DEFINIÇÃO
Sabe-se que n r ‾ = n ( n − 1 ) ( n − 2 ) ⋅ ⋅ ⋅ ( n − r + 1 ) n^{\underline{r}} = n(n-1)(n-2)\cdot\cdot\cdot(n-r+1) n r = n ( n − 1 ) ( n − 2 ) ⋅ ⋅ ⋅ ( n − r + 1 )
n r ‾ = ∑ k = 0 r [ r k ] n k n^{\underline{r}} = \sum_{k=0}^{r} \begin{bmatrix}
r \\
k
\end{bmatrix}n^k n r = k = 0 ∑ r [ r k ] n k Ou seja, os números de Stirling de primeira espécie dão-nos os coeficientes da expansão de um polinómio fatorial.
DE NOTAR
O sinal dos números de Stirling de primeira espécie depende da paridade de n + k n+k n + k s ( n , k ) s(n,k) s ( n , k )
s ( n , k ) = ( − 1 ) n + k ∣ [ n k ] ∣ s(n,k) = (-1)^{n+k}\left|\begin{bmatrix}n \\ k \end{bmatrix}\right| s ( n , k ) = ( − 1 ) n + k [ n k ] Exemplo É possível relacionar os polinómios usuais com os fatoriais através da primeira espécie, como feito no exemplo seguinte:
n 3 ‾ = n ( n − 1 ) ( n − 2 ) = ( n 2 − n ) ( n − 2 ) = n 3 − 2 n 2 − n 2 + 2 n = 0 n 0 + 2 n 1 + ( − 3 ) n 2 + n 3 = [ 3 0 ] n 0 + [ 3 1 ] n 1 + [ 3 2 ] n 2 + [ 3 3 ] n 3 \begin{aligned}
n^{\underline{3}} & =n( n-1)( n-2)\\
& =\left( n^{2} -n\right)( n-2)\\
& =n^{3} -2n^{2} -n^{2} +2n\\
& =0n^{0} +2n^{1} +( -3) n^{2} +n^{3}\\
& =\begin{bmatrix}
3\\
0
\end{bmatrix} n^{0} +\begin{bmatrix}
3\\
1
\end{bmatrix} n^{1} +\begin{bmatrix}
3\\
2
\end{bmatrix} n^{2} +\begin{bmatrix}
3\\
3
\end{bmatrix} n^{3}
\end{aligned} n 3 = n ( n − 1 ) ( n − 2 ) = ( n 2 − n ) ( n − 2 ) = n 3 − 2 n 2 − n 2 + 2 n = 0 n 0 + 2 n 1 + ( − 3 ) n 2 + n 3 = [ 3 0 ] n 0 + [ 3 1 ] n 1 + [ 3 2 ] n 2 + [ 3 3 ] n 3 É possível calcular todos os números de Stirling de primeira espécie através da construção de uma tabela , de forma
semelhante à construção do Triângulo de Pascal.
Abaixo está a tabela das primeiras 6 linhas.
0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 ‾ 1 0 0 0 0 0 1 k 1 ‾ 0 1 0 0 0 0 2 k 2 ‾ 0 − 1 1 0 0 0 3 k 3 ‾ 0 2 − 3 1 0 0 4 k 4 ‾ 0 − 6 11 − 6 1 0 5 k 5 ‾ 0 24 − 50 35 − 10 1 \begin{array}{ c c c c c c c c }
& & 0 & 1 & 2 & 3 & 4 & 5\\
& & k^{0} & k^{1} & k^{2} & k^{3} & k^{4} & k^{5}\\
0 & k^{\underline{0}} & 1 & 0 & 0 & 0 & 0 & 0\\
1 & k^{\underline{1}} & 0 & 1 & 0 & 0 & 0 & 0\\
2 & k^{\underline{2}} & 0 & -1 & 1 & 0 & 0 & 0\\
3 & k^{\underline{3}} & 0 & 2 & -3 & 1 & 0 & 0\\
4 & k^{\underline{4}} & 0 & -6 & 11 & -6 & 1 & 0\\
5 & k^{\underline{5}} & 0 & 24 & -50 & 35 & -10 & 1
\end{array} 0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 1 0 0 0 0 0 1 k 1 0 1 − 1 2 − 6 24 2 k 2 0 0 1 − 3 11 − 50 3 k 3 0 0 0 1 − 6 35 4 k 4 0 0 0 0 1 − 10 5 k 5 0 0 0 0 0 1 A diagonal da tabela é composta por 1 1 1 0 0 0 0 0 0 1 1 1
Por fim, todos os outros números se geram através de um algoritmo , representado abaixo.
