February 2011 - mathfun@mailman.xmission.com

[math-fun] Aperiodic tiling of hyperbolic plane?
by Mike Stay 21 Feb '11

21 Feb '11

Has anyone worked out aperiodic tiles for the hyperbolic plane? Is there something analogous to Penrose tiles? Penrose tiles are a 2-d projection of a 5-d hypercube lattice; in 3d, there's a "cubic" tiling of hyperbolic space with dodecahedra where eight dodecahedra meet around a central vertex like cubes do in Euclidean space: http://reperiendi.files.wordpress.com/2011/01/dodecahedral.jpg If we took some "hypercubic" tiling of 5d hyperbolic space and projected down to 2d hyperbolic space, what would we get? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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[math-fun] Spherical polygon area
by Bill Gosper 21 Feb '11

21 Feb '11

Belatedly answering D.Wilson, given vertices as a list of {latitude,longitude} pairs, the (unit sphere) area is: xyz[{\[Phi]_,\[Theta]_}]:={Cos[\[Theta]]*Cos[\[Phi]],Sin[\[Theta]]*Cos[\[Phi]],Sin[\[Phi]]} Spharea[latlongs_List] := Total[MapThread[ Function[{\[Phi]\[Theta]1, \[Phi]\[Theta]2, \[Phi]\[Theta]3}, ArcCos[normcross[\[Phi]\[Theta]1, \[Phi]\[Theta]2].normcross[\ \[Phi]\[Theta]3, \[Phi]\[Theta]2]]], {RotateRight[latlongs], latlongs, RotateLeft[latlongs]}]] - Pi*(Length[latlongs] - 2) normcross[\[Phi]\[Theta]1_, \[Phi]\[Theta]2_] := Normalize[Cross[xyz[\[Phi]\[Theta]1], xyz[\[Phi]\[Theta]2]]] which coincides with a plan suggested by "Ask Dr. Math", if you Google the subject line. Testing on David's equator-based trapezoid of width a and height b, paired with the rest of the polar triangle of width a, so that their areas sum to a: In[121]:= FullSimplify[{Spharea[{{0, 0}, {0, a}, {b, a}, {b, 0}}], Spharea[{{b, 0}, {b, a}, {Pi/2, foobar}}]}, 0 < a < Pi && 0 < b < Pi/2] where foobar, as everyone knows, is the longitude of the North Pole, Out[121]= {ArcSin[(2*Sin[a/2]*Sin[b])/Sqrt[3 + Cos[a] + (-1 + Cos[a])*Cos[2*b]]] - ArcSin[((-1 + Cos[a])*Sin[b])/ Sqrt[Cos[b]^2*Sin[a]^2 - 2*(-1 + Cos[a])*Sin[b]^2]], a - ArcCos[-((2*Sin[a/2]*Sin[b])/ Sqrt[3 + Cos[a] + (-1 + Cos[a])*Cos[2*b]])] + ArcCos[(2*Sin[a/2]^2*Sin[b])/ Sqrt[Cos[b]^2*Sin[a]^2 - 2*(-1 + Cos[a])*Sin[b]^2]]} Unfortunately, Mma chose to write -1 + Cos[a] for the trapezoid but 2*Sin[a/2]^2 for the triangle! And it is completely incapable of simplifying down to a unless we help it: In[133]:= {#, TrigReduce[#]}&[2*Sin[a/2]^2]*Sin[b] Out[133]= {2*Sin[a/2]^2*Sin[b], (1 - Cos[a])*Sin[b]} In[143]:= FullSimplify[Total[%121] /. Rule @@ %133] Out[143]= a --rwg

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Re: [math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients
by Bill Gosper 21 Feb '11

