I've been interested in non-standard analysis recently. I was reading up on it and noticed the following interesting comment on the Wikipedia page about hyperreal numbers, right after giving an example of a nonstandard differentiation:
The use of the standard part in the definition of the derivative is a rigorous alternative to the traditional practice of neglecting the square of an infinitesimal quantity... the typical method from Newton through the 19th century would have been simply to discard the $dx^2$ term.
I've never heard anything like this before, and really find it fascinating that Newton's method was to define the relation $dx^2 = 0$. If we actually formalize the above structure by taking $\mathbb{R}$ and adjoining an element $dx^2 = 0$ to it, we get the "dual numbers," isomorphic to the quotient ring $\mathbb{R}[x]/x^2$. I'd seen some things about how this algebra plays into automated differentiation algorithms for some computer software systems, but I've never heard anything about Newton directly working in this algebra. So I have a few questions:
- Does anyone have more historical information on the way that Newton performed differentiation, and its relation to the dual numbers?
- Does anyone know how effectively real analysis can be formalized with the dual numbers? Does the resulting system play nice enough to develop all of the important modern results?
- If we start with $\mathbb{C}[x]/x^2$ instead, can we likewise develop complex analysis?
Since this idea is so simple, I'm very curious how powerful it is. I'm also curious if it has any major drawbacks too, since I'm not sure why anyone would mess with the foundational baggage involved in defining the hyperreals if this simple 2-dimensional real algebra could really do the trick.
$\endgroup$ 28 Answers
$\begingroup$The biggest draw back (and it's a big one) is that the ring of dual numbers is not a field. It has plenty of zero divisors. So, Newton, or any of the mathematicians of the early days of calculus, certainly did not work directly in the ring of dual numbers. They of course did not consider the ring to exist (as rings did not exist at all yet), but from their writing it is clear they envisaged a field of real numbers with, somehow, some notions of infinitesimals. Their work is of course very vague, but correct. Much more on that can be found in math history books. Many interesting discussions can be found in the recent book "Adventures in Formalism", also related to the early days of calculus and how things developed.
Some (rather unsatisfactory) portions of analysis can be developed in the ring of dual numbers, but it does not go too far. The idea, as you say, is very simple, perhaps too simple. One immediately gets into trouble when trying to define the derivative as the quotient of the infinitesimal $f(x+h)-f(x)$ divided by $h$, where $h$ is infinitesimal. The difficulty is that the non-zero infinitesimals in the ring of dual numbers are not invertible. So, it's the end of the party. (As you say though, some aspects of the party remain with automatic differentiation). In some sense, the dual numbers form a first order approximation to actual infinitesimals: The square of an infinitesimal is of an order of magnitude smaller than the infinitesimal you started with, but in the ring of dual numbers, the square of an 'infinitesimal' is precisely $0$. So, in a nonstandard model of the reals you have whole layers of infinitesimals. In the dual numbers there is only one layer, nothing in it is invertible, and they all square to $0$.
The book Models for smooth infinitesimal analysis explores many different models for analysis with infinitesimals. None of them is particularly simple.
$\endgroup$ 16 $\begingroup$No for 1. and 3., this ring is not really useful in analysis. But it is quite important for analytical considerations in algebraic geometry, the main reason being that the scheme $\mathrm{Spec}(k[\varepsilon]/\varepsilon^2)$ classifies tangent vectors. This makes it possible to define the tangent space of arbitrary functors $F : \mathsf{CRing} \to \mathsf{Set}$ at some $x \in F(k)$, namely as the fiber of $F(k[\varepsilon]/\varepsilon^2) \to F(k)$ at $x$. There is no manifold which represents tangent vectors for manifolds, so this is the main difference.
$\endgroup$ 4 $\begingroup$The answers by Ittay Weiss and Martin Brandenburg are helpful. I would like to point out a more direct shorcoming of the dual numbers as far as analysis (and even calculus) is concerned is that it is not clear how to extend a generic real function to the dual numbers, even say a $C^\infty$ smooth function. Thus, if one wishes to form a ratio of infinitesimals involved in the definition of the derivative, it is not clear what should appear in the numerator. Over the hyperreals, one has a systematic way of extending every real function to the wider hyperreal domain, and the transfer principle (which is arguably a formalisation of the Leibnizian Law of Continuity) ensures that such an extension is meaningful.
