February 2011 - mathfun@mailman.xmission.com

Re: [math-fun] eta triples again
by Bill Gosper 03 Mar '11

03 Mar '11

Joerg>A collection of (currently) 5416 such identities by Michael Somos is available at http://eta.math.georgetown.edu/ Enjoy! * Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Oct 01. 2010 09:35]: > [...] These identities seem to fail for complex values. E.g., for the simplest, good old aequatio satis abstrusa, In[318]:= Equal[0, 16*(DedekindEta[-((Log[q]*I)/(2*Pi))])^8*(DedekindEta[-((Log[q^4]* I)/(2*Pi))])^16 + (DedekindEta[-((Log[q]*I)/(2* Pi))])^16*(DedekindEta[-((Log[q^4]*I)/(2* Pi))])^8 - (DedekindEta[-((Log[q^2]*I)/(2*Pi))])^24] In[319]:= %318 /. q -> Exp[-\[Pi]*(I + Sqrt[5])] Out[319]= False (*yet*) In[320]:= %318 /. q -> Exp[-\[Pi]*(+Sqrt[5])] Out[320]= 0 == -DedekindEta[I Sqrt[5]]^24 + DedekindEta[(I Sqrt[5])/2]^16 DedekindEta[2 I Sqrt[5]]^8 + 16 DedekindEta[(I Sqrt[5])/2]^8 DedekindEta[2 I Sqrt[5]]^16 Can they be restated to hold more broadly? --rwg BtW, DedekindEta[1/2 + (I Sqrt[5])/2] == ((-1)^(1/24) Sqrt[ Gamma[1/20] Gamma[9/20]])/(2 5^(5/16) GoldenRatio^(1/8) \[Pi]^(3/4))

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[math-fun] eta triple 1,2,4 isn't good old aequatio!
by Bill Gosper 28 Feb '11

28 Feb '11

Except for positive q. For negative q, empirically, triple 1,2,4 becomes (c67) (ETA(-Q),SUBST((-1)^-(1/6)*%%,%%,SUBST(-Q,Q,D34))); 8 16 4 16 eta (- q) eta (q ) 2/3 16 8 4 24 2 (d67) - ---------------------- - %i eta (- q) eta (q ) = eta (q ) 2/3 %i I.e., we need to multiply eta(-q) by e^(-i pi/6) in the usual aequatio, which was d34. Testing numerically, (c68) EXPAND(BFLOAT(SUBST(69/105,Q,%))); (d68) 6.1178890272559794661b-22 %i + 1.1762322592495258282b-10 = 1.1762322592637729936b-10 Testing symbolically, (c64) (ETA(-%E^-%PI),SUBST((-1)^-(1/6)*%%,%%,D56)); 2/3 8 - 4 %pi 16 - %pi (d64) - %i eta (%e ) eta (- %e ) 16 - 4 %pi 8 - %pi 16 eta (%e ) eta (- %e ) 24 - 2 %pi - ----------------------------------- = eta (%e ) 2/3 %i (c66) SUBST([D46,D48,D50],D64); 24 1 24 1 gamma (-) gamma (-) 4 4 (d66) -------------- = -------------- 18 18 16777216 %pi 16777216 %pi where (c69) [D45,D46,D48,D50]; 1 1 gamma(-) gamma(-) - %pi 4 - 2 %pi 4 (d69) [eta(%e ) = -----------, eta(%e ) = --------, 7/8 3/4 3/4 2 %pi 2 %pi 1 1/4 1/12 1 gamma(-) 2 %i gamma(-) - 4 %pi 4 - %pi 4 eta(%e ) = -------------, eta(- %e ) = --------------------] 3/8 3/4 3/4 2 2 %pi 2 %pi The obvious suggestion is that eta(-q) is off by i^(-1/3), both numerically and symbolically. But here are four different expressions for eta(-e^-pi): Out[100]= {((-1)^(1/24)*Gamma[1/4])/(2*Pi)^(3/4), (-1)^(1/24 + I/24)* QPochhammer[-E^(-Pi), -E^(-Pi)], DedekindEta[1/2 + I/2], EllipticTheta[1, Pi/3, (-1)^(1/6 + I/6)]/Sqrt[3]} In[101]:= N[%] Out[101]= {0.9057633419831963 + 0.11924600619519435*I, 0.9057633419831964 + 0.11924600619519436*I, 0.9057633419831964 + 0.11924600619519445*I, 0.9057633419831963 + 0.11924600619519422*I} which seem to refute the i^(-1/3) theory. Could eta be defined wrong? --rwg

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[math-fun] Kevin Brown?
by Christian Boyer 28 Feb '11

28 Feb '11

If somebody knows how to reach Kevin Brown of www.mathpages.com (excellent pages!), please send me a PRIVATE message. The @gte.net address given by www.who.is is not working. Thanks. Christian.

