June 2010 - mathfun@mailman.xmission.com

Re: [math-fun] And Beano was his name-0
by Dan Asimov 20 Jun '10

20 Jun '10

<< The web tells me that a Carl Leffler, a Columbia math prof, was contracted around 1930 to produce 6000 cards with all with unique ³Bingos², and that the effort drove him insane, despite his ultimately billing at a rate of $100/card (eg http://www.strangelife.com/bingodoc/bingohist.html) >> I haven't found any reliable documentation about the claim he went insane (which could have a lot of meanings). One site said the price per card gradually rose, eventually to $100 as Leffler found it increasingly hard to generate new cards, presumably because it was increasingly hard to verify that a new card wasn't the same as an old card. Which suggests he didn't bother to come up with an algorithm before proceeding, or else his algorithm was flawed. Which would make anyone grumpy, but not necessarily insane. He was reportedly elderly at the time, and senile dementia may have been just as common in those days as now. So he may well have "gone insane", but it seems unlikely to me that this was caused by the bingo project. --Dan Even though kleptomaniacs can't help themselves, they do.

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[math-fun] And Beano was his name-0
by Marc LeBrun 20 Jun '10

20 Jun '10

The web tells me that a Carl Leffler, a Columbia math prof, was contracted around 1930 to produce 6000 cards with all with unique ³Bingos², and that the effort drove him insane, despite his ultimately billing at a rate of $100/card (eg http://www.strangelife.com/bingodoc/bingohist.html) But the web says a lot of things, and my Googling only seems to churn up the same stuffis any of it true?

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Re: [math-fun] Complex projective spaces
by Dan Asimov 19 Jun '10

19 Jun '10

Gene wrote: << Can someone give a geometrical or topological description of the shape of the complex projective space CP^n ? While CP^1 is the 2-sphere, some searching reveals a few results. CP^n is simply connected (in contrast to RP^n). CP^n for n>1 is not homeomorphic to the sphere S^2n. The quotient of CP^2 by complex conjugation is S^4. I'm writing a paper on the optics of polarized light, and I need to understand CP^2 and CP^3. >> In general, CP^n is the union of closed 2k-balls for 0 <= k <= n: CP^n = e_0 u e_2 u e_4 u ... e_2n. The e_2k are glued to one another in order with the glueing maps independent of which CP^n we're talking about. So to get CP^n from CP^(n-1), we take a CP^(n-1) and and e_2n, whose boundary is S^(2n-1). We then glue this S^(2n-1) to CP^(n-1) by thinking of S^(2n-1) as {(z_1,...,z_n) in C^n | |z_1|^2 + ... + |z_n|^2 = 1}. Then one definition of CP^(n-1) is to quotient out S^(n-1) by the action of the circle: (z_1,...,z_n) ~ (c z_1,...,c z_n) where c is an arbitrary complex number with |c| = 1. The equivalence classes constitute CP^(n-1). Then, to construct CP^n from CP^(n-1) you glue on the e_2n by this very same quotient map from S^(2n-1) = bd(e_2n) -> CP^(n-1). So to get CP^2 from CP^1 = S^2, you take a 4-ball e_4 and glue it to S^2 by mapping its boundary S^3 to S^2 via the Hopf fibration. If you start out from any point of the (real) 4-manifold CP^2 and march out in all directions (there is an S^3 of them), then you just get a larger and larger 4-ball, until suddenly at one stage you just get an S^2 (and that's the whole thing), with the sphere of directions at a certain radius being identified by that circle action, i.e., the Hopf fibration. CP^3, and all higher CP^n's, are described analogously. Despite this apparently asymmetric definition, it's clear from the quotient description that in fact each point of CP^n is like any other; there is an self-isometry carrying every point to any other. It's not isotropic as a real manifold, but as a complex manifold, all complex "directions" at any point look like every other. --Dan Even though kleptomaniacs can't help themselves, they do.

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[math-fun] Complex projective spaces
by Eugene Salamin 19 Jun '10

19 Jun '10

Can someone give a geometrical or topological description of the shape of the complex projective space CP^n ? While CP^1 is the 2-sphere, some searching reveals a few results. CP^n is simply connected (in contrast to RP^n). CP^n for n>1 is not homeomorphic to the sphere S^2n. The quotient of CP^2 by complex conjugation is S^4. I'm writing a paper on the optics of polarized light, and I need to understand CP^2 and CP^3. -- Gene

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[math-fun] Replicating Life pattern
by Bernie Cosell 16 Jun '10

