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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 $

financial_physician
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  • Can you provide more context? It is not clear from your question what $dX^2$ is supposed to be. – Math1000 Dec 05 '19 at 03:29
  • Ahhh I think it makes sense now. We're using a heuristic that dt^(some power greater than one) is zero. So I think dX must be the same thing here. Thank you! – financial_physician Dec 05 '19 at 21:10
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    If $dX$ is a Wiener process, it should be for Ito calculus $(dX)^2\sim dt$ and not a higher power of $dt$. – Jon Dec 05 '19 at 23:29
  • You're right, that's one of the identities we learned for stochastic processes. I'm still trying to figure it out then... I've updated the question. Could you flesh out the details my professor omitted? It's not standard calculus for me haha. – financial_physician Dec 06 '19 at 03:17

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