May 2020 - mathfun@mailman.xmission.com

[math-fun] prob gen fcn for *product* of Bernoulli random vars?
by Henry Baker 29 May '20

29 May '20

If I have a Bernoulli random variable with a probability p of value 2 and a probability q=(1-p) of value 98, I can represent its PGF as G(z,p,q,1), see below. Is G(z,p,q,n) below correct for the PGF of the *product* of n independent such random samples? (%i1) G(z,p,q,1); 98 2 (%o1) q z + p z (%i2) G(z,p,q,2); 2 9604 196 2 4 (%o2) q z + 2 p q z + p z (%i3) G(z,p,q,n); n ==== i n - i \ i n - i 2 98 (%o3) > binomial(n, i) p q z / ==== i = 0 --- An aside: trying to *factor* such a polynomial as in %o2 above causes mucho pain! Are there any poly factoring algs that can factor polys like these (or prove they aren't factorable) ?

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[math-fun] Polynomial torus knots redux
by Dan Asimov 29 May '20

29 May '20

Say p, q are positive integers with gcd(p,q) = 1 and p > q. Then a (p,q) torus knot in S^3 in R^4 = C^2 = {(z,w)} is given by the intersection of the 3-sphere S^3 |z|^2 + |w|^2 = 1 and the variety z^p + w^q = 0. Via stereographic projection S^3 - {(0,0,0,1)} —> R^3 = {(X,Y,Z)}, this gives the complex polynomial equation P(X,Y,Z) = 0 where P(X,Y,Z) = 2^p (X+iY)^p + (Q+1)^(p-q) (2Z + i(Q-1))^q and Q stands for X^2 + Y^2 + Z^2, which is equivalent to the two real polynomial equations Re(P(X,Y,Z)) = 0 and Im(P(X,Y,Z) = 0, each of degree = 2p. This is the lowest degree for two real polynomials that give the (p,q) torus knot in R^3 (p > q). Of course we could write (Re("))^2 + (Im("))^2 = 0 and get the same knot as the zero-set of just one real polynomial of degree 4p. —Dan

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[math-fun] New Minimal Trefoil with a Period ODE
by Brad Klee 28 May '20

28 May '20

Proof Certificate: https://0x0.st/ifNB.txt . Meta-Data, by (line #): (1) Parametric Eqs, (2) Triangular-Symmetric Surface, degree 3, (3) Triangular-Symmetric Surface, degree 4, (4) dt/d\phi, (6) Annihilator of dt/d\phi, an ODE in k, (7) Annihilator Certificate. It can be checked by Mma script: https://0x0.st/ifNL.txt Prints 8 zeros. I'm guessing Ed Pegg will want to publish this data, with a little slider for viewing the knot: https://0x0.st/ifNv.png . Woohoo! --Brad

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[math-fun] The eraser game
by Éric Angelini 28 May '20

28 May '20

Hello Math-Fun, (I'm sorry to propose this here – but SeqFan didn't publish my last three posts, dunno why) The idea is to always extend the sequence E with the sum of the last two terms – and to apply the "eraser rule" when needed: E = 1, 2, 3, 5, 8, 13, 2(1), 15, (1)7, ... The eraser rule says that the last term of E and the coming sum cannot share any digit. The duplicate digit (parenthesized) are thus erased from the sum. (When needed, the remaining digits are concatenated to form a new integer). Example: We see that after 8 + 13 = 21 we must erase the 1 from 21 and proceed with 2; this 2 will be added to 13 to produce 15; but as 2 + 15 = 17 we will erase 1 again from 17 and extend E with 7; etc. The above start stops quickly: E = 1, 2, 3, 5, 8, 13, 2(1), 15, (1)7, 22, (2)9, 31, 40, 71, (111) stop. What about leading zeros in the "erased" sum? Let's see: E = 598, 414, (1)0(1)2,... Well, let's erase any leading zero – we will thus proceed here with: E = 598, 414, 2, 416, (41)8, 424, etc. Questions: What is the lexicographically earliest start pushing the sequence into a loop? Is there a start leading to an infinite sequence? Best, É. As usual, the same text as above is here -- with colors and pictures: http://cinquantesignes.blogspot.com/2020/05/the-eraser-game.html

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Re: [math-fun] Fibred knots
by Dan Asimov 28 May '20

