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Re: [math-fun] Random slice of a cube
by Keith F. Lynch 23 Oct '17

23 Oct '17

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron. I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes. > Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons). I meant regular polyhedrons, not polygons. > * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides. I meant the maximum distance from the center of the cube to the closest point on the plane. > Unfortunately, my program cannot easily be adapted to other regular > polygons. Again, I meant polyhedrons, not polygons. > I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested. In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices. > The average number of sides is 4.0004. I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4. This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square. I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4. I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried. Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Dan Asimov 23 Oct '17

23 Oct '17

Henry got that right — in fact, something quite interesting happens, as Wikipedia says: ----- Prince Rupert's Drops (also known as Dutch or Batavian tears) are toughened glass beads created by dripping molten glass into cold water, which causes it to solidify into a tadpole-shaped droplet with a long, thin tail. These droplets are characterized internally by very high residual stresses, which give rise to counter-intuitive properties, such as the ability to withstand a blow from a hammer or a bullet on the bulbous end without breaking, while exhibiting explosive disintegration if the tail end is even slightly damaged. In nature, similar structures are produced under certain conditions in volcanic lava. ----- more at https://en.wikipedia.org/wiki/Prince_Rupert's_Drop. —Dan Henry Baker wrote: ----- 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. ----- I wrote: ----- ... 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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[math-fun] accelerated convergence
by François Mendzina Essomba 23 Oct '17

23 Oct '17

Hello, After reading some articles on the Brent-Salamin algorithm based on the AGM and algorithms of the Borweins brothers arising from the famous Ramanujan formulas on the Pi number. I wondered if it was possible to find an algorithm for any convergence speed m considered. Finally I came to establish the following general formula which is of convergence (2 * m): x(k+1)=x(k)+sum(2*((m!)^2)*sin(n*x(k))/(n*((m+n)!)*((m-n)!)),n=1..m); x(0) = 3 ; k --- infinity ; x(k) ---- Pi ; If we take m = 5, the convergence is (2 * m) = 10; we are getting: x(k+1)= x(k)+(5/3)*sin(x(k))+(10/21)*sin(2*x(k))+(5/42)*sin(3*x(k))+(5/252)*sin(4*x(k))+(1/630)*sin(5*x(k)); x(0)=3; x(1)=3.141592653589628958434997482316410876703079409530919198836399516902375075829274562410138150533960715 ; x(2)=3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068 ; So if I want to write a 2000000 convergence formula, then I would simply have: x(k+1)=x(k)+sum(2*((1000000!)^2)*sin(n*x(k))/(n*((1000000+n)!)*((1000000-n)!)),n=1..1000000); Unfortunately, this formula can not be useful for the calculation of Pi given the difficulty related to the calculation of sinuses. Best regards...

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Re: [math-fun] math-fun Digest, Vol 176, Issue 31
by françois mendzina essomba2 23 Oct '17

23 Oct '17

Gosper's identities (Bill Gosper); Hello, Thank you dear Professor Bill Gosper, for the details related to my recurrence, I did not verify it experimentally. The negative terms (-2) ^ n or (-3) ^ n of this identities seem to me so rich and so profound ... It took me longer to understand Best regards. Le Samedi 21 octobre 2017 20h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. 31 lines of zero (mod 15) puzzle. (Ed Pegg Jr) 2. Gosper's identities (Bill Gosper) 3. Gosper's nonidentities (Bill Gosper) ---------------------------------------------------------------------- Message: 1 Date: Fri, 20 Oct 2017 23:00:36 -0500 From: Ed Pegg Jr <ed(a)mathpuzzle.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: [math-fun] 31 lines of zero (mod 15) puzzle. Message-ID: <CAGog0+Y90bRpf2mWS9eZ_Ho3DbAdt6UHVWvNfGCRev2CbLqm7g(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" 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 ------------------------------ Message: 2 Date: Sat, 21 Oct 2017 02:22:35 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Gosper's identities Message-ID: <CAA-4O0E0cmKVasfuY65rGjf496FDtY8anutimpgecOeLyngZgw(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" Dear Dr. Essomba, Thank you for catching my D111 mistake. Your claim is immediately substantiated by the limit ??0, which gives 1 = ?. And confirmed by rederivation, simply multiplying D110 by itself with ???3?. 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 ?, f for ?,...) 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 ??. --rwg Date: 2017-10-16 05:35 From: Fran?ois 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: Fran?ois 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 ------------------------------ Message: 3 Date: Sat, 21 Oct 2017 10:15:19 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Gosper's nonidentities Message-ID: <CAA-4O0ER2aMjEHX2+SdqNzGc0rpkfXApmnGrx4jX9mWAzuNT0g(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" 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 ???? out of anything this weird. Help? --rwg ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 31 *****************************************

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

22 Oct '17

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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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 20 Oct '17

20 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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