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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 d

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

mlochbaum 4y ago
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 https://en.wikipedia.org/wiki/Operational_calculus
Koshkin 4y ago
> Can you multiply by dx? Maybe. But can you multiply by ∂x?
- Blammar 4y ago
  "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.
  - Koshkin 4y ago
    Actually, “dx” is rigorously defined in the standard calculus, too, and yes, you can multiply (and divide) by it.
    - oolveea 4y ago
      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!
      - bmitc 4y ago
        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.
        Koshkin 4y ago
        > 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.
        oolveea 4y ago
        Can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.
thrown_22 4y ago
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.

mlochbaum 4y ago

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 https://en.wikipedia.org/wiki/Operational_calculus

Koshkin 4y ago

> Can you multiply by dx? Maybe. But can you multiply by ∂x?

Blammar 4y ago
"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.
- Koshkin 4y ago
  Actually, “dx” is rigorously defined in the standard calculus, too, and yes, you can multiply (and divide) by it.
  - oolveea 4y ago
    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!
    - bmitc 4y ago
      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.
      - Koshkin 4y ago
        > 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.
        oolveea 4y ago
        Can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.

thrown_22 4y ago

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.