Indefinite sum

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

${\displaystyle \mathrm{\Delta }\sum _{x}f(x)=f(x).}$

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

${\displaystyle F(x+1)-F(x)=f(x).}$

The solution is not unique; it is determined only up to an additive periodic function with period 1. Therefore, each indefinite sum represents a family of functions.

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

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

The inverse forward difference operator, ${\displaystyle {\mathrm{\Delta }}^{-1}}$, extends the summation up to ${\displaystyle x-1}$:

${\displaystyle {\displaystyle \sum _{k=0}^{x-1}}f(k).}$

Some authors analytically extend summation for which the upper limit is the argument without a shift:^[5][6][7]

${\displaystyle {\displaystyle \sum _{k=1}^x}f(k).}$

In this case, a closed-form expression ${\displaystyle F(x)}$ for the sum is a solution of

${\displaystyle F(x+1)-F(x)=f(x+1),}$

which is called the telescoping equation.^[8] It is the inverse of the backward difference operator ${\displaystyle \mathrm{\nabla }}$, ${\displaystyle {\mathrm{\nabla }}^{-1}}$:

${\displaystyle F(x)-F(x-1)=f(x),}$

It is related to the forward antidifference operator using the fundamental theorem of the calculus of finite differences.

The functional equation ${\displaystyle F(x+1)-F(x)=f(x)}$ does not have a unique solution. If ${\displaystyle F_1(x)}$ is a particular solution, then for any function ${\displaystyle C(x)}$ satisfying ${\displaystyle C(x+1)=C(x)}$ (i.e., any 1-periodic function), the function ${\displaystyle F_2(x)=F_1(x)+C(x)}$ is also a solution. Therefore, the indefinite sum operator defines a family of functions differing by an arbitrary 1-periodic component, ${\displaystyle C(x)}$.

To select the unique principal solution (German: Hauptlösung) up to an additive constant ${\displaystyle C}$ (instead of up to the additive 1-periodic function ${\displaystyle C(x)}$) one must impose additional constraints.

Suppose ${\displaystyle f(z)}$ is analytic in a vertical strip containing the real axis, and let ${\displaystyle F(z)}$ be an analytic solution of ${\displaystyle F(z+1)-F(z)=f(z)}$ in that strip. To ensure uniqueness, we require ${\displaystyle F(z)}$ to be of minimal growth, specifically to be of exponential type less than ${\displaystyle 2\pi }$ in the imaginary direction. That is, there exist constants ${\displaystyle M>0}$ and ${\displaystyle ϵ>0}$ such that $${\displaystyle |F(z)|\le M{e}^{(2\pi -ϵ)|\mathrm{\Im }(z)|}}$$ as ${\displaystyle |\mathrm{\Im }(z)|\to \mathrm{\infty }}$.^[9][10]

Now let ${\displaystyle F_1(z)}$ and ${\displaystyle F_2(z)}$ be two analytic solutions satisfying this growth condition. Their difference ${\displaystyle C(z)=F_1(z)-F_2(z)}$ is then analytic, 1‑periodic (i.e., ${\displaystyle C(z+1)=C(z)}$), and inherits the same exponential type less than ${\displaystyle 2\pi }$.

A fundamental result in complex analysis states that a non‑constant 1‑periodic entire function must have exponential type at least ${\displaystyle 2\pi }$. This follows from its Fourier series expansion: if ${\displaystyle C(z)}$ is non‑constant, its Fourier series contains a term ${\displaystyle a_n{e}^{2\pi inz}}$ with ${\displaystyle n\ne 0}$, which has type ${\displaystyle 2\pi |n|\ge 2\pi }$. Since ${\displaystyle C(z)}$ has type strictly less than ${\displaystyle 2\pi }$, it cannot contain any such term and therefore must be constant.

Consequently, under this minimal growth condition, any two solutions differ by at most a constant. Hence ${\displaystyle F(z)}$ is unique up to an additive constant ${\displaystyle C}$.

