May 2019 - mathfun@mailman.xmission.com

[math-fun] Three piece Golden Rectangle to square dissection
by Bill Gosper 09 Jun '19

09 Jun '19

Golden rectangle↔︎square <http://gosper.org/gold2square.gif> I forget where I stole this, but can you find the superfluous step I just noticed? (Not the retrace, which was to tell you a dimension.) —rwg (Rebuild in progress)

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[math-fun] Stirlingoid figure of merit
by Bill Gosper 07 Jun '19

07 Jun '19

n!/√(2πn)e^n/n^n ~ 1 + O(1/n), whose infinite product over n blows up, giving an infinite figure of demerit. For convergence, we can improve Stirling's approximation: n! ~ √(2π(n+1/6)) n^n/e^n ~ n! (1+1/(144 n^2)+. . .). Then the infinite product of all the relative errors is In[86]:= Product[n!/√(2 π (n + 1/6)) n^n/E^n), {n, ∞}] Out[87]= E^( 2 Zeta'[-1] - 1/12) √(⅙! √(2π))) Out[88]= 1.00781097654253 (BtW, Stirling's contribution was "only" the √(2π); de Moivre had already found the rest. This must have been nearly as stunning as Euler's 𝜁(2) = π²/6.) Stirling's approximation can be improved no end. Perhaps the next better is E^-z √(2 π z) (1/(12 z) + z)^z ~ z! (1-1/(1440 z^3)+O[1/z]^4) whose "figure of merit" is Product[z!/(E^-z √(2 π z) (1/(12 z) + z)^z),{z,∞}] == E^(2 Zeta'[-1]) √√(2π)/(BarnesG[1 - I/(2√3)] BarnesG[1 + I/(2√3)]) ~ 1.001178221812, which seems to cry out for the BarnesG reflection formula. Which, unfortunately, only applies to G(z)/G(-z), not G(z)G(-z): E.g. BarnesG[1 - I/(2 √3)]/BarnesG[1 + I/(2 √3)] == (-1)^(1/24) E^(-((I PolyLog[2, E^(-(π/√3))])/(2 π))) (π Csch[π/(2 √3)])^(-(I/(2 √3))) —rwg

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[math-fun] Mathematical flimflam
by James Propp 01 Jun '19

01 Jun '19

As a teenager I loved "How to Lie With Statistics"; does it have any successors, especially ones that focus on non-statistical forms of mathematical flimflam? I'd also be interested in egregious and/or amusing examples of math abuse, and in particular, examples of misuse of Lagrange interpolation. We mathematicians all know that it should always be used for interpolation, never for extrapolation. But do practitioners of client disciplines know this? I hope to include some examples of mathematical flimflam in my June 17 Mathematical Enchantments essay. As always, all contributions will be gratefully acknowledged in the essay. Jim Propp

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[math-fun] 170
by James Propp 01 Jun '19

01 Jun '19

I’ve agreed to host Carnival of Mathematics #170 next month, and I’m wondering whether the number 170 has any especially nice properties that might deserve mention. The number 170 has its own Wikipedia page, but none of the mathematical properties listed there appealed to me. I might say a little about writing numbers as sums of two squares, but I’d prefer something a little more exciting. Jim Propp

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[math-fun] The first valence 7 graceful graph is unique.
by Ed Pegg Jr 31 May '19

31 May '19

I was looking at high valence graceful graphs. I was surprised to find the valence 7 e=35 graceful graph is unique. https://math.stackexchange.com/questions/3246000/graceful-graphs-with-minim… --Ed Pegg Jr

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[math-fun] Fwd: about bugs in mathematica and pari-gp
by Bill Gosper 30 May '19

30 May '19

On 2019-05-27 05:42, Hans Havermann wrote: > SP: "type Prime[8200000000] or Prime[93000000000]" > > note the eight zeros after 82, nine zeros after 93 > OOPS! See below. > RWG: "My old Macsyma gets > (c71) PRIME(8200000000); > (d71) 205021987301 > (c72) PRIME(9300000000); > (d72) 233746743437" > > note only eight zeros after 93 > > VM: "I think that macsyma is wrong about 93*10^9: > ./primecount -n 93000000000 > 2560462462787 > ./primecount 2560462462786 > 92999999999" > > RWG: "? It agrees with (somewhat cajoled) Mathematica: > In[159]:= NextPrime@Prime[9299999999] > Out[159]= 233746743437" > > :) In[456]:= Prime[93 10^9] // tim During evaluation of In[456]:= 36292.726051,0 (ten hrs +. Clearly unintended. Yet correct.) Out[456]= 2560462462787 —rwg

