July 2007 - mathfun@mailman.xmission.com

[math-fun] solid angles yet again
by Bill Gosper 17 Jul '07

17 Jul '07

Circa Xmas '02 I sent Subj: more solid angles In my version of Macsyma's Geofuncs package is now (c1) solid_angle(a,b,c) 2 (cos(c) + cos(b) + cos(a) + 1) (d1) acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) where a, b, and c are the ordinary angles at the trihedral vertex. Further simplified to cos(c) + cos(b) + cos(a) + 1 (d1) 2 acos ---------------------------- . a b c 4 cos(-) cos(-) cos(-) 2 2 2 It seems too nice to be new, but an hour or so of (low bandwidth) Googling failed to turn it up. Here are some playings with it. A sector with wedge angle v will have solid angle v/(2 pi) 4 pi = 2 v. But we can split it perpendicular to its straight edge to make two trihedrals with angles pi/2, pi/2, and v: (c2) 2*solid_angle(%pi/2,%pi/2,v) (d2) 2 v We can dissect a regular n-gonal pyramid of height h and base circumradius r into n congruent tetrahedra with trihedral components v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., half the apex angle. This gives, after much simplification, two expressions for the solid angle of the apex: 2 %pi 2 %pi 2 %pi 2 cos (---) (2 sin (---) cos(v) + cos(-----)) n n n (d3) n acos(--------------------------------------------- 2 %pi 2 2 %pi sin (---) cos (v) + cos (---) n n 2 %pi - cos(-----)) n n = acos((- 1) n - 1 /===\ 2 %pi 2 | | %pi %pi k 2 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) n | | n n k = 1 (1 - ------------------------------------------------------)) 2 2 %pi n (cos (v) + cot (---)) n Shame on me. The product over a half-period of a polynomial in trigs of angles in arithmetic progression always simplifies, in this case, mondomiraculously: n 2 %pi acos((- 1) (1 - 2 sin (n acot(tan(---) cos(v))))) (!) n The (-1)^n looks very strange. But it seems to work, and the succeeding 1-2sin^2 expression oscillates without it. Letting n -> oo, the solid angle of a cone is 2 pi (1 - cos v), where, again, tan v = (base radius)/height. Is there a polyhedral analog of n-gon angle sum = (n-2) pi, presumably involving vertex angles, dihedrals, and maybe vertex solid angles? Disappointingly, for the regular tetrahedron, (c683) (%pi/3,solid_angle(%%,%%,%%)) 5 (d683) 2 acos(---------) 3 sqrt(3) (c684) dfloat(4*%) (d684) 2.20514239373012d0 while for the tetrahedral corner of a cube, (c692) (%pi/2,solid_angle(%%,%%,%%)+3*solid_angle(%pi/4,%pi/4,%pi/3)) 2 3 ------- + - sqrt(2) 2 %pi (d692) 6 acos(-------------------) + --- 2 %pi 2 2 sqrt(3) cos (---) 8 (c693) dfloat(%) (d693) 2.59030705515727d0 I.e., the solid angles of the vertices don't even sum to a constant. --rwg steradian dentarias

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[math-fun] solid conclusion(?)
by Bill Gosper 16 Jul '07

16 Jul '07

>Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is 2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2 I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2 Clearer might have been the difference between two isosceles trihedrals: solid_angle(d,d,b)-solid_angle(d-a,d-a,c) with d chosen to equalize the dihedrals, using dihedral(a,b,c):=acos(csc(a)*csc(b)*cos(c)-cot(a)*cot(b)), which gives sin(a) d = atan(----------------------), b c cos(a) - csc(-) sin(-) 2 2 and eventually the same solid angle formula. But the max technique generalizes from trapezoidal pyramids to arbitrary quadrilateral pyramids with apex angles a,b,c,d, whose maximal solid angle is cos(d) + cos(c) + cos(b) + cos(a) a b c d --------------------------------- - tan(-) tan(-) tan(-) tan(-) a b c d 2 2 2 2 4 cos(-) cos(-) cos(-) cos(-) 2 2 2 2 Note we get the old formula when d=0. Note also the symmetry, giving the same result for a,b,d,c. In this maximal case, the "diagonal angles", i.e. the apex angles exposed by splitting into two triangular pyramids, are a d b c b d a c 2 (sin(-) sin(-) + sin(-) sin(-)) (sin(-) sin(-) + sin(-) sin(-)) 2 2 2 2 2 2 2 2 acos(1 - -----------------------------------------------------------------) c d a b sin(-) sin(-) + sin(-) sin(-) 2 2 2 2 and same(a<->c). This is analogous to the maximal area of a quadrilateral with sides a,b,c,d (or a,b,d,c): sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 with diagonals (a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c Whose formula is this? (Superhero of Alexandria?) And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? The formulas must be doozies. --rwg IMPORTUNATE PERMUTATION

