October 2017 - mathfun@mailman.xmission.com

Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Dan Asimov 22 Oct '17

22 Oct '17

[Without having read any other replies] I'd try dropping molten glass some distance through the air to make it form into a sphere, and have it land in water to make it rigid. —Dan Henry Baker wrote: ----- It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. How would you fabricate it? ... ... -----

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Henry Baker 22 Oct '17

22 Oct '17

If you don't remove the air in such a column, you're likely to get shot that is more raindrop-shaped than sphere-shaped. BTW, I believe an analogous method has been used to fabricate spherical items in space orbit. So *that's* what Galileo was really doing on the Tower of Pisa: making musket balls and cannon balls! Unfortunately, Galileo was about 100 years later -- almost 1600. So we have a very interesting question: is it possible that da Vinci considered the same idea for making spherical glass objects? The problem with glass is that unless it is extremely hot, it will take quite a while to "relax" into a spherical shape during the fall. A slightly more modern Galileo might have used an elevator, part of whose travel was in free fall, because an elevator would eliminate the airflow problem. BTW, now that we have really, really tall buildings, where are the *free fall* elevators? Why do we have to go to Magic Mountain (N. of Los Angeles) to find a free fall ride? At 10:19 AM 10/22/2017, you wrote: >The usual method for producing spherical lead shot was as follows: > >1. Position a large container of cold water at the bottom of a tower. > >2. Stand at the top of the tower. > >3. Drop a volume of molten lead, in freefall, from the tower. > >4. Appeal to the isoperimetric inequality. > >Best wishes, > >Adam P. Goucher > >> Sent: Sunday, October 22, 2017 at 5:03 PM >> From: "Henry Baker" <hbaker1(a)pipeline.com> >> To: math-fun(a)mailman.xmission.com >> Subject: [math-fun] Fabricating a glass/crystal sphere circa 1500 >> >> It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. >> >> How would you fabricate it? >> >> I don't know how they actually did it, but here are some of my speculations: >> >> 1. To make a solid glass sphere, one might make a spherical mold, melt glass into it, and then polish the result. >> >> 2. To make a rock crystal sphere, one might build a lathe, and then utilize a semicircle guide to cut a rough crystal block into nearly spherical shape, and then polish the result. >> >> One might even take the glass from the mold & mount it onto a lathe for the polishing step. >> >> Clearly (!), one could continually reposition the sphere in the lathe to work on different axes to attempt to make the sphere as precisely spherical as possible. >> >> Are there more elegant methods? >> >> Using just the technology available in 1500, what kind of precision could be achieved? >> >> [BTW, a similar problem arose in the 19th Century when *ball bearings* became important. Apparently, the longevity of various kinds of engines depended critically on making the bearings as spherical as possible. I don't know enough about how this precision was achieved, but during WWII, bearing manufacturing plants were heavily targeted for bombing.]

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[math-fun] Gosper's nonidentities
by Bill Gosper 21 Oct '17

21 Oct '17

While collecting reasons why Mathematica's StirlingS1 and StirlingS2 should, like Macsyma, admit negative and fractional arguments, I found, to my discomfort, >From rwg(a)NEWTON.Macsyma.COM Wed Jul 24 03:00:00 1996 Received: from optima.CS.Arizona.EDU by cheltenham.cs.arizona.edu; Wed, 24 Jul 1996 03:03:19 MST Received: from NEWTON.Macsyma.COM by optima.cs.arizona.edu (5.65c/15) via SMTP id AA07538; Wed, 24 Jul 1996 03:03:16 MST Received: from SWEATHOUSE.macsyma.com ([192.233.166.105]) by NEWTON.Macsyma.COM via INTERNET with SMTP id 363715; 24 Jul 1996 06:01:03-0400 Date: Wed, 24 Jul 1996 03:00-0700 From: Bill Gosper <rwg(a)NEWTON.macsyma.com> Reply-To: rwg(a)NEWTON.macsyma.com Subject: funny-looking sum To: math-fun(a)cs.arizona.edu Message-Id: <19960724100013.6.RWG(a)SWEATHOUSE.macsyma.com> (Using Knuthian (nonnegative) Stirlings) inf n - 2 ==== ==== x k + 1 \ \ (%e - x - 1) stirling_s1(n, n - k) - x n > ( > ----------------------------------------) %e = log(x + 1) / / (k + 1)! ==== ==== n = 1 k = 0 apparently for x>=0, anyway. (Convergence rate is useless.) ----------------------------------- It doesn't even slightly work, presumably mistranscribed. To my active distress, Eric Weisstein has faithfully quoted this bogon as eqn (26) in http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html, (except my S is his |S|). I am so far unable to guess its correct form, which presumably existed because I would have Taylor-expanded the ⦓Я∀π out of anything this weird. Help? --rwg

