July 2014 - mathfun@mailman.xmission.com

Re: [math-fun] analytic function for new Airbnb logo?
by Bill Gosper 06 Aug '14

06 Aug '14

HGB>I wanted an approximation that was entire over the complex plane, so the straight lines and discontinuities would have to be fudged. If I take the image, convert it to black-and-white, and erode the boundaries to a single pixel width, I will get a "continuous" curve that goes around the origin twice. If I consider this curve in (r,theta) coordinates, I can now consider the Fourier transform. I'd like to reset the origin in such a way that the "fundamental" is zero -- i.e., the lowest frequency (which takes 2 revolutions) has a coefficient of zero. (There are other possible locations for the origin: the balancing point, the center of gyration, etc.) That having been done, I could alternatively use (log(r),theta) coordinates, which have some nicer properties, because it is the actual complex log function. Fourier analysis can approximate the discontinuous changes in curvature; hopefully modifying these coefficients a little will remove these discontinuities with a minimum of ringing. <HGB Ringing only happens at true discontinuities (e.g. http://www.tweedledum.com/rwg/gibbs.htm ) A nice example of going around twice, mixing circular arcs and straight segments, is to draw, with a single set of amplitudes and phases (for which I have closed forms) alternately an equilateral triangle and its inscribed circle: gosper.org/Incirc_1.mp4 (The file isn't corrupt--download it.) --rwg The sides of your rounded edge squares are impressively flat. But not as flat as those tan(sin())-sin(tan()) games we played. At 03:48 PM 7/30/2014, Bill Gosper wrote: HGB> My approach would be to take the Fourier transform of the polar plot, rwg>Defined as piecewise linear|circular? and then simplify to a small number of coefficients. Of course, it will take 2 circuits around the origin to complete the figure. The sides are perhaps a little too flat, which will blow up the number of coefficients required. I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. At 02:56 PM 7/29/2014, Whitfield Diffie wrote: HGB>Anyone game for an analytic function to approximate Airbnb's new logo? They should manufacture a paper clip and give it out at trade shows. Whithttp://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.… rwg> You can get some impressively straight segments with theta fcns: gosper.org/halph.pdf (10MB?!) or see http://gosper.org/thetpak.html (scrolled 69%). (.3281... is âˆš(one ninth constant), which I claim should be named Halphen's. It's the largest q for which the curve is convex.)) But a problem with this logo is that it switches curvatures discontinuously. <rwg Another possibility: modulate the arcspeed (slow down for curves, speed up for straightaways) before taking the Fourier.

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Re: [math-fun] analytic function for new Airbnb logo?
by Henry Baker 30 Jul '14

30 Jul '14

I wanted an approximation that was entire over the complex plane, so the straight lines and discontinuities would have to be fudged. If I take the image, convert it to black-and-white, and erode the boundaries to a single pixel width, I will get a "continuous" curve that goes around the origin twice. If I consider this curve in (r,theta) coordinates, I can now consider the Fourier transform. I'd like to reset the origin in such a way that the "fundamental" is zero -- i.e., the lowest frequency (which takes 2 revolutions) has a coefficient of zero. (There are other possible locations for the origin: the balancing point, the center of gyration, etc.) That having been done, I could alternatively use (log(r),theta) coordinates, which have some nicer properties, because it is the actual complex log function. Fourier analysis can approximate the discontinuous changes in curvature; hopefully modifying these coefficients a little will remove these discontinuities with a minimum of ringing. The sides of your rounded edge squares are impressively flat. At 03:48 PM 7/30/2014, Bill Gosper wrote: >HGB> My approach would be to take the Fourier transform of the polar plot, > >rwg>Defined as piecewise linear|circular? > >and then simplify to a small number of coefficients. > >Of course, it will take 2 circuits around the origin to complete the figure. > >The sides are perhaps a little too flat, which will blow up the number of coefficients required. > >I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. > >It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. > >At 02:56 PM 7/29/2014, Whitfield Diffie wrote: >HGB>Anyone game for an analytic function to approximate Airbnb's new logo? > They should manufacture a paper clip and give it out at trade shows. > Whit >http://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.png > >rwg> You can get some impressively straight segments with theta fcns: gosper.org/halph.pdf (10MB?!) or see http://gosper.org/thetpak.html (scrolled 69%). (.3281... is â(one ninth constant), which I claim should be named Halphen's. > >It's the largest q for which the curve is convex.)) > >But a problem with this logo is that it switches curvatures discontinuously. <rwg > >Another possibility: > modulate the arcspeed (slow down for curves, speed up for straightaways) before taking the Fourier.

