October 2017 - mathfun@mailman.xmission.com

[math-fun] Great Pyramid angle ~ atan(4/pi)
by Bill Gosper 06 Oct '17

06 Oct '17

Henry, I can't figure out what you mean by the angle of a pyramid, nor the aesthetic criteria motivating the numerology. FWIW, the pyramid whose apex is the center of the cube on its base (and thus apex solid angle 4π/6) has its four apex angles = ArcTan[√8] ~ 70.5º. Is this too steep? Too squat? --rwg On 2017-10-03 09:08, Henry Baker wrote: > The Great Pyramid angle is actually atan(14/11). > > Why 14:11 ? > > cf(14/11) = [1,3,1,2] > > atan(14/11) = 51.843 degrees. > = asin(14/sqrt(317)) > = acos(11/sqrt(317)) > > ["317" really weird coincidence: 22/7 = 3 1/7] > > Apparently, 14/11 ~ 4/pi, if you believe pi=22/7. > > cf(4/pi) = [1,3,1,1,1,15,...] > > So, the angle the Egyptians really desired was > > atan(4/pi) = 51.854 degrees. > > Supposedly, the circumference of the base of > the Great Pyramid ~ the circumference of a circle > whose radius is the height of the Great Pyramid. > (Note that the base corners would protrude outside > the sphere of radius = height, so this sphere would > NOT enclose the Pyramid; the Roman Pantheon is more > mathematically interesting, as it perfectly > illustrates Archimedes's "Hatbox Theorem".) > > So if r is the height of the pyramid, the circle > circumference is 2*pi*r ~ 2*(22/7)*r or (44/7)*r. > > So each side of the Pyramid base is (11/7)*r, and > the distance from the center of the base to the > center of one of the base sides is (11/14)*r. > > So we have a right triangle > r, (11/14)*r, (sqrt(317)/14)*r > through the center of the Pyramid for a face > angle of atan(14/11). > > David Ian Lightbody. "Biography of a Great > Pyramid Casing Stone", The Journal of Ancient > Egyptian Architecture vol. 1, 2016. > > http://www.egyptian-architecture.com/JAEA1/article3/JAEA1_Lightbody.pdf >

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[math-fun] Complex dynamics
by Dan Asimov 05 Oct '17

05 Oct '17

Complex dynamics is the field that has brought us the Mandelbrot set, Julia sets, and lots of beautiful and mysterious phenomena. In 2006 topologist John Milnor published the latest edition (3rd) of his book "Dynamics in One Complex Variable". The original (1990) notes this book is based on are freely downloadable, but the book goes into more detail in many, many ways, and of course is more up-to-date. I finally got my hands on it, and it looks to be another Milnor tour de force. There were 3 published editions after these notes: https://arxiv.org/abs/math/9201272 so the 3rd edition is in that sense the 4th. If you have any interest in this subject, book is highly recommended. —Dan

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[math-fun] Erik Verlinde lecture
by Fred Lunnon 05 Oct '17

05 Oct '17

"A New View on Gravity and the Dark Side of the Cosmos" https://www.youtube.com/watch?v=rWW3S5fNf5c WFL

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[math-fun] Fwd: Re: [ROBBINS] A 7-cycle
by Bill Gosper 04 Oct '17

