September 2012 - mathfun@mailman.xmission.com

[math-fun] That fractal pentagram again
by Bill Gosper 09 Apr '15

09 Apr '15

For some pictorial. mildly trick questions: http://gosper.org/foupent.pdf (1.6M) --rwg

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[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 19 Jun '14

19 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

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[math-fun] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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[math-fun] is your Hurwitz # unlisted?
by Bill Gosper 28 Sep '12

28 Sep '12

Can someone name a single (uncontrived) constant (e.g., π, e^3, 2^(1/3), parity number,...) that is provably nonHurwitz? And if Hurwitz, is not homographically equivalent to a single- mover, linear? Boy, are we ignorant. --rwg

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[math-fun] 18kt golden ratio
by Bill Gosper 26 Sep '12

26 Sep '12

An infinite continued fraction with fractional terms approaching 1. Out[123]= 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 {-, -, -, -, -, -, -, -, --, --, --, --, --, --, --, --, --, --, --, --, 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 57 58 59 60 61 62 63 64 65 66 67 68 69 --, --, --, --, --, --, --, --, --, --, --, --, --} == 58 59 60 61 62 63 64 65 66 67 68 69 70 1 Sqrt[5] 1 1 1 -(-) + ------- + - Hypergeometric2F1[1, 1, 2 - -------, -- (5 - Sqrt[5])] 2 2 2 Sqrt[5] 10 Check: In[124]:= N[FromContinuedFraction[%[[1]]] -> %[[2]], 33] Out[124]= 1.23225970948789169611305410061559 -> 1.23225970948789169611305410135170 Elementwise reciprocating the terms, In[125]:= 1 ------------- == Out[95][[2]] Out[123][[1]] Out[125]= 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 {2, -, -, -, -, -, -, -, --, --, --, --, --, --, --, --, --, --, --, --, 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, --, 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 58 59 60 61 62 63 64 65 66 67 68 69 70 --, --, --, --, --, --, --, --, --, --, --, --, --} == 57 58 59 60 61 62 63 64 65 66 67 68 69 (5 + 3 Sqrt[5]) / 1 (5 + Sqrt[5] - Sqrt[5] Hypergeometric2F1[1, 1, 2 + -------, Sqrt[5] 1 -- (5 - Sqrt[5])]) 10 In[126]:= N[FromContinuedFraction[%[[1]]] -> %[[2]], 33] Out[126]= 2.49245974602128660339685183239050 -> 2.49245974602128660339685183239151 Larger terms -> more digits. --rwg

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[math-fun] Do you have an unidentified Hurwitz #?
by Bill Gosper 24 Sep '12

