August 2013 - mathfun@mailman.xmission.com

[math-fun] Q about a (planar-)partition bijection
by Joerg Arndt 16 Aug '13

16 Aug '13

Hesitatingly, I'll dare ask a question of rather specific nature: The generating function of planar partitions is 1 / prod(n>=1, (1 - q^n)^n ) Now that is also the g.f. of (the usual) partitions with one sort of 1, two sorts of 2, three of 3, etc. I am the only person on this planet that cannot figure out a (natural, constructive) bijection? Best, jj

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Re: [math-fun] rationalize exp(%i*x), for floating point x.
by Henry Baker 15 Aug '13

15 Aug '13

What a _great_ picture! It looks like the Moon or some other moon, pitted with meteor craters. At 07:21 PM 8/15/2013, Fred Lunnon wrote: > Tracked it down: plot shows rational points [x/w, y/w, z/w] >on unit sphere with w^2 + x^2 + y^2 + z^2 < 2 s^2 , >where s = 199 . See > > https://www.dropbox.com/s/7u73fm17o5av7fd/rat_pts_sph.gif > > Note the ominously large "no-man's land" disc centred on an axis >(I presume); and the intriguing pattern of (orthogonal?) circles traced >by the points surrounding its boundary. > >Fred Lunnon > >On 8/16/13, rcs(a)xmission.com <rcs(a)xmission.com> wrote: >> >> 2-phi (= 1/phi^2 = .382-) is an example. >> The typical convergent is Fib(n)/Fib(n+2), e.g. 34/89. >> Fib(odd) is the sum of two squares, so at least half of >> the convergents for 2-phi fill the bill. >> >> Rich >> >> --------- >> Quoting Victor Miller <victorsmiller(a)gmail.com>: >> >>> In this paper by Glyn Harman: >>> http://matwbn.icm.edu.pl/ksiazki/aa/aa53/aa5323.pdf he proves (on page 210) >>> that almost all real numbers have infinitely many convergents from their >>> continued fraction expansion whose numerators and denominators are both >>> sums of two integral squares. >>> >>> However, this theorem appears not to be able to address any individual real >>> number! >>> >>> Victor >>> >>> On Thu, Aug 15, 2013 at 3:53 PM, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: >>> >>>> On 8/15/13, Olivier Gerard <olivier.gerard(a)gmail.com> wrote: >>>> > On Tue, Aug 13, 2013 at 2:31 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >>>> > >>>> >> It occurred to me that this is a classical problem of factoring with Gaussian integers. >>>> >> >>>> >> The best 'denominators' are the most composite of Gaussian integers, so that you have the widest >>>> >> variety of angles represented. >>>> >> >>>> >> In particular, pick denominators which are the product of 2 with all of >>>> >> the (rational integer) primes of the form 4n+1. >>>> >> >>>> >> Turn on a light at the origin of the complex plane, put up thin barriers at every Gaussian integral point, >>>> >> and distribute some photograhic film around the circumference of a circle of radius r. Wherever the >>>> >> light shines through the most, those are angles that aren't well >>>> >> represented. In particular, I'm interested >>>> >> in the largest _gap_ in this circle for a given radius r; the size of that >>>> >> gap indicates the worst case for that radius r. >>>> > >>>> > I did a few things close to what you describe. It can be downloaded here >>>> > >>>> > http://list.seqfan.eu/extfiles/Gaussian%20Shadow.pdf >>>> > >>>> > I draw a thin blue line through every Gaussian integer (here with absolute value less than 120) >>>> > and I display a corona between two concentric circle (here 50 and 35) so >>>> > that one can appreciate visually the resulting density. >>>> > >>>> > http://list.seqfan.eu/extfiles/Gaussian%20Prime%20Shadow.pdf >>>> > >>>> > is the same idea, restricted to Gaussian primes. >>>> > >>>> > The angles less represented are the ones close to exact rational approximation, >>>> > it reminds of the noman's land around a potent bacterial colony. You find similar >>>> > isolation around algebraic integers with small coefficients. >>>> > >>>> > Olivier >>>> >>>> I did the same thing for the sphere quite recently: the "no-mans lands" are >>>> circular discs instead. [Unfortunately, I have no recollection of why I did it; >>>> so cannot track down the --- rather pretty --- pictorial results.] >>>> >>>> It does emphasise --- relevantly to what Henry is doing, perhaps --- >>>> that points equidistant from the centre and circumference of such a >>>> disc are very badly approximated by rational points! >>>> >>>> WFL

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Re: [math-fun] pythagorean rationals & approximations by them
by Henry Baker 15 Aug '13

