Indefinite sum

Short description: Inverse of a finite difference

In the calculus of finite differences, the indefinite sum (or antidifference operator), denoted by $\sum _x$ or ${\mathrm{\Delta }}^{-1}$, is the linear operator that inverts the forward difference operator $\mathrm{\Delta }f(x)=f(x+1)-f(x).$ That is, if $\sum _{x}f(x)=F(x)$, then $F$ satisfies the functional equation

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

so that applying the forward difference recovers the original function: $\mathrm{\Delta }\sum _{x}f(x)=f(x).$ The operator thus plays the same role for finite differences that the indefinite integral plays for the derivative.

An indefinite sum is not unique: adding any 1-periodic function $C(x)$ (satisfying $C(x+1)=C(x)$), the function $F(x)+C(x)$ is also a solution. Therefore, an indefinite sum is unique up to a 1-periodic function $C(x)$ instead of up to a constant $C$ as the indefinite integral is.

To obtain the unique solution up to a constant $C$, one must impose additional analytic constraints. The Nørlund principal solution is the unique analytic solution that has the minimal possible exponential type (that is, its growth in the imaginary direction on the complex plane is the minimal possible), filtering out any non-constant periodic component. Other methods include higher-order convexity conditions in real analysis, or using axioms and complex analysis to step back the function's behavior from a neighborhood of infinity in which it behaves polynomially.

For integer arguments, the indefinite sum naturally extends ordinary summation, turning a discrete sum into a continuous function. Many such extensions are well-known special functions.

Forward and backward difference conventions

File:Indefinite Sum Forward vs Backward Comparison.png

A comparison of the indefinite sum operators to their discrete counterparts. The inverse backward difference of x is shown in yellow, and the inverse forward difference of x is shown in blue (both with respect to x).

The inverse forward difference operator, ${\mathrm{\Delta }}^{-1}$ ($F(x+1)-F(x)=f(x)$), extends the summation up to $x-1$, typically starting with the iterator at $0$:

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

Some authors analytically extend summation for which the upper limit is the argument without a shift, typically starting the iterator at $1$:

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

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

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

Which follows from the discrete counterpart:

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

Some authors use the equivalent form called the telescoping equation:

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

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 $C$.

Fundamental theorem of the calculus of finite differences

Indefinite sums can be used to calculate definite sums with the formula:

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

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

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

Examples

Main article: List of indefinite sums

The following basic indefinite sums follow from the fundamental properties of the difference operator, where $C(x)$ represents an arbitrary 1-periodic function (or a constant if the Nørlund principal solution is assumed):

Constant:

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

Exponential:

$\sum _x{a}^x=\frac{a^x}{a-1}+C(x)a\ne 1$

Logarithm:

$\sum _x\ln x=\ln \mathrm{\Gamma }(x)+C(x)$

Powers:

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

where $B_a(x)$ are the Bernoulli polynomials (via Abel-Plana, Hurwitz zeta, or as defined by their recurrence; not the definition by generating functions), $\zeta (s,a)$ is the Hurwitz zeta function, and $\psi (z)$ is the digamma function. This is related to the generalized harmonic numbers. Combined with series expansions (such as Taylor series) or partial fraction decomposition, the power formula allows the indefinite summation of many analytic functions and rational functions (term-wise, through the linearity of the operator).

Falling factorials

Main article: Falling and rising factorials

Falling factorials provide the discrete analog of the power rule from differential calculus. In infinitesimal calculus, $\frac{d}{dx}{x}^n=n{x}^{n-1}$. In the calculus of finite differences, the falling factorial

$(x{)}_n=x^{\underset{\_}{n}}=x(x-1)(x-2)\cdots (x-n+1)=\frac{\mathrm{\Gamma }(x+1)}{\mathrm{\Gamma }(x-n+1)}$

plays the role of $x^n$, and the forward difference operator satisfies

$\mathrm{\Delta }(x{)}_n=n(x{)}_{n-1}.$

The indefinite sum of a falling factorial is given by the discrete analog of the power rule for integration:

$\sum _x(x{)}_n=\frac{(x{)}_{n+1}}{n+1}+C(x),n\ne -1.$

Equivalently, using the Gamma function:

$\sum _x\frac{\mathrm{\Gamma }(x+1)}{\mathrm{\Gamma }(x-n+1)}=\frac{\mathrm{\Gamma }(x+1)}{(n+1)\mathrm{\Gamma }(x-n)}+C(x),n\ne -1.$

For the case where $n=-1$, the solution is the digamma function with a shift, $\psi (x+1)+C(x)$, which naturally extends the harmonic numbers.

Example: Sum of the first $x$ squares. Using $k^2=(k{)}_2+(k{)}_1$ and the indefinite sum formula above,

$\sum _k{k}^2=\frac{(k{)}_3}{3}+\frac{(k{)}_2}{2}+C(k).$

Applying the fundamental theorem of the calculus of finite differences,

$\begin{array}{rl}\sum _{k=0}^x{k}^2& ={(\frac{(k{)}_3}{3}+\frac{(k{)}_2}{2})|}_0^{x+1}\\ & =(\frac{(x+1{)}_3}{3}+\frac{(x+1{)}_2}{2})-(\frac{(0{)}_3}{3}+\frac{(0{)}_2}{2})\\ & =\frac{(x+1{)}_3}{3}+\frac{(x+1{)}_2}{2}.\end{array}$

Expanding the falling factorials,

$(x+1{)}_3=(x+1)x(x-1),(x+1{)}_2=(x+1)x,$

and simplifying yields the formula

$\sum _{k=0}^x{k}^2=\frac{x(x+\frac{1}{2})(x+1)}{3}.$

Summation by parts

Main article: Summation by parts

Indefinite summation by parts is the discrete analog of integration by parts. It is derived from the product rule for the forward difference operator.

Product rule. For two functions $u(x)$ and $v(x)$, the product rule for the forward difference is:

$\mathrm{\Delta }(u(x)v(x))=u(x)\mathrm{\Delta }v(x)+v(x+1)\mathrm{\Delta }u(x).$

Introducing the shift operator $\mathrm{E}$, defined by $\mathrm{E}f(x)=f(x+1)$, this can be written more compactly as:

$\mathrm{\Delta }(uv)=u\mathrm{\Delta }v+\mathrm{E}v\mathrm{\Delta }u.$

Summation by parts. Rearranging the product rule gives:

$u(x)\mathrm{\Delta }v(x)=\mathrm{\Delta }(u(x)v(x))-v(x+1)\mathrm{\Delta }u(x).$

Taking the indefinite sum of both sides and using the fact that $\sum _x\mathrm{\Delta }F(x)=F(x)+C(x)$ (where $C(x)$ is an arbitrary 1‑periodic function) yields the formula for summation by parts:

$\sum _{x}u(x)\mathrm{\Delta }v(x)=u(x)v(x)-\sum _{x}v(x+1)\mathrm{\Delta }u(x)+C(x).$

A symmetrical form, also obtained from the product rule, is:

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

Definite summation by parts. For definite sums from $a$ to $b$, the formula becomes:

$\sum _{k=a}^{b}u(k)\mathrm{\Delta }v(k)=[u(b+1)v(b+1)-u(a)v(a)]-\sum _{k=a}^{b}v(k+1)\mathrm{\Delta }u(k).$

Example: product of a polynomial and an exponential

Summation by parts is effective for functions like $k{2}^k$. To find the indefinite sum $\sum _{k}k{2}^k$, let $u(k)=k$ and $\mathrm{\Delta }v(k)=2^k$. Then:

• $\mathrm{\Delta }u(k)=(k+1)-k=1$
• $v(k)=\sum _k{2}^k=\frac{2^k}{2-1}=2^k$
• $\mathrm{E}v(k)=v(k+1)=2^{k+1}$

Applying the summation by parts formula:

$\sum _{k}k{2}^k=k\cdot 2^k-\sum _k{2}^{k+1}\cdot 1+C(k).$

The remaining sum is elementary:

$\sum _k{2}^{k+1}=2\sum _k{2}^k=2\cdot 2^k=2^{k+1}.$

Hence the indefinite sum (antidifference) is

$F(k):=\sum _{k}k{2}^k=k{2}^k-2^{k+1}+C(k)=(k-2)2^k+C(k).$

To evaluate the definite sum from $0$ to $x$, we use the fundamental theorem with the forward difference inverse:

$\sum _{k=0}^{x}k{2}^k=F(x+1)-F(0).$

Substituting the expression for $F$:

$\begin{array}{rl}\sum _{k=0}^{x}k{2}^k& =[(x+1-2)2^{x+1}]-[(0-2)2^0]\\ & =(x-1)2^{x+1}-(-2)\\ & =(x-1)2^{x+1}+2.\end{array}$

Thus, for any non‑negative integer $x$,

$\sum _{k=0}^{x}k{2}^k=(x-1)2^{x+1}+2.$

Uniqueness of the principal solution

File:Finite Difference Visualisation One Periodic Vanishing.webm

Visualization of the property ΔC(x)=0 for a 1-periodic function C(x). Because C(x+1)=C(x), the forward difference C(x+1)-C(x) and backward difference C(x)-C(x-1) both vanish.

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

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

Complex analysis (exponential type)

Following the theory developed by Niels Erik Nørlund, 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.

File:Inverse Finite Difference Partitioning.webm

Partitioning the complex plane for the inverse finite difference of $1/(x^2+1)$.

The usual formulation assumes that the summand $f(z)$ is analytic in a vertical strip containing a portion of the real line. However, when $f(z)$ has singularities (including those extending into the imaginary direction), a single vertical strip cannot contain the entire real axis. Instead, these singularities create vertical boundaries that split the domain into disjoint connected components. For example, poles at $\pm i$ prevent a single strip from crossing the imaginary axis, splitting the domain at $\mathrm{\Re }(z)=0$ into disjoint connected half-planes.

Nørlund’s theory provides a principal solution in each connected component that contains a segment of the real line. While these infinite vertical strips can be shifted horizontally to evaluate the function, they cannot cross the singularities without the recurrence relation causing the singularities to repeat (e.g. digamma). Thus, each connected component's principal solution contains no singularities in its respective connective component, but contains singularities that recur outwards into outer disjoint connected components.

The solution that contains the largest defined portion of the discrete sum being extended is then considered the disjoint connected component defining the canonical principal solution. This usually becomes, in practice, the right half-plane.

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

Let $F_1(z)$ and $F_2(z)$ be two analytic solutions satisfying this growth condition on the same connected component. Their difference $C(z)=F_1(z)-F_2(z)$ is then analytic, 1-periodic (i.e., $C(z+1)=C(z)$), and inherits the same exponential type less than $2\pi$.

Nørlund uses a fundamental result in complex analysis (related to Carlson's theorem, the Phragmén–Lindelöf principle, and the Paley–Wiener theorem) which states that a non-constant periodic entire function must have exponential type at least $2\pi$. This follows from its Fourier series expansion: if $C(z)$ is non-constant, its Fourier series contains a term $a_n{e}^{2\pi inz}$ with $n\ne 0$, which has type $2\pi |n|\ge 2\pi$. Since $C(z)$ has type strictly less than $2\pi$, it cannot contain any such term and therefore must be constant. Hence, on any fixed connected component where the growth condition holds, the solution is unique up to a constant.

The exponential type less than $2\pi$ in the imaginary direction on $f$ condition is sufficient but not strictly necessary. Nørlund's general definition of the principal solution is the analytic solution $F$ having Fourier components of the minimal possible exponential type for the given $f$ ($F$ of slowest possible growth in the complex plane). If $f$ has exponential type $k$ in imaginary direction, then the principal solution $F(z)$ will also have type $k$ in that strip, provided it converges. For example, $f(z)=\sin (7z)$ has exponential type $7$; its principal solution exists and has type $7$, even though $7>2\pi$.

