Wikipedia:Reference desk/Archives/Mathematics/2021 March 27

= March 27 =

How do you convert from something like (10^78)^(10^32) to 'approximately 10^10^34', and what's the technical term for the latter notation '10^10^x'?
Those numbers come from an episode of PBS Spacetime 'How Many Universes Are There?', https://www.youtube.com/watch?v=XglOw2_lozc - but unfortunately, they don't explain how to convert from one form to the other. If there isn't already a term for numbers of the form 10^10^x, I suggest 'cosmological notation', because the term in double-quotes gets very few hits on Google, so it doesn't seem to have another meaning yet, and that's one of the few fields where ordinary scientific notation is sometimes inadequate. MathewMunro (talk) 07:30, 27 March 2021 (UTC)


 * Well, the conversion is simple: (10^78)^(10^32) is just 10^(78&times;10^32), and if you're just looking at the order of magnitude, you will say that 78&times;10^32 is approximately 10^34. --184.147.181.129 (talk) 08:13, 27 March 2021 (UTC)


 * (E.C.) The calculation appears at 4:27 in that video.
 * We know (x^a)^b = x^(a·b); see Exponentiation.
 * (Note that when we omit parentheses and write x^a^b, by convention we mean x^(a^b) as that can't be reduced to a simpler form.)
 * So (10^78)^(10^32) = 10^(78·10^32). Since 78 ~ 100 = 10^2, 78·10^32 ~ 10^2·10^32 = 10^(2+32) = 10^34.
 * More precisely, since log1078 = 1.892 (to four significant figures), then 78 = 10^1.892 and (10^78)^(10^32) = 10^10^33.892.
 * I'm not familiar with a term specifically for numbers of the form 10^10^x.
 * Hopefully another editor can help with that. -- ToE 08:18, 27 March 2021 (UTC)
 * careful there-- a^b^c^d associates to the right; think a^(b^(c^d)) as shown below — See tetration. --Ancheta Wis    (talk  &#124; contribs) 11:49, 27 March 2021 (UTC)


 * An expression with repeated nested exponentiation, such as $$x^{y^z}$$, is often referred to as a "tower of powers". This is not specific to the use of $$10$$ as the base. The value $$10^{10^{10^2}}$$ is called googolplex, but this is a name just for this one value. The use of the terminology "or around" in saying "$$(10^{78})^{10^{32}}$$, or around $$10^{10^{34}}$$", is rather loose. The second value is larger than the first one by 2,200,000,000,000,000,000,000,000,000,000,000 orders of magnitude, so they are not quite in the same ballpark. For comparison, the radius of the proton and that of the observable universe differ by a mere 42 orders of magnitude. --Lambiam 11:18, 27 March 2021 (UTC)
 * Thanks everyone, I consider this answered MathewMunro (talk) — Preceding undated comment added 16:36, 27 March 2021 (UTC)

Possible Cross Sections of the Regular polyhedra.
The possible cross sections of a cube are triangle, quadrilateral, pentagon and hexagon. Is there any way to gauge which of those types are possible. For example, I believe that *every* triangle is possible, but for a Pentagon, I don't think that those with a significantly obtuse angle (say > 150 degrees) or equal to a regular pentagon are possible. Is there a way to tell which ones are?Naraht (talk) 15:40, 27 March 2021 (UTC)

Other polyhedra
Are all convex quadrilateral possible cross sections of a tetrahedron?Naraht (talk) 15:40, 27 March 2021 (UTC)


 * Assuming that "type" means an equivalence class of similar shapes, the short answer to gauging which types are possible is "yes, it is possible". I don't know if there is a simple way of classifying these types. All acute triangles are possible as a cross section of a cube, but I think that only acute triangles are possible. I have not proved this, but I expect that likewise not all convex quadrilaterals are possible as a cross section of a regular tetrahedron. --Lambiam 21:54, 27 March 2021 (UTC)


 * You're right, only acute triangles are possible cross sections of a cube. Let the cube be [0,1]3 and wlog let the vertex cut off be (0,0,0). Then the three vertices of the triangle are A=(a,0,0), B=(0,b,0), and C=(0,0,c), with 0<a,b,c<1. The angle at A is the angle formed by the vectors (-a,b,0) and (-a,0,c), and the cosine of angle A is then
 * $$\frac{a^2}{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}$$
 * It' not hard to see that this must be between 0 and 1 implying that the angle is acute.


