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[math-fun] A word sequence
by Éric Angelini 03 Dec '20

03 Dec '20

Hello Math-Fun, [and J.-M. Falcoz for a possible French version], here is an English word sequence, for a change: S=7,5,7,4,2,-16,1,-22,-22,-22,-26,0,0,-31,-31,-6,-3,-37,-38,-37,-7,-8,... If you now compute k= |n+a(n)| the result k is the number of letters used in English to write the integers « n » and « a(n) ». Example for n=1: k = |1 + a(1)| = |1+7| = 8 = {ONE,SEVEN} For n=2 we have: k = |2 + a(2)| = |2+5| = 7 = {TWO,FIVE} For n = 3: k = |3+a(3)| = |3+7| = 10 = {THREE,SEVEN} ... For n=6: k = |6+a(6)| = |6-16| = 10 = {SIX,SIXTEEN} etc. When a(n) = 0, there is (apparently) no solution [see a(12) and a(13)]. We’d like S to be the lexicographically earliest seq with this property. Best, É.

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[math-fun] Is 17 trying to tell us something?
by Bill Gosper 03 Dec '20

03 Dec '20

{Cos[π/17]==1/16 (1-√17+√(34-2 √17)+2 √(17+3 √17+√(170+38 √17))), Sin[π/34]==1/16 (-1+√17+√(34-2 √17)-2 √(17+3 √17-√(170+38 √17)))} or even 2 √(17 + 3 √17 - √(170 + 38 √17)) == √( 2 (3 + √17) (2 √17 - √(2 (17 - √17)))) et al. https://mathworld.wolfram.com/TrigonometryAnglesPi17.html includes a messier version of Cos[π/17]. (People keep forgetting that Sin[π/5] is messier than Sin[π/10].) —rwg

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[math-fun] Kids: An errorless way to freely expand and contract the continued fraction of an irrational.
by Bill Gosper 02 Dec '20

02 Dec '20

The function and its inverse are named longer and shorter. For example, expand Cot[1/4]-1 to one, two, three, and four terms: In[258]:= NestList[longer,Cot[1/4]-1,4] Out[258]= {-1+Cot[1/4], 2+1/1/(-3+Cot[1/4]), 2+1/(1+1/(3-Cot[1/4])/(-4+Cot[1/4])), 2+1/(1+1/(10+1/(4-Cot[1/4])/(-43+11 Cot[1/4]))), 2+1/(1+1/(10+1/(1+1/(43-11 Cot[1/4])/(-47+12 Cot[1/4]))))} Shorten the 3rd expansion by two: In[260]:= shorter[%[[4]],2] Out[260]= 2+1/1/(-3+Cot[1/4]) (In a notebook, these partial expansions display traditionally, but for a messy last term.) And now, ladies and gentlemen, here they are, Miss Jayne ... oops, no, here they are, the definitions of longer and shorter: Clear[longer]; longer[cf_, 0] := cf longer[x_?NumericQ, n_ : 1] := longer[HoldForm @@ {x}, n] longer[cf_, n_ : 1] := longer[cf /.HoldForm[r : Except[_Integer]] :> Floor[r] + 1/HoldForm @@ {Simplify@Together[1/Mod[r, 1]]}, n - 1] Clear[shorter]; shorter[cf_, 0] := cf shorter[cf_, n_ : 1] := shorter[cf /. (a_ : 0) + 1/HoldForm[r_] :> HoldForm @@ {1/Simplify[1/(a + 1/r)]}, n - 1] This continued fraction begins 2, 1, 10, 1, 18, 1, ... . Does it continue to grow by {8, 0}? Yes indeed, In[254]:= fromcf[{8 # + 2, 1}] Out[254]= -1 + Cot[1/4] where fromcf provides closed forms for a class of Hurwitz numbers: fromcf[pre_List : {}, {a___, (c1_ : 1)*# + c0_ : 0, b__}] := fromcf[Join[pre, {a, c0}], {b, a, c1*# + c0 + c1}] fromcf[pre_List : {}, {a___, (c1_ : 1) # + c0_ : 0}] := Simplify[FunctionExpand[ 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)])/( A c1 ((B + C + A c0)/(A c1))! Hypergeometric0F1Regularized[( B + C + A c0)/(A c1), det/(A^2 c1^2)]))]]]]] —rwg