The indefinite product operator, denoted by ${\displaystyle \prod _x}$, is the multiplicative analogue of the indefinite sum. If ${\displaystyle \prod _{x}f(x)=F(x)}$, then:

${\displaystyle \frac{F(x+1)}{F(x)}=f(x).}$

Its common discrete analog is ${\displaystyle {\displaystyle \prod _{k=1}^{x-1}}f(k)}$. The two operators are related by:

${\displaystyle \prod _{x}f(x)=\exp (\sum _x\ln f(x)),}$

${\displaystyle \sum _{x}f(x)=\ln (\prod _x\exp (f(x))).}$

The Laplace summation formula is a formal asymptotic expansion (generally convergent only for polynomials) of the inverse forward difference ${\displaystyle {\mathrm{\Delta }}^{-1}f(x)}$:^[11][12]

${\displaystyle \sum _{x}f(x)={\displaystyle \underset{0}{\overset{x}{\int }}}f(t)dt-{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{c_k{\mathrm{\Delta }}^{k-1}f(x)}{k!}+C}$

where ${\displaystyle c_k={\displaystyle \underset{0}{\overset{1}{\int }}}(x{)}_{k}dx}$ are the Cauchy numbers of the first kind.

${\displaystyle (x{)}_k=\frac{\mathrm{\Gamma }(x+1)}{\mathrm{\Gamma }(x-k+1)}}$ is the falling factorial.

The inverse forward difference operator, ${\displaystyle {\mathrm{\Delta }}^{-1}f(x)}$, can be expressed formally (generally convergent only for polynomials) by its Newton series expansion:

${\displaystyle \sum _{x}f(x)={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle (\genfrac{}{}{0ex}{}{x}{k})}{\mathrm{\Delta }}^{k-1}f\left(0\right)+C={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{{\mathrm{\Delta }}^{k-1}f(0)}{k!}(x{)}_k+C,}$

Given that ${\displaystyle f(x)}$ can be represented by its Maclaurin Series expansion, the Taylor series about ${\displaystyle 0}$, it is sometimes possible to represent the indefinite sum using Bernoulli polynomials because ${\displaystyle \sum _x{x}^a=\frac{B_{a+1}(x)}{a+1}+C}$:

${\displaystyle \sum _{x}f(x)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{f^{(n-1)}(0)}{n!}{B}_n(x)+C.}$

If ${\displaystyle f(x)}$ is analytic on the right half-plane and satisfies the decay condition ${\displaystyle \underset{x\to +\mathrm{\infty }}{\lim }f(x)=0}$, the analytic continuation of ${\displaystyle {\mathrm{\nabla }}^{-1}f(x)={\displaystyle \sum _{k=1}^x}f(k)}$ is given by:^[5]

${\displaystyle {\mathrm{\nabla }}^{-1}f(x)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(f(n)-f(n+x))+C.}$

This formula is derived from axioms presented in the paper based on fractional sums, which uniquely extends the definition of the summation to complex limits. The decay condition ${\displaystyle \underset{x\to +\mathrm{\infty }}{\lim }f(x)=0}$ represents the simplest case of the general asymptotic requirements for the function ${\displaystyle f(x)}$.

The Euler–Maclaurin formula extends ${\displaystyle {\mathrm{\nabla }}^{-1}f(x)={\displaystyle \sum _{k=1}^x}f(k)}$:^[6][9] $${\displaystyle \begin{array}{rl}{\mathrm{\nabla }}^{-1}f(x)& ={\displaystyle \underset{1}{\overset{x}{\int }}}f(t)dt+\frac{f(1)+f(x)}{2}\\ & +{\displaystyle \sum _{k=1}^p}\frac{B_{2k}}{(2k)!}(f^{(2k-1)}(x)-f^{(2k-1)}(1))+R_p+C\end{array}}$$ where ${\displaystyle B_{2k}}$ are the even Bernoulli numbers, ${\displaystyle p}$ is an arbitrary positive integer, and ${\displaystyle R_p}$ is the remainder term given by:

${\displaystyle R_p=(-1{)}^{p+1}{\displaystyle \underset{1}{\overset{x}{\int }}}{f}^{(p)}(t)\frac{P_p(t)}{p!}dt,}$

with ${\displaystyle P_p(t)=B_p(t-\lfloor t\rfloor )}$ being the periodized Bernoulli function related to the Bernoulli polynomials.