Para obter o número (d d d a a a b b b c c c d d d a a a b b b c c c
0 1 2 3 k 0 k 1 k 2 k 3 0 k 0 ‾ 1 0 0 0 c k 1 ‾ a b 0 0 2 k 2 ‾ 0 d 1 0 3 k 3 ‾ 0 2 − 3 1 \begin{array}{ c c c c c c }
& & 0 & 1 & 2 & 3\\
& & k^{0} & k^{1} & k^{2} & k^{3}\\
0 & k^{\underline{0}} & 1 & 0 & 0 & 0\\
\smartcolor{orange}{\mathbf{c}} & k^{\underline{1}} & \smartcolor{orange}{\mathbf{a}} & \smartcolor{orange}{\mathbf{b}} & 0 & 0\\
2 & k^{\underline{2}} & 0 & \smartcolor{orange}{\mathbf{d}} & 1 & 0\\
3 & k^{\underline{3}} & 0 & 2 & -3 & 1
\end{array} 0 c 2 3 k 0 k 1 k 2 k 3 0 k 0 1 a 0 0 1 k 1 0 b d 2 2 k 2 0 0 1 − 3 3 k 3 0 0 0 1 Os números de Stirling de segunda espécie servem para relacionar polinómios (usuais) com os polinómios fatoriais. São sempre positivos.
n 3 = n 3 ‾ + 3 n 2 ‾ + n 1 ‾ = 0 n 0 ‾ + 1 n 1 ‾ + 3 n 2 ‾ + 1 n 3 ‾ = { 3 0 } n 0 ‾ + { 3 1 } n 1 ‾ + { 3 2 } n 2 ‾ + { 3 3 } n 3 ‾ \begin{aligned}
n^{3} & =n^{\underline{3}} +3n^{\underline{2}} +n^{\underline{1}}\\
& =0n^{\underline{0}} +1n^{\underline{1}} +3n^{\underline{2}} +1n^{\underline{3}}\\
& =\begin{Bmatrix}
3\\
0
\end{Bmatrix} n^{\underline{0}} +\begin{Bmatrix}
3\\
1
\end{Bmatrix} n^{\underline{1}} +\begin{Bmatrix}
3\\
2
\end{Bmatrix} n^{\underline{2}} +\begin{Bmatrix}
3\\
3
\end{Bmatrix} n^{\underline{3}}
\end{aligned} n 3 = n 3 + 3 n 2 + n 1 = 0 n 0 + 1 n 1 + 3 n 2 + 1 n 3 = { 3 0 } n 0 + { 3 1 } n 1 + { 3 2 } n 2 + { 3 3 } n 3 o que facilita imenso a avaliação da soma de n 3 n^{3} n 3
DEFINIÇÃO
∑ k = 0 m n p = ∑ k = 0 m ( ∑ k = 0 p { p k } n k ‾ ) \sum_{k=0}^{m}n^{p} = \sum_{k=0}^{m}\left( \sum_{k=0}^{p} \begin{Bmatrix}p\\k\end{Bmatrix}n^{\underline{k}}\right) k = 0 ∑ m n p = k = 0 ∑ m ( k = 0 ∑ p { p k } n k ) onde { m k } \begin{Bmatrix}m\\k\end{Bmatrix} { m k }
Tal como com os números de primeira espécie, é possível calcular, através de uma tabela, os números de segunda espécie.
