Indefinite sum

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

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

If ${\textstyle \sum _{x}f(x)=F(x)}$, 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.

The Nørlund principal solution represents the analytic solution without any such non-constant periodic terms. Two conventions exist, one for the forward difference, ${\displaystyle \Delta }$, and one for the backward difference, ${\displaystyle \nabla }$. The inverse forward difference, denoted ${\displaystyle \Delta ^{-1}}$, naturally extends the summation up to ${\displaystyle x-1}$. The inverse backward difference, denoted ${\displaystyle \nabla ^{-1}}$, naturally extends the summation up to ${\displaystyle x}$.

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

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

Alternatively, using the inverse backward difference operator, the relation is:

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

The following basic indefinite sums follow from the fundamental properties of the difference operator:^[5]

Constant:

${\displaystyle \sum _{x}c=cx+C\left(x\right)}$

Falling factorial:

${\displaystyle \sum _{x}(x)_{n}={\frac {(x)_{n+1}}{n+1}}+C\left(x\right)}$

Exponential:

${\displaystyle \sum _{x}a^{x}={\frac {a^{x}}{a-1}}+C\left(x\right)\quad a\neq 1}$

Logarithm:

${\displaystyle \sum _{x}\ln x=\ln \Gamma (x)+C\left(x\right)}$

In the above identities, ${\displaystyle C(x)}$ represents an arbitrary 1-periodic function (or a constant if the Nørlund principal solution is assumed).

Indefinite summation by parts is the discrete analog of integration by parts. It is used to find the indefinite sum of a product of two functions:^[6][5]

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

Using the identity ${\displaystyle g(x+1)=g(x)+\Delta g(x)}$, this is often written more compactly as:

${\displaystyle \sum _{x}f(x)\Delta g(x)=f(x)g(x)-\sum _{x}g(x+1)\Delta f(x)}$

A symmetrical form derived from the discrete product rule is:

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

Definite summation by parts is defined as:

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

Example product of a polynomial and exponential

Summation by parts is effective for functions like ${\displaystyle x2^{x}}$. To find the indefinite sum, let ${\displaystyle f(x)=x}$ and ${\displaystyle \Delta g(x)=2^{x}}$:^[7]

1. Find the components:

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

   ${\displaystyle g(x)=\sum _{x}2^{x}={\frac {2^{x}}{2-1}}=2^{x}}$

   ${\displaystyle g(x+1)=2^{x+1}}$

2. Apply the summation by parts formula:

   ${\displaystyle \sum _{x}x2^{x}=x\cdot 2^{x}-\sum _{x}2^{x+1}\cdot 1}$

3. Evaluate the remaining sum:

   ${\displaystyle \sum _{x}x2^{x}=x2^{x}-2^{x+1}+C\left(x\right)}$

4. Result:

   ${\displaystyle \sum _{x}x2^{x}=(x-2)2^{x}+C\left(x\right)}$

The inverse forward difference operator, ${\displaystyle \Delta ^{-1}}$, extends the summation up to ${\displaystyle x-1}$, typically starting with the iterator at ${\displaystyle 0}$:

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

Some authors analytically extend summation for which the upper limit is the argument without a shift, typically starting the iterator at ${\displaystyle 1}$:^[8][9][10]

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

In this case, the analytic continuation, ${\displaystyle F(x)}$, for the sum is a solution of ${\displaystyle \nabla ^{-1}f(x)}$. Stated explicitly, that is:

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

Some authors use the equivalent form called the telescoping equation:^[11]

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

The lower bounds of the discrete analog for both inverse forward difference and inverse backward difference can be an arbitrary constant other than those listed here, as it is absorbed into the height of the 1-periodic or constant term ${\displaystyle C}$.

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)^[12] up to an additive constant ${\displaystyle C}$ (instead of up to the additive 1-periodic function ${\displaystyle C(x)}$) one must impose additional constraints.

Following the theory developed by Niels Erik Nørlund,^[12] the indefinite sum can be uniquely determined for analytic functions by imposing restriction on their growth in the complex plane. Specifically, by imposing minimal growth, the non-constant periodic terms can be filtered out.

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, 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 ${\displaystyle \epsilon >0}$ such that ${\displaystyle |F(z)|\leq Me^{(2\pi -\epsilon )|\Im (z)|}}$ as ${\displaystyle |\Im (z)|\to \infty }$.^[13][14]

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 ${\displaystyle 2\pi }$.

Nørlund uses a fundamental result in complex analysis (related to Carlson's theorem and the Paley–Wiener theorem) which states that a non‑constant periodic entire function must have exponential type at least ${\displaystyle 2\pi }$.^[12] 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 ${\displaystyle n\neq 0}$, which has type ${\displaystyle 2\pi |n|\geq 2\pi }$. Since ${\displaystyle C(z)}$ has type strictly less than ${\displaystyle 2\pi }$, it cannot contain any such term and therefore must be constant.

