代數/本書課文/求和/差分變換

組合數求和中由$p(k)=\sum _{j=0}^m{C}_j^{k-a}{\mathrm{\Delta }}^{j}p(a)$得到多項式的求和結果。

Template:Robox 設$p(k)=\sum _{j=0}^m{a}_j{C}_j^{k-a}=a_0+a_1{C}_1^{k-a}+a_2{C}_2^{k-a}+\dots +a_m{C}_m^{k-a}$

$p(a)=a_0$

$\mathrm{\Delta }{C}_l^k=C_l^{k+1}-C_l^k=C_{l-1}^k$

${\mathrm{\Delta }}^{j}p(k)=a_j+a_{j+1}{C}_1^{k-a}+a_{j+2}{C}_2^{k-a}+\dots +a_m{C}_{m-j}^{k-a}$

${\mathrm{\Delta }}^{j}p(a)=a_j$

$p(k)=\sum _{j=0}^m{C}_j^{k-a}{\mathrm{\Delta }}^{j}p(a)$

除此之外，還可以利用變換$\sum _{k=0}^{\mathrm{\infty }}{u}_k{v}_k{x}^k=\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{k!}\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)$求和，可設$u_k=p(k)$為多項式使和式的項數有限。（${\mathrm{\Delta }}^{m+1}{u}_k=0$）

$\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)=\sum _{l=k}^{\mathrm{\infty }}l(l-1)\dots (l-k+1)v_l{x}^{l-k}=\sum _{l=k}^{\mathrm{\infty }}\frac{l!}{(l-k)!}{v}_l{x}^{l-k}$

$\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{k!}\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)=\sum _{k=0}^{\mathrm{\infty }}\sum _{l=k}^{\mathrm{\infty }}{C}_k^l{\mathrm{\Delta }}^k{u}_0{v}_l{x}^l=\sum _{l=0}^{\mathrm{\infty }}\sum _{k=0}^l{C}_k^l{\mathrm{\Delta }}^k{u}_0{v}_l{x}^l=\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l(\sum _{k=0}^l{C}_k^l{\mathrm{\Delta }}^k{u}_0)=\sum _{l=0}^{\mathrm{\infty }}{u}_l{v}_l{x}^l=\sum _{k=0}^{\mathrm{\infty }}{u}_k{v}_k{x}^k$

$v_k=1,\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)=\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{x}^l)=\frac{d^k}{d{x}^k}(\frac{1}{1-x})=\frac{k!}{(1-x{)}^{k+1}}$

$\sum _{k=0}^{\mathrm{\infty }}{u}_k{x}^k=\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{k!}\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)=\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{(1-x{)}^{k+1}}$

$v_k=\frac{1}{k!},\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}{v}_l{x}^l)=\frac{d^k}{d{x}^k}(\sum _{l=0}^{\mathrm{\infty }}\frac{x^l}{l!})=\frac{d^k}{d{x}^k}{e}^x=e^x$

$\sum _{k=0}^{\mathrm{\infty }}\frac{u_k{x}^k}{k!}=\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{k!}{e}^x=e^x(\sum _{k=0}^{\mathrm{\infty }}\frac{{\mathrm{\Delta }}^k{u}_0{x}^k}{k!})$ Template:Robox/Close