28 May '20

So I wanted to know what is this glueing map h : T^2 - D^2 —> T^2 - D^2 which for the trefoil knot K determines its complement S^3 - K via S^3 - K = (T^2 - D^2) x [0,1] / (x,1) ~ (h(x),1) ??? I finally figured it out, and it's kind of pretty. (No doubt this was known long, long ago to knot theorists.) Consider the hexagonal torus: a regular hexagon with each pair of its opposite sides identified by a translation across the hexagon. Now remove a small disk D^2 centered at the center of the hexagon. Finally, the glueing map h just takes this holey hexagonal torus and rotates it 120º into itself. That's it. That's S^3 - K when K is the trefoil. If you're partial to knots in R^3 instead, just remove another point from that. (Something similar happens with all (p,q) torus knot complements. —Dan ----- There's an interesting feature of many knots called "fibring", and such knots are called "fibred knots". A knot can be thought of as living in R^3 but it can be more useful to think of it as living in the 3-sphere S^3 (which is topologically just R^3 with a point added at infinity). To distinguish knots — since all knots are topologically circles — people have studied not the knot itself but what becomes of S^3 after the knot is removed. S^3 - K is called the knot complement. For many knots K in S^3, it turns out that there is a lovely surjective map π : S^3 - K —> S^1 to the circle, such that the inverse image of each point of S^1 is always a certain M (with boundary) in a continuous manner. If we removed one of these surfaces, say π^(-1)(1), what remained of S^3 - K would be topologically M x (0, 1), the product of the surface with an open interval. If we like we can glue on the missing end-surfaces to get M x [0, 1]. Then knot complement is completely determined by the way the end-surfaces are glued back together: S^3 - K = M x [0,1] / (x,1) ~ (h(x),0) where h is some self-homeomorphism h : M —> M of M. ----

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Re: [math-fun] Knot Annihilator
by Dan Asimov 27 May '20

27 May '20

We need three constraints to accomplish what? —Dan ----- Version 1.0 in Mathematica: https://0x0.st/ip7E.txt Analyzing outputs, I found the following for case (3,4): {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], 2 Sin[4 t]} 0 = -81 + 117 X^2 - 40 X^4 + 4 X^6 + 117 Y^2 - 112 X^2 Y^2 + 12 X^4 Y^2 - 40 Y^4 + 12 X^2 Y^4 + 4 Y^6; 0 = -8 X^3 Y + 8 X Y^3 + 27 Z - 24 X^2 Z + 4 X^4 Z - 24 Y^2 Z + 8 X^2 Y^2 Z + 4 Y^4 Z; 0 = 81 - 81 X^2 - 81 Y^2 + 32 X^2 Y^2 + 16 X^2 Z^2 + 16 Y^2 Z^2 Contrary to expectation, we need not two but three constraints. The first two equations define the knot + 8 vertical lines at crossing points. The third constraint filters out vertical lines. -----

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[math-fun] Fibred knots
by Dan Asimov 27 May '20

27 May '20

There's an interesting feature of many knots called "fibring", and such knots are called "fibred knots". A knot can be thought of as living in R^3 but it can be more useful to think of it as living in the 3-sphere S^3 (which is topologically just R^3 with a point added at infinity). To distinguish knots — since all knots are topologically circles — people have studied not the knot itself but what becomes of S^3 after the knot is removed. S^3 - K is called the knot complement. For many knots K in S^3, it turns out that there is a lovely surjective map π : S^3 - K —> S^1 to the circle, such that the inverse image of each point of S^1 is always a certain M (with boundary) in a continuous manner. If we removed one of these surfaces, say π^(-1)(1), what remained of S^3 - K would be topologically M x (0, 1), the product of the surface with an open interval. If we like we can glue on the missing end-surfaces to get M x [0, 1]. Then knot complement is completely determined by the way the end-surfaces are glued back together: S^3 - K = M x [0,1] / (x,1) ~ (h(x),0) where h is some self-homeomorphism h : M —> M of M. The trefoil can be thought of a the intersection of the zero-locus of the polynomial P : C^2 —> C (C = complex numbers) defined by P(z,w) = z^3 + w^2 with a small 3-sphere S^3 about the origin (0,0) of C^2. Then on S^3 - K, that polynomial is always non-zero. So we can define a mapping π : S^3 - K —> S^1 via π(z,w) = P(z,w) / |P(z,w)| and that defines the fibring as above. For the trefoil, it turns out that the fibre M is just a torus with a disk removed. Even the unknot (a geometric circle) has a fibring: The fibre is just a disk. But some knots have no such fibring. —Dan

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Re: [math-fun] Knot Annihilator
by Dan Asimov 27 May '20

27 May '20

Actually, I'm just trying to understand the comment "we need three constraints". I'm not asking about the details of your calculation. I'm asking about what is the final goal that requires three constraints? I ask because, if the final goal is to represent the (3,4) torus knot in 3-space as the intersection of the zero-loci of each of several real polynomials, I would have expected that only two are required. —Dan ----- Again, I think it my last message was fine as stated, but I will explain a few finer points. Call the three polynomials S(X,Y), H(X,Y,Z), and F(X,Y,Z). They all equal zero for the parametric curve {X(t),Y(t),Z(t)} after reducing terms. The linear height function, H, can be factored as H = L(X,Y)+C(X,Y)*Z with L the product of four linear factors, and C the product of two circular factors. Eight solutions of L=0 & C=0, are a solutions of H=0 & S=0 regardless of Z, see: https://0x0.st/iph1.png So the algebraic varietry { (X,Y,Z) : H=0 & S=0 } accidentally contains 8 vertical lines. If (X_0,Y_0) is any one of the eight singular points, then F(X_0,Y_0,Z)=0 implies either: 48 Z^2-45 = 0, or 144 Z^2 - 567 = 0. On the set of singularities--which corresponds to the set of knot crossings--the quadratic filter function F chooses just two points from the singular, vertical line, one on the overpass, and one on the underpass. Thus the variety: { (X,Y,Z) : H=0 & S=0 & F=0 } Is the (4,2) torus knot in the strict sense, where it contains no extra points or mirror images. -----

