July 2019 - mathfun@mailman.xmission.com

[math-fun] Direct proof of the Pentagonal Number Theorem
by Bill Gosper 16 Jul '19

16 Jul '19

In[150]:= EllipticTheta[1, π/3, q^(1/6)]/√3/q^(1/24) // theta2product Out[150]= Product[(1 - q^(n/3))*(1 + q^(n/3) + q^((2*n)/3)), {n, ∞}] (I.e., (q;q)_∞.) In[149]:= EllipticTheta[1, π/3, q^(1/6)]/√3/q^(1/24) // theta2sum Out[149]= (2*Sum[(-1)^n*q^((1/6)*(1/2 + n)^2)*Sin[(1/3)*(1 + 2*n)*Pi], {n, 0, ∞}])/ (√3*q^(1/24)) Group terms in threes: In[152]:= gruple[%149, 3] Out[152]= (1/(Sqrt[3]*q^(1/24)))*2* Sum[ (-1)^(-3 + 3*n)*q^((1/6)*(-(5/2) + 3*n)^2)*Sin[(1/3)*(1 + 2*(-3 + 3*n))*π] + (-1)^(-2 + 3*n)*q^((1/6)*(-(3/2) + 3*n)^2)*Sin[(1/3)*(1 + 2*(-2 + 3*n))*π] + (-1)^(-1 + 3*n)*q^((1/6)*(-(1/2) + 3*n)^2)*Sin[(1/3)*(1 + 2*(-1 + 3*n))*π], {n,∞}] Remind Mathematica that n is an integer: In[154]:= Assuming[n \[Element] Integers, MapAt[Simplify /@ # &, %, {-1, 1}]] Out[154]= (2*Sum[(1/2)*(-1)^n*√3*q^((1/24)*(1 - 6*n)^2) - (1/2)*(-1)^n*√3*q^((1/24)*(5 - 6*n)^2), {n, ∞}])/ (√3*q^(1/24)) In[160]:= ExpandAll@intosum@%154 Out[160]= Sum[(-1)^(1 + n)*q^(1 - (5*n)/2 + (3*n^2)/2) + (-1)^n*q^(-(n/2) + (3*n^2)/2), {n, ∞}] —rwg Anybody want theta2product, theta2sum, gruple, intosum? Let me know. It's actually a little disappointing that Mathematica doesn't automatically revert back to EllipticTheta! When that day comes we'll need to Inactive the Sum and Product.

1 0

[math-fun] Apollo Guidance Computer: 10 secs/SHA256 hash
by Henry Baker 16 Jul '19

16 Jul '19

FYI -- 50 years of Moore's Law at (proof of) work: Mining Bitcoins on the Apollo Guidance Computer (see below). --- I've recently been learning about the detailed calculations of satellite GPS computations, so I'm much more interested in how quickly the AGC could do today's GPS calculations. BTW, Einstein's special relativity (satellite speed adjustment to the clock) and Einstein's general relativity (satellite altitude from the Earth's center) are both important: higher altitude clocks tick faster by ~45usecs/day, while fast orbiting satellites tick slower by ~7usecs/day, resulting in a ~38usec/day (11.4km/day=7.1miles/day) difference. Q: Did the Apollo Guidance Computer have to deal with corrections for relativity? --- http://www.righto.com/2019/07/bitcoin-mining-on-apollo-guidance.html Ken Shirriff's blog Bitcoin mining on an Apollo Guidance Computer: 10.3 seconds per hash We've been restoring an Apollo Guidance Computer. Now that we have the world's only working AGC, I decided to write some code for it. Trying to mine Bitcoin on this 1960s computer seemed both pointless and anachronistic, so I had to give it a shot. Implementing the Bitcoin hash algorithm in assembly code on this 15-bit computer was challenging, but I got it to work. Unfortunately, the computer is so slow that it would take about a million times the age of the universe to successfully mine a Bitcoin block. ... https://arstechnica.com/tech-policy/2019/07/computer-from-nasas-apollo-prog… Computer from NASA's Apollo program reprogrammed to mine bitcoin It takes the Apollo Guidance Computer 10 seconds to compute a single hash value. Timothy B. Lee - Jul 9, 2019 10:15 pm UTC

4 3

[math-fun] Mitchell Feigenbaum
by Keith F. Lynch 15 Jul '19

15 Jul '19

Today I learned that Mitchell Feigenbaum died on June 30th at age 74. We should commemorate with 4.6692016... minutes of silence.

