Timeline for Is the $x$ in $\frac{\mathrm{d}}{\mathrm{d}x}$ a symbol in the sense of Harper's PFPL?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2017 at 9:26 | comment | added | Michael Bächtold | This comment section is to short to explain better what I mean, but maybe you can get a glimpse of the historical development of this change of perspective by reading Frege's 1904 "What is a function?" and by looking at the discussion here. | |
| Nov 24, 2017 at 9:26 | comment | added | Michael Bächtold | Thanks. I can fully understand where this answer comes from. It seems to be in line with what Derek Elkins and Andrej Bauer say in the comments. My suspicion though, is that this modern perspective (which became popular among mathematicians and logicians only around 1900) ignores a useful interpretation of the concepts of "variable" and "function", that was predominant prior to 1900 (and seems still to be awaiting a proper direct formalisation). | |
| Nov 22, 2017 at 17:50 | history | edited | D.W.♦ | CC BY-SA 3.0 |
added 66 characters in body
|
| Nov 22, 2017 at 17:44 | comment | added | D.W.♦ | @MichaelBächtold, good points. I've edited my answer extensively to try to clarify these points. Can you read through my answer a second time and see what you think now? | |
| Nov 22, 2017 at 17:44 | history | edited | D.W.♦ | CC BY-SA 3.0 |
added 857 characters in body
|
| Nov 22, 2017 at 16:26 | history | edited | D.W.♦ | CC BY-SA 3.0 |
added 21 characters in body
|
| Nov 22, 2017 at 8:14 | comment | added | Michael Bächtold | Also when you say "these operators are symbols" are you using the term symbol in the same technical sense Harper does? | |
| Nov 22, 2017 at 8:13 | comment | added | Michael Bächtold | There are several things in your answer that I don't understand. "since $x$ is a bound variable of $f(x)$"? Then in one paragraph you say "interpret $\frac{d}{dx}f(x)$ as $Df(x)$", in the next you say "Interpret $\frac{dy}{dx}$ as $\frac{d}{dx}f(x)$ ... in particular interpret this as $Df$". Isn't that a contradiction? Or maybe you equate $f$ with $f(x)$? | |
| Nov 22, 2017 at 1:21 | history | answered | D.W.♦ | CC BY-SA 3.0 |