September 2016 - mathfun@mailman.xmission.com

[math-fun] Fwd: Bar bet
by Bill Gosper 26 Sep '16

26 Sep '16

By "opposite" I think Julian means reflected in a horizontal plane. --rwg ---------- Forwarded message ---------- From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Date: Sun, Sep 25, 2016 at 7:55 PM Subject: Re: [math-fun] Bar bet To: Bill Gosper <billgosper(a)gmail.com> But don't we need to prove that it _is_ tetrahedral, i.e. the tetrahedron > and octahedron > dihedrals are supplementary? Easier to just intersect two tetrahedra and note that this gives an octahedron with all the symmetries of the regular octahedron. How do we know that the intersection is an octahedron? What we're actually doing is taking a regular tetrahedron and cutting off the vertices to get an octahedron, then noting that the planes we're using to cut them off have the same symmetries hence also form a regular tetrahedron, which by inspection is opposite to the original (so that you can swap them). Julian

1 1

Re: [math-fun] Question about base 3/2
by Don Reble 26 Sep '16

26 Sep '16

> Is it possible to write 1/3 as an infinite sum of the form a_1 (2/3)^1 + > a_2 (2/3)^2 + ... where the sequence of nonnegative integer coefficients > a_1, a_2, ... is eventually periodic? No, I figure. Here's why http://www.graysage.com/djr/propp.pdf -- Don Reble djr(a)nk.ca

1 0

Re: [math-fun] Bar bet
by Bill Gosper 26 Sep '16

26 Sep '16

On 2016-09-25 16:18, James Propp wrote: > Kate Fractal should be invited to join math-fun! > > Jim Propp > > On Sunday, September 25, 2016, Andy Latto <andy.latto(a)pobox.com> wrote: > >> On Sun, Sep 25, 2016 at 12:01 PM, Hans Havermann <gladhobo(a)bell.net >> <javascript:;>> wrote: >> >> "The volume of a regular tetrahedron (triangular pyramid with unit >> edges) >> >> is exactly half the volume of a square pyramid with unit edges." >> > >> > This is (I think) equivalent to stating that the volume of an octahedron >> is four times the volume of a tetrahedron (unit edges assumed). >> >> Yes because you can slice an octahedron into two square pyramids. >> >> The geometrical "proof" of this is that you can assemble 4 unit >> tetrahedrons and a unit octahedron to produce a tetrahedron with edge >> length two, and therefore area equal to 8 tetrahedrons. >> >> I don't have the ability to draw a picture of this here, but start >> with an equilateral triangle of side 2, and divide it into 4 >> equilateral triangles of side 1. Place tetrahedrons on the 3 of these >> that are in the same orientation, and then add as a second layer a >> fourth tetrahedron whose bottom three vertices are the points of the >> first 3 tetrahedrons. These four tetrahedrons outline the shape of a >> side-2 tetrahedron, and the space in the large tetrahedron that isn't >> in any of the small ones is exactly a unit octahedron. >> >> Thanks to Kate Fractal for the idea of this construction. >> >> Andy https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb#/media/File:… (Stick a tetrahedron onto the foremost face to complete a double size tetrahedron.) But don't we need to prove that it _is_ tetrahedral, i.e. the tetrahedron and octahedron dihedrals are supplementary? ("Behold!") Hans Moravec designed a version of hashlife based on this recursive partitioning of space. Time ran vertically, and the "initial" conditions were the lower faces, which completely determined the interiors and upper faces. I think he even got some unfortunate grad student to implement it. It was awkward because you normally start with a horizontal square grid initial condition. I think their scheme was to treat this as a "waffle iron" of minimal octahedra and then fill in the next layer of tetrahedra to make an "endo waffle iron", which then determined the next layer of octahedra. But this needs to be done cleverly, so as to halve the spatial frequency at each step, to get the crucial geometric growth of cell sizes, and hopefully, running speed. The hash table records upper faces as results of lower ones. I have no idea by how much this beats my inelegant scheme of overlapping frusta: https://www.lri.fr/~filliatr/m1/gol/gosper-84.pdf --rwg

1 0

[math-fun] Bar bet
by Bill Gosper 25 Sep '16

25 Sep '16

"The volume of a regular tetrahedron (triangular pyramid with unit edges) is exactly half the volume of a square pyramid with unit edges." True or false? --rwg

