November 2018 - mathfun@mailman.xmission.com

[math-fun] http://mathworld.wolfram.com/EllipticAlphaFunction.html α(11) is missing.
by Bill Gosper 15 Nov '18

15 Nov '18

Numerically, I get 1/12 (2 + 6 Sqrt[11] - 2 (-5 + Sqrt[33]) (17 + 3 Sqrt[33])^(1/3) + (-7 + Sqrt[33]) (17 + 3 Sqrt[33])^(2/3)) . —rwg

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[math-fun] MathWorld EllipticK bogon?
by Bill Gosper 13 Nov '18

13 Nov '18

http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html Eqn (18) gives Out[290]= K[Sqrt[1 - k^2]] == E[k]/(k (1 - k^2)) - K[k]/k (modulus convention) which translates to Mathematica (parameter convention) Out[284]= EllipticK[1 - k^2] == EllipticE[k^2]/(k (1 - k^2)) - EllipticK[k^2]/k which doesn't work. Could there be any such relation between K, K', and only one of {E, E'}, in light of Legendre's relation? —rwg

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Re: [math-fun] Dandelin hyperbola proof
by Henry Baker 12 Nov '18

12 Nov '18

Animated gif's work just fine, and have been supported since the middle ages (of the Internet). FFMPEG can convert any movie or image sequence into an animated gif, although if you want decent color rendition, the process is currently more complicated than one might want. But if color accuracy isn't an issue -- e.g., animated mathematical objects -- then it isn't difficult. Also, the symbolic algebra system Maxima can generate animated gif's, and I believe that other CAS systems can, as well. At 07:04 AM 11/12/2018, Fred Lunnon wrote: >An interactive fly-through demo with fancy lighting? > >On Wikipedia? > >It might just be feasible, if XML is up to this sort of job ... current user PC's have plenty of power to run what is effectively a simple 3-D game. > >This might make a nice student project. > >But I never even got around to uploading the static paper to ArXiV yet, let alone getting it past Math. Intell. --- so don't hold yer breath! > >WFL > >On 11/12/18, Bill Gosper <billgosper(a)gmail.com> wrote: >> On 2018-11-11 17:21, Fred Lunnon wrote: >>> Sadly, technological progress has relegated our early (and usually >>> laboriously crafted) efforts to museum exhibits --- compare the demo at >>> https://en.wikipedia.org/wiki/Dandelin_spheres >>> >>> WFL >> And where might we find a similarly technologically ultraprogressive >> Wikipedient obsoletion of your laboriously crafted gosper.org/binomial.pdf ? >> Ârwg >>> >> >> >> >> On 11/11/18, Billl Gosper <billgosper(a)gmail.com> wrote: >>>> Eons ago, trying to prove that locus-of-constant-difference is a conic section, >>>> I reinvented Dandelin spheres, with a possibly simpler proof that does not >>>> refer to their centers nor radii. gosper.org/hyperb.GIF >>>> uses instead the lemma that all tangents to a sphere from a point are >>>> equally long. The drawing probably needs stereo vision to be convincing. >>>> I'm not sure I realized at the time that the spheres could be unequal. >> Ârwg

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Re: [math-fun] Dandelin hyperbola proof
by Bill Gosper 12 Nov '18

12 Nov '18

On 2018-11-11 17:21, Fred Lunnon wrote: > Sadly, technological progress has relegated our early (and usually > laboriously crafted) efforts to museum exhibits --- compare the demo > at > https://en.wikipedia.org/wiki/Dandelin_spheres > > WFL And where might we find a similarly technologically ultraprogressive Wikipedient obsoletion of your laboriously crafted gosper.org/binomial.pdf ? —rwg > > > > > On 11/11/18, Bill Gosper <billgosper(a)gmail.com> wrote: >> Eons ago, trying to prove that locus-of-constant-difference is a conic >> section, >> I reinvented Dandelin spheres, with a possibly simpler proof that does not >> refer to their centers nor radii. gosper.org/hyperb.GIF >> uses instead the lemma that all tangents to a sphere from a point are >> equally long. The drawing probably needs stereo vision to be convincing. >> I'm not sure I realized at the time that the spheres could be unequal. —rwg

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[math-fun] Some conic thoughts (long)
by Dan Asimov 12 Nov '18

