May 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Embedding with integer coordinates
by Adam P. Goucher 26 May '13

26 May '13

Yeah, that's subsumed by my inner product argument. Also, the vertices of a pentagon are a subset of those of the dodecahedron and icosahedron, so they follow as trivial corollaries. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Warren D Smith > Sent: 05/26/13 07:23 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Embedding with integer coordinates > > If your thing has got two interpoint distances A,B such that (A/B)^2 > is irrational, > then its points cannot be be embedded in any finite dimension with > integer coordinates. > > Regular pentagon, (A/B)^2=2.6180339887498948482 = g+1 > where g=golden ratio > where B=side, A=2-apart vertex sep > > icosahedron (including center as point) > (A/B)^2 = sin(2*pi/5)^2 = (g/2)^2 + 1/4 =0.9045084971874737120 > > dodecahedron (ditto) > (A/B)^2 = g^2 * (3/4) = 1.96352549156242113610 > where A=radius, B=side > > all easily seen to be irrational, so none can be embedded. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Embedding with integer coordinates
by Warren D Smith 26 May '13

26 May '13

If your thing has got two interpoint distances A,B such that (A/B)^2 is irrational, then its points cannot be be embedded in any finite dimension with integer coordinates. Regular pentagon, (A/B)^2=2.6180339887498948482 = g+1 where g=golden ratio where B=side, A=2-apart vertex sep icosahedron (including center as point) (A/B)^2 = sin(2*pi/5)^2 = (g/2)^2 + 1/4 =0.9045084971874737120 dodecahedron (ditto) (A/B)^2 = g^2 * (3/4) = 1.96352549156242113610 where A=radius, B=side all easily seen to be irrational, so none can be embedded.

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[math-fun] Peano midpoints curve
by Bill Gosper 26 May '13

26 May '13

The usual discrete-sampled Peano curve self contacts, so you can't follow the path. The standard fix is to average consecutive vertices to form the median curve. Ever seen Peano's? Here's why not: http://gosper.org/fylfy-mind.png . A really pleasing design, if you're young enough. --rwg

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Re: [math-fun] Platonic regular polyhedra with integer vertices ?
by Bill Gosper 26 May '13

26 May '13

Cris Moore>What is the smallest number of dimensions in which we can embed a pentagon with integer vertices? More generally, for an n-gon, the argument below suggests (I think) that we need at least p-1 dimensions where p is the largest prime factor of n. We can do a hexagon in three dimensions, but not, I think, in two. Cris On May 25, 2013, at 11:09 PM, Bill Gosper wrote: Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center. Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix. No, That was DanA. http://www.tweedledum.com/rwg/rectarith12.pdf (end) illustrates hexagons in R^3, followed immediately by "How many dimensions before we see (regular) octagons, pentagons, dodecagons, ...? We won't! Even in an infinite dimensional grid, the only regular polygons of finite size are triangles, hexagons, and squares. We could probably prescribe the nth coordinate of the kth vertex of a regular pentagon, say, but infinitely many coordinates would be nonzero, resulting in infinite size. And it would only approach regularity as we consider higher and higher dimensions." --rwg

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Re: [math-fun] Platonic regular polyhedra with integer vertices ?
by Bill Gosper 26 May '13

26 May '13

Henry> Interesting! Would your proof indicate that the dodecahedron _could_ be embedded in 4D? Or would 4D require a different proof of non-embeddability? At 09:05 PM 5/25/2013, Dan Asimov wrote: Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center. Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix. This excludes the dodecahedron and icosahedron, and the other three are pretty easy to embed with integer vertices. --Dan On 2013-05-25, at 7:21 PM, Henry Baker wrote: Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices? What are the coordinates? (Here I'm talking about normal square Cartesian coordinates.) Obviously the cube has the trivial embedding with the unit cube. http://www.tweedledum.com/rwg/rectarith12.pdf claims without proof that you can't even embed a regular pentagon in R^n. I recall it was fairly obvious. --rwg

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[math-fun] Platonic regular polyhedra with integer vertices ?
by Henry Baker 26 May '13

26 May '13

Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices? What are the coordinates? (Here I'm talking about normal square Cartesian coordinates.) Obviously the cube has the trivial embedding with the unit cube.

