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While this may be a fruitless pursuit of anecdotes, I still ask: what is the strangest (or most blatantly wrong (at least in the eyes of common notation)) mathematical notation you have ever seen?

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closed as primarily opinion-based by Zev Chonoles, Shuchang, M Turgeon, Claude Leibovici, Paramanand Singh Mar 1 at 7:06

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

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There is a thread about this on MathOverflow also. There, as here, the top scorer is Mazur's $\overline{\Xi}\over\Xi$. –  MJD Mar 1 at 4:39
    
This is not an answer, but $dx$ and $dy$ are the best notation for confusing students. –  Sawarnik Mar 1 at 5:02
    
@Sawarnik do you mean contrary to using $\Delta x$ or $\Delta y$? –  user76061 Mar 1 at 5:05
    
@user76061 I meant that unlike $f'(x)$, the imprecise $dx$.. confuses people. Why the derivative became a quotient, what is a differential, why did $dx$ come in the integral, among other things. –  Sawarnik Mar 1 at 5:11
    
"A woman stepped forward and asked, / What is the strangest day? // Tuesday, the Master replied." —Kehlog Albran, The Profit –  Rahul Mar 3 at 22:39

9 Answers 9

There is an old story about Lang and Mazur, Mazur tried to get Lang attention by using the worst notation possible. He wrote Xi conjugated over Xi, which looks like:

$$\frac{\overline{\Xi}}{\Xi}$$

P.S. You can read the story, narrated by Paul Vojta, in the AMS Notices issue dedicated to Lang: AMS Nottices Lang

It is on pages 546-547.

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Objective achieved! –  user76061 Mar 1 at 3:21
    
Just curious: What did Mazur use the notation for (except for getting Lang's attention, of course)? –  SDevalapurkar Mar 1 at 3:25
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@SanathDevalapurkar Just for getting Lang's attention. I posted the link to the story, on short Lang was criticizing a lot the notations Mazur was using, so they prepared a t-shirt for Lang, and they were waiting for Lang to say the magic words... But for some reason, that particular day Lang was quiet... –  N. S. Mar 1 at 3:29

The Landau big-$O$ notation is extremely strange.

  1. One writes $$f(x) = O(g(x))$$ which looks like $f$ is the composition of $O$ and $g$, but it is nothing of the sort. Is $O()$ an operator that can be applied to any term? Can I write $$O(x^2) = O(x^3)$$ or $O(x^2) = 2x^2$? Not normally.

  2. It is easily confused with a whole family of similar notations for similar notions; computer programmers regularly talk about $O(n)$ algorithms when they mean $\Omega(n)$ algorithms, for example. This is exacerbated because someone decided that instead of using mnemonic abbreviations, it would be a good idea arbitrarily assign every possible variant of the letter ‘o’ in naming them. Then when they ran out of letter O’s they used $\Theta$, seemingly because it looks enough like an O that you might confuse it with one.

  3. It is written with an $=$ even though the relation is asymmetric! We have both $x=O(x^2)$ and $x=O(x^3)$ although $O(x^2)$ and $O(x^3)$ are not the same, and we have both $1 = O(x)$ and $x = O(x)$ even though $1\ne x$.

