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Such discussions show that teaching of calculus often tends to be overly algebraic, for no good reason. Clearly, Leibniz notation does not intrinsically contai

Such discussions show that teaching of calculus often tends to be overly algebraic, for no good reason. Clearly, Leibniz notation does not intrinsically contain any deep insights since at one point Leibniz himself was erroneously induced by it to think that d(xy)= d(x)d(y), which is false. Newton on the other hand thought in terms of simple geometric concepts (areas), which make it crystal clear that d(xy) = x d(y) + y d(x). On the topic of history of early analysis I cannot recommend enough this collection of anecdotes from Arnold: https://archive.org/details/huygensbarrownew0000arno https://archive.org/details/huygensbarrownew0000arno

Q6T46nT668w6i3m 3y ago
Ultimately, algebra _is_ important and calculus is a useful setting to practice algebraic manipulation. Nevertheless, the trend in modern mathematics education in the United States is increasingly towards a more balanced understanding (e.g., greater emphasis on geometric interpretations).
impossiblefork 3y ago
But you do have 1/(dy/dx) = dx/dy and you can get the insight deep if you follow on with actually using differentials properly, as in the paper linked here in this very thread by LudwigNagasena. But then of course, you see that d^2 f/dx^2 is a bad expression for the second derivative, and that you should actually use differentials properly and write it as (d^2 f)/(dx)^2 - df/dx (d^2 f)/(df)^2.
- bikenaga 3y ago
  The second derivative notation makes sense when you think of d/dx as an operator on functions. (As someone noted below, d/dx is essentially a tangent vector, which is something that operates on functions.) d/dx takes the derivative of what follows with respect to x, so (d/dx)(f) = df/dx is the derivative function of f with respect to x. Now you want to differentiate df/dx with respect to x, so you do "d/dx" to "df/dx": (d/dx) (df/dx) (This looks better if you write it with "real" fractions.) On the top you have "d d f" with is naturally written "d^2f". On the bottom, you have "dx dx", which is naturally written "dx^2". So the result should be written d^2 f/dx^2.
  - impossiblefork 3y ago
    The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.
    - bikenaga 3y ago
      We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.
      - impossiblefork 3y ago
        I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
        bikenaga 3y ago
        > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]
bikenaga 3y ago
Leibniz and the product/quotient rules: "Although at the outset Leibniz was uncertain about his method and hesitated as to whether or not d(x y) is the same as dx dy and whether d(x/y) is equal to dx/dy, he in the end answered these questions correctly, determining that d(x y) = x dy + y dx and d(x/y) = (y dx - x dy)/y^2. These values he found by allowing x and y to become x + dx and y + dy respectively. Upon subtracting the original value of the function from the new one and observing that dx dy is infinitely small in comparison with the terms x dy and y dx, the results are obtained." [1] Newton's algebraic vs. geometric approaches to calculus: Boyer [1] notes that in Principia (1687) "Newton presented them [his propositions] in the form of synthetic geometrical demonstrations with an almost complete lack of analytical calculations" - hence the geometrical approach you point out. But he actually wrote up accounts of his calculus in three papers (De analysi ... (1669), Methodus fluxionum ... (1671), De quadratura ... (1676)), all of which were published after 1700. In those earlier papers, his approach is algebraic, often using the binomial theorem. In one incredible result from De analysi, he shows that the area under y = a x^(m/n) is given by z = (m/(m + n)) a x^((m + n)/n). To do this, he increments z and x: z + o y = (m/(m + n)) a (x + o)^((m + n)/n) Expand the right side using the binomial theorem, subtract the z-expression from both sides, divide both sides by o, then drop any terms on the right containing o ... and you get y = a x^(m/n). That's pretty algebraic, right? But the incredible thing is that he's basically showing that (roughly) "the rate of change of the area under the y-curve is y", which is the Fundamental Theorem of Calculus! He's figured out that derivatives (rates of change) and integrals (areas under curves) are somehow inverses (at least in this case). Newton was pretty amazing! [1] Carl Boyer, "The History of the Calculus and Its Conceptual Development". Dover Publications, 1949.

Q6T46nT668w6i3m 3y ago

Ultimately, algebra _is_ important and calculus is a useful setting to practice algebraic manipulation. Nevertheless, the trend in modern mathematics education in the United States is increasingly towards a more balanced understanding (e.g., greater emphasis on geometric interpretations).

impossiblefork 3y ago

But you do have 1/(dy/dx) = dx/dy and you can get the insight deep if you follow on with actually using differentials properly, as in the paper linked here in this very thread by LudwigNagasena. But then of course, you see that d^2 f/dx^2 is a bad expression for the second derivative, and that you should actually use differentials properly and write it as (d^2 f)/(dx)^2 - df/dx (d^2 f)/(df)^2.

bikenaga 3y ago
The second derivative notation makes sense when you think of d/dx as an operator on functions. (As someone noted below, d/dx is essentially a tangent vector, which is something that operates on functions.) d/dx takes the derivative of what follows with respect to x, so (d/dx)(f) = df/dx is the derivative function of f with respect to x. Now you want to differentiate df/dx with respect to x, so you do "d/dx" to "df/dx": (d/dx) (df/dx) (This looks better if you write it with "real" fractions.) On the top you have "d d f" with is naturally written "d^2f". On the bottom, you have "dx dx", which is naturally written "dx^2". So the result should be written d^2 f/dx^2.
- impossiblefork 3y ago
  The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.
  - bikenaga 3y ago
    We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.
    - impossiblefork 3y ago
      I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
      - bikenaga 3y ago
        > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]

bikenaga 3y ago

Leibniz and the product/quotient rules: "Although at the outset Leibniz was uncertain about his method and hesitated as to whether or not d(x y) is the same as dx dy and whether d(x/y) is equal to dx/dy, he in the end answered these questions correctly, determining that d(x y) = x dy + y dx and d(x/y) = (y dx - x dy)/y^2. These values he found by allowing x and y to become x + dx and y + dy respectively. Upon subtracting the original value of the function from the new one and observing that dx dy is infinitely small in comparison with the terms x dy and y dx, the results are obtained." [1] Newton's algebraic vs. geometric approaches to calculus: Boyer [1] notes that in Principia (1687) "Newton presented them [his propositions] in the form of synthetic geometrical demonstrations with an almost complete lack of analytical calculations" - hence the geometrical approach you point out. But he actually wrote up accounts of his calculus in three papers (De analysi ... (1669), Methodus fluxionum ... (1671), De quadratura ... (1676)), all of which were published after 1700. In those earlier papers, his approach is algebraic, often using the binomial theorem. In one incredible result from De analysi, he shows that the area under y = a x^(m/n) is given by z = (m/(m + n)) a x^((m + n)/n). To do this, he increments z and x: z + o y = (m/(m + n)) a (x + o)^((m + n)/n) Expand the right side using the binomial theorem, subtract the z-expression from both sides, divide both sides by o, then drop any terms on the right containing o ... and you get y = a x^(m/n). That's pretty algebraic, right? But the incredible thing is that he's basically showing that (roughly) "the rate of change of the area under the y-curve is y", which is the Fundamental Theorem of Calculus! He's figured out that derivatives (rates of change) and integrals (areas under curves) are somehow inverses (at least in this case). Newton was pretty amazing! [1] Carl Boyer, "The History of the Calculus and Its Conceptual Development". Dover Publications, 1949.