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Re: [math-fun] exact inches per meter
by Bill Gosper 26 May '13

26 May '13

ACW>In general, regardless of base, the period of 1/n divides phi(n). So perhaps it would be interesting to provide phi(n)/period. On Sat, May 25, 2013 at 6:32 PM, Hans Havermann <gladhobo(a)teksavvy.com>wrote: Bill Gosper: In[209]:= RealDigits[1/499999] Out[209]= A very large output was generated. Here is a sample of it: {{{2,0,0,0,0,0,4,0,0,0,0,0,8,0,0,0,0,1,6,0,0,0,0,3,2,0,0,0,0,6,4,0,0,0,1,2,8,0,0,0,2,5, <<26587>>, 5,1,5,6,2,5,0,3,1,2,5,0,0,6,2,5,0,0,1,2,5,0,0,0,2,5,0,0,0,0,5,0,0,0,0,1,0,0,0,0,0}},<<1>>} HH> Mathematica 9 just gives: A very large output was generated. Here is a sample of it: <<1>> RWG>I would have expected shorter periods. HH>I wanted to chart the period lengths for numbers of the form 49..9: Do[Print[Length[RealDigits[1/(10^n/2-1)][[1,1]]]],{n,2,16}] 42 498 357 641 26670 2499999 10248414 27737540 7001968 $Aborted This is better: Do[Print[MultiplicativeOrder[10,10^n/2-1]],{n,2,32}] 42 498 357 [...] -------------- FI[n_Integer] := Times @@ (HoldForm @@ {#[[1]]}^#[[2]] &) /@ FactorInteger[n] n 10^n/2-1 phi/period period In[230]:= TableForm[Table[Join[{n},FI/@{10^n/2-1,"\t",EulerPhi[10^n/2-1]/#,"\t",#}&(a)MultiplicativeOrder[10,10^n/2-1]/.FI->(#&)],{n,2,32}]] Out[230]//TableForm= 2 7^2 1 2 3 7 3 499 1 2 3 83 4 4999 2 7 3 7 17 5 49999 2 3 13 641 6 31 127^2 2 3^2 2 3 5 7 127 7 4999999 2 3 191 4363 8 7 23 310559 2^2 2 3 11 31 5009 9 691 723589 2 3^2 2^2 5 17 23 3547 10 17 14033 20959 2^5 3 7 2^4 499 877 11 29 1724137931 2 2^2 5 7 139 1240387 12 2969 168406871 2^4 5 7 53 3499 4813 13 491 10183299389 2 2^2 5 7^2 43 3041 19469 14 7 7142857142857 2 3 2^3 3 19 191 4363 18797 15 499999999999999 2 3 307 271444082519 16 17737 281896600327 2 3 739 2^3 3 11 739 5779649 17 19 2631578947368421 2 3 2^2 3^2 5 7 23 739 1187 310559 18 223 208513 10753058401 2^6 3^2 5 37 2^7 3^2 5 7 37 181 17299 19 1871 3061 873037477229 2^3 3 5 17 2^2 3 5 7^2 11 17 4454272843 20 7 782209 9131647862473 2^5 3^3 7 2^6 3 17^2 97 947 1390241 21 31 8999 1792313841322871 2^3 5 3 5 11 409 1789 57503 1742261 22 1049 2137 2230437170146223 2^4 2^3 3 89 131 1601 501187 1389853 23 1699 2069 3121 1382819 3295771 2^7 3^2 5 2^2 3 5 11 13 47 283 109859 691409 24 41717706103 11985318626233 2 3 2^3 3^3 7 587 1559 1316099 45760861 25 59 88369891 958989105371471 2^2 3 5 2 5 7 29 420809 95898910537147 26 7 17 420168067226890756302521 2^4 2^4 3 5 43 244283760015634160641 27 229 367021 409453351 14529115561 2^6 3^3 5^2 2^2 3^2 5^2 19 61 73 613 2039 121075963 28 4728835511159 1057342761912761 2^2 2^2 5 11 17 19 47 73 653 1109 3323 2130536717 29 71 76863049 9162079325173218881 2^6 3 5 2^4 5 7 47 4973 5741 12967 68141 77339 30 23 47 457 3691153 274199124029771599 2^6 3^2 2^4 3^2 7^2 11 19 23 25633 16760311 55646347 31 761 1295159 1798241 44167979 63871459 2^7 5 7 487 2^4 3 5 19 71 137 331 11239 21419 647579 32 7 7142857142857142857142857142857 2 3 2^3 3 19 23 389 739 1187 128509 310559 50010001 --rwg

