# Approximating a curve near a point

Close to a point, a smooth curve is described by its value there, its slope and its curvature. Dropping the rest needs a reason: how small the step is, and how fast the dropped terms shrink with it.

Written for this edition, not translated from Einstein.

## Why can a curve be replaced, close to one point, by a few simple terms?

What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller.

The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:

$$
\begin{aligned}f(x+\Delta)&=f(x)+\Delta f'(x)\\&\quad+\frac{\Delta^2}{2}f''(x)+\ldots\end{aligned}
$$

The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms.

The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small.

## Worked example: What the dropped term costs

$$
(x+\Delta)^2=x^2+2x\Delta+\Delta^2
$$

For the square function, the local expansion is exact through the quadratic term.

Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:

1. Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.
2. Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.
3. Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.
4. Halving the step quarters the error, because the dropped term is Δ².

## Where this lesson stops

An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.

## This lesson builds on

- [Squares and square roots](/foundations/bridge-squaring-square-roots/)