In the calculus of finite differences, the indefinite sum operator (also known as the antidifference operator), denoted by ∑ x {\textstyle \sum _{x}} or Δ − 1 } , is the linear operator that is the inverse of the forward difference operator, Δ : Δ f ( x )
Dalam matematika, operator jumlah taktentu atau operator antiselisih, dilambangkan sebagai ∑ x } atau Δ − 1 } , adalah operator linear, yang kebalikan dari operator selisih (atau selisih tentu) Δ .
If ∑ x f ( x )F ( x ) f(x)=F(x)} , then F satisfies the functional equation F ( x + 1 ) − F ( x )f ( x ) .
Lebih eksplisit lagi, jika ∑ x f ( x )F ( x ) f(x)=F(x)} , kemudian F ( x + 1 ) − F ( x )
The solution F ( x ) is not unique: if F ( x ) is one solution, then for any 1‑periodic function C ( x ) (i.e., C ( x + 1 )
f ( x ) Jika F ( x ) adalah solusi untuk persamaan fungsional ini untuk fungsi f ( x ) , maka F ( x ) + C ( x ) untuk setiap fungsi periodik C ( x ) dengan periode 1.
In this case, the analytic continuation, F ( x ) , for the sum is a solution of ∇ − 1 f ( x ) f(x)} .
Lebih eksplisit lagi, jika ∑ x f ( x )F ( x ) f(x)=F(x)} , kemudian F ( x + 1 ) − F ( x )
f ( x ) , Which follows from the discrete counterpart: ∑ k1 x f ( k ) − ∑ k1 x − 1 f ( k )f ( x ) .
Lebih eksplisit lagi, jika ∑ x f ( x )F ( x ) f(x)=F(x)} , kemudian F ( x + 1 ) − F ( x )
− F ( x )f ( x + 1 ) .
− f ( n + x ) ) + C .
^{b}f(k)=\Delta ^{-1}f(b+1)-\Delta ^{-1}f(a).}
Δ − 1 f ( b + 1 ) − Δ − 1 f ( a ) ^{b}f(k)=\Delta ^{-1}f(b+1)-\Delta ^{-1}f(a)}
Γ ( x − n + 1 )x^{\underline {n}}=x(x-1)(x-2)\cdots (x-n+1)={\frac {\Gamma \left(x+1\right)}{\Gamma \left(x-n+1\right)}}} plays the role of x n } , and the forward difference operator satisfies Δ ( x ) nn ( x ) n − 1 .
Γ ( x − i a ) ) + C \arctan ax={\frac {i}{2}}\ln \left({\frac {\Gamma (x+{\frac {i}{a}})}{\Gamma (x-{\frac {i}{a}})}}\right)+C} ∑ x ψ ( x )( x − 1 ) ψ ( x )
Γ ( x − n + 1 )Γ ( x + 1 ) ( n + 1 ) Γ ( x − n ) + C ( x ) , n ≠ − 1. {\frac {\Gamma \left(x+1\right)}{\Gamma \left(x-n+1\right)}}={\frac {\Gamma \left(x+1\right)}{\left(n+1\right)\Gamma \left(x-n\right)}}+C(x),\quad n\neq -1.}
Γ ( x − i a ) ) + C \arctan ax={\frac {i}{2}}\ln \left({\frac {\Gamma (x+{\frac {i}{a}})}{\Gamma (x-{\frac {i}{a}})}}\right)+C} ∑ x ψ ( x )( x − 1 ) ψ ( x )
u ( x ) Δ v ( x ) + v ( x + 1 ) Δ u ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
Δ ( u ( x ) v ( x ) ) − v ( x + 1 ) Δ u ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
− ∑ x v ( x + 1 ) Δ u ( x ) + C ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
− ∑ x Δ f ( x ) Δ g ( x ) + C ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
f(x)\Delta g(x)+\sum _{x}g(x)\Delta f(x)=f(x)g(x)-\sum _{x}\Delta f(x)\Delta g(x)+C(x).} Definite summation by parts.
− ∑ x Δ f ( x ) Δ g ( x ) f(x)\Delta g(x)+\sum _{x}g(x)\Delta f(x)=f(x)g(x)-\sum _{x}\Delta f(x)\Delta g(x)} Penjumlahan tentu berdasarkan bagian, yaitu: ∑ ia b f ( i )
[ u ( b + 1 ) v ( b + 1 )
f ( b + 1 ) g ( b + 1 ) − f ( a ) g ( a )
^{b}u(k)\Delta v(k)={\bigl [}u(b+1)v(b+1)-u(a)v(a){\bigr ]}-\sum _{k=a}^{b}v(k+1)\Delta u(k).}
Δ − 1 f ( b + 1 ) − Δ − 1 f ( a ) ^{b}f(k)=\Delta ^{-1}f(b+1)-\Delta ^{-1}f(a)}
If F 1 ( x ) (x)} is a particular solution, then for any function C ( x ) satisfying C ( x + 1 )C ( x ) (i.e., any 1-periodic function), the function F 2 ( x )F 1 ( x ) + C ( x ) (x)=F_{1}(x)+C(x)} is also a solution.
f ( x ) Jika F ( x ) adalah solusi untuk persamaan fungsional ini untuk fungsi f ( x ) , maka F ( x ) + C ( x ) untuk setiap fungsi periodik C ( x ) dengan periode 1.
f ( x ) F ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
ln ⁡ ( ∏ x exp ⁡ ( f ( x ) ) ) + C ( x ) .
