组合数求和中由 p ( k ) = ∑ j = 0 m C j k − a Δ j p ( a ) {\displaystyle p(k)=\sum _{j=0}^{m}C_{j}^{k-a}\Delta ^{j}p(a)} 得到多项式的求和结果。[1]
设 p ( k ) = ∑ j = 0 m a j C j k − a = a 0 + a 1 C 1 k − a + a 2 C 2 k − a + ⋯ + a m C m k − a {\displaystyle p(k)=\sum _{j=0}^{m}a_{j}C_{j}^{k-a}=a_{0}+a_{1}C_{1}^{k-a}+a_{2}C_{2}^{k-a}+\dots +a_{m}C_{m}^{k-a}}
p ( a ) = a 0 {\displaystyle p(a)=a_{0}}
Δ C l k = C l k + 1 − C l k = C l − 1 k {\displaystyle \Delta C_{l}^{k}=C_{l}^{k+1}-C_{l}^{k}=C_{l-1}^{k}}
Δ j p ( k ) = a j + a j + 1 C 1 k − a + a j + 2 C 2 k − a + ⋯ + a m C m − j k − a {\displaystyle \Delta ^{j}p(k)=a_{j}+a_{j+1}C_{1}^{k-a}+a_{j+2}C_{2}^{k-a}+\dots +a_{m}C_{m-j}^{k-a}}
Δ j p ( a ) = a j {\displaystyle \Delta ^{j}p(a)=a_{j}}
p ( k ) = ∑ j = 0 m C j k − a Δ j p ( a ) {\displaystyle p(k)=\sum _{j=0}^{m}C_{j}^{k-a}\Delta ^{j}p(a)}
除此之外,还可以利用变换 ∑ k = 0 ∞ u k v k x k = ∑ k = 0 ∞ Δ k u 0 x k k ! d k d x k ( ∑ l = 0 ∞ v l x l ) {\displaystyle \sum _{k=0}^{\infty }u_{k}v_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})} 求和,可设 u k = p ( k ) {\displaystyle u_{k}=p(k)} 为多项式使和式的项数有限。( Δ m + 1 u k = 0 {\displaystyle \Delta ^{m+1}u_{k}=0} )[2]
d k d x k ( ∑ l = 0 ∞ v l x l ) = ∑ l = k ∞ l ( l − 1 ) … ( l − k + 1 ) v l x l − k = ∑ l = k ∞ l ! ( l − k ) ! v l x l − k {\displaystyle {\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{l=k}^{\infty }l(l-1)\dots (l-k+1)v_{l}x^{l-k}=\sum _{l=k}^{\infty }{\frac {l!}{(l-k)!}}v_{l}x^{l-k}}
∑ k = 0 ∞ Δ k u 0 x k k ! d k d x k ( ∑ l = 0 ∞ v l x l ) = ∑ k = 0 ∞ ∑ l = k ∞ C k l Δ k u 0 v l x l = ∑ l = 0 ∞ ∑ k = 0 l C k l Δ k u 0 v l x l = ∑ l = 0 ∞ v l x l ( ∑ k = 0 l C k l Δ k u 0 ) = ∑ l = 0 ∞ u l v l x l = ∑ k = 0 ∞ u k v k x k {\displaystyle \sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{k=0}^{\infty }\sum _{l=k}^{\infty }C_{k}^{l}\Delta ^{k}u_{0}v_{l}x^{l}=\sum _{l=0}^{\infty }\sum _{k=0}^{l}C_{k}^{l}\Delta ^{k}u_{0}v_{l}x^{l}=\sum _{l=0}^{\infty }v_{l}x^{l}(\sum _{k=0}^{l}C_{k}^{l}\Delta ^{k}u_{0})=\sum _{l=0}^{\infty }u_{l}v_{l}x^{l}=\sum _{k=0}^{\infty }u_{k}v_{k}x^{k}}
v k = 1 , d k d x k ( ∑ l = 0 ∞ v l x l ) = d k d x k ( ∑ l = 0 ∞ x l ) = d k d x k ( 1 1 − x ) = k ! ( 1 − x ) k + 1 {\displaystyle v_{k}=1,{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})={\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }x^{l})={\frac {d^{k}}{dx^{k}}}({\frac {1}{1-x}})={\frac {k!}{(1-x)^{k+1}}}}
∑ k = 0 ∞ u k x k = ∑ k = 0 ∞ Δ k u 0 x k k ! d k d x k ( ∑ l = 0 ∞ v l x l ) = ∑ k = 0 ∞ Δ k u 0 x k ( 1 − x ) k + 1 {\displaystyle \sum _{k=0}^{\infty }u_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{(1-x)^{k+1}}}}
v k = 1 k ! , d k d x k ( ∑ l = 0 ∞ v l x l ) = d k d x k ( ∑ l = 0 ∞ x l l ! ) = d k d x k e x = e x {\displaystyle v_{k}={\frac {1}{k!}},{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})={\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }{\frac {x^{l}}{l!}})={\frac {d^{k}}{dx^{k}}}e^{x}=e^{x}}
∑ k = 0 ∞ u k x k k ! = ∑ k = 0 ∞ Δ k u 0 x k k ! e x = e x ( ∑ k = 0 ∞ Δ k u 0 x k k ! ) {\displaystyle \sum _{k=0}^{\infty }{\frac {u_{k}x^{k}}{k!}}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}e^{x}=e^{x}(\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}})}