# indefinite sum

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

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

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 $f:\mathbb{R}\to\mathbb{R}$ is the set of functions

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

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

Remark. Like the indefinite integral, the indefinite sum $\Delta^{-1}$ is shift invariant. This means that for any $F\in\Delta^{-1}f$, then $F+c\in\Delta^{-1}f$ for any $c\in\mathbb{R}$. 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. 1.

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

2. 2.

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

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

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

3. 3.

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

4. 4.

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

 $\displaystyle F(a+1)-F(a)$ $\displaystyle=$ $\displaystyle f(a),$ $\displaystyle F(a+2)-F(a+1)$ $\displaystyle=$ $\displaystyle f(a+1),$ $\displaystyle\vdots$ $\displaystyle F(x)-F(x-1)$ $\displaystyle=$ $\displaystyle f(x-1).$

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

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

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

$f(x)$ $\Delta^{-1}f(x)$ Comment
$r\in\mathbb{R}$ $rx+C$
$x$ $\displaystyle{\frac{x(x-1)}{2}+C}$
$x^{2}$ $\displaystyle{\frac{x(x-1)(2x-1)}{6}+C}$
$x^{3}$ $\displaystyle{\frac{x^{2}(x-1)^{2}}{4}+C}$
$x^{n}$ $T_{n}(x)+C$ See this link (http://planetmath.org/SumOfPowers) for detail
$a^{x}$ $\displaystyle{\frac{a^{x}}{a-1}+C}$ $a\neq 1$
$(x)_{n}$ $\displaystyle{\frac{(x)_{n}}{n+1}+C}$ $(x)_{n}$ is the falling factorial  of degree $n$
$\displaystyle{\binom{x}{n}}$ $\displaystyle{\binom{x}{n+1}+C}$ $\displaystyle{\binom{x}{n}:=\frac{(x)_{n}}{n!}}$
$\displaystyle{\frac{1}{x}}$ $\psi(x)+C$ $\psi(x)$ is the digamma function  $\ln{x}$ $\ln{\Gamma(x)}+C$ $\Gamma(x)$ is the gamma function    $\sin{x}$ $\displaystyle{-\frac{\cos(x-1/2)}{2\sin(1/2)}+C}$
$\cos{x}$ $\displaystyle{\frac{\sin(x-1/2)}{2\sin(1/2)}+C}$

${{{{}\end{center}\inner@par\thebibliography\bibitem{cj}C.Jordan.\emph{Calculus of % Finite Differences},thirdedition.Chelsea,NewYork(1965)\endthebibliography% \begin{flushright}\begin{tabular}[]{|ll|}\hline Title&indefinite sum\\ Canonical name&IndefiniteSum\\ Date of creation&2013-03-22 17:35:14\\ Last modified on&2013-03-22 17:35:14\\ Owner&CWoo (3771)\\ Last modified by&CWoo (3771)\\ Numerical id&20\\ Author&CWoo (3771)\\ Entry type&Definition\\ Classification&msc 39A99\\ Related topic&FiniteDifference\\ \hline}\end{tabular}}}\end{flushright}\end{document}$