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[math-fun] Gosper's identities
by Bill Gosper 21 Oct '17

21 Oct '17

Dear Dr. Essomba, Thank you for catching my D111 mistake. Your claim is immediately substantiated by the limit 𝜃→0, which gives 1 = ¼. And confirmed by rederivation, simply multiplying D110 by itself with 𝜃→√3𝜃. There are several errors on that page (which needs to be done over in Mathematica) due to faulty conversion to html, (e.g., p for π, f for 𝜙,...) but this isn't one of them, since the error is repeated in C111. Puzzling. Also, here is experimental confirmation of your 2017-10-13 07:51 recurrence: In[502]:= NestList[(Print[√√(2^(4 #[[1]] + 5) (3 - 2 √(2 + 2 #[[2]]) + #[[2]]))]; {#[[1]] + 1, √(2 + 2 #[[2]])/2}) &, {0, -1.}, 9] During evaluation of In[502]:= 2.82842712474619 During evaluation of In[502]:= 3.061467458920717 During evaluation of In[502]:= 3.12144515225806 During evaluation of In[502]:= 3.136548490545999 During evaluation of In[502]:= 3.140331156951562 During evaluation of In[502]:= 3.141277250930965 During evaluation of In[502]:= 3.141513800680598 During evaluation of In[502]:= 3.141572952943204 During evaluation of In[502]:= 3.141587832038716 Out[502]= {{0, -1.}, {1, 0.}, {2, 0.7071067811865476}, {3, 0.9238795325112867}, {4, 0.9807852804032304}, {5, 0.9951847266721969}, {6, 0.9987954562051724}, {7, 0.9996988186962042}, {8, 0.9999247018391445}, {9, 0.9999811752826011}} where u_n (= #[[2]]) converges to 1 while a_n converges to π⁴. --rwg Date: 2017-10-16 05:35 From: François Mendzina Essomba via math-fun <math-fun(a)mailman.xmission.com> To: math-fun(a)mailman.xmission.com, m_essob(a)yahoo.fr Reply-To: François Mendzina Essomba <m_essob(a)yahoo.fr>, math-fun < math-fun(a)mailman.xmission.com> Hello, I found in the page http://www.tweedledum.com/rwg/idents.htm strange identities, as incredible as each other ... I have observed at length some of the identities ranging from C105 to C115. And this is what I deduced from it: (9) D106 product(2*cos(theta/3^(n/m))-1,n=1..infinity)=product(cos(3^(i/m)*theta/2),i=0...m-1) ; (10) D105 product(2*cos(theta/2^(n/m))-1,n=1..infinity)=product((1+2*cos(2^(i/m)*theta))/3,i=0..m-1); (11) D110 product((2*cos(theta/3^(n/m))+1)/3,n=1..infinity)=(2^m/(theta)^m)*product(sin(3^(i/m)*theta/2)/3^(i/m),i=0..m-1); However I am a little perplexed as to the last formula that differs from Gosper's D111 identity by a factor of 4. The result I get is this: (12) D111 product((2*cos(theta/3^(n/2))+1)/3,n=1..infinity)=(2^2/(theta)^2)*product(sin(3^(i/2)*theta/2)/3^(i/2),i=0..2-1); I suppose it must be the same for this small series, I have yet to check. Best regards

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[math-fun] 31 lines of zero (mod 15) puzzle.
by Ed Pegg Jr 21 Oct '17

21 Oct '17

Here's an orchard of 31 lines of three on 15 points. That might be maximal according to A003035. https://i.imgur.com/w18qs7R.jpg Arrange the fifteen numbers -7 to 7 in the disks so that the (mod 15) sum of all lines is zero. --Ed Pegg Jr

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[math-fun] Fwd: Random slice of a cube
by James Propp 20 Oct '17

20 Oct '17

Apropos of the question "Why is the valency 4 on average?" (In Warren's argument), here's a mental picture of the generic situation: An edge in the graph is the intersection of the oblique plane with a face of a cube in the tessellation. A vertex A vertex in the graph is the intersection of the oblique plane with an edge of a cube in the tessellation. Four faces surround each edge in the tessellation, so four edges surround each vertex in the graph. Jim Propp

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[math-fun] AlphaGo Zero
by James Propp 19 Oct '17

19 Oct '17

Just because AlphaGo Zero can beat its predecessor, which in turn can beat expert humans, does it follow that AlphaGo Zero can beat expert humans? https://www.quantamagazine.org/artificial-intelligence-learns-to-learn-enti… It seems possible to me that AlphaGo Zero could be exploiting weaknesses in AlphaGo that humans might not share. It's not obvious to me that transitivity applies in this kind of situation. I was surprised that Kevin Hartnett didn't address this question. (Or am I missing something?) Jim Propp

