May 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Hilbert(Hilbert)
by Bill Gosper 23 May '13

23 May '13

NeilB> Once we got the code working, we soon found out that it would be necessary to go down to level 9 to get a good result. [...] >Lastly, some things about the final picture could be improved. [...] Julian uncovered a bug in Neil's code, whose repair enabled Neil to reach level 10. I hope he doesn't mind my leaking http://gosper.org/hilbert-10.png . --rwg

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Re: [math-fun] Twin Prime conjecture
by Henry Baker 22 May '13

22 May '13

Re 'Twin Prime': I spent some time as a kid in Detroit, MI, where they had a 'Twin Pines' dairy that supported the Soupy Sales TV program. Every time I hear/see 'Twin Prime', I hear/see 'Twin Pine' instead. Math-fun content: I also watched Mr. Wizard on TV! http://www.detroitkidshow.com/twin_pines_truck.htm http://en.wikipedia.org/wiki/Soupy_Sales http://en.wikipedia.org/wiki/Watch_Mr._Wizard

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Re: [math-fun] Twin Prime conjecture
by Warren D Smith 22 May '13

22 May '13

>i don't have access. could somebody, who feels comfortable doing so, email me a copy? >bob baillie rjbaillie(a)frii.com --me too! warren.wds at gmail.com

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[math-fun] planar duals, Maxwell, circuits
by Warren D Smith 21 May '13

21 May '13

Sorry for my stupidity, but I cannot figure out what you are talking about. First, for electrical circuits, whatever you wanted to do with a planar dual is useless if the graph is nonplanar (which for a complicated circuit, it always will be). Second, for trusses, if we have a V-vertex planar graph it has E edges where E<=3*V-6. Now seems to me, if you demand zero vector force at each vertex (2D universe) that is 2 equations per vertex, but you've got E unknown forces. So if E>2*V, such as E=3*V-6, then how are you going to solve for the forces? No comprende. Maybe you have in mind an "outerplanar" graph (all vertices on one face). Then looks like can do. Seems like the solution for general planar graph is going to be highly non-unique but by minimizing elastic energy over all solutions, you'd uniquify. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Maxwell/graphs/reciprocal/dual/polygon projection/trusses
by Henry Baker 19 May '13

19 May '13

In preparation for doing some 3D printing & designing, I've been learning about structural engineering & trusses, and tripped over a very beautiful part of mathematics. As an electrical engineering student at MIT, I learned about planar & dual graphs for circuit diagrams, but the duality is even stronger & prettier for the 'trusses' found in civil & mechanical engineering. For example, if you have a planar system of rods (of fixed lengths) and (pin) joints which make up a 'truss', you can analyze this diagram to compute internal forces when external forces are applied to this structure. The cool thing about these truss diagrams is that the _dual diagram_ (in the graph theory sense) represents precisely the directions and magnitudes (lengths of the arrows) of the forces in the primal diagram (and vice versa). This duality was discovered by Maxwell (of Maxwell's equations). Thus, whereas in electrical engineering the dual diagram was purely conceptual; in trusses, the dual diagram has the actual x,y components of the forces. Structural engineers for about 50 years after Maxwell used these dual diagrams -- called 'graphic statics' -- to design many bridges, buildings, etc. (I would even speculate that Archimedes may have intuitively understood this duality, but we don't have enough extant material from Archimedes to know for sure.) http://en.wikipedia.org/wiki/Cremona_diagram Michell, following up on Maxwell, came out with 'Michell trusses' which optimized the amount of material required to carry a particular load. Some of Michell's structures look like solutions of Maxwell's (electromagnetic) Equations, and indeed they are! One of Michell's solution form logarithmic spirals in the complex plane; one can generate some of these discrete spirals using Gaussian integer multiplication -- including Gaussian prime numbers. Even cooler, the discrete form of some of Michell's trusses are _self-reciprocal_ (self-dual), in the sense that the force diagram and the truss diagram are identical (although the 'force' edge doesn't usually correspond with itself, but with another edge in another part of the same diagram). But wait, there's more! AI vision folks in the 1960's & 1970's were trying to 'see' & interpret simple objects like polyhedra using video cameras, and rediscovered another portion of Maxwell's theory: that _projections_ of polyhedra onto planes had certain combinatorial constraints; and Maxwell found that these constraints were identical to those involving stresses in planar 2D truss networks. Here's the basic intuition, which has be formalized into a proof. Consider a (3D) tetrahedron and its projection onto the plane (assume that no face is perpendicular to the plane, and no vertices or edges coincide). You get a 2D diagram with 4 vertices, 6 edges, and 4 'windows' -- including the region outside the diagram. Consider that the tetrahedron as a set of 6 bars/edges connecting the 4 vertices in 3D. If you _stand on_ the tetrahedron, the bottom 3 edges/bars will be in _tension_, while the 3 edges/bars holding up the 4 vertex will be in _compression_. Alternatively, if you look at the 2D projection of the tetrahedron, and attempt to _pick it up using the center vertex_, this process will induce _tension_ in the middle 3 edges/bars, and _compression_ in the outer 3 edges/bars. Thus, there is a perfect correspondence between 2D diagrams which can support 'self-stress' (through tension & compression of the bars/edges) and projections of 3D polyhedra. Interestingly, Maxwell knew about this 3D polyhedra/2D truss correspondence, too! --- I believe that there may be even more reciprocal/dual beauty hidden in the control points of Bezier splines and 'forces' on the splines, but I haven't gotten that far yet. For example, a typical suspension bridge provides _piers_, cables and suspenders to keep a roadway horizontal in the presence of loads, but if you want a _curved_ roadway, you need multiple piers, cables & suspenders to hold that roadway in the shape of the curve. The 'piers' in this case act like 'control points' for the (curved) roadway. http://en.wikipedia.org/wiki/De_Casteljau's_algorithm --- Besides Maxwell & Michell, there are many, many other names: Huffmann, Laman, Crapo, Whiteley, Sugihara, etc.