In real analysis, the uniqueness condition can be given using higher‑order convexity, generalizing the Bohr-Mollerup theorem. For an integer ${\displaystyle p\geq 0}$, a function is called ${\displaystyle p}$-convex if its divided differences of order ${\displaystyle p}$ are non‑negative, and ${\displaystyle p}$-concave if those divided differences are non-positive. A function is called eventually ${\displaystyle p}$-convex (resp. eventually ${\displaystyle p}$-concave) if there exists ${\displaystyle M>0}$ such that it is ${\displaystyle p}$-convex (resp. ${\displaystyle p}$-concave) on the interval ${\displaystyle (M,\infty )}$.

Marichal and Zenaïdi proved the following uniqueness theorem, their method requiring the solution to be eventually ${\displaystyle p}$-convex or ${\displaystyle p}$-concave.^[15][16]

Theorem. Let ${\displaystyle p\geq 0}$ be an integer and let ${\displaystyle g:\mathbb {R} _{+}\to \mathbb {R} }$ satisfy ${\displaystyle \lim _{n\to \infty }\Delta ^{p}g(n)=0}$. If ${\displaystyle f:\mathbb {R} _{+}\to \mathbb {R} }$ is an eventually ${\displaystyle p}$-convex or eventually ${\displaystyle p}$-concave solution of ${\displaystyle \Delta f=g}$, then ${\displaystyle f}$ is uniquely determined up to an additive constant. Moreover, for any ${\displaystyle x>0}$,

${\displaystyle f(x)=f(1)+\lim _{n\to \infty }\left(\sum _{k=1}^{n-1}g(k)-\sum _{k=0}^{n-1}g(x+k)+\sum _{j=1}^{p}{\binom {x}{j}}\Delta ^{\,j-1}g(n)\right),}$

and the convergence is uniform on bounded subsets of ${\displaystyle \mathbb {R} _{+}}$.

In their paper How to Add a Noninteger Number of Terms^[8], Müller and Schleicher introduced an axiomatic approach to fractional summation with a real or complex number of terms. Their method extends the classical discrete sum

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

to non-integer and complex upper limits ${\displaystyle x}$. The definition is built upon six natural axioms that uniquely determine the extension of fractional sums for functions that grow polynomially, S1 through S6 are as follows:

1. Continued Summation: ${\displaystyle \sum _{\nu =x}^{y}f(\nu )+\sum _{\nu =y+1}^{z}f(\nu )=\sum _{\nu =x}^{z}f(\nu )}$.
2. Translation Invariance: ${\displaystyle \sum _{\nu =x+s}^{y+s}f(\nu )=\sum _{\nu =x}^{y}f(\nu +s)}$.
3. Linearity: ${\displaystyle \sum _{\nu =x}^{y}(\lambda f(\nu )+\mu g(\nu ))=\lambda \sum _{\nu =x}^{y}f(\nu )+\mu \sum _{\nu =x}^{y}g(\nu )}$.
4. Empty Sum Condition: ${\displaystyle \sum _{\nu =1}^{1}f(\nu )=f(1)}$ (equivalent to the empty sum condition).
5. Holomorphy for Monomials: for each ${\displaystyle d\in \mathbb {N} }$, ${\displaystyle z\mapsto \sum _{\nu =1}^{z}\nu ^{d}}$ is holomorphic in ${\displaystyle \mathbb {C} }$.
6. Right-Shift Continuity: if ${\displaystyle f(z+n)\to 0}$ pointwise as ${\displaystyle n\to +\infty }$, then ${\displaystyle \sum _{\nu =x}^{y}f(\nu +n)\to 0}$; more generally, if ${\displaystyle f(z+n)}$ can be approximated by polynomials ${\displaystyle p_{n}(z+n)}$ of fixed degree with ${\displaystyle |f(z+n)-p_{n}(z+n)|\to 0}$, then:

${\displaystyle \left|\sum _{\nu =x}^{y}f(\nu +n)-\sum _{\nu =x}^{y}p_{n}(\nu +n)\right|\to 0}$.

This axiomatic framework assumes that for sufficiently large inputs, the indefinite sum behaves like a polynomial (S5); this asymptotic behavior is then "stepped back" to the rest of the complex plane using translation invariance (S2) and continuity (S6) to uniquely determine the analytic continuation.

