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[math-fun] Brain hemisphere spacefills likely with sextuple points.
by Bill Gosper 02 Dec '19

02 Dec '19

Julian & I (again mostly Julian) found exactly the two continuous functions schematized by this old Macsyma plot <http://gosper.org/hemis.gif> of a closed and unclosed curve. So now we can sample the functions at arbitrary rationals. E.g., triple frequency <http://gosper.org/semihexes3x.png>. But actually the median curve <http://gosper.org/semihexesmedian.png> most resembles the Macsyma plot. Since Julian recently found a sampling of Heighway's Dragon equal to its median curve, it's likely that there are samplings of these semihexes coincident with their median curves. This will be easy to test when we manage to invert the semihex functions. Which will also let us directly exhibit the sextuple points. Failing that, it should be possible to tediously guess and check a few. (Actually, since the Dragon truly spacefills, there is obviously some way to hit all the vertices on the median curve. But "luckily" not crazy and disordered.) Heretofore, the only spacefill dense with sextuple points I'd seen was Julian's huge Fourier triangle (@http://howwords.com/triangles/artcircle, which seems to be down), made from 25 copies of itself. Unfortunately, sextuple points are visited very furtively, so the Fourier plot is unconvincing, never (in six tries) coming very close to the actual sextuple points. For the exact definition, see my 21 Aug 2016 math-fun, which gives the six preimages of In[381]:= untrifil25[1/2 + I Sqrt[3]/10] Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} (In arithmetic progression. Always?) —rwg

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[math-fun] Did René Descartes say "Cogito, ergo sum"?
by Bill Gosper 02 Dec '19

02 Dec '19

Wikipedia: *Cogito, ergo sum*[a] <https://en.wikipedia.org/wiki/Cogito,_ergo_sum#cite_note-capitalization_and…> is a Latin <https://en.wikipedia.org/wiki/Latin> philosophical <https://en.wikipedia.org/wiki/Philosophy> proposition <https://en.wikipedia.org/wiki/Proposition> by René Descartes <https://en.wikipedia.org/wiki/Ren%C3%A9_Descartes> usually translated into English as "*I think, therefore I am*".[b] <https://en.wikipedia.org/wiki/Cogito,_ergo_sum#cite_note-better_translation…> The phrase originally appeared in French <https://en.wikipedia.org/wiki/French_language> as *je pense**, donc je suis* in his *Discourse on the Method <https://en.wikipedia.org/wiki/Discourse_on_the_Method>*, so as to reach a wider audience than Latin would have allowed. I heard that the real reason was that Descartes was lousy at Latin and paid someone for the translation! Using Google Translate for the French pronunciation, I am baffled by the sound of mysterious consonants in "je suis". ACW, Simon, is that authentic? How is it written phonetically? I have no confidence in Google Translate's pronunciations after hearing Joe, The Pizza Guy say "Cogito, ergo sum" in "Latin". —rwg

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[math-fun] History: A memoir on the theory of matrices (1858).
by Brad Klee 01 Dec '19

01 Dec '19

Arthur Cayley (1821-1895) was a pioneer of invariant theory and among the first to work out the details of the characteristic equation for 3x3 matrices in general. The Cayley-Hamilton theorem is given on page 24 of "A memoir on the theory of matrices" [1], and quickly followed by the great sensibility of an old-style mathematician: "I have not thought it necessary to undertake the labour of a formal proof of the theorem in the general case of a matrix of any degree." ( though nowadays we automate the next few cases on a computer ) Coefficients of the characteristic equation are homogeneous polynomials in the matrix elements, and may be obtained by summing over matrix minor determinants. The characteristic equation itself holds for any similar matrix, thus we may construct a zero sum for every coefficient by subtracting the most general polynomial from a polynomial in a set of distinct variables, with all off-diagonal symbols set to zero. If the dimension of the matrix is N, then we obtain N distinct polynomials z_i, such that z_i = 0. Over field K, the null ring RZ=K[[z_1,z_2,...,z_N]] is a subring of R = K[[a,b,...,L1,L2...,L_N]] with lowercase alphabet for generic matrix elements and L_i for eigenvalues. Two expressions p1 and p2 from R are equivalent whenever a third expression p3 from RZ exists such that p1 + p3 = p2. Without the normalizing factors in the denominator, the squared elements of the eigenvectors are obtained from the diagonal of projector matrices, also guaranteed to exist as corollary to the Cayley-Hamilton theorem. The real problem with Terence Tao's proposal to obtain "Eigenvectors from Eigenvalues" is that it engages in revisionist history. The authors expect to find proof an old, fundamental results in a post-Hilbert language that didn't exist during the days of Cayley and Hamilton, when the question of finding "Eigenvectors from Eigenvalues" was originally settled. Were the old masters alive today, we think that they would be more impressed with the performance of modern computer algebra systems, which easily find and prove identities within well-defined polynomial rings. Some applause is deserved by subsequent authors for finding particularly nice polynomials in R; however, the amount of success needs to be gauged relative to other theorems in the field. In our perspective, a "theorem" of the sort "p1 + p3 = p2", is not really that great compared to the discovery of the null ring KZ and the original proof of the Cayley-Hamilton theorem. In fact, infinitely many equivalent functions p2 are guaranteed to exist as a corollary. In our opinion, that is how they should be viewed. All the preceding text was written as notes for an article titled "Inside the quantum bikeshed", which planned to discuss Parkinson's law of triviality, as it often manifests in the sciences. However, Predrag C. convinced me that it would be rude and inconsiderate to write such an article. Counterpoint: If instead, we were living in a world of science fiction, and suffering from hard-scrabble existence after the fall of the empire, of course we would : * * demand the return of the Beskar Iron to its rightful owners! * * --Brad [1] https://royalsocietypublishing.org/doi/pdf/10.1098/rstl.1858.0002