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[math-fun] new repunit (probable) prime [10^270343/9]
by Schroeppel, Richard 16 Jul '07

16 Jul '07

Maksym Voznyy and Anton Budnyy report (on NMBRTHRY) the probable prime repunit number R(270343) (consisting of 270343 1s, base 10). Known repunit primes are R(2), R(19), R(23), R(317), and R(1031). Known PRP repunits are R(49081), R(86453), and R(109297). Nothing is said about intermediate values. I believe I saw a recent report that all P<250K have been examined. Rich

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[math-fun] remorselessly solid angles
by Bill Gosper 15 Jul '07

15 Jul '07

WFL> normalised so that \sum_j (a_ij)^2 = 0 for i = 1,...,n, Now that was a normalization liberals can only dream of. My turn: > So, fresh start: the solid angle at the apex of a regular n-gonal > pyramid with slant edges 1 and height h > ... > = 2 pi - 2 n atan(h t), > > where t = tan pi/n, but explement the middle term (2 pi - acos) > when x< 1/(1+sec pi/n). Should say h < 1/(1+sec pi/n). rwg> There must be some obvious insight for the incredibly simple rhs. WFL>This looks suspiciously like the area formula for the dual spherical polygon: this has angles and sides corresponding to the sides and angles of the of the original polygon (whose angles corrspond to dihedral angles, and sides to apical angles). See Todhunter. Pending a jaunt to the Stanford Math Library, the formula says the following: from the unit southern hemisphere slice off enough of the polar cap to leave a segment of thickness h. Inscribe in-phase regular n-gons on the two circular faces. (The lower n-gon and the sphere center form the pyramid in question.) Then one of the isosceles trapezoidal pyramids with apex at the center and base defined by one side of an n-gon and the corresponding side of the other has apical solid angle 2 atan(h tan pi/n). Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is 2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2 I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2 Plugging into d79 the pyramid's particulars: (c80) subst([a = asin(h),b = 2*%pi/n,c = 2*asin(sqrt(1-h^2)*sin(%pi/n))],%) (d80) <big mess> (c81) trigsimp(factor(trigsimp(factor(rootscontract(tan(d80/2)))))) | %pi | abs(h) |sin(---)| | n | (d81) ----------------- %pi cos(---) n I.e., h tan pi/n. QED. Is someone out there who can say if Todhunter has the solid angle from vertex angles formula cos(c) + cos(b) + cos(a) + 1 (d1) 2 acos ---------------------------- ? a b c 4 cos(-) cos(-) cos(-) 2 2 2 Odditum: the height of a square pyramid with unit slant edges and apex angles v is just sqrt(cos(v)). Apropos the question of a 3D analog of the (n-2) pi formula, there's Descartes' total angular defect formula in which a sum over all the 2D angles of a convex polyhedron gives 4 pi. Not a solid result. (But could it test for convexity?) --rwg

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[math-fun] Correction to solid angles yet again
by Bill Gosper 15 Jul '07