04 Oct '17

On 2017-10-02 13:08, rcs(a)xmission.com wrote: > This seems worth passing along to the Math-Fun crowd. > Read the bottom half first. > My apologies to Propp & Speyer. -- Rich > > ----- Forwarded message from speyer(a)UMICH.EDU ----- > Date: Mon, 2 Oct 2017 12:34:10 -0400 > From: David Speyer <speyer(a)UMICH.EDU> > Reply-To: The Robbins Forum <ROBBINS(a)LISTSERV.UML.EDU> > Subject: Re: [ROBBINS] A 7-cycle > To: ROBBINS(a)LISTSERV.UML.EDU > > This provides no insight but, yes, the composition has order 7. Define the > seven cones > > 1: 0 < x,y > 2: 0 < -x < y > 3: 0 < y < -x > 4: x<y<0 > 5: y<x<0 > 6: 0 < x < -y > 7: 0 < y < -x > > Then one can check that h and v are linear on each cone, with image another > cone. Specifically, h(k) = 4-k mod 7 and v(k) = -k mod 7. So hv permutes > the cones by k+4 mod 7 and (hv)^7 is the identity on the set of cones; it > is then easy to check that it is the identity on each cone. > > I'm curious about the tropicalization question -- seven isn't a number > which tends to come out of cluster algebras. > READ THIS FIRST > On Mon, Oct 2, 2017 at 12:20 PM, James Propp <jamespropp(a)gmail.com> wrote: > >> Apologies for the double posting to the DAC listserv and Robbins listserv; >> both seem equally appropriate. >> >> If we define piecewise-linear continuous maps h and v from R^2 to itself >> with >> h((x,y)) = (-x-max(y,0),y) >> and >> v((x,y)) = (x,-y-max(x,0)), >> then the composition of h and v appears to be of order 7. >> >> Is this true? >> >> Usually the way I prove a PL identity like this is by finding a birational >> identity that tropicalizes to it (and having Mathematica do the heavy >> lifting), but in this case that strategy failed; I tried the obvious >> de-tropicalization >> H((x,y)) = (1/(x (1 + y)), y) >> and >> V((x,y)) = (x, 1/(y (1 + x))) >> but the composition of these maps is not of order 7. >> >> Is the composition of h and v the tropicalization of some birational map >> of order 7? >> >> (If you're wondering where h and v come from: they're the horizontal and >> vertical fliber-flipping maps for the polygon bounded by the lines x=-1, >> x=1, y=-1, y=1, and x+y=1; the composition of h and v leaves the origin >> fixed and cyclically permutes the 7 lattice points on the boundary. If the >> term "fiber-flipping" is unfamiliar, see the bottom of page 3 and the top >> of page 4 of https://arxiv.org/pdf/1404.3455.pdf, where the >> fiber-flipping maps are called piecewise-linear toggles.) >> >> Jim Propp >> Mathematica comes bizarrely close to proving order 7 symbolically, in ℂ rather than ℝ² : h = I Im@# - Re@# - Max[0, Im@#] & v = Re@# - I (Im@# + Max[0, Re@#]) & The 1st iterate: In[29]:= hv = Function @@ {FullSimplify[Composition[h, v]@#]} Out[29]= -I (Im[#1] - I Max[0, -Im[#1] - Max[0, Re[#1]]] + Max[0, Re[#1]]) - Re[#1] & The 2nd iterate: In[7]:= Function[x, Evaluate[FullSimplify[Composition[%, %]@x]]] // tim During evaluation of In[7]:= 1.486366,2 Out[7]= Function[x, I Im[x] + Max[0, -Im[x] - Max[0, Re[x]]] - Max[0, Im[x] - Max[0, -Max[0, -Im[x] - Max[0, Re[x]]] - Re[x]] + Max[0, Re[x]]] - I Max[0, -Max[0, -Im[x] - Max[0, Re[x]]] - Re[x]] + I Max[0, Re[x]] + Re[x]] The 4th iterate: In[8]:= f4=Function[y, Evaluate[FullSimplify[Composition[%, %]@y]]] // tim During evaluation of In[8]:= 8034.496115,2 Out[8]= Function[y, Piecewise[ {{I y - Re[y], y == 0 || (Re[y] > 0 && ((Im[y] < 0 && Im[y] + Re[y] != 0) || Im[y] > 0))}, {(-1 + I) Re[y], (Im[y] == Re[y] && Re[y] < 0) || (y == Re[y] && Re[y] > 0)}, {I y, (Re[y] < 0 && ((Im[y] + Re[y] < 0 && Im[y] >= 0) || (Im[y] < 0 && Im[y] != Re[y]) || (Im[y] > 0 && Im[y] + Re[y] > 0))) || (Re[y] == 0 && y != 0)}, {y + 2 I Re[y], Im[y] + Re[y] == 0 && Im[y] > 0}}, (1 + I) y - (2 - I) Re[y]]] The 3rd