When $f$ has exponential type exactly $2\pi n$ for some non-zero integer $n$ in every strip where it is analytic (e.g. $f(z)=\sin (2\pi nz)$ has type $2\pi n$; its antidifference contains $\sin (\pi n)=0$ in the denominator) the principal solution fails to exist (or is undefined everywhere) because it resonates with the kernel of the difference operator: $${\mathrm{\Delta }}^{-1}=\frac{1}{e^D-1}.$$ In all other cases (i.e., when $f$ is meromorphic and on some vertical strip that contains a segment of the real line and its exponential type is not an integer multiple of $2\pi$) the principal solution exists and is uniquely determined (up to a constant) on that connected component. Different components may give distinct branches; the canonical branch is the one analytic on the component containing the positive integers.

Example: Partitioning disjoint connected components (${\mathrm{\nabla }}^{-1}$)
Consider the meromorphic function $f(x)=1/(x^2+1)$. Its poles at $x=\pm i$ split the complex plane into two maximal vertical strips that each contain a segment of the real line: the right half-plane $\mathrm{\Re }(x)>0$ and the left half-plane $\mathrm{\Re }(x)<0$. We construct the Nørlund principal solution of the backward difference equation $F(x)-F(x-1)=f(x)$ with empty-sum normalization $F(0)=0$ on each strip.

Right half-plane ($\mathrm{\Re }(x)>0$). Partial fractions give

$\frac{1}{x^2+1}=\frac{1}{2i}(\frac{1}{x-i}-\frac{1}{x+i}).$

The digamma function satisfies $\psi (z+1)-\psi (z)=1/z$ for all $z\notin \{0,-1,-2,\dots \}$. For $\mathrm{\Re }(x)>0$ the arguments $x+1\pm i$ and $1\pm i$ never hit a non-positive integer, so the identity is valid at every term of the sum:

$\sum _{k=1}^x\frac{1}{k^2+1}=\frac{1}{2i}\left[\right(\psi (x+1-i)-\psi (1-i))-(\psi (x+1+i)-\psi (1+i)\left)\right].$

Because $\psi (\bar{z})=\overline{\psi (z)}$, the two terms are complex conjugates, and the expression simplifies to a real function:

$F_{\text{right}}(x)=\mathrm{Im}\psi (1+i)-\mathrm{Im}\psi (x+1+i),\mathrm{\Re }(x)>0.$

The non-simplified function is analytic on the whole right half-plane, and $F_{\text{right}}(0)=0$. Within this strip the difference equation $F(x)-F(x-1)=1/(x^2+1)$ holds for every $x$ in the respective connected component (half-plane). If one attempts to continue $F_{\text{right}}$ across the imaginary axis by the recurrence $F(x+1)=F(x)+f(x+1)$, poles appear at $x=\pm i-1,\pm i-2,\dots$-integer shifts of the original singularities that lie in the left half-plane.

Left half-plane ($\mathrm{\Re }(x)<0$). Using the reflection $x\mapsto -x$, the analogous solution on the left half-plane is

$F_{\text{left}}(x)=\mathrm{Im}\psi (-x+i)-\mathrm{Im}\psi (i),\mathrm{\Re }(x)<0.$

For $\mathrm{\Re }(x)<0$ the argument $-x+i$ has a positive real part, so it never equals $0,-1,-2,\dots$; the digamma identity applies and the difference equation is satisfied for all $x$ in the strip. Again the non-simplified function is analytic on the whole left half-plane. Extending this solution to the right via the recurrence would introduce poles at $x=\pm i,\pm i+1,\pm i+2,\dots$, which lie in the right half-plane outside the original strip.