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Re: [math-fun] Euler's crazy pi product
by Bill Gosper 28 Feb '11

28 Feb '11

interesting how if you start with 1 you get a line of slope 1. you can also include all integers. π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ... if I've got that right. For the n-1/2 cases, round up and round down produce divergent products. Round to even gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: > http://math.ucr.edu/home/baez/week127.html finds in > Lennart Berggren, Jonathan Borwein and Peter Borwein, > π: A Source Book, Springer-Verlag, New York, 1997, > "the following weirdly beautiful formula due to Euler, > which unfortunately is not explained:" > (Then how can they call it a sourcebook?--rwg) > > 3 5 7 11 13 17 19 > π/2 = - × - × - × -- × -- × -- × -- × ... > 2 6 6 10 14 18 18 > > "Here the numerators are the odd primes, and the denominators are the > closest numbers of the form 4n+2." > E.g., for 10, 100, ..., 10^6 terms, > In[209]:= > > Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2,10^n]],{n,6}] > > Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, > 3.14157195} > > Why the heck?? Was Euler a preincarnation of Ramanujan? > --rwg

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Re: [math-fun] strange approximations for F(x) = partition function. regarding A000041.
by Bill Gosper 24 Feb '11

24 Feb '11

SP>Hello, thank you for the reference, yes, that talks to me, in other words, some values of Gamma[a/b], Pi^(c/d), Exp[Pi*e/f] and 2^(p/q), a,b,c,d,e,f, p and q integers are well approximated by algebraic numbers, this is the point, this should be explored more isn't ? I am not certain that a table ? a compendium of values like that are all known ? Actualy, considering the simplest possible a,b,c,d,... what is the best example or the most striking evaluation that could exist ? (rwg)> I don't recall anything like these. SP> Also, about a certain algorithm for denesting radicals, is there a solution to this ? this is a very difficult problem, mathematica is quite good for extracting these automatically but it does not solve the problem completely by far. I tried many times to solve this : niet, I do not see how it can be done. best regards, simon plouffe People have claimed knowing how to do this for at least 18 years, yet no one seems to have actually done it. See http://math.stackexchange.com/questions/4680/simplification-of-expressions-… which is my thinly veiled attempt to reestablish contact with oldtime funster Bill Dubuque, who seems to have been driven into reclusiveness by a crank/troll named Archimedes Plutonium. --rwg

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Re: [math-fun] strange approximations for F(x) = partition function. regarding A000041.
by Bill Gosper 24 Feb '11

24 Feb '11

Hi Simon, I would bet that your values are related to this: http://www.mathstat.carleton.ca/~williams/papers/pdf/220.pdf Victor (Miller) YOW! So evaluating etas (and thetas and lambdas and ...) is reduced to evaluating L series! Presumably, a solved problem. I had no idea. Simon, your F*(x) := x^(25/24)/eta[x] . Your F*(e^(-2*Pi/5)) result is equivalent to Out[214]= (4 E^(-5 \[Pi]/6) GoldenRatio*Pi^(3/2))/Gamma[1/4]^2 In[215]:= N[%, 22] Out[215]= 0.2000000000000090844043 which really needs to be explained. Your F*(e^(-2*Pi/5)) radical easily denests, and the approximation is Out[246]= 1/10 (5 + 3 Sqrt[5] - 5^(3/4) (1 + Sqrt[5])) -> (E^(-5 Pi/6) Pi^(3/4))/(2^(1/8) Sqrt[2 - 5^(1/4) + 5^(3/4)] Gamma[5/4]) In[247]:= N[%, 33] Out[247]= 0.0887758501511156259651860669196022 -> 0.0887758501511156259651860669653919 which *really* needs to be explained. (The sqrt(trinomial) was reduced from a quadrinomial^(1/8) by Corey's (still unfinished) denester.) --rwg On Wed, Feb 23, 2011 at 12:11 PM, Simon Plouffe <simon.plouffe(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=simon.plouffe%40gmail.com>>wrote: > Hello, > > I stumbled on these 2 approximations regarding F(x), > the partition function, where p(n) = the number > of partitions of n, as usual. aka A000041. > > F(x) = sum(p(n)*x^n, n=0..infinity): > > Instead I use F*(x) = sum(p(n)*x^(n+1), n=0..infinity): > > Then here is the strange thing, for x = exp(-2*Pi/5) then > the value is 1/sqrt(5), well almost ; the precision is > 13 digits. > > For x = exp(-4*Pi/5) the value is 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) > the > precision is 28 decimal digits. I find this quite surprising. > I was sure it was exact, it is NOT. I verified with large values. > > Also, apparently these are the only 2 examples I have found > within F60 : the Farey fractions up to denominators = 60. > Also when x = exp(-Pi/5) = apparently nothing algebraic of > a low degree. > > caution : do not mistake these values for the standard > F(x) which goes 0 for the exponent too, it is not the > same. > > I added these 2 values in the formulas of A000041 of course. > > Does anybody have an idea why these values just pop out > like that and apparently no other ???! > > Bonne journée. > Simon Plouffe > >