5 4

Re: [math-fun] Question about base 3/2
by Dan Asimov 25 Sep '16

25 Sep '16

As for periodic expansions in the most obvious sense: Let S = (2/3)^n + (2/3)^(n+k) + (2/3)^(n+2k) + ... = (2/3)^n / (1 - (2/3)^k) = 2^n / (3^k - 2^k). Not much thought went into this, but can S = 1/3 be achieved by some choice of n and k positive integral? We'd need 3^k = 2^k + 3*(2^n) which is impossible with n, k in Z+. —Dan Allan wrote: ----- ... It is entirely conceivable that there exists a *non-canonical* but *periodic* expansion. Jim provided one for 4/5 -- I'm sure the canonical expansion is completely chaotic in this case too. Jim was asking whether there was *any* periodic expansion. -----

2 1

[math-fun] Tightly coloring K_n?
by Michael Kleber 25 Sep '16

25 Sep '16

Suppose you color the edges of K_n with C colors. Say the "spread" of a color = how many of the n vertices are an endpoint of an edge of that color. How small can you make the spread of the most-spread color? The pigeon-hole principle says you can't get below n/sqrt(k), since you'll have at least n-choose-2 / k edges of some color, and even if you pack them all into a complete subgraph, that's how big it would have to be. But surely you can't force *all* the colors into a really tight complete-graph-like arrangement, so seems like you should be able to improve on that substantially. Is there terminology for this question? My naive Google searchings are clobbered by the usual kind of edge coloring, where edges of the same color aren't allowed to meet at a vertex, so maybe I'm just looking for the right names to find prior results. --Michael -- Forewarned is worth an octopus in the bush.

2 3

[math-fun] Question about base 3/2
by James Propp 25 Sep '16

25 Sep '16

Is it possible to write 1/3 as an infinite sum of the form a_1 (2/3)^1 + a_2 (2/3)^2 + ... where the sequence of nonnegative integer coefficients a_1, a_2, ... is eventually periodic? Jim Propp

5 8

Re: [math-fun] Bar bet
by Dan Asimov 25 Sep '16

25 Sep '16

----- On Sun, Sep 25, 2016 at 12:01 PM, Hans Havermann wrote: >> "The volume of a regular tetrahedron (triangular pyramid with unit edges) >> is exactly half the volume of a square pyramid with unit edges." > > This is (I think) equivalent to stating that the volume of an octahedron is four times the volume of a tetrahedron (unit edges assumed). . . . At some point after the above time Andy Latto wrote: The geometrical "proof" of this is that you can assemble 4 unit tetrahedrons and a unit octahedron to produce a tetrahedron with edge length two, and therefore area equal to 8 tetrahedrons. ----- Lovely proof, Andy. Much better than doing the calculation (though it was fun doing it in my head). —Dan

1 0

Re: [math-fun] Bar bet
by Bill Gosper 24 Sep '16

24 Sep '16

> From: Don Reble <djr(a)nk.ca> > To: math-fun(a)mailman.xmission.com > Sent: Friday, September 23, 2016 1:52 PM > Subject: Re: [math-fun] Bar bet > >> "The volume of a regular tetrahedron (triangular pyramid with unit edges) >> is exactly half the volume of a square pyramid with unit edges." >> True or false? --rwg > > Trick puzzle: prove that the volume of _every_ such pyramid > (unit edges, regular polygon base) is a quadratic surd. On 2016-09-23 13:59, Eugene Salamin via math-fun wrote: > The only remaining nontrivial case is the pentagonal base, and it's > clear that all the needed heights are no more involved than stuff > containing sqrt(5). In[445]:= PolyhedronData[{"Pyramid", 5}, "Volume"] Out[445]= 1/24 (5 + Sqrt[5]) > For the hexagonal base, the pyramid is flat, zero > volume, and for higher order polygons, the lateral edges cannot reach > the vertex while remaining of unit length. > > -- Gene In[587]:= PolyhedronData[{"Pyramid", 6}, "Volume"] During evaluation of In[587]:= PolyhedronData::notdef: PolyhedronData has no value associated with the specified argument(s). >> Why, do you suppose?-) --rwg

1 0

[math-fun] Tenure-Track CS Position
by Michael Collins 24 Sep '16

24 Sep '16

My department is again advertising a full-time tenure-track position in computer science, in any area of specialization. Christopher Newport's CS department is an excellent place to work and I am happy to chat with anyone who might have questions. Mike Collins <http://cnu.edu/titleix-eo/pdf/8530%20assistant%20professor% 20of%20computer%20science.pdf>

1 0