December 2020 - mathfun@mailman.xmission.com

[math-fun] sin(x/3)
by Bill Gosper 09 Dec '20

09 Dec '20

Do they even teach the halfangle formulas in Trig anymore? They're not even in Mathematica 12.2! Macsyma had them forever. And they were useful. So here I was thinking the kids already knew them, and planning to ask them why the Trig books never mention "thirdangle" formulas, and why don't the kids derive them for themselves. But if they try to invert the triple angle formulas, they hit the casus irreducibilis bizarreness, where they need to solve a cubic whose real roots cannot be expressed in real terms. Kids should meet this while still young enough to struggle against it. And then they'd know why the Trig books never bring it up. But the books have an excuse: Bringing it up requires kids to already know complex numbers. Lots of prerequisites is not a good way to sell textbooks. But if the booksellers would accept the complex numbers prerequisite, instead of explaining about the casus pain-in-the-asus, they could actually provide a nifty thirdangle formula! Sin[x/3] == 1/2 Re[(1 + I Sqrt[3]) (I Cos[x] + Sin[x])^(1/3)], -π/2 ≤ x < 3π/2 E.g., In[212]:= %/. x -> π/6 Out[212]= Sin[π/18] == 1/2 Re[(1/2 + (I Sqrt[3])/2)^(1/3) (1 + I Sqrt[3])] In[213]:= %% /. x ->π/3 Out[213]= Sin[π/9] == 1/2 Re[(1 + I Sqrt[3]) (I/2 + Sqrt[3]/2)^(1/3)] In[214]:= %%%/. x -> π/4 (* Sin[π/12] == *) Out[214]= (-1 + Sqrt[3])/(2 Sqrt[2]) == 1/2 Re[((1 + I)^(1/3) (1 + I Sqrt[3]))/2^(1/6)] —rwg

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[math-fun] 'Column operations' & 'Gram-Schmidt' on quaternions
by Henry Baker 06 Dec '20

06 Dec '20

'Column operations' and 'Gram-Schmidt' on quaternion sums A classic problem with quaternion equations is how to 'simplify' *linear* sums such as A1 Z B1 + A2 Z B2 + A3 Z B3 + ..., where Ai, Bi, Z are all general quaternions. The game here is to avoid converting quaternion equations into sets of linear equations by appealing to an existing 4D coordinate system -- e.g., 1,I,J,K. If these letters Ai, Bi, Z all denoted complex numbers, then the solution is trivial: A1 Z B1 + A2 Z B2 + A3 Z B3 + ... = (A1 B1 + A2 B2 + A3 B3 + ...) Z Unfortunately, the quaternions