April 2016 - mathfun@mailman.xmission.com

Re: [math-fun] Simple finite group problem
by Keith F. Lynch 21 Apr '16

21 Apr '16

Allan Wechsler <acwacw(a)gmail.com> wrote: > Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form > a cyclic group under multiplication, generated by 3 (or 7). This > group has only two subgroups, {1} and {1,9}. These are the three > groups Dan enumerated. .... > Wait, I think I just realized. Consider, for instance, {2,4,6,8}. > Under multiplication, these do form a group with identity 6. I > didn't think of it because it doesn't "cohere" with the ring of > integers mod 10. {5} is another example. I'm not sure how to > enumerate these. {6} and {4,6} are two more. Right. Dan Asimov <dasimov(a)earthlink.net> wrote: > I think it would also fair to say {0} is a group under > multiplication modulo 10. Right. And that's all eight of them. Mod N, the identity elements are those numbers I such that I^2 = I mod N. For N=10, those are 0, 1, 5, and 6. The numbers of identity elements mod N appear to be given by A034444. Mod 30, there are 8 of them, and 28 groups. Every number 0...N-1 "belongs to" at most one identity element. For N=10, 0 belongs to 0; 1,3,5,7 belong to 1; 5 belongs to 5; and 2,4,6,8 belong to 6. That's all 10 numbers. They each appear in at least one group. Mod N, the number of numbers that *don't* appear in any multiplicative group appears to be given by A055654. It appears to be always 0 for square-free N, and never 0 if N is not square free. Those groupless numbers are always divisors of N. I think the order of any of these groups always divides phi(N). Except for very small N, you'll always have {0}, {1}, {1, N-1}, and the phi(N) numbers relatively prime to N. The last may have other subgroups. If M divides N, I think you'll always also have, for each group in M, the same group in N only with each term multiplied by N/M. For instance {2,4,6,8} where N=10 is an echo of {1,2,3,4} where N=5. And {5} where N=10 is an echo of {1} where N=2. So there's probably some simple recurrence relation which can be used to find how many multiplicative groups there are mod N. How many are there? 1,2,3,3,4,6,5,6,5,8,5,10,7,10,14,9,6,10,7,14,17,10,5,24,7,14,7,17,7,28,9,12,17,12,24,17,... Not in OEIS. I plan to add it. This discussion thread makes it obvious that I need help in phrasing my description of what this is about, so someone please help. Thanks.

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[math-fun] Failure to Rupert the regular icosahedron
by Warren D Smith 21 Apr '16

21 Apr '16

Icosahedron of circumradius=1, inradius=0.79465447, midradius=0.85065080: (1) Viewed face-on, cross section is a regular hexagon with circumradius=sqrt((5+sqrt(5))*2/15)=0.982246946 and inradius=sqrt((5+sqrt(5))/10)=0.850651. (2) Viewed vertex-on, cross section is a regular 10-gon with circumradius=2/sqrt(5)=0.89442719 and inradius=cos(18 degrees)*circumradius i.e. inradius=sqrt((5+sqrt(5)/10)=0.850651. (3) Viewed edge-on, cross section is a hexagon (which is non-regular, with sym group of order 4 not 12) with 4 "far" vertices at radius=1, forming a "golden rectangle" inscribable in the icosahedron, plus 2 "close" vertices at radius=sqrt((5+sqrt(5))/10)=0.850651. These facts hopefully can be seen from https://en.wikipedia.org/wiki/Regular_icosahedron#Orthogonal_projections and doing some trig. (Assuming I have not made an error) these facts fail to show that an icosahedron can be Ruperted (i.e. passed thru a hole drilled in a copy) since none of these 3 orthogonal projections fits strictly inside any other. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Simple finite group problem
by Keith F. Lynch 21 Apr '16

21 Apr '16

This is a simple finite group problem, not to be confused with a finite simple group problem. It's a simple problem about finite groups, not a difficult problem about finite simple groups. List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another. I noticed that the number of multiplicative groups of integers mod N is not in OEIS. I plan to add it.