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Re: [math-fun] exact inches per meter
by Bill Gosper 25 May '13

25 May '13

David Wilson>Also related and interesting is 499999 = 31 x 127^2 ----------------- In[209]:= RealDigits[1/499999] Out[209]= A very large output was generated. Here is a sample of it: {{{2,0,0,0,0,0,4,0,0,0,0,0,8,0,0,0,0,1,6,0,0,0,0,3,2,0,0,0,0,6,4,0,0,0,1,2,8,0,0,0,2,5, <<26587>>, 5,1,5,6,2,5,0,3,1,2,5,0,0,6,2,5,0,0,1,2,5,0,0,0,2,5,0,0,0,0,5,0,0,0,0,1,0,0,0,0,0}},<<1>>} In[213]:= Length[%209[[1,1]]] Out[213]= 26670 In[214]:= %209[[2]] Out[214]= -5 It "abbreviated" -5 to <<1>> !! In[212]:= Length[RealDigits[1/499999,2][[1,1]]] Out[212]= 4445 I would have expected shorter periods. --rwg -----Original Message----- From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun- bounces(a)mailman.xmission.com] On Behalf Of Bill Gosper Sent: Thursday, May 23, 2013 4:55 PM To: math-fun(a)mailman.xmission.com Subject: [math-fun] exact inches per meter has a long repeating decimal: In[170]:= RealDigits[100/(254/100)] Out[170]= {{{3, 9, 3, 7, 0, 0, 7, 8, 7, 4, 0, 1, 5, 7, 4, 8, 0, 3, 1, 4, 9, 6, 0, 6, 2, 9, 9, 2, 1, 2, 5, 9, 8, 4, 2, 5, 1, 9, 6, 8, 5, 0}}, 2} I.e. 39.3700... But the binary is very short: In[171]:= RealDigits[100/(254/100), 2] Out[171]= {{1, 0, 0, {1, 1, 1, 0, 1, 0, 1}}, 6} i.e. 100111.0101... Likewise the octal: In[172]:= RealDigits[100/(254/100), 8] Out[172]= {{4, {7, 2, 7, 5, 3, 6, 5}}, 2} Reason: 254 = 2*(2^7 - 1) . --rwg

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[math-fun] exact inches per meter
by Bill Gosper 25 May '13

25 May '13

has a long repeating decimal: In[170]:= RealDigits[100/(254/100)] Out[170]= {{{3, 9, 3, 7, 0, 0, 7, 8, 7, 4, 0, 1, 5, 7, 4, 8, 0, 3, 1, 4, 9, 6, 0, 6, 2, 9, 9, 2, 1, 2, 5, 9, 8, 4, 2, 5, 1, 9, 6, 8, 5, 0}}, 2} I.e. 39.3700... But the binary is very short: In[171]:= RealDigits[100/(254/100), 2] Out[171]= {{1, 0, 0, {1, 1, 1, 0, 1, 0, 1}}, 6} i.e. 100111.0101... Likewise the octal: In[172]:= RealDigits[100/(254/100), 8] Out[172]= {{4, {7, 2, 7, 5, 3, 6, 5}}, 2} Reason: 254 = 2*(2^7 - 1) . --rwg

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[math-fun] group theory
by Bill Gosper 24 May '13

24 May '13

Yasmine wanted to divide 57 buttons into 13 groups. How many groups will Yasmine have? Will there be any buttons left over? (Saxon Math, Intermediate 4) --rwg

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