May 2020 - mathfun@mailman.xmission.com

[math-fun] R1 & Plateau functions
by Brad Klee 21 May '20

21 May '20

pictures: https://0x0.st/i9kU.png Start with infection curve p0(t), and spawn pn(t)=R1^n*p0(t-n) from n = 0,1,...,N. When R1 approaches 1, the total of all pn(t) has a plateau rather than a peak. The slope of the Plateau depends on R1, while its duration depends on N. During the peak, R0 is not apparent. See pics above, N=20. When successive pn(t) are assigned to nodes of a graph corresponding to different geographic locations, the above calculation is a naive estimate of what a PDE would look like assuming SIR holds locally. --Brad

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[math-fun] Hilbert space puzzle
by Dan Asimov 21 May '20

21 May '20

Let H be the Hilbert space of square-summable sequences of reals: H = 𝓵^2(Z+; R). Then the vectors e_n, n in Z+, that are 1 in the nth place and 0 elsewhere form an orthonormal Hilbert basis. For any subset X of H, the closed convex hull CC(X) of X is defined as the intersection of all closed half-spaces L^(-1)([c,oo)) that contain X, where L is a continuous nonzero linear map L : H —> R and c is a real number. Let the Hilbert (regular) simplex Δ be defined as the closed convex hull of that orthonormal basis: Δ = CC({e_n}). This is by analogy with the n-dimensional regular simplex Δ_n (n >= 0) in H, defined as the closed convex hull of the basis of R^(n+1) in H: CC({e_1,...,e_n}) (though of course the convex hull of a finite set is automatically closed). Now suppose we want to walk from the centroid e_1 of Δ_0 to the centroid of Δ_1 to the centroid of Δ_2 to ... to the centroid of Δ_n to ..., hopefully ending up at the centroid of Δ. 1) Assuming these steps for n = 1, 2, 3, ... are taken in straight segments, what is the net distance traveled in the limit? 2) What is the total distance traveled, in the limit? 3) If all these centroids were projected radially onto the unit sphere S of H, obtaining points d_0, d_1, d_2, ..., d_n, ... on S, what would be the net distance traveled from d_0 to d_1, then from d_1 to d_2, etc., in the limit? Again assume each step is in a straight line. —Dan

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[math-fun] The triadic dragon triple points
by Bill Gosper 20 May '20

20 May '20

seem all <http://gosper.org/tridragtrips.png> to be merely triadic rationals. —rwg

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[math-fun] The Sudoku Miracle by Mitchell Lee
by Hans Havermann 20 May '20

20 May '20

https://www.youtube.com/watch?v=yKf9aUIxdb4 To his own amazement, Simon Anthony solves it!

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[math-fun] The Seven+ operation
by Éric Angelini 20 May '20

20 May '20

Hello Math-Fun, The Seven+ operation transforms a digit 7 into a plus sign. The integer 170 becomes 1+0 which is 1. We want that a(n), transformed by the Seven+ operation, divides a(n+1). The term a(n+1) has to contain at least one digit 7. No two identical terms in the sequence S – which should be the lexico-first of its kind. For a(1) = 170, we get: S = 170, 171, 172, 174, 175, 270, 176, 273, 275, 371, 272, 276, 376, 279, 374, 378, 473, 476, 370,... We see indeed that: 170 = 1+0 = 1 divides 171; 171 = 1+1 = 2 divides 172; 172 = 1+2 = 3 divides 174; 174 = 1+4 = 5 divides 175; 175 = 1+5 = 6 divides 270; etc. Numbers we don't want to see in S: — no term starting with a 7 (ex. 754 or 7574); — no term ending with a 7 (ex. 127 or 1727); — no term with 2 or more consecutive 7 (ex. 1778); — between two 7, no string with a leading 0 (ex. 170578 – but 17078 is ok as this term would produce 9). Where does S (S for "seven") go? As usual, please forgive my hand mistakes. Best, É. Nothing more on my personal web page: http://cinquantesignes.blogspot.com/2020/05/seven-plus.html

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Re: [math-fun] Graph question
by Dan Asimov 20 May '20