aren't commutative, so this plan fails. Nevertheless, we can still perform certain types of simplification -- e.g., I want to 'eliminate' the 'scalar part' of A in the first term of sum AZB+CZD. If s(C)=0, then simply exchange the terms: CZD+AZB and we're done. On the other hand, if s(C)/=0, then let R=A-C*s(A)/s(C); clearly s(R)=0. Then A=R+C*s(A)/s(C), so AZB+CZD = (R+C*s(A)/s(C))ZB+CZD = RZB + CZB*s(A)/s(C) + CZD = RZB + CZ(D+B*S(A)/s(C)) We now have a 2-term sum with s(R)=0. We can perform an analogous operation to eliminate the scalar part of B in the first term of the sum AZB+CZD. If s(D)=0, then we exchange terms as before. If s(D)/=0, then let S=B-D*s(B)/s(D); clearly s(S)=0. AZB+CZD = AZ(S+D*s(B)/s(D))+CZD = AZS + (s(B)/s(D))AZD + CZD = AZS + (C+A*s(B)/s(D))ZD We now have a 2-term sum with s(S)=0 (and since A didn't change, it will also preserve s(A)=0 if we previously assured that property). But we're just warming up. What if R=A-(A.C)/(C.C)*C ? (where (A.C) denotes the 4D dot product: (A.C)=s(AC').) Then A=R+(A.C)/(C.C)*C, and AZB+CZD = (R+(A.C)/(C.C)*C)ZB+CZD = RZB + CZB*(A.C)/(C.C) + CZD = RZB + CZ(D+B*(A.C)/(C.C)) So now R.C=0, and we have R orthogonal to C. We can do the same thing with B,D: Let S=B-(B.D)/(D.D)*D. Then B=S+(B.D)/(D.D)*D, and AZB+CZD = AZ(S+(B.D)/(D.D)*D)+CZD = AZS + (B.D)/(D.D)*AZD + CZD = AZS + (C+(B.D)/(D.D)*A)ZD So S.D=0, and we have S orthogonal to D. So we have a type of 'Gram-Schmidt' orthogonalization process. What can we do with these types of operations? 1. Given a sum of n terms of the form Ai Z Bi, we can assure that at most one of the Ai's and at most one of the Bi's (not necessarily for the same i) has a non-zero scalar part. 2. Given a sum of n terms of the form Ai Z Bi, we can assure that all of the Ai's are orthogonal, OR all of the Bi's are orthogonal (making Bi's orthgonal messes up the orthogonality of the Ai's, and vice versa). Since the dimension of the quaternion space is 4, these operations produce sums having at most 4 non-zero terms. This is messy for symbolic algebra, but works just fine for numerical quaternions. 3. These operations extend to quaternions over other fields -- e.g., Zp -- even when the characteristic is 2.

