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Re: [math-fun] The Hippasus Integers
by Dan Asimov 19 Nov '16

19 Nov '16

Thanks. Yes, I have that book and agree it is a fine book to have. —Dan >From: Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> >You were right the first time. If h=1 mod 4, the ring is Z[(1 + sqrt(h))/2]. Otherwise, the ring is Z[sqrt(h)]. It is understood that square factors are removed from h, so that h is squarefree. This is all derived in Harvey Cohn, "Advanced Number Theory", Dover, p. 45, a wonderful book.

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Re: [math-fun] The Hippasus Integers
by Dan Asimov 19 Nov '16

19 Nov '16

[By the way, nifty use of the blackboard font ℚ!] I didn't know the discover of the irrationality of sqrt(2) was known by name! I'd like to see a reference for that, since I always heard the history was unclear. ----- The Heegner numbers are 1, 2, 3, 7, 11, 19, 43, 67, 163. The ℚ(√−1) numbers are known as Gaussian integers. The ℚ(√−3) numbers are known as Eisenstein integers. The ℚ(√−7) numbers are known as Kleinian integers. It's bugged me for awhile that ℚ(√−2) wasn't named. Today I decided that the obvious name was ... The ℚ(√−2) numbers are now known as Hippasus integers. Hippasus proved √2 was irrational. He was then murdered. Seemed like an apt name to use. I calculated the nine types of Heegner primes and plotted them out. http://community.wolfram.com/groups/-/m/t/965609 <http://community.wolfram.com/groups/-/m/t/965609> Anyone agree / disagree about calling ℚ(√−2) the Hippasus integers? ----- I might be misinterpreting what is meant by "The ℚ(√[whatever]) numbers" are known as [whoever] integers." Usually ℚ(√[whatever]) is labeled an example of a "number field", which I think is any field of algebraic numbers with a finite index over ℚ. But to use the word "integers" implies that they are the roots of monic polynomials, with integer coefficients, that lie in the number field. For the number field ℚ(√-h) where h is a Heegner number the set of algebraic integers is as follows: Let O_h be the ring of algebraic integers of ℚ(√-h), Then: O_1 = Z[i] O_2 = Z[sqrt(-2)] O_h = Z[(1 + sqrt(h))/2] for h = -3, -7, -11, -19, -43, -67, -163. —Dan

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Re: [math-fun] The Hippasus Integers
by Dan Asimov 18 Nov '16

18 Nov '16

As a very kind colleague pointed out: There is still a typo: the complete absence of any correction. As this mundane finder (apply S_6*, S_5 respectively) pointed out: > SHOULD READ: ----- O_h = Z[(1 + sqrt(-h))/2] for -h = -3, -7, -11, -19, -43, -67, -163 ----- ((( And NOT: >> O_h = Z[(1 + sqrt(h))/2] for -h = -3, -7, -11, -19, -43, -67, -163 ))) As my maternal grandmother used to say: Oy!!!!!!!!!!!!!!!!!!!!!!! I can say this: The descent into total senility is at least interesting! —Dan ———— * But notice that Out(S_6) is different. > On 11/18/16, Dan Asimov <asimov(a)msri.org> wrote: >> Correction: >> >>> On Nov 18, 2016, at 10:30 AM, Dan Asimov <asimov(a)msri.org> wrote: >>> >>> Let O_h be the ring of algebraic integers of ℚ(√-h), Then: >>> >>> O_1 = Z[i] >>> >>> O_2 = Z[sqrt(-2)] >>> >>> O_h = Z[(1 + sqrt(h))/2] for h = -3, -7, -11, -19, -43, -67, -163. >> >> >> >> The last line should read: >> >> ----- >> O_h = Z[(1 + sqrt(h))/2] for -h = -3, -7, -11, -19, -43, -67, -163. >> ----- >> >> (and in that e-mail "discover" should be "discoverer". Argggh.

