August 2020 - mathfun@mailman.xmission.com

[math-fun] What do you mean, "Fake news"?
by Bill Gosper 09 Aug '20

09 Aug '20

Define hilbert:[0,1] → [0,1] ⨉[0, i] be the Hilbert Curve function. E.g., In[58]:= #[[1]] & /@ hilbert /@ Range[0, 1, 1/6] Out[58]= {0, 1/2 + I/2, I, 1/2 + I/2, 1 + I, 1/2 + I/2, 1} (corner,middle,corner,middle,corner,middle,corner) (But the Hilbert function is dense with *quadruple* points! In[233]:= unbert[1/2 + I/4] Out[233]= {5/48, 7/48, 41/48, 43/48} In[234]:= hilbert /@ % Out[234]= {{1/2 + I/4}, {1/2 + I/4}, {1/2 + I/4}, {1/2 + I/4}} ) Then, empirically, Out[38]= hilbert[1/2 EllipticTheta[3,0,1/4]] == 2+I/2-1/2 EllipticTheta[3,0,1/2] (Hilbert meets Jacobi.) In[49]:= $RecursionLimit=2500;%38/.R:EllipticTheta[3,0,_]:>Round[R,2^-999] Out[49]= 46727909977607706913365982377862004313596030844679844841051801456313857598476681679840123330654943383021969653600824586099255880151054521703858175/49947976805055875702105555676690660891977570282639538413746511354005947821116249921924897649015871538557230897942505966327167610868612564900642816+I/2== 18234098153986631008405986432405557662078277564803117560901382754419566447242273596263185647274799844525865896201189350917120311482141620852228639269436148149700137184604665725070440348534903185396966774064837413388299772999715897332003677006757173294606789913563827485196972130849897054207/19490628022799998160706764775750376621752453715190015053735812914425897381532852204931230131764020518450609832462817336366918339730406188093155974592625306839062555399912946059741579310980107296705599186958436757747371195850789749891492727230937931225655477606208555094163657179983828221952+I/2 In[50]:= N[Subtract@@%,9] Out[50]= 5.69026240*10^-160 In[51]:= N[%49,160] Out[51]= 0.9355315863940614206652707257275243376937417300295609535554944039715198668204215933408693598375308515525975283883147938845745807397312533384005722661851704410260+0.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000I== 0.9355315863940614206652707257275243376937417300295609535554944039715198668204215933408693598375308515525975283883147938845745807397312533384005722661851704410254+0.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 I

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[math-fun] Fwd: Discrete dynamical system, interacting particles with mass, spin and charge
by James Propp 09 Aug '20

