On 2019-04-14 18:18, Adam P. Goucher wrote:
By tweaking the circumradius code, I get an inradius of:
r = 0.43089181059646955564926306618266477789
Whee, gosper.org/inradius.png ! Scalable hi-res: https://www.wolframcloud.com/objects/b2a91ed1-e02e-4961-96ee-fcc151e61f9d Your browser may cop out and make you download.
which is slightly smaller than (sqrt(3) - 1)R. Again, I don't get anything exciting from RIES (from trying each of R, r, and R/r).
FindIntegerNullVector and LatticeReduce likewise come up empty on a sextic(R,r). —rwg
Best wishes,
Adam P. Goucher
Sent: Sunday, April 14, 2019 at 11:09 AM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] That curve I regrettably named flowsnake
Warren Smith privately objected to my reasoning here and requested higher-res evidence. Boosting the sampling frequency 9-fold and zooming <http://gosper.org/In&Out9.png> appears to bear him out! —rwg
On Sat, Apr 13, 2019 at 7:36 PM Bill Gosper <billgosper@gmail.com> wrote:
Adam, your winning candidate was c ~ 0.553742984573881 - 0.220041001470351 i . In view of the tessellation <http://gosper.org/In&Out.png>, have you not simultaneously solved the inradius question? r = (√3-1) |c| = .4361*999*1440836026564516021182720360465 —rwg
On Fri, Apr 12, 2019 at 4:48 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2019-04-11 11:05, Adam P. Goucher wrote:
Bill,
The first few decimal places of the circumradius are:
0.595860164210417909024063032937823238789
Amazing!
That's enough to see that RIES can't find a simple closed form.
Also amazing! How can such a simple construction compute an entirely new constant? Do we get another one from the same base but the "triskelion" digits triskelion <http://gosper.org/tril7image2.gif>, gosper.org/tril7image3.gif, gosper.org/tril7image4.gif? At least the inradius is easy!
And therefore the circumradius! —rwg
—rwg
Here's some Wolfram Mathematica code to obtain accurate lower bounds:
omega = Simplify[(-1)^(2/3)]; base = 3 + omega; candidates = {1}; unit = base; Do[ unit = Simplify[base unit]; candidates = DeleteDuplicates@ Flatten@Table[{1, omega, omega^2, -1, -omega, -omega^2} + base e, {e, candidates}]; candidates = Map[Simplify, candidates]; maxnorm = Max[Map[Norm, candidates]]; candidates = Select[candidates, (Norm[#] + 1 >= maxnorm) &]; Print[{Length[candidates], N[maxnorm/Norm[unit], 50]}]; , {i, 1, 100}];
It does this by a tree search on sequences of digits (from the set of sixth roots of unity, because 0 can't be in the digit expansion of an extremal point). It prunes branches whose upper bound is smaller than the currently established lower bound. Fortunately, the number of candidates seems to be O(1), independent of the iteration, so this runs relatively quickly.
(Also, the first digit is wlog assumed to be a 1.)
Best wishes,
Adam P. Goucher
On Thu, Apr 11, 2019 at 7:29 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2019-04-10 06:10, Joerg Arndt wrote:
Pics or it didn't happen!
Agreed: https://www.wolframcloud.com/objects/797a1110-f0cd-47a5-90f6-f0b3d1d8ce4b
Joking aside: I'd like to understand, but don't.
Best regards, jj
My mail to arndt@jjj.de is bouncing. Correction?
* Bill Gosper <billgosper@gmail.com> [Apr 08. 2019 15:50]: > On Sat, Apr 6, 2019 at 9:20 AM Bill Gosper <billgosper@gmail.com>
wrote:
> > > Mandelbrot detested the name. Then he discovered a spacefill of Koch's > > Snowflake. > > The worthy claimant to the Flowsnake name! Which he still detested. > > I recently realized how little I know about my own curve, which fills the > > FranceFlake, the fundamental region for the positional number system > > Base: 2 + i^(2/3) = √7 e^(i arccos(11/14)/2) > > and the (necessarily) seven digits 0 and the 6th roots of 1. > > > > SPOILERS > > > If FranceFill(0)= 0+0i, FranceFill(1) = 1 + 0i: > > 1) What is the area of a FranceFlake? > > > Only √3/2. > > > 2) Where is its centroid? > > > 1/2+i/2/√3 > > > 3) Is its circumradius √(52/147) ? > > > No. ≥ √(6140317/3)/2401. (What the heck is this approaching?) > The same as the FranceFlake, duh. > > > 4) Ironically, it was popular partly for how completely its canonical > > sampling (at 0/7, 1/7,...,1) self-avoided. But like all true spacefills, > > it is *dense* with *triple* points. > > E.g., what are the preimages of (9 + i √3)/21? > > > {5/42, 11/42, 17/42} > Are there quadruple points? How can we prove "No."? > > > —rwg
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