A function ${\displaystyle f}$ is called fractional summable of degree ${\displaystyle \sigma }$ if, for large ${\displaystyle n}$, the values ${\displaystyle f(n+z)}$ can be approximated by a sequence of polynomials ${\displaystyle p_{n}(n+z)}$ of fixed degree ${\displaystyle \sigma }$, with the error tending to zero. For such functions, the fractional sum is uniquely given by the limit:

${\displaystyle \sum _{\nu =x}^{y}f(\nu )=\lim _{n\to \infty }\left(\sum _{\nu =1}^{n}{\bigl (}f(\nu +x-1)-f(\nu +y){\bigr )}+\sum _{\nu =n+x}^{n+y}p_{n}(\nu )\right),}$

where the sum over polynomials is evaluated via the polynomial antidifference formula.

In the simplest case when ${\displaystyle f(t)\to 0}$ as ${\displaystyle t\to \infty }$ (i.e., the approximating polynomials are zero), this reduces to:

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

Following directly from uniqueness, if ${\displaystyle f(z)}$ is a meromorphic function, one can define a unique analytic solution of the backward difference sum, by imposing the conditions that:

• Difference Equation: ${\displaystyle F(x)-F(x-1)=f(x)}$
• Normalization: ${\displaystyle F(0)=0}$ (empty sum boundary condition).
• Growth constraint: ${\displaystyle F(z)}$ has exponential type less than ${\displaystyle 2\pi }$ in the imaginary direction.

Under these conditions, ${\displaystyle F(z)}$ satisfies a reflection formula (referred to by Nørlund as Ergänzungssatz, a complementary theorem to uniqueness of the principal solution [Hauptlösung], presenting it as ${\displaystyle G\left(x-\omega |-\omega \right)=G\left(x|\omega \right),}$${\displaystyle F\left(x-\omega |-\omega \right)=F\left(x|\omega \right)}$ where ${\displaystyle \omega }$ is the span).^[17]

If ${\displaystyle f(z)}$ is an odd function (${\displaystyle f(-z)=-f(z)}$), the unique analytic solution ${\displaystyle F(z)}$ satisfies:^[17]

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

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 solution ${\displaystyle F(z)}$ satisfies:^[17]

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

In the symbolic method developed by L. M. Milne-Thomson, the indefinite product operator ${\displaystyle \prod _{x}}$ serves as the multiplicative analog to the indefinite sum. It is defined by the first order homogeneous equation ${\displaystyle F(x+1)=f(x)F(x).}$

By taking the logarithm of the product formula, one obtains the telescoping identity ${\displaystyle \Delta \ln F(x)=\ln f(x)}$.^[18] This allows any indefinite product to be expressed through an indefinite sum:

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

where ${\displaystyle \varpi (x)}$ is an arbitrary periodic function of period 1.^[19] Conversely, an indefinite sum may be represented as the logarithm of an indefinite product:

${\displaystyle \sum _{x}f(x)=\ln \left(\prod _{x}\exp(f(x))\right)+C(x).}$

For an entire function of exponential type less than ${\displaystyle \ln \left(2\right)}$^[20] the inverse forward difference operator, ${\displaystyle \Delta ^{-1}f(x)}$, can be expressed by its Newton series expansion: ^[21][22]

${\displaystyle \sum _{x}f(x)=\sum _{k=1}^{\infty }{\binom {x}{k}}\Delta ^{k-1}f\left(0\right)+C\left(x\right)=\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}f(0)}{k!}}(x)_{k}+C\left(x\right).}$

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

Formally, the inverse forward difference operator can be expressed in terms of the derivative operator ${\displaystyle D={\frac {d}{dx}}}$ using the exponential generating function of the Bernoulli numbers:^[23][24][25]

$${\displaystyle \Delta ^{-1}={\frac {1}{e^{D}-1}}=\sum _{v=0}^{\infty }{\frac {B_{v}}{v!}}D^{\,v-1},}$$

where ${\displaystyle B_{v}}$ are the Bernoulli numbers defined by the generating function ${\displaystyle {\frac {t}{e^{t}-1}}=\sum _{v=0}^{\infty }B_{v}{\frac {t^{v}}{v!}}}$. Under this convention ${\displaystyle B_{1}=-{\tfrac {1}{2}}}$.

If ${\displaystyle f}$ is a polynomial, only finitely many terms of the series are non-zero as the finite difference of a monomial is a polynomial of one degree lower (following by induction, finitely many terms are required). For ${\displaystyle f(x)=x^{n}}$ one obtains the antidifference:^[24]

$${\displaystyle \sum _{x}x^{n}={\frac {B_{n+1}(x)}{n+1}}+C\left(x\right),}$$

where ${\displaystyle B_{n}(x)}$ are the Bernoulli polynomials of the first order.^[24]

If ${\displaystyle f}$ admits a Maclaurin series expansion ${\displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}}$, the antidifference of monomials in the series expansion yields the formal series:^[25]

$${\displaystyle \sum _{x}f(x)=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}B_{n}(x)+C\left(x\right).}$$

For non‑polynomials this expansion is generally asymptotic.