12 Nov '18

I don't know much about this, but that never stopped me before. >From things I've heard, I get the feeling that the best way to think of conic sections in the plane is as if they belonged to the projective plane P^2. P^2 is a surface that is not easy to visualize, but one way to describe it is: P^2 = the ordinary plane R^2 plus the "circle at infinity". The circle at infinity (aka P^1) consists of one point for each family of parallel lines in R^2. This is equivalent to the set of antipodal pairs {exp(it, -exp(it)} for 0 <= t <= π, except we have to identify the pairs for t = 0 and t = π; hence a "circle". Then ellipses are hyperbolas and hyperbolas are ellipses (when you include the points at infinity that belong in the locus). For example, the hyperbola (x/4)^2 - (y/3)^2 = 1 includes which points at infinity? Plug in the equation of a line through 0: y = m x to get ((1/4)^2 - (m/3)^2) = 1 / x^2 Now let x —> oo to determine the values m = ±3/4, which tell us just which points on the circle at infinity we should add to our hyperbola to "complete" it. Of course, these are the same points in the circle at infinity determined by the asymptotes. Now consider conic sections, what you get in 3-space by cutting a cone with a plane. The projective wisdom seems to apply here, too. By adding new points to R^3 we get projective 3-space P^3. The new points this time will be one point for each complete family of mutually parallel planes in R^3, or in other words one point for each antipodal pair of points on the unit sphere S^2 in R^3. Now instead of a *circle* at infinity, we have to add a copy of P^2 at infinity. Let's see what we get when we try to complete a quadratic surface Q in R^3. For any x in Q, let T(x) denote the tangent plane to Q at the point x. Since T(x) is a plane in R^3 it belongs to just one point of the P^2 at infinity. What happens to T(x), as a point of P^2, as |x| —> oo ? If Q is a hyperboloid of one sheet, it's easy to see from a picture that the directions of the tangent planes T(x) approach two simple closed curves in the sphere S^2, each curve the antipodal image of the other. In P^2 itself this is just one simple closed curve at infinity to add to Q. Thus Q gets "joined to itself at infinity". But wait - things are seldom what they seem (cream is milk and milk is cream). In order to add this simple closed curve at infinity, you have to *twist* the hyperboloid, because look at which points have parallel tangent planes! Wait, that's no ellipsoid at all, that's topologically ... ... a torus! And while we're at it, what is a cone once it's been "completed" this way? It's a ... a ... I have no idea what it's called. Its a torus with a narrow waist (circumference zero). And going back to the hyperbola in the plane, we should have "twisted" that, too, when completing it. The strangeness we are detecting all (OK, mostly) goes away in the setting that algebraic geometers have decided is most appropriate to look at this kind of thing: Complex projective space and homogeneous polynomial equations. Say P(X,Y,Z) is a complex quadric polynomial in complex variables X, Y, Z such that every monomial is the same degree. E.g. P(X,Y,Z) = X^2 + YZ. Then whenever X, Y, Z make the equation P(X,Y,Z) = 0 true, then so do cX, cY, cZ for any nonzero complex number c, since P(cX,cY,cZ) = c^2 P(X,Y,Z) = 0. Ignoring the triple X = Y = Z = 0, we may as well take equivalence classes in the space C^3 - {(0,0,0)} via (X,Y,Z) ~ c * (X,Y,Z). When we do this, the set of equivalence classes is called "the complex projective plane and denoted CP^2. And now I have to eat dinner. —Dan

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[math-fun] Dandelin hyperbola proof
by Bill Gosper 12 Nov '18

12 Nov '18

Eons ago, trying to prove that locus-of-constant-difference is a conic section, I reinvented Dandelin spheres, with a possibly simpler proof that does not refer to their centers nor radii. gosper.org/hyperb.GIF uses instead the lemma that all tangents to a sphere from a point are equally long. The drawing probably needs stereo vision to be convincing. I'm not sure I realized at the time that the spheres could be unequal. —rwg

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[math-fun] Words in a graph puzzles
by Victor Miller 09 Nov '18

09 Nov '18

There are a bunch of puzzles I've encountered of the following form: You're given an undirected (finite) graph G, and a list of words. The problem is to label the vertices of the graph G with letters so that each of the words can be "spelled out" by following a path in the graph. The specific versions of this puzzle are the following: The list of words is the names of the nine (before Pluto got demoted) planets. The graphs are "rectangle graphs" -- i.e. an (m,n) rectangle graph has vertices labeled by (i,j) for 1 <= i <= m, 1 <= j <= n. Two vertices (i,j) and (i',j') are adjacent if you can transform one to the other by choosing one coordinate and adding or subtracting 1 -- i.e. Manhattan distance 1. The challenges are for the (5,7) and (4,8) rectangle graphs with the above words. A bigger challenge is to prove that you *can't* do it for the (5,6) graph. I know of no reasonable proof for the impossibility for the (5,6) graph other than running some intelligent exhaustive program for a long time (over 1 day). Can anyone do better? Victor