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[math-fun] continued fractions (2)
by François Mendzina Essomba 19 Oct '17

19 Oct '17

Formula 26 a := (sqrt(exp(2*Pi)-exp(Pi)+1)*sqrt(exp(2*Pi)+exp(Pi)+1)*sqrt(exp(8*Pi)+exp(2*Pi)+1)*sqrt(exp(12*Pi)+exp(10*Pi)+exp(8*Pi)+exp(6*Pi)+2*exp(4*Pi)-2*exp(2*Pi)+1))/(sqrt(5)*exp(4*Pi)+sqrt(5)*exp(2*Pi)+sqrt(5)); b:= -(a/2)+(exp(12*Pi)+exp(10*Pi)+exp(8*Pi)+exp(6*Pi)-1)/(2*(exp(4*Pi)+exp(2*Pi)+1)); Golden ratio phi:=a/(b+1/(1+exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+ exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+…)))))))))))))); Formula 27 a:=(sqrt(exp(2*Pi)+exp(Pi)+1)*sqrt(exp(3*Pi)+exp(2*Pi)+1)*sqrt(exp(5*Pi)-exp(4*Pi)+exp(2*Pi)-exp(Pi)+1)*sqrt(exp(10*Pi)+exp(9*Pi)+exp(7*Pi)+exp(5*Pi)+exp(4*Pi)+exp(3*Pi)-2*exp(2*Pi)+1))/(sqrt(5)*exp(3*Pi)+sqrt(5)*exp(2*Pi)+sqrt(5)); b:= -(a/2)+(exp(10*Pi)+exp(9*Pi)+exp(7*Pi)+exp(5*Pi)+ exp(4*Pi)-exp(3*Pi)-1)/(2*(exp(3*Pi)+exp(2*Pi)+1)); Golden ratio phi:=a/(b+1/(1+exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+ exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+…)))))))))))))); Formula 28 a:=(sqrt(24*exp(6*Pi)+12*exp(5*Pi)+4*exp(4*Pi)+2*exp(3*Pi)+7*exp(2*Pi)-2*exp(Pi)+1)*sqrt(24*exp(6*Pi)+12*exp(5*Pi)+4*exp (4*Pi)+2*exp(3*Pi)+7*exp(2*Pi)+2*exp(Pi)+1))/(6*sqrt(5)*exp(2*Pi)+3*sqrt(5)*exp(Pi)+sqrt(5)) ; b:= -(a/2)+ (24*exp(6*Pi)+12*exp(5*Pi)+4*exp(4*Pi)+2*exp(3*Pi)-5*exp(2*Pi)-1)/(2*(6*exp(2*Pi)+3*exp(Pi)+1)); Golden ratio phi:=a/(b+1/(1+exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+ exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+…)))))))))))))); Formula 29 cos(k*exp(x))*cos(k*exp(-x))=(sum((-1)^(n)*k^(2*n)*cosh(2*n*x)/(2*n)!,n=0..infinity))^2-(sum((-1)^(n)*k^(2*n)*sinh(2*n*x)/(2*n)!,n=0..infinity))^2;

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[math-fun] continued fractions
by François Mendzina Essomba 19 Oct '17