Summary:

Domain	Principal solution (inverse backward difference, $F(0)=0$)
$\mathrm{\Re }(x)>0$	$\mathrm{Im}\psi (1+i)-\mathrm{Im}\psi (x+1+i)$
$\mathrm{\Re }(x)<0$	$\mathrm{Im}\psi (-x+i)-\mathrm{Im}\psi (i)$

The two expressions are analytic on their respective strips and give distinct principal solutions. The poles of the digamma function, which would violate the identity $\psi (z+1)-\psi (z)=1/z$, are never reached inside the respective domains. However, the recurrence propagates the original singularities of $f$ by integer steps, so any attempt to analytically continue one branch into the other component introduces poles.

Real analysis (higher‑order convexity)

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

Marichal and Zenaïdi proved the following uniqueness theorem, their method requiring the solution to be eventually $p$-convex or $p$-concave.

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

$f(x)=f(1)+\underset{n\to \mathrm{\infty }}{\lim }(\sum _{k=1}^{n-1}g(k)-\sum _{k=0}^{n-1}g(x+k)+\sum _{j=1}^p(\genfrac{}{}{0ex}{}{x}{j}){\mathrm{\Delta }}^{j-1}g(n)),$

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

Müller–Schleicher axiomatic method

In their paper How to Add a Noninteger Number of Terms, 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

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

to non-integer and complex upper limits $x$. The definition is built upon six natural axioms:

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

$|\sum _{\nu =x}^{y}f(\nu +n)-\sum _{\nu =x}^y{p}_n(\nu +n)|\to 0$.

Axioms S1–S4 force the sum to align with the ordinary finite sum when the limits are integers. Axiom S5 forces monomials to behave the same way under the generalization of fractional sums. Axiom S6 is the crucial axiom which allows one to "step back" the asymptotic region to determine the fractional sum in a finite interval. The exact conditions for the method to work are, as stated in the Definition 1.2 of the paper:

Let $U\subset \mathbb{ℂ}$ and $\sigma \in \mathbb{\mathbb{N}}\cup \{-\mathrm{\infty }\}$. A function $f:U\to \mathbb{ℂ}$ will be called fractional summable of degree $\sigma$ if the following conditions are satisfied:

• $x+1\in U$ for all $x\in U;$
• there exists a sequence of polynomials $(p_n{)}_{n\in \mathbb{\mathbb{N}}}$ of fixed degree $\sigma$ such that for all $x\in U$

$|f(n+x)-p_n(n+x)|\to 0$ as $n\to +\mathrm{\infty }$

• for every $x,y+1\in U,$ the limit

$\underset{n\to \mathrm{\infty }}{\lim }(\sum _{\nu =n+x}^{n+y}{p}_n(\nu )+\sum _{\nu =1}^n(f(\nu +x-1)-f(\nu +y)\left)\right),$

exists.

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

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

Symmetry of the principal solution

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

• Difference Equation: $F(x)-F(x-1)=f(x)$
• Normalization: $F(0)=0$ (empty sum boundary condition).
• Growth constraint: $F(z)$ has the minimal possible exponential type in the imaginary direction.

Under these conditions, $F(z)$ satisfies a reflection formula (referred to by Nørlund as Ergänzungssatz, a complementary theorem to uniqueness of the principal solution [Hauptlösung], presenting it as $G(x-\omega |-\omega )=G(x|\omega ),$$F(x-\omega |-\omega )=F(x|\omega )$ where $\omega$ is the span). From Nørlund’s Ergänzungssatz for the principal solution, one obtains the following symmetry for the inverse backward difference when the summand is odd or even under the condition $F(0)=0$ via direct application (setting $\omega =1$).

File:Odd Symmetry Nörlund Principal Solution.png

Image of the inverse backward difference of x, where f(x)=x is a simple example odd function. In the real plane, the point symmetry appears as a line symmetry about negative 1 half.