24 Sep '12

Suppose we have a mysterious quantity foo. We can quickly detect its Hurwitz signature: In[44]:= hursig[ContinuedFraction[foo, 69]] Out[44]= {1, 1, 1, 4 k, 3, 2 + 4 k} This is not a single mover, so we can't use fromcf. Propose successively larger homographic determinants 2, 3, ... In[45]:= hursigs[foo, 69, 2] During evaluation of In[45]:= 0.151464 Out[45]= {3, 2 -2 18 {1, 1, 1, 4 + 6 k, 1, 4, 1, 5 + 6 k, 1, 1, 1, 29 + 24 k, } -2 3 3, 8 + 6 k, 6, 9 + 6 k, 3, 41 + 24 k} 1 0 18 {1, 1, 1, 6 k, 1, 4, 1, 1 + 6 k, 1, 1, 1, 13 + 24 k, 3, -1 2 4 + 6 k, 6, 5 + 6 k, 3, 25 + 24 k} 1 -1 18 {1, 1, 1, 2 + 6 k, 1, 4, 1, 3 + 6 k, 1, 1, 1, 21 + 24 k, -2 4 3, 6 + 6 k, 6, 7 + 6 k, 3, 33 + 24 k} No single movers--fail. In[46]:= hursigs[foo, 69, 3] During evaluation of In[46]:= 0.27556 Out[46]= {6, 3 -3 22 {1, 1, 1 + 4 k, 9, 2 + 4 k, 4, 1, 1, 2 + 4 k, 2, 3, 1, -3 4 2 + 4 k, 1, 7, 1, 3 + 4 k, 2, 3, 1, 4 + 4 k, 4} 1 3 16 {1, 1, 4, 9 + 3 k, 2 + k, 3, 1 - k, 9 + 2 k, 1 - 3 k, 1, 1 6 4 + 3 k, 10 + 3 k, 1, 3, 2, 11 + 3 k} 1 1 2 {1, 1 + 6 k} -1 2 1 -1 20 {1, 1, 4, 4 k, 9, 4 k, 1, 7, 1, 4 k, 1, 3, 2, 1 + 4 k, 1, -3 6 7, 1, 2 + 4 k, 9, 3 + 4 k} 1 -3 16 {1, 3, 2, 8 + 3 k, 1 + 3 k, 1, 4, 9 + 3 k, 2 + k, 3, 3 -6 1 - k, 9 + 2 k, 1 - 3 k, 1, 4 + 3 k, 10 + 3 k} 1 -3 12 {1, 1 - 2 k, 4 + 2 k, 10 + 2 k, 1, 3 + 2 k, 2 - 2 k, 2 -3 11 + 2 k, 4 + 2 k, 1 - 2 k, 1, 11 + 2 k} } Bingo in the third row. In[48]:= %46[[2, 1, 3]] Out[48]= {1 1 , 2, {1, 1 + 6 k}} -1 2 In[50]:= Simplify[mat2fn[Inverse[%%[[1, 1]]]]@fromcf[%%[[3]], k]] Out[50]= (1 + Tan[1/3])/(2 - Tan[1/3]) In[51]:= ContinuedFraction[#, 22] & /@ {%, foo} Out[51]= {{0, 1, 4, 2, 1, 1, 1, 4, 3, 6, 1, 1, 1, 8, 3, 10, 1, 1, 1, 12, 3, 14}, {1, 4, 2, 1, 1, 1, 4, 3, 6, 1, 1, 1, 8, 3, 10, 1, 1, 1, 12, 3, 14, 1}} So foo == (2*Cos[1/3] - Sin[1/3])/(Cos[1/3] + Sin[1/3]) I've yet to see a "naturally occurring" Hurwitz number that this scheme won't catch, i.e., reduce to a single mover. But if need be, the hursigs function could be generalized to handle Hurwitzes with quadratic progressors, or even higher. I have no idea how to generalize fromcf. --rwg

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[math-fun] Was: Continued fraction for exp(3)
by Bill Gosper 24 Sep '12