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[math-fun] Very weak Skolem-Langford numbers
by Éric Angelini 06 Dec '20

06 Dec '20

Hello Math-Fun, I've considered this lately ("Base 10 weak Skolem-Langford numbers"): https://oeis.org/A108116 The idea would be to have a "lighter constraint" on the digits. We would say: « Numbers such that for any digit "d", the _closest_ duplicate of "d" is at exactly "d" digits ». The new sequence would then admit numbers like 2002002, or 2312131 because we would drop the "unique pair (of digits)" obligation. Best, É. P.-S. My friend Carole and I are currently working on a sequence related to this post, visible here (in French), where the constraint « For any digit "d", the _closest_ duplicate of "d" is at exactly "d" digits » is applied to a sequence -- not to integers themselves: http://cinquantesignes.blogspot.com/2020/11/une-suite-facon-skolem-langford… Yes, this is a backtracking nightmare!

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[math-fun] Magic zero-sum hexagons
by ed pegg 05 Dec '20

05 Dec '20

Let k be three times a Triangular number. k=3 T(n), 3, 9, 18, 30, 45, ... Arrange numbers -k to k in a hexagon so that all hex lines have sum zero. From Tomas Rokicki, here are all the solutions for hex size 3. {{{-9,1,8},{3,4,-5,-2},{6,-7,0,7,-6},{2,5,-4,-3},{-8,-1,9}},{{-9,1,8},{6,5,-4,-7},{3,-8,-3,9,-1},{2,4,0,-6},{-5,-2,7}},{{-9,3,6},{7,-8,5,-4},{2,1,0,-1,-2},{4,-5,8,-7},{-6,-3,9}},{{-9,4,5},{7,3,-6,-4},{2,-8,-2,9,-1},{1,6,0,-7},{-3,-5,8}},{{-9,5,4},{6,2,-7,-1},{3,-8,0,8,-3},{1,7,-2,-6},{-4,-5,9}},{{-8,1,7},{6,3,-4,-5},{2,-9,0,9,-2},{5,4,-3,-6},{-7,-1,8}},{{-8,2,6},{9,-5,3,-7},{-1,-4,0,4,1},{7,-3,5,-9},{-6,-2,8}},{{-8,3,5},{7,-9,6,-4},{1,2,0,-2,-1},{4,-6,9,-7},{-5,-3,8}},{{-8,3,5},{7,4,-7,-4},{1,2,0,-2,-1},{-9,-6,9,6},{8,-3,-5}},{{-8,3,5},{9,-7,4,-6},{-1,-2,0,2,1},{6,-4,7,-9},{-5,-3,8}},{{-7,1,6},{9,-4,3,-8},{-2,-5,0,5,2},{8,-3,4,-9},{-6,-1,7}},{{-7,2,5},{3,6,-8,-1},{4,-9,0,9,-4},{1,8,-6,-3},{-5,-2,7}},{{-7,2,5},{6,8,-5,-9},{1,-2,-6,3,4},{-8,-1,9,0},{7,-3,-4}},{{-7,3,4},{9,-8,5,-6},{-2,-1,0,1,2},{6,-5,8,-9},{-4,-3,7}},{{-6,7,-1},{9,-7,2,-4},{-3,-8,6,0,5},{8,-2,3,-9},{-5,1,4}},{{-5,2,3},{9,-8,6,-7},{-4,-1,0,1,4},{7,-6,8,-9},{-3,-2,5}},{{-5,2,3},{9,-7,6,-8},{-4,-2,0,1,5},{7,-6,8,-9},{-3,-1,4}},{{-5,3,2},{9,-8,6,-7},{-4,-2,1,0,5},{7,-6,8,-9},{-3,-1,4}},{{-5,8,-3},{9,-2,-6,-1},{-4,-7,0,7,4},{1,6,2,-9},{3,-8,5}}} Here's a hex size 4 solution. {{-12,-11,15,8},{-7,-10,-3,4,16},{13,-2,17,-9,-1,-18},{6,5,-14,0,14,-5,-6},{18,1,9,-17,2,-13},{-16,-4,3,10,7},{-8,-15,11,12}}, Here's a hex size 5 solution. {{-27,15,2,-13,23}, {22,11,-17,28,-19,-25}, {4,7,-18,30,-21,24,-26}, {9,1,-6,-29,-10,12,3,20}, {-8,-14,16,-5,0,5,-16,14,8}, {-20,-3,-12,10,29,6,-1,-9}, {26,-24,21,-30,18,-7,-4}, {25,19,-28,17,-11,-22}, {-23,13,-2,-15,27}} At Wikipedia's entry on magic hexagons there is an order 8 solution by Louis K. Hoelbling. Above the size 2 hexagon, do solutions always exist? --Ed Pegg Jr

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Re: [math-fun] An alternative math model for COVID
by Steve Witham 04 Dec '20

04 Dec '20

> From: Henry Baker <hbaker1(a)pipeline.com> > Date: 12/1/20, 12:21 PM Hi, Henry, what does your model look like mathematically if not a differential equation? One reason I ask is that the classic predator-prey model is a differential equation (modulo the granularity of population numbers). --Steve

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Re: [math-fun] Prime-encoding constants/functions
by Steve Witham 04 Dec '20