16 Jun '10

<http://tinyurl.com/272pyfx> IF YOU found a self-replicating organism living inside your computer, your first instinct might be to reach for the antivirus software. If, however, you are Andrew Wade, an avid player in the two-dimensional, mathematical universe known as the Game of Life, such a discovery is nothing short of an epiphany. When Wade posted his self-replicating mathematical organism on a Life community website on 18 May, it sparked a wave of excitement. "This is truly ground-breaking work," wrote a fellow Life enthusiast, Adam Goucher, on the website Game of Life News. "In fact, this is arguably the single most impressive and important pattern ever devised." [...] /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] "Utilizing Infinite Loops to Compute an Approximate Value of Infinity"
by Henry Baker 15 Jun '10

15 Jun '10

I've commented elsewhere that access to terabyte (& beyond) storage poses a problem: Simply accessing every single byte of a terabyte disk is a non-trivial exercise. Also, many storage & network devices consider an error rate of 1 in 10^12 quite acceptable, so you can't be sure that you've actually read all of these bytes correctly. Re incrementing integers: The usual method of incrementing bignums in Lisp (& many other interpreted languages) is terribly expensive, because it copies the entire bignum on each increment & then causes lots of garbage collections. This is one of the reasons why I investigated implementing bignums which were forced to have a reference count of one -- i.e., guaranteed to be unshared. Then you can "increment in place" & avoid copying. At 01:13 PM 6/15/2010, Paul Reiners wrote: >Kind of funny: > >http://www-user.slac.stanford.edu/jimstan/Infinity.htm > >The author should probably have tried using an Amiga 500, rather than a >Commodore 64, though.

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[math-fun] "Utilizing Infinite Loops to Compute an Approximate Value of Infinity"
by Paul Reiners 15 Jun '10

15 Jun '10

Kind of funny: http://www-user.slac.stanford.edu/jimstan/Infinity.htm The author should probably have tried using an Amiga 500, rather than a Commodore 64, though.

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Re: [math-fun] Ramanujan's Hardy-stumping CFs
by Bill Gosper 14 Jun '10

