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Timeline for Suggestions for good notation

Current License: CC BY-SA 3.0

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Apr 10, 2023 at 14:52 answer added David E Speyer timeline score: 3
Apr 10, 2023 at 3:26 answer added Michael Hardy timeline score: 0
Dec 22, 2021 at 4:58 comment added Michael Hardy If you want an undergraduate to understand why you call a function $x\mapsto x\sin x$ instead of just calling it $x\sin x,$ point out that $x\mapsto(x+2y)^2$ and $y\mapsto(x+2y)^2$ are two different functions. $\qquad$
Dec 19, 2021 at 9:51 comment added François Brunault The notation $\delta_P$ is also used for a formula $P$, and if $A$ is a set then $\delta_A$ is the characteristic function of $A$. So $[n]=\delta_{n=0}=\delta_0(n)$. In comparison to $[n]$, this makes it easier to view this as a function. This is also more general than the traditional $\delta$.
Dec 19, 2021 at 5:40 answer added Daniel Sebald timeline score: 1
Dec 19, 2021 at 5:08 answer added Dan Ramras timeline score: 2
Dec 17, 2021 at 18:28 answer added Mozibur Ullah timeline score: 0
Dec 17, 2021 at 0:16 answer added Michael Hardy timeline score: 4
Jun 18, 2021 at 0:24 answer added Tom Copeland timeline score: 1
May 19, 2021 at 10:03 answer added Ben McKay timeline score: 3
Feb 9, 2021 at 15:15 answer added Ben McKay timeline score: 6
Sep 4, 2020 at 7:05 answer added Qiaochu Yuan timeline score: 8
Nov 20, 2019 at 14:43 comment added firtree I also don't have enough reputation on this one to post an answer, so I'll do mine as a comment: In basic linear algebra, N.Vavilov suggests $[u\leadsto v]$ for the transition matrix from basis $u$ to basis $v$. That makes much easier and more natural and more memorizable formulas like $u[u\leadsto v]=v$ and for coordinates columns $x,y$: $a=ux=vy=(u[u\leadsto v])y=u([u\leadsto v]y)=ux$. Sure $[v\leadsto u]=[u\leadsto v]^{-1}$, and one never needs a transpose matrix.
May 13, 2019 at 18:59 answer added Dirk timeline score: 10
May 13, 2019 at 16:49 answer added k.stm timeline score: 9
Mar 5, 2018 at 14:55 comment added johnnyb I don't have enough reputation on this one to post an answer, so I'll do mine as a comment: 1) The second derivative written so that it can be used as a fraction: $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. See arxiv.org/abs/1801.09553 for more details on this approach. 2) Partial differentials written so that they can be used as a fraction: The partial derivative of $u$ with respect to $x$ where $x$ is the only variable allowed to independently vary written as $\frac{\partial_x u}{dx}$. Both of these allow differentials to more freely be used as fractions.
Jan 27, 2018 at 22:46 answer added Itai Bar-Natan timeline score: 3
Jan 27, 2018 at 21:48 answer added Mr Pie timeline score: -3
Jan 27, 2018 at 21:11 comment added Mr Pie I invented my own function $\delta(n, k)$ which simply means the sum of the digits of $n$ in base $k$ since $\delta$ looks like an $S$ in english but corresponds to a $d$ in greek ..... until I read this post and found out there was already a similar delta function (sigh).
Jan 13, 2018 at 20:32 answer added echinodermata timeline score: 4
Oct 26, 2017 at 5:53 review Close votes
Oct 26, 2017 at 9:23
Jul 15, 2017 at 11:29 comment added Peter Heinig Re the comments on $B^A$: like you are likely to know, a useful and usual alternative notation for $B^A$, coming from category theory, is $[A,B]$, often referred to as the "internal-hom". In the category $\mathsf{C}$ of sets, we have $B^A = \mathsf{C}(A,B) = [A,B]$. The notation somewhat clashes with Iverson brackets, of course; and then again...one can perhaps reconcile the two with a suitable interpretation.
Jul 15, 2017 at 3:28 answer added ಠ_ಠ timeline score: 3
Jul 15, 2017 at 1:45 answer added k.stm timeline score: 5
Jul 18, 2016 at 12:39 answer added Matt Majic timeline score: 6
Dec 16, 2014 at 0:57 answer added echinodermata timeline score: 52
Dec 13, 2014 at 14:22 answer added Sylvain JULIEN timeline score: -2
Dec 13, 2014 at 12:24 answer added Rasmus timeline score: 0
Dec 13, 2014 at 11:48 answer added Basil timeline score: 13
Dec 1, 2013 at 2:32 history edited Tom LaGatta CC BY-SA 3.0
switched B -> A to A -> B
Jul 9, 2013 at 16:41 history protected François G. Dorais
Feb 2, 2013 at 23:15 answer added Douglas Lind timeline score: 11
Jan 24, 2013 at 22:28 answer added Hans Schoutens timeline score: 0
Jan 24, 2013 at 21:04 answer added Anton Petrunin timeline score: 1
Jan 24, 2013 at 5:25 answer added Anton Petrunin timeline score: 23
Dec 10, 2012 at 1:13 comment added Daniel McLaury Well, if X is a set, then the Cartesian product $X^n$ is the set of functions from $[n] = {1, 2, \ldots, n}$ to X, so naturally $X^A$ is the set of functions $A \to X$.