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[math-fun] Why R0 is a useless concept
by Henry Baker 27 May '20

27 May '20

I think that we are all in violent agreement that these epidemiological models are 'ill-conditioned', hence *any* noise in the input can be dramatically *amplified* in such a way that it can often overwhelm any 'answer'. Analogy: those screeching noises that are often heard from audio public address systems that have positive feedback; the screeches often overwhelm the person speaking. Re network-simulation Monte Carlo models, e.g., the Imperial model: Monte Carlo models require enough iterations/runs in order to *average out* the sampling noise, *and to fully "explore" the nether/tail regions of the particular probability density function*. The most trivial Monte Carlo model is that of estimating the *mean* of a distribution by computing statistics from N samples. How many samples are required in order to assure a reasonable estimate of the mean? Answer: N ~ O(distribution variance). OK. Let's take an oversimplified 'superspreader' model for R0: 99% of the time, R0=2, and 1% of the time, R0=98. The mathematical mean of this bimodal distribution is 2.96, and the mathematical variance of this distribution is ~91. But I just ran this experimental model and it takes at least 15,000 random samples of this distribution just to get a decent approximation to one number -- its mean! The reason why so many samples are required is that the relatively rare event where R0=98 has to occur often enough to average out against the vastly more probable R0=2 events. But we're only getting started. R0 appears as the *base* of an exponential in various epidemic models -- e.g., (R0)^(a*t), for some constant a. The most elementary statistics classes don't deal with *products* of random variables, much less *exponentials* of random variables. One simple way to understand such products and exponentials uses *lognormal* distributions. If X=L(m,v) is a lognormal distribution with parameters m,v, then the distribution for the exponential X^n is L(n*m,n*v). The mean of L(n*m,n*v) is (exp(m+v/2))^n; the variance of L(n*m,n*v) is exp(2*m+v)^n*(exp(v)^n-1). If we choose m,v to match the mean and variance of the bimodal distribution above, then m~-0.1322 and v~2.4348, so the mean of X^n is (2.96)^n and the variance of X^n is (2.96)^(2n)*(11.414^n-1) ~ 100^n. So what if we have to sample, e.g., (R0)^10, i.e., a*t=10 -- to compute its mean ? How many samples will we need to get a decent approximation ? (Note that this is the 10-fold product of independently chosen R0's, so we can't simply average numbers like sample^(1/10).) Since variance of (R0)^10 is ~100^10 = 10 *billion*, it could take O(10 billion) random samples to get a decent approximation to the mean of (R0)^10. I'd be willing to bet that the Imperial model was not run 10 billion times! But this is merely one positive feedback loop in such a Monte Carlo network simulation. What happens when there are multiple feedback loops ? How many runs might then be required ? The problem here is that our samples have to explore an incredibly wide and incredibly shallow distribution, and then accumulate enough weight for each sample to guarantee some reasonable accuracy for our result. But even if we performed such a computation, what would it mean when the *variance* of the distribution is so wide -- hence the weight of any particular value is so tiny -- of what practical use is *any* particular value -- e.g., the "mean" ? This is the reason why "R0" models make no sense in the presence of superspreaders -- there is no single 'R0' that captures any useful aspect of the behavior.

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[math-fun] Knot Annihilator
by Brad Klee 27 May '20

27 May '20

Version 1.0 in Mathematica: https://0x0.st/ip7E.txt Analyzing outputs, I found the following for case (3,4): {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], 2 Sin[4 t]} 0 = -81 + 117 X^2 - 40 X^4 + 4 X^6 + 117 Y^2 - 112 X^2 Y^2 + 12 X^4 Y^2 - 40 Y^4 + 12 X^2 Y^4 + 4 Y^6; 0 = -8 X^3 Y + 8 X Y^3 + 27 Z - 24 X^2 Z + 4 X^4 Z - 24 Y^2 Z + 8 X^2 Y^2 Z + 4 Y^4 Z; 0 = 81 - 81 X^2 - 81 Y^2 + 32 X^2 Y^2 + 16 X^2 Z^2 + 16 Y^2 Z^2 Contrary to expectation, we need not two but three constraints. The first two equations define the knot + 8 vertical lines at crossing points. The third constraint filters out vertical lines. --Brad

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