14 Jun '10

> > > (Let's try this with Dr. Stay's BadRapBane. I hope there was plenty of > blood spilled at the the GMail design meetings where the current behavior > was > defaulted.) > Weird. The Courier font went way tiny. > > Theorem 4.5 in Bruce's http://www.math.uiuc.edu/~berndt/articles/rrcf.pdf > indeed reduces the problem of evaluating the four Pochhammers in f(q)/g(q) > to two ratios of "second differences" of the form > eta(q^2)^2/eta(q)/eta(q^4), > q:=e^(pi alpha). These are even easier to evaluate than plain etas--their > gamma components have been divided out, and they have a lower algebraic > order. > > This cleverly dodges the problem of evaluating naked (q;q)_oo, let alone f > and g, > which are of the form (q^r,q^(1-r);q)_oo for rational r. I can get these > (even > quadratic surd r?) by guessing plus lattice reduction, but is there a > theory? > And does the theory preclude closed forms for naked (q^r;q)_oo, 2r > noninteger? > It resembles the problem of evaluating Gamma(r). > --rwg > > Ah, here's what I was after--an expression of theta_1(pi/5,q) in terms of > theta constants: > > EllipticTheta[1, \[Pi]/5, E^(\[Pi]^2/(5 Log[q]))]==(Sqrt[5] r^( > 1/4) (11 - 11 r + 5 r^2 - > r^3 + (5 + (-4 + r) r) Sqrt[5 + (-2 + r) r])^(1/10) > EllipticTheta[4, 0, q^5] Sqrt[-Log[q]])/(2^(1/10) > Sqrt[\[Pi]] (-1 + r)^(3/5)) > > (after an unbelievable struggle against FullSimplify, which can't even > confirm > algebraic equivalence with my starting expression) with > > {Rule[s, Sqrt[r^3 - 2*r^2 + 5*r]], > Rule[r, ((EllipticTheta[2, 0, q]* > EllipticTheta[3, 0, q])/(EllipticTheta[2, 0, q^5]* > EllipticTheta[3, 0, q^5]))]} > > This reduces evaluating the R-R CF to evaluating etas. Theoretically, just > two, > since, if r is rational, eta(q^r) is a (probably brutal) algebraic function > of > eta(q) and eta(q^2). e.g., > 256*eta[q]^2*eta[q^10]^24 == > 5*eta[q^5]^18*eta[q^10]^8 + 10*eta[q]^6*eta[q^5]^12*eta[q^10]^8 + > eta[q]^12*eta[q^5]^6*eta[q^10]^8 - eta[q]^2*eta[q^5]^24 . > (This is a *tiny* one.) > > Mma's woeful Theta machinery was of no help on this, except to speed up > some of > the failed brute force approaches. The solution was in my old q-Trig > paper, whose > abstract prophesied: > > We introduce a q-generalization of the sine and cosine functions, related > to the θ functions, but (as revealed by computer experiments) possessing > addition and multiplication formulas more analogous to those of ordinary sin > and cos. These formulas then contribute identities to θ theory, and hint of > a more natural formulation of θ functions as outgrowths of elementary > functions. Nevertheless, this paper can be read without knowledge of θ > functions--it was certainly written that way. > > The paper was soundly rejected by the august SYMSAC '76 Computer Algebra > seance, but > later won high praise from Dyson. > > Three by-products: > Equal[EllipticTheta[1, Pi/10, > E^((Pi^2)/(5*Log[q]))], ((EllipticTheta[4, 0, q^5]*Sqrt[5]* > Sqrt[-Log[q]]*(s*Sqrt[s^2 - 4*r] - s^2 + 2*r + 2)^(1/5))/(2^(1/ > 5)*(r - 1)^(1/5)*(s^2 - r^2 - 2*r - 1)^(1/5)*Sqrt[Pi]))] > > Equal[ > EllipticTheta[2, Pi/10, > E^((Pi^2)/(5*Log[q]))], ((EllipticTheta[4, 0, q^5]*Sqrt[5]* > Sqrt[-Log[q]]*(s*Sqrt[s^2 - 4*r] - s^2 + 2*r + 2)^(1/5)* > Sqrt[((s*Sqrt[s^2 - 4*r] + (1 - 2*r)*s^2 + 2*r^3 + > 2*r^2)/((s^2 - r^2 - r)*Sqrt[s^2 - 4*r] - > s^3 + (r^2 + r + 2)*s))])/(2^(1/5)*(r - 1)^(1/5)*(s^2 - > r^2 - 2*r - 1)^(1/5)*Sqrt[Pi]))] > > Equal[ > EllipticTheta[2, Pi/5, > E^((Pi^2)/(5*Log[q]))], ((2*EllipticTheta[4, 0, q^5]*Sqrt[5]* > Sqrt[-Log[ > q]]*((-s^4 + (2*r^2 + 3*r)*s^2 - r^4 - 2*r^3 - r^2)* > Sqrt[s^2 - 4*r] + > s^5 + (-2*r^2 - 5*r)*s^3 + (r^4 + 6*r^3 + 5*r^2)*s)^(3/ > 5))/((s^2 - r^2 - 2*r - 1)^(1/ > 5)*((r - 1)^(1/5)*(2^(3/5)*s^2 - 2^(3/5)*r^2 - 2^(3/5)*r)* > Sqrt[s^2 - > 4*r] + (r - 1)^(1/5)*((2^(3/5)*r^2 + 3*2^(3/5)*r)*s - > 2^(3/5)*s^3))*Sqrt[Pi]))] > > Besides its more serious failing on [EllipticTheta[2, Pi, q] , 0 < q < 1], > FullSimplify mysteriously converts Sqrt[-5*Loq[q]] to Sqrt[Log[1/q^5]] . > --rwg > > > On Fri, Jun 11, 2010 at 3:55 PM, <berndt(a)math.uiuc.edu> wrote: > >> In his notebooks and lost notebook, Ramanujan records many special values >> for the Rogers-Ramanujan continued fraction. Perhaps the best source for >> Ramanujan's contributions is Chapter 2 of Ramanujan's Lost Notebook, Part >> I, by George Andrews and myself. In this chapter you will find a general >> formula that Ramanujan might have used as well as proofs of all the values >> of the c.f. found in his lost notebook. In a paper I published with Heng >> Huat Chan and Liang-Cheng Zhang in Crelle in 1996, we presented a general >> method for evaluating the Rogers-Ramanujan continued fraction. We also >> prove a theorem showing that the values at a certain class of arguments >> containing all those mentioned above, up to a possible small fractional >> power of 2, are units in the algebraic number fields in which they lie. >> My paper with Chan and Zhang as well as some other papers on the >> Rogers-Ramanujan continued fraction are available at my web-site. >> >> Best wishes, >> Bruce Berndt >> >> >> >> >> > I hadn't realized how directly connected they are to the >> Rogers-Ramanujan >> > identities: >> > If f(q):=(q,q^4;q^5)_oo and g(g):=(q^2,q^3;q^5)_OO, the reciprocals of >> the >> > g.f.s for the >> > +-1 (resp +-2) mod 5 partitions, then >> > 5 5 >> > f(q) 1 2 (q ;q )_oo >> > ---- = -------- = f(q) --------- >> > g(q) q (q;q)_oo >> > 1 + -------- >> > 2 >> > q >> > 1 + ------- >> > 3 >> > q >> > 1 + ----- >> > ... >> > >> > which appears to come out an <algebraic #>/q^(1/5) when q = e^(pi r), r >> a >> > quadratic surd (in the >> > left halfplane) "by an appeal to the theory of elliptic theta >> functions" >> > (G. Andrews: The >> > Theory of Partitions, p105 :-). Ramanujan sent Hardy the cases q = >> e^(-2 >> > pi) and - e^-pi, >> > which presumably give the nicest algebraics, >> > >> > f(e^-2 pi) pi 3 pi 2 pi/5 >> > ---------- = 4 sin -- sin ---- e , >> > g(e^-2 pi) 20 20 >> > >> > whereas the nicest cf is presumably >> > >> > f(e^-pi) sqrt(40 + 18 sqrt(5) - 25 5^(1/4) - 11 5^(3/4)) - sqrt(36 + >> > 18 >> > sqrt(5) - 25 5^(1/4) - 11 5^(3/4)) pi/5 >> > -------- = >> > ------------------------------ ------------------------------ >> ------------------------------------- >> > e . >> > g(e^-pi) 2 >> > >> > ~ 0.958650181596. >> > >> > Now we know why Ramanujan didn't send Hardy this one. >> > >> > I'm really interested in how to get these theta special values (other >> than >> > by numerical searching), but >> > Andrews seems to cite only combinatoric references rather than (theta) >> > number-theoretic. In particular, >> > where do I learn that f and g can be evaluated separately, e.g. here in >> > terms of e^(pi/60) because >> > >> > -pi -pi - pi/6 10 >> > f(e ) g(e ) = e sqrt ------------------ . >> > 5^3/4 - 5^(1/4) +2 >> > >> > Other forms of f and g are >> n >> > (5n-3) n/2 >> > Sum (-) >> > q >> > 1 Theta_4(7 i log(q)/4, q^(5/2)) n >> > f(q) = ----------- = - q ------------------------------ = >> > -------------------- , >> > n^2 (q^5;q^5)_oo >> > (q^5;q^5)_oo >> > q >> > Sum ------- >> > n>0 (q;q)_n >> > >> > the "gofigurate number theorem", and >> > n >> > (5n-1) n/2 >> > Sum >> > (-) >> > q >> > 1 2 Theta_4(9 i log(q)/4, q^(5/2)) n >> > g(q) = ------------ = - q ------------------------------ = >> > -------------------- . >> > (n+1) n (q^5;q^5)_oo >> > (q^5;q^5)_oo >> > q >> > Sum ------- >> > n>0 (q;q)_n >> > >> > f(q) and g(q) become Theta_1(pi/5, Q) under Jacobi's transformation. If >> > we >> > can express these in terms of >> > Theta_s(0,p) ("Theta constants") we'd have a nearly automatic source of >> > special values because the Theta >> > constants all come out as etas, at which we're pretty good. I can do >> > Theta_1(pi/10, Q), but so far no >> > luck with pi/5. >> > --rwg >> > >> >> >> -- >> Bruce Berndt >> Dept. of Mathematics >> University of Illinois >> 1409 West Green St. >> Urbana, IL 61801 >> phone: 217-333-3970 (office) >> fax: 217-333-9576 >> >> My web-page was updated on August 28, 2009. >> http://www.math.uiuc.edu/~berndt/ >> Please inform me of any mistakes or problems. >> >> My favorite composers (in no particular order): Bach, Purcell, Vaughan >> Williams, Sibelius, Ravel, Schubert >> >> >> >