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[math-fun] An alternative math model for COVID
by Henry Baker 02 Dec '20

02 Dec '20

The 100-year-old differential equation model for infectious diseases -- the one with R/R0 -- isn't the only possibility. I'm going to suggest another one -- which I'm not claiming is necessarily correct, but which has some of the same characteristics as the R/R0 model. My model is more akin to 'predator-prey' models, only in this case, the predator is microscopic. The diffeq model suggests that 'herd immunity' is reached when the % of immune/already-had-its is large enough that there are too few uninfected people left to sustain a chain reaction. According to most estimates using this type of model, herd immunity % for COVID will be in the 70-80% range. But there is an alternative model where a kind of 'herd immunity' is reached at a far lower %. Consider a predator-prey model of lions and antelopes. The lions don't have to *win* the lion-v-antelope races; they just have to not come in *last* in order to eat. Suppose that a lion has a top speed of 20mph for 1/4 mile (I have no idea what a lion's actual speed is). All the antelopes have to do is to stay ahead of the lion for that period of time in order to survive. At time t=0, some % of the antelope herd can't achieve 20mph, and are bound to be eaten. But as time goes on, more and more of these 'slow' antelopes are eaten and no longer remain in the herd. At some point, *all* of these 'slow' antelopes will be gone, and since a famished lion isn't likely to get any faster, he will be forced to find another herd of antelopes to chase which might still contain some 'slow' antelopes. Note that in my model *everyone* who is exposed gets the disease -- i.e., R/R0 is exceedingly high, but only some % -- e.g., those with 'co-morbidities' -- will succumb. Note that while the herd immunity % in the diffeq model is dependent upon R/R0, the herd immunity % in my model is dependent on a 'critical velocity', and is independent of chain reaction dynamics. Thus, if 10% of the antelope population can't achieve 20mph, then the 'herd immunity' will be achieved at 10%. Yes, I have mixed up to some extent the difference between unexposed and immune, but in reality, most people only care if they survive, not whether they catch an asymptomatic case of the disease.

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[math-fun] Klein bottle puzzles
by Dan Asimov 28 Nov '20

28 Nov '20

Let K denote a Klein bottle. 1. The cartesian product S^1 x S^2 of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold. Puzzle: Does this manifold contain a Klein bottle K ⊂ S^1 x S^2 as a subset ??? 2. Call two simple closed curves C_0, C_1 on the Klein bottle "equivalent" if there is a continuous family {C_t ⊂ K | 0 ≤ t ≤ 1} of simple closed curves on K. How many inequivalent simple closed curves are there on K, and what is an example of each one ??? —Dan Appendix -------- A surface is called "nonorientable" if it has a subset that is a Möbius band. Otherwise it is called "orientable". The (compact) orientable and nonorientable surfaces (without boundary) were first classified up to topological equivalence by August Ferdinand Möbius in 1861, but this was not proved rigorously until Henry Roy Brahana did so in 1921. Any two surfaces can be combined topologically by the operation called "connected sum", which means to remove the interior of a disk from each surface and then connect the resulting two circular boundaries by a cylinder. Each orientable surface M is the connected sum of g tori (where the connected sum of 0 tori is defined to be the sphere). Each nonorientable surface M is the connected sum of g projective planes for some integer g ≥ 1, also called the "genus" of M. If S denotes the sphere, T denotes the torus and P denotes the projective plane, then every surface (compact and without boundary) can be expressed as the connected sum of S with finitely many copies of T and finitely many copies of P. This operation, denoted by #, has S as the identity element and is commutative with just one relation: P # P # P = P + T. It turns out that each compact surface M can be "triangulated" — expressed as the union of finitely many 2D triangles any two of which intersect in a common edge or vertex. Then if V, E, F denote the number of vertices, edges, and faces (triangles) of the surface, the computation 𝜒(M) = V - E + F is a topological invariant called the Euler characteristic of M. If M is orientable its genus satisfies g = 2 - 2 𝜒(M), and if M is nonorientable its genus satisfies g = 2 - 𝜒(M). Thus Euler characteristic and orientability determine the surface up to topological equivalence. The Klein bottle K shows up in this classification as K = P # P, the connected sum of two projective planes. It can also be viewed as the result of starting with a square and identifying the top and bottom sides in the same direction, but identifying the left and right sides in the reverse direction: |—————➞————| | | ↑ // ↓ | | |—————➞————|

