Wikipedia:Reference desk/Archives/Mathematics/2024 February 12

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February 12

Slicing a 5-simplex hexateron in half.

A Tetrahedron (3-simplex) if cut in half with a plane for which two vertices are equally above and two equally below gives a square. What 4-D shape do you get if you slice a 5-simplex in half with a 4-space with three vertices equally above and three equally below? there will be 9 of the edges that cross the 4-space cut, but I'm not quite sure what relationship they have. If the three vertices "above" are A, B & C and those below are X, Y & Z, the crossing point (midpoint) of AX is the same distance from AY, AZ, BX and CX, but are a different equal distance to BY, BZ, CY & CZ. Ideas? Naraht (talk) 05:29, 12 February 2024 (UTC)

Computations are easier if you work in the 5-dimensional affine subspace p+q+r+s+t+u=1 in R⁶. The coordinates of the 5 simplex are then (1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0), (0, 0, 0, 0, 0, 1). If you scale everything up by a factor of 2, the coordinates of the vertices of the slice are (1, 0, 0, 1, 0, 0), (1, 0, 0, 0, 1, 0), (1, 0, 0, 0, 0, 1), (0, 1, 0, 1, 0, 0), (0, 1, 0, 0, 1, 0), (0, 1, 0, 0, 0, 1), (0, 0, 1, 1, 0, 0), (0, 0, 1, 0, 1, 0), (0, 0, 1, 0, 0, 1). The equation of the slice is given by p+q+r=s+t+u, and there are six faces given by p=0, q=0, ... u=0. You can describe this as the Cartesian product of a triangle with itself. We have an article on it: 3-3 duoprism. I don't know if I can add anything that's not in the article. --RDBury (talk) 07:08, 12 February 2024 (UTC)

PS. Here's something that's not mentioned in the article: It's the dual of the Birkhoff polytope or order 3, B₃. I don't have a reference, but the computation isn't that hard so I assume it's well known. Perhaps it's just not considered noteworthy. --RDBury (talk) 07:26, 12 February 2024 (UTC)

Extending

RDBuryThis would theoreticaly be a series of the 2n dimensional cuts of the 2n+1 dimensional simplexes. Not sure how many of them would reach notability though.

Simple Limit Question

Why Lim x→0+ (1⁄x) ≠ 1⁄0 if both are equal to infinity (or that is both are not defined). Note that this is not my HW question, but question of curiosity (so don't tag it unnecessarily). Thank You. --ExclusiveEditor _{Notify Me!} 12:07, 12 February 2024 (UTC)

Considering the function 1/x, it has an essential discontinuity when x=0. Therefore there can be no limit as x approaches zero. Dolphin (t) 12:21, 12 February 2024 (UTC)

@Dolphin51: Thank you, but I am learning calculus, so I don't get those terms. (I read the joint article's section). Okay I get it a bit, but my question roots to basics that doesn't (1/0⁺) also have essential discontinuity in a sense that it is conceptually defined to be just ahead of zero, thus on graph (like 0 itself) have it too and approach infinity/ not defined status and hence get equaled to 1/0? By the way, thanks for essential discontinuity. ExclusiveEditor _{Notify Me!} 15:04, 12 February 2024 (UTC)

X approaching 0+ doesn’t mean x is slightly greater than 0. It means x approaches 0 from the positive side (x is moving left towards 0.) Conversely, x approaching 0- means x approaches 0 from the negative side (x is moving right towards 0.) Students of calculus are mostly asked to find the limit at discontinuities at which 0+ and 0- are the same and so there is only one value of the limit. For these functions you will be asked to find the limit at a removable discontinuity. Dolphin (t) 16:12, 12 February 2024 (UTC)

1/0 hasn't a defined value, 1/x could mean plus or minus infinity when approached from the positive or negative side. A simple example of a removable disconuity is the value of x/x. This is undefined when x=0 but when approached from either side the value is 1. NadVolum (talk) 16:48, 12 February 2024 (UTC)

The notation

$\lim _{x\to 0^{+}}f(x)=\infty$

is somewhat misleading. It is not an ordinary equality in which the two sides denote real numbers, but a convenient abbreviation for

$\forall y>0~\exists \,\delta >0~\forall x\in \mathbb {R} _{>0}~(x<\delta \implies f(x)>y).$

(See Limit of a function § Infinite limits.) It can be interpreted as an equality in the domain of the extended real number system.

Accepting $\lim _{x\to 0^{+}}{\frac {1}{x}}={\frac {1}{0}}$ as an equality in the extended reals is problematic, because, by symmetry, also $\lim _{x\to 0^{-}}{\frac {1}{x}}={\frac {1}{0}}$ should then hold. But $\lim _{x\to 0^{+}}{\frac {1}{x}}$ and $\lim _{x\to 0^{-}}{\frac {1}{x}}$ are as apart as it gets. (They meet on the projectively extended real line, but this is no longer an ordered set, so then we get into a very different game.)  --Lambiam 19:26, 12 February 2024 (UTC)

A good explanation. I'd go a bit further and say that since since 1/0 is not defined, it's not equal or unequal to anything. Part of the problem is that when it comes to limits the '=' sign often used in ways that don't signify actual equality, and the $\lim _{x\to 0^{+}}f(x)=\infty$ statement is a case in point. Another example is Big O notation; in particular see the section "Matters of notation". This abuse of notation is fine as long as the intended audience can interpret the intended meaning, but it's confusing for people who haven't mastered calculus. It's all part of the rather notorious mathematical jargon. But jargon seems to to arise naturally in any specialized area of knowledge; even Fortnite has it's own jargon. As an aside, when I taught calculus we made a distinction between $\infty$ and $+\infty$; one tells you about the behavior of |f(x)| and the other about f(x). Apparently it depends on which textbook you use. RDBury --RDBury (talk) 03:01, 13 February 2024 (UTC)

The original Intel IEEE floating point chips supported that, see IEEE 754-1985#Extending the real numbers. NadVolum (talk) 09:33, 13 February 2024 (UTC)