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I have been wondering for some time what the limits of Leibniz notation is, and what exactly its meaning is. I learned limits and later learned (to some extent) infinitesimals, but there are some oddities which have me befuzzled. The one person I know who could answer the question gave me a reference so dense I couldn't make heads or tails of it.

In any case, let's say you have a function $y = f(x)$. Now, the derivative is $\frac{dy}{dx} = f'(x)$ and the second derivative is $\frac{d^2y}{dx^2} = f''(x)$. Anyway, if you play around with these a bit, you can see that $\frac{dx}{dx} = 1$, which means that $x$ always changes in unity with itself. However, a very odd result happens if you look at the second derivative. Since $\frac{dx}{dx} = 1$, and 1 is a constant, that means that the second derivative, $\frac{d^2x}{dx^2} = 0$, which means that x never has any acceleration with respect to itself.

However, algebraically, what this seems to mean to me is that $d^2x$ is always zero, but this is obviously not the case, as it could be put in ratio with $dy^2$ to produce a real-valued function. However, this seems to be at odds with an infinitesimal definition of $d^2x$ (or any other definition I have seen). It seems to imply that that $dx$ is more of a relational quantity than an infinitesimal or even a limit.

I did not know if anyone had any specific knowledge about this, or knew of any books that dealt with this topic. I have a hard time finding any at all that approach this subject.

On a side note (but related), I would also be interested in any books which discussed any possible meaning of quantities like $\frac{d^2y}{d^2x}$ (note that this is different from the Leibniz second derivative which is $\frac{d^2y}{dx^2}$). Anyway, if anyone has ideas or references, I would love to investigate this topic further.

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  • $\begingroup$ Please be careful about using $d/dx$ as a fraction. There have been so many questions about this on this site, but here is one of the most popular. $\endgroup$ – Eff Oct 11 '16 at 20:13
  • $\begingroup$ Dr. MV - I don't see how you think I thought they were the same. $\endgroup$ – johnnyb Oct 11 '16 at 20:17
  • $\begingroup$ Read about "Non-Standard Analysis," a branch of mathematics pioneered by Abraham Robinson. $\endgroup$ – Mark Viola Oct 11 '16 at 20:18
  • $\begingroup$ I have read a few books on the subject, such as Henle's Infinitesimal Calculus. I may have missed it, but I did not find the answer to my question. $\endgroup$ – johnnyb Oct 11 '16 at 20:23
  • $\begingroup$ In general, $d^2f$ should be the part coming from the second-degree part of the Taylor polynomial of $f$. It wants to be fed two "infinitesimals" (or tangent vectors, in the multivariable case) to give the second-order increment in $f$. $\endgroup$ – Ted Shifrin Oct 11 '16 at 20:37
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$\frac{dx}{dx} = 1$ because the infinitesimal $dx$ changes at the same rate as $dx$.

$\frac{d}{dx}\frac{dx}{dx} = 0$ because $\frac{dx}{dx}$ changes infinitely slower than $dx$. One way to represent this is $\frac{d}{dx}\frac{dx}{dx} = lim_{n \rightarrow \infty} \frac{(\frac{1}{n})}{1} = 0$

The same holds true for $\frac{d^2 x}{dx^2}$. It's not that $d^2 x = 0$, rather, it is that $dx^2$ changes infinitely faster than $d^2 x$. Both are infinitesimals, so on their own they both approach zero. But the ratio approaches zero.


Some good reading is found on another page: Is There a Difference Between $d^2x$ and $(dx)^2$?

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  • $\begingroup$ So, what I think you are saying is that $\frac{d^2x}{dx}$ is not producing zero but rather an infinitesimal. Is that correct? And it is just the standard part that is zero (at least when using infinitesimal terminology). $\endgroup$ – johnnyb Oct 11 '16 at 20:36
  • $\begingroup$ Yes, it produces an infinitesimal. The operator $d$ is taking an infinitesimal of $x$. $\frac{dx}{dx}$ is taking an infinitesimal of the top and bottom, yielding 1. $\frac{d^2x}{dx^2}$ is taking an infinitesimal of an infinitesimal on the top, which is much smaller than an infinitesimal squared. $\endgroup$ – Paul Terwilliger Oct 11 '16 at 20:42
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The key term here is second difference. If $y=f(x)$ then the meaning of the symbol $d^2y$ is the infinitesimal version of the second difference $\Delta^2y=f(x)-2f(x+\Delta x)+f(x+2\Delta x)$. If $f\in C^2$ then the second derivative $f''$ can be calculated in terms of the second difference by taking the limit of $\frac{\Delta^2 y}{\Delta x^2}$. This explains Leibniz's notation for second derivative that the OP mentioned.

