In the derivation for the diffusion equation on the wikipedia article for Brownian motion, they have these equations:

enter image description here

I can't figure out how $\rho(x+\Delta,t)$ gets expanded, though. For a Taylor series, we have

$f(x) = f(\Delta) + \frac{df}{dx}(\Delta) (x-\Delta) + \frac{d^2 f}{d x^2}(\Delta) (x-\Delta)^2 + ...$

and therefore

$f(x+\Delta) = f(\Delta) + \frac{df}{dx}(\Delta) \Delta + \frac{d^2 f}{d x^2}(\Delta) \Delta^2 + ...$

but the derivatives are then defined at $\Delta$, not $x$. Can anyone tell me what step I'm missing in this expansion?

  • $\begingroup$ I can't speak specifically to this, but you'll often find that expansions in Physics equations resolve when terms become increasingly insignificant (i.e. the terms only impact decimal points that are way beyond sensible experimental limits). Another approach might be that derivatives of polynomials will disappear if there are only positive integer powers. $\endgroup$ – Dancrumb Dec 21 '15 at 19:57
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    $\begingroup$ $f(x+\Delta)=f(x)+f'(x) \Delta + ..$ ? Just do the expansion at $x$ $\endgroup$ – Bort Dec 21 '15 at 19:58
  • $\begingroup$ Hi Bort, could you go into more detail? Is this a variation of a Taylor expansion? $\endgroup$ – user1704042 Dec 21 '15 at 20:02
  • $\begingroup$ You should adress other users like this @user1704042 (so that they get notified). Its the same as your formula with $x$ and $\Delta$ interchanged. You don't have to expand in $x$ around $\Delta$, you could just as well expand in $\Delta$ around $x$ $\endgroup$ – Bort Dec 21 '15 at 20:06
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    $\begingroup$ the derivative is with respect to the argument of f, so more clearly you could write $f(x+\Delta)=f(x)+\left.\frac{df(y)}{dy}\right|_{y=x} \Delta + \ldots$ and $f(x+\Delta)=f(\Delta)+\left.\frac{df(y)}{dy}\right|_{y=\Delta } x+ \ldots$ $\endgroup$ – Bort Dec 21 '15 at 20:10

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