why does $dX^2 \rightarrow 0$?

55 Views Asked by At

In my textbook, we're talking a bit about Brownian Motion and Ito's Lemma. A short note that's in the book that I don't understand is that the Taylor series expansion for our higher order terms (2 or higher) go to zero. I'm just not sure why $dX^2$ goes to zero.


Here's the section of the text in question:

Ito's Lemma

$ f(X+\Delta) = f(X) + \Delta f_x(X) + \frac{1}{2} \Delta^2 f_{XX}(X) + \frac{1}{6}\Delta^3f_{XXX}(X) + ... $

As $\Delta$ approaches $dX$ standard calculus tells us that the second order and higher terms vanish $(dX^2 \rightarrow 0).$ In standard calculus,

$ f(X+dX) = f(X) + \Delta f_x(X)dX $