15 Aug '13

If I understand you correctly, you're recommending the following sequence of operations: To find a Pythagorean approximation for floating point (x+iy)/|x+iy|. Denote atan2(y,x) by phi. So tan(phi)=y/x, cos(phi)=x/sqrt(x^2+y^2), sin(phi)=y/sqrt(x^2+y^2). What we really want is a rational approximation to tan(phi/2). tan(phi/2) = sin(phi)/(1+cos(phi)) = y/(sqrt(x^2+y^2)+x), or = (1-cos(phi))/sin(phi) = (sqrt(x^2+y^2)-x)/y, depending upon the sign of x, so we get the most accurate answer. We then use continued fractions to produce a '1/2-length' rational approximation to t=tan(phi/2). The Pythagorean result is ((t^2-1)+2it)/(t^2+1), which is 'full length'. --- I ran this procedure on the test case: cos(phi)+i*sin(phi), where phi=2*%pi/100000. The best my previous procedure was able to achieve was 506,605,917 + 31,831*i, which is _not_ Pythagorean. However, the above procedure produces (using one of the rational continued fraction approximations for tan(phi/2)): 506,606,280 + 31,831*i, which _is_ Pythagorean, with hypotenuse 506,606,281 !! So maybe there is hope for a 'Pythagorean number system' after all ! I now need to check all 100,000 points -- phi=2pik/100000, k=0..99999 -- to make sure that they can all be represented with reasonable size denominators. At 01:46 PM 8/14/2013, Warren D Smith wrote: >Henry Baker wanted to approximate reals x by "pythagorean rationals" > x =approx= a/b >with squareroot(a^2+b^2)=integer. > >Obviously, yes, any such pythagorean (a,b) does have a continued fraction >(namely just use the CF for a/b); and as Asimov remarked the set of >rational points on the unit circle is dense, hence any x can be approximated >arbitrarily closely by a pythagorean rational. > >The pythagorean rationals form a group in the sense that any two rational points >on the unit circle can be multiplied (viewed as complex numbers). >That is, if v+iw has v,w rational with v^2+w^1, then view that (v,w) >as the rational > (v*LCM(vdenominator,wdenominator)) / (w*LCM(vdenominator,wdenominator)). > >Euclid found all pythagorean triples (a,b,c) namely > a = (m-n)*(m+n), b=2*m*n, c=m^2+n^2 >which enables easily finding arbitrarily many arbitrarily close >pythagorean rational approximations to x by this method: >1. use CFs to find large a,b so x =approx= a/b >2. solve for m,n satisfying above two simultaneous quadratic equations >(as real numbers) >3. round off m,n to nearest integers >4. find resulting pythagorean a,b.

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[math-fun] pythagorean rationals & approximations by them
by Warren D Smith 15 Aug '13

15 Aug '13

The procedure I suggested for approximating real x by rational a/b so that sqrt(a^2+b^2)=integer, was one of many possible and I do not claim my procedure is "best." It is merely good enough to produce an infinite set of approximations a/b which obey the constraint and are arbitrarily good. It is clear this procedure will cause approximation error=|x - a/b| which will be upper bounded by some negative power of |b|, say error<PositiveConstant*|b|^(-P), for an infinite set of (a,b). A natural question would be: what is the greatest P that can be achieved? For the non-pythagorean rational approximation problem P=2 is best possible and achieved by the usual continued fraction algorithm with the hardest-to-approximate number being the Golden number. Now that the Carsten Elsner: Fibonacci Quarterly (May 2003) 98-104 paper has been pointed out, it appears to have pretty much the same ideas that I used, and he demonstrates in coro 1.1 that P>=1 is valid for almost all irrational x, and also demonstrates in coro 1.2 that P=1 is best possible i.e that there exist quadratic-irrational x such that always error>PositiveConstant*|b|^(-1). I think the procedure I had described achieved P=1 in which case it actually was best possible in that weak sense (although you should check this numerically please).

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[math-fun] totalperiodic redux
by Bill Gosper 15 Aug '13

15 Aug '13

(This was for the Life list, but you might find it interesting.) I tightened it so you can watch the first couple of deletions. #CXRLE #C Total periodic #C #C Without the block, every cell in the universe is aperiodic. #C #C The block starts to delete the nth shuttling glider at time #C 14 - 35 (-1)^n - 2 n + (1 + 12 n) 3^(2 + 5 n) #C = 28478, 13286000, 4778186074, 1537671920812,... #C each time lurching slightly northeast. Thus every cell in #C the universe is eventually permanently vacant. #C #C It would be nice to "shew that" no shuttling glider escapes #C the reflecting walls, although all that is required is that #C infinitely many hit the block. Which would also be nice to shew. x = 56, y = 60, rule = B3/S23 38bo$37b3o$36b2obo5bo$36b3o5b3o$37b2o4bo2b2o3b3o$43b3o4bo2bo$53bo$49bo 3bo$53bo$37b3o10bobo$37bo2bo$37bo$37bo$38bobo7$35b3o$35bo2bo11bo$35bo 13b3o$35bo3bo8b2obo$35bo12b3o$36bobo10b2o3$32b3o$27b2o2b5o$27b2o2b3ob 2o7b2o5bo2bo$34b2o7b4o8bo$43b2ob2o3bo3bo$45b2o5b4o5$18bo32b3o$16bo3bo 32bo$21bo31bo$b2o13bo4bo30bo$4o13b5o8bo2bo$2ob2o29bo$2b2o26bo3bo$31b4o 15b2o$47b3ob2o$21b2o24b5o$22b2o24b3o$21b2o$21bo3$22b2o$13b4o4b4o$12bo 3bo4b2ob2o$b5o10bo6b2o$o4bo6bo2bo$5bo$o3bo! --rwg

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[math-fun] number theory: continued fractions with constraints
by Henry Baker 14 Aug '13

14 Aug '13

Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint. In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral. Suppose I used a continued fraction process to generate better & better approximations. Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ? I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral. So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions? Or perhaps this situation is exceedingly rare. If so, how rare is it?

