If f(x) is a function on integers, the forward difference operator is defined by

For example, say *f*(*x*) = *x*^{2}. The forward difference of the sequence of squares 1, 4, 9, 16, … is the sequence of odd numbers 3, 5, 7, …

There are many identities for the forward difference operator that resemble analogous formulas for derivatives. For example, the forward difference operator has its own product rule, quotient rule, etc. These rules are called the calculus of finite differences. The finite results are often much easier to prove than their continuous counterparts.

The calculus of finite differences makes it possible to solve some discrete problems systematically, analogous to the way one would solve continuous problems with ordinary calculus. For example, there is a “summation by parts” technique for computing sums analogous to integration by parts for integrals. This is quite useful technique and I intend to write a separate post on it.

The product rule for forward differences looks a little odd:

The left hand side is symmetric in f and g though the right side is not. There is also a symmetric version:

Here is the quotient rule for forward differences.

One of the first things you learn in calculus is how to take the derivative of powers of *x*. If *f(x*) = *x ^{n}* then

*f*‘(

*x*) =

*n*

*x*

^{n-1}. There is an analogous formula in the calculus of finite differences, but with a different kind of power of

*x*. For positive integers

*n*, define the

*n*

^{th}falling power of

*x*by

Then

Falling powers can be generalized to non-integer exponents by defining

The formula for finite difference of falling powers given above remains valid when using the more general definition of falling powers. Falling powers arise in many areas: generating functions, power series solutions to differential equations, hypergeometric functions, etc.

The function 2* ^{x}* is its own forward difference, i.e.

analogous to fact that *e ^{x}* is its own derivative.

Here are a couple more identities showing a connection between the gamma function and finite differences. First,

Also,

I plan to write posts in the future about summation by parts and the connection between harmonic numbers and logs.

To read more about the calculus of finite differences, see Concrete Mathematics.

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John,

Thanks for posting this: I would be interested to see some applications. Would you believe that in my four year degree course on Mathematics we never covered this subject!

Sam

Sam, I got a PhD in math without ever seeing this. I worked with finite differences as approximations, but I never saw this elaborate calculus for working with finite differences until sometime after I was out of school.

Notice there’s no approximation going on here, only exact results. These tools can be used to solve discrete problems with no reference to ordinary calculus.

I’ve listened some older teachers telling about this but I never had the chance to study with details. I’ve been using foward and backward operators but didn’t know there were calculus-like rules. Thanks for the post.

John,

“Discrete Calculus” is indeed very interesting. I only discovered its existence last year, and I was immediately intrigued by it.

You mentioned the “summation by parts” technique. Another cool technique of discrete calculus that bears resemblance with continuous calculus is the “discrete l’Hôpital’s rule”, i.e., the Stolz-Cesàro theorem.

So which function should be considered the discrete analogue (no pun) of the logarithm function? Log base 2 , or H(n), the nth harmonic number?

Jack, it depends on which analogy you want to draw. Log base 2 is analogous to the natural logarithm in that it is the inverse of 2

^{x}. But the harmonic numbers are analogous to the natural logarithm in that Δ H_{n}= 1/n, just as the derivative of ln(x) is 1/x.Please give us a short course in summation by parts. Thank You.

@John: I would also very much appreciate a post about summation by parts! Thank you!

Thanks, great article!

Thanks muchly for this informative post. whilst I look forward to your followup on summation by parts, I jumped the gun a little and derived the formula from the product rule (since that’s the easiest way in infinitesimal calculus), with a bit of effort I used it to glean a summation formula for [k^n from k = 1 to x] with any chosen n in terms of the summation formulae for lower values of n.

Very satisfying result for me, feel like I’m much more apt with sums now that I have this ‘disctrete calculus’ in my toolkit and some practice under my belt.