09 Aug '20

I just got this email from someone I never heard of. Any thoughts? (I thought some of the videos were pretty.) Jim Propp ---------- Forwarded message --------- From: Wolfhard Hövel <Wolfhard.Hoevel(a)th-nuernberg.de> Date: Sun, Aug 9, 2020 at 4:30 AM Subject: Discrete dynamical system, interacting particles with mass, spin and charge To: <jamespropp(a)gmail.com> Dear Professor Propp, I would like to propose an algorithm that simulates spontaneously interacting particles with mass, spin and charge, using only position and displacement vectors. Both the momentum and the energy conservation law are always fulfilled. You can see some results of the approach on YouTube. You can find the YouTube videos „Experiments with vectors“ under the following link: https://www.youtube.com/playlist?list=PLBda4QvOnxQHVsqa3l2ixuqD7xlRlYWVx Also you can view various examples interactively on the web page of Ulrich Schwebinghaus: http://www.fraktalwelt.de/myhome/selforg_2.htm I have attached a description of the calculation model below. There I tried to describe the algorithm “Experiments with Vectors” as far as possible with words. I would be happy if you could take a look at them. Sincerely yours, Wolfhard Hövel D 92268 Etzelwang Gerhardsberg 7 Email: wolfhard.hoevel(a)th-nuernberg.de Telephone: +491735737624 Experiments with vectors A digital computer is a wonderful tool for performing mathematical experiments. I would like to introduce an algorithm that spontaneously generates particles that move and interact in Euclidean space. This only happens with the help of position and displacement vectors. First of all I have to define some rules according to which the algorithm should run. These rules are not entirely random and when experimenting quickly show whether the rules produce something interesting. I believe that the proposed algorithm actually gives amazing results. The program describes a discrete dynamical system. In a loop calculation, the states of the system change in equidistant time steps. With each iteration, the states are always recalculated according to specified procedures. A general rule is introduced that a quantity G is made up of the integer multiples of a constant base quantity G0 (G = n G0, n is a natural number). All quantities are finite. Now I can define an important quantity in this way: time. Each cycle in the calculation loop specifies the shortest time step dt0. The time is then granular and corresponds to the number n of iterations t = n dt0. You also need objects with which something happens. Let's start with a point. The point has a certain position in space, which is determined by the position vector r (vectors are printed in bold). This position vector points from a freely chosen zero vector to the specified point. The point is by definition always in motion, I achieve this with a displacement vector dr, which is added to the position of the point with every iteration in the calculation loop. Since dr has by definition a constant length, the step width of the point is constant and is equal to the magnitude of the displacement vector |dr| = dr0. The point now moves straight ahead and has the constant speed v = dr / dt0. According to Newton's laws, this is a physical statement. After a while the point disappears from the screen and that would be a boring event. I have to constrain the movement of the point to a defined area. I achieve this by providing the point with a second point as a partner. These two points are supposed to dance around each other. For this, the dancers need certain rules. The first rule is: move with the velocities v and -v respectively exactly in opposite direction on two parallel lines with the distance s. The second rule is: If your distance becomes longer than the elongation e, turn around and run on two new parallel lines with the distance s. This distance s and the direction of rotation remaining always the same. A third rule should take into account the principle that all quantities can only change step by step, including the parameters s and e. They are integer multiples of the step width dr0. The two points form a system whose center of mass remains constant. The angular momentum of the points is constant and given by v and s. In addition, a maximum distance is set, which the dancing points should not exceed if possible. They don't quite succeed because they move step by step. The limit is not sharp, it can be exceeded by a maximum of dr0. There is still something missing to make the algorithm physical: it is the mass. So I assign a smallest possible mass m0 to each point. This assumption must not appear artificial or forced. A common center of mass can also be specified for a point cloud. Now the constant momentum for a point is given by p = m0 v and the kinetic energy by E = ½ m0 v². The described pair of points ("bivector") is the most important basic building block of the calculation model. It is the smallest possible particle of the simulation. From a geometric point of view, a parallelogram is spanned in space by the distance vector r12 = r2 - r1 between the two points and the displacement vector dr, whose area is equal to s · dr0. The parallelogram is an oriented plane and has a sense of rotation. The bivector exists in every dimension > 1. The bouncing points describe a circular disc, the edge of which is not very sharp. For example, if e >> s the points are oscillating in an almost star shape around the circle with the diameter s and for the special case e = s the points are moving unsharp on a circle with the diameter e. The bivector contains two points and therefore its mass is 2 m0. How can the bivector be represented in a program? At the beginning of each iteration, it is checked whether the