Timeline for If d/dx is an operator, on what does it operate?
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| Feb 23, 2023 at 23:09 | comment | added | Zach Teitler | Even without the differentiation $\dfrac{d}{dt}$, I'm not really sure how I would approach the integral $\displaystyle \int_t^x (t^2+x^3) dt$. Using $t$ both as a limit of integration and also as the variable of integration is a little confusing. | |
| Sep 6, 2018 at 18:25 | comment | added | peter | this last comment by @DonuArapura is all this is about. d/dx can be applied to expressions (hopefully containing x), but without abuse of notation (or fancy definitions with coordinate functions) not to functions. since some people have a tendency to identify f with f(x) and thereby functions with expressions, they can apply d/dx to f. | |
| Dec 7, 2012 at 14:58 | comment | added | Donu Arapura | I agree with this answer. While professional mathematician differentiate functions, in basic calculus we differentiate expressions. | |
| Dec 6, 2012 at 19:35 | vote | accept | Jason Howald | ||
| Dec 6, 2012 at 19:35 | comment | added | Jason Howald | I am grateful to Joel for his support of the question, including this interesting answer. Certainly $\frac{d}{dx}$ is similar to a quantifier: It "shields" occurrences of the variable $x$ in its scope from direct substitution. It is defined in terms of the limit, which also binds a variable, as a quantifier could. It is a very strange quantifier, though, as $x$ once again occurs free in the ("bound"?) expression $\frac{d}{dx} x^3$ since $\frac{d}{dx} x^3 = 3x^2$. | |
| Dec 6, 2012 at 17:01 | comment | added | Joel David Hamkins | To use the $\lambda$-calculus notation, the multivariate approach is dealing with the binary function $\lambda x,t.x^3$, and the scheme-of-unary-functions approach involves the function-valued function $\lambda x.(\lambda t.x^3)$. The introduced constant concept is the role played by $G$, when one wants to prove a theorem about all groups, and begins, "Let $G$ be a group". | |
| Dec 6, 2012 at 16:24 | comment | added | Joel David Hamkins | The function to be integrated in your case, as I am sure you know very well (and so it is hard to take you seriously here), is the function with constant value $x^3$, constant as a function of $t$. One can think multivariately, but this is not actually necessary, for it suffices to have a separate function specific to the value of $x$, which in effect here is an introduced constant, as with the coefficients in $ax^2+bx+c$. The subsequent differentiation with respect to $x$ is with respect to the function $x\mapsto x^4$, on your set-up. But I don't think you are confused about it... | |
| Dec 6, 2012 at 16:20 | comment | added | S. Carnahan♦ | In a now-deleted answer, "none" says, "The issue of switching variables and messing up the answer has been called "perturbation confusion" in automatic differentiation (AD) and it's discussed at some length here (pdf): bcl.hamilton.ie/~qobi/nesting/papers/ifl2005.pdf and here (sigfpe blog): blog.sigfpe.com/2011/04/perturbation-confusion-confusion.html . It is apparently sometimes a (pun not intended) confusing issue. AD itself is discussed by sigfpe here: blog.sigfpe.com/2005/07/automatic-differentiation.html ." | |
| Dec 6, 2012 at 16:07 | comment | added | user9072 | I did not mean anything fancy. The situation I envision is just $x^3$ does not depend on $t$ for a result of $x^4$ and $7x$ in the latter case. But what is the precise nature of the finction in the integeral. In particular, please make it so that the $d/dx$ application you give makes still sense. | |
| Dec 6, 2012 at 15:48 | comment | added | Joel David Hamkins | Well, of course I would give the answer explaining precisely which function I had meant; the notation is ambiguous without doing so, and this is the point of the question and my answer. One can imagine a variety of perfectly reasonable answers that would correspond to different intended meanings here. The discussion of whether $x$ in the integrand was meant to depend on $t$ or be an independent variable from $t$ and so on was the discussion I alluded to at the end of my answer. The part of the question interesting me is the precise nature of this particular ambiguity. | |
| Dec 6, 2012 at 15:31 | comment | added | user9072 | I assume you still define integeration as an operation on a function, as opposed to a purely syntactic process (which for calculus would mostly be possible). So, if you write $\int_{0}^x x^3 dt$ and a student ask: please explain precisely which function is meant (with domain of definition and all that) by $x^3$ what do you reply? Same question for $7$ in $\int_{0}^x 7 dt$. | |
| Dec 6, 2012 at 6:58 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |