Two of the ways of doing calculus with algebra are non-standard analysis NSA and smooth infinitesimal analysis SIA. NSA has a technique called 'taking the standard part' which neglects incremental (or infinitesimal) terms at the end of derivations, whereas SIA has the nilsquare rule which neglects higher power incremental terms during derivations. For example if we differentiate $y = x^2$ from first principles we have: $$(x + h)^2 = x^2 + hx^{2'}$$ $$x^2 + 2hx + h^2 = x^2 + hx^{2'}$$ $$x^{2'} = 2x + h$$ $$x^{2'} = 2x$$ in NSA, whilst in SIA we have: $$(x + h)^2 = x^2 + hx^{2'}$$ $$x^2 + 2hx + h^2 = x^2 + hx^{2'}$$ $$2hx = hx^{2'}$$ $$x^{2'} = 2x$$ It's tempting to think the two methods are equivalent. However, consider the formula for secant length: $$s^2 = h^2 + (hy')^2$$ $$s = h \sqrt{1 + y'^2}$$ Taking the standard part of $y'$ while evaluating $s$ would result in the first RHS term being set to zero, rendering the equation useless, whereas the nilsquare rule doesn't have this effect. What does this tell us, if anything, about the validity and/or usefulness of NSA and SIA relative to eachother?


Taking the standard part is not really a technique for neglecting infinitesimal terms at the end of a derivation. Instead, non-standard analysis has a bona fide standard part function $st: \mathbb{R}^* \rightarrow \mathbb{R}$, a map that assigns an (extended) real number to each hyper-real number.

With this in mind, your first derivation, where you claim to differentiate the squaring function using NSA, is invalid. In particular, the claimed equality $(x + h)^2 = x^2 + h(x^2)'$ does not hold at all; even if it did, you are not allowed just neglect infinitesimals like you do when you pass from $(x^2)' = 2x + h$ to $(x^2)' = 2x$. None of this works.

What you should say instead is that the squaring function has derivative $2x$ because $$ st\left( \frac{(x+h)^2 - x^2}{h} \right) = 2x $$ for each real number $x \in \mathbb{R}$ and infinitesimal hyperreal $h \in \mathbb{R}^*$.

The same thing goes for your third derivation: you cannot just take the standard part in the middle of an equality and expect the resulting equation to hold. As an aside: in its present form, that third derivation does not hold up in Synthetic Differential Geometry (Smooth Infinitesimal Analysis) either - it amounts to inferring $s = h\sqrt{1 + (y')^2}$ from $s^2 = 0$.

This should answer your question: these arguments don't tell us anything about the validity and/or usefulness of NSA and SDG relative to each other, simply because they are not correct.

That said, the methods of NSA and SDG are far from equivalent: NSA strictly includes usual (standard) analysis, while SDG is specifically a way of doing differential calculus in a smooth topos - which excludes many mathematical artifacts that analysts are concerned with, but greatly simplifies a lot of differential-geometric concepts or computations. Then there are models of SDG in which both nilsquare SDG-style and proper NSA-style infinitesimals coexist. I'll stop here: a detailed look at these would be best suited as an answer to another (softer) question.

  • $\begingroup$ Firstly, if your take on NSA were correct then it's useless as a guide to calculus, which I don't believe even though on balance I prefer SIA (or SDG). Secondly, the equation for the secant $s$ is true by definition if $y'$ is the finite difference quotient and also for $s = 0$. The question is - is it true if $h$ is infinitesimal? Well SDG has nilpotent infinitesimals, so of course $s^2 = 0$; this observation doesn't tell us anything about $s$ itself. $\endgroup$ – selfawareuser1 Mar 1 at 20:23
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    $\begingroup$ The conclusion of your third argument is used in SDG when deriving arc length (with $y'$ denoting the derivative of the function $y$ - what else would it denote?!) but that inference is not valid. You can find a valid derivation somewhere in Bell's A Primer of Infinitesimal Analysis. As for NSA: it's the topic of my PhD thesis, so I'm fairly unlikely to be wrong about it, especially about such simple facts. None of those equations hold. $\endgroup$ – Z. A. K. Mar 1 at 20:45
  • $\begingroup$ It's on page 45 in this form: $s'(x) = (1 + f'(x)^2)^{1/2}$ ("Cancelling [h] yields the familiar equation"). So the answer is Yes - it does apply to infinitesimals, just by changing $s$ to $ds$ and dividing by $dh$. Also, if NSA can't even provide simple rules for differentiation then it's useless - why would you advocate for that? $\endgroup$ – selfawareuser1 Mar 1 at 21:09

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