indefinite sum

Recall that the finite difference operator $\mathrm{\Delta }$ defined on the set of functions $ℝ\to ℝ$ is given by

$$\mathrm{\Delta }f(x):=f(x+1)-f(x).$$

The difference operator can be thought of as the discrete version of the derivative operator sending a function to its derivative (if it exists). With the derivative operation, there corresponds an inverse operation called the antiderivative, which, given a function $f$, finds its antiderivative $F$ so that the derivative of $F$ gives $f$. There is also a discrete analog of this inverse operation, and it is called the indefinite sum.

The indefinite sum of a function $f:ℝ\to ℝ$ is the set of functions

$$\{F:ℝ\to ℝ\mid \mathrm{\Delta }F=f\}.$$

This set is often denoted by ${\mathrm{\Delta }}^{-1}f$ or $\mathrm{\Sigma }f$, and any element in ${\mathrm{\Delta }}^{-1}f$ is called an indefinite sum of $f$.

Remark. Like the indefinite integral, the indefinite sum ${\mathrm{\Delta }}^{-1}$ is shift invariant. This means that for any $F\in {\mathrm{\Delta }}^{-1}f$, then $F+c\in {\mathrm{\Delta }}^{-1}f$ for any $c\in ℝ$. But, unlike the indefinite integral, the indefinite sum is also invariant by a shift of a periodic real function of period $1$. Conversely, the difference of two indefinite sums of a function $f$ is a periodic real function of period $1$.

In the following discussion, we consider the indefinite sum of a function as a function.

Basic Properties

1. $\mathrm{\Delta }{\mathrm{\Delta }}^{-1}f=f$, and ${\mathrm{\Delta }}^{-1}\mathrm{\Delta }f=f$ modulo a real function of period $1$.

2. Modulo a real number, and treating ${\mathrm{\Delta }}^{-1}$ as an operator taking a function into a function, we see that ${\mathrm{\Delta }}^{-1}$ is linear, that is,

   • ${\mathrm{\Delta }}^{-1}(rf)=r{\mathrm{\Delta }}^{-1}f$ for any $r\in ℝ$, and

   • ${\mathrm{\Delta }}^{-1}(f+g)={\mathrm{\Delta }}^{-1}f+{\mathrm{\Delta }}^{-1}g$.

3. If $F(x)={\mathrm{\Delta }}^{-1}f(x)$, then $F(x+a)={\mathrm{\Delta }}^{-1}f(x+a)$.

4. If $F={\mathrm{\Delta }}^{-1}f$, then we see that

   $F(a+1)-F(a)$$=$$f(a),$

   $F(a+2)-F(a+1)$$=$$f(a+1),$

   $\mathrm{⋮}$

   $F(x)-F(x-1)$$=$$f(x-1).$

   where $x-a$ is a positive integer. Summing these expressions, we get

   $$F(x)-F(a)=\sum _{i=1}^{x-a}f(a+i-1).$$

   This is the discrete version of the fundamental theorem of calculus.

Below is a table of some basic functions and their indefinite sums ($C$ is a real-valued periodic function with period $1$):