04 Dec '20

First of all, this pursuit reminds me of a quote from the Gnostic Gospels: Anyone who attempts to generate random numbers by deterministic means is, of course, living in a state of sin. --John von Neumann. As already pointed out, the notion of restricting oneself to a "simple" language is arbitrary and can't be made non-arbitrary non-arbitrarily. Patterns of certain sorts can be interesting-- for instance, long arithmetic or quadratic series of primes, or Simon Plouffe's f(n) = round(exp(f(n-1)) serieses. A program to produce all the primes, and only the primes, is easy to write in finite space, as long as one's language is Turing complete. That's the ultimate compression ratio, one for the Guinness Book of Records that will never be surpassed. But there's another way that seems to me not-as- ill-defined and not as pointless. Take a finite, plausible but imperfect model of the probability that a given number is prime, and then see what happens when you entropy-encode (e.g. arithmetic- code) the actual primes with it. (One cute thing about arithmetic coding is that it's already part of the idea to interpret the encoded form of an infinite sequence as a single real number.) For instance, there's this formula for prime density: P(is_prime(n)) = 1 / log(n). What's interesting with this (I think) is to ask how unexpected the primes are (i.e. how much info each number's prime-or-not-ness represents in light of the assumption) which amounts to a function (already mentioned in this thread) from how many primes to cumulative bits to encode them. One could complicate one's model by adding "not divisible by two", "not divisible by three", ad infinitum. What's to stop us from doing that isn't complexity exactly, it's that *that would be boring*, at least without some sort of interesting result coming out of it. I think it's right to say that what compels this sort of quest is interestingness as a whole, not simplicity per se. What is the curve of information-- the curve of mismatch between the 1/log(n) model and reality-- per n as n gets higher? One could guess at it and have a model of the inaccuracy of the 1/log(n) model. (For sure the information per n gets *spikier* as n goes up.) Does the naive model generally do what a reasonable mathematician would expect? Is the natural guess way off? Is the curve just generally higher, generally *lower*?? does it have unexpected long deep wobbles? Sharp cliffs? What's the frequency spectrum of the error? Is it brownian noise? 1/f noise? What is the expected and measured fractal dimension of the distribution of spikes in a given neighborhood? Also, we know the 1/log(n) model is wrong in specific ways! With this sort of meta-error perspective, eliminating even numbers, numbers divisible by three, etc. is more interesting: what happens to the curve of model inaccuracy as you improve the model by making it reject numbers with low prime divisors? A quality-improvement curve. Does *that* curve go the way one would expect? Also, there are probably (!) more refined, general- seeming ideas about prime density that I don't know about. How much improvement over -log2(1/log(n)) bits should we expect each of them to give? The thing is, maybe there are ideas that are both more complicated and less accurate than the definition of "prime," but more interesting or seemingly likely to be fruitful. The real meta question here is what makes a math investigation interesting. I believe there's such a thing as interesting, and math has it. --Steve

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[math-fun] A word sequence
by Éric Angelini 03 Dec '20

03 Dec '20

Hello Math-Fun, [and J.-M. Falcoz for a possible French version], here is an English word sequence, for a change: S=7,5,7,4,2,-16,1,-22,-22,-22,-26,0,0,-31,-31,-6,-3,-37,-38,-37,-7,-8,... If you now compute k= |n+a(n)| the result k is the number of letters used in English to write the integers « n » and « a(n) ». Example for n=1: k = |1 + a(1)| = |1+7| = 8 = {ONE,SEVEN} For n=2 we have: k = |2 + a(2)| = |2+5| = 7 = {TWO,FIVE} For n = 3: k = |3+a(3)| = |3+7| = 10 = {THREE,SEVEN} ... For n=6: k = |6+a(6)| = |6-16| = 10 = {SIX,SIXTEEN} etc. When a(n) = 0, there is (apparently) no solution [see a(12) and a(13)]. We’d like S to be the lexicographically earliest seq with this property. Best, É.

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[math-fun] Is 17 trying to tell us something?
by Bill Gosper 03 Dec '20

03 Dec '20

{Cos[π/17]==1/16 (1-√17+√(34-2 √17)+2 √(17+3 √17+√(170+38 √17))), Sin[π/34]==1/16 (-1+√17+√(34-2 √17)-2 √(17+3 √17-√(170+38 √17)))} or even 2 √(17 + 3 √17 - √(170 + 38 √17)) == √( 2 (3 + √17) (2 √17 - √(2 (17 - √17)))) et al. https://mathworld.wolfram.com/TrigonometryAnglesPi17.html includes a messier version of Cos[π/17]. (People keep forgetting that Sin[π/5] is messier than Sin[π/10].) —rwg

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[math-fun] Kids: An errorless way to freely expand and contract the continued fraction of an irrational.
by Bill Gosper 03 Dec '20