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[math-fun] strange approximations for F(x) = partition function. regarding A000041.
by Simon Plouffe 23 Feb '11

23 Feb '11

Hello, I stumbled on these 2 approximations regarding F(x), the partition function, where p(n) = the number of partitions of n, as usual. aka A000041. F(x) = sum(p(n)*x^n, n=0..infinity): Instead I use F*(x) = sum(p(n)*x^(n+1), n=0..infinity): Then here is the strange thing, for x = exp(-2*Pi/5) then the value is 1/sqrt(5), well almost ; the precision is 13 digits. For x = exp(-4*Pi/5) the value is 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) the precision is 28 decimal digits. I find this quite surprising. I was sure it was exact, it is NOT. I verified with large values. Also, apparently these are the only 2 examples I have found within F60 : the Farey fractions up to denominators = 60. Also when x = exp(-Pi/5) = apparently nothing algebraic of a low degree. caution : do not mistake these values for the standard F(x) which goes 0 for the exponent too, it is not the same. I added these 2 values in the formulas of A000041 of course. Does anybody have an idea why these values just pop out like that and apparently no other ???! Bonne journÃ©e. Simon Plouffe

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[math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients
by Fred lunnon 23 Feb '11

23 Feb '11

Everybody can solve these, can't they --- just like quadratic equations ... Here's a simple example: solve for x,y,z, b z - c y = f , c x - a z = g , a y - b x = h . Oops --- the determinant vanishes: | c 0 -a | | -b a 0 | = 0 . | 0 -c b | No solution, says our CAS equation solver, firmly. Except of course, if the RHS also vanishes there is obviously a 1-dimensional homogeneous solution; but simply substituting f,g,h -> 0 should encourage our CAS to disgorge some version of that too. End of story? Not quite: you see, adding a equn1 + b equn2 + c equn3 eliminates all variables, leaving a f + b g + c h = 0 ; and in my particular application, this constraint happens to be satisfied by the coefficients. Now some head-scratching unearths a 3-dimensional (in general) solution spanned by the symmetric linear basis [ x -> -h/b, y -> 0, z -> +f/b ] , [ x -> +g/c, y -> -f/c, z -> 0 ] , [ x -> 0, y -> +h/a, z -> -g/a ] . In particular, summing eventually delivers the symmetric solution x -> (g/c - h/b)/3 , y -> (h/a - f/c)/3 , z -> (f/b - g/a)/3 , which I knew ought to exist by the symmetry of the equations. I have encountered this type of problem --- solving SLE's on a manifold --- on a number of occasions, without ever finding an effective solution algorithm. Surely it must have been considered before --- any pointers, anyone? Is it reasonable to expect that the next generation of CAS solvers might inform us about constraints on our coefficients under which such equations might possess extra solutions? Fred Lunnon

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[math-fun] Time-reversed laser absorbs coherent light
by Henry Baker 22 Feb '11

