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I began reading a book on mathematical finance out of curiosity. It is called "Paul Wilmott introduces Quantitative Finance" (2007) and is meant to show quantitative finance with the least amount of math (just enough to apply the concepts). In chapter 1 p. 15, the book gives some example:

$$M(t+dt) - M(t) \approx \frac{dM}{dt}dt + ...$$

This somehow doesn't completely match with my understanding, which is mostly math for engineering (not 100% rigorous). Although I can see that this is clearly a Taylor Series centered around t, some apparent contradictions/issues popped up:

  1. From what I recall, the Taylor series when taken as an infinite sum is supposed to be exactly equal to the function M(t).
  2. The fundamental theorem of calculus says (?): $$M(t+dt) - M(t) = \int_{t}^{t+dt}{\frac{dM}{dt}dt}$$ So does that mean that $$\int_{t}^{t+dt}{\frac{dM}{dt}dt} = \frac{dM}{dt}dt + \frac{1}{2}\frac{d^2M}{dt^2}dt^2...$$ and if so how do all those higher order terms add up that cleanly? (Or do they vanish?)
  3. I have also seen the definition of the derivative written as: $$\lim_{h \to 0}\frac{M(t+h) - M(t)}{h} = \frac{dM}{dt}$$ If dt is an infinitesimal, then what is mathematically wrong with writing something like: $$\frac{M(t+dt) - M(t)}{dt} = \frac{dM}{dt}$$ which would make the original expression in the book exactly equal (not approximate), and would make the higher order terms in the Taylor expansion be exactly zero?

My question is what is wrong with my assessment of the situation (in any or all of the sections I enumerated 1 to 3)?

If my logic is somehow correct for all 3 things I have shown above, then there still is the issue of showing that all higher order terms equal zero exactly. In engineering usually this is hand waved but I would like to know how they would equal zero exactly (not approximately zero).

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  • $\begingroup$ The final expression is only correct if you take the limit as $dt$ approaches zero. Before that happens, it is an approximation $\endgroup$ – Alex Pavellas Dec 10 '16 at 4:14
  • $\begingroup$ @AlexPavellas It is exactly correct for a secant. It is an approximation for the tangent. $\endgroup$ – user301988 Dec 10 '16 at 13:10

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