03 Dec '20

The function and its inverse are named longer and shorter. For example, expand Cot[1/4]-1 to one, two, three, and four terms: In[258]:= NestList[longer,Cot[1/4]-1,4] Out[258]= {-1+Cot[1/4], 2+1/1/(-3+Cot[1/4]), 2+1/(1+1/(3-Cot[1/4])/(-4+Cot[1/4])), 2+1/(1+1/(10+1/(4-Cot[1/4])/(-43+11 Cot[1/4]))), 2+1/(1+1/(10+1/(1+1/(43-11 Cot[1/4])/(-47+12 Cot[1/4]))))} Shorten the 3rd expansion by two: In[260]:= shorter[%[[4]],2] Out[260]= 2+1/1/(-3+Cot[1/4]) (In a notebook, these partial expansions display traditionally, but for a messy last term.) And now, ladies and gentlemen, here they are, Miss Jayne ... oops, no, here they are, the definitions of longer and shorter: Clear[longer]; longer[cf_, 0] := cf longer[x_?NumericQ, n_ : 1] := longer[HoldForm @@ {x}, n] longer[cf_, n_ : 1] := longer[cf /.HoldForm[r : Except[_Integer]] :> Floor[r] + 1/HoldForm @@ {Simplify@Together[1/Mod[r, 1]]}, n - 1] Clear[shorter]; shorter[cf_, 0] := cf shorter[cf_, n_ : 1] := shorter[cf /. (a_ : 0) + 1/HoldForm[r_] :> HoldForm @@ {1/Simplify[1/(a + 1/r)]}, n - 1] This continued fraction begins 2, 1, 10, 1, 18, 1, ... . Does it continue to grow by {8, 0}? Yes indeed, In[254]:= fromcf[{8 # + 2, 1}] Out[254]= -1 + Cot[1/4] where fromcf provides closed forms for a class of Hurwitz numbers: fromcf[pre_List : {}, {a___, (c1_ : 1)*# + c0_ : 0, b__}] := fromcf[Join[pre, {a, c0}], {b, a, c1*# + c0 + c1}] fromcf[pre_List : {}, {a___, (c1_ : 1) # + c0_ : 0}] := Simplify[FunctionExpand[ Block[{M = cfmat[{a}], A, B, C, D, det}, det = Det[{{A, B}, {C, D}} = M]; FromContinuedFraction[ Append[pre, A/(C + (det Hypergeometric0F1[1 + (B + C + A c0)/(A c1), det/( A^2 c1^2)])/( A c1 ((B + C + A c0)/(A c1))! Hypergeometric0F1Regularized[( B + C + A c0)/(A c1), det/(A^2 c1^2)]))]]]]] —rwg

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[math-fun] An alternative math model for COVID
by Henry Baker 02 Dec '20

02 Dec '20

The 100-year-old differential equation model for infectious diseases -- the one with R/R0 -- isn't the only possibility. I'm going to suggest another one -- which I'm not claiming is necessarily correct, but which has some of the same characteristics as the R/R0 model. My model is more akin to 'predator-prey' models, only in this case, the predator is microscopic. The diffeq model suggests that 'herd immunity' is reached when the % of immune/already-had-its is large enough that there are too few uninfected people left to sustain a chain reaction. According to most estimates using this type of model, herd immunity % for COVID will be in the 70-80% range. But there is an alternative model where a kind of 'herd immunity' is reached at a far lower %. Consider a predator-prey model of lions and antelopes. The lions don't have to *win* the lion-v-antelope races; they just have to not come in *last* in order to eat. Suppose that a lion has a top speed of 20mph for 1/4 mile (I have no idea what a lion's actual speed is). All the antelopes have to do is to stay ahead of the lion for that period of time in order to survive. At time t=0, some % of the antelope herd can't achieve 20mph, and are bound to be eaten. But as time goes on, more and more of these 'slow' antelopes are eaten and no longer remain in the herd. At some point, *all* of these 'slow' antelopes will be gone, and since a famished lion isn't likely to get any faster, he will be forced to find another herd of antelopes to chase which might still contain some 'slow' antelopes. Note that in my model *everyone* who is exposed gets the disease -- i.e., R/R0 is exceedingly high, but only some % -- e.g., those with 'co-morbidities' -- will succumb. Note that while the herd immunity % in the diffeq model is dependent upon R/R0, the herd immunity % in my model is dependent on a 'critical velocity', and is independent of chain reaction dynamics. Thus, if 10% of the antelope population can't achieve 20mph, then the 'herd immunity' will be achieved at 10%. Yes, I have mixed up to some extent the difference between unexposed and immune, but in reality, most people only care if they survive, not whether they catch an asymptomatic case of the disease.

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