15 Jul '07

I blathered: We can dissect a regular n-gonal pyramid of height h and base circumradius r into n congruent tetrahedra with trihedral components v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., half the apex angle. This gives, after much simplification, two expressions for the solid angle of the apex: 2 %pi 2 %pi 2 %pi 2 cos (---) (2 sin (---) cos(v) + cos(-----)) n n n (d3) n acos(--------------------------------------------- 2 %pi 2 2 %pi sin (---) cos (v) + cos (---) n n 2 %pi - cos(-----)) n n = acos((- 1) n - 1 /===\ 2 %pi 2 | | %pi %pi k 2 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) n | | n n k = 1 (1 - ------------------------------------------------------)) 2 2 %pi n (cos (v) + cot (---)) n Shame on me. The product over a half-period of a polynomial in trigs of angles in arithmetic progression always simplifies, in this case, mondomiraculously: n 2 %pi acos((- 1) (1 - 2 sin (n acot(tan(---) cos(v))))) (!) n The (-1)^n looks very strange. But only because I'm losing it. 1-2 sin^2 x = cos 2x, and simplification continues to an incredible 2 n atan (cos(v) tan(pi/n)), which is wrong, because the rhs(d3) takes the wrong branch when cos v < 1/(1+sec pi/n) = (1-tan^2 2pi/n)/2. It's obviously wrong, since the answer needs to approach 2 pi for small cos v, but acos is normally limited to pi. So, fresh start: the solid angle at the apex of a regular n-gonal pyramid with slant edges 1 and height h is 2 2 (h t - h t + t + 1) (h t + h t - t + 1) n acos(-----------------------------------------) 2 2 2 (t + 1) (h t + 1) n - 1 pi k /===\ cot ---- 2 | | n 2 2 n | | (--------- + h) | | t n k = 1 = acos((- 1) (1 - -----------------------------)) 1 2 n 2 (-- + h ) t 2 t = 2 pi - 2 n atan(h t), where t = tan pi/n, but explement the middle term (2 pi - acos) when x< 1/(1+sec pi/n). There must be some obvious insight for the incredibly simple rhs. --rwg <explement deleted>, but there is a brief CO2 remark at gosper.org/co2.txt

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[math-fun] Letter insertions
by Steve Witham 12 Jul '07

12 Jul '07

"Insert letters to form words" is a nice highway passtime with license plates. Similar lateral feeling. Different section of the brain than those used for "what does this acronym stand for" and "what is this vanity plate about." Maybe that would make a cool FMRI study. Trivial Pursuit, too, when you really don't know, but get the right answer out of the blue. Once I did $ grep inel /usr/dict/words and got this scary poem: inelastic inelegant ineligible ineluctable machinelike sentinel trichinella --Steve p.s. what the hec are hectorems? > > .*H.*R.*E.*M.*S.* > >Dan A: HREMS -> harems, theorems, and ? >Mike S: hectorems. > >From: Bill Gosper <rwmgosper(a)yahoo.com> > >scHwaRtzEneggerisMS? higHeR mathEMaticS! > >hypergeometrics, hectorems, hordeolums, >hypernatremias. > >Some reasons to kick yourself: >theremins, thermochemist(ry), threadworms, threesomes, >chrysanthemums,

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Re: [math-fun] Letter insertions
by Eugene Salamin 12 Jul '07

12 Jul '07

----- Original Message ---- From: Steve Witham <sw(a)tiac.net> To: math-fun(a)mailman.xmission.com Sent: Thursday, July 12, 2007 9:40:56 AM Subject: [math-fun] Letter insertions "Insert letters to form words" is a nice highway passtime with license plates. Similar lateral feeling. Different section of the brain than those used for "what does this acronym stand for" and "what is this vanity plate about." Maybe that would make a cool FMRI study. Trivial Pursuit, too, when you really don't know, but get the right answer out of the blue. Once I did $ grep inel /usr/dict/words and got this scary poem: inelastic inelegant ineligible ineluctable machinelike sentinel trichinella --Steve p.s. what the hec are hectorems? > > .*H.*R.*E.*M.*S.* > >Dan A: HREMS -> harems, theorems, and ? >Mike S: hectorems. > >From: Bill Gosper <rwmgosper(a)yahoo.com> > >scHwaRtzEneggerisMS? higHeR mathEMaticS! > >hypergeometrics, hectorems, hordeolums, >hypernatremias. > >Some reasons to kick yourself: >theremins, thermochemist(ry), threadworms, threesomes, >chrysanthemums, 1 hectorem = 100 rems. A rem is a unit of ionizing radiation dose, equal to the absorption of 1 Joule of energy per kilogram of tissue, and corrected for the biological effectiveness of the type of radiation. Is it just coincidence that this is associated with inel = Idaho National Engineering Laboratory (now called Idaho National Laboratory), a nuclear reactor R&D facility? Gene ____________________________________________________________________________________ Fussy? Opinionated? Impossible to please? Perfect. Join Yahoo!'s user panel and lay it on us. http://surveylink.yahoo.com/gmrs/yahoo_panel_invite.asp?a=7

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Re: [math-fun] Stairway to Heaven?
by Eugene Salamin 11 Jul '07