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[math-fun] Our state of ignorance regarding multiply-perfect numbers
by Allan Wechsler 30 May '19

30 May '19

It's astonishing how little we know about multiply-perfect numbers.. Reminder: let sigma(n) be the sum of all divisors of n, including 1 and n. A number n is multiply-perfect (of order k) if sigma(n) = kn. The classical perfect numbers are exactly the multiply-perfect numbers of order 2. We do not know if there are any odd multiply-perfect numbers, other than 1. Let me restate this in a slightly weird way: we do not know if there are any multiply-perfect numbers that are multiples of 1 but not of 2, outside the set {1} But our ignorance is much, much more profound than that. To my knowledge, we can't prove any theorem of the form: "There are no multiply-perfect numbers that are multiples of N but not of M, outside the set S." In particular, for example, we do not know if there are any multiply-perfect numbers (other than 1) whose prime factors are all larger than 100. It seems staggeringly unlikely that there would be any, and it is easy to put astronomical lower bounds on such numbers, but we still have no proofs. I'm quite sure this was all true about 30 years ago. Is it still true?

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[math-fun] 170 addendum
by Bill Gosper 30 May '19

30 May '19

A point on the unit circle: In[471]:= (((Sqrt[5] + 1)*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1)/ 64) - ((Sqrt[5 - Sqrt[5]]* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/16))* I + (((Sqrt[5] + 1)* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/(16* Sqrt[2])) + ((Sqrt[ 5 - Sqrt[5]]*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1))/(32*Sqrt[2])) // Abs // FullSimplify // tim During evaluation of In[471]:= 127967.849905,0 (coupla days) Out[471]= 1 Shoulda said AbsArg. 12 more seconds, In[472]:= (((Sqrt[5] + 1)*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1)/ 64) - ((Sqrt[5 - Sqrt[5]]* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/16))* I + (((Sqrt[5] + 1)* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/(16* Sqrt[2])) + ((Sqrt[ 5 - Sqrt[5]]*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1))/(32*Sqrt[2])) // Arg // FullSimplify // tim During evaluation of In[472]:= 12.379624,2 Out[472]= π/170 Extra cores and RAM. Nice hand- and lapwarmer. It's cold in California these days. —rwg (Kids: How does it get the Arg ten thousand times faster than the Abs?) (Support: Note this (from Macsyma) is simpler than the cis π/170 from Developer`TrigToRadicals. Hey, have you considered (trivially) providing CiS?)

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Re: [math-fun] Another pic of Hilbert's spacefill
by Bill Gosper 30 May '19

30 May '19

Nice. style=rect is the "median curve" of style=basic. I.e. join the midpoints of consecutive line segments of the latter. I suspect these are images of equally spaced smallish rationals in the preimage. But how do you spot a triple point? (I should plot a Dragon connecting only small-rational triple points.) Note https://oeis.org/A260482 & its many friends. Of course, all spacefills are provably dense with triple points. You may recall Julian's triangle filler dense with sextuple points: *http://howwords.com/triangles/?a_=sexttri&radiusPct=50&nRotors=1476&dtPct=20&rotorSort=1&cmn_=32&cmt_=highcontrast&cmabs_=false&stepsPerCycle_=5&delayMS_=0 <http://howwords.com/triangles/?a_=sexttri&radiusPct=50&nRotors=1476&dtPct=2…>* ------------------------------ © 2019 by Howard I. Cannon —rwg On 2019-05-28 23:07, Tom Karzes wrote: > Bill, you aren't kidding about the Heighway Dragon being dense with > triple points! Here's a self-avoiding rendering that makes them > really easy to spot: > > http://www.karzes.com/dragon/dragon.html?style=rect&form=h > > Tom > > Bill Gosper writes: > > to accompany should've been "sampHilbert" > > <http://www.tweedledum.com/rwg/sampeano.htm> > > Hilbert's [spacefilling] function is dense with quadruple points, i.e. > > points in [0,*i*] with four distinct preimage points in [0,1]: > > gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, > > 41/48, 43/48}. This picture connects, in the order they were swept, all the > > points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All > > quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost > > vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and > > 43/12288. The white segments are retraced boundary, retroflexed in the > > middle, (making quadruple points). (Spacefilling functions map closed > > intervals to closed sets.) > > > > The Heighway Dragon is dense with mere triple points. Here's one: Hi res > > Heighway <http://gosper.org/dragtrip!.png> . > > —rwg