21 Feb '11

(Neil B. beat me to this but is too shy...) In Mathematica 8 and 7 and probably back to 5, In[63]:= Reduce[-c y + b z == f && c x - a z == g && -b x + a y == h, {x, y, z}] Out[63]= (f != 0 && a == (-b g - c h)/f && a != 0 && y == (h + b x)/a && z == (-g + c x)/a) || (f == 0 && g != 0 && b == -((c h)/g) && a != 0 && y == (h + b x)/a && z == (-g + c x)/a) || (h == 0 && g == 0 && f == 0 && c == 0 && b == 0 && a == 0) || (h == 0 && g == 0 && f == 0 && a != 0 && y == (b x)/a && z == (c x)/a) || (f == 0 && g != 0 && b == -((c h)/g) && a == 0 && b != 0 && x == -(h/b) && z == (c y)/b) || (g == 0 && c == 0 && a == 0 && b != 0 && x == -(h/b) && z == f/b && f h != 0) || (g == 0 && f == 0 && h != 0 && c == 0 && a != 0 && y == (h + b x)/a && z == 0) || (h == 0 && b == 0 && a == 0 && c != 0 && x == g/c && y == -(f/c) && f g != 0) || (h == 0 && g == 0 && a == 0 && x == 0 && b != 0 && z == (f + c y)/b && f != 0) || (h == 0 && g == 0 && f == 0 && a == 0 && b != 0 && x == 0 && z == (c y)/b) || (g != 0 && b == -((c h)/g) && a == 0 && b != 0 && x == -(h/b) && z == (f + c y)/b && f != 0) || (g == 0 && f == 0 && c == 0 && a == 0 && b != 0 && x == -(h/b) && z == 0 && h != 0) || (h == 0 && f == 0 && b == 0 && a == 0 && c != 0 && x == g/c && y == 0 && g != 0) || (h == 0 && g == 0 && b == 0 && a == 0 && x == 0 && c != 0 && y == -(f/c) && f != 0) || (h == 0 && g == 0 && f == 0 && b == 0 && a == 0 && c != 0 && x == 0 && y == 0) The first "Orand" contains your a f + b g + c h =0 constraint. The future is now? --rwg From: Fred lunnon <fred.lunnon(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=fred.lunnon%40gmail.com>> To: math-fun <math-fun(a)mailman.xmission.com <http://gosper.org/webmail/src/compose.php?send_to=math-fun%40mailman.xmissi…>> Sent: Sat, February 19, 2011 6:53:26 PM Subject: [math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients Everybody can solve these, can't they --- just like quadratic equations ... Here's a simple example: solve for x,y,z, b z - c y = f , c x - a z = g , a y - b x = h . Oops --- the determinant vanishes: | c 0 -a | | -b a 0 | = 0 . | 0 -c b | No solution, says our CAS equation solver, firmly. Except of course, if the RHS also vanishes there is obviously a 1-dimensional homogeneous solution; but simply substituting f,g,h -> 0 should encourage our CAS to disgorge some version of that too. End of story? Not quite: you see, adding a equn1 + b equn2 + c equn3 eliminates all variables, leaving a f + b g + c h = 0 ; and in my particular application, this constraint happens to be satisfied by the coefficients. Now some head-scratching unearths a 3-dimensional (in general) solution spanned by the symmetric linear basis [ x -> -h/b, y -> 0, z -> +f/b ] , [ x -> +g/c, y -> -f/c, z -> 0 ] , [ x -> 0, y -> +h/a, z -> -g/a ] . In particular, summing eventually delivers the symmetric solution x -> (g/c - h/b)/3 , y -> (h/a - f/c)/3 , z -> (f/b - g/a)/3 , which I knew ought to exist by the symmetry of the equations. I have encountered this type of problem --- solving SLE's on a manifold --- on a number of occasions, without ever finding an effective solution algorithm. Surely it must have been considered before --- any pointers, anyone? Is it reasonable to expect that the next generation of CAS solvers might inform us about constraints on our coefficients under which such equations might possess extra solutions? Fred Lunnon

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Re: [math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients
by Eugene Salamin 20 Feb '11