Odd functions

If $f$ is an odd function ($f(-z)=-f(z)$) and that a principal solution $F$ exists. Define $H(x)=F(x)-F(-1-x)$. Using the difference equation $F(x)-F(x-1)=f(x)$ and oddness,

$\begin{array}{rl}H(x)-H(x-1)& =[F(x)-F(x-1)]-[F(-1-x)-F(-x)]\\ & =f(x)-[-f(-x)]=f(x)+f(-x)=0,\end{array}$

so $H$ is 1-periodic. Because $F$ has minimal exponential type, $H$ does as well; by Nørlund’s uniqueness theorem, a non-constant 1-periodic function of type $<2\pi$ must be constant. Hence $H$ is constant. Evaluating at $x=0$ with $F(0)=0$ and $f(0)=0$ (oddness) gives $F(-1)=0$, so $H(0)=0$. Therefore $H\equiv 0$, yielding $0=F(x)-F(-1-x),$

$F(z)=F(-1-z),$

a point symmetry about $z=-1/2$. For example, $f(z)=z$ gives $F(z)=\frac{z(z+1)}{2}$.

Even functions

If $f$ is an even function ($f(-z)=f(z)$) with a principal solution $F$, define $H(x)=F(x)+F(-1-x)$. Then

$\begin{array}{rl}H(x)-H(x-1)& =[F(x)-F(x-1)]+[F(-1-x)-F(-x)]\\ & =f(x)+[-f(-x)]=f(x)-f(-x)=0,\end{array}$

so again $H$ is a constant 1-periodic function. Setting $x=-1$ gives the constant: $H(-1)=F(-1)+F(0)=F(-1)$. Consequently,

$F(z)+F(-1-z)=F(-1).$

Relationship to indefinite products

Main article: Indefinite product

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

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

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

where $\varpi (x)$ is an arbitrary periodic function of period 1. This representation is valid provided a branch of the logarithm can be chosen so that $\ln \left(f\left(x\right)\right)$ is single-valued and its indefinite sum exists. Conversely, an indefinite sum may be represented as the logarithm of an indefinite product:

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

Expansions and definitions

Newton series

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

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

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

Bernoulli‑operator series expansion

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

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

where $B_v$ are the Bernoulli numbers defined by the generating function $\frac{t}{e^t-1}=\sum _{v=0}^{\mathrm{\infty }}{B}_v\frac{t^v}{v!}$. Under this convention $B_1=-\frac{1}{2}$.

If $f$ 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 $f(x)=x^n$ one obtains the antidifference:

$$\sum _x{x}^n=\frac{B_{n+1}(x)}{n+1}+C(x),$$

where $B_n(x)$ are the Bernoulli polynomials of the first order.

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

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

For non‑polynomials this expansion is generally asymptotic.

Relation to the inverse backward difference

If one instead expands the inverse backward difference operator, ${\mathrm{\nabla }}^{-1}=\frac{e^D}{e^D-1}$ (which extends $\sum _{k=1}^{x}f(k)$), it admits to the same expansion, but with $B_1=+\frac{1}{2}$ in place of $B_1=-\frac{1}{2}$.

Euler–Maclaurin formula

Main article: Euler–Maclaurin formula

The Euler–Maclaurin formula provides an asymptotic expansion for the inverse backward difference ${\mathrm{\nabla }}^{-1}f(x)=\sum _{k=1}^{x}f(k)$ when the function is sufficiently smooth. For any positive integer $m$, one has:

$$\begin{array}{rl}{\mathrm{\nabla }}^{-1}f(x)& =\underset{1}{\overset{x}{\int }}f(t)dt+\frac{f(1)+f(x)}{2}\\ & +\sum _{k=2}^m\frac{(-1{)}^k{B}_k}{k!}(f^{(k-1)}(x)-f^{(k-1)}(1))+R_m(x)+C(x),\end{array}$$

where $B_k$ are the Bernoulli numbers ($B_1=-\frac{1}{2}$, $B_3=B_5=\cdots =0$), and the remainder term is

$$R_m(x)=(-1{)}^{m+1}\underset{1}{\overset{x}{\int }}\frac{b_m(t)}{m!}{f}^{(m)}(t)dt,$$

with $b_m(t)=B_m(t-\lfloor t\rfloor )$ the periodized Bernoulli polynomial. The terms with odd $k>1$ vanish, so the sum effectively runs only over even indices. Choosing $m=2p$ gives the form

