brownian motion

The equation explained

f(x + \Delta, t) = f(x, t) + \Delta \frac{\partial f(x, t)}{\partial x} + \frac{\Delta^2}{2!} \frac{\partial^2 f(x, t)}{\partial x^2} \ldots \text{ in inf.}

In wordsThe number of particles per unit volume a displacement away, at the same time, written as its value here plus the displacement times its slope, plus half the displacement squared times the curvature, and so on without end.OverviewThe second expansion: the density a small distance away, written as a series in that distance. It runs on indefinitely, and the paper prints it so.Full explanationThe paper expands the density in powers of

\Delta

and then says the expansion may be carried out under the integral, because only very small values of

\Delta

contribute to it. That is where the second condition on

\varphi

earns its keep: the series in

\Delta

need not converge usefully for large displacements, since the frequency law gives those no weight. The series is printed as running on without end, and the truncation comes later and is argued separately. Each power of

\Delta

will meet the frequency law under the integral and turn into a moment of the step law, which is how the physics of the stepping enters an equation about the density.Every step

f(x + \Delta, t) = f(x, t) + \Delta \frac{\partial f(x, t)}{\partial x} + \frac{\Delta^2}{2!} \frac{\partial^2 f(x, t)}{\partial x^2} \ldots \text{ in inf.}

The Taylor expansion of the density in the position, about the point

x

, at fixed time. The paper prints it as continuing indefinitely and does not truncate it here.

\Delta^2 \ll \Delta

For the small displacements the frequency law admits, each power is far smaller than the one before, so the later terms of the series matter less and less. This is the ground the paper gives for expanding under the integral, not a separate assumption.AlsoOverviewFull explanationEvery step

Read it in the paper