distance between the points exeeds e. If this is not the case, the points can continue to move freely in a straight line. If the permitted limit is exceeded, the points must be reflected so that their distance is reduced again. The rules given above must be observed. From a geometric point of view, this is easy to illustrate. The distance between the points is halved and a circle with the diameter s is drawn around the center found, which is also the center of mass of the bivector. Then, starting from the two points, tangents are placed on the circle, whereby the sense of rotation must be the same as before. The points continue to move on these newly found tangents. One can imagine that the two points are connected by a chain with the length e. As long as the chain is still loose, the points can move freely. However, if the chain is suddenly tensioned after a step, the points must reverse abruptly. The chain can be spanned by a maximum of one chain link. This methode of reflection has the advantage that only internal quantities of the bivector are used and no impulse is exchanged to the outside. The bivectors can be moved freely in the space. This fact will become important later. Now we add more points. Each point must be at least a member of one bivector to give it a momentum. However, one point can be a partner of several bivectors and then the situation quickly becomes confusing. Therefore, it makes sense to represent the relationship between the points using a graph of graph theory. The points are the vertices and the bivectors are the edges of the graph. If the point cloud consists of N points, a maximum of N(N-1)/2 bivectors can be inserted between the points. If a point is involved in several bivectors, there is a strong coupling between these linked bivectors. This should be illustrated using a simple example. Suppose we connect three points P1, P2 and P3 with two bivectors with the pairing 1-2 and 2-3. With each iteration, each bivector is treated as before, regardless of the others. The associated displacement vectors dr12 and dr23 result from this. Finally, the displacements are superposed. Point P1 shifts by dr12, point P2 by (-dr12 + dr23) and point P3 by -dr23. Imagine that dancer P2 has to perform the steps of two dance pairs at the same time! The two dance couples will have to dance completely new figures, they are closely linked. This example describes another possible particle. Since it contains three points, its mass is 3 m0. If you look at the energy situation, it gets a little more complicated. Two displacement vectors -dr12 and dr23 are added to point P2. In the case that both vectors have the same direction, the point P2 has the speed 2v. Its kinetic energy is therefore 2 m0 v². On the other hand, if the two displacement vectors are exactly opposite, the kinetic energy is zero. So that the theorem of energy conservation remains valid for the simulation model, the energy of point P2 could now be interpreted as potential energy. As the directions of the displacement vectors will change, the potential energy is converted back into kinetic energy. Nothing is lost, the number of displacement vectors is constant. This interplay is also known from a tensioned and relaxed metal spring. If you experiment with the calculation model described, you will experience a surprise. Under certain conditions, self-organization creates stable structures in the space. These attractors can change spontaneously, they can attract and repel, they behave like particles with mass and obey the momentum conservation law. To do this, however, the algorithm still needs to be expanded. A weak interaction must be added. Let's start again with a point cloud. The points are linked by bivectors. When points are members of several bivectors, there are strongly coupled sets of points, these are our particles. Several isolated graphs and thus particles can be defined. The particles do not yet interact, their graphs are isolated networks in space. How can I couple the isolated particles? To do this, I assign a plus or minus sign to some points. The additional rule is: if the points have opposite signs, the points move iteratively radially towards each other with the step size drL, if they are like-signed, they repel each other. These shifts are added to the other displacements. The momentum conservation law is still observed, the sum of the momentum vectors remains equal to zero. With drL you can control the weak coupling between the particles, often I set drL = dr0 / 150. The algorithm can be programmed on the basis of real or integer numbers. Further information on this particle simulation can be found on the Internet under the keywords „search opus4 hoevel“ . Appendix You can make a bivector geometrically clear with a drawing. Draw two points P1 and P2 16 cm apart. Connect the two points with a straight line and halve them. The point M found in this way is the center of mass of the pair of points. Draw two circles around M with the radius s/2 = 2 cm and e/2 = 4 cm. Place two parallel tangent lines starting from points P1 and P2 to the s-circle. Starting from point P1, move P1 step by step with the step width dr0 = 2 cm along the tangent. The point now jumps to the e-circle and runs through it. As soon as P1 has left the e-circle, a new tangent line from P1 is placed to the s-circle, keeping the direction of rotation, and the game is repeated. The point P1 is now captured and circles the s-circle. Point P2 is always moved in the opposite direction at the same time. This system is a bivector, it is characterized by a constant angular momentum (~ s) and an elongation e. The bivector can be moved freely in space, its position is only determined by the two points P1 and P2.

