代数/本书课文/求和/差分变换

组合数求和中由$p(k)=\sum _{j=0}^{m}C_{j}^{k-a}\Delta ^{j}p(a)$得到多项式的求和结果。

证明：$p(k)=\sum _{j=0}^{m}C_{j}^{k-a}\Delta ^{j}p(a)$ 设$p(k)=\sum _{j=0}^{m}a_{j}C_{j}^{k-a}=a_{0}+a_{1}C_{1}^{k-a}+a_{2}C_{2}^{k-a}+\dots +a_{m}C_{m}^{k-a}$ $p(a)=a_{0}$ $\Delta C_{l}^{k}=C_{l}^{k+1}-C_{l}^{k}=C_{l-1}^{k}$ $\Delta ^{j}p(k)=a_{j}+a_{j+1}C_{1}^{k-a}+a_{j+2}C_{2}^{k-a}+\dots +a_{m}C_{m-j}^{k-a}$ $\Delta ^{j}p(a)=a_{j}$ $p(k)=\sum _{j=0}^{m}C_{j}^{k-a}\Delta ^{j}p(a)$

除此之外，还可以利用变换$\sum _{k=0}^{\infty }u_{k}v_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})$求和，可设$u_{k}=p(k)$为多项式使和式的项数有限。（$\Delta ^{m+1}u_{k}=0$）

证明：$\sum _{k=0}^{\infty }u_{k}v_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})$ ${\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{l=k}^{\infty }l(l-1)\dots (l-k+1)v_{l}x^{l-k}=\sum _{l=k}^{\infty }{\frac {l!}{(l-k)!}}v_{l}x^{l-k}$ $\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{k=0}^{\infty }\sum _{l=k}^{\infty }C_{k}^{l}\Delta ^{k}u_{0}v_{l}x^{l}=\sum _{l=0}^{\infty }\sum _{k=0}^{l}C_{k}^{l}\Delta ^{k}u_{0}v_{l}x^{l}=\sum _{l=0}^{\infty }v_{l}x^{l}(\sum _{k=0}^{l}C_{k}^{l}\Delta ^{k}u_{0})=\sum _{l=0}^{\infty }u_{l}v_{l}x^{l}=\sum _{k=0}^{\infty }u_{k}v_{k}x^{k}$

例子：$\sum _{k=0}^{\infty }u_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{(1-x)^{k+1}}}$ $v_{k}=1,{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})={\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }x^{l})={\frac {d^{k}}{dx^{k}}}({\frac {1}{1-x}})={\frac {k!}{(1-x)^{k+1}}}$ $\sum _{k=0}^{\infty }u_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{(1-x)^{k+1}}}$

例子：$\sum _{k=0}^{\infty }{\frac {u_{k}x^{k}}{k!}}=e^{x}(\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}})$ $v_{k}={\frac {1}{k!}},{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})={\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }{\frac {x^{l}}{l!}})={\frac {d^{k}}{dx^{k}}}e^{x}=e^{x}$ $\sum _{k=0}^{\infty }{\frac {u_{k}x^{k}}{k!}}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}e^{x}=e^{x}(\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}})$