The indefinite sum ${\displaystyle {\mathrm{\nabla }}^{-1}f(x)={\displaystyle \sum _{k=1}^x}f(k)}$ can be analytically continued by applying the standard Abel-Plana formula to the finite sum ${\displaystyle {\displaystyle \sum _{k=1}^n}f(k)}$ and then analytically continuing the integer limit ${\displaystyle n}$ to the variable ${\displaystyle x}$. This yields the formula:^[7] $${\displaystyle \begin{array}{rl}{\mathrm{\nabla }}^{-1}f(x)& ={\displaystyle \underset{1}{\overset{x}{\int }}}f(t)dt+\frac{f(1)+f(x)}{2}\\ & +i{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}\frac{(f(x-it)-f(1-it))-(f(x+it)-f(1+it))}{e^{2\pi t}-1}dt+C\end{array}}$$

This analytic continuation is valid when the conditions for the original formula are met. The sufficient conditions are:^[9][10]

1. Analyticity: ${\displaystyle f(z)}$ must be analytic in the closed vertical strip between ${\displaystyle \mathrm{\Re }(z)=1}$ and ${\displaystyle \mathrm{\Re }(z)=\mathrm{\Re }(x)}$. The formula provides analytic continuation up to, but not beyond, the nearest singularities of ${\displaystyle f}$ to the line ${\displaystyle \mathrm{\Re }(z)=1}$.
2. Growth: ${\displaystyle f(z)}$ must be of exponential type less than ${\displaystyle 2\pi }$ in this strip, satisfying ${\displaystyle |f(z)|\le M{e}^{(2\pi -ϵ)|\mathrm{\Im }(z)|}}$ for some ${\displaystyle M>0}$, ${\displaystyle ϵ>0}$ as ${\displaystyle |\mathrm{\Im }(z)|\to \mathrm{\infty }}$.

The constant term ${\displaystyle C}$, in the context of indefinite sums naturally extending the discrete summation, is often defined based on the respective empty sum.

For the inverse forward difference, ${\displaystyle {\mathrm{\Delta }}^{-1}f(x)}$, the typical summation equivalent is ${\displaystyle {\displaystyle \sum _{k=0}^{x-1}}f(k)}$ so the empty sum is when ${\displaystyle {\mathrm{\Delta }}^{-1}f(0)=0}$ as it correlates to ${\displaystyle {\displaystyle \sum _{k=0}^{-1}}f(k).}$

For the inverse backward difference, ${\displaystyle {\mathrm{\nabla }}^{-1}f(x)}$, the typical summation equivalent is ${\displaystyle {\displaystyle \sum _{k=1}^x}f(k)}$ so the empty sum is when ${\displaystyle {\mathrm{\nabla }}^{-1}f(0)=0}$ as it correlates to ${\displaystyle {\displaystyle \sum _{k=1}^0}f(k).}$

In older texts relating to Bernoulli polynomials (predating more modern analytic techniques) the constant ${\displaystyle C}$ was often fixed using integral conditions.

Let

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

Then the constant ${\displaystyle C}$ is fixed from the condition

${\displaystyle {\displaystyle \underset{0}{\overset{1}{\int }}}F(x)dx=0}$

or

${\displaystyle {\displaystyle \underset{1}{\overset{2}{\int }}}F(x)dx=0}$

Alternatively, Ramanujan summation can be used:

${\displaystyle {\displaystyle \sum _{x\ge 1}^{\mathrm{\Re }}}f(x)=-f(0)-F(0)}$

or at 1

${\displaystyle {\displaystyle \sum _{x\ge 1}^{\mathrm{\Re }}}f(x)=-F(1)}$

respectively.^[13][14]

Indefinite summation by parts:^[15]

${\displaystyle \sum _{x}f(x)\mathrm{\Delta }g(x)=f(x)g(x)-\sum _x(g(x)+\mathrm{\Delta }g(x))\mathrm{\Delta }f(x)}$

${\displaystyle \sum _{x}f(x)\mathrm{\Delta }g(x)+\sum _{x}g(x)\mathrm{\Delta }f(x)=f(x)g(x)-\sum _x\mathrm{\Delta }f(x)\mathrm{\Delta }g(x)}$

Definite summation by parts:

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

If ${\displaystyle T}$ is a period of function ${\displaystyle f(x)}$ then

${\displaystyle \sum _{x}f(Tx)=xf(Tx)+C.}$

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

${\displaystyle \sum _{x}f(Tx)=-\frac{1}{2}f(Tx)+C.}$

The unique analytic continuation of ${\displaystyle F(z)={\mathrm{\nabla }}^{-1}f(z)}$ defined as ${\displaystyle F(x)-F(x-1)=f(x)}$ with exponential type less than ${\displaystyle 2\pi }$ in the imaginary direction where ${\displaystyle f(z)}$ is entire and the constant term ${\displaystyle C}$ is chosen such that ${\displaystyle F(0)=0}$ (the empty sum condition), ${\displaystyle F(z)}$ satisfies a reflection formula.