20 May '20

At the risk of sounding stupid, I'd like to see a proof of the claim (that the isomorphism of the subgraphs taking 1—>1, 2—>2, 4—>5, 5—>6 does not extend to an isomorphism G —> G). Andy Latto wrote: ----- > ... the graph that links each > integer n to every other integer except n-1 and n+1 would be a solution. I don't think this works. The induced subgraph on {1,2,4,5} is isomorphic to the induced subgraph on {1,2,5,6}, using the isomorphism that fixes 1 and 2, sends 4 to 5 and 5 to 6. But this does not extend to an isomorphism of G to G. -----

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[math-fun] 9,9 ends the sequence
by Éric Angelini 20 May '20

20 May '20

Hello Math-Fun, This seq S is finite — but how many terms in it? And which is the last one? S = 1, 3, 4, 2, 41, 42, 21, 43, 44, 22, 45, etc. We are only interested in the cumulative sum Q which must be prime at every stage. The first term of Q is 13 — and this prime comes from the digits squeezing the 1st comma; the 2nd term of Q is 13 + 34 = 47 (prime) and 34 comes from the digits squeezing the 2nd comma; the 3rd term of Q is 47 + 42 = 89 (prime) and 42 comes from the digits squeezing the 3rd comma; the 4th term of Q is 89 + 24 = 113 (prime) and 24 comes from the digits squeezing the 4th comma; the 5th term of Q is 113 + 14 = 127 (prime) and 14 comes from the digits squeezing the 5th comma; etc. We don’t accept terms of S ending in 0 (zero). Best, É.

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Re: [math-fun] Fwd: Pufferfish creates art
by Dan Asimov 19 May '20

19 May '20

Male bowerbirds also spend a long time creating elaborate displays to impress the females. The pufferfish certain have some kind of spatial savvy, but I don't know that I'd call it "reasoning" exactly. Dogs are said to have poor eyesight, but it's amazing how some can run after a thrown Frisbee and leap to catch it in mid-air. Re pufferfish, I'd like to know the vantage point of the female surveying the male's finwork: Is it from way above, as in the video, showing the macro pattern, or something else? —Dan Brent Meeker wrote: ----- I wonder if we can infer some simple algorithm that might be encoded in the little fellow such that he produces this. LC wrote: -------- Forwarded Message -------- https://www.youtube.com/watch?v=p1PID91sEW8 The thing that is remarkable is there is a high measure of spatial reasoning in this fish in order to do this. I would ordinarily consider a dog more intelligent than a fish, but as far as I can see dogs have almost no spatial reasoning. -----

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[math-fun] Regular polyhedra question
by Dan Asimov 19 May '20

19 May '20

Suppose T, Q, O, D, I represent the *vertex sets* of the 5 regular polyhedra (Platonic solids), assumed to be on the unit sphere S^2 centered at the origin. For any two of these, i.e., one of {TT, TQ, TO, TD, TI, QQ, QO, QD, QI, OO, OD, OI, DD, DI, II}, I want to know how far one of each pair can be rotated from the other one. So for the pair PP' (i.e., P and P'), consider any rotation of g of the sphere S^2 and define: f_g(x,y) = minimum_{x in P, y in P'} dist(x, g(y)) where dist is arclength over S^2, i.e., central angle. Now set F(P, P') = maximum_{g in SO(3)} f_g(x,y) where SO(3) is the set of all rotations of S^2. Note this is unchanged if we had used dist(g(x), y) instead of dist(x, g(y)). Note: In many cases (e.g., if P and P' are dual) the points P' can be rotated to (or already are) a subset of the face centers of P — or vice versa — so for these cases the value of F(P, P') is easy to determine. Namely, the cases PP' = TT, TO, TI, QO, QI, DI. That leaves the following 9 less-obvious cases: {TQ, TD, QQ, QD, OO, OD, OI, DD, II}. I hope I didn't misstate something, but if so I hope someone will point it out. —Dan

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[math-fun] Fwd: Pufferfish creates art
by Brent Meeker 19 May '20

19 May '20

I wonder if we can infer some simple algorithm that might be encoded in the little fellow such that he produces this? Brent -------- Forwarded Message -------- https://www.youtube.com/watch?v=p1PID91sEW8 The thing that is remarkable is there is a high measure of spatial reasoning in this fish in order to do this. I would ordinarily consider a dog more intelligent than a fish, but as far as I can see dogs have almost no spatial reasoning. LC

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