22 Feb '11

FYI -- If you used a laser to "push"/accelerate such a device in outer space, it would continue accelerating until the doppler frequency matched the absorption frequency, at which point it would stop accelerating & explode ? http://www.nytimes.com/2011/02/22/science/22laser.html February 21, 2011 A Dazzling Show Inside a Laser, but a Vacuum of Light Outside By HENRY FOUNTAIN In an elegant melding of theoretical and experimental physics, scientists at Yale University have taken the basic function of a laser and flipped it around Â producing a device that absorbs, rather than emits, a beam of light. The device, which the scientists call a Âcoherent perfect absorberÂ or, more popularly, an anti-laser, may lead to the development of new kinds of switches, filters and other components that could be useful in hybrid optical-electronic computers under development, among other applications. A. Douglas Stone, a theoretical physicist at Yale, developed the concept of a backward-running laser in a paper in Physical Review Letters last spring. The actual device, described in a paper published last week in Science, was created in the laboratory of a laser physicist, Hui Cao. In a laser, energy is pumped into a medium Â which can be a solid, liquid or gas Â between two mirrors, stimulating the emission of photons that are coherent, or of the same frequency and phase. The photons reflect back and forth between the mirrors, resulting in amplification of the light. ÂYou put energy into it, and some of that energy gets converted into that beautiful coherent light beam,Â Dr. Stone said. In his theoretical work, Dr. Stone said, he made use of the fact that the equations that describe how a laser works have certain symmetrical properties. ÂIf you can make a laser of a certain type, the equations say you can make a reverse device as well,Â he said. An anti-laser uses mirrors, too, but the other components are the reverse of a laser. The medium that provides amplification is replaced with one that provides absorption, and the outgoing light beam is replaced with an incoming one. (This light needs to be coherent, so it takes a laser to make an anti-laser.) The incoming beam is split in two, and hits the medium from two sides. The photons bounce around between the mirrors and interfere with one another, eventually wiping themselves out in a flurry of electrons and heat. The experimental device absorbed about 99.4 percent of the light. In theory, an anti-laser should be able to absorb 100 percent. ÂItÂs a one-way trap for light,Â Dr. Stone said. Dr. Cao said the device they built was relatively simple, using silicon as the absorptive medium and a couple of ÂbadÂ mirrors. ÂBut we should be able to get coherent perfect absorption in more complicated systems,Â she said. Eventually it may even be possible to make an ÂantiÂ version of a so-called random laser, in which the medium is highly disordered and there are no mirrors. The experimental device works in the near-infrared, outside of the visible spectrum. But Dr. Stone said that in principle anti-lasers would not be limited in terms of frequency. ÂWe could move it into the visible, or the farther infrared,Â he said. ÂItÂs definitely possible to engineer this across the whole range.Â Stefano Longhi, a physicist at the Polytechnic Institute of Milan in Italy who was not involved in the work, said the anti-laser was an Âimportant achievementÂ that was Âexciting and surprising to the scientific community.Â He said one important characteristic of the device is that the absorption could be turned on or off. This might make anti-lasers extremely useful as optical switching devices. A device that absorbs light perfectly might be considered ideal for solar energy applications, but Dr. Longhi said this is not the case. Sunlight is not coherent, and an anti-laser will not work with incoherent light, he said. A physicist would describe the device as a Âtime-reversedÂ laser, since the symmetrical properties are related to the concept of time reversal. But Dr. Stone said he thought the term anti-laser was a better description for nonscientists, so that no one would think the device had anything to do with time travel. But even Âanti-laserÂ is problematic, he noted. ÂI donÂt want people to think this is some kind of laser shield,Â he said. ÂIf R2-D2 had our anti-lasers, it would be melted into a puddle.Â

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Re: [math-fun] Aperiodic tiling of hyperbolic plane?
by Dan Asimov 21 Feb '11

21 Feb '11

According to Wikipedia: << There are no finite-faceted regular tessellations of hyperbolic space of dimension 5 or higher. There are 5 regular honeycombs in H5 with infinite (Euclidean) facets or vertex figures: {3,4,3,3,3}, {3,3,4,3,3}, {3,3,3,4,3}, {3,4,3,3,4}, {4,3,3,4,3}. >> So it appears that the choices are severely limited. --Dan Mike Stay wrote: << Has anyone worked out aperiodic tiles for the hyperbolic plane? Is there something analogous to Penrose tiles? Penrose tiles are a 2-d projection of a 5-d hypercube lattice; in 3d, there's a "cubic" tiling of hyperbolic space with dodecahedra where eight dodecahedra meet around a central vertex like cubes do in Euclidean space: http://reperiendi.files.wordpress.com/2011/01/dodecahedral.jpg If we took some "hypercubic" tiling of 5d hyperbolic space and projected down to 2d hyperbolic space, what would we get? >> Those who sleep faster get more rest.

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