iterate: f3 = Function[{x}, Evaluate[FullSimplify[Composition[hv,hv,hv]@x]]] Function[{x}, Piecewise[ {{Im[x], (Re[x] == 0 && Im[x] < 0) || x == 0}, {-I x, Im[x] < 0 && ((Re[x] < 0 && Im[x] != Re[x]) || (Re[x] > 0 && Im[x] + Re[x] != 0))}, {(1 - I) Im[x], Im[x] > 0 && Re[x] == 0}, {(1 - I) Re[x], Im[x] == Re[x] && Im[x] < 0}, {(-1 - I) x + Re[x], Im[x] >= 0 && ((Re[x] < 0 && Im[x] + Re[x] != 0) || Re[x] > 0)}, {-Re[x], Re[x] < 0 && Im[x] + Re[x] == 0}}, (-1 - I) Re[x]]]; And for the grand finale 7th iterate: In[14]:= Function[z, Evaluate[FullSimplify[Composition[f4, f3]@z]]] // tim During evaluation of In[14]:= 46.694108,2 Out[14]= Function[z, Piecewise[ {{z, z == 0 || z == 0 || ((1 + I) z == Re[z] && Re[z] < 0 && Im[z] + Re[z] != 0 && Im[z] >= 0) || ((1 + I) z == Re[z] && Im[z] >= 0 && Re[z] > 0) || (Im[z] + Re[z] < 0 && Im[z] > 0) || (Re[z] < 0 && Im[z] + Re[z] > 0) || (Re[z] > 0 && Im[z] > 0) || (Im[z] < 0 && Re[z] != 0 && Im[z] + Re[z] != 0 && ((Re[z] < 0 && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0) && Im[z] != Re[z]) || (Re[z] > 0 && z + (1 - I) Im[z] != 0)))}, {I Im[z], Re[z] == 0 && z != 0}, {(1 - I) Re[z], Im[z] + Re[z] == 0 && (Im[z] < 0 || Re[z] < 0)}, {(2 - I) Re[z], (z == Re[z] && (z == 0 || Re[z] == 0 || ((1 + I) z == Re[z] && (Re[z] >= 0 || Im[z] + Re[z] != 0)))) || (Im[z] >= 0 && Re[z] != 0 && (((1 + I) z == Re[z] && (Im[z] + Re[z] != 0 || Re[z] >= 0)) || (Im[z] > 0 && Im[z] + Re[z] != 0)))}, {(1 + I) Re[z], Re[z] < 0 && (Im[z] == Re[z] || (z - (1 + I) Re[z] == 0 && -z + (1 + I) Im[z] >= 0 && Im[z] < 0 && -z + (1 + I) Im[z] <= 0))}, {(1 - I) z + I Re[z], (z == Re[z] && (1 + I) z != Re[z] && ((Re[z] < 0 && Im[z] + Re[z] != 0) || Re[z] > 0)) || (Im[z] < 0 && ((z + (1 - I) Im[z] != 0 && Im[z] + Re[z] != 0 && Re[z] > 0) || (Re[z] < 0 && Im[z] != Re[z] && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0)) || Re[z] == 0))}, {(-1 - 3 I) z, (Im[z] == Re[z] || z - (1 + I) Re[z] == 0) && Im[z] < 0 && Re[z] < 0}, {0, Re[z] == 0 && Im[z] < 0}, {(1 - I) z - (2 - I) Im[z], Im[z] < 0 && ((Re[z] < 0 && Im[z] != Re[z]) || (Re[z] > 0 && Im[z] + Re[z] != 0))}, {z - I Re[z], Im[z] >= 0 && ((Re[z] < 0 && Im[z] + Re[z] != 0) || Re[z] > 0)}}, (2 - 3 I) z]] The 4th iterate took 2 hrs, This 7th took 46 secs. Notice the first branch is z, the identity! Is that crazy condition an unsimplified tautology? Anything but! Out[22]= z == 0 || z == 0 || ((1 + I) z == Re[z] && Re[z] < 0 && Im[z] + Re[z] != 0 && Im[z] >= 0) || ((1 + I) z == Re[z] && Im[z] >= 0 && Re[z] > 0) || (Im[z] + Re[z] < 0 && Im[z] > 0) || (Re[z] < 0 && Im[z] + Re[z] > 0) || (Re[z] > 0 && Im[z] > 0) || (Im[z] < 0 && Re[z] != 0 && Im[z] + Re[z] != 0 && ((Re[z] < 0 && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0) && Im[z] != Re[z]) || (Re[z] > 0 && z + (1 - I) Im[z] != 0))) Out[24]= {{z -> 1}, {z -> 1 + I}, {z -> 2 I}, {z -> -2 + 2 I}, {z -> -4}, {z -> -4 - 4 I}, {z -> -8 I}, {z -> 8 - 8 I}, {z -> 16}} In[25]:= %22 /. % Out[25]= {False, True, False, False, False, False, False, False, False} So Mathematica blew it and the 7th iterate isn't really the identity -- NOT!! In[17]:= (4/5 - 3 I/5)^Range[-5, 5] Out[17]= {-(3116/3125) - (237 I)/3125, -(527/625) + (336 I)/625, -(44/125) + (117 I)/125, 7/25 + (24 I)/25, 4/5 + (3 I)/5, 1, 4/5 - (3 I)/5,7/25 - (24 I)/25, -(44/125) - (117 I)/125, -(527/625) - (336 I)/625, -(3116/3125) + (237 I)/3125} In[19]:= %14 /@ %17 Out[19]= {-(3116/3125) - (237 I)/3125, -(527/625) + (336 I)/625, -(44/125) + (117 I)/125, 7/25 + (24 I)/25, 4/5 + (3 I)/5, 1, 4/5 - (3 I)/5,7/25 - (24 I)/25, -(44/125) - (117 I)/125, -(527/625) - (336 I)/625, -(3116/3125) + (237 I)/3125} It's just a bizarrely unsimplified identity. Maybe someday FullSimplifyInTheSky will say so. Something for WRI's To Do List. --rwg