24 Sep '12

Again, the relevant property isn't really interleaved polynomials, but rather eventual periodicity mod any integer. But the only Hurwitz numbers we have formulæ for are "single mover", linear, or equivalent thereto under homographic maps. Julian habilitated my nonworking attempt at this, but before doing so correctly predicted that, for any Hurwitz number h, there would be exactly σ(|ad-bc|) equivalence classes of patterns among the continued fractions for (a h +b)/(c h + d), where σ is DivisorSigma. E.g., there are seven possible patterns when Δ := |ad-bc| =4, regardless of h. For h := Coth[1/3], e.g., In[75]:= hursigs[Coth[1/3], 69, 4] During evaluation of In[75]:= 0.606005 (secs) Out[75]= {7, 4 -4 10 {1, 3, 8 + 9 k, 156 + 144 k, 11 + 9 k, 3, 1, 12 + 9 k, } 0 1 228 + 144 k, 15 + 9 k} 2 -3 18 {1, 1, 1 + 9 k, 1, 3, 3 + 9 k, 1, 1, 20 + 36 k, 1, 1, 0 2 6 + 9 k, 3, 1, 7 + 9 k, 1, 1, 38 + 36 k} 2 -4 1 {3 + 6 k} 0 2 1 0 10 {1, 3, 2 + 9 k, 60 + 144 k, 5 + 9 k, 3, 1, 6 + 9 k, -1 4 132 + 144 k, 9 + 9 k} 1 -1 18 {1, 1, 3 + 9 k, 3, 1, 4 + 9 k, 1, 1, 26 + 36 k, 1, 1, -1 5 7 + 9 k, 1, 3, 9 + 9 k, 1, 1, 44 + 36 k} 1 -2 18 {1, 1, 9 k, 3, 1, 1 + 9 k, 1, 1, 14 + 36 k, 1, 1, -3 10 4 + 9 k, 1, 3, 6 + 9 k, 1, 1, 32 + 36 k} 1 -3 10 {1, 3, 5 + 9 k, 108 + 144 k, 8 + 9 k, 3, 1, 9 + 9 k, -4 16 180 + 144 k, 12 + 9 k} showing quasiperiods 10, 18, and 1. The argument 69 is how many continued fraction terms to use, and should exceed thrice the expected longest quasiperiod. If h fails to be Hurwitz, the function hursig (below) will just find longer and longer quasiperiods for increasing values of 69. We yet lack a proven bound for how many matrices to try, and are just using a,|b|,|c| ≤ Δ, but we'll know if we miss any because Julian proved his prediction. Here is our code: Clear[hurcanon]; hurcanon[L_List] := Block[{v = Variables[L][[1]]}, First[Sort[ NestList[ Expand[If[ FreeQ[# /. k -> -1, _?(# < 0 &)], # /. k -> k - 1, #] &[ kRotateLeft[#, 1]]] &, #, Length[#] - 1]]] &[ Expand[L /. v -> v + Max[Ceiling[ Solve[# == 0][[1, 1, 2]] & /@ Select[L, Not[NumberQ[#]] &]]]]]] kRotateLeft[L_, n_] := Join[Drop[L, n], Take[L, n] /. k -> k + 1] trailing0s[L_List] := If[(Equal @@ L && (L == {} || L[[1]] == 0)), Length[L], Length[L] - Position[L, Except[0], 1][[-1, 1]]] Clear[hurperiod]; hurperiod[L_List] := Ordering[Table[ trailing0s[ Drop[L, 2*k] - 2*Drop[Drop[L, k], -k] + Drop[L, -2*k]], {k, Floor[Length[L]/3]}], -1][[1]] Clear[hursig]; hursig[L_List] := Block[{per = hurperiod[L]}, hurcanon[k*(Take[L, {-per - 1, -2}] - Take[L, {-2*per - 1, -2 - per}]) + Take[L, {-per - 1, -2}]]] hursigs[h_?NumericQ, terms_Integer, det_?NumericQ, rang_: Hold[]] := (Print[#1]; #2) & @@ Timing[Block[{sigs = {}, range = {ReleaseHold[rang], Abs[det]}[[1]]}, Do[If[IntegerQ[(b*c + det)/a] && (a*h + b)/(c*h + (b*c + det)/a) > 0, Block[{sig = hursig[ContinuedFraction[(a*h + b)/(c*h + (b*c + det)/a), terms]]}, If[Not[MemberQ[sigs, {_, _, sig}]], PrependTo[ sigs, {MatrixForm[{{a, b}, {c, (b*c + det)/a}}], Length[sig], sig}]]]], {a, 1, range}, {b, -range, range}, {c, -range, range}]; {Length[sigs], MatrixForm[sigs]}]] Here is Julian' proof of σ(|Δ|) distinct signatures for det = Δ We first note that we can make the lower-left entry smaller by adding a term: In[217]:= {{a, b}, {c, d}} == {{b, r}, {d, s}}.