By far the clearest discussion of this is in Keisler's textbook Elementary Calculus, page 94.

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After a lot of searching, I realized that the answer to this question depends on which variable is the independent variable. $d^2u$ reduces to $0$ when $u$ is the independent variable, but, otherwise, it should stay.

The problem is that the second derivative as it is generally taught is actually a special case. The real notation for the second derivative of $x$ with respect to $y$ should be $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. If (and only if!) $x$ is the independent variable, then $d^2x$ reduces to $0$ and the second term goes away. However, if you leave it in the expanded form, you can do a lot of cool tricks with it, like algebraically manipulate it into the second derivative of $x$ with respect to $y$.

Anyway, I wrote a paper on my investigations into this, and you can see it here:

https://arxiv.org/abs/1801.09553

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You seem to be adding some context to the notation that was never meant to be there.

$\frac{\mathrm{d}}{\mathrm{d}x}$ is just an operator.$\frac{\mathrm{d}^2}{\mathrm{d}x^2}$ is just the notation used when operating twice.

In the language of infinitesimals, you can consider

$\frac{\mathrm{d}y}{\mathrm{d}x}$ to be the infinitesimal change in $y$ and how it compares to $x$ (in a ratio).

I took a course in my second year of University which addressed the differential, differential forms (one-forms, two-forms, etc) and wedge products. It's a slippery slope, but you might want to look at a textbook from a Calculus III course. I can't share mine, as my Professor was the author (I'm not sure if I'm able to distribute this book).

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  • $\begingroup$ Be warned, though, that second derivatives are not built out of differential forms, because they have a symmetry (rather than skew-symmetry) to them. Another way to say this is that when you change coordinates, the product rule messes you up and you get an extra term: $$\frac{d^2y}{dx^2} = \frac d{dx}\left(\frac{dy}{dx}\right) = \frac d{dx}\left(\frac{dy}{du}\frac{du}{dx}\right) = \frac{d^2y}{du^2}\left(\frac{du}{dx}\right)^2 + \frac{dy}{du}\frac{d^2u}{dx^2}.$$ The second term means things aren't transforming the way a tensor should ... unless you're at a critical point of $y$. $\endgroup$ – Ted Shifrin Oct 11 '16 at 20:32
  • $\begingroup$ Ted - these are exactly the kinds of considerations I have been thinking about. What books or kinds of books do I need to look for to find discussions like this. $\endgroup$ – johnnyb Oct 11 '16 at 20:41
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$d^2 x$ is not "always" zero. It depends on the context. With respect to itself, sure $\frac{d^2 x}{dx^2} = \frac{d}{dx}\frac{dx}{dx} = 0$ but suppose with respect to $y$ it isn't 0. Then $\frac{d^2 x}{dy^2} = \frac{d}{dy}\frac{dx}{dy} \ne 0$ then how could $d^2x = 0$ ?

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  • $\begingroup$ math.stackexchange.com/questions/1550600/… this might help $\endgroup$ – Joe Dirt Oct 11 '16 at 20:28
  • $\begingroup$ Right, that's my question. How is it working to produce this. How should we conceive of a differential properly in order to expect this. $\endgroup$ – johnnyb Oct 11 '16 at 20:34
  • $\begingroup$ Use the limit definition of a derivative. $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{y(x+\Delta x) - y(x)}{x + \Delta x - x}$ observe the function at these scales to develop an intuition for how the function changes with respect to the variable. So, $x(x) = x$ by definition, then: $\frac{dx}{dx} = \lim_{\Delta x \to 0} \frac{x(x+\Delta x) - x(x)}{x + \Delta x - x} = \frac{x + \Delta x - x}{x + \Delta x - x} = 1$ But suppose $y(x) = x^2$ then you input the function into the definition and get something else. $\endgroup$ – Joe Dirt Oct 11 '16 at 20:39

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