If ${\displaystyle f(z)}$ is an odd function (${\displaystyle f(-z)=-f(z)}$), the unique analytic continuation ${\displaystyle F(z)}$ satisfies:

${\displaystyle F(z)=F(-1-z).}$

This represents a point symmetry about ${\displaystyle z=-1/2}$.

If ${\displaystyle f(z)}$ is an even function (${\displaystyle f(-z)=f(z)}$), the unique analytic continuation ${\displaystyle F(z)}$ satisfies:

${\displaystyle F(z)+F(-1-z)=F(-1)}$.

For positive integer exponents, Faulhaber's formula can be used. Note that ${\displaystyle x}$ in the result of Faulhaber's formula must be replaced with ${\displaystyle x-1}$ due to the offset, as Faulhaber's formula finds ${\displaystyle {\mathrm{\nabla }}^{-1}}$ rather than ${\displaystyle {\mathrm{\Delta }}^{-1}}$.

For negative integer exponents, the indefinite sum is closely related to the polygamma function:

${\displaystyle \sum _x\frac{1}{x^a}=\frac{(-1{)}^{a-1}{\psi }^{(a-1)}(x)}{(a-1)!}+C,a\in \mathbb{\mathbb{N}}}$

For fractions not listed in this section, one may use the polygamma function with partial fraction decomposition. More generally,

${\displaystyle \sum _x{x}^a=\{\begin{array}{ll}\frac{B_{a+1}(x)}{a+1}+C,& \text{if }a\ne -1\\ \psi (x)+C,& \text{if }a=-1\end{array}=\{\begin{array}{ll}-\zeta (-a,x)+C,& \text{if }a\ne -1\\ \psi (x)+C,& \text{if }a=-1\end{array}}$

where ${\displaystyle B_a(x)}$ are the Bernoulli polynomials, ${\displaystyle \zeta (s,a)}$ is the Hurwitz zeta function, and ${\displaystyle \psi (z)}$ is the digamma function. This is related to the generalized harmonic numbers.

As the generalized harmonic numbers use reciprocal powers, ${\displaystyle a}$ must be substituted for ${\displaystyle -a}$, and the most common form uses the inverse of the backward difference offset:

${\displaystyle {\mathrm{\nabla }}^{-1}{x}^a=H_x^{(-a)}=\zeta (-a)-\zeta (-a,x+1).}$

Here, ${\displaystyle \zeta (-a)}$ is the constant ${\displaystyle C}$.

The Bernoulli polynomials are also related via a partial derivative with respect to ${\displaystyle x}$:

${\displaystyle \frac{\partial }{\partial x}\left(\sum _x{x}^a\right)=B_a(x)=-a\zeta (1-a,x).}$

Similarly, using the inverse of the backwards difference operator may be considered more natural, as:

${\displaystyle \frac{\partial }{\partial x}\left({\mathrm{\nabla }}^{-1}{x}^a\right){|}_{x=0}=-a\zeta (1-a,x+1){|}_{x=0}=-a\zeta (1-a)=B_a.}$

Further generalization comes from use of the Lerch transcendent:

${\displaystyle \sum _x\frac{z^x}{(x+a{)}^s}=-z^x\mathrm{\Phi }(z,s,x+a)+C,}$

which generalizes the generalized harmonic numbers as ${\displaystyle z\mathrm{\Phi }(z,s,a+1)-z^{x+1}\mathrm{\Phi }(z,s,x+1+a)}$ when taking ${\displaystyle {\mathrm{\nabla }}^{-1}}$. Additionally, the partial derivative is given by

${\displaystyle \frac{\partial }{\partial x}(-z^x\mathrm{\Phi }(z,s,x+a))=z^x(s\mathrm{\Phi }(z,s+1,x+a)-\ln (z)\mathrm{\Phi }(z,s,x+a)).}$

${\displaystyle \sum _x{B}_a(x)=(x-1)B_a(x)-\frac{a}{a+1}{B}_{a+1}(x)+C}$

${\displaystyle \sum _x{a}^x=\frac{a^x}{a-1}+C}$

${\displaystyle \sum _x{\log }_{b}x={\log }_b\mathrm{\Gamma }(x)+C}$

${\displaystyle \sum _x{\log }_{b}ax={\log }_b(a^{x-1}\mathrm{\Gamma }(x))+C}$

${\displaystyle \sum _x\sinh ax=\frac{1}{2}\mathrm{csch}\left(\frac{a}{2}\right)\cosh (\frac{a}{2}-ax)+C}$

${\displaystyle \sum _x\cosh ax=\frac{1}{2}\mathrm{csch}\left(\frac{a}{2}\right)\sinh (ax-\frac{a}{2})+C}$

${\displaystyle \sum _x\tanh ax=\frac{1}{a}{\psi }_{e^a}(x-\frac{i\pi }{2a})+\frac{1}{a}{\psi }_{e^a}(x+\frac{i\pi }{2a})-x+C}$

where ${\displaystyle {\psi }_q(x)}$ is the q-digamma function.