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[math-fun] "MALAYALAM" isn't just the longest palindrome in my length 400000 wordlist;
by Bill Gosper 27 Nov '20

27 Nov '20

It's the longest word whose reverse is a word: In[282]:= MaximalBy[DeleteCases[words, "MALAYALAM"], If[StringQ@# && MemberQ[words, Evaluate[StringReverse@#]], StringLength@#, 0] &] /@Range@6// tim During evaluation of In[282]:= 4166.771102,6 Out[282]= {"DESSERTS", "DIORAMAS", "REDRAWER", "REWARDER", "SAMAROID", "STRESSED"} W/o the Evaluate this took over 2.5 hrs. In[295]:= WordDefinition@"samaroid" Out[295]= {"pain caused by maple seeds at or inside the anal sphincter. Relieved by Preparation S."} —Bill

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[math-fun] NestList. For when you get bored and keep pushing the same buttons on your calculator.
by Bill Gosper 27 Nov '20

27 Nov '20

In[290]:= Integrate[Cos@x/(x^2+1),{x,-∞,∞}] Out[290]= π/*e* In[291]:= ContinuedFraction[%,105] Out[291]= {1,6,2,2,1,2,6,8,2,1,1,1,4,3,1,1,66,2,1,1,2,4,2,7,46,10,2,1,18,1,23,10,14,1,1,4,3,1,6,2,5,1,1,1,18,355, 1,1,5,3,4,1,28,5,2,1,3,16,3,25,4,7,1,11,2,2,1,9,2,31,1,20,1,2,1,1,6,13,1,3,1,3,4,125,1,7,2,2,1,3,5,1,1,6,1,1,1,9,1,1,4,6,4,1,13} (Prove it's endless; Win a Fields Medal.) In[292]:= {GeometricMean@%} (*In[295]:= N@Khinchin Out[295]= 2.68545200106531*) Out[292]= {2^(8/15) 3^(23/105) 5^(13/105) 7^(1/21) 2201^(1/105) 3289^(2/105)} In[293]:= NestList[(Print[N@#];GeometricMean@ContinuedFraction [#,105])&,%[[-1]],9] During evaluation of In[293]:= 3.09506434681568 During evaluation of In[293]:= 2.41520362449092 During evaluation of In[293]:= 2.9335577403418 During evaluation of In[293]:= 2.44930689361668 During evaluation of In[293]:= 2.71218585275826 During evaluation of In[293]:= 2.6450717647845 During evaluation of In[293]:= 2.77752000086026 During evaluation of In[293]:= 2.74022410644692 During evaluation of In[293]:= 2.59352990789751 Out[293]= {2^(8/15) 3^(23/105) 5^(13/105) 7^(1/21) 2201^(1/105) 3289^(2/105),2^(18/35) 3^(29/105) 5^(1/35) 104977873^(1/105),2^(19/35) 3^(22/105) 5^(3/35) 7^(2/21) 11^(2/105) 39497^(1/105),2^(7/15) 3^(4/21) 5^(8/105) 7^(2/35) 11^(1/35) 589^(1/105),2^(7/15) 3^(2/7) 5^(8/105) 7^(2/35) 19^(2/105) 1639^(1/105),2^(64/105) 3^(5/21) 5^(8/105) 209^(2/105) 851^(1/105),2^(53/105) 3^(4/21) 5^(11/105) 13^(4/105) 77^(1/35) 1919^(1/105),2^(47/105) 3^(6/35) 5^(11/105) 11^(1/35) 91^(2/105) 316238983^(1/105),2^(3/5) 3^(8/35) 5^(2/35) 7^(4/105) 