# 代數/本書課文/求和/差分算子求逆法

${\displaystyle S(n)=\sum _{k=1}^{n}p(k)\Rightarrow \Delta S(n)=p(n+1)\Rightarrow S(n)=\Delta ^{-1}p(n+1)}$

## 多項式公比求和

${\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}$

${\displaystyle f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)}$[1]

 證明多項式公比求和有表達式：${\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}$其中${\displaystyle deg(f)=deg(p)=m}$ ${\displaystyle \sum _{k=1}^{n}q^{k-1}={\frac {q^{n}-1}{q-1}}}$ 設${\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}+C}$ 當${\displaystyle m=deg(p)=1}$時，${\displaystyle deg(f)=m=1}$ 兩邊逐項求導：${\displaystyle \sum _{k=1}^{n}(k-1)p(k)q^{k-2}=[{\frac {df(n)}{dq}}+nf(n)q^{-1}]q^{n}+{\frac {dC}{dq}}}$ 可見兩邊多項式的階數同時增加1 ${\displaystyle \sum _{k=0}^{n}p(k)q^{k-1}=f(n)q^{n}+C+p(0)q^{-1}}$ 代入${\displaystyle n=0}$得到${\displaystyle p(0)q^{-1}=f(0)+C+p(0)q^{-1}\Rightarrow C=-f(0)}$
 證明：${\displaystyle f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)}$ 其中${\displaystyle \Delta p(n)=p(n+1)-p(n),m=deg(p)}$ 對${\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}$兩邊差分 ${\displaystyle p(n+1)q^{n}=f(n+1)q^{n+1}-f(n)q^{n}}$ ${\displaystyle p(n+1)=qf(n+1)-f(n)}$ ${\displaystyle p(n+1)=p(n)+\Delta p(n)=(I+\Delta )p(n)}$ ${\displaystyle (I+\Delta )p(n)=q(I+\Delta )f(n)-f(n)=[(q-1)I+q\Delta ]f(n)}$ ${\displaystyle f(n)={\frac {I+\Delta }{(q-1)I+q\Delta }}p(n)={\frac {1}{(q-1)I+q\Delta }}p(n+1)}$ 由於f,p都是m次多項式，以下可進行滿足條件${\displaystyle \Delta ^{m+1}=0}$的求逆運算： ${\displaystyle {\frac {I+\Delta }{(q-1)I+q\Delta }}={\frac {I+\Delta }{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}}$ ${\displaystyle ={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k+1}]={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}]}$ ${\displaystyle ={\frac {1}{q-1}}I+{\frac {1}{q-1}}\sum _{k=1}^{m}[({\frac {-q}{q-1}})^{k}+({\frac {-q}{q-1}})^{k-1}]\Delta ^{k}={\frac {1}{q-1}}I-{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}}$ ${\displaystyle ={\frac {1}{q-1}}I+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}}$ 最後把算子表達式作用到p上得證。
 例子：差比數列求和 計算各階差分：${\displaystyle p(n)=a+(n-1)d,\Delta p(n)=d}$ ${\displaystyle \displaystyle f(n)={\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}}$ ${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}$
 例子：求${\displaystyle \displaystyle \sum _{k=1}^{n}(A+Bk+Ck^{2})q^{k-1}}$ 計算各階差分：${\displaystyle p(n)=A+Bn+Cn^{2},\Delta p(n)=B+C(2n+1)=B+C+2Cn,\Delta ^{2}p(n)=2C}$ ${\displaystyle \displaystyle f(n)={\frac {A+Bn+Cn^{2}}{q-1}}-{\frac {B+C+2Cn}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q}$ ${\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=[{\frac {A+Bn+Cn^{2}}{q-1}}-{\frac {B+C+2Cn}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]q^{n}-[{\frac {A}{q-1}}-{\frac {B+C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]}$

${\displaystyle \sum _{k=1}^{\infty }p(k)q^{k-1}={\frac {p(0)}{1-q}}+{\frac {1}{(1-q)^{2}}}\sum _{k=1}^{m}{\frac {q^{k-1}}{(1-q)^{k-1}}}\Delta ^{k}(p(0)),~|q|<1}$

 例子：求${\displaystyle \displaystyle \sum _{k=1}^{\infty }(A+Bk+Ck^{2})q^{k-1},~|q|<1}$ ${\displaystyle \displaystyle \sum _{k=1}^{\infty }(A+Bk+Ck^{2})q^{k-1}={\frac {A}{1-q}}+{\frac {B+C}{(1-q)^{2}}}+{\frac {2C}{(1-q)^{3}}}q}$

## 參考資料

1. 黄嘉威. 方幂和及其推广和式. 数学学习与研究. 2016, (7).