For this reason, the answer to the original question would be: No, dual numbers are insufficient to capture "Newton's historical method for doing calculus". The hyperreals provide a framework where the procedures of 17th century infinitesimal calculus can be successfully formalized.
$\endgroup$ 4 $\begingroup$I think no one mentioned Synthetic differential geometry, there you have nonzero quantities with $dx^2=0$. For a very readable introduction I suggest:
Bell, A primer of infinitesimal Analysis
$\endgroup$ 6 $\begingroup$In the dual numbers for any differentiable function holds $ f(x+\epsilon) = f(x) + \epsilon f^\prime (x)$. This is enough to handle computationally 1st derivatives. Of course it is not enough for the conventional definition of second derivative. So you can consider the duals as a computational model. Of course, on of the drawbacks is that the purely imaginary numbers are not invertible.
$\endgroup$ 12 $\begingroup$Duals Numbers, attributed to Eduard Study, are already
practically used, for example here:
From my point of view division seems not so much a problem, since it fails when ordinary division would also fail. Here is a set of arithmetic operations defined for duals, I am writing (x,y) instead of x+εy:
$$-(x,y) = (-x,-y)$$
$$(x,y)+(z,t) = (x+z,y+t)$$
$$(x,y)*(z,t) = (x*z,x*t+y*z)$$
$$\frac{(x,y)}{(z,t)} = (\frac{x}{z}, \frac{y*z-x*t}{z^2})$$
I guess the claim that duals cannot be used to define derivative, stems from a confusion with Jerome Keislers standard part. He writes translated to dual equations the following, and division is exactly the problem:
f((x,h)) - f((x,0))
------------------- = (f'(x), e) /* doesn't work */ (0,h)But if we use the hypothesis:
f((x,y)) = (f(x), f'(x)*y)We then find the following by using this hypothesis and the aforementioned arithmetic operations:
f((x,h)) - f((x,0))
------------------- = (0, f'(x)) /* works */ (h,0)And if this isn't convincing enough, we can also use the hypothesis to show, that duals reflect the chain rule:
f(g( (x,1) )) = f( (g(x), g'(x)) ) = ( f(g(x)), f'(g(x))*g'(x) )Hyperduals are an extension of duals where second or higher order derivatives can be also calculated.
But currently I rather would wish for duals that can compute f(x+) and f(x-) for me, i.e. left and right derivative. Currently experimenting.
$\endgroup$ $\begingroup$Yes... and no....
On the one hand, the dual numbers $\mathbb{R}[\epsilon] / (\epsilon^2)$ are a topological ring, and the projection $\mathbb{R}[\epsilon] / (\epsilon^2) \to \mathbb{R}$ is a vector bundle over the real line. In fact, it is isomorphic (as a vector bundle) to the tangent bundle $T\mathbb{R}$ and to the cotangent bundle $T^*\mathbb{R}$ in a rather suggestive way.
On the other hand, aside from being a neat way to differentiate polynomials, I'm not sure it actually does anything for you. e.g. while it's interesting to organize derivative information such as $\log(x+\epsilon y) = \log(x) + \epsilon \frac{y}{x}$, I don't know if it actually does anything to help you derive such formula.
On the cotangent side, matters are worse — I'm not aware of anything the dual numbers could do for you that the exterior algebra doesn't do as well or better.
$\endgroup$ 1 $\begingroup$This seems to be generating a lot of confusion.
It is simply that f(x+he)=f(x)+he df/dx where ee=0 ; algebraically and precisley for Taylor expandable differentiable functions.(ie analytic functions)
You can go through the table of all the elementary functions and write out the answer using this algebraic definition. Also proove all properties of derivatives. It is no better nor worse than the Lagrange definition of derivative, of whatever is sitting on the second term of the expansion.
Probably good for educational purposes, no torture using limit concept, the derivative is simply there. Of course analyists may misunderstand the algebra part. It is a ring, not a field. The he part is an ideal of the ring.
$\endgroup$
