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I am yet to see the following disambiguated:

Is it true that $ds^{2}$ as in $\frac{d^{2}a(s)}{ds^2}$ is such that $ds^{2}=(ds)^{2}$?

This comes into play when dealing with SDE's as $(dW_{t})^{2}=dt$ which can be shown if it assumed $(dW_{t})^{2}=dW_{t}^{2}$.

BayesIsBae
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1 Answers1

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No, $\dfrac{ {\rm d}^2a(s)}{{\rm d}s^2}$ is an abbreviation for $\dfrac{{\rm d}\left( \dfrac{ {\rm d} a(s) }{{\rm d} s}\right)} {{\rm d} s}$. You can multiply out one of the ${\rm d}s$ but not the other.

See for example https://math.stackexchange.com/a/765077/97045 and other answers in that thread for reasons why treating ${\rm d}s^2$ as a product leads to contradictions.

DanielV
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  • Interesting. Outside of the context of $\dfrac{{\rm d}\left( \dfrac{ {\rm d} a(s) }{{\rm d} s}\right)} {{\rm d} s}$, does the differential $ds^{2}=(ds)^{2}$, ever? – BayesIsBae Apr 03 '19 at 00:39
  • I tentatively would say you could do that when you are multiplying 2 first derivatives together, but to claim something never leads to a contradiction is a really heavy claim. – DanielV Apr 03 '19 at 00:59
  • Perhaps then I will make a follow-up question w.r.t. the generality of it. – BayesIsBae Apr 03 '19 at 02:11