19 Oct '17

hello, just to leave some formulas. Formula 26 a := (sqrt(exp(2*Pi)-exp(Pi)+1)*sqrt(exp(2*Pi)+exp(Pi)+1)*sqrt(exp(8*Pi)+exp(2*Pi)+1)*sqrt(exp(12*Pi)+exp(10*Pi)+exp(8*Pi)+exp(6*Pi)+2*exp(4*Pi)-2*exp(2*Pi)+1))/(sqrt(5)*exp(4*Pi)+sqrt(5)*exp(2*Pi)+sqrt(5)); b:= -(a/2)+(exp(12*Pi)+exp(10*Pi)+exp(8*Pi)+exp(6*Pi)-1)/(2*(exp(4*Pi)+exp(2*Pi)+1)); phi:=a/(b+1/(1+exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+ exp(-2*Pi)/(1+exp(-4*Pi)/(1+exp(-6*Pi)/(1+exp(-8*Pi)/(1+…)))))))))))))); Formula 27 a:=(sqrt(exp(2*Pi)+exp(Pi)+1)*sqrt(exp(3*Pi)+exp(2*Pi)+1)*sqrt(exp(5*Pi)-exp(4*Pi)+exp(2*Pi)-exp(Pi)+1)*sqrt(exp(10*Pi)+exp(9*Pi)+exp(7*Pi)+exp(5*Pi)+exp(4*Pi)+exp(3*Pi)-2*exp(2*Pi)+1))/(sqrt(5)*exp(3*Pi)+sqrt(5)*exp(2*Pi)+sqrt(5)); b:= -(a/2)+(exp(10*Pi)+exp(9*Pi)+exp(7*Pi)+exp(5*Pi)+ exp(4*Pi)-exp(3*Pi)-1)/(2*(exp(3*Pi)+exp(2*Pi)+1)); phi:=a/(b+1/(1+exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+ exp(-Pi)/(1+exp(-3*Pi)/(1+exp(-5*Pi)/(1+exp(-7*Pi)/(1+…)))))))))))))); Formula 28 a:=(sqrt(24*exp(6*Pi)+12*exp(5*Pi)+4*exp(4*Pi)+2*exp(3*Pi)+7*exp(2*Pi)-2*exp(Pi)+1)*sqrt(24*exp(6*Pi)+12*exp(5*Pi)+4*exp (4*Pi)+2*exp(3*Pi)+7*exp(2*Pi)+2*exp(Pi)+1))/(6*sqrt(5)*exp(2*Pi)+3*sqrt(5)*exp(Pi)+sqrt(5)) ; b:= -(a/2)+ (24*exp(6*Pi)+12*exp(5*Pi)+4*exp(4*Pi)+2*exp(3*Pi)-5*exp(2*Pi)-1)/(2*(6*exp(2*Pi)+3*exp(Pi)+1)); phi:=a/(b+1/(1+exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+ exp(-Pi)/(2+exp(-2*Pi)/(3+exp(-3*Pi)/(4+exp(-4*Pi)/(1+…)))))))))))))); Best regards François...

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Re: [math-fun] prime targets for Coppersmith hacking
by Don Reble 17 Oct '17

17 Oct '17

> What does this tell us about how the keys were poorly generated, and > what does it tell us about how to factor them? > Here's a count of "bad" remainders for each prime. Very odd. > 3 1 > 5 1 > 7 1 > 11 9 > 13 7 > 17 9 > 19 10 > ... That's an enormous hint. Each set of remainders has the Nth-order residues modulo the prime, for some N. This python program tests and proves it. #!/usr/bin/python Primes = ( \ 3, 5, 7, 11, 13, 17, 19, 23, \ 29, 31, 37, 41, 43, 47, 53, 59, \ 61, 67, 71, 73, 79, 83, 89, 97, \ 101, 103, 107, 109, 113, 127, 131, 137, \ 139, 149, 151, 157, 163, 167 ) # all odd primes from 3 to 167 Generators = ( \ 2, 2, 3, 2, 2, 3, 2, 5, \ 2, 3, 2, 6, 3, 5, 2, 2, \ 2, 2, 7, 5, 3, 2, 3, 5, \ 2, 5, 2, 6, 3, 3, 2, 3, \ 2, 2, 6, 5, 2, 5 ) Rems = ( \ 6, 30, 126, 1026, 5658, 107286, 199410, 8388606, 536870910, 2147483646, 67109890, 2199023255550, 8796093022206, \ 140737488355326, 5310023542746834, 576460752303423486, \ 1455791217086302986, 147573952589676412926, 20052041432995567486, \ 6041388139249378920330, 207530445072488465666, \ 9671406556917033397649406, 618970019642690137449562110, \ 79228162521181866724264247298, 2535301200456458802993406410750, \ 1760368345969468176824550810518, 50079290986288516948354744811034, \ 473022961816146413042658758988474, \ 10384593717069655257060992658440190, \ 144390480366845522447407333004847678774, \ 2722258935367507707706996859454145691646, \ 174224571863520493293247799005065324265470, \ 696898287454081973172991196020261297061886, \ 713623846352979940529142984724747568191373310, \ 1800793591454480341970779146165214289059119882, \ 126304807362733370595828809000324029340048915994, \ 11692013098647223345629478661730264157247460343806, \ 187072209578355573530071658587684226515959365500926 ) def countBits(x): qu = 0 while x != 0: qu += 1 x = x & (x-1) return qu def computeResidues(pr, gen, order): power = 1 res = 1 << power gen = (gen ** order) % pr for kk in range(order, pr, order): power = (power * gen) % pr res |= 1 << power return res for pr,gen,rem in zip(Primes,Generators,Rems): qu = countBits(rem) if ((pr-1) % qu) != 0: print pr, qu, hex(rem) else: order = (pr-1) / qu residues = computeResidues(pr, gen, order) if residues == rem: print pr, qu, str(order) + "-order residues" else: print pr, qu, hex(rem), hex(residues) -- Don Reble djr(a)nk.ca

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