Abaixo está a tabela dos números de Stirling de segunda espécie:
0 1 2 3 4 5 k 0 ‾ k 1 ‾ k 2 ‾ k 3 ‾ k 4 ‾ k 5 ‾ 0 k 0 1 0 0 0 0 0 1 k 1 0 1 0 0 0 0 2 k 2 0 1 1 0 0 0 3 k 3 0 1 3 1 0 0 4 k 4 0 1 7 6 1 0 5 k 5 0 1 15 25 10 1 \begin{array}{ c c c c c c c c }
& & 0 & 1 & 2 & 3 & 4 & 5\\
& & k^{\underline{0}} & k^{\underline{1}} & k^{\underline{2}} & k^{\underline{3}} & k^{\underline{4}} & k^{\underline{5}}\\
0 & k^{0} & 1 & 0 & 0 & 0 & 0 & 0\\
1 & k^{1} & 0 & 1 & 0 & 0 & 0 & 0\\
2 & k^{2} & 0 & 1 & 1 & 0 & 0 & 0\\
3 & k^{3} & 0 & 1 & 3 & 1 & 0 & 0\\
4 & k^{4} & 0 & 1 & 7 & 6 & 1 & 0\\
5 & k^{5} & 0 & 1 & 15 & 25 & 10 & 1
\end{array} 0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 1 0 0 0 0 0 1 k 1 0 1 1 1 1 1 2 k 2 0 0 1 3 7 15 3 k 3 0 0 0 1 6 25 4 k 4 0 0 0 0 1 10 5 k 5 0 0 0 0 0 1 O algoritmo para encontrar um número de Stirling da segunda espécie consiste em multiplicar o que se encontra diretamente acima (a a a b b b c c c d d d
0 1 b 3 k 0 ‾ k 1 ‾ k 2 ‾ k 3 ‾ 0 k 0 1 0 0 0 1 k 1 0 1 0 0 2 k 2 0 c a 0 3 k 3 0 1 d 1 \begin{array}{ c c c c c c }
& & 0 & 1 & \smartcolor{orange}{\mathbf{b}} & 3\\
& & k^{\underline{0}} & k^{\underline{1}} & k^{\underline{2}} & k^{\underline{3}}\\
0 & k^{0} & 1 & 0 & 0 & 0\\
1 & k^{1} & 0 & 1 & 0 & 0\\
2 & k^{2} & 0 & \smartcolor{orange}{\mathbf{c}} & \smartcolor{orange}{\mathbf{a}} & 0\\
3 & k^{3} & 0 & 1 & \smartcolor{orange}{\mathbf{d}} & 1
\end{array} 0 1 2 3 k 0 k 1 k 2 k 3 0 k 0 1 0 0 0 1 k 1 0 1 c 1 b k 2 0 0 a d 3 k 3 0 0 0 1 Os números de Stirling de segunda espécie também podem ser calculados através do método de Horner (Ruffini).n 3 n^{3} n 3 n n n n − 1 n-1 n − 1 n − 2 n-2 n − 2
1 0 0 0 0 ↓ 0 0 0 1 0 0 0 1 0 0 1 ↓ 1 1 1 1 1 1 1 2 ↓ 2 1 3 n 3 = n ( n − 1 ( ( n − 2 ) + 3 ) + 1 ) + 0 n 3 = 1 n 3 ‾ + 3 n 2 ‾ + 1 n 1 ‾ + 0 n 0 ‾ \begin{array}{ c|c c c c }
& 1 & 0 & 0 & 0\\
0 & \downarrow & 0 & 0 & 0\\
\hline
& 1 & 0 & 0 & \smartcolor{blue}{0}
\end{array}\\
\\
\begin{array}{ c|c c c }
& 1 & 0 & 0\\
1 & \downarrow & 1 & 1\\
\hline
& 1 & 1 & \smartcolor{orange}{1}
\end{array}\\
\\
\begin{array}{ c|c c }
& 1 & 1\\
2 & \downarrow & 2\\
\hline
& 1 & \smartcolor{green}{3}
\end{array}\\
\\
n^{3} =n( n-1(( n-2) +3) +\smartcolor{orange}{1}) +\smartcolor{blue}{0}\\
n^{3} =1n^{\underline{3}} +\smartcolor{green}{3}n^{\underline{2}} +\smartcolor{orange}{1}n^{\underline{1}} +\smartcolor{blue}{0}n^{\underline{0}} 0 1 ↓ 1 0 0 0 0 0 0 0 0 0 1 1 ↓ 1 0 1 1 0 1 1 2 1 ↓ 1 1 2 3 n 3 = n ( n − 1 (( n − 2 ) + 3 ) + 1 ) + 0 n 3 = 1 n 3 + 3 n 2 + 1 n 1 + 0 n 0 Assim, é fácil encontrar a expressão para a soma de qualquer polinomial.