{{q, 1}, {1, 0}} /. {q -> Floor[c/d], r -> a - b*Floor[c/d], s -> c - d*Floor[c/d]} Out[217]= True so by induction {{a,b},{c,d}}=={{A,B},{C,0}}.<CF matrices>, and we need only consider numbers of the form (a*x+b)/c. Note that this allows negative terms, so we need the lemma that a negative term can be assumed to be the first term without changing the tail. Next, a only matters mod b: In[283]:= {{a, b}, {c, 0}}.{{0, 1}, {1, 0}}.{{x, 1}, {1, 0}} Out[283]= {{a + b x, b}, {c, 0}} Then, for a given determinant ∆, the number of possible tails is at most the number of {{a,b},{c,0}} with determinant ±∆ (since the CF terms contribute a ±1) and b>0 (negating the matrix does not affect what the transformation is), where a is considered mod b. Since bc=±∆ b divides ∆, and since each b can be paired with b different a's and two different c's, the number is at most twice the sum of the divisors of ∆. But we also have that In[284]:= {{a, b}, {c, 0}}.{{-1, 1}, {1, 0}}.{{1, 1}, {1, 0}}.{{-1, 1}, {1, 0}} Out[284]= {{-a, b}, {-c, 0}} so (a,b,c) is the same as (-a,b,-c), and we can assume c positive, for a bound of σ(∆). But are these all different? Suppose that two were the same, so that there would be a unimodular matrix relating them: In[289]:= {{p, q}, {r, s}} /. Solve[{{a, b}, {c, 0}}.{{p, q}, {r, s}} == {{d, e}, {f, 0}}, {p, q, r, s}][[1]] Out[289]= {{f/c, 0}, {-((-c d + a f)/(b c)), e/b}} Since e/b and f/c have to be integers, and ef=±bc (determinants), e=±b and f=±c. Then (af-cd)/(bc) being an integer means that either a+d is divisible by b (if f=-c) or a-d is (if f=c). Thus, taking d mod b, we find that {{d,e},{f,0}}={{εa,±b},{εc,0}}, which is an equivalence we have already dealt with. Thus there are exactly σ(∆) different tails for continued fractions of the numbers (ax+b)/(cx+d), with ad-bc=∆ and x is neither rational nor a quadratic irrational (this last comes from eliminating possible identities in which algebraic conditions on x mean that the matrices can be non-obviously equivalent (and equivalent "only" for that particular x)). QED --rwg ACW> For some reason I find this remarkable. It suggests the question: how does one go about searching for interleaved polynomial sequences? On Thu, Sep 6, 2012 at 3:53 PM, Jeffrey Shallit <shallit(a)uwaterloo.ca <http://gosper.org/webmail/src/compose.php?send_to=shallit%40uwaterloo.ca>>wrote: > As far as I know, nothing is known. I computed thousands of partial> quotients 30 years ago, but there was no simple pattern I could detect.>>> On 9/5/12 3:17 PM, Allan Wechsler wrote:>>> What is known about the simple continued fraction expansion of exp(3)?>> Obviously we can generate it fairly easily, but I want to know if there>> are>> any known patterns or periodicities in the terms. I thought exp(3) was a>> known Hurwitz number (continued fraction consists of interleaved>> polynomials) but I can't see any obvious patterns.>> ______________________________**_________________