$$\begin{array}{rl}{\mathrm{\nabla }}^{-1}f(x)& =\underset{1}{\overset{x}{\int }}f(t)dt+\frac{f(1)+f(x)}{2}\\ & +\sum _{k=1}^p\frac{B_{2k}}{(2k)!}(f^{(2k-1)}(x)-f^{(2k-1)}(1))+R_{2p}(x)+C(x),\end{array}$$

with the remainder

$$R_{2p}(x)=-\underset{1}{\overset{x}{\int }}\frac{b_{2p}(t)}{(2p)!}{f}^{(2p)}(t)dt.$$

The formula gives the analytic continuation of the discrete sum.

Laplace summation (Gregory summation formula)

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 ${\mathrm{\Delta }}^{-1}f(x)$:

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

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

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

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

$$\begin{array}{rl}\sum _{x=a}^{z}f(x)& =\underset{a}{\overset{z}{\int }}f(x)dx\\ & -\sum _{m=1}^n{b}_m[{\mathrm{\Delta }}^{m-1}f(z)-{\mathrm{\Delta }}^{m-1}f(a)]\\ & -b_n(z-a){\mathrm{\Delta }}^{n}f(\xi ),a<\xi <z,\end{array}$$ where the coefficients $b_m$ are the Bernoulli numbers of the second kind. Note the argument is without a shift, aligning with the inverse backward difference.

Abel–Plana formula

Main article: Abel–Plana formula

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

This analytic continuation is valid when the conditions for the original formula are met. The sufficient conditions are:

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

Example: Step size generalization
Let $S>0$ be a real step size and suppose $f(z)$ satisfies the standard Abel–Plana conditions on the appropriate strips. Apply the Abel–Plana formula to the function $g(t)=f(St)$ with upper limit $n=x/S+1$: $$\begin{array}{rl}\sum _{k=1}^{x/S+1}f(Sk)& =\underset{1}{\overset{x/S+1}{\int }}f(St)dt+\frac{f(S)+f(x+S)}{2}\\ & +i\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(x+S-iSt)-f(S-iSt)-f(x+S+iSt)+f(S+iSt)}{e^{2\pi t}-1}dt.\end{array}$$ Now subtract the last term $f(x+S)$ from both sides, because $\sum _{k=1}^{x/S+1}f(Sk)=\sum _{k=1}^{x/S}f(Sk)+f(x+S)$: $$\begin{array}{rl}\sum _{k=1}^{x/S}f(Sk)& =\underset{1}{\overset{x/S+1}{\int }}f(St)dt+\frac{f(S)+f(x+S)}{2}-f(x+S)\\ & +i\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(x+S-iSt)-f(S-iSt)-f(x+S+iSt)+f(S+iSt)}{e^{2\pi t}-1}dt.\end{array}$$ Simplify the boundary terms: $\frac{f(S)+f(x+S)}{2}-f(x+S)=\frac{f(S)-f(x+S)}{2}.$ In the real integral, substitute $u=St$, $dt=du/S$, limits $t=1\to u=S$, $t=x/S+1\to u=x+S$: $$\underset{1}{\overset{x/S+1}{\int }}f(St)dt=\frac{1}{S}\underset{S}{\overset{x+S}{\int }}f(u)du.$$ The imaginary part is already in a convenient form; by reordering the terms it becomes: $$i\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{(f(S+iSt)-f(S-iSt))-(f(x+S+iSt)-f(x+S-iSt))}{e^{2\pi t}-1}dt.$$ Thus we obtain the step size generalization: $$\begin{array}{rl}\underset{S}{\overset{-1}{\mathrm{\nabla }}}f(x)& =\frac{1}{S}\underset{S}{\overset{x+S}{\int }}f(u)du+\frac{f(S)-f(x+S)}{2}\\ & +i\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{(f(S+iSt)-f(S-iSt))-(f(x+S+iSt)-f(x+S-iSt))}{e^{2\pi t}-1}dt\\ & +C(x),\end{array}$$ where $C(x)$ is a $S$-periodic function. The expression satisfies $F(x)-F(x-S)=f(x)$ and, with the empty sum convention $F(0)=0$ (or up to another constant convention; $C(x)$ being a constant function), defines the Nørlund principal solution where the growth condition on $f$ becomes type $<2\pi /S$ after the scaling.