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Re: [math-fun] High schoolers: simplify this expression
by Axel Vogt 09 Aug '20

09 Aug '20

a,b := (sqrt(108) +- 10)^(1/3) = (6*sqrt(3) +- 10)^(1/3), 0 < b < a One has a*b = (a^3*b^3)^(1/3) = 8^(1/3) = 2, which is elementary. Moreover a-b = a - 2/a is *guessed* to be 2 , using some calculator. The quadratic equation x - 2/x = 2 has solutions 1 +- sqrt(3), by hand. Taking 3rd powers one sees that a = 1 + sqrt(3). The assertion follows. > Date: Sun, 26 Jul 2020 14:55:27 -0700 > From: Henry Baker <hbaker1(a)pipeline.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] High schoolers: simplify this expression > Message-ID: <E1jzocv-0007m4-Tx(a)elasmtp-masked.atl.sa.earthlink.net> > Content-Type: text/plain; charset="iso-8859-1" >

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[math-fun] Heighway meets Jacobi
by Bill Gosper 09 Aug '20

09 Aug '20

Out[157]= Im[dragun[EllipticTheta[3, 0, 1/4] - 1]] == 1/2 (EllipticTheta[3, 0, 1/2] - 1) In[159]:= N[%157 /.EllipticTheta[3, 0, q_] :> Round[EllipticTheta[3, 0, q], 2^-691], 99] Out[159]= 0.564468413605938579334729274272475662306258269970439046444505596028480133179578406659130640162469148 == 0.564468413605938579334729274272475662306258269970439046444505596028480133179578406659130640162469148 Out[139]= EllipticTheta[3, 0, 1/2]/2 - Im[dragun[EllipticTheta[3, 0, 1/4]/2]] + Re[dragun[EllipticTheta[3, 0, 1/4]/2]] == 3/2 In[146]:= $RecursionLimit = 5333; N[%139[[1]] /. T : EllipticTheta[3, 0, _] :> Round[T, 2^-999], 99] Out[146]= 1.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 —rwg NonMathematickers should wonder why I goosed $RecursionLimit simply to replace the 𝜗₃ by a rational approximation. The recursion is in the definition of dragun, iff it receives a rational argument. So its definition begins dragun@0 = 0; dragun@1 = 1; dragun[t_Real | t_Rational] := piecewiserecursivefractal[ . . .

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Re: [math-fun] Paradoxical decomposition of the plane or sphere?
by Dan Asimov 08 Aug '20

08 Aug '20

I'll have to think about Andy's post when it's not so close to my bedtime. But meanwhile I noticed that Hausdorff proved that there *does* exist a paradoxical decomposition of the sphere S^2 - D where D is countable, using the Axiom of Choice: S^2 - D = A + B + C such that A, B, C are all disjoint and isometric to each other, and also A is isometric to B + C. —Dan Andy Latto wrote: ----- No, this is not possible. The key fact here is that it's possible to assign a nonnegative real number, the "content" of the set, to every subset of the plane, such that this agrees with the Lebesgue measure on all Lebesgue-measurable sets, and is finitely additive. It cannot of course be countably additive, because there exist non-measureable sets. Any isometry of the plane can be decomposed into a translation, a rotation, and possibly a reflection. Consider this decomposition for the isometries that take A-> B, B->C, and A+B -> C. Take a disk containing the three points of rotation of these three isometries, that is very large compared to both the distance between the three points of rotation and the lengths of the three translations. Looking at the content of the intersections of this disk with A, B, and C will produce a contradiction. Andy Intersecting A, B, and C with a disk that is very large compared to the length of the On Fri, Aug 7, 2020 at 2:50 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > Does there exist a decomposition of the plane into three disjoint sets > A, B, C such that all three are isometric to each other but also A is > isometric to the union B + C ??? > > Ideally each set would be connected. Also measurable, but I would be > surprised if measurable is possible. > > And ideally, without needing the Axiom of Choice. > > How about the same thing but with some points omitted: > > A + B + C = R^2 - Z > > where Z is a set of measure 0 ??? > > And if not the plane, how about the sphere S^2 ??? > > —Dan -----