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Re: [math-fun] Simple finite group problem
by Keith F. Lynch 21 Apr '16

21 Apr '16

Dan Asimov <asimov(a)msri.org> wrote: > Keith F. Lynch <kfl(a)KeithLynch.net> wrote: >> List all multiplicative groups of integers mod ten. For instance >> {1,9} is one of them, and {1} is another. > If I understand the problem, it's to list all subgroups > of the group of invertible elements of the ring Z/10. > The invertible elements can be chosen as {-3, -1, 1, 3}. > Since there is an element of order 4, this is isomorphic to > the abelian group Z/4, so it has: .... > {-3, -1, 1, 3}, {-1, 1}, {1}. Not quite what I meant. Yes, those are solutions, but those are not the only solutions. I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36. How can I phrase that better? Thanks.

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Re: [math-fun] Minsky "starfish"
by Bill Gosper 20 Apr '16

20 Apr '16

Regardless of whether the center is a true vertex, the odd number of spokes would require retracing somehow. But, apologies. My experimental procedure to produce starfish43.png must have been flawed. Running backwards (red) and forwards (green) 10^7 points and plotting every 101st: gosper.org/starfish43rg.png shows distinct inbound and outbound spokes, no problem. --rwg On 2016-04-14 17:53, Allan Wechsler wrote: > I'm guessing that the apparent order-5 "vertex" is formed from two effects. > First, that the scale of the picture is large -- a zoom on the center would > reveal a void around the origin where the spokes seem to converge. Second, > the points that are close in the plot are actually separated by a small > multiple of 5, so adjacency in the plot is not the same as adjacency in the > dynamic system. > > On Thu, Apr 14, 2016 at 7:20 PM, rwg <rwg(a)sdf.org> wrote: > >> (Hopefully) I'd've said "puzzle" if I knew the answer. My only guess is >> that >> it retraces the spokes. Needs to be checked. -rwg >> >> On 2016-04-14 12:06, Allan Wechsler wrote: >> >>> Was your last question a puzzle, or an honest question? I'm withholding my >>> answer in case it was a puzzle. >>> >>> On Thu, Apr 14, 2016 at 2:48 PM, Bill Gosper <billgosper(a)gmail.com> >>> wrote: >>> >>> For x₀ = 34, y₀ = 55, δ > 0, and ε = 4 (sin π/5]^2/δ, without the Floor >>>> operation, the Minsky iteration simply repeats five points on an ellipse >>>> whose shape depends on δ. If we include the Floors, we get a variety of >>>> fivish orbits. >>>> δ = 1 produces those intriguing fractal pentagrams >>>> gosper.org/rastar.pdf discovered >>>> by Corey and Julian, and perhaps earlier by Steve Witham. A portion of >>>> the >>>> x-y rug plot is Figure 31 (p17) in >>>> http://www.blurb.com/books/2172660-minskys-trinskys-3rd-edition . A few >>>> months ago, Adam Goucher made the startling observation that this plot >>>> defines a legal kites-and-darts Penrose tiling. (See the last three >>>> images >>>> in gosper.org/g12a.pdf (102MB!), the stills from my (and Rohan >>>> Ridenour's) >>>> recent G4G12 talk. (Rohan did the animations on a separate screen.)) >>>> Varying δ gives a variety of orbits. δ = 33/100 simply loops over ten >>>> points in five close pairs. But δ = 17/50 makes five fuzzy patches of >>>> 100 points each. And δ = 17/50 + 1/googol makes a pretty doily with >>>> 30071 >>>> points! gosper.org/mnskplt.png >>>> Similarly, δ = 14/25 gives a 5794 point starfish, while δ = 14/25 + >>>> 1/googol gives a 72721 point obese starfish. gosper.org/mnskplt1.png . >>>> Various other starfish are in gosper.org/mnskplt2.png . Warning: The >>>> third one is actually five sea-cucumbers having a little fun at your >>>> expense. The last two (δ =.62 and .63) are just printouts of merely >>>> period >>>> 5 loops illustrating the experimental fact that every perturbation of δ >>>> by >>>> as little as .01 seems to produce a different orbit. >>>> The first image (δ = 43/100) is the first 10 million points of a 31 >>>> billion >>>> point orbit! Here is every 5001st point of it: >>>> gosper.org/starfish43.png . Note the radius grew from 55 to > 200000. >>>> But something is (star)fishy. Finite Minsky orbits are unicursal. How >>>> can >>>> there be a graph vertex of degree 5 in the center?? --rwg >>>> _______________________________________________ >>>> >>>