Choice of the constant term

Because the indefinite sum is defined only up to an arbitrary 1-periodic function, the constant $C$ must be fixed by an additional condition. Three common choices are the empty sum condition, an integral mean condition that identifies the result with the classical Bernoulli polynomials, and Ramanujan summation.

Empty sum boundary condition

The most direct method forces the indefinite sum to extend the usual discrete sum and to satisfy the empty sum convention. This is the same as $\underset{x\to 0^{-}}{\lim }F(x)=0\text{or}\underset{x\to 0^{+}}{\lim }F(x)=0.$

File:Analytic Continuation Harmonic.png

Inverse backward difference of 1/x with respect to x, showing a shifted digamma function.

Inverse backward difference

${\mathrm{\nabla }}^{-1}f(x)$ corresponds to ${\sum }_{k=1}^{x}f(k)$. The convention ${{\mathrm{\nabla }}^{-1}f(x)|}_{x=0}=0$ makes the sum over an empty interval zero.

Inverse forward difference

${\mathrm{\Delta }}^{-1}f(x)$ corresponds to ${\sum }_{k=0}^{x-1}f(k)$. The same convention yields ${{\mathrm{\Delta }}^{-1}f(x)|}_{x=0}=0$.

These conditions determine the solution uniquely up to an additive constant. For example,

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

Here, $\zeta (-a)$ is the constant $C$ such that $F(0)=0$.

Integral mean condition

In the study of Faulhaber's formula and the Euler–Maclaurin formula, it is convenient to identify the indefinite sum of a monomial with the corresponding Bernoulli polynomial. The Bernoulli polynomials $B_n(x)$ are defined by the generating function

$\frac{t{e}^{xt}}{e^t-1}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x)\frac{t^n}{n!}$

together with the normalization

$\underset{0}{\overset{1}{\int }}{B}_n(x)dx=0(n\ge 1).$

This property follows from the difference equation $\mathrm{\Delta }{B}_{\nu }(x)=\nu x^{\nu -1}$ and the integration formula $\underset{x}{\overset{x+1}{\int }}{B}_{\nu }(z)dz=x^{\nu }$, derived by Nørlund and found in standard references.

To match this convention, the constant is fixed by requiring that the solution have zero mean over a unit interval. For the inverse backward difference one may use

$\underset{-1}{\overset{0}{\int }}({\mathrm{\nabla }}^{-1}f(x)+C)dx=0\text{or}\underset{0}{\overset{1}{\int }}({\mathrm{\nabla }}^{-1}f(x)+C)dx=0,$

and for the inverse forward difference

$\underset{0}{\overset{1}{\int }}({\mathrm{\Delta }}^{-1}f(x)+C)dx=0\text{or}\underset{1}{\overset{2}{\int }}({\mathrm{\Delta }}^{-1}f(x)+C)dx=0.$

Example. For ${\mathrm{\Delta }}^{-1}x=\frac{x(x-1)}{2}+C$, the condition $\underset{0}{\overset{1}{\int }}({\mathrm{\Delta }}^{-1}x)dx=0$ gives $C=\frac{1}{12}$. Hence ${\mathrm{\Delta }}^{-1}x=\frac{x(x-1)}{2}+\frac{1}{12}=\frac{1}{2}{B}_2(x)$ with $B_2(x)=x^2-x+\frac{1}{6}$, consistent with the Bernoulli normalization.

This normalization is not mandatory; in modern treatments the empty sum condition is usually preferred. This is usually used in context of Bernoulli polynomials, the Hurwitz or Riemann zeta functions, generalized harmonic number function, or when dealing with monomials.

See also

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

Further reading

• "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

External links

• 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.