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It probably makes more sense if you give $O$ a variable to bind, e.g. $Ox(x^2)$ could denote the equivalence class of functions associated with the mapping $x \in \mathbb{R} \mapsto x^2 \in \mathbb{R}$. –  goblin Mar 1 at 5:28
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There is still no excuse for the $=$ sign. If it were instead a $\in$ sign, or some other asymmetric sign like $\prec$, instead, my objection would only be a minor quibble. –  MJD Mar 3 at 21:50
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@MJD I've always treated $O$ as a function which when given a function $f$ returns the set of functions asymptotically upper-bounded by $f$, because the = in that context is absolutely appalling! Makes much more sense to say $\lambda x.4x^2-3 \in O(\lambda x. x^2)$. Regarding your point 2, I do believe $\Theta$ is gaining much more popularity in computer science circles these days, at least. –  Ray Toal Mar 4 at 6:08
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On the equals sign for O notation, Knuth indicates that (at least) he thinks of (say) $O(x^2)$ as "something that is at most a constant times $x^2$"—see his letter "Teach Calculus with Big O" (blog repost, PDF of AMS) — and of the equals sign as the English "is": "Aristotle is a man, but a man isn’t necessarily Aristotle". He's channelling de Bruijn: see pp. 5–7 of Asymptotic Methods in Analysis. –  ShreevatsaR Mar 4 at 20:14
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In particular, Knuth (and de Bruijn before him, and perhaps even Bachmann and Landau onwards) would write $O(x^2) = O(x^3)$ (for $x \to \infty$, not $x \to 0$ of course), with the meaning that "something that is at most a constant times $x^2$ is also at most a constant times $x^3$". E.g. de Bruijn writes, on page 6 of his book, the equation $O(x) + O(x^2) = O(x) \quad (x \to 0).$ At least de Bruijn agrees that it is abuse of notation and that the equals sign is a poor choice because it "suggests symmetry, and there is no such symmetry". But it's customary. –  ShreevatsaR Mar 4 at 20:25

The single worst use of mathematical notation I have ever seen was in a set of lecture notes in which the author wanted to construct a sequence of equivalence relations, each one ($\equiv_n$) derived from the previous one ($\equiv_{n-1}$). After $i_0$ iterations of this procedure, the construction has no more work to do, and the sequence has converged to a certain equivalence relation $\equiv$ with desirable properties. The notes contained this formula: $$\equiv_{i_0+1}=\equiv_{i_0}=\equiv$$

I regret that I did not make a note of the source.

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Could this actually be considered correct? –  user76061 Mar 1 at 4:31
    
I believe it is correct. –  MJD Mar 1 at 4:31
    
So it treats the equivalence like a variable? –  user76061 Mar 1 at 4:32
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I'm not sure what you mean. An equivalence relation, as any relation, is a set. The $=$ signs assert that the sets are equal. –  MJD Mar 1 at 4:33
    
Wait. It just clicked, I misread. Sorry –  user76061 Mar 1 at 4:34

The usage of pi:

$\pi$ is a constant. $\pi(x)$ is the prime counting function. $\prod(x)$ is a product of a sequence.

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$\pi$ is often a homomorphism or projection map as well. –  Steven Gubkin Mar 1 at 4:21
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$\phi = \frac{1 + \sqrt 5}{2}$, $\phi (n)$ is the totient function, $\Phi = \frac{1 - \sqrt5}{2}$, $\Phi_n$ is the $n$th cyclotomic polynomial, $\phi$ is often used in logic (e.g. $\psi \implies \phi$)... –  Soke Mar 1 at 4:50
    
$\phi(\alpha)$ is also used in NF literature as a special operation on cardinals. If you're writing about the proof theory surrounding NF's big cardinals, it gets ugly. That's why I've pledged to overwork $\xi$ in my own writing... –  Malice Vidrine Mar 1 at 7:44
    
$\pi$ is also often a permutation. –  ShreevatsaR Mar 4 at 19:45
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isn't used for the fundamental group as well? –  MphLee Mar 6 at 7:25

I took a long time to get used to derivative of integrals like this $$\frac{\partial}{\partial x}\int_{x_0}^x f(x,y) \ dx$$

It's just too much $x'$s in the same formula, and each one has a different meaning. Nevertheless, its common to see people writing down this way.

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I think this is actually meaningless. You cannot have the same variable in the integrand as in one of the limits of integration. –  Steven Gubkin Mar 1 at 4:21
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I couldn't agree more. –  Integral Mar 1 at 4:25
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If you are careful, you can, but there is no excuse to require the care. The $x$ in $f(x,y)$ is bound by the integral, while the $x$ upper limit is free, so that is what the derivative is taken with respect to. The result is then $f(x,y)$ with $x$ free. –  Ross Millikan Mar 1 at 5:07
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The integration variable could be named anything; why on earth would you pick the one letter that would cause the most confusion. –  mjqxxxx Mar 3 at 22:33
    
