Timeline for Examples of bad notation and its consequences
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 2 at 20:54 | comment | added | Joe Lamond | @LSpice: I'm quite late to this post, but I believe that one correct way to write that $D(g\circ f)(x)=Dg(f(x))\circ Df(x)$ for all $x$ is $D(g\circ f)=\operatorname{comp}\circ\,(Dg\circ f,Df)$, where $\operatorname{comp}$ is the function given by $\operatorname{comp}(\phi,\psi)=\phi\circ \psi$ for linear maps $\phi$ and $\psi$. It can be shown that $\operatorname{comp}$ is a smooth function, which is an ingredient in the proof in Dieudonne's Foundations of Modern Analysis that the composition of smooth functions (between Banach spaces) is smooth (Chapter 12, p. 183). | |
| Apr 28, 2023 at 0:34 | comment | added | darij grinberg | @HumbertoJoséBortolossi: Theorem 2.1 (b) in arxiv.org/abs/math/0005260v1 is an example where it means "$a\leq x$ and $y \leq b$". In matrix algebra, it usually means "both $x$ and $y$ belong to $[a, b]$". | |
| Apr 27, 2023 at 23:47 | comment | added | Humberto José Bortolossi | @darij grinberg, POLEASE COULD YOU GIVE ONE OR TWO REFERENCES? | |
| Apr 21, 2023 at 17:37 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
edited body
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| Apr 21, 2023 at 6:04 | comment | added | Carl-Fredrik Nyberg Brodda | @R.vanDobbendeBruyn ]Thanks a lot, now I'm going to spend all day with negative unresolved tension. | |
| Apr 21, 2023 at 1:10 | comment | added | Timothy Chow | @LoïcTeyssier I can see how we can have some fun with that notation if we also allow multiplication of an interval by a scalar (e.g., $3]a,b] = ]3a,3b] = ]a,b]3$). Then if I refer to $[a,b[c,d]e,f]$ and $[g,h]$, am I referring to three intervals $[a,b[c = [ac,bc[$ and $d]e,f] = ]de,df]$ and $[g,h]$, or am I referring to a 3-tuple and a 2-tuple, where the 3-tuple consists of $a$, the interval $b[c,d]e = [bce,bde]$, and $f$? | |
| Apr 21, 2023 at 0:17 | comment | added | Steve Costenoble | As a topologist, I do think of the superscript in $\mathbb{S}^2$ as an exponent, as $\mathbb{S}^2 = \mathbb{S}^1\wedge\mathbb{S}^1$ is the smash product of two 1-spheres, and the $n$-sphere is the smash product of $n$ 1-spheres. This is very useful. | |
| Apr 20, 2023 at 21:57 | comment | added | R. van Dobben de Bruyn |
@LoïcTeyssier again stating the obvious: there is a difference between not closing every bracket with the same type or haphazardly opening and closing brackets everywhere. A syntax where ( and [ always increase the parenthesis state by one and ) and ] decrease it still has a semantics-independent parity check ($\mathbf N$-valued, not $\mathbf Z/2$-valued). This also serves a role in readability: opening a bracket makes you look ahead, whereas closing one makes you look back at what came before it.
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| Apr 20, 2023 at 8:27 | comment | added | Loïc Teyssier | @R.vanDobbendeBruyn: I don't see how $(1,2]$ is less ill-formed in bracket-pairing than $]1,2]$. You just can't parse expressions containing intervals in the usual way, that's all. | |
| Apr 20, 2023 at 7:04 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
terminolgy corrected
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| Apr 20, 2023 at 0:56 | comment | added | LSpice | @wlad, re, I think that $D(g \circ f) = (Dg \circ f) \circ Df$ comes from thinking of the derivatives as linear operators (to be composed) rather than as matrices (to be multiplied). | |
| Apr 20, 2023 at 0:21 | comment | added | darij grinberg | In the same vein, "$a \leq x,y \leq b$" is a horror show. Both possible meanings are widespread in the literature. | |
| Apr 19, 2023 at 22:42 | comment | added | R. van Dobben de Bruyn | The obvious problem with the French notation for open intervals is bracket pairing — it looks syntactically ill-formed. Your sentence "(I do prefer the French notation $]a,b[$)" highlights this issue: it looks like the matching sets of brackets are of the form ($\ldots$] and [$\ldots$), which is not what is intended. ]Even worse is a negative number of opened brackets.[ | |
| Apr 19, 2023 at 22:09 | comment | added | wlad | $D(g \circ f) = (Dg \circ f) \circ Df$. WTF? That should be $D(g \circ f) = (Dg \circ f) \cdot Df$, then there's nothing wrong with it. | |
| S Apr 19, 2023 at 21:54 | history | answered | Christophe Leuridan | CC BY-SA 4.0 | |
| S Apr 19, 2023 at 21:54 | history | made wiki | Post Made Community Wiki by Christophe Leuridan |