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[math-fun] The Hippasus Integers
by Ed Pegg Jr 18 Nov '16

18 Nov '16

The Heegner numbers are 1, 2, 3, 7, 11, 19, 43, 67, 163. The ℚ(√−1) numbers are known as Gaussian integers. The ℚ(√−3) numbers are known as Eisenstein integers. The ℚ(√−7) numbers are known as Kleinian integers. It's bugged me for awhile that ℚ(√−2) wasn't named. Today I decided that the obvious name was ... The ℚ(√−2) numbers are now known as Hippasus integers. Hippasus proved √2 was irrational. He was then murdered. Seemed like an apt name to use. I calculated the nine types of Heegner primes and plotted them out. http://community.wolfram.com/groups/-/m/t/965609 Anyone agree / disagree about calling ℚ(√−2) the Hippasus integers? Ed Pegg Jr

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[math-fun] The Hippasus Integers
by Bill Gosper 17 Nov '16

17 Nov '16

Date: 2016-11-17 18:43 From: Ed Pegg Jr <ed(a)mathpuzzle.com> The Heegner numbers are 1, 2, 3, 7, 11, 19, 43, 67, 163. The ℚ(√−1) numbers are known as Gaussian integers. The ℚ(√−3) numbers are known as Eisenstein integers. The ℚ(√−7) numbers are known as Kleinian integers. It's bugged me for awhile that ℚ(√−2) wasn't named. Today I decided that the obvious name was ... The ℚ(√−2) numbers are now known as Hippasus integers. Hippasus proved √2 was irrational. He was then murdered. Seemed like an apt name to use. I calculated the nine types of Heegner primes and plotted them out. http://community.wolfram.com/groups/-/m/t/965609 Anyone agree / disagree about calling ℚ(√−2) the Hippasus integers? Ed Pegg Jr --------------- His assailants probably didn't kill him for proving √2 irrational, but rather for taunting them for being unable to state the numerator and denominator. --rwg

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[math-fun] Hyperbolic paraboloid
by James Propp 17 Nov '16

17 Nov '16

John Partridge asked me about the hyperbolic paraboloid shown in http://pyrigan.com/wp-content/uploads/2016/10/CelebrationOfMind.pdf Is the 2D projection of the hole a true circle? See page 12. Jim Propp

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[math-fun] Mapping problem
by David Wilson 17 Nov '16

17 Nov '16

Let S(n) be the largest subset of Z(n) fixed by the mapping n -> n^2, and let f(n) = |Z(n)|. For example, S(25) = {0, 1, 6, 11, 16, 21} is the largest set of residues modulo 25 fixed by the mapping n -> n^2, so f(25) = |S(25)| = 6. Can you find a formula for f(n) in terms of n?

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[math-fun] Landau's function of 2^16
by Allan Wechsler 17 Nov '16

17 Nov '16

I got an email from Claus Michael Ringel at the University of Bielefeld, asking me if I knew a reasonable way to evaluate Landau's function (A000793) at 2^16 = 65536. This is the maximum least common multiple of any partition of 65536; equivalently it is the maximum order of any element of Sym(65536). I am sure there are lots of shortcuts, that save one from enumerating all the partitions of 65536. Can anyone suggest one? (Of all the code snippets in the relevant page at A000793, all of them except one are clearly just enumerating partitions and comparing LCM's. But Charles Greathouse's PARI code is not doing anything like that, and I don't have a clue what it *is* doing.)

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[math-fun] Not particularly math, possibly fun, particularly particles
by Bill Gosper 16 Nov '16