@mjqxxxx: Because it's consistent with how one usually notates antidifferentiation; in fact, the integral above can be viewed not as intending to be a definite integral, but as intending to be an anti-derivative with a particular constant of integration determined by the lower bound $x_0$. –  Hurkyl May 5 at 21:52

From a proof that convergence a.e. implies convergence in measure for $\mu(\Omega)<\infty$:$$\bigcup_{r\geq 1}\bigcap_{n\geq 1}\bigcup_{j\geq n}\{|{f_j-f}|>\frac{1}{r}\}=\{\omega:f_j(\omega) \not \to f(\omega)\}$$

Also, labeling graphs of functions as $f(x)$ (which I end up still doing to my undergraduates, who are bored when I mention my reservations about it), $\coprod$, "Random Variable," calling a domain the preimage but switching it to a connected open set in complex talk, etc. etc. etc.

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When I first learnt trigonometry (at the age of $6\frac{1}{2}$), I encountered the $\sin,\cos,\tan,\sec,\csc,\cot$ notation. This confused me since I believed that $\sin x$ was $s\cdot i\cdot n\cdot x$. (However, I got over it and now I'm doing category theory.) One more strange notation that I've encountered is $\partial\Sigma$, used to denote the boundary of a space $\Sigma$.

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Wow, you learned trig at 6.5? Very impressive! –  user76061 Mar 1 at 3:19
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@user76061 I should also add that $\sin^{-1}$ is also pretty ambiguous. –  SDevalapurkar Mar 1 at 3:20
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$6\frac{1}{2}=6.5\ne 3$ is pretty strange too –  user80551 Mar 1 at 3:21
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Yeah, I have no clue why arcsin went out of style. Every semester some student will ask why I don't just write cosecant. –  Darrin Mar 1 at 4:06
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@Darrin I don't blame them. The notation $\sin^{-1} x$ meaning inverse is fine, but it's completely messed up by the fact that $\sin^n x = (\sin x)^n$. Granted, there's really no reason why the latter would ever mean $\sin(\sin(\sin \dots \sin(\sin x) \dots ))$, but it just makes the rest confusing and inconsistent. –  Soke Mar 1 at 4:09

How about using pairs of letters like $r,s$ or $u,v$ , or $m,n$ when writing on a blackboard? Unless you're extremely careful, the two in any pair get very easily confused with each other. Or, when you're told you have two collections of objects ( with maybe some additional propreties ) , say $S,X$ , and then you have that $a$, or worse $x$ is an element in $S$. Isn't it so much better to just say $s$ is in $S$, and $x$ is in $X$ ; isn't an element $s$ in $S$ better than any other letter?

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$$\large{\prod_{n = 1}^3 \mathbb{R} = \mathbb{R}^3}$$

Edit: Apparently this is common notation. MJD suggests a better example:

$$\large{\prod_{n = 1}^3 S \neq S^3}$$

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Where has that been used? I've never seen this before (I've seen $(\mathbb{R}^1)^3=\mathbb{R}^3$), though I can understand the logic behind it. –  SDevalapurkar Mar 1 at 3:43
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Why is this strange? $\mathbb{R}^3 = \mathbb{R}\times\mathbb{R}\times\mathbb{R}$. It's is a cartesian product, but still a product. Its quite natural to write like the way you showed. –  Integral Mar 1 at 3:47
    
Maybe I just don't work with this a lot so it seems weird to me. Usually I see it as $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ -- the use of product notation of $n = 1$ to $3$ just seems a bit unnatural to me. –  Soke Mar 1 at 4:05
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This is quite natural. The one you should have mentioned is $$\prod_{n=1}^3 S\color{red}{\ne} S^3.$$ –  MJD Mar 1 at 4:23
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$S$ is the circle, $S^2$ is the sphere (that is, the two-dimensional manifold that is the boundary of the ball in $\Bbb R^3$), $S^3$ is the 3-sphere, which is the 3-dimensional boundary of the ball in $\Bbb R^4$. But unfortunately $S^2 \ne S\times S $; the latter is the torus, sometimes written $T^2$, just to really confuse matters. –  MJD Mar 1 at 4:44

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