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Comment by kjellsbells | Hacker News Reader

Does the notion of suggestability extend to the notion of "creative breakage"? The latter is an important feature of a mathematical notation.

For example the dy/dx notation in calculus naturally leads to inquisitive thinking. Can you multiply by dx? It looks like a division but it isnt, really, except that a soon as you get past Calc 101 you are spraying dys and dxs all over. the notation is just incredibly suggestive.

Another example is exponents where m^n is introduced naively for natural n but instantly prompts the question around non integer n.

i think that having a notation that is suggestive and can be creatively broken to use in new ways is v important

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Iverson usually expresses the derivative as an operator, which ties into Heaviside's operational calculus[0] but doesn't have multiplication rules like dx/dy (I'd probably be more on your side here, in favor of dx and dy). I don't think he was too comfortable with mathematical concepts that don't resolve into specific calculations. Which may be why his notation ended up being so easy to turn into a programming language. To me, Iverson's suggestivity would be more about unifying or making analogies between established concepts. For example, APL uses * for exponentiation and ⍣ for repeated function application.

[0] https://en.wikipedia.org/wiki/Operational_calculus

Leibniz's notation for calculus didn't start out as creative breakage, it meant what it said on the tin, dy/dx is an infinitesimal fraction. By comparison the modern notion of a derivative is that of a higher order function of type (R -> R) -> (R -> R). Something that was only made clear with the advent of types in the turn of the last century.

Creative breakage is a bug which tells you to think harder about what you're doing.

> Can you multiply by dx?

Maybe. But can you multiply by ∂x?

"dx" is a term rigorously defined in infinitesimal calculus (IMO a much easier to understand approach to differentiation.) So, yes, you can multiply by dx (even if x is a more complicated function.)

∂x would have a problematic definition, as it would require selecting from its components based on information not supplied. E.g., let x = y+z. Then dx = dy + dz. But ∂x (by extension) = ∂y + ∂z, but at least one of these terms on the right is identically zero, depending on whether y or z is held constant. So ∂x doesn't have a meaning.

Actually, “dx” is rigorously defined in the standard calculus, too, and yes, you can multiply (and divide) by it.

For the love of anything sacred, can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.

Like, I get “dx”, but I cannot put my finger to it!!! This might be because the Precise Definition of the Limit phrases it as “x approaches a”; it is as though we are “sent” to the land of dx, but not told what it is as an atomic concept!

I am not entirely sure what the above commenter means that dx is rigorously defined in "infinitesimal calculus" because I don't know what they necessarily mean by "infinitesimal calculus". As far as I am aware, there is standard calculus, non-standard analysis by Robinson, and smooth infinitesimal analysis that uses intuitionistic logic. The three are very different. dx has no meaning in standard calculus. It is simply there for notation. It is given meaning by the theory of smooth manifolds and differential forms. In that setting, differentials such as dx are given explicit meaning: they are functions that operate on tangent vectors. For example, apply dx to the unit vector d/dx + d/dy to get dx(d/dx+ d/dy) = d/dx(x) + d/dy(x) = 1 + 0 = 1.

> dx has no meaning in standard calculus

Sure it does. There is no need to know about smooth manifolds or differential forms to understand the differential of a function of one variable at a point and the meaning of dy = f’(x)dx.

Can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.

Any introductory calculus book worth the paper it’s printed on would gladly tell you that the differential of the function y = x at a point x0 is nothing more than x - x0 and that you do not have to think about it as something that is “infinitely small” or anything equally mysterious. (Some would even go as far as saying that “the differential of a function of one variable is a linear map of the increment of the argument.”) So, with dx = x - x0, you can do with it anything you want, even divide by it (assuming that dx stays non-zero).

What is the meaning then?

dy = f'(x)dx is just a definition for notional convenience, primarily employed when doing u or u-v substitution. My point is that dx in single variable calculus is notation. It is not an intrinsic object. dx is an intrinsic object as a differential form on a smooth manifold. Of course, the real line R is a 1-manifold, so dx does have that meaning, but you need to understand what a differential form is to know that.

One doesn't necessarily need the full generality of smooth manifolds though. Harold Edwards' Advanced Calculus: A Differential Forms Approach and Advanced Calculus: A Geometric View teach differential forms for Euclidean manifolds.

The second book is by James Callahan. I accidentally left that off.

I think I get your point, but at the same time I disagree.

1) The differential of a function (at a point), dy, is not notation, it is a concept.

2) The differential of the function y = x, dx, is not, then, a notation, either; and, since the derivative is 1, dx = 1 Δx = Δx = x - x0.

3) You can argue, of course, that using dx instead of Δx in dy = f'(x)dx is "notation," but I think the above shows that it is more than that.

Thank you for asking this. This plagued me for years as an undergrad physics student.

This is one book on infinitesimal calculus: https://www.amazon.com/Infinitesimal-Calculus-Dover-Books-Ma... .

However, and this is very amusing to me, it turns out that the process of automatic differentiation (see https://en.wikipedia.org/wiki/Automatic_differentiation, the section on dual numbers) works in exactly the same way. Just replace all of the primed symbols (u', v', etc.) with du, dv, etc. and dual numbers are isomorphic to infinitesimals (if I'm using that term correctly.)

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