Dec 9, 2012 at 23:36 answer added Qfwfq timeline score: 8
Oct 16, 2012 at 14:34 comment added Hans-Peter Stricker (1) I only know the Kronecker delta with two indices, yielding $\delta_{ij} = [i = j]$ - but this is better anyhow! (2) I always found the notation $A^B$ useful since the number of functions from $B$ to $A$ is - in the finite case - just $|A|^{|B|}$ (3) I believe it's a matter of taste: for me $A\rightarrow B$ indicates one morphism from $A$ to $B$, not all of them.
Oct 15, 2012 at 19:32 answer added Jonathan Beardsley timeline score: 9
Oct 15, 2012 at 19:27 answer added Todd Trimble timeline score: 5
Oct 15, 2012 at 19:07 answer added David E Speyer timeline score: 41
Sep 20, 2012 at 5:29 answer added 36min timeline score: 24
Jun 14, 2012 at 18:48 comment added LSpice @Peter LeFanu Lumsdaine, there's no need to do that spacing manually; stmaryrd (ctan.org/tex-archive/fonts/stmaryrd) includes \llbracket and \rrbracket.
Feb 5, 2012 at 17:50 comment added timur I would say the second bullet is completely standard.
Dec 19, 2011 at 1:27 answer added Zeeshan Mahmud timeline score: 0
Jul 2, 2011 at 21:35 comment added Noam D. Elkies How about the notation $x^n$ itself? Initially for positive integers $n$ (and even then Euler still wrote $xx$ and $xxx$ on occasion) but suggesting fruitful generalization where $n$ might be negative, fractional, etc. and eventually even $x$ might "live" somewhere other than the real or complex numbers.
Dec 18, 2010 at 22:12 answer added Andreas Blass timeline score: -3
Dec 18, 2010 at 13:49 answer added Zhen Lin timeline score: 8
Nov 16, 2010 at 5:56 answer added Alan Guo timeline score: 89
Nov 16, 2010 at 1:54 answer added Jérôme JEAN-CHARLES timeline score: 10
Nov 9, 2010 at 0:06 answer added Gareth timeline score: 2
Nov 8, 2010 at 23:00 answer added David MJC timeline score: 18
Nov 8, 2010 at 12:41 answer added Zsbán Ambrus timeline score: 13
Nov 2, 2010 at 1:41 answer added John D. Cook timeline score: 2
Nov 2, 2010 at 0:13 answer added darij grinberg timeline score: 12
Nov 1, 2010 at 23:19 answer added Chris Hardin timeline score: 3
Oct 29, 2010 at 5:21 answer added Sándor Kovács timeline score: 15
Oct 29, 2010 at 3:26 answer added Chua KS timeline score: 1
Oct 27, 2010 at 8:13 answer added Darsh Ranjan timeline score: 27
Oct 22, 2010 at 21:20 answer added David MJC timeline score: 31
Oct 22, 2010 at 13:24 answer added Qfwfq timeline score: 5
Oct 22, 2010 at 8:13 answer added Roy Maclean timeline score: 2
Oct 22, 2010 at 8:00 answer added Suvrit timeline score: 7
Oct 22, 2010 at 4:59 answer added Cristi Stoica timeline score: 15
Oct 22, 2010 at 4:59 history edited timur CC BY-SA 2.5
Hindu-Arabic numerals
Oct 22, 2010 at 4:52 answer added Cristi Stoica timeline score: 31
Oct 21, 2010 at 17:52 answer added Willie Wong timeline score: 28
Oct 21, 2010 at 17:27 answer added Terry Tao timeline score: 23
Oct 21, 2010 at 17:05 answer added Terry Tao timeline score: 49
Oct 21, 2010 at 16:59 answer added Ryan Reich timeline score: 14
Oct 21, 2010 at 13:12 history made wiki Post Made Community Wiki by Ben Webster
Oct 21, 2010 at 12:00 answer added Qfwfq timeline score: 42
Oct 21, 2010 at 11:55 answer added Qfwfq timeline score: 48
Oct 21, 2010 at 11:15 answer added John Bentin timeline score: 2
Oct 21, 2010 at 8:18 comment added Neel Krishnaswami @Peter: I've always heard them called "Oxford brackets" -- I've never heard the name Scott brackets before. I wouldn't mind attributing them to Scott though, if he invented them.
