February 2007 - mathfun@mailman.xmission.com

Re: [math-fun] Re: Simplest Ovals
by Marc LeBrun 13 Mar '07

13 Mar '07

Fascinating and surprisingly long-lived thread! My "natural" choice would be to make ovals by interpolating or blending two ellipses (perhaps from too much computer graphics): Standardize the graphs of the ellipses by aligning them with their major axis coincident with the X axis and scale them so their ends are at x=0 and x=1. These graphs form a family indexed by "eccentricity parameter" [e]: y = ellipse[e](x) Now define oval[e0,e1] := (1-x) ellipse[e0](x) + x ellipse[e1](x) (having "eccentricity vector" [e0,e1]) So if ellipse(x) is quadratic in x then oval(x) will be some cubic. Alternatively, we might define oval[e0,e1] := ellipse[(1-x) e0 + x e1](x) For what parameterizations of eccentricity are these the same?

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[math-fun] irrational eggs
by James Buddenhagen 28 Feb '07

28 Feb '07

Since eggs are aux currant, let me say that my favorite eggs are the eggs of real elliptic curves. Many years ago I asked Peter Montgomery if the rational points on a real elliptic curve (of positive rank over Q) are dense in the real curve. If I remember correctly, his answer was that they are dense in the connected component containing the identity but that the egg might have no rational points. So, since this list obviously has readers who know much more than I do, let me ask: suppose y^2 = cubic(x) models an elliptic curve of positive rank over Q, and suppose that the cubic has 3 real roots. How can we tell (say from the coefficients of the cubic) whether or not the real egg has no rational points or is dense with rational points? Specific examples would be really nice. Jim Buddenhagen

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[math-fun] late follow-up
by Michael Reid 28 Feb '07

28 Feb '07

some time ago, i wrote: > > Before people go too far looking for such charts, I have a question: is > > it possible to have 4 points in a plane, no 3 colinear, so that all 6 > > distances are integers? And with the additional constraint that no 2 > > distances are equal? > > it is possible to have arbitrarily many points in the plane, > no three collinear, such that all distances between them are > integers, and no two such distances are equal. there hasn't been any response, except for dan's query "why is this true?". since no one has answered, here is my solution. rational points on the unit circle x^2 + y^2 = 1 are (rationally) parametrized by (x, y) = P(t) = ((1 - t^2) / (1 + t^2), 2t / (1 + t^2)) . let A = P(0) = (1, 0), B = P(oo) = (-1, 0) and O = (0, 0) ; then the parameter t is tan(<PBA) = tan(<POA / 2) . it is easy to show that the distances PA and PB are rational if and only if t^2 + 1 is the square of a rational number, which is the case if and only if s = tan(<PBA / 2) is rational. we get a new parametrization of the unit circle by this parameter s : P(s) = ((1 - 6s^2 + s^4) / (1 + s^2)^2, 4s(1 - s^2) / (1 + s^2)^2) for rational s , such points have rational distance to both A and B , and the points obtained are dense in the unit circle. moreover, by ptolemy's theorem, any two such have rational distance to each other. this gives an infinite subset of the plane such that all the pairwise distances are rational, and one can also include the center of the circle. i do not know if this is a maximal subset of the plane with this property, although i think that it is likely. it remains to get a set with all distances different. this can be done in several ways. one way is to choose an infinite sequence {P_j} of points in the above set such that each point is on the top half of the circle, and such that <P_(j+1)OA < (<P_jOA) / 2 . for an explicit example, choose s = 2^(-j) , j = 0, 1, 2, ... to get the points P_j = ((2^(4j) + 6(2^(2j)) + 1) / (2^(2j) + 1)^2 , 4(2^j)(2^(2j) - 1) / (2^(2j) + 1)^2) . another way is to choose z to be the complex number corresponding to P(s) , and then to take the complex numbers z^(2^k) , k = 1, 2, 3, ... . (this will work as long as s is not 0, 1, -1 or oo , so that z is not a root of unity.) for a *finite* set, we can get all distances to be integers simply by scaling. i do not know if it's possible to have an infinite set, not all on a line, such that all pairwise distances are integers. i also do not know about dan's follow-up question: | And what about the higher-dimensional analogues: | | Q3: What is the maximum number of points in R^n, no hyperplane | (i.e., n-1 dimensional plane) containing n+1 of them, such that all | interpoint distances are integers? but i note that even with the weaker condition that no hyperplane contains all of the points, the question is still non-trivial! mike

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Re: [math-fun] Re: Simplest Ovals
by Daniel Asimov 28 Feb '07

28 Feb '07

Franklin writes: << Actually, I would think that two degrees of freedom is the ideal: one for how long the oval is compared to its width (eccentricity), and one for relative size of big end vs. small end (ovalicity?). >> I agree. In fact, here's a somewhat imprecise physical basis for determining such an oval (as I mentioned to WFL): Let kmin and kmax, 0 < kmin < kmax, be the curvatures at the two local maxima our oval will have. Now choose a radius r intermediate between rmin = 1/kmax and rmax = 1/kmin. (A natural choice is the arithmetic mean of rmin and rmax, i.e., r = 1/harmonic_mean(kmin, kmax), since 1/curvature = radius of osculating circle.) Now consider a perfect circle C of radius r made a thin strip of spring steel (an idealized such circle would be ideal). Constrain one point of C to have curvature kmin, and the antipodal point to have curvature kmax. Then let C relax into its natural shape under these constraints. That's my ideal oval. --Dan

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[math-fun] Re: Simplest Ovals
by Henry Baker 27 Feb '07

27 Feb '07

After doing some plotting with Maxima/gnuplot on some ovals produced from parabolae, I concluded that the ovals produced by parabolae from isosceles triangles were a little bumpy for my taste. The parabolic top lobe looks pretty good, but the two bottom lobes are too pointy. There appear to be no degrees of freedom to adjust things. Rules: 1. The apex of each of the three parabolae goes through one of the vertices. 2. At the points of intersection of the parabolae (in pairs), the slopes also match. This is kind of cool, as it appears that each triangle has a unique circumparabolic path. If one were to make an amusement park ride based on this path, and trace it appropriately, one would successively undergo uniform acceleration in three different directions, with a discontinuous change in acceleration in between. I wonder if there are rides in amusement parks with this sort of characteristic. I did find after a Google search that mechanical cams are sometimes shaped to impart uniform acceleration at various portions of their turning. At 08:52 PM 2/25/2007, Henry Baker wrote: >I did a little more research on this topic & found that the answer appears to be very simple. > >Consider a moving particle going around the outside of a triangle as fast as it can. It obviously has to change direction in order to complete a circuit of the triangle -- the simplest way to do this is obviously the circumcircle. However, going around the circumcircle may take longer than utilizing this same potential for acceleration & deceleration to navigate closer to the triangle. For example, if the triangle is very long and thin -- approximating a line segment -- then the fastest way to circumnavigate it would be to utilize all of the acceleration to gain as much speed as possible up to the midpoint, and then to utilize all of the acceleration to slow down towards the end. The reason is to avoid the spike in acceleration that would be required to turn the 180 degree corner at the end (we have limited the maximum magnitude of acceleration for the entire round trip). > >It would appear that the fastest circuits around triangles utilize pieces of _parabolas_ to get around the triangle faster -- three pieces to be exact. During each of the three parabolic pieces, the entire ability to accelerate is used to create a uniform acceleration field much like that studied in high school physics by the acceleration due to gravity, except that the uniform acceleration is in the plane of the triangle, and used to slow down the particle just enough to clear each vertex before accelerating to the next. > >One way to think about this curve is to follow Hamilton and consider its "hodograph" -- take all of the _velocity_ vectors and bunch them at the origin. Then the hodograph for the fastest curve can be constructed by taking each side of the triangle and considering it as a vector in the direction that the particle will "trace" it (i.e., stay "close" to it); place each of these three vectors at the origin. Then the acceleration vectors for the fastest curve will trace out the perimeter of the triangle formed by these three vectors (suitably scaled). Since the _magnitude_ of the acceleration is at all times the same, the perimeter will be traced out at constant speed (we aren't charged for the change in direction of the acceleration at the corners of this derived triangle). Around the original triangle, however, the particle will speed up & slow down as it goes around the parabolic path pieces. > >(The intuitive idea is that the longer sides of the triangle give the particle more distance to speed up and more distance to slow down, so the particle can achieve a higher velocity "along" those sides.) > >If the original triangle is isosceles, then the overall shape of the 3 parabolic pieces will look egg-shaped. > >I haven't yet worked out what triangle shape will produce a pleasing egg shape. > >If someone already has a particle simulation applet, then perhaps these data could be put into it. > >At 08:09 AM 2/22/2007, Henry Baker wrote: >>In bicycle/car races, the vehicles have a maximum acceleration potential. For simplicity's sake, let's assume that the max acceleration in any direction is capped at a certain constant. When going around corners, these vehicles "go wide", so that they don't go beyond their maximum acceleration when turning the corner. On the other hand, their ability to "go wide" is limited by the width of the road/track. >> >>So, one way to produce an egg-shape is to consider a racing track of a certain width, whose overall shape is in the form of an isosceles triangle. If the triangle is equilateral, and the width is wide enough, then the curve is circular. However, when we start reducing the width, we get a curve which starts getting pinched in 3 places. If we then lengthen the altitude of the triangle so that it is no longer equilateral, then the curve starts looking egg-shaped. >> >>I assume that curves such as these have been studied. Does anyone know what their equations are? Can anyone point me to any references?

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[math-fun] More oval pix up
by Steve Witham 26 Feb '07

26 Feb '07

Fred supplied pictures of ovals of Greg Fee's and Bill Gosper's, which I've added to http://www.tiac.net/~sw/2007/02/Steve_Gray_oval --Steve

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[math-fun] Re: Simplest Ovals
by Henry Baker 26 Feb '07

26 Feb '07

I did a little more research on this topic & found that the answer appears to be very simple. Consider a moving particle going around the outside of a triangle as fast as it can. It obviously has to change direction in order to complete a circuit of the triangle -- the simplest way to do this is obviously the circumcircle. However, going around the circumcircle may take longer than utilizing this same potential for acceleration & deceleration to navigate closer to the triangle. For example, if the triangle is very long and thin -- approximating a line segment -- then the fastest way to circumnavigate it would be to utilize all of the acceleration to gain as much speed as possible up to the midpoint, and then to utilize all of the acceleration to slow down towards the end. The reason is to avoid the spike in acceleration that would be required to turn the 180 degree corner at the end (we have limited the maximum magnitude of acceleration for the entire round trip). It would appear that the fastest circuits around triangles utilize pieces of _parabolas_ to get around the triangle faster -- three pieces to be exact. During each of the three parabolic pieces, the entire ability to accelerate is used to create a uniform acceleration field much like that studied in high school physics by the acceleration due to gravity, except that the uniform acceleration is in the plane of the triangle, and used to slow down the particle just enough to clear each vertex before accelerating to the next. One way to think about this curve is to follow Hamilton and consider its "hodograph" -- take all of the _velocity_ vectors and bunch them at the origin. Then the hodograph for the fastest curve can be constructed by taking each side of the triangle and considering it as a vector in the direction that the particle will "trace" it (i.e., stay "close" to it); place each of these three vectors at the origin. Then the acceleration vectors for the fastest curve will trace out the perimeter of the triangle formed by these three vectors (suitably scaled). Since the _magnitude_ of the acceleration is at all times the same, the perimeter will be traced out at constant speed (we aren't charged for the change in direction of the acceleration at the corners of this derived triangle). Around the original triangle, however, the particle will speed up & slow down as it goes around the parabolic path pieces. (The intuitive idea is that the longer sides of the triangle give the particle more distance to speed up and more distance to slow down, so the particle can achieve a higher velocity "along" those sides.) If the original triangle is isosceles, then the overall shape of the 3 parabolic pieces will look egg-shaped. I haven't yet worked out what triangle shape will produce a pleasing egg shape. If someone already has a particle simulation applet, then perhaps these data could be put into it. At 08:09 AM 2/22/2007, Henry Baker wrote: >In bicycle/car races, the vehicles have a maximum acceleration potential. For simplicity's sake, let's assume that the max acceleration in any direction is capped at a certain constant. When going around corners, these vehicles "go wide", so that they don't go beyond their maximum acceleration when turning the corner. On the other hand, their ability to "go wide" is limited by the width of the road/track. > >So, one way to produce an egg-shape is to consider a racing track of a certain width, whose overall shape is in the form of an isosceles triangle. If the triangle is equilateral, and the width is wide enough, then the curve is circular. However, when we start reducing the width, we get a curve which starts getting pinched in 3 places. If we then lengthen the altitude of the triangle so that it is no longer equilateral, then the curve starts looking egg-shaped. > >I assume that curves such as these have been studied. Does anyone know what their equations are? Can anyone point me to any references?

