In the calculus of finite differences, the indefinite sum (or antidifference operator), denoted by ∑ x {\textstyle \sum _{x}} or Δ − 1 {\displaystyle \Delta ^{-1}} , is the linear operator that inverts the forward difference operator Δ f ( x ) = f ( x + 1 ) − f ( x ) . {\displaystyle \Delta f(x)=f(x+1)-f(x).}
That is, if ∑ x f ( x ) = F ( x ) {\textstyle \sum _{x}f(x)=F(x)} , then F {\displaystyle F} satisfies the functional equation F ( x + 1 ) − F ( x ) {\displaystyle F(x+1)-F(x)}
= f ( x ) {\displaystyle =f(x)} so that applying the forward difference recovers the original function: Δ ∑ x f ( x ) = f ( x ) . {\textstyle \Delta \sum _{x}f(x)=f(x).} The operator thus plays the same role for finite differences that the indefinite integral plays for the derivative. An indefinite sum is not unique: adding any 1-periodic function C ( x ) {\displaystyle C(x)} (satisfying C ( x + 1 ) = C ( x ) {\displaystyle C(x+1)=C(x)} ), the function F ( x ) + C ( x ) {\displaystyle F(x)+C(x)} is also a solution. Therefore, an indefinite sum is unique up to a 1-periodic function C ( x ) {\displaystyle C(x)} instead of up to a constant C {\displaystyle C} as the indefinite integral is. To obtain the unique solution up to a constant C {\displaystyle C} , one must impose additional analytic constraints. The Nørlund principal solution is the unique analytic solution that has the minimal possible exponential type (that is, its growth in the imaginary direction on the complex plane is the minimal possible), filtering out any non-constant periodic component. Other methods include higher-order convexity or concavity conditions in real analysis, or using axioms and complex analysis to step back the function's behavior from a neighborhood of infinity in which it behaves polynomially. For integer arguments, the indefinite sum naturally extends ordinary summation, turning a discrete sum into a continuous function. Many such extensions are well-known special functions.
Forward and backward difference conventions The inverse forward difference operator, Δ − 1 {\displaystyle \Delta ^{-1}} ( F ( x + 1 ) − F ( x ) = f ( x ) {\displaystyle F(x+1)-F(x)=f(x)} ), extends the summation up to x − 1 {\displaystyle x-1} , typically starting with the iterator at 0 {\displaystyle 0} :
∑ k = 0 x − 1 f ( k ) . {\displaystyle \,\sum _{k=0}^{x-1}f(k).}
Some authors analytically extend summation for which the upper limit is the argument without a shift, typically starting the iterator at 1 {\displaystyle 1} :
∑ k = 1 x f ( k ) . {\displaystyle \,\sum _{k=1}^{x}f(k).}
In this case, the analytic continuation, F ( x ) {\displaystyle F(x)} , for the sum is a solution of ∇ − 1 f ( x ) {\displaystyle \nabla ^{-1}f(x)} . Stated explicitly, that is:
F ( x ) − F ( x − 1 ) = f ( x ) , {\displaystyle \ F(x)-F(x-1)=f(x),}
Which follows from the discrete counterpart:
∑ k = 1 x f ( k ) − ∑ k = 1 x − 1 f ( k ) = f ( x ) . {\displaystyle \sum _{k=1}^{x}f(k)-\sum _{k=1}^{x-1}f(k)=f(x).}
Some authors use the equivalent form called the telescoping equation:
F ( x + 1 ) − F ( x ) = f ( x + 1 ) . {\displaystyle F(x+1)-F(x)=f(x+1).}
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