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[math-fun] All Factors of n
by Eric Angelini 19 Jul '07

19 Jul '07

Hello Math-Fun, does someone know of an on-line factorization applet showing ALL divisors of n (not only the prime ones) Not this applet, for instance : http://www.merriampark.com/factor.htm Neither this famous one : http://www.alpertron.com.ar/ECM.HTM Thanks, Best, É.

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RE : [math-fun] All Factors of n
by Eric Angelini 19 Jul '07

19 Jul '07

Thanks to Edwin, Neil and Mike -- this is beautiful ! Best, É. ________________________________ De: math-fun-bounces+eric.angelini=kntv.be(a)mailman.xmission.com de la part de Edwin Clark Date: jeu. 19/07/2007 21:25 À: math-fun Objet : Re: [math-fun] All Factors of n On Thu, 19 Jul 2007, Eric Angelini wrote: > > does someone know of an on-line factorization applet showing > ALL divisors of n (not only the prime ones) Use for example the command Divisors(12345678); at this site http://magma.maths.usyd.edu.au/calc/ and click on evaluate. You will get [ 1, 2, 3, 6, 9, 18, 47, 94, 141, 282, 423, 846, 14593, 29186, 43779, 87558, 131337, 262674, 685871, 1371742, 2057613, 4115226, 6172839, 12345678 ] Total time: 0.350 seconds, Total memory usage: 6.46MB _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Re: solid conclusion(?)
by Bill Gosper 18 Jul '07

18 Jul '07

> 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? Brahmagupta's (cyclic case). His formula makes it clear that the area is maximized when the vertices lie on a circle. >And what about arbitrary pentagons, etc? Are the maximal areas fixed >w.r.t. permuting the sides? If the area is maximized when the vertices lie on a circle, we can freely swap adjacent sides, preserving the chord (and circumscribing) arc, and thus area. --rwg

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Re: [math-fun] ENIAC and maple : a simple test
by Eugene Salamin 18 Jul '07

18 Jul '07

Bill Gosper asks: Why does everybody ignore Macsyma? Perhaps it's because, thanks to Dick Petti and the other former Macsyma Inc. stockholders, Macsyma has self-destructed. You can't buy the software (macsyma.com is now an entertainment site), and it is being maintained and upgraded only by unofficial programmers, and yet it cannot be open sourced. What would be the legal status of using a Macsyma that was not purchased years ago from Macsyma Inc? If a project discloses its use of Macsyma, will its members be subject to legal action? Could funding be found to create an open source computer algebra system? Extract the best ideas out of Macsyma, Maple, Mathematica, Mathcad, etc., chuck the ancient excess baggage, and make a fresh start. Gene ____________________________________________________________________________________ Be a better Heartthrob. Get better relationship answers from someone who knows. Yahoo! Answers - Check it out. http://answers.yahoo.com/dir/?link=list&sid=396545433

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Re:[math-fun] ENIAC and maple : a simple test
by Bill Gosper 18 Jul '07

18 Jul '07

Simon Plouffe wrote: > I was reading an article on the ENIAC, that monster could add 100,000 10 digits numbers in 1 second. At the time it was impressive in 1946 of course. I made this simple test, a naive test. I took maple 11, with a 3.0 Ghz PC pentium. ##################################### Digits:=10: som:=0: for i from 1 to 1e99 do: som:=som+1.0/i: if i mod 100000 = 0 then lprint(i,som) fi: od: ##################################### It displays the partial sum of the harmonic series every 100000 terms. this is not optimized, a lazy program. >Well, this is the bad news. That loop made on a modern computer with one of the top programs goes at the SAME speed. It takes 1 second to display each step. Mma seems comparable: Timing[Block[{s = 0}, For[{n = 1}, n <= 100000, s += 1/N[n++, 10]]; s]] {0.906 Second, 12.09014613} Macsyma (using 16 digit doublefloats instead of 10 digit bigfloats): (c4) (showtime:true,block([s:0],for n thru 100000 do s:s+1.0d0/n,s)) Time= 1594 msec. (d4) 12.0901461298633d0 (c5) harm(n):=block([s:0.0d0],mode_declare([n,k,s],float),for k thru n do s:s+1.0d0/k,s) Time= 0 msec. (d5) harm(n):=block([s:0.0d0],mode_declare([n,k,s],float), 1.0d0 for k thru n do s : s + -----, s) k (c6) harm(100000.0) Time= 1672 msec. (d6) 12.0901461298633d0 BUT (c7) compile(harm) time= 0 msec. (d7) [harm] (c8) harm(100000.0) Time= 78 msec. (d8) 12.0901461298633d0 Why does everybody ignore Macsyma? It should be mentioned that these were run in conflict with a second, rogue Mma process trying to FactorInteger repunit 71. Pausing continued to hog cpu, and subsession won't enter, so I just whacked the process priority. --rwg

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[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 15 Jul '07

15 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 14 Jul '07

14 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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