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[math-fun] The astroid is a trisectrix
by Bill Gosper 19 May '13

19 May '13

and a rather handy one. Wikipedia mentions this in neither its astroid article nor its list of trisectrices. It is obvious from a diagram. Will Wikipedia, as usual, reject this as "original research"? --rwg Here is Julian's animation, with the comment Using the quadricuspid cycloids to trisect and pentasect an arbitrary angle: The angle to be tri/pentasected is between the short green segment and horizontal. Note the hypocycloid gives the three correct angles (w.r.t. horizontal, cyan [rigid asterisk]) and five [magenta] bogons, while the epicycloid gives the five correct angles (dark blue [rigid asterisk]) and three [red] bogons. We've never understood the geometric significance of the bogons. --Bill and Julian [Somebody should export this as an mgif or something.] Block[{three = ParametricPlot[{3/4*Cos[t] + 1/4*Cos[-3*t], 3/4*Sin[t] + 1/4*Sin[-3*t]}, {t, -\[Pi], \[Pi]}, Axes -> None], five = ParametricPlot[{5/4*Cos[t] - 1/4*Cos[5*t], 5/4*Sin[t] - 1/4*Sin[5*t]}, {t, -\[Pi], \[Pi]}, Axes -> None]}, ListAnimate[ Table[Show[ Graphics[{AbsolutePointSize[10], AbsoluteThickness[2], Circle[{Cos[t], Sin[t]}/4, 5/4], Black, Point[{0, 0}], Red, Point[{Cos[t], Sin[t]}/4], Green, Line[{{0, 0}, {Cos[t], Sin[t]}}/4], Red, Line[Transpose[{Table[{Cos[t], Sin[t]}/4, {Length[#]}], #}] &@ Select[{x, y} /. (NSolve[{(x - Cos[t]/4)^2 + (y - Sin[t]/4)^2 == 25/16, 5/4*Cos[\[Theta]] - 1/4*Cos[5*\[Theta]] == x, 5/4*Sin[\[Theta]] - 1/4*Sin[5*\[Theta]] == y}, {x, y, \[Theta]}, WorkingPrecision -> MachinePrecision] /. {C[1] -> 0, Or -> List, And -> List, Equal -> Rule}), Re[#] == # &]], Blue, Line[ Transpose[{Table[{Cos[t], Sin[t]}/4, {5}], Table[{Cos[t], Sin[t]}/4 + 5/4*{Cos[t/5 + 2*k*\[Pi]/5 + \[Pi]], Sin[t/5 + 2*k*\[Pi]/5 + \[Pi]]}, {k, 0, 4}]}]]}, PlotRange -> {{-3/2, 3/2}, {-3/2, 3/2}}, ImageSize -> 1000], Graphics[{AbsolutePointSize[10], AbsoluteThickness[2], Circle[{Cos[t], Sin[t]}/4, 3/4], Black, Point[{0, 0}], Red, Point[{Cos[t], Sin[t]}/4], Green, Line[{{0, 0}, {Cos[t], Sin[t]}}/4], Magenta, Line[Transpose[{Table[{Cos[t], Sin[t]}/4, {Length[#]}], #}] &@ Select[{x, y} /. (NSolve[{(x - Cos[t]/4)^2 + (y - Sin[t]/4)^2 == 9/16, 3/4*Cos[\[Theta]] + 1/4*Cos[-3*\[Theta]] == x, 3/4*Sin[\[Theta]] + 1/4*Sin[-3*\[Theta]] == y}, {x, y, \[Theta]}, WorkingPrecision -> MachinePrecision] /. {C[1] -> 0, Or -> List, And -> List, Equal -> Rule}), Re[#] == # &]], Cyan, Line[ Transpose[{Table[{Cos[t], Sin[t]}/4, {3}], Table[{Cos[t], Sin[t]}/4 + 3/4*{Cos[-t/3 + 2*k*\[Pi]/3], Sin[-t/3 + 2*k*\[Pi]/3]}, {k, 0, 2}]}]]}, PlotRange -> {{-3/2, 3/2}, {-3/2, 3/2}}, ImageSize -> 900], three, five], {t, 0, 2 \[Pi], \[Pi]/60.}], 30, AnimationRunning -> False]]