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Re: [math-fun] 3-coloring the rationals
by Keith F. Lynch 20 Apr '16

20 Apr '16

James Propp <jamespropp(a)gmail.com> wrote: > Is there a standard term for this tripartition? I don't know. I discovered it independently. I think I was trying to find a parity-related proof that there was no point that has rational distances from all four corners of the unit square. I didn't succeed, and as far as I know the problem is still unsolved. No point within the plane, I mean. Outside the plane, it's easy to find such a point. But that would be a pyramid scheme. :-) > But I think there's another coloring that satisfies the conditions I > stated: -r is blue if r is green and vice versa. Maybe so. But I'm still not convinced that there isn't some completely unrelated coloring that would also solve your problem. > Keith F. Lynch <kfl(a)keithlynch.net> wrote: >> Extra credit: Is there any color or pairs of colors that form a group >> under multiplication? Is there any color or any pair of colors that >> forms a group under addition? > Yes and yes. Do tell. > Maybe all but finitely many convergents to pi are the same color? Sure, and maybe all but finitely many decimal digits of pi are fours and sevens. It's never been disproven, but it seems kind of unlikely. > Maybe some quadratic irrational has the property in question, or > maybe not; that might be my fall-asleep problem for tonight. The convergents to sqrt(2) don't, but I noticed that if you cheat by starting with 2/1 instead of 1/1, and then, as always, repeatedly replacing p/q with (p+2q)/(p+q), all these pseudo-convergents will be red (even/odd), and will still converge to sqrt(2). But I'd be surprised if it's not possible to find pseudo-convergents of any color you please to converge to any real number you please. For instance one simple formula for pi is 4/1 - 4/3 + 4/5 - 4/7 + ... The partial sums of that series are all red. And the standard formula for e, the sum of the reciprocals of the factorials, gives nothing but green (odd/even) partial sums after the four four terms. Your fall-asleep problem for tonight is to find a sequence for pi whose partial sums are either all blue or all green, or a sequence for e whose partial sums are all red or all blue. :-) Is there some natural coloring for algebraic numbers? I no longer think convergents are a way to get anything interesting. Maybe something to do with the parity of the coefficients, e.g. what patterns of even and odd A, B, and C have when Ax^2 + Bx + C = 0. What are the rules for adding, subtracting, multiplying, and dividing numbers expressed in that form? I never learned it. I do know you'd typically get higher-order polynomials, e.g. the set of all quadratic numbers isn't closed under any interesting operations, similarly with the set of all cubic numbers, etc.

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Re: [math-fun] "starshot"
by Keith F. Lynch 19 Apr '16

19 Apr '16

This sounds almost identical to the late Robert Forward's Starwisp proposal, except that that would have used microwaves rather than laser light for power and propulsion. Wikipedia is a good starting point for learning about Starwisp.