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[math-fun] Great Pyramid angle ~ atan(4/pi)
by Henry Baker 03 Oct '17

03 Oct '17

The Great Pyramid angle is actually atan(14/11). Why 14:11 ? cf(14/11) = [1,3,1,2] atan(14/11) = 51.843 degrees. = asin(14/sqrt(317)) = acos(11/sqrt(317)) ["317" really weird coincidence: 22/7 = 3 1/7] Apparently, 14/11 ~ 4/pi, if you believe pi=22/7. cf(4/pi) = [1,3,1,1,1,15,...] So, the angle the Egyptians really desired was atan(4/pi) = 51.854 degrees. Supposedly, the circumference of the base of the Great Pyramid ~ the circumference of a circle whose radius is the height of the Great Pyramid. (Note that the base corners would protrude outside the sphere of radius = height, so this sphere would NOT enclose the Pyramid; the Roman Pantheon is more mathematically interesting, as it perfectly illustrates Archimedes's "Hatbox Theorem".) So if r is the height of the pyramid, the circle circumference is 2*pi*r ~ 2*(22/7)*r or (44/7)*r. So each side of the Pyramid base is (11/7)*r, and the distance from the center of the base to the center of one of the base sides is (11/14)*r. So we have a right triangle r, (11/14)*r, (sqrt(317)/14)*r through the center of the Pyramid for a face angle of atan(14/11). David Ian Lightbody. "Biography of a Great Pyramid Casing Stone", The Journal of Ancient Egyptian Architecture vol. 1, 2016. http://www.egyptian-architecture.com/JAEA1/article3/JAEA1_Lightbody.pdf

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[math-fun] Fwd: Re: [ROBBINS] A 7-cycle
by rcs＠xmission.com 02 Oct '17