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[math-fun] the tsunami function
by Bill Gosper 18 May '13

18 May '13

Minus the 3rd derivative of a bell curve: In[234]:= D[-E^-t^2, {t, 3}] Out[234]= -12 E^-t^2 t + 8 E^-t^2 t^3 In[235]:= Plot[%, {t, -3, 3}] looks remarkably like the water depth variation along the Tohoku coast during the 11-3-11 tsunami: Drawback, huge flood, huge backwash+drawback, lesser flood. Coincidence? I can supply YouTube URLs. --rwg Challenge: Combine %234 and %235 into a single "command". Macsyma has a similar problem.

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Re: [math-fun] the tsunami function
by Bill Gosper 18 May '13

18 May '13

?!? That was one of the first things I tried! OK, reality rot marches on. Tnx, --rwg On Fri, May 17, 2013 at 6:57 PM, Erich Friedman <efriedma(a)stetson.edu> wrote: > i can't seem to reply to math-fun (why not?) but i think what you want is: > > Plot[Evaluate[D[-E^-t^2, {t, 3}]], {t, -3, 3}] > > erich > > Minus the 3rd derivative of a bell curve: > In[234]:= D[-E^-t^2, {t, 3}] > > Out[234]= -12 E^-t^2 t + 8 E^-t^2 t^3 > > In[235]:= Plot[%, {t, -3, 3}] > > looks remarkably like the water depth variation along the Tohoku coast > during the 11-3-11 tsunami: > Drawback, huge flood, huge backwash+drawback, lesser flood. > Coincidence? I can supply YouTube URLs. > --rwg > Challenge: Combine %234 and %235 into a single "command". Macsyma > has a similar problem. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > >

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Re: [math-fun] Hilbert(Hilbert)
by Stuart Anderson 17 May '13

17 May '13

Using this site http://image.online-convert.com/convert-to-svg I was able to convert the black and white Hilbert to svg, ( the red Hilbert came out black) The compressed file is here; http://www.squaring.net/downloads/hilbert_10.svg.zip uncompressed it is 20M, and is very slow to load in any browser or image viewer (only part of the image was viewable in Chrome and Firefox). Stuart

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Re: [math-fun] Hilbert(Hilbert)
by Henry Baker 17 May '13

17 May '13

'.svg' format is a 'vector' format, which is why I asked for a .svg file. .svg files are displayable by most browsers today without any plugins. At 03:41 PM 5/16/2013, quad wrote: >PNG is not a vector format. It is just a lossless format. > >I do also see loops. > >On Thu, May 16, 2013 at 8:14 AM, Fred lunnon <fred.lunnon(a)gmail.com> wrote: > >> I don't see David's loops; however, near the centre top of hat-crown is >> the hair jutting right which he mentioned. At the centre lower edge of >> hat-brim are two T-junctions, and there are several T-junctions elsewhere >> in the light regions. >> >> All these remain distinct on zooming x8 --- presumably .PNG is a vector >> format permitting indefinite zooming, but nowhere to hide? >> >> My earlier remarks may have been off-target: on more careful inspection, >> it becomes clear that Neil is (as hinted somewhere?) representing pixels >> by lattice nodes, whereas I conceived them as 4-node (black) or 0-node (white) >> walk segments. Furthermore, I had failed to take into account that a sequence >> of 0-node segments creates large rectangles, of which both spare corners >> P,Q might apparently overlap distant segments of the path. >> >> [Retires wiping customary egg from face.] >> >> Fred Lunnon >> >> On 5/16/13, David Wilson <davidwwilson(a)comcast.net> wrote: >> > Very impressive. >> > >> > I do, however, notice a few hairs and loops. >> > >> > One example is a loop extending down from the top of the at just left of the >> > hat's center, which has a hair at its bottom jutting out to the right. >> > >> > I found a few other hairs and closed loops on inspection. >> > >> > I realize these might be artifacts of resolution, but they might also be >> > bugs, so thought you would like to know. >> > >> >> -----Original Message----- >> >> From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun- >> >> bounces(a)mailman.xmission.com] On Behalf Of Bill Gosper >> >> Sent: Thursday, May 16, 2013 7:12 AM >> >> To: math-fun(a)mailman.xmission.com >> >> Subject: Re: [math-fun] Hilbert(Hilbert) >> >> >> >> NeilB> >> >> >> >> Once we got the code working, we soon found out that it would be >> >> necessary to >> >> go down to level 9 to get a good result. [...] >> >> >> >> >Lastly, some things about the final picture could be improved. [...] >> >> >> >> Julian uncovered a bug in Neil's code, whose repair enabled Neil to >> >> reach level 10. I hope he doesn't mind my leaking >> >> http://gosper.org/hilbert-10.png . >> >> >> >> --rwg

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