16 Nov '16

Right. Thanks--that explains everything. But it would be so cool to have exact reversibility. To (perhaps unnecessarily) repeat: Might it be possible to update the particles sequentially a la the (two skew) Minsky "circle" algorithm: newx -> x - floor(d y), newy -> y + floor(e newx) which is exactly reversible, no matter how batsh*t the floor function is. The penalty for this is (very remediable) non-circularity. Could we update the particles with sequential (and thus reversible) skews? If so, there are probably remedies for any visible imperfections in the forward simulation. --rwg On 2016-11-15 11:06, Mike Speciner wrote: > Bill, I'm assuming you created this by time-reversing an initial > configuration. If so, .... > The three balls that end up apparently motionless and linearly > together at the lower left aren't really quite as motionless or > linearly together as they appear, presumably due to tiny floating > point "errors" that accumulate in both the reverse and the forward > simulation. (You can examine the actual positions and velocities by > mousing over the balls with the simulation paused.) In the forward > simulation, there are probably no simultaneous collisions. In any > case, only one collision (the earliest) is processed at a time, which > might occur at a small fraction of the time step. Then all the > positions and velocities are updated to the collision time post > collision, and the process is repeated until all the collisions within > the stepsize have been processed. > > --ms > > On 14-Nov-16 19:39, Bill Gosper wrote: >> Simple experiment; negative result: I can't see a difference between >> >> title:gas2 >> clearwalls >> clearballs >> time:0 >> speed:4 >> wall:{x:[[-256,-256],[-256,256]]} >> wall:{x:[[-256,256],[256,256]]} >> wall:{x:[[256,256],[256,-256]]} >> wall:{x:[[256,-256],[-256,-256]]} >> wall:{x:[[0,56],[-56,0]]} >> ball:{x:[138.95544361180137,-53.80397594446178],v:[1.510335795838575,25.481978458903548],m:1,r:10,c:"gray"} >> ball:{x:[-134.9335319061281,-186.86010684600106],v:[21.85076947267741,9.512509711816417],m:1,r:10,c:"gray"} >> ball:{x:[54.22470051433805,245.51158342487696],v:[-24.566342384761167,-19.008502444414795],m:1,r:10,c:"gray"} >> ball:{x:[-10.77118714505185,226.1293232034858],v:[-22.55619118918868,-2.3002862909202832],m:1,r:10,c:"gray"} >> ball:{x:[-75.57273059907199,-225.01762417294833],v:[-1.2330104812166782,-0.14817511072452252],m:1,r:10,c:"gray"} >> >> and the same text with the last five lines reversed. The continuation from >> time 400 should have one ball shuttling >> across the bottom for many periods. Do you presort the balls? Take pains >> to simulate parallel update? Either >> would foil my plan. >> --rwg >> One non-surprise: speed -2 is a poor slo-mo time-reversal of speed 4. >> >> >> >> On Sun, Nov 13, 2016 at 10:01 PM, Bill Gosper <billgosper(a)gmail.com> wrote: >> >>> Maybe we can at least get Minsky-style reversibility by updating the >>> particles in reverse order if >>> time runs negative. --rwg >>> I somehow missed the Load from file button! My file is small enough to >>> fit in a url. >>> Is there a way for it to override the default initial configuration? >>> >>> Date: 2016-11-12 13:44 >>> From: Mike Speciner <ms(a)alum.mit.edu> >>> To: math-fun <math-fun(a)mailman.xmission.com> >>> Reply-To: math-fun <math-fun(a)mailman.xmission.com> >>> >>> If you SHOW AS TEXT, you can cut and paste to save in a file of your >>> choosing (and edit it as you choose). You can load such a file with the >>> "Load from file" function. (The LOAD and SAVE functions use cookies, with a >>> nonstandard encoding to reduce their size.) >>> >>> Good question about perfect computation with rationals. There are an awful >>> lot of square roots in the calculations; I'm not sure the conditions under >>> which they don't produce irrationals. I suspect one dimensional scenarios >>> are more likely to be able to produce exact results. (It's been a long time >>> since I looked at this code. I do remember deliberately not adding in >>> gravity because that made reversibility harder. The PDP-6 version had >>> gravity. And I vaguely remember having some code to make sure energy was >>> [nearly] conserved, but I'm not sure it's in this new version.) >>> >>> --ms >>> >>> On 12-Nov-16 01:44, Bill Gosper wrote: >>>>> Date: 2016-11-05 09:51 >>>>> From: Mike Speciner <ms(a)alum.mit.edu> >>>>> https://ms0.github.io/gas.html >>>> This is a great toy! I delete everything except six identical balls (and >>>> maybe one wall), set one ball >>>> to bounce around for a while before disrupting the others, run it half a >>>> minute into the disruption, save >>>> it out, and show it time-reversed to some unsuspecting person. Or if I'm >>>> afraid they'll notice the >>>> negative time passage, I manually negate all the velocities before >>> saving. >>>> (Is there a way to load >>>> one of those text files?) >>>> >>>> Unfortunately, neither of these reversal stunts is perfect, due to >>>> accumulated numerical errors, >>>> which I assume you've worked hard to minimize. >>>> >>>> So let's get particularly math-y and ask: Given rational positions and >>>> velocities, can we simulate >>>> one of these six-ball scenarios exactly and indefinitely, without runaway >>>> denominator buildup, >>>> nor irrational collision results? --rwg >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Mapping problem
by Dan Asimov 16 Nov '16

16 Nov '16

I'm not feeling particularly patient at the moment. Can you kindly say what the notation "Z(n)" stands for? I can try to guess, of course, but I would rather not. —Dan ----- Let S(n) be the largest subset of Z(n) fixed by the mapping n -> n^2, and let f(n) = |Z(n)|. For example, S(25) = {0, 1, 6, 11, 16, 21} is the largest set of residues modulo 25 fixed by the mapping n -> n^2, so f(25) = |S(25)| = 6. Can you find a formula for f(n) in terms of n? -----

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