Wikipedia:Reference desk/Archives/Mathematics/2022 January 2

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January 2

Most notable mathematical discovery by an amateur since WWII

After noticing a quick addition (and then appropriate reversion and comment about Original Research) to the Collatz Problem, I was wondering what the most significant mathematical discovery since WWII by an amateur mathematician. The only one that comes to mind is Marjorie Rice and the Pentagonal tilings. Are there others at this level (geometrical constructions) or beyond? I understand that amateur mathematician is a nebulous term, I'm particularly excluding whoever happened to have the computer that found specific entries for GIMPS.Naraht (talk) 00:42, 2 January 2022 (UTC)

We have List of amateur mathematicians with notable contributions. Greg Egan probably also deserves to be on the list; see the Quanta article on superpermutations. --{{u|Mark viking}} {Talk} 00:54, 2 January 2022 (UTC)

There is an MO thread[1] about this too. 2601:648:8202:350:0:0:0:9435 (talk) 00:51, 4 January 2022 (UTC)

Don't really know how notable it is but "A lower bound on the length of the shortest superpattern" coauthored by an anonymous 4chan user is a nice story. 2A01:E34:EF5E:4640:6D7C:A7E:28B7:1036 (talk) 19:13, 4 January 2022 (UTC)

Notation

Does $-{\sqrt {x^{2}}}$ have two solutions? For example, for $-2$, I could just $-2^{2}=4$, extract the root and end up with $-2$. Or, I could cancel the square root with the exponent, and end up with $-(-2)=2$ --Bumptump (talk) 01:44, 2 January 2022 (UTC)

The term "solution" does not apply, since there is no equation. There is a connection between square roots and equations: $x={\sqrt {a}}$ is a solution of the equation $x^{2}=a$ in the unknown $x$. When $a>0$, there are two solutions, but only one is (by definition) the square root of $a$: the positive one. It follows that ${\sqrt {x^{2}}}=|x|,$ the absolute value of $x$. So then $-{\sqrt {x^{2}}}=-|x|.$ There is no ambiguity of any kind. This assumes we are working in the world of real numbers. Some mathematicians treat the complex logarithm as a "multivalued function", which (depending on how one defines everything) may rub off on the square root function. Others find this unnecessarily confusing.  --Lambiam 02:32, 2 January 2022 (UTC)

So, if I have $f(x)=x^{2}$ and $g(x)=-sqrt(x)$, no one will see $f\circ g=g\circ f$ in general? Would any restriction, like $x\leq 0forf(x)$ make the composed functions equivalent? --Bumptump (talk) 14:13, 2 January 2022 (UTC)

If we're talking about real valued functions, then $g\circ f=-{\sqrt {x^{2}}}=-|x|$ for all x, and $f\circ g=(-{\sqrt {x}})^{2}=x$ for x ≥ 0 and undefined for x < 0. The functions agree for x ≥ 0, but the domains are different so they are different functions. If you want do as Lambiam suggested for complex valued functions, and talk about multivalued functions and branch cuts then it depends on what conventions you're using. But if you're using multivalued functions then $g\circ f=-{\sqrt {x^{2}}}=\pm x$ and $f\circ g=(-{\sqrt {x}})^{2}=x$; one is single valued and one is multi(two)valued so again, they are different functions. In general, composition of functions does not commute, but you can demonstrate that with much simpler functions. (I've used the convention that composition follows the reverse of the order the functions are evaluated. I've always thought that was somewhat counterintuitive but it seems to be the prevailing, but perhaps not the universal standard.) --RDBury (talk) 17:07, 2 January 2022 (UTC)

I too dislike function composition order, but we're stuck with it because of Path dependence.  Card Zero  (talk) 01:36, 3 January 2022 (UTC)

The ordering of the function operands is the same as they appear in stepwise application: $(f\circ g)(x)=f(g(x)).$ So the "backwardness" starts already with the ordering in function application, which may derive from the grammatical ordering in Germanic languages. Leibniz used "eine Function von $x$", which Euler notated as $f(x).$ If they had spoken Turkish and Gottfried Bey had used "$x$'in bir fonksiyonu", Leonhard Bey might have used the notation $(x)f,$ and function composition could have been defined by $(x)(f\,;g)=((x)f)g.$  --Lambiam

I thought you might say that. :) I reserve the right to dislike it the other way round, also. Actually I hadn't anticipated as far as that comparison to the grammar of spoken languages. I wonder what the grammar would be in Polish?  Card Zero  (talk) 03:37, 3 January 2022 (UTC)

The same as in Łukasiewicz notation, "funkcją zmiennej $x$". So with Polish genitors we might have dropped some of the brackets but still be stuck with $\circ fg\,x=f(gx).$ There is a reason why the reverse notation is called "reverse" (which is not that it is the notation in Reverse Polish). I could not use more generally "Indo-European grammar", though, because in Latin and Ancient Greek, for example, the order of possessor and possessee can freely vary.  --Lambiam 10:48, 3 January 2022 (UTC)