Timeline for If d/dx is an operator, on what does it operate?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2019 at 20:45 | comment | added | Michael Bächtold | @Valerio yes, your suggestion clashes with 300 years of mathematical usage. And a question arises if we adopt it: why should we use a notation ($d/dx$) that contains an irrelevant letter ($x$), when we already have a perfectly fine notation for what you want to say? The prime notation $(x\mapsto x^2)'=(t\mapsto 2t)$ is almost as old as Leibniz $d/dx$: hsm.stackexchange.com/questions/6206. | |
| Jul 7, 2019 at 20:18 | comment | added | Valerio_xula | The way I look at it, $d/dx$ is the operator that takes the derivative of the one-variable function that follows it, and the symbol $x$ is (or should be) irrelevant. So it just operates on the space of all differentiable one variable functions. I think that when we write $\frac{d}{dx} x^2$ the confusing part is the $x^2$, that should be written as $x\mapsto x^2$. Then $\frac{d}{dx} (x\mapsto x^2)$ and $\frac{d}{dt}(x\mapsto x^2)$ give the same answer, $(x\mapsto 2x)$, while $\frac{d}{dx}(t\mapsto x^2)$ and $\frac{d}{dt}(t\mapsto x^2)$ are both $0$. But I know this clashes with common usage... | |
| Aug 23, 2018 at 13:54 | comment | added | Toby Bartels | True, although I like to say that if $y=f(x)$, then $dy=f'(x)\,dx$, as long as $f$ and $x$ are differentiable (so $dy=z\,dx$, where $z=f'(x)$), regardless of whether $x$ is constant. But we need $dx\ne0$ for even $f$ (or $z$) to be unique (even locally). | |
| Aug 22, 2018 at 11:33 | comment | added | Michael Bächtold | @TobyBartels thanks! I also forgot the fine print that $dx$ needs to be non-zero, otherwise $z$ is not unique and we would be allowed to say things like "3 is a function of 5". One might interpret the condition $dx\neq 0$ as saying that $x$ can 'truly vary' or is an 'independent variable', and that this is a necessary condition for some other variable quantity to change with $x$. | |
| Aug 21, 2018 at 22:18 | comment | added | Toby Bartels | [I edited the answer yesterday to require $x$ and $y$ to be differentiable maps on $M$, which is just a technicality to make $dx$ and $dy$ exist. But at the time, I didn't notice that $f$ has to be differentiable too to make $z$ exist, and that's more than just fine print in this context, so I made it a comment.] | |
| Aug 21, 2018 at 22:15 | comment | added | Toby Bartels | To prove the existence of $z$ such that $dy=z\,dx$, you not only need that $y$ is a function of $x$ but that $y$ is a differentiable function of $x$. This is basic fine print, of course; my real point is that one can (and people did) use language exactly in that way: treating $x$ and $y$ formally as maps on $M$, $y$ is a differentiable function of $x$ iff there exists a differentiable map $f$ such that $y = f \circ x$; or treating $x$ and $y$ informally as variable quantities, $y$ is a differentiable function of $x$ iff there exists a differentiable map $f$ such that $y = f(x)$. | |
| S Aug 20, 2018 at 14:33 | history | suggested | Toby Bartels | CC BY-SA 4.0 |
$x$ and $y$, interpreted as maps on $M$, have to be differentiable if you're to prove the existence of $z$ (or even of $dx$ and $dy$).
|
| Aug 20, 2018 at 5:34 | comment | added | Toby Bartels | Note that the requirement to avoid dividing by zero is also what stops you from substituting $5$ for $x$ in $d(x^2)/dx = 2x$. It would be fine to substitute, say, $5y$ for $x$ instead; or to substitute $5$ for $x$ in the nearly equivalent equation $d(x^2) = 2x\,dx$. | |
| Aug 20, 2018 at 5:32 | review | Suggested edits | |||
| S Aug 20, 2018 at 14:33 | |||||
| Aug 13, 2018 at 19:14 | comment | added | Michael Bächtold | @MikeShulman: I just re-read you're answer on the other question and have a better picture of what's going on now. From your perspective $dy/dx$ will always be a partial function on $TX$, but only when $y$ is (locally) a function of $x$ will $dy/dx$ be the pullback of a function on $X$. So only then will $dy/dx$ be an observable quantity on the same state space as $y$ and $x$. | |
| Aug 13, 2018 at 9:02 | comment | added | Mike Shulman | Possibly a more faithful way to deal with the worry about whether ${\rm d}x=0$ is to consider all functions to be partial (as one generally does in calculus anyway). Then $\frac{{\rm d}y}{{\rm d}x}$ just has its domain restricted to the points of the tangent bundle where ${\rm d}x \neq 0$. Your calculation of $0=1$ is then perfectly valid as long as you keep track of domains, which are empty in this case -- the two functions $\emptyset \to \mathbb{R}$ constant at 0 and 1 are in fact equal! | |
| Aug 13, 2018 at 8:59 | comment | added | Mike Shulman | In that case the problem is that ${\rm d}x = 0$, so you can't divide by it. The state-space perspective is that there is no distinguished operator to call "d/dx", it really is literally taking the differential $\rm d$ followed by dividing pointwise by the differential of $x$, ${\rm d}x$, so there is nothing to "allow" except to worry about whether ${\rm d}x = 0$. | |
| Aug 13, 2018 at 8:43 | comment | added | Michael Bächtold | @MikeShulman Hmm, I'm not so sure if one can run into trouble with allowing arbitrary applications of d/dx. Consider the following example: take the equation $x=1$ and derive both sides with respect to $x$ to arrive at $0=1$. | |
| Aug 13, 2018 at 8:37 | comment | added | Mike Shulman | Before seeing this answer, I just added essentially this proposal as an answer to the "very similar question" you linked to. (-: I don't think it's necessary to restrict this notation to act only on "functions of $x$", however. For instance, if $z = x^2+y^2$ for another "independent" variable $y$, then ${\rm d}z = 2 x \,{\rm d}x + 2 y\,{\rm d}y$, so that $\frac{{\rm d}z}{{\rm d}x} = 2 x + 2 y \frac{{\rm d}y}{{\rm d}x}$, which makes perfect sense. | |
| Aug 12, 2018 at 10:50 | history | answered | Michael Bächtold | CC BY-SA 4.0 |