Indefinite sum

In mathematics the indefinite sum operator (also known as the antidifference operator), denoted by $\sum _{x}\,$ or $\Delta ^{{-1}}\,$ ,^[1]^[2]^[3] is the linear operator, inverse of the forward difference operator $\Delta \,$ . It relates to the forward difference operator as the indefinite integral relates to the derivative. Thus

\Delta \sum _{x}f(x)=f(x)\,.

More explicitly, if $\sum _{x}f(x)=F(x)\,$ , then

F(x+1)-F(x)=f(x)\,.

If F(x) is a solution of this functional equation for a given f(x), then so is F(x)+C(x) for any periodic function C(x) with period 1. Therefore each indefinite sum actually represents a family of functions. However the solution equal to its Newton series expansion is unique up to an additive constant C.

Fundamental theorem of discrete calculus

Indefinite sums can be used to calculate definite sums with the formula:^[4]

\sum _{k=a}^{b}f(k)=\Delta ^{-1}f(b+1)-\Delta ^{-1}f(a)

Definitions

Laplace summation formula

\sum _{x}f(x)=\int _{0}^{x}f(t)dt+\sum _{{k=1}}^{\infty }{\frac {c_{k}\Delta ^{{k-1}}f(x)}{k!}}+C

where

c_{k}=\int _{0}^{1}{\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}dx

are the Bernoulli numbers of the second kind.^[5]

Newton's formula

\sum _{x}f(x)=-\sum _{{k=1}}^{{\infty }}{\frac {\Delta ^{{k-1}}f(x)}{k!}}(-x)_{k}+C

where

(x)_{k}={\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}

is the falling factorial.

Faulhaber's formula

\sum _{x}f(x)=\sum _{{n=1}}^{{\infty }}{\frac {f^{{(n-1)}}(0)}{n!}}B_{n}(x)+C\,,

provided that the right-hand side of the equation converges.

Mueller's formula

If $\lim _{{x\to {+\infty }}}f(x)=0,$ then^[6]

\sum _{x}f(x)=\sum _{{n=0}}^{\infty }\left(f(n)-f(n+x)\right)+C.

Euler–Maclaurin formula

\sum _{x}f(x)=\int _{0}^{x}f(t)dt-{\frac 12}f(x)+\sum _{{k=1}}^{{\infty }}{\frac {B_{{2k}}}{(2k)!}}f^{{(2k-1)}}(x)+C

Choice of the constant term

Often the constant C in indefinite sum is fixed from the following condition.

Let

F(x)=\sum _{x}f(x)+C

Then the constant C is fixed from the condition

\int _{0}^{1}F(x)dx=0

\int _{1}^{2}F(x)dx=0

Alternatively, Ramanujan's sum can be used:

\sum _{{x\geq 1}}^{{\Re }}f(x)=-f(0)-F(0)

or at 1

\sum _{{x\geq 1}}^{{\Re }}f(x)=-F(1)

respectively^[7]^[8]

Summation by parts

Main article: Summation by parts

Indefinite summation by parts:

\sum _{x}f(x)\Delta g(x)=f(x)g(x)-\sum _{x}(g(x)+\Delta g(x))\Delta f(x)\,

\sum _{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)\,

Definite summation by parts:

\sum _{{i=a}}^{b}f(i)\Delta g(i)=f(b+1)g(b+1)-f(a)g(a)-\sum _{{i=a}}^{b}g(i+1)\Delta f(i)

Period rules

If $T\,$ is a period of function $f(x)\,$ then

\sum _{x}f(Tx)=xf(Tx)+C\,

If $T\,$ is an antiperiod of function $f(x)\,$ , that is $f(x+T)=-f(x)$ then

\sum _{x}f(Tx)=-{\frac 12}f(Tx)+C\,

Alternative usage

Some authors use the phrase "indefinite sum" to describe a sum in which the numerical value of the upper limit is not given. e.g.

\sum _{{k=1}}^{n}f(k)

In this case a closed form expression F(k) for the sum is a solution of

F(x+1)-F(x)=f(x+1)\,

which is called the telescoping equation.^[9] It is inverse to backward difference

\nabla

operator.

It is related to the forward antidifference operator using the fundamental theorem of discrete calculus described earlier.

List of indefinite sums

This is a list of indefinite sums of various functions. Not every function has an indefinite sum that can be expressed in terms of elementary functions.