 * A plane cuts two parallel planes in parallel lines, and a plane which cuts through more than three sides of a cube must cut at least one pair of opposite sides, so at least two of the side of the resulting cross-section must be parallel. Therefore a quadrilateral cross section of a cube must be a trapezoid. A hexagonal cross section of a cube must have opposite sides parallel. --RDBury (talk) 00:56, 28 March 2021 (UTC)


 * It seems that each of the four angles of a quadrilateral cross section of a regular tetrahedron is larger than 60°, but that all convex quadrilaterals obeying this constraint are possible. --Lambiam 10:08, 28 March 2021 (UTC)
 * According to my calculations, any angle greater than arccos(1/√3) ≈ 54.7° is possible. Take the vertices of the tetrahedron to be (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1); we'll only consider points (w,x,y,z) with w+x+y+z=1. As the cutting plane, use qw-qx+py-pz=0 where p,q>0, p+q=1. This intersects the tetrahedron in a kite shaped quadrilateral with vertices A=(1/2,1/2,0,0), B=(0,p,q,0), C=(0,0,1/2,1/2), D=(p,0,0,q). In the limiting case of p=0, C is the midpoint of B and D and the angle at B is arccos(1/√3). By taking p to be small but >0, the angle at B can be made as close as desired to this value.


 * I'm not sure what the most convenient description of "quadrilateral space" would be, but it seems clear that it would be 5-dimensional. But the possible slices of a (regular) tetrahedron can be described with four parameters: One for the size of the tetrahedron, and three for the positions of three of the vertices of the quadrilateral along three edges of the tetrahedron; since three points determine a plane, the remaining vertex is determined by the other three. So the space of "quadrilateral slices" is 4-dimensional, implying there is some constraint other than just limits on the angles. I don't know if it would even be feasible to work out what the constraint would be, but I think the dimension argument shows that one exists. --RDBury (talk) 13:41, 28 March 2021 (UTC)


 * PS. I looked a special case of this and I believe the following is true: If the cross section of a regular tetrahedron is a parallelogram, then it is a rectangle. Also, the sum of the dimensions of the rectangle is equal to the side of the tetrahedron. --RDBury (talk) 22:50, 28 March 2021 (UTC)


 * PPS: GeoGebra has this interactive gadget related to this question. --RDBury (talk) 23:04, 28 March 2021 (UTC)


 * Take the unit tetrahedron with its vertices in 4-space as above. Just any four points on different edges, such as one on AB, one on BC, one on CD and one on DA, can be described with four parameters in the unit interval. Their coordinates are an easy function of these parameters, and therefore so are the six distances between these points. To be the vertices of a quadrilateral, they have to be coplanar, which is equivalent to the Cayley–Menger determinant vanishing. This requirement reduces the dimension of the parameter space by 1. The determinant is, I think, a sextic polynomial in the parameters. --Lambiam 01:05, 29 March 2021 (UTC)


 * Cayley–Menger gives a description of quadrilateral space as degree six, dimension 5 variety in 6-space. But not every (irregular) tetrahedron can be inscribed in a regular tetrahedron in the required way. That would produce a condition independent of Cayley–Menger, and the combination of the two conditions would give a dimension 4 variety in 6-space. The condition that a general tetrahedron be inscribable in a regular tetrahedron is equivalent to saying that, up to isometry, it's coordinates can be given as (a,0,a), (0,b,b), (c,s,s-c), (s,s-d,d). There are only 5 parameters used here and you need six parameters to describe a general tetrahedron up to isometry, so there must be an extra condition needed on the tetrahedron. But this should reduce the problem somewhat, since now what is required to so find a single condition in the space of general tetrahedra, which would seem to have a much simpler description than the space of quadrilaterals. --RDBury (talk) 11:48, 29 March 2021 (UTC)


 * The original question (see the heading) was about regular polyhedra. Using the (regular) 4-space unit tetrahedron, assigning parameters $$w,x,y,z$$ so that the points are at $$(w,1{-}w,0,0)$$, $$(0,x,1{-}x,0)$$, $$(0,0,y,1{-}y)$$ and $$(1{-}z,0,0,z)$$, the Cayley–Menger determinant equals
 * $$32\left(wxyz-(1{-}w)(1{-}x)(1{-}y)(1{-}z)\right)^2.$$
 * So, effectively, the quadrilateral subspace of the parameter space is given by a very symmetric cubic polynomial. --Lambiam 13:33, 29 March 2021 (UTC)


 * There's a miscommunication here. The question was, given a quadrilateral, can it be described as the cross section of a regular tetrahedron? How are we to be given a quadrilateral? One way is by giving the lengths of its sides and diagonals. If these lengths satisfy Cayley–Menger then it's a quadrilateral. Another way is by giving the coordinates in the plane and allowing for translation and rotation. In the first method you're given six parameters and one constraint giving a 5 dimensional space. In the second method you're given 8 parameters, the coordinates of the points. But because of translation you can assume the coordinates of the first point are (0,0), and because of rotation you can assume that the coordinates of the second point are (a,0) for some a. That, again, reduces the number of parameters to 5. You're starting with the assumption that the quadrilateral is given a certain way that would guarantee that it's the cross section of a tetrahedron. But that's not how a quadrilateral would be given. The quadrilateral would have to be given in terms of its intrinsic properties: the lengths of sides and diagonal, angles at the vertices, etc. But the w,x,y,z parameters you're using aren't intrinsic properties, they're properties of an assumed embedding in 4-space which you haven't been given. The problem is: Given an intrinsic description of a quadrilateral, find w,x,y,z so that the above points fit that description. You starting with w,x,y,z and trying to show it's a quadrilateral.