11^(1/35) 221^(1/105),2^(43/105) 3^(13/105) 35^(8/105) 242280181^(1/105)} In[294]:= NestList[(Print[N@#];GeometricMean@ContinuedFraction [#,105])&,%[[-1]],42] During evaluation of In[294]:= 2.39797710382867 During evaluation of In[294]:= 2.69893228647732 During evaluation of In[294]:= 3.29061256432359 During evaluation of In[294]:= 2.03162608678672 During evaluation of In[294]:= 3.01906752481969 During evaluation of In[294]:= 2.59538767809429 During evaluation of In[294]:= 2.68778297494002 During evaluation of In[294]:= 3.11804540780474 During evaluation of In[294]:= 2.98027331397385 During evaluation of In[294]:= 2.23842419582078 During evaluation of In[294]:= 2.73195233677031 During evaluation of In[294]:= 2.32112558219677 During evaluation of In[294]:= 2.65266653473585 During evaluation of In[294]:= 2.45995266400843 During evaluation of In[294]:= 1.94020851734991 During evaluation of In[294]:= 2.13798184316968 During evaluation of In[294]:= 2.75100265543326 During evaluation of In[294]:= 3.56350709479461 During evaluation of In[294]:= 2.59571502601519 During evaluation of In[294]:= 2.4740594814946 During evaluation of In[294]:= 2.38638422095896 During evaluation of In[294]:= 2.94605839370828 During evaluation of In[294]:= 2.82204016953752 During evaluation of In[294]:= 2.25028325915228 During evaluation of In[294]:= 2.36248747987146 During evaluation of In[294]:= 2.39582513918321 During evaluation of In[294]:= 2.79821884430297 During evaluation of In[294]:= 2.51021506839255 During evaluation of In[294]:= 2.52467515174716 During evaluation of In[294]:= 2.40158175513891 During evaluation of In[294]:= 2.54575104382523 During evaluation of In[294]:= 2.70499590456546 During evaluation of In[294]:= 2.36456538650387 During evaluation of In[294]:= 2.82338870219823 During evaluation of In[294]:= 2.33636891461969 During evaluation of In[294]:= 2.4610503767373 During evaluation of In[294]:= 2.50128322134356 During evaluation of In[294]:= 2.43209873810393 During evaluation of In[294]:= 2.83712666091056 During evaluation of In[294]:= 2.92217685976039 During evaluation of In[294]:= 2.58135001107047 During evaluation of In[294]:= 2.97981675199117 Out[294]= {2^(43/105) 3^(13/105) 35^(8/105) 242280181^(1/105),2^(7/15) 3^(16/105) 5^(1/21) 77^(1/35) 54503796054221^(1/105),2^(58/105) 3^(41/105) 5^(1/15) 17^(2/105) 77^(4/105) 247^(1/105),2^(43/105) 3^(17/105) 5^(2/35) 133^(2/105) 671^(1/105),2^(11/21) 3^(4/21) 5^(8/105) 7^(2/35) 