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[math-fun] Most transcendental number?
by Kerry Mitchell 14 Jun '10

14 Jun '10

I've read that phi (~ 1.618, (1+sqrt(5))/2) is the "most irrational" number because of how poorly it is approximated by rational numbers. I assume that this is because it's continued fraction expansion is all 1's. Is there a sense in which some number is the "most transcendental" number? If so, what would that number and that meaning be? Kerry Mitchell -- lkmitch(a)gmail.com www.kerrymitchellart.com http://spacefilling.blogspot.com/

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Re: [math-fun] Intuitively simple trig functions
by Dan Asimov 12 Jun '10

12 Jun '10

Joshua wrote: << On Sat, Jun 12, 2010 at 1:04 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > I recently gave a talk to a bunch of smart youngsters about complex numbers, > and was unable to find a truly graceful way to explain why > > exp(ix) = cos(x) + i sin(x), > > (without deriving their Taylor series). > > So if anyone knows a way to see this, I'd love to know it. > > But of course that would require giving exp a meaning on the > imaginaries. You need some physics, so that may lose some youngsters, but I find the first chapter of Tristan Needham's Visual Complex Analysis to give a very nice way of thinking about complex numbers addition, multiplication, and exponentiation, and it naturally yields this identity. >> If you can sketch the derivation, that would be great. If on the other hand it requires reading the first chapter of that book, then I'd just as soon derive the Taylor series. Also: A physics derivation would surprise the heck out of me, though it would certainly be interesting. Still, what I'm really looking for is a math derivation. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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