 * (I'm now switching to using w,x,y,z as coordinates instead of parameters, so I can talk about, say, the 3-space of points satisfying aw+bx+cy+dz=e in 4-space.) It's actually not that hard to determine if four points lying in a given 3-subpsace w+x+y+z=1 of 4-space are coplaner without using the Cayley–Menger determinant. They will be coplanar iff they satisfy some equation of the form aw+bx+cy+dz=e which is independent of w+x+y+z=1. Subtract e times the second equation from the first, to get (a-e)w+(b-e)x+(c-e)y+(d-e)z=0 and relabel to get aw+bx+cy+dz=0. The two equations are to be independent, so not all of a,b,c,d are 0. So the points are coplanar iff they satisfy aw+bx+cy+dz=0 for some a,b,c,d not all 0. In other words they are coplanar iff they are linearly dependent if taken as vectors. But they are linearly dependent iff the determinant is 0. In the case where the points are (p,1-p,0,0), (0,q,1-q,0), (0,0,r,1-r), (1-s,0,0,s), this determinant is pqrs-(1-p)(1-q)(1-s)(1-t). Another method would be it add a point that's not on w+x+y+z=1, say (0,0,0,0), and say that the original four points are coplanar if the five points including the new one are in the same affine 3-subspace. Again this is easy to check using a determinant and the result turns out the same after a bit of simplification. I'm not sure how you get your expression from the Cayley–Menger determinant, but it matches the condition and has the right degree so it seems correct.


 * In any case, look that the result I mentioned earlier that if a cross section of a regular tetrahedron is a parallelogram then it must be a rectangle. I don't think that can be derived from the Cayley–Menger determinant, since it can only tell you that the points are coplanar, and you already know the points of a parallelogram are coplanar. I was able to prove it by showing that if the quadrilateral is a parallelogram then a=c and b=d in the equation, aw+bx+cy+dz=0, of the cutting plane. --RDBury (talk) 20:38, 29 March 2021 (UTC)


 * PS. Another special case: If a cross section of a regular tetrahedron is a trapezoid but not a parallelogram, then the non-parallel opposite sides are equal and the diagonals are equal, in other words it's an isosceles trapezoid. This is proved using the same idea as the result with parallelograms. --RDBury (talk) 21:10, 29 March 2021 (UTC)


 * I may have misunderstood your previous contribution. You are absolutely right that it is not possible to see from the CM determinant that parallelogram solutions are rectangular. Still, might one not hope that insight in the structure of the coplanarity constraint on 4D parameter space could be of some help in finding general answers to the question? --Lambiam 22:31, 29 March 2021 (UTC)


 * Yes, I do think there is some insight to be gained there. My current thinking it there is some algebraic relation on the distances between the points that's independent of the Cayley–Menger determinant being 0. Using the model you suggested seems like the most efficient way of finding it. It's still not easy though, not that it can't be done with a computer algebra package and sufficient computing power. As I see it, the problem is similar to finding the equation of a curve when it's given parametrically. For example from x=t2, y=(1-t)2, eliminate t to get x2-2xy+y2-2x-2y+1=0. In this case though, there are 5 parameters and the equation is in six variables. --RDBury (talk) 13:06, 30 March 2021 (UTC)

Let $$(\kappa,\lambda,\mu,\nu)$$ be any quadruple of positive numbers. Then
 * $$(w,x,y,z) = (\frac{\kappa}{\kappa+\mu},\frac{\lambda}{\lambda+\kappa},\frac{\mu}{\mu+\nu},\frac{\nu}{\nu+\lambda})$$

satisfies the constraints
 * $$0<w,x,y,z<1;wxyz=(1{-}w)(1{-}x)(1{-}y)(1{-}z)$$

and thus determines a quadrilateral cut of a regular tetrahedron. Conversely, given values for such $$w{,}x{,}y{,}z$$, values for $$\kappa{,}\lambda{,}\mu{,}\nu$$ that reproduce them are given by:
 * $$(\kappa,\lambda,\mu,\nu)=((1{-}x)(1{-}y)(1{-}z),x(1{-}y)(1{-}z),xyz,x(1{-}y)z).$$

So this is another complete parametrization. Since only the ratio $$\kappa:\lambda:\mu:\nu$$ is relevant, and there is no equational constraint, it is possibly more convenient for understanding the space of these quadrilateral cuts. --Lambiam 12:07, 31 March 2021 (UTC)