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[math-fun] Taylor series "inverter"?
by Steve Witham 25 Feb '07

25 Feb '07

Is there a machine out there, analogous to Plouffe's Inverter ( http://pi.lacim.uqam.ca/eng/ ), or the sequences database ( http://www.research.att.com/~njas/sequences/ ), that can take the first n coefficients of a Taylor series and find a function close to that Taylor series? Better yet (of course) take *any* expression for a function and return the simplest expression of the function based on well- known functions. --Steve p.s. I am behind in my egg-collecting.

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[math-fun] Re: Interesting ovoids
by dean＠math.ucdavis.edu 24 Feb '07

24 Feb '07

Franklin T. Adams-Watters wrote: > I asked, what happens if the 1 in that equation is replaced by some > other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us > period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, > x/y, and then loops. ... > When you graph a(n) vs. a(n+1), with suitable parameters the result is > an ovoid. ... > The obvious next step is to find a non-trivial function f_c which > satisfies the functional equation f_c(x,y) = f_c(y,(y+c)/x). ... > This is where I got stuck - I wasn't able to find such an f_c. Maybe > somebody with better analytic skills than I have will be able to. Let f_c(x,y) = (c + (c+1)(x+y) + x^2 + y^2 + x^2 y + x y^2)/(x y). I wish I could claim that I used great analytic skills, but all I did was look at some special cases, assume that the curves satisfied polynomial equations of low degree, and use Mathematica to solve for the coefficients. After a few cases, the pattern was obvious, and led to the invariant f_c. I did notice that this recurrence is related to a 7-th order Somos recurrence: Define s(0), ..., s(4) arbitrarily and then define s(n) for n>=5 so that a(n) = s(n) s(n+5) / (s(n+2) s(n+3)). Then s satisfies the Somos recurrence s(n) s(n-7) = s(n-1) s(n-6) + c s(n-3) s(n-4). I'm sure there must be an easy derivation of f_c from known facts about Somos sequences, but I couldn't find one. Both Franklin's recurrence and Somos's have been studied by many people, so this is probably all well-known to someone. David W. Cantrell wrote: > Here's an example. Suppose a(1) = 1 and a(2) = 2 and c = 1/2. The > sequence itself is not periodic. The continuous function, of which the > sequence is a restriction, has a period of roughly 5.248; a graph of > one cycle is shown at > <http://img453.imageshack.us/img453/744/recursioncu0.gif>. > Its maximum and minimum are roughly 2.549 and 0.741, resp. The max and min are roots of 4x^4 - 80x^3 + 228x^2 - 128x + 1 = 0 Numerically, they're about 2.549193378506151504365717868828 and 0.741012926112203188665166488474 Dean Hickerson dean(a)math.ucdavis.edu

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[math-fun] Picture of Steve Gray's oval
by Steve Witham 24 Feb '07

24 Feb '07

Fred Lunnon sent me a picture of an oval based on Steve Gray's suggestion. >What about the locus of points such that d(A)+kd(B) is constant, I pulled out a bunch of quotes from the list to put it in context: http://www.tiac.net/~sw/2007/02/Steve_Gray_oval/index.html --Steve

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