17^(1/35) 437^(1/105) 4433^(2/105),2^(58/105) 3^(29/105) 13^(4/105) 17^(1/35) 11165^(1/105),2^(7/15) 3^(19/105) 5^(3/35) 19^(1/35) 2639^(2/105) 19987^(1/105),2^(3/5) 3^(8/35) 17^(1/35) 35^(1/21) 209^(2/105) 245401^(1/105),2^(58/105) 3^(4/15) 11^(1/35) 35^(4/105) 1417^(2/105) 2369^(1/105),2^(41/105) 3^(16/105) 5^(1/15) 7^(2/105) 11^(1/35) 11497861^(1/105),2^(46/105) 3^(4/15) 5^(13/105) 7^(4/105) 1431859^(1/105),2^(17/35) 3^(1/5) 5^(1/21) 91^(2/105) 410533^(1/105),2^(58/105) 3^(19/105) 11^(1/35) 35^(2/35) 374153^(1/105),2^(59/105) 3^(23/105) 35^(1/21) 39121^(1/105),2^(18/35) 3^(2/21) 5^(2/35) 17^(1/105) 77^(2/105),2^(13/35) 3^(4/21) 5^(8/105) 11^(1/35) 44863^(1/105),2^(19/35) 3^(8/35) 5^(2/35) 7^(4/105) 11^(2/105) 13^(1/35) 34523^(1/105),2^(22/35) 3^(8/35) 5^(1/15) 7^(2/35) 17^(1/35) 143^(2/105) 460112447^(1/105),2^(31/105) 3^(32/105) 5^(13/105) 7^(1/21) 11^(1/35) 17^(2/105),2^(11/21) 3^(19/105) 5^(1/15) 19^(2/105) 77^(1/35) 377^(1/105),2^(41/105) 3^(9/35) 5^(4/105) 17^(1/35) 77^(2/105) 15067^(1/105),2^(22/35) 3^(26/105) 5^(1/15) 7^(4/105) 11^(2/35) 299^(1/105),2^(18/35) 3^(23/105) 5^(13/105) 17^(1/35) 77^(2/105) 3379^(1/105),2^(3/7) 3^(23/105) 5^(1/21) 7^(1/15) 1133^(1/105),2^(16/35) 3^(6/35) 11^(1/35) 13^(2/105) 35^(1/21) 1241^(1/105),2^(3/7) 3^(4/21) 5^(1/21) 11^(1/35) 13^(2/105) 80741659^(1/105),2^(3/7) 3^(16/105) 31^(2/105) 35^(1/15) 221^(1/35) 82777^(1/105),2^(8/21) 3^(9/35) 5^(1/21) 11^(1/35) 26739058073^(1/105),2^(16/35) 3^(5/21) 7^(1/15) 11^(1/35) 115^(2/105) 493^(1/105),2^(18/35) 3^(23/105) 5^(1/21) 7^(1/35) 19^(2/105) 13651^(1/105),2^(7/15) 3^(23/105) 11^(1/35) 35^(1/15) 899^(1/105),2^(62/105) 3^(19/105) 5^(4/105) 7^(1/21) 247^(2/105) 694331^(1/105),2^(41/105) 3^(2/15) 5^(2/21) 7^(1/21) 13^(4/105) 35563^(1/105),2^(18/35) 3^(6/35) 5^(8/105) 7^(2/105) 377^(1/35) 29932163^(1/105),2^(37/105) 3^(19/105) 5^(1/15) 7^(2/35) 11^(2/105) 2800733^(1/105),2^(41/105) 3^(1/7) 5^(3/35) 7^(1/21) 112819455463^(1/105),2^(10/21) 3^(6/35) 5^(1/15) 7^(1/35) 221^(2/105) 1123441^(1/105),2^(8/15) 3^(1/7) 5^(1/15) 7^(1/21) 24845483^(1/105),2^(53/105) 3^(26/105) 5^(8/105) 7^(2/35) 338981071^(1/105),2^(47/105) 3^(29/105) 5^(4/35) 7^(3/35) 13^(2/105) 493^(1/105),2^(46/105) 3^(19/105) 5^(8/105) 77^(1/35) 1202287073^(1/105),2^(64/105) 3^(5/21) 7^(4/105) 55^(1/15) 1079^(1/105),2^(61/105) 3^(6/35) 5^(1/15) 7^(1/21) 13^(2/105) 3023603^(1/105)} Methinks this could have a largish period. Limit? LUB? GLB? Tighter for larger values of "105"? —rwg

