k\end{array}\right\},\quad k=2,3,\ldots,n\tag{15}\\\left\{\begin{array}{c}n+1\\k+1\end{array}\right\}&=\left\{\begin{array}{c}n\\n\end{array}\right\}+(n+1)\cdot\left\{\begin{array}{c}n\\n+1\end{array}\right\}.\tag{16} \end{align}
• $(14),$ $(15)$ and $(16)$ can be written as a single recurrence relation \begin{align} \left\{\begin{array}{c}n+1\\r\end{array}\right\}&=\left\{\begin{array}{c}n\\r-1\end{array}\right\}+r\cdot \left\{\begin{array}{c}n\\r\end{array}\right\},\nonumber\\ \left\{\begin{array}{c}n\\n\end{array}\right\}&:=1,\quad \text{ for }n\ge 1\nonumber\\ \left\{\begin{array}{c}n\\r\end{array}\right\}&:=0,\quad \text{ for }r=0 < n\text{ or }r > n.\tag{17} \end{align}
q.e.d

| | | | created: 2020-04-08 10:22:55 | modified: 2020-04-08 10:22:55 | by: | references: [8404]