02 Oct '17

This seems worth passing along to the Math-Fun crowd. Read the bottom half first. My apologies to Propp & Speyer. -- Rich ----- Forwarded message from speyer(a)UMICH.EDU ----- Date: Mon, 2 Oct 2017 12:34:10 -0400 From: David Speyer <speyer(a)UMICH.EDU> Reply-To: The Robbins Forum <ROBBINS(a)LISTSERV.UML.EDU> Subject: Re: [ROBBINS] A 7-cycle To: ROBBINS(a)LISTSERV.UML.EDU This provides no insight but, yes, the composition has order 7. Define the seven cones 1: 0 < x,y 2: 0 < -x < y 3: 0 < y < -x 4: x<y<0 5: y<x<0 6: 0 < x < -y 7: 0 < y < -x Then one can check that h and v are linear on each cone, with image another cone. Specifically, h(k) = 4-k mod 7 and v(k) = -k mod 7. So hv permutes the cones by k+4 mod 7 and (hv)^7 is the identity on the set of cones; it is then easy to check that it is the identity on each cone. I'm curious about the tropicalization question -- seven isn't a number which tends to come out of cluster algebras. On Mon, Oct 2, 2017 at 12:20 PM, James Propp <jamespropp(a)gmail.com> wrote: > Apologies for the double posting to the DAC listserv and Robbins listserv; > both seem equally appropriate. > > If we define piecewise-linear continuous maps h and v from R^2 to itself > with > h((x,y)) = (-x-max(y,0),y) > and > v((x,y)) = (x,-y-max(x,0)), > then the composition of h and v appears to be of order 7. > > Is this true? > > Usually the way I prove a PL identity like this is by finding a birational > identity that tropicalizes to it (and having Mathematica do the heavy > lifting), but in this case that strategy failed; I tried the obvious > de-tropicalization > H((x,y)) = (1/(x (1 + y)), y) > and > V((x,y)) = (x, 1/(y (1 + x))) > but the composition of these maps is not of order 7. > > Is the composition of h and v the tropicalization of some birational map > of order 7? > > (If you're wondering where h and v come from: they're the horizontal and > vertical fliber-flipping maps for the polygon bounded by the lines x=-1, > x=1, y=-1, y=1, and x+y=1; the composition of h and v leaves the origin > fixed and cyclically permutes the 7 lattice points on the boundary. If the > term "fiber-flipping" is unfamiliar, see the bottom of page 3 and the top > of page 4 of https://arxiv.org/pdf/1404.3455.pdf, where the > fiber-flipping maps are called piecewise-linear toggles.) > > Jim Propp > > ----- End forwarded message ----- This provides no insight but, yes, the composition has order 7. Define the seven cones 1: 0 < x,y 2: 0 < -x < y 3: 0 < y < -x 4: x<y<0 5: y<x<0 6: 0 < x < -y 7: 0 < y < -x Then one can check that h and v are linear on each cone, with image another cone. Specifically, h(k) = 4-k mod 7 and v(k) = -k mod 7. So hv permutes the cones by k+4 mod 7 and (hv)^7 is the identity on the set of cones; it is then easy to check that it is the identity on each cone. I'm curious about the tropicalization question -- seven isn't a number which tends to come out of cluster algebras. On Mon, Oct 2, 2017 at 12:20 PM, James Propp <jamespropp(a)gmail.com> wrote: > Apologies for the double posting to the DAC listserv and Robbins listserv; > both seem equally appropriate. > > If we define piecewise-linear continuous maps h and v from R^2 to itself > with > h((x,y)) = (-x-max(y,0),y) > and > v((x,y)) = (x,-y-max(x,0)), > then the composition of h and v appears to be of order 7. > > Is this true? > > Usually the way I prove a PL identity like this is by finding a birational > identity that tropicalizes to it (and having Mathematica do the heavy > lifting), but in this case that strategy failed; I tried the obvious > de-tropicalization > H((x,y)) = (1/(x (1 + y)), y) > and > V((x,y)) = (x, 1/(y (1 + x))) > but the composition of these maps is not of order 7. > > Is the composition of h and v the tropicalization of some birational map > of order 7? > > (If you're wondering where h and v come from: they're the horizontal and > vertical fliber-flipping maps for the polygon bounded by the lines x=-1, > x=1, y=-1, y=1, and x+y=1; the composition of h and v leaves the origin > fixed and cyclically permutes the 7 lattice points on the boundary. If the > term "fiber-flipping" is unfamiliar, see the bottom of page 3 and the top > of page 4 of https://arxiv.org/pdf/1404.3455.pdf, where the > fiber-flipping maps are called piecewise-linear toggles.) > > Jim Propp > >

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