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Re: [math-fun] mysterious arguments of the sum of 2 complex numbers / Maple
by Axel Vogt 14 Aug '13

14 Aug '13

For 1/5^(1/2+12*I) and the above commands one gets -12*ln(5)+6*Pi as argument. Essentially I agree. But you add 2 vectors in the plane (in polar coordinates), which can be "done" by ruler and pencile. And you ask for the angle (Maple ...) r*exp(I*phi) + s*exp(I*psi); eval(%, s=1); # some scaling to unit circle: assume s=1 argument(%); convert(%, arctan); evalc(%); arctan(r*sin(phi)+sin(psi),r*cos(phi)+cos(psi)); This is arctan with certain "sign rule" Without further assumptions I would not expect more "structure" in a result: One adds 2 vectors, their resulting angle is expressed in terms of trigonometrics.

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Re: [math-fun] mysterious arguments of the sum of 2 complex numbers / Maple
by Axel Vogt 14 Aug '13

14 Aug '13

... > For example, arg(1/2^(1/2+12*I) , if you use Maple or Mathematica > or by hand you can expand this BUT, actually the expression is quite > cumbersome and tricky. > > The proper answer is 2*Pi - 12*log(2). But that answer does not > appear at fist hand, the <12> comes from an operation with { } and [ ]. ... Using Maple 17: (1/2^(1/2+12*I)); convert(%, exp); argument(%); -12*ln(2)+2*Pi

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[math-fun] pythagorean rationals & approximations by them
by Warren D Smith 14 Aug '13

14 Aug '13

Henry Baker wanted to approximate reals x by "pythagorean rationals" x =approx= a/b with squareroot(a^2+b^2)=integer. Obviously, yes, any such pythagorean (a,b) does have a continued fraction (namely just use the CF for a/b); and as Asimov remarked the set of rational points on the unit circle is dense, hence any x can be approximated arbitrarily closely by a pythagorean rational. The pythagorean rationals form a group in the sense that any two rational points on the unit circle can be multiplied (viewed as complex numbers). That is, if v+iw has v,w rational with v^2+w^1, then view that (v,w) as the rational (v*LCM(vdenominator,wdenominator)) / (w*LCM(vdenominator,wdenominator)). Euclid found all pythagorean triples (a,b,c) namely a = (m-n)*(m+n), b=2*m*n, c=m^2+n^2 which enables easily finding arbitrarily many arbitrarily close pythagorean rational approximations to x by this method: 1. use CFs to find large a,b so x =approx= a/b 2. solve for m,n satisfying above two simultaneous quadratic equations (as real numbers) 3. round off m,n to nearest integers 4. find resulting pythagorean a,b.

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Re: [math-fun] "firewall paradox" about black holes
by Henry Baker 14 Aug '13

14 Aug '13

At 07:22 PM 8/13/2013, Rowan Hamilton wrote: >Really smart guys are confused by this issue. And even Ed Witten is staying >silent. To me, that means that the jury is still out while the experts >digest things. Yea, these are the same 'really smart guys' who built their physics on _infinitely precise_ real numbers. These are the same 'really smart guys' who have (mostly) ignored the concept of 'information' for the past 60+ years. They've now swept too many infinities under the rug, and those bulges in the rug refuse to go away. The situation in physics today is analogous to that of Ptolemaic astronomy prior to Copernicus: it isn't right, but it gives such great answers that they refuse to give it up. Maxwell formulated the main problem with Maxwell's Demon. No one since has really made that much progress, and the subsequent development of quantum mechanics made the situation considerably stranger. The field of quantum computation started out on the right track, but was hijacked by more traditional physicists before being able to provide new insights. I believe that the right answer is tied up with quantum computation, but someone will have to first come up with a new name for the field, so that all of the assumptions can be re-examined, before further progress will be made. The lack of progress in solving these problems, which I generally put under the heading of 'Second Law' problems, is related to the fact that there are relatively few people working on them. Whatever progress has been made is mostly afterthoughts and stolen moments from more 'pressing' research: building bombs, building larger colliders, getting Nobel prizes, etc. Bottom line: novelty doesn't come from large organizations with billions of dollars at stake; novelty comes from individuals willing to question every assumption -- e.g., some random patent clerk in Switzerland.

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