Tal como já foi visto anteriormente , sabemos que:
Δ u n r ‾ = u n r − 1 ‾ ( u n + 1 − u n − r + 1 ) \Delta u_n^{\underline r} = u_n^{\underline{r-1}}\left(u_{n+1}-u_{n-r+1}\right) Δ u n r = u n r − 1 ( u n + 1 − u n − r + 1 ) Trocando isto agora por uma sucessão aritmética, isto é, u n = a n + b u_n = an+b u n = an + b
Δ ( a n + b ) r ‾ = u n r − 1 ‾ ( a ( n + 1 ) + b − ( a ( n − r + 1 ) + b ) ) = a ⋅ r ⋅ u n r − 1 ‾ \begin{aligned}
\Delta (an+b)^{\underline r} &= u_n^{\underline{r-1}}\left(a(n+1)+b-(a(n-r+1)+b)\right)\\
&= a\cdot r\cdot u_n^{\underline{r-1}}
\end{aligned} Δ ( an + b ) r = u n r − 1 ( a ( n + 1 ) + b − ( a ( n − r + 1 ) + b ) ) = a ⋅ r ⋅ u n r − 1 Escrevendo em ordem a ( a n + b ) r ‾ (an+b)^{\underline{r}} ( an + b ) r
( a n + b ) r ‾ = Δ 1 a ( r + 1 ) ( a n + b ) r + 1 ‾ (an+b)^{\underline{r}} = \Delta\frac{1}{a(r+1)}(an+b)^{\underline{r+1}} ( an + b ) r = Δ a ( r + 1 ) 1 ( an + b ) r + 1 Exemplo 1 Através desta fórmula é possível achar a soma fechada para expressões do tipo:
∑ k = 0 n ( 2 k + 3 ) 2 ‾ \sum_{k=0}^{n}(2k+3)^{\underline{2}} k = 0 ∑ n ( 2 k + 3 ) 2 ∑ k = 0 n ( 2 k + 3 ) 2 ‾ = ∑ k = 0 n Δ ( 2 k + 3 ) 3 ‾ 3 × 2 = [ ( 2 k + 3 ) 3 ‾ 6 ] 0 n = ( 2 n + 3 ) ( 2 n + 1 ) ( 2 n − 1 ) 6 − 2 × 0 + 3 6 = ( 2 n + 3 ) ( 2 n + 1 ) ( 2 n − 1 ) − 3 6 = ( n + 1 ) ( 4 n 2 + 2 n − 3 ) 3 \begin{aligned}
\sum ^{n}_{k=0}( 2k+3)^{\underline{2}} & =\sum ^{n}_{k=0} \Delta \frac{(2k+3)^{\underline{3}}}{3\times 2}\\
& =\left[\frac{( 2k+3)^{\underline{3}}}{6}\right]^{n}_{0}\\
& =\frac{( 2n+3)( 2n+1)( 2n-1)}{6} -\frac{2\times 0+3}{6}\\
& =\frac{( 2n+3)( 2n+1)( 2n-1) -3}{6}\\
& =\frac{( n+1)\left( 4n^{2} +2n-3\right)}{3}
\end{aligned} k = 0 ∑ n ( 2 k + 3 ) 2 = k = 0 ∑ n Δ 3 × 2 ( 2 k + 3 ) 3 = [ 6 ( 2 k + 3 ) 3 ] 0 n = 6 ( 2 n + 3 ) ( 2 n + 1 ) ( 2 n − 1 ) − 6 2 × 0 + 3 = 6 ( 2 n + 3 ) ( 2 n + 1 ) ( 2 n − 1 ) − 3 = 3 ( n + 1 ) ( 4 n 2 + 2 n − 3 ) Exemplo 2 Um comboio de mercadorias viaja durante 5 h 5\ h 5 h 3 k m h − 1 3\ km\ h^{-1} 3 km h − 1 7 h 7\ h 7 h 5 k m h − 1 5\ km\ h^{-1} 5 km h − 1 9 h 9\ h 9 h 7 k m h − 1 7\ km\ h^{−1} 7 km h − 1 25 h 25\ h 25 h 23 k m h − 1 23\ km\ h^{−1} 23 km h − 1
∑ k = 0 n ( 2 k + 5 ) 3 ‾ = [ 1 4 × 2 ( 2 k + 5 ) 4 ‾ ] 0 n + 1 = [ 1 8 ( 2 k + 5 ) ( 2 k + 3 ) ( 2 k + 1 ) ( 2 k − 1 ) ] 0 n + 1 = ( 2 n + 7 ) ( 2 n + 5 ) ( 2 n + 3 ) ( 2 n + 1 ) + 15 8 . \begin{aligned}
\sum_{k=0}^{n}(2 k+5)^{\underline 3} &=\left[\frac{1}{4 \times 2}(2 k+5)^{\underline 4}\right]_{0}^{n+1} \\
&=\left[\frac{1}{8}(2 k+5)(2 k+3)(2 k+1)(2 k-1)\right]_{0}^{n+1} \\
&=\frac{(2n+7)(2 n+5)(2 n+3)(2 n+1)+15}{8} .