20 Feb '11

From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Sat, February 19, 2011 6:53:26 PM Subject: [math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients Everybody can solve these, can't they --- just like quadratic equations ... Here's a simple example: solve for x,y,z, b z - c y = f , c x - a z = g , a y - b x = h . Oops --- the determinant vanishes: | c 0 -a | | -b a 0 | = 0 . | 0 -c b | No solution, says our CAS equation solver, firmly. Except of course, if the RHS also vanishes there is obviously a 1-dimensional homogeneous solution; but simply substituting f,g,h -> 0 should encourage our CAS to disgorge some version of that too. End of story? Not quite: you see, adding a equn1 + b equn2 + c equn3 eliminates all variables, leaving a f + b g + c h = 0 ; and in my particular application, this constraint happens to be satisfied by the coefficients. Now some head-scratching unearths a 3-dimensional (in general) solution spanned by the symmetric linear basis [ x -> -h/b, y -> 0, z -> +f/b ] , [ x -> +g/c, y -> -f/c, z -> 0 ] , [ x -> 0, y -> +h/a, z -> -g/a ] . In particular, summing eventually delivers the symmetric solution x -> (g/c - h/b)/3 , y -> (h/a - f/c)/3 , z -> (f/b - g/a)/3 , which I knew ought to exist by the symmetry of the equations. I have encountered this type of problem --- solving SLE's on a manifold --- on a number of occasions, without ever finding an effective solution algorithm. Surely it must have been considered before --- any pointers, anyone? Is it reasonable to expect that the next generation of CAS solvers might inform us about constraints on our coefficients under which such equations might possess extra solutions? Fred Lunnon _______________________________________________ We have vectors A = (a,b,c), Z = (x,y,z), and F = (f,g,h). The linear equations become a vector cross product AxZ = F. A solution exists only when A.F = 0, and then the most general solution consists of the sum of any one particular solution P with an arbitrary solution of the homogeneous equation AxZ = 0. The latter has the 1-dimensional solution space consisting of multiples of A. So the general solution is Z = P + kA. If A' is any vector in the plane normal to F, and linearly independent of A, then AxA' lies along F, and so we have a solution Z = ((F.F)/[F,A,A']) A', [...] denoting the scalar triple product. Any such A' is of the form FxB with B arbitrary. Then [F,A,FxB] = ((FxB)xF).A = ((F.F)B - (B.F)F).A = (F.F)(A.B), and Z = (FxB)/(A.B) is a general solution. If A' = FxB is linearly dependent on A, we have 0 = Ax(FxB) = (A.B)F - (A.F)B = (A.B)F, so B is restricted to A.B nonzero. The solution found in the original post is of this form with B = (1/a, 1/b, 1/c) and A.B = 3. -- Gene

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[math-fun] Digital question
by David Wilson 20 Feb '11

20 Feb '11

Let d > 0 divide 99. I can then show that d divides n iff d divides the base-10 reverse of n. I believe the converse is true as well, that these are the only d with this property. In general, I believe that ( d | n <=> d | base-b reverse of n ) <=> d | b^2-1 Supposing this is indeed true, it can't be too hard to prove, but I'm at a loss. ----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1204 / Virus Database: 1435/3454 - Release Date: 02/19/11

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[math-fun] House numbers ...
by Guy Haworth 19 Feb '11

19 Feb '11

I'm at 33 (close but no cigar) but a son IS at 31 ... cf 31, M31 and M(M31) ... I was asked at a door, security-coded with the number '1066', if all the codes were year numbers. I said 'Yes, but they haven't all happened yet'. g

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Re: [math-fun] Kevin Brown?
by Dan Asimov 19 Feb '11

19 Feb '11

Michael wrote: << I suppose it's possible that a true math fiend might spend years of his [life] waiting for Taxicab Number on Hardy Street to come on the market. >> I once jumped at the chance to choose a telephone number ending in 1729, keeping it until I moved 5 years later. (In the days when you couldn't keep a number after moving, alas.) Dan P.S. Definition from < www.m-w.com >: PRIVATE: intended for or restricted to the use of a particular person, group, or class Those who sleep faster get more rest.

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Re: [math-fun] Kevin Brown?
by Dan Asimov 19 Feb '11

19 Feb '11

A certain mathematician, whom I know well, lived at street number 841 for 40 years without realizing that it's 29^2. I hope this isn't OCD, but I've got into the habit of trying to factor every street address and [last 4 digits of] every phone number (is there a word for this?) that I come in close contact with. --Dan Thane wrote: << I have a mathematician friend who lived in a house with street address 2401 for over ten years before he realized it was (and in fact still is) 7^4, causing him significant mental distress >> Those who sleep faster get more rest.

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Re: [math-fun] Kevin Brown?
by Dan Asimov 18 Feb '11

18 Feb '11

Mine might be prime, but because it ends in an even digit the probability of this is 0. Tom wrote: << My social security number is prime. . . . >> Those who sleep faster get more rest.

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[math-fun] Ian Porteous
by Fred lunnon 18 Feb '11

18 Feb '11

Ian R. Porteous, author of the standard textbook "Clifford Algebras and the Classical Groups": RIP. http://www.guardian.co.uk/education/2011/feb/17/ian-porteous-obituary WFL

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