What are the applications of finite calculus

Solution 1:

As I didn't really get a satisfactory answer (although thanks to user35071 for his link, and to those of you that commented), I have decided to answer my own question as best as I can (if I've missed anything important, please let me know in the comments):

1. Calculating closed-form summations from known expressions.
2. Approximation of infinite-calculus (forward-difference methods, numerical solutions to PDEs and ODEs, etc.)

It is often the case that we are trying to express the summation of some expression $f(x)$ in closed form, and this can be tricky, using standard-techniques.

For instance, computation of $\sum{x^{2}\:\delta x}$ can be done by using two known formulae:

$$x^{n}=\sum_{k}{\left\{n \atop k\right\}x^{\underline{k}}} \text{ and } \sum{x^{\underline{m}}\:\delta x}=\frac{x^{\underline{m+1}}}{(m+1)}\tag{1}$$

Where $\left\{n\atop k\right\}$ is read "$n$ subset $k$", and is a Stirling number of the second kind. And $x^{\underline{k}}=x(x-1)\cdots(x-k+1)$ is the falling-factorial function (note this is also written as the Pochhammer symbol: $(x)_{k}$).

Expanding $x^{2}$ in terms of falling factorials, we get:

$$x^{2}=x^{\underline{2}}+x^{\underline{1}}\implies\sum{x^{2}\:\delta x}=\sum{x^{\underline{2}}\:\delta x}+\sum{x^{\underline{1}}\:\delta x}$$

Using our known formulae in $(1)$, we get:

$$\sum{x^{2}\:\delta x}=\frac{x^{\underline{3}}}{3}+\frac{x^{\underline{2}}}{2}=\frac{x(x-1)(x-2)}{3}+\frac{x(x-1)}{2}=\frac{x(x-1)(2x-1)}{6}$$

Which corresponds to our standard closed form for $\sum_{k=1}^{n}{k^{2}}$.

If we don't restrict ourselves to integral finite-differences, then we can also use it to numerically approximate derivatives and integrals of continuous functions. There are several types of difference methods often used in numerical approximation to calculus,

$$\Delta_{h}[f](x)=f(x+h)-f(x) \tag{2}$$ $$\nabla_{h}[f](x)=f(x-h)-f(x) \tag{3}$$ $$\delta_{h}[f](x)=f\left(x+\frac{h}{2}\right)-f\left(x-\frac{h}{2}\right)\tag{4}$$

Where $(2)$ is called the forward-difference, $(3)$ is called the backwards-difference, and $(4)$ is called the central-difference. These can be used to approximate the derivative using the following formulae:

$$f'(x)\approx\frac{\Delta_{h}[f](x)}{h}\approx\frac{\nabla_{h}[f](x)}{h}\approx\frac{\delta_{h}[f](x)}{h} \tag{5}$$

Which is often used when a non-analytic evaluation of a derivative is required.

Solution 2:

Here is a link. There are some applications at the bottom!

Solution 3:

Solving ("integrating") difference equations.

Formally, of course, this is the same as finding closed forms of sums.

But only finite calculus itself unveils this equivalence, in the same way as differential calculus relates solving a differential equation to integrating an area.