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[math-fun] denesting paradox?
by Bill Gosper 26 Nov '20

26 Nov '20

Are we still looking for the fabled case where neither √A nor √B denests, but √A+√B does? Unfortunately, this isn't a clear cut case, because the denesters go off the rails instead of declaring impossibility. Even the numerics is shaky. Out[19]= Sqrt[5] + Sqrt[7] == Sqrt[4 + (-54 + 35 I Sqrt[3])^(1/3)/3^(2/3) + 13/(3 (-54 + 35 I Sqrt[3]))^(1/3)] + \[Sqrt]((1/(162 I + 105 Sqrt[3]))(1296 I + 840 Sqrt[3] - 35 3^(5/6) (-54 + 35 I Sqrt[3])^(1/3) - 54 I (3 (-54 + 35 I Sqrt[3]))^(1/3) + 13 I (3 (-54 + 35 I Sqrt[3]))^(2/3))) Try to denest this RHS with Swastik's new package: In[20]:= ResourceFunction["RadicalDenest"] /@ %[[2]] During evaluation of In[20]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. During evaluation of In[20]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[20]= Sqrt[ 4 + (-54 + 35 I Sqrt[3])^(1/3)/3^(2/3) + 13/(3 (-54 + 35 I Sqrt[3]))^( 1/3)] + \[Sqrt]((1/( 162 I + 105 Sqrt[3]))(1296 I + 840 Sqrt[3] - 35 3^(5/6) (-54 + 35 I Sqrt[3])^(1/3) - 54 I (3 (-54 + 35 I Sqrt[3]))^(1/3) + 13 I (3 (-54 + 35 I Sqrt[3]))^(2/3))) And with Corey's original package: In[49]:= $MaxExtraPrecision = 999; Strad /@ %20 During evaluation of In[49]:= N::meprec: Internal precision limit $MaxExtraPrecision = 999.` reached while evaluating -(105/13) Cos[2/3 (\[Pi]-ArcTan[Times[<<2>>]])]+3 Sqrt[13] Sin[1/3 (\[Pi]-ArcTan[Times[<<2>>]])]-54/13 Sqrt[3] Sin[2/3 (\[Pi]-ArcTan[Times[<<2>>]])]. During evaluation of In[49]:= MinimalPolynomial::nalg: (1/(Sqrt[6] 13^(1/4)))Cos[1/6 (\[Pi]+2 ArcTan[Times[<<2>>]])]^(\[Pi]/2+ArcTan[35/(18 Sqrt[3])]) Sqrt[(26 Sqrt[13]+<<1>>+(67+35 I Power[<<2>>]) Sin[1/3 <<6>>[<<1>>]]) <<1>>] is not an explicit algebraic number. What's all this trigamarole? In[50]:= %20 == \[Sqrt]5 + \[Sqrt]7 Out[50]= Sqrt[ 4 + (-54 + 35 I Sqrt[3])^(1/3)/3^(2/3) + 13/(3 (-54 + 35 I Sqrt[3]))^( 1/3)] + \[Sqrt]((1/( 162 I + 105 Sqrt[3]))(1296 I + 840 Sqrt[3] - 35 3^(5/6) (-54 + 35 I Sqrt[3])^(1/3) - 54 I (3 (-54 + 35 I Sqrt[3]))^(1/3) + 13 I (3 (-54 + 35 I Sqrt[3]))^(2/3))) == Sqrt[5] + Sqrt[7] In[51]:= N@% Out[51]= False Well, no wonder it didn't denest! Wait a minute, In[52]:= N[%%, 22] Out[52]= True That was a false negative! —rwg

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[math-fun] Hits when concatenating?
by Éric Angelini 24 Nov '20

24 Nov '20

Hello Math-Fun Say we add the last odd digit of S to a(n): S=1,2,3,6,9,18,19,28,37,44,51,52,57,64,71,72, 79,88,97,104,105,110,111,112,113,116,117,124,... Say we have a hit when the concatenation of two or more initial terms of S reappear in S; will we ever find a hit here (the hit « 123 » was missed by a single unit with « 124 », in the above example; the next possibilities would be with 1236, 12369, 1236918,... but the margin of this post, etc. Best, É.

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Re: [math-fun] "MALAYALAM" isn't just the longest palindrome
by Don Reble 23 Nov '20

23 Nov '20

> ... > 15 dermatoglyphics > 14 hydromagnetics > 14 hydropneumatic * > ... > * Plans for a gun with hydropneumatic brakes figured in the > Dreyfus Affair. So... the other words are uncopyrightables? -- Don Reble djr(a)nk.ca

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