1 0

[math-fun] A Coloring Technique
by Brad Klee 15 Jul '19

15 Jul '19

Recently Beth Burgess exhibited some wire-sculpture artwork in Fayetteville. This inspired me to spend a colorful few hours over the weekend, wire-bending and thread turning: https://0x0.st/z99S.JPG https://0x0.st/z99u.JPG Crochet thread is cheap and will also work for coloring 3D-printed models. If anyone is interested, I can also get pictures of this technique applied to Schwarz's D-surface, or to a toric cross section of the cubic oscillator's phase space geometry. Of course, when the Mobius strip hovers over a cyclic plane curve it characterizes 4*Pi periodicity. After only one turn by 2*Pi radians, the surface normal inverts, and then another 2*Pi cycle brings the vector back to its initial condition. This odd type of transformation sometimes occurs in Quantum Mechanics, under the heading of "geometric phase". Cheers, Brad

1 0

[math-fun] Alan Turing on 50 pounds bank notes
by Simon Plouffe 15 Jul '19

15 Jul '19

Hello, these are the new 50 pounds bank note : https://www.bbc.com/news/business-48962557 many things on it, his brithdate in binary, a page of formulas and a citation. At last, some recognition from the British gov. Best regards, Simon Plouffe

1 0

[math-fun] A true Mac Mahon ruler?
by Éric Angelini 15 Jul '19

15 Jul '19

Hello Math-Fun (posted here as SeqFans didn't publish my mail) The first 8 terms of A002049 are: 1,3,7,12,20,30,44 and 59. I see that I can measure 29 in two ways with this ruler (I thought only one way was authorized): 30-1=29 59-30=29 I am obviously missing smthg as this is an ancient seq. signed by Neil himself. Anyway -- is it possible to build a wooden ruler with as few as possible vertical marks such that all integers measures between 1 and 200 can be materialized only once? I guess this sort of question is as old as the hat I'm wearing right now to avoid the killer sun we currently have in Brussels. Forgive me. Best, É.

6 8

[math-fun] A046693 and the Sparse Ruler Conjecture
by ed pegg 15 Jul '19

15 Jul '19

I've verified the Sparse Ruler conjecture to at least 1017. Sparse Ruler Conjecture: If a minimal sparse ruler of length n has m marks, then m − round(sqrt(3×n+9/4))∈(0,1). Anyone that would like my list of best known sparse rulers to length 1200, please contact me. The only possible counterexample under 1200 is at 1018 at the moment. https://math.stackexchange.com/questions/3279766/sparse-ruler-conjecture --Ed Pegg Jr

1 0

[math-fun] Draft of short essay on Scott Kim's motley dissections
by James Propp 13 Jul '19

13 Jul '19

At http://faculty.uml.edu/jpropp/mathenchant/motley-draft3.pdf you'll find a draft of an essay that will be published on July 11th. Comments are welcome, but even more welcome would be a good picture of the "spherical pseudo-cuboctahedron"! In principle I know how to make such a picture, sort of: take the motley dissection of the surface of the cube (shown in the essay), push all the vertices out onto a sphere, and draw spherical arcs joining up those vertices. This seems like a job for Mathematica, but my mastery of 3D graphics isn't up to the job. I'll gratefully acknowledge anyone who can create a picture I can use! Thanks, Jim Propp

5 15

[math-fun] rhopalic sentences
by Cris Moore 13 Jul '19

13 Jul '19

From a friend: > https://richardelwes.co.uk/2013/04/04/a-rhopalic-sentence/ <https://richardelwes.co.uk/2013/04/04/a-rhopalic-sentence/> And my entry: a la GEB, self- refer: create Quinier rhopalic sentences, imitations shamelessly triangularly incrementally arithmetically Hofstadteresque. - Cris

1 0

Re: [math-fun] Draft of my July 2019 essay
by Brad Klee 12 Jul '19

12 Jul '19

Maybe it is also okay to say: The commutator identity [ d/dt, f(t) ] = 0 is true if and only if f(t)=f(0), i.e. if and only if f(t) is a constant function. Usually [ d/dt, f(t) ] = f'(t), which recovers the earlier statement ( with "only if" added, is this missing from the essay? ). Aside from the relation to QM maths, there is a reason to write the statement as such: The commutation property [d/dt,c]=0 is used all the time when solving problems in integral and differential calculus. But do we yet need a delta-epsilon proof that [d/dt,0]=0? Arg! ( I am sure this is one of the reasons C. Huneke gave me a B in Vector Calculus so many years ago. ) --Brad On Thu, Jul 11, 2019 at 2:01 PM <bradklee(a)gmail.com> wrote: > I still have nightmares about delta-epsilon proofs

1 0