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Re: [math-fun] conic sections
by Dan Asimov 09 Nov '18

09 Nov '18

So I was idly wondering what the locus is for the sum of distances to *three* points to be constant looks like. For convenience I took (-1,0), (1,0) and (0, sqrt(3)) as the vertices of an equilateral triangle, and the sum-of-distances equal to 4. Anyone care to guess what this looks like before graphing it? —Dan ----- On Nov 8, 2018, at 10:06 AM, Mike Speciner <ms(a)alum.mit.edu> wrote: > > And the focus to focus reflection of ellipses is easy to see when using the constant sum of distances to foci property. I guess I need to sketch a proof that at each point x on the ellipse, the curve is perpendicular to the bisector of the lines connecting x to the two foci. -----

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Re: [math-fun] conic sections
by Bill Gosper 09 Nov '18

09 Nov '18

You might mention that any old cone contains *all* the conic sections, e.g., ellipses of every size and eccentricity. Thus, through any ellipse fixed in space there are infinitely many cones. It isn't well known, but their apices all lie on a hyperbola through the two foci. And the apices of all the cones through this hyperbola lie on the original ellipse! gosper.org/conethm.png (Somebody should animate this.) —rwg On 2018-11-08 10:35, Henry Baker wrote: > You should definitely show them the "Dandelion" (actually, Dandelin!) Theorem, > which I call the *ice cream cone* theorem. It proves the sum-of-the-distances > theorem about ellipses w/o requiring analytic geometry. > > When my plane geometry teacher showed it to us, it completely blew my mind! > > In particular, I couldn't get over the fact that the ellipse was symmetrical, > even though the spheres were of different sizes! > > Even if the kids haven't yet had 3D geometry, they can still follow > this easy theorem. > > https://en.wikipedia.org/wiki/Dandelin_spheres > > At 05:43 AM 11/8/2018, Cris Moore wrote: >>I'm going to talk to my daughter's 8th grade class about conic sections. >> >>I want to focus on foci (ha), and how curves with beautiful geometric descriptions also have nice algebraic descriptions in Cartesian geometry. >> >>But I found it surprisingly tricky to work out examples. >> >>Consider an ellipse with foci at (-1,0) and (+1,0), and define the set of points where the sum of its distances from these two is 4. >> >>Using Pythagoras' theorem produces an equation with a bunch of square roots. >> >>Squaring both sides eventually turns this into 3x^2 + 4y^2 = 12 but this takes a bunch of steps of algebra, and mysterious cancellations of 4th-order terms. >> >>Similarly, it takes a fair amount of work to get from the hyperbola with foci at (+2,+2) and (-2,-2), where the difference in distances is 4, to the simple equation xy = 2. >> >>Am I doing something wrong? >> >>Is there an easier way to get from foci and distances to these simple quadratic equations - without recourse to canonical forms, linear transformations, polar coordinates etc.? >> >>Of course, I then want to talk about light waves bouncing from one focus to another… >> >>I'm not sure how to justify this without a little calculus. >> >>- Cris > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] conic sections
by Cris Moore 09 Nov '18

09 Nov '18

I’m going to talk to my daughter’s 8th grade class about conic sections. I want to focus on foci (ha), and how curves with beautiful geometric descriptions also have nice algebraic descriptions in Cartesian geometry. But I found it surprisingly tricky to work out examples. Consider an ellipse with foci at (-1,0) and (+1,0), and define the set of points where the sum of its distances from these two is 4. Using Pythagoras’ theorem produces an equation with a bunch of square roots. Squaring both sides eventually turns this into 3x^2 + 4y^2 = 12 but this takes a bunch of steps of algebra, and mysterious cancellations of 4th-order terms. Similarly, it takes a fair amount of work to get from the hyperbola with foci at (+2,+2) and (-2,-2), where the difference in distances is 4, to the simple equation xy = 2. Am I doing something wrong? Is there an easier way to get from foci and distances to these simple quadratic equations - without recourse to canonical forms, linear transformations, polar coordinates etc.? Of course, I then want to talk about light waves bouncing from one focus to another… I’m not sure how to justify this without a little calculus. - Cris

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