Oct 21, 2010 at 5:17 answer added Harry Altman timeline score: 2
Oct 21, 2010 at 4:56 comment added J. M. isn't a mathematician For those who missed the point of Chandan's comment: it's "Hindu-Arabic numerals". ;)
Oct 21, 2010 at 4:33 answer added Quadrescence timeline score: 42
Oct 21, 2010 at 3:38 comment added Aleksei Averchenko Isn't $x \mapsto f(x)$ commonplace? As for homomorphisms, they are not simply maps, and $\mathrm{Hom}(A, B)$ denotes the whole class, while $A \to B$ denotes a single mapping.
Oct 21, 2010 at 3:37 answer added Andrey Rekalo timeline score: 88
Oct 21, 2010 at 3:29 comment added Chandan Singh Dalawat Arabic numerals ? Ah yes, they were transmitted to Europe by the Arabs.
Oct 21, 2010 at 3:21 answer added Jason Morton timeline score: 13
Oct 21, 2010 at 3:05 comment added Peter LeFanu Lumsdaine (a version of) the Iverson bracket notation is common in categorical logic: we write them like $[\![ \mathrm{this} ]\!]$, i.e. [\![ … ]\!], and call them “Scott brackets”. Besides the examples already given, it extends beautifully to more general models with some other lattice of truth-values; but I guess that's of less general interest :-)
Oct 21, 2010 at 2:50 answer added Richard Stanley timeline score: 105
Oct 21, 2010 at 2:38 answer added Allen Knutson timeline score: 19
Oct 21, 2010 at 2:18 answer added Hsien-Chih Chang 張顯之 timeline score: 23
Oct 21, 2010 at 2:06 answer added Karl Schwede timeline score: 0
Oct 21, 2010 at 1:30 answer added Dick Palais timeline score: 20
Oct 21, 2010 at 0:59 comment added J. M. isn't a mathematician I still find it terribly annoying that I have to write things like "where $[Q]$ is an Iversonian bracket" every time I use it. I wish it reaches the point where it should only have to be explained in introductory work. On the other hand, since brackets get used a lot for other things as well, maybe someone should develop specific enclosures like what was done for floor and ceiling I suppose...
Oct 20, 2010 at 23:25 answer added Qfwfq timeline score: 9
Oct 20, 2010 at 23:20 comment added darij grinberg Yes, among other things. Also $A^B\times A^C=A^{B+C}$, where $+$ is disjoint union. But all the great reasons for it don't help for our mind thinking that maps start with the source and end with the image, not the other way round.
Oct 20, 2010 at 23:19 comment added Kevin H. Lin I guess, in terms of lambda calculus, this is called "currying" ...
Oct 20, 2010 at 23:18 comment added Kevin H. Lin I've always assumed that the notation $A^B$ is because of the "exponential law" $(A^B)^C = A^{B\times C}$ ...
Oct 20, 2010 at 23:13 answer added Qfwfq timeline score: 24
Oct 20, 2010 at 23:08 answer added Qfwfq timeline score: 1
Oct 20, 2010 at 22:51 answer added Qfwfq timeline score: 21
Oct 20, 2010 at 22:29 answer added Mark timeline score: 73
Oct 20, 2010 at 21:17 comment added lhf Knuth made an argument for Iverson's notation in arxiv.org/abs/math/9205211 .
Oct 20, 2010 at 21:02 answer added Bjørn Kjos-Hanssen timeline score: 43
Oct 20, 2010 at 20:54 comment added Andrés E. Caicedo In set theory we write ${}^B A$ for the set of functions from $B$ to $A$.
Oct 20, 2010 at 20:52 answer added Mark Grant timeline score: 50
Oct 20, 2010 at 20:48 answer added José Figueroa-O'Farrill timeline score: 42
Oct 20, 2010 at 20:44 answer added Michael Lugo timeline score: 46
Oct 20, 2010 at 20:39 answer added Joel David Hamkins timeline score: 6
Oct 20, 2010 at 20:19 answer added Pietro Majer timeline score: 121
Oct 20, 2010 at 20:14 answer added Faisal timeline score: 19
Oct 20, 2010 at 20:13 answer added Jan Weidner timeline score: 59
Oct 20, 2010 at 20:13 history edited Harald Hanche-Olsen CC BY-SA 2.5
Replaced → by ↦
Oct 20, 2010 at 20:12 answer added Suvrit timeline score: 31
Oct 20, 2010 at 20:00 history edited Alison Miller
edited tags
Oct 20, 2010 at 19:58 history asked Richard Borcherds CC BY-SA 2.5