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Re: [math-fun] analytic function for new Airbnb logo?
by Bill Gosper 30 Jul '14

30 Jul '14

HGB> My approach would be to take the Fourier transform of the polar plot, rwg>Defined as piecewise linear|circular? and then simplify to a small number of coefficients. Of course, it will take 2 circuits around the origin to complete the figure. The sides are perhaps a little too flat, which will blow up the number of coefficients required. I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. At 02:56 PM 7/29/2014, Whitfield Diffie wrote: HGB>Anyone game for an analytic function to approximate Airbnb's new logo? They should manufacture a paper clip and give it out at trade shows. Whit http://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.png rwg> You can get some impressively straight segments with theta fcns: gosper.org/halph.pdf (10MB?!) or see http://gosper.org/thetpak.html (scrolled 69%). (.3281... is √(one ninth constant), which I claim should be named Halphen's. It's the largest q for which the curve is convex.)) But a problem with this logo is that it switches curvatures discontinuously. <rwg Another possibility: modulate the arcspeed (slow down for curves, speed up for straightaways) before taking the Fourier.

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[math-fun] quantum cheshire cat
by Warren D Smith 30 Jul '14

30 Jul '14

Don't know what to make of this: http://www.nature.com/ncomms/2014/140729/ncomms5492/full/ncomms5492.html

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[math-fun] Shegar: Macro-scale quantum entanglement
by Henry Baker 30 Jul '14

30 Jul '14

FYI -- There are probably analogous examples of quantum superposition... http://www.bbc.co.uk/programmes/p023c98x Yemens Swap Marriages Duration: 25 minutes First broadcast: Tuesday 29 July 2014 Ill marry your sister if you marry mine. And if you divorce my sister, Ill divorce yours. That is Yemens Shegar, or swap marriage, an agreement between two men to marry each others sisters, thereby removing the need for expensive dowry payments. But the agreement also entails that if one marriage fails, the other couple must separate as well, even if they are happy. BBC Arabics Mai Noman returns to her native Yemen and hears the stories of two women who have loved and lost because of Shegar. Nadia lives in the village of Sawan on the outskirts of the capital Sanaa with her family. She was married off at the age of 22 and has three children. Nadia had to pay a high emotional price because of her familys decision to marry her off in the Shegar tradition. She was forced to divorce and now she and her mother have to live with the stigma that brings. Nora and her brother Waleed had little say in marrying their cousins through Shegar. But what happens when one siblings happiness depends on ending the marriage of the other? Could you choose your siblings happiness over your own? Stay in an unhappy marriage so your sibling can stay in a happy one? Mai asks why an old tradition that forces you to love only to force you to part, is still practised in Yemen. What do the religious authorities think of it? And is it tied to Islam? In Yemen, the heart of Arabia, ancient traditions and values have kept the fabric of society unchanged. They helped preserve Yemens unique charm and character, but also imprisoned Yemens people in the past. Shegar marriage is one such tradition. It helps poorer families to marry, but at what price?

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[math-fun] pi/2 + 2/pi yet again
by Bill Gosper 30 Jul '14

30 Jul '14

Adam's blog http://cp4space.wordpress.com/ currently has "<s, t : s^10 = t^3 = (st)^2 = 1> This is a surprisingly exciting group. Firstly, it is straightforward to verify from its presentation that the group is a hyperbolic *Dyck group*, *i.e.* the orientation-preserving symmetries of the kaleidoscopic tiling of the hyperbolic plane by triangles with interior angles (π/2, π/3, π/10). [...] In other words, it is the group generated by the following Möbius transformations: - s(z) = z exp(i pi/5); - any t with the property that t(t(t(z))) = s(t(s(t(z)))) = z for all z. Now, this is a ring-theoretic definition, so we can apply a ring automorphism and replace exp(pi/5) with exp(3pi/5) in the above definition without changing the resulting abstract group. However, this is entirely different from a topological perspective, since we have a group of spherical symmetries — rotations — rather than hyperbolic symmetries. So we can regard this as a dense subgroup of SO(3) generated by two rotations." Question: Does this have a "group volume", and can it show the expected magnitude of an element of SO(3) is pi/2 + 2/pi? --rwg

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Re: [math-fun] analytic function for new Airbnb logo?
by Henry Baker 30 Jul '14

