I’m not sure why, but high school calculus traditionally jumps straight into the continuous setting completely ignoring the fact that most of the computational methods have clear discrete analogues.

Discrete Derivatives and Integrals

For discrete functions \(f: \mathbb{Z} \to \mathbb{R}\) the derivative is pointwise defined as

\[\Delta f(x) = f(x) - f(x-1)\]

Like \(\frac{d}{dx}\) the \(\Delta\) is linear, \(\Delta(af + bg) = a\,\Delta f + b\,\Delta g\), and kills constants, \(\Delta c = 0\).

Looking at monomials, we see that \(\Delta x^n = x^n - (x-1)^n\)

expands out into an ugly sum of terms through binomial expansion. This isn’t that nice to work with, and so instead we can look at a better set of monomials that behave more like the continuous versions. These are the rising factorials:

\[x^{\overline{n}} = x(x+1)(x+2)\cdots(x+n-1) \quad (n \text{ factors})\]

which when we apply \(\Delta\) we get

\[\Delta x^{\overline{n}} = x^{\overline{n}} - (x-1)^{\overline{n}} = \big[x(x+1)\cdots(x+n-2)\big]\big[(x+n-1) - (x-1)\big] = n\, x^{\overline{n-1}}\]

just like with \(\frac{d}{dx}\). There’s a product rule too, with one discrete wrinkle, just requiring a shift on the second factor: \(\Delta(uv) = v\,\Delta u + u(x{-}1)\,\Delta v\). Summing both sides gives summation by parts, the analogue of integration by parts: \(\sum_{x=a}^{b} v\,\Delta u = uv\big\rvert_{a-1}^{b} - \sum_{x=a}^{b} u(x{-}1)\,\Delta v\), which resolves terms like \(\sum x\,2^x\) exactly the way \(\int x e^x\,dx\) yields to parts.

The chain rule is the one place the analogy breaks: \(\Delta f(g(x)) = f(g(x)) - f(g(x-1))\) won’t factor unless \(g\) is affine (e.g., shifting \(g(x) = x+c\) just reindexes, \(\Delta f(x+c) = (\Delta f)(x+c)\)) since differences care about integer steps and most \(g\) don’t preserve them.

Integration Is Just Summation

Suppose we want \(\sum_{x=a}^{b} g(x)\). The trick is to find an antidifference, a function \(F\) with \(\Delta F = g\), just as we’d hunt for an antiderivative. The sum telescopes:

\[\sum_{x=a}^{b} \Delta F(x) = \sum_{x=a}^{b} \big[F(x) - F(x-1)\big] = F(b) - F(a-1)\]

Every interior term is added once and subtracted once; only the ends survive, an analogue of the fundamental theorem of discrete calculus:

\[\sum_{x=a}^{b} g(x) = F(b) - F(a-1), \qquad \text{where } \Delta F = g\]

And we already have the antidifference of a rising factorial. Reading the power rule backwards:

\[\Delta\!\left(\frac{x^{\overline{n+1}}}{n+1}\right) = x^{\overline{n}}\]

which is the discrete twin of \(\int x^n\,dx = \frac{x^{n+1}}{n+1}\).

Quick Applications

Once you have the power rule and the fundamental theorem, a surprising amount falls out with no cleverness required.

Power sums. For \(\sum_{x=1}^N x\) we have \(g(x) = x = x^{\overline{1}}\), so the antidifference is \(x^{\overline{2}}/2 = x(x+1)/2\) and

\[\sum_{x=1}^{N} x = \frac{x(x+1)}{2}\Big|_{x=0}^{N} = \frac{N(N+1)}{2}\]

For higher powers, we just rewrite the monomial in the rising-factorial basis. e.g. for \(k=2\) we get \(x^{\overline{2}} = x^2 + x\), and \(x^2 = x^{\overline{2}} - x^{\overline{1}}\), and antidifferencing each piece gives

\[\sum_{x=1}^{N} x^2 = \left(\frac{x^{\overline{3}}}{3} - \frac{x^{\overline{2}}}{2}\right)\Bigg|_{x=0}^{N} = \frac{N(N+1)(N+2)}{3} - \frac{N(N+1)}{2} = \frac{N(N+1)(2N+1)}{6}\]

The same recipe handles \(\sum x^k\) for any \(k\): expand \(x^k\) in rising factorials (the coefficients turn out to be the Stirling numbers) and integrate term by term.

Geometric series. In ordinary calculus the exponential is special because it’s nearly its own derivative; the same is true here. For a constant \(c\), \(\Delta c^x = c^x - c^{x-1} = c^x\!\left(\frac{c-1}{c}\right)\), so \(c^x\) is its own difference up to a constant, with antidifference \(c^{x+1}/(c-1)\). The fundamental theorem then hands you the geometric series for free:

\[\sum_{x=1}^{N} c^x = \frac{c^{x+1}}{c-1}\Bigg|_{x=0}^{N} = \frac{c^{N+1} - c}{c-1} = \frac{c\,(c^N - 1)}{c-1}\]

Recurrences, and solving Fibonacci. Recurrence formulas are the discrete analogues of ODEs: \(\Delta a_n = g(n) \quad\Longleftrightarrow\quad a_n = a_{n-1} + g(n),\)

To solve a constant-coefficient ODE in school you’re taught to guess \(e^{\lambda x}\) and solve a characteristic polynomial; since \(c^x\) is the discrete exponential, the discrete move is to guess \(a_n = r^n\). Take Fibonacci, \(F_n = F_{n-1} + F_{n-2}\). Substituting \(r^n\) and dividing by \(r^{n-2}\) leaves the characteristic equation

\[r^2 = r + 1 \quad\Longrightarrow\quad r = \frac{1 \pm \sqrt{5}}{2},\]

the golden ratio \(\varphi\) and its conjugate \(\psi\). The general solution is \(F_n = A\varphi^n + B\psi^n\), and the initial conditions \(F_0 = 0\), \(F_1 = 1\) pin down \(A = -B = 1/\sqrt{5}\) to give Binet’s formula

\[F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}\]