+ ∑ x g ( x ) Δ f ( x )f ( x )
f(x)=\ln \left(\prod _{x}\exp(f(x))\right)+C(x).}
− x + C \psi (x)=(x-1)\psi (x)-x+C} ∑ x Γ ( x )
∑ k1 ∞ ( x k ) Δ k − 1 f ( 0 ) + C ( x )
∑ x f ( x )∑ k1 ∞ ( x k ) Δ k − 1 [ f ] ( 0 ) + C∑ k1 ∞ Δ k − 1 [ f ] ( 0 )
∑ k1 ∞ Δ k − 1 f ( 0 ) k ! ( x ) k + C ( x ) .
∑ x f ( x )∑ k1 ∞ ( x k ) Δ k − 1 [ f ] ( 0 ) + C∑ k1 ∞ Δ k − 1 [ f ] ( 0 )
f(x)=\sum _{k=1}^{\infty }{\binom {x}{k}}\Delta ^{k-1}f\left(0\right)+C(x)=\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}f(0)}{k!}}(x)_{k}+C(x).} ( x )
k + C f(x)=\sum _{k=1}^{\infty }{\binom {x}{k}}\Delta ^{k-1}[f]\left(0\right)+C=\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}[f](0)}{k!}}(x)_{k}+C} di mana ( x ) kΓ ( x + 1 ) Γ ( x − k + 1 ){\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}} adalah faktorial menurun.
For f ( x )x n } one obtains the antidifference: ∑ x x nB n + 1 ( x ) n + 1 + C ( x ) , x^{n}={\frac {B_{n+1}(x)}{n+1}}+C(x),} where B n ( x ) (x)} are the Bernoulli polynomials of the first order.
∑ x aa x + C a=ax+C} ∑ x xx 2 2 − x 2 + C x={\frac {x^{2}}{2}}-{\frac {x}{2}}+C} ∑ x x aB a + 1 ( x ) a + 1 + C , a ∉ Z − x^{a}={\frac {B_{a+1}(x)}{a+1}}+C,\,a\notin \mathbb {Z} ^{-}} di mana B a ( x )− a ζ ( − a + 1 , x ) (x)=-a\zeta (-a+1,x)} , yang digeneralisasikan ke orde polinomial Bernoulli yang sebenarnya.
∑ n1 ∞ f ( n − 1 ) ( 0 ) n ! B n ( x )
∑ x f ( x )∑ k1 ∞ ( x k ) Δ k − 1 [ f ] ( 0 ) + C∑ k1 ∞ Δ k − 1 [ f ] ( 0 )
Δ k − 1 f ( x ) + C ( x ) f(x)=\int _{0}^{x}f(t)dt-\sum _{k=1}^{\infty }{\frac {c_{k}}{k!}}\Delta ^{k-1}f(x)+C(x)} where c k∫ 0 1 ( x ) k d x\int _{0}^{1}(x)_{k}dx} are the Cauchy numbers of the first kind.
+ C f(x)=\int _{0}^{x}f(t)dt-\sum _{k=1}^{\infty }{\frac {c_{k}\Delta ^{k-1}f(x)}{k!}}+C} di mana c k∫ 0 1 Γ ( x + 1 )
− f ( 1 − i t ) ) − ( f ( x + i t )
− f ( n + x ) ) + C .
Then, the constant C is fixed from the condition ∫ − 1 0 F ( x )
∑ x f ( x ) + C f(x)+C} Maka, konstanta C diperbaiki dengan kondisi ∫ 0 1 F ( x )
d x0 ^{0}F(x)\,dx=0} or ∫ 0 1 F ( x )
d x0 ^{1}F(x)dx=0} atau ∫ 1 2 F ( x )
d x0 ^{1}F(x)\,dx=0} .
d x0 ^{1}F(x)dx=0} atau ∫ 1 2 F ( x )
Then, the constant C is fixed from the condition ∫ 0 1 F ( x )
∑ x f ( x ) + C f(x)+C} Maka, konstanta C diperbaiki dengan kondisi ∫ 0 1 F ( x )
d x0 ^{1}F(x)\,dx=0} or ∫ 1 2 F ( x )
d x0 ^{1}F(x)dx=0} atau ∫ 1 2 F ( x )
d x0. ^{2}F(x)\,dx=0.} Alternatively, Ramanujan summation can be used: ∑ x ≥ 1 ℜ f ( x )
d x0 ^{2}F(x)dx=0} Secara alternatif, penjumlahan Ramanujan digunakan sebagaiː ∑ x ≥ 1 ℜ f ( x )− f ( 0 ) − F ( 0 ) ^{\Re }f(x)=-f(0)-F(0)} atau dengan 1 ∑ x ≥ 1 ℜ f ( x )