\end{aligned} k = 0 ∑ n ( 2 k + 5 ) 3 = [ 4 × 2 1 ( 2 k + 5 ) 4 ] 0 n + 1 = [ 8 1 ( 2 k + 5 ) ( 2 k + 3 ) ( 2 k + 1 ) ( 2 k − 1 ) ] 0 n + 1 = 8 ( 2 n + 7 ) ( 2 n + 5 ) ( 2 n + 3 ) ( 2 n + 1 ) + 15 . Como n = 10 n = 10 n = 10 40755 k m 40 755\ km 40755 km
DEFINIÇÃO
Define-se um polinomial fatorial com valor negativo ( n − r ‾ ) (n^{\underline{-r}}) ( n − r )
u n − r ‾ = 1 u n + 1 ⋅ ⋅ ⋅ u n + r u_{n}^{\underline{-r}} = \frac{1}{u_{n+1} \cdot \cdot \cdot u_{n+r}} u n − r = u n + 1 ⋅ ⋅ ⋅ u n + r 1 Como mnemónica, escreve-se o último termo no denominador primeiro porque coincide com o expoente fatorial e depois conta-se o número de fatores (existem r r r
Exemplos u n − 2 ‾ = 1 ( n + 1 ) ( n + 2 ) u_{n}^{\underline{-2}} = \frac{1}{(n+1)(n+2)} u n − 2 = ( n + 1 ) ( n + 2 ) 1 1 ( n + 2 ) ( n + 3 ) ( n + 4 ) = ( n + 1 ) − 3 ‾ \frac{1}{(n+2)(n+3)(n+4)} = (n+1)^{\underline{-3}} ( n + 2 ) ( n + 3 ) ( n + 4 ) 1 = ( n + 1 ) − 3 1 ( 2 n + 1 ) ( 2 n + 3 ) ( 2 n + 5 ) = 1 ( 2 ( n + 1 ) − 1 ) ( 2 ( n + 2 ) − 1 ) ( 2 ( n + 3 ) − 1 ) = ( 2 n − 1 ) − 3 ‾ \frac{1}{(2n+1)(2n+3)(2n+5)} = \frac{1}{(2(n+1)-1)(2(n+2)-1)(2(n+3)-1)} = (2n-1)^{\underline{-3}} ( 2 n + 1 ) ( 2 n + 3 ) ( 2 n + 5 ) 1 = ( 2 ( n + 1 ) − 1 ) ( 2 ( n + 2 ) − 1 ) ( 2 ( n + 3 ) − 1 ) 1 = ( 2 n − 1 ) − 3 A fórmula anteriormente conhecida para polinómios elevados a fatoriais positivos pode-se prolongar, funcionando da mesma maneira para os de expoente negativo, ou seja:
DEFINIÇÃO
∑ k = 0 n − 1 k r ‾ = [ k r + 1 ‾ r + 1 ] 0 n , r ∈ Z \ { − 1 } \sum_{k=0}^{n-1}k^{\underline{r}} = \left[\frac{k^{\underline{r+1}}}{r+1}\right]_{0}^{n}\quad,\quad r \in \mathbb{Z}\backslash \{-1\} k = 0 ∑ n − 1 k r = [ r + 1 k r + 1 ] 0 n , r ∈ Z \ { − 1 } ∑ k = 0 n − 1 u k r ‾ = [ u k r + 1 ‾ a ( r + 1 ) ] 0 n , r ∈ Z \ { − 1 } , u k = a k + b \sum_{k=0}^{n-1}u_{k}^{\underline{r}} = \left[\frac{u_{k}^{\underline{r+1}}}{a(r+1)}\right]_{0}^{n}\quad,\quad r \in \mathbb{Z}\backslash \{-1\}\quad,\quad u_k=ak+b k = 0 ∑ n − 1 u k r = [ a ( r + 1 ) u k r + 1 ] 0 n , r ∈ Z \ { − 1 } , u k = ak + b Exemplo ∑ k = 0 n 1 k 2 + 3 k + 2 = ∑ k = 0 n 1 ( k + 1 ) ( k + 2 ) = ∑ k = 0 n k − 2 ‾ = [ 1 − 2 + 1 k − 2 + 1 ‾ ] 0 n + 1 = [ − 1 k + 1 ] 0 n + 1 = − ( 1 n + 2 − 1 ) = n + 1 n + 2 \begin{aligned}