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[math-fun] Find the matrix
by Fred Lunnon 30 Nov '19

30 Nov '19

Given two (unordered) sets of (binary) integer vectors. Considerations of symmetry lead to the conjecture that there exists a linear bijection (indeed, a permutation of the basis) relating them. Design an (efficient) algorithm to construct some such matrix. In the case to hand, the dimension equals 20 (reducible to nearer 10 after some finagling), and vector sets have size 120 and full rank. Fred Lunnon

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[math-fun] nonstandard Heighway Dragon recipe
by Bill Gosper 26 Nov '19

26 Nov '19

Huh? <http://gosper.org/symdrag.gif> —rwg

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[math-fun] Fwd: Re: Hot tea sucks.
by Bill Gosper 25 Nov '19

25 Nov '19

Duct / Duck (sic) tape comes in handy in such situations. Or you could just drink it hot with milk, a la Brit. WFL OMG—I shoulda stood in bed! Next to which sits a huge roll of Gaffer/Gopher (sic) tape! (Way better than duct tape.) —rwg I glossed over the hassle factor of the trip to Draeger's, which has an excessively level parking lot. You might not perceive this as a disadvantage until I reveal that reverse gear on my car is broken. On 11/24/19, Bill Gosper <billgosper(a)gmail.com> wrote: > It just seized again for the first time since this discussion. I ran > out > of nerve trying to > repressurize it in the 𝜇wave, and drove to Draeger's to bum a big > rubber > band from > the produce dept. But their's were too small. Trudging back to my > car, I > found a > bigger one in the parking lot! Which, along with more 𝜇wave courage, > did > the trick. > —rwg > > On Fri, Apr 26, 2019 at 3:54 PM Bill Gosper <billgosper(a)gmail.com> > wrote: > >> I BCCed a lot of people and got a lot of nontrivial suggestions. >> (Restaurateur) Mike Wang's rubber band suggestion >> was probably best. Just encircle the lid with a big one from a >> broccoli >> bunch. It's amazing if you've never tried it. >> But that's not trivial enough! (I forget where I stashed my rubber >> band.) Someone suggested that if I have a mountain >> handy, I could drive up it. Nobody suggested the dicey solution of >> sticking the jar back in the 𝜇wave. Really hot tapwater >> should work, but our water heater is set way low. C'mon >> people—*trivial*! >> >> SPOILER: >> In[532]:= StringReverse@"!no ti pals uoy sa dil eht rednu ria )looc >> ylevitaler( fo llufgnul kciuq a wolb tsuJ" >> —rwg >> >> On Fri, Apr 26, 2019 at 8:12 AM Bill Gosper <billgosper(a)gmail.com> >> wrote: >> >>> I brew it in a 46oz pickle jar that easily tips over, and leaks if >>> the >>> lid isn't snug. >>> But if I forget to momentarily loosen the lid within a few minutes, a >>> powerful vacuum >>> develops that makes it virtually impossible to unscrew. What is the >>> trivial solution? >>> —Bill >>> >> > _______________________________________________ > 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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[math-fun] Hot tea sucks.
by Bill Gosper 24 Nov '19

24 Nov '19

I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? —Bill

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[math-fun] Nice video NASA
by Brad Klee 23 Nov '19

23 Nov '19

https://m.youtube.com/watch?v=0yNzSwlnQ2Q In case you didn’t see... it takes a lot of math to make this possible, and even more to explain what it is we are seeing. No, mercury isn’t riding a perfect ellipse. Elliptic functions do better to account for precession, but still, nature is messy. I like the music too, great! And no astronauts were harmed, much less involved, another plus. Enjoy —Brad

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[math-fun] Brad's recent interest in sn|cn double angles
by Bill Gosper 21 Nov '19

21 Nov '19

prompts me to t(rot )out these four old 𝜗 half-angle formulas: Out[419]= {EllipticTheta[1, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[2, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[3, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[4, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4))} Check: In[432]:= Assuming[0 < x < \[Pi]/2 && 0 < q < 1, FullSimplify@Series[%419, {q, 0, 22}, {x, 0, 22}]] // tim During evaluation of In[432]:= 419.850403,4 (*Seven minutes*). Out[432]= {True, True, True, True} —rwg

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[math-fun] Primes with strokes
by Éric Angelini 19 Nov '19

19 Nov '19

Hello Math-Fun, I was wondering if this seq S was sound: "Put a stroke behind every prime digit. Any integer formed by the concatenated digits between two strokes is prime." S = 1,3,2,4,7,5,6,8,9,10,11,14,16,18,83,13,17,,... The same with the strokes: S = 1,3,\2,\4,7\,5,\6,8,9,10,11,14,16,18,83\,13\,17\,... If I'm not wrong, S is the lexicographically first seq of distinct terms with this property. The successive primes are: 13, 2, 47, 5, 689101114161883, 13, 17,... What do you think? Best, É.

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