Relation to the inverse backward difference

If one instead expands the inverse backward difference operator, ${\displaystyle \nabla ^{-1}={\frac {e^{D}}{e^{D}-1}}}$ (which extends ${\displaystyle \sum _{k=1}^{x}f(k)}$), it admits to the same expansion, but with ${\displaystyle B_{1}=+{\tfrac {1}{2}}}$ in place of ${\displaystyle B_{1}=-{\tfrac {1}{2}}}$.

The Euler–Maclaurin formula extends ${\displaystyle \nabla ^{-1}f(x)=\sum _{k=1}^{x}f(k)}$:^[9][13] $${\displaystyle {\begin{aligned}\nabla ^{-1}f(x)&=\int _{1}^{x}f(t)dt+{\frac {f(1)+f(x)}{2}}\\&\quad +\sum _{k=1}^{p}{\frac {B_{2k}}{(2k)!}}\left(f^{(2k-1)}(x)-f^{(2k-1)}(1)\right)+R_{p}+C\left(x\right)\end{aligned}}}$$ 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:

${\displaystyle R_{p}=(-1)^{p+1}\int _{1}^{x}f^{(p)}(t){\frac {P_{p}(t)}{p!}}\,dt,}$

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

Laplace's summation formula, closely related to the Gregory summation formula, can be seen as the discrete counterpart to the Euler–Maclaurin formula. The inverse forward difference ${\displaystyle \Delta ^{-1}f(x)}$:^{[26][27][7][28]}

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

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

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

Truncating the series after ${\displaystyle n}$ terms leaves a remainder that can be expressed as an integral of ${\displaystyle f^{(n)}}$ times a periodic Bernoulli polynomial.^[7][28] In the notation of Charles Jordan, Gregory's formula is:^[7]

$${\displaystyle {\begin{aligned}\sum _{x=a}^{z}f(x)&=\int _{a}^{z}f(x)\,dx\\&\quad -\sum _{m=1}^{n}b_{m}{\bigl [}\Delta ^{m-1}f(z)-\Delta ^{m-1}f(a){\bigr ]}\\&\quad -b_{n}\,(z-a)\,\Delta ^{n}f(\xi ),\quad a<\xi <z,\end{aligned}}}$$ where the coefficients ${\displaystyle b_{m}}$ are the Bernoulli numbers of the second kind. Note the argument is without a shift, aligning with the inverse backward difference.

The indefinite sum ${\displaystyle \nabla ^{-1}f(x)=\sum _{k=1}^{x}f(k)}$ can be analytically continued by applying the standard Abel-Plana formula to the finite sum ${\displaystyle \sum _{k=1}^{n}f(k)}$ and then analytically continuing the integer limit ${\displaystyle n}$ to the variable ${\displaystyle x}$. This yields the formula:^[10] $${\displaystyle {\begin{aligned}\nabla ^{-1}f(x)&=\int _{1}^{x}f(t)dt+{\frac {f(1)+f(x)}{2}}\\&\quad +i\int _{0}^{\infty }{\frac {\left(f(x-it)-f(1-it)\right)-\left(f(x+it)-f(1+it)\right)}{e^{2\pi t}-1}}dt+C\left(x\right)\end{aligned}}}$$

This analytic continuation is valid when the conditions for the original formula are met. The sufficient conditions are:^[13][14]

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

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

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

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

Let ${\displaystyle F(x)=\nabla ^{-1}f(x)+C}$. Then, the constant ${\displaystyle C}$ is fixed from the condition ${\displaystyle \int _{-1}^{0}F(x)\,dx=0}$ or ${\displaystyle \int _{0}^{1}F(x)\,dx=0}$.

Let ${\displaystyle F(x)=\Delta ^{-1}f(x)+C}$. Then, the constant ${\displaystyle C}$ is fixed from the condition ${\displaystyle \int _{0}^{1}F(x)\,dx=0}$ or ${\displaystyle \int _{1}^{2}F(x)\,dx=0}$.

Alternatively, Ramanujan summation can be used: ${\displaystyle \sum _{x\geq 1}^{\Re }f(x)=-f(0)-F(0).}$ Or at 1: ${\displaystyle \sum _{x\geq 1}^{\Re }f(x)=-F(1)}$ respectively.^[29][30]

• 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

• Brian Hamrick: Discrete Calculus (PDF, 70 kB)
• Interactive visualization of the Nörlund principal solution for inverse backward differences. Implements Candelpergher's analytic continuation (Abel-Plana formula with recurrence) for visualizing Nörlund's principal solution.