Editing Summation (section)

==Calculus of finite differences==

Given a function {{mvar|f}} that is defined over the integers in the [[interval (mathematics)|interval]] {{math|[''m'', ''n'']}}, the following equation holds:
:<math>f(n)-f(m)= \sum_{i=m}^{n-1} (f(i+1)-f(i)).</math>

This is known as a [[telescoping series]] and is the analogue of the [[fundamental theorem of calculus]] in [[calculus of finite differences]], which states that:
:<math>f(n)-f(m)=\int_m^n f'(x)\,dx,</math>
where
:<math>f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}</math>
is the [[derivative]] of {{mvar|f}}.

An example of application of the above equation is the following: 
:<math>n^k=\sum_{i=0}^{n-1} \left((i+1)^k-i^k\right).</math>
Using [[binomial theorem]], this may be rewritten as: 
:<math>n^k=\sum_{i=0}^{n-1} \biggl(\sum_{j=0}^{k-1} \binom{k}{j} i^j\biggr).</math>

The above formula is more commonly used for inverting of the [[difference operator]] <math>\Delta</math>, defined by:
:<math>\Delta(f)(n)=f(n+1)-f(n), </math>
where {{mvar|f}} is a function defined on the nonnegative integers.
Thus, given such a function {{mvar|f}}, the problem is to compute the [[antidifference]] of {{mvar|f}}, a function <math>F=\Delta^{-1}f</math>  such that <math>\Delta F=f</math>. That is, <math>F(n+1)-F(n)=f(n).</math>
This function is defined up to the addition of a constant, and may be chosen as<ref name=CRC>''Handbook of Discrete and Combinatorial Mathematics'', Kenneth H. Rosen, John G. Michaels, CRC Press, 1999, {{isbn|0-8493-0149-1}}.</ref>
:<math>F(n)=\sum_{i=0}^{n-1} f(i).</math>

There is not always a [[closed-form expression]] for such a summation, but [[Faulhaber's formula]] provides a closed form in the case where <math>f(n)=n^k</math> and, by [[linearity]], for every [[polynomial function]] of {{mvar|n}}.