${\displaystyle \sum _x\sin ax=-\frac{1}{2}\csc \left(\frac{a}{2}\right)\cos (\frac{a}{2}-ax)+C,a\ne 2n\pi }$

${\displaystyle \sum _x\cos ax=\frac{1}{2}\csc \left(\frac{a}{2}\right)\sin (ax-\frac{a}{2})+C,a\ne 2n\pi }$

${\displaystyle \sum _x{\sin }^{2}ax=\frac{x}{2}+\frac{1}{4}\csc (a)\sin (a-2ax)+C,a\ne n\pi }$

${\displaystyle \sum _x{\cos }^{2}ax=\frac{x}{2}-\frac{1}{4}\csc (a)\sin (a-2ax)+C,a\ne n\pi }$

${\displaystyle \sum _x\tan ax=ix-\frac{1}{a}{\psi }_{e^{2ia}}(x-\frac{\pi }{2a})+C,a\ne \frac{n\pi }{2}}$

where ${\displaystyle {\psi }_q(x)}$ is the q-digamma function.

$${\displaystyle \begin{array}{rl}\sum _x\tan x& =ix-{\psi }_{e^{2i}}(x+\frac{\pi }{2})+C\\ & =-{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}(\psi (k\pi -\frac{\pi }{2}+1-x)+\psi (k\pi -\frac{\pi }{2}+x)\\ & -\psi (k\pi -\frac{\pi }{2}+1)-\psi (k\pi -\frac{\pi }{2}))+C\end{array}}$$

${\displaystyle \sum _x\cot ax=-ix-\frac{i{\psi }_{e^{2ia}}(x)}{a}+C,a\ne \frac{n\pi }{2}}$

$${\displaystyle \begin{array}{rl}\sum _x\mathrm{sinc}x& =\mathrm{sinc}(x-1)(\frac{1}{2}+(x-1)\\ & \times (\ln (2)+\frac{\psi (\frac{x-1}{2})+\psi (\frac{1-x}{2})}{2}\\ & -\frac{\psi (x-1)+\psi (1-x)}{2}))+C\end{array}}$$

where ${\displaystyle \mathrm{sinc}(x)}$ is the normalized sinc function.

${\displaystyle \sum _x\psi (x)=(x-1)\psi (x)-x+C}$

${\displaystyle \sum _x\mathrm{\Gamma }(x)=(-1{)}^{x+1}\mathrm{\Gamma }(x)\frac{\mathrm{\Gamma }(1-x,-1)}{e}+C}$

where ${\displaystyle \mathrm{\Gamma }(s,x)}$ is the incomplete gamma function.

${\displaystyle \sum _x(x{)}_a=\frac{(x{)}_{a+1}}{a+1}+C}$

where ${\displaystyle (x{)}_a}$ is the falling factorial.

${\displaystyle \sum _x{\mathrm{sexp}}_a(x)={\ln }_a\frac{({\mathrm{sexp}}_a(x){)}^{\prime }}{(\ln a{)}^x}+C}$

(see super-exponential function)

• Indefinite product
• Time scale calculus
• List of derivatives and integrals in alternative calculi

• "Difference Equations: An Introduction with Applications", Walter G. Kelley, Allan C. Peterson, Academic Press, 2001, ISBN 0-12-403330-X
• Markus Müller. How to Add a Non-Integer Number of Terms, and How to Produce Unusual Infinite Summations
• Markus Mueller, Dierk Schleicher. Fractional Sums and Euler-like Identities
• S. P. Polyakov. Indefinite summation of rational functions with additional minimization of the summable part. Programmirovanie, 2008, Vol. 34, No. 2.
• "Finite-Difference Equations And Simulations", Francis B. Hildebrand, Prenctice-Hall, 1968

0.00 (0 votes) Original source: https://en.wikipedia.org/wiki/Indefinite sum. Read more