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Re: [math-fun] 3-coloring the rationals
by Bill Gosper 18 Apr '16

18 Apr '16

In terms of the Newman enumeration of the rationals In[596]:= NestList[1/(2 Floor@# - # + 1) &, 0, 42] Out[596]= {0, 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8, 8/5, 5/7, 7/2, 2/7, 7/5, 5/8, 8/3, 3/7, 7/4, 4/5, 5, 1/6, 6/5, 5/9, 9/4, 4/11, 11/7, 7/10, 10/3, 3/11, 11/8, 8/13} In[595]:= color[r_?NumberQ] := col[{0, R}, r, {ComplexInfinity, B}] In[582]:= col[L_, r_, H_] := Block[{m}, Which[L[[1]] == r, L[[2]], H[[1]] == r, H[[2]], r < (m = {Mediant[L[[1]], H[[1]]], Complement[{R, G, B}, {L[[2]], H[[2]]}][[1]]})[[1]], col[L, r, m], True, col[m, r, H]]] In[597]:= color /@ %596 Out[597]= {R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R, G, B, R} Taking every other term (i.e., restricting to [0/1,1/1]) still gives a 3-coloring. (Mediant works given ComplexInfinity because In[591]:= Denominator@ComplexInfinity Out[591]= 0 Yay!) --rwg On 2016-04-17 19:53, Tom Karzes wrote: > This statement is true for rationals between 0 and 1. You cannot > obtain rationals greater than 1 if you only consider the half of the > s-b tree between 0/1 and 1/1. > > For rationals greater than 1, it still works if you throw (1/0) into > the mix (i.e., treat it as a rational) and color it green. But if you > don't consider (1/0) to be a rational, then you could break the > pattern by coloring both (0/1) and (1/0) the same color, e.g. red. > And you could then choose any color for (1/1), but if you force it to > blue, then you get the same color for any a/b and b/a, as opposed to > red/green reversal if you allow (1/0) as a rational. > > Tom > > James Propp writes: > > For uniqueness, one may cite the theory of Farey fractions and the > > Stern-Brocot tree. Every rational can be obtained by an iterative process > > of inserting mediants, starting from 0/1 and 1/1. This means that at each > > stage one has no choice (that is to say, that one has 1 choice).

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Re: [math-fun] 3-coloring the rationals
by Keith F. Lynch 18 Apr '16

18 Apr '16

James Propp <jamespropp(a)gmail.com> > There's a unique way to 3-color the rationals in [0,1] using the colors > red, blue, and green so that 0 is red, 1 is blue, and the fractions a/b, > c/d, and (a+c)/(b+d) all have distinct colors whenever ad-bc=1. > Puzzle (no spoilers till Sunday please!): What color is 355/113? > Argh: 355/113 isn't in [0,1]; replace it by 113/355. Either way, it's blue. Odd/odd is blue. Even/odd is red. Odd/even is green. Even/even can be reduced to one of the above. I have no idea how to prove that this coloring is unique, however. Do you have a proof? I noticed, years ago, that the rationals have these three parities, so that was the first thing I tried. I agree with RCS that this is unambiguous over the whole of the rationals. I don't understand why you think there's any ambiguity with negatives. Just as negative integers have the same parity as if they were positive, so do negative rationals. Extra credit: Is there any color or pairs of colors that form a group under multiplication? Is there any color or any pair of colors that forms a group under addition? Can this coloring be extended in any natural way to any irrationals? You mention 355/113, which is best known as one of the convergents to pi. Are there any irrational numbers whose convergents are all the same color? (Pi isn't one.) If so, it might make sense to regard the number as having that color. Especially if it wouldn't break any group of rationals with that color.

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Re: [math-fun] object stably resting on tabletop
by Warren D Smith 18 Apr '16

18 Apr '16

The conditions for stability If object rests on 1 point, then that point must lie vertically below center of mass, and the radii of curvature of the surface at that point, must exceed the distance to the center of mass. If object rests on 2 points A,B (or line segment AB) then the center of mass must be vertically above the line segment, and the max radius of curvature of the surface at the contact points in directions perp to the line AB, must exceed the distance to the center of mass. If object rests on 3 or more noncollinear points then the center of mass must be vertically above a point interior to their convex hull (if above a hull-boundary point then use the 2-point condition). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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