30 Jul '14

My approach would be to take the Fourier transform of the polar plot, and then simplify to a small number of coefficients. Of course, it will take 2 circuits around the origin to complete the figure. The sides are perhaps a little too flat, which will blow up the number of coefficients required. I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. At 02:56 PM 7/29/2014, Whitfield Diffie wrote: >> Anyone game for an analytic function to approximate Airbnb's new logo? > > They should manufacture a paper clip and give it out at trade shows. > > Whit http://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.png

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[math-fun] analytic function for new Airbnb logo?
by Henry Baker 29 Jul '14

29 Jul '14

Anyone game for an analytic function to approximate Airbnb's new logo? http://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.png

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[math-fun] Experimental multiple choice question
by Bill Gosper 29 Jul '14

29 Jul '14

gosper.org/problem9.png Feel free to spoil it, else reply privately. --rwg

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Re: [math-fun] Asimov's rolling ball (I was wrong; partial repair)
by Warren D Smith 29 Jul '14

29 Jul '14

AW rightly objected to my "foobar" solution. Sorry, my brain malfunctioned. I now reveal the reasoning I had inside the brain... but with error now removed... and considerable more details... View the curve C in the plane as described by curvature=F(arclength). The curve C' of sphere-plane contact drawn on the sphere surface, has the same F in its GeodesicCurvature=F(GeodesicArcelength) description. For polygonal C, we get polygonal (with geodesic edges) C', with the same bend-angles at corners. We want C' to join the N & S poles. We want C to be "closed" which implies integral curvature darclength=2*pi, or for polygonal C the bend angles at corners sum to 2*pi. (The reverse implication would also be true except that you might close a subsegment of C, and there's issues at the two endpoints, and you might need to add straight sections to C...) So anyhow, if C' is smooth in a neighborhood of its start & end, and C is smooth except that there is a corner of interior angle theta>=0 at its start=end point, then C' needs to have integral curvature darclength = (pi+theta). This would be accomplished if C' encloses sphere surface area A and winds round it W times, where W*A=pi+theta and W>0 is integer. (In view of the "angle defect = area" theorem for non-Euclidean geometry.) In view of the isoperimetric theorem on surface of sphere (valid for all areas A less than 2*pi) the shortest closed Jordan curve on sphere surface, which encloses area=pi+theta, is a (non-great) circle with (by Archimedes hat box theorem) circumference = 2*pi*sinR where R is the angular radius of the circle and 2*pi*(1-cosR)=pi+theta=area. Incidentally since theta>=0 we find circle circumference>=sqrt(3)*pi=5.441398... So, consider in the plane, differentiable C of the following form: 1. line segment of length pi/2, then 2. circular arc of arclength Q, subtending angle pi+theta [hence radius = Q/(pi+theta) and curvature=(pi+theta)/Q], then 3. another line segment of length pi/2, closing the curve at its only corner, of interior angle theta<pi. This will yield C' on sphere joining N to S pole, consisting of geodesic segment of length pi/2 to reach equator, then non-great circle of arclength=Q, then finally geodesic pi/2 to reach S pole. The total arclength of C' or of C is pi+Q. And as we saw, Q>=sqrt(3)*pi=5.441398... To make this work: i. The surface area inside the circle on the sphere, needs to equal pi+theta to get the right amount of total curvature. ii. Hence area = pi+theta = 2*pi*(1-cosR), Q = 2*pi*sinR, where R is angular radius of that circle. iii. sin(theta/2)*(pi/2)=cos(theta/2)*Q/(pi+theta) to make C close. Trying to solve these 3 equations for the 3 unknowns Q,R,theta... from iii we have Q = tan(theta/2)*pi*(pi+theta)/2 from iib we have R = arcsin(Q/(2*pi)) from iia we have pi+theta = 2*pi*(1-cosR) so t=pi+theta is expressed as a function of Q, and conversely Q as a function of t, with R eliminated: t=2*pi*(1-Cos[Arcsin[Q/(2*pi)]]), Q=Tan[(t-pi)/2]*pi*t/2 = -Tan[(pi-t)/2]*pi*t/2 = -Cot[t/2]*pi*t/2 hence we can eliminate t: Q+Cot[pi*(1-Cos[Arcsin[Q/(2*pi)]])]*pi*(pi*(1-Cos[Arcsin[Q/(2*pi)]]) = 0 and... there seem to be no solutions for |Q|<=2*pi except for Q=multiple of 2*pi. So... the arclength is Q+pi=3*pi... which is the same solution as Asimov's. This is assuming I have not messed up in all that algebra, and unfortunately it is quite possible I have messed up; little attempt to check. Also, this merely finds an upper bound on the minimum length, not necessarily tight (although it might be; it would be minimum among curves of my class).

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