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Re: [math-fun] Find a polyhedron with all vertices on its circumsphere, but without all faces touching its insphere
by Bill Gosper 30 May '19

30 May '19

And don't forget poor bastard "Archimedean 14" In[208]:= PolyhedronData[{"Johnson", 37}, "Name"] Out[208]= "elongated square gyrobicupola" (Why should I have to retype this?) In[202]:= Norm@Mean@# &@ PolyhedronData["ElongatedSquareGyrobicupola", "Vertices"][[#]] & /@ PolyhedronData["ElongatedSquareGyrobicupola", "Faces"] // FullSimplify // DeleteDuplicates Out[202]= {1/2 + 1/Sqrt[2], Sqrt[11/12 + 1/Sqrt[2]]} In[203]:= Min@% Out[203]= 1/2 + 1/Sqrt[2] In[200]:= PolyhedronData["ElongatedSquareGyrobicupola", "Inradius"] Out[200]= Missing["NotAvailable"] (Yoohoo, wri support!) —rwg On 2019-05-27 14:55, Tom Karzes wrote: > Actually, I believe *all* Archimedean solids satisty the condition > (note that they exclude the Platonic solids). > > All vertices in an Archimedean solid are the same distance from the > center of the polyhedron. There are at least two different polygons > among the faces, and only the largest of them will touch the insphere. > > Tom > > Tom Karzes writes: > > There are lots of solutions to this. Any elongated prism will work. > > The insphere is not unique, but it can be made unique by tapering one > > end, forcing the insphere to the other end. The smallest is a > > triangular prism, with 5 faces. You can taper one end if you want to > > force a unique insphere. The face at the smaller end doesn't touch > > it. > > > > A specific instance of this is a tetrahedron with one truncated > > vertex. Just snip a tiny piece off of one vertex, leaving a little > > triangle that's parallel to the opposite face. The result is a > > 5-sided tapered triangular prism that satisfies the condition. > > > > More symmetrical, albeit larger, solutions can also be found. For > > instance, almost any Archimedean solid will work. All vertices are > > the same, so they all touch the circumsphere. But the faces differ, > > and in most cases some faces have their center closer to the center of > > the polyhedron than others, so only those faces touch the insphere. > > > > Example: a truncated tetrahedron (3.6.6). This has 8 faces (4 > > hexagons and 4 triangles). Only the hexagons touch the insphere: > > > > http://www.karzes.com/polyhedra/polyhedron.html?ph=3.6.6 > > > > As I said, I believe most of the other Archimedean solids work > > as well. > > > > Tom > > > > > > Andy Latto writes: > > > Or simpler, just take any rectangular prism that is not a cube. Works in > > > two dimensions too. > > > > > > Andy > > > > > > On Mon, May 27, 2019, 03:52 Dan Asimov <dasimov(a)earthlink.net> wrote: > > > > > > > There is one with 6 faces, anyhow. > > > > > > > > (Far below if you want to try to find one without spoilage.) > > > > > > > > —Dan > > > > > > > > ... > > > > > > > > Let T be a skinny isosceles triangle with apex A. > > > > > > > > Let the polyhedron P be the result of truncating > > > > [the product T x [0, 1] of T with the unit interval] > > > > by cutting off a neighborhood N of {A} x [0, 1], thereby > > > > creating a new face — a skinny rectangle R. > > > > > > > > Then P has all 8 of its vertices on its circumsphere C. > > > > But clearly C would not touch the rectangle R if N is > > > > small enough. > > > > > > > > —Dan > > > > > > >

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