11 Jul '07

----- Original Message ---- From: Steve Witham <sw(a)tiac.net> To: math-fun(a)mailman.xmission.com Sent: Wednesday, July 11, 2007 10:32:57 AM Subject: [math-fun] Stairway to Heaven? I wrote to the webmaster of the HIPAS site http://www.hipas.alaska.edu/hipasweb/hipas.htm and they have improved the description of their transmitters: The Heater system consists of 8 transmitters capable of conducting amplitude modulation of 100 Hz - 20 kHz and phase modulation of 0 -20 kHz. Each transmitter can transmit up to 150kW at 2.85 or 4.53 MHz on CW mode. The HAARP transmitter is also in the HF band (3 - 30mHz), so neither is anywhere close to 17 Hz! I think a 17 Hz transmitter and antenna need to be quite different from HF far-cousins. In fact I'm having a hard time imagining a 17 Hz directional antenna in Alaska since the wavelength is 18,000 kM. CO2 is only .04% of the atmosphere by volume. Maybe the laser can be tuned to heat CO2. And then, maybe a heated column of CO2 can be a major absorber of 17 Hz radio waves even with an omnidirectionnal transmitter? I only watch EE from the sidelines. --Steve _______________________________________________ The Navy once had Project ELF that transmitted at 76 Hz to deep submarines. Perhaps some of that equipment could be recycled for this new purpose. The 17 Hz is the cyclotron resonance frequency, qB/(2 pi m), for singly charged CO2 molecules in a field of 50 µT. However, as the molecules are pushed to higher altitude (which is a pretty dubious assumption), the field becomes smaller and the resonance frequency drops. It would be nice if someone who has confidence in this project would post some quantitative performance calculations. Gene ____________________________________________________________________________________ Fussy? Opinionated? Impossible to please? Perfect. Join Yahoo!'s user panel and lay it on us. http://surveylink.yahoo.com/gmrs/yahoo_panel_invite.asp?a=7

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[math-fun] Stairway to Heaven?
by Steve Witham 11 Jul '07

11 Jul '07

I wrote to the webmaster of the HIPAS site http://www.hipas.alaska.edu/hipasweb/hipas.htm and they have improved the description of their transmitters: The Heater system consists of 8 transmitters capable of conducting amplitude modulation of 100 Hz - 20 kHz and phase modulation of 0 -20 kHz. Each transmitter can transmit up to 150kW at 2.85 or 4.53 MHz on CW mode. The HAARP transmitter is also in the HF band (3 - 30mHz), so neither is anywhere close to 17 Hz! I think a 17 Hz transmitter and antenna need to be quite different from HF far-cousins. In fact I'm having a hard time imagining a 17 Hz directional antenna in Alaska since the wavelength is 18,000 kM. CO2 is only .04% of the atmosphere by volume. Maybe the laser can be tuned to heat CO2. And then, maybe a heated column of CO2 can be a major absorber of 17 Hz radio waves even with an omnidirectionnal transmitter? I only watch EE from the sidelines. --Steve

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Re: [math-fun] Superduperghost
by Bill Gosper 11 Jul '07

11 Jul '07

Dan A: HREMS -> harems, theorems, and ? Mike S: hectorems. scHwaRtzEneggerisMS? higHeR mathEMaticS! > (The grep search would be something like > > .*H.*R.*E.*M.*S.* > > but no fair doing one!) I did one anyway. Merriam-Webster finds 258, excluding hypergeometrics, hectorems, hordeolums, hypernatremias. Some reasons to kick yourself: theremins, thermochemist(ry), threadworms, threesomes, chrysanthemums, de|enthronements, disheartenments, enshrinements, hemorrhoidectomies|mise|mizes, horsemanship, hysterectomies||. More math: three-dimensional, Khwarezmians! Reasons for medics to kick themselves: thyroidectomies|mise|mizes, nephrectomies||, thromboembolism, choleresteremias, hemispherectomies, hyperglycemias. Mensa specials: whoremaster|monger, Christadelphianisms, harlequinisms, headforemost, headstreams, mithraeums (an anagram of rheumatism), phrasemakers|mongers, shrewmouse. Interesting illegals: white sumpremacist,northern|southern hemisphere, charlie mccarthys, hebrewisms HUH?: schnurkeramiks --rwg ____________________________________________________________________________________ Pinpoint customers who are looking for what you sell. http://searchmarketing.yahoo.com/

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