\sum ^{n}_{k=0}\frac{1}{k^{2} +3k+2} & =\sum ^{n}_{k=0}\frac{1}{( k+1)( k+2)}\\
& =\sum ^{n}_{k=0} k^{\underline{-2}}\\
& =\left[\frac{1}{-2+1} k^{\underline{-2+1}}\right]^{n+1}_{0}\\
& =\left[ -\frac{1}{k+1}\right]^{n+1}_{0}\\
& =-\left(\frac{1}{n+2} -1\right)\\
& =\frac{n+1}{n+2}
\end{aligned} k = 0 ∑ n k 2 + 3 k + 2 1 = k = 0 ∑ n ( k + 1 ) ( k + 2 ) 1 = k = 0 ∑ n k − 2 = [ − 2 + 1 1 k − 2 + 1 ] 0 n + 1 = [ − k + 1 1 ] 0 n + 1 = − ( n + 2 1 − 1 ) = n + 2 n + 1 DEFINIÇÃO
Δ n − r ‾ = − r ⋅ n − r − 1 ‾ , r > 0 \Delta n^{\underline{-r}}=-r\cdot n^{\underline{-r-1}}\quad,\quad r>0 Δ n − r = − r ⋅ n − r − 1 , r > 0 Demonstração Δ n − r ‾ = ( n + 1 ) − r ‾ − n − r ‾ = 1 ( n + 2 ) … ( n + r ) ( n + r + 1 ) − 1 ( n + 1 ) ( n + 2 ) … ( n + r ) = 1 ( n + 2 ) … ( n + r ) ( 1 n + r + 1 − 1 n + 1 ) = 1 ( n + 2 ) … ( n + r ) × ( n + 1 ) − ( n + r + 1 ) ( n + 1 ) ( n + r + 1 ) = − r ( n + 1 ) ( n + 2 ) … ( n + r ) ( n + r + 1 ) = − r × n − ( r + 1 ) ‾ = − r × n − r − 1 ‾ \begin{aligned}
\Delta n^{\underline{-r}} & =( n+1)^{\underline{-r}} -n^{\underline{-r}}\\
& =\frac{1}{( n+2) \dotsc ( n+r)( n+r+1)} -\frac{1}{( n+1)( n+2) \dotsc ( n+r)}\\
& =\frac{1}{( n+2) \dotsc ( n+r)}\left(\frac{1}{n+r+1} -\frac{1}{n+1}\right)\\
& =\frac{1}{( n+2) \dotsc ( n+r)} \times \frac{( n+1) -( n+r+1)}{( n+1)( n+r+1)}\\
& =\frac{-r}{( n+1)( n+2) \dotsc ( n+r)( n+r+1)}\\
& =-r\times n^{\underline{-( r+1)}}\\
& =-r\times n^{\underline{-r-1}}
\end{aligned} Δ n − r = ( n + 1 ) − r − n − r = ( n + 2 ) … ( n + r ) ( n + r + 1 ) 1 − ( n + 1 ) ( n + 2 ) … ( n + r ) 1 = ( n + 2 ) … ( n + r ) 1 ( n + r + 1 1 − n + 1 1 ) = ( n + 2 ) … ( n + r ) 1 × ( n + 1 ) ( n + r + 1 ) ( n + 1 ) − ( n + r + 1 ) = ( n + 1 ) ( n + 2 ) … ( n + r ) ( n + r + 1 ) − r = − r × n − ( r + 1 ) = − r × n − r − 1 Podemos também obter a derivada em função do polinómio fatorial, tal como feito anteriormente:
n − r ‾ = 1 − r + 1 Δ n − r + 1 ‾ , r ≠ 1 n^{\underline{-r}}=\frac{1}{-r+1} \Delta n^{\underline{-r+1}}\quad,\quad r\ne 1 n − r = − r + 1 1 Δ n − r + 1 , r = 1