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[math-fun] Paradoxical decomposition of the plane or sphere?
by Dan Asimov 07 Aug '20

07 Aug '20

Does there exist a decomposition of the plane into three disjoint sets A, B, C such that all three are isometric to each other but also A is isometric to the union B + C ??? Ideally each set would be connected. Also measurable, but I would be surprised if measurable is possible. And ideally, without needing the Axiom of Choice. How about the same thing but with some points omitted: A + B + C = R^2 - Z where Z is a set of measure 0 ??? And if not the plane, how about the sphere S^2 ??? —Dan

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Re: [math-fun] Paradoxical decomposition of the plane or sphere?
by Dan Asimov 07 Aug '20

07 Aug '20

For the sphere, yes. But for the plane, oo + oo = oo, so it's not immediately obvious why measurability must be ruled out. —Dan Allan Wechsler wrote: ----- Maybe I'm not understanding, but I think we can prove right away that the measurability requirement is not attainable. Two congruent measurable sets, after all, must have the same measure. And I think that if A is congruent to the union of the disjoint sets B and C, its measure must be at least |B| + |C|. But maybe I am forgetting some horrible pathological case. -----

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[math-fun] What do you mean, "Fake news"?
by Bill Gosper 06 Aug '20

06 Aug '20

Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16) In[372]:= N[%364,105] Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 ==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg

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[math-fun] Minimizing products of differences of powers in the same base
by Andres Valloud 05 Aug '20

05 Aug '20

Hi, I'm looking for some references to the problem of minimizing products of the form \prod_{j=0 \neq k}^d |b^k - b^j| where k is chosen between 0 and j, for a fixed base b. It looks like, for sufficiently large b > 1, minimizing this product requires choosing k = d, unless b gets close to 1 in which case eventually k must be chosen to be e.g. floor(d/2). If 0 < b < 1, the reverse process occurs: k must be chosen to be 0, unless b is really close to 1 in which case k must become e.g. floor(d/2). I'd like to believe there's literature on this subject, however searching for things like "minimize the product of differences of powers" returns material that is only related to the search query by the fact that the terms appear in the search results. Looking at the expression with Wolfram, like this, https://www.wolframalpha.com/input/?i=%5Cprod_%7Bj%3D0%7D%5Ed+%28b%5Ek+-+b%… returns a reference to the q-Pochhammer symbol. Digging through that, I was not able to immediately find a way to minimize the product or show which factor has to be removed from the product. I also tried making b = e, applying ln to the product, then approximating the sum by the integral. The resulting integral (courtesy of Wolfram Alpha) relies on Spence's dilogarithm function Li_2. After some approximations evaluating the integral, it looks like the product as a whole is about e^O(d). However, I do not immediately see how to minimize the product by choice of k depending on b. Do you have any suggestions on where to look? TIA! Andres.

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[math-fun] Sum of consecutive terms = shortcut
by Éric Angelini 01 Aug '20

01 Aug '20

Hello Math-Fun Instead of writing: 1+2+3+4+5+6+7+8+9+10+11+12+13 +14+15+16+17+18+19+20+21+22+23 +24+25+... I could’ve written the « shortcut »: 3+7+11+24+46+62+37+41+94+... as 3=1+2 7=3+4 11=5+6 24=7+8+9 46=10+11+12+13 62=14+15+16+17 37=18+19 41=20+21 94=22+23+24+25 ... The terms 3,7,11,24,46,62,37,41,94,... are numbers that can be expressed as the sum of k>1 consecutive integers _in only one way_ (they belong to A038550). The terms were always chosen to be the smallest one that could belong to the « shortcut » sequence. One needs to backtrack sometimes as some chunks of consecutive integers cannot be « shortcut » (like the chunk that starts with 16+17+18... if I’m right). This « backtrack » remark applies to the lexicographically earliest « shortcut » tentative: 3,7,11,24,46,29,... which seems impossible to extend. Best, É. Catapulté de mon aPhone

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