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[math-fun] single mover Hurwitz numbers
by Bill Gosper 23 Sep '12

23 Sep '12

Sometime around 18 Nov 2010, Julian derived the general closed form for the continued fraction a,c,d,...,z,a+b,c,d,...,z,a+2b,..., but we (or a disk crash) seem to have misplaced it, despite my nagging false(?) memory that I reported it here. This, derived in a possibly different way, makes a nice toy. fromcf[optional preamble,quasiperiod,variable or {variable, nonzero starting value}] In[481]:= fromcf[{1, 2*k, 1}, k] Out[481]= E In[487]:= fromcf[{2}, {1, 2*k, 1}, {k, 1}] Out[487]= E In[473]:= fromcf[{2*n + 1}, n] Out[473]= Coth[1] In[484]:= fromcf[{4*k, 1}, k] Out[484]= -1 + Cot[1/2] In[486]:= fromcf[{Sqrt[2] - 1, -1/Sqrt[2], Sqrt[2] - 1}, {4*n/3 + 2/3}, n] Out[486]= E^3 Code: In[106]:= cfmat[cf_List] := {Numerator /@ #, Denominator /@ #} &@ Convergents[cf][[{-1, -2}]] In[107]:= cfmat[{a_}] := {{a, 1}, {1, 0}} In[108]:= cfmat[{}] = IdentityMatrix[2] Out[108]= {{1, 0}, {0, 1}} Clear[fromcf]; fromcf[pre_List: {}, {a___, (c1_: 1)*v_ + c0_: 0}, {v_Symbol, lo_: 0}] := fromcf[pre, {a, c1*v + c1*lo + c0}, v] /; FreeQ[{a}, v] In[103]:= fromcf[pre_List: {}, {a___, (c1_: 1)*v_ + c0_: 0, b__}, {v_Symbol, lo_: 0}] := fromcf[Join[pre, {a, c1*lo + c0}], {b, a, c1*v + c0 + c1}, {v, lo}] /; FreeQ[{a, b}, v] In[104]:= fromcf[pre_List: {}, {a___, (c1_: 1)*v_ + c0_: 0, b__}, v_Symbol] := fromcf[Join[pre, {a, c0}], {b, a, c1*v + c0 + c1}, v] /; FreeQ[{a, b}, v] fromcf[pre_List, {a___, (c1_: 1)*v_ + c0_: 0}, v_] := Block[{M = cfmat[{a}], A, B, C, D, det}, det = Det[{{A, B}, {C, D}} = M]; FromContinuedFraction[ Append[pre, A/(C + (det Hypergeometric0F1[1 + (B + C + A c0)/(A c1), det/( A^2 c1^2)])/((B + C + A c0) Hypergeometric0F1[( B + C + A c0)/(A c1), det/(A^2 c1^2)]))]]] /; FreeQ[{a}, v] --rwg

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[math-fun] ABC conjecture math colloquium at UCSC
by Eugene Salamin 22 Sep '12

22 Sep '12

This is at UC Santa Cruz in Santa Cruz CA. Complete list of math colloquia: http://www.math.ucsc.edu/seminars/index.html October 9, 2012 The ABC Conjecture Professor Brian Conrad Stanford University The ABC Conjecture, formulated in the mid-1980's by Oesterle and Masser, is one of the most important conjectures in number theory. It has many deep consequences, but its basic formulation can be given in entirely elementary terms. Earlier this month, a 500-page solution was announced by Shinichi Mochizuki (building on several thousand pages of work he has carried out over the last 20 years). In this talk, I will explain what the conjecture asserts, some evidence that supports it, and a few of its corollaries. McHenry Building - Room 4130 Tuesday - 4:00 p.m. Refreshments served at 3:30 p.m. For further information please contact the Mathematics Department at (831) 459-2969.

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[math-fun] New dice problem, also kill-joy for old magic permutation-generating dice problem
by Warren Smith 22 Sep '12

22 Sep '12

> Neat idea! Synopsis for people who didn't click through: a set of four > 12-sided dice, faces numbered from 1 to 48 with each number appearing > once, such that if you roll all four of them, all four permutations of the > dice are equally likely. > > With three people, it looks to me like you can do it with six-sided dice. > Here are the 11 solutions, according to a little Mma, though I'd be happy > to have someone else confirm that I didn't mess things up: > > {{1, 5, 9, 12, 14, 16}, {2, 6, 7, 11, 13, 18}, {3, 4, 8, 10, 15, 17}},... --yes but, you could just roll ONE die with the 6 permutations written on its 6 faces. And for 4-item perms use an icosahedral die plus a coinflip to get the 24 possibilities... or a cube & tetrahedron... So I don't see the point. (Although perhaps this has some other application.) --Here's a different problem: Devise a convex polyhedron with 5 faces, such that the probability it lands on each face, is 1/5. I have a solution in mind which in fact works for any number N>=4 of faces (here N=5), although I'm not sure whether we should accept my solution. The question inside the question is: "what is the right probabilistic model?" And the answer inside the answer is not so obvious to me. --Warren D. Smith

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