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[math-fun] The True Tau Day(s)
by Greg Huber 17 Dec '20

17 Dec '20

Dear Bill - I agree with you 100% that the Greek letter *tau *should honor Gamma(1/4) = 3.625... (though I confess to having a fondness for *tau*^4, maybe because it's easier -- for me -- to remember that it's near 173). Should some enlightened individuals start a campaign to celebrate a true *Tau Day *in honor of Gamma(1/4)? (2π and June 28 are OK, but NOT *tau* = 2π -- c'mon people, seriously, grow up -- there's no need to throw away another Greek letter, 'cause they ain't making any new ones!) But when to kick off the new, true Tau Day?? Perhaps on 3/6/25, which could mean either March 6, 2025 or 3 June 2025 depending on where you are located? That would give two different target dates to aim for, in case only 3.6... people show up for the party on March 6. Or should it be 10 days after Star Wars Day, with the slogan "May One-Four ^th Be With You"? *Or, perhaps the funniest idea so far, should the annual celebrations commence in Israel, in conjunction with 1 April, since "Gam ma 1/4" means "Also from April 1" *(Heb./Eng. translation: G.Huber)*.* This latter possibility could help to alleviate the worries of those stereotypical Jewish parents who spent undue time and effort sending their kids money in foreign currency, all because of some joker named Rothberg, according to "The Schmooze" section of *The Forward:* https://forward.com/schmooze/154158/israeli-april-fools-jokes-went-too-far/ . Instead, those anxious parents would know that on April Fools' Day their children are studying the properties of the Gamma function, not to mention the many shekels said parents and kids would save because Artin's book is available as a Dover reprint. Furthermore, to the list of leg-pulling utterances such as "I saw your car being towed" and "A beautiful lady asked me to give this phone number to you" (both suggested here https://www.hebrewpod101.com/blog/2018/03/19/how-to-celebrate-april-fools-d… ) one could add remarks especially designed for a dual-purpose April-Fools'/Tau-Day, such as: "I saw your keys around the first zero of the Gamma function" (oh boy! imagine the belly laughs while the remarkee looks around for that), or "Let's do a Zoom call around Gamma of 0 'clock" (they wait centuries in the Zoom room, as you barely suppress guffaws) So, Bill, it seems a true *tau*-inspired Tau Day would truly be a multifaceted laughfest, whenever it happens to happen. -- Greg. ==================== G Huber Theory, Biohub San Francisco, CA On Wed, Dec 16, 2020 at 2:22 AM Bill Gosper <billgosper(a)gmail.com> wrote: > The arclength of one period of sin x = the circumference of an ellipse > with semiaxes 1 and √2. > > In[112]:= #1 == #2 == FunctionExpand@#1== N@# &[ArcLength[Circle[{0, 0}, > {√2, 1}]], ArcLength[Sin@x, {x, 0, 2 π]}]] > > Out[112]= 4 EllipticE[-1] == 4 √2 EllipticE[1/2] == 4 √2 > π^(3/2)/Gamma[1/4]^2 + Gamma[1/4]^2/√(2π) == 7.64039557805542 > > 𝚪(¼) is the rightful value of the symbol 𝛕. And someone should write > Beckmann II: A History of 𝛕. > —rwg >

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[math-fun] Prime-encoding constants/functions
by Leo Broukhis 14 Dec '20

14 Dec '20

Apropos https://www.youtube.com/watch?v=_gCKX6VMvmU There are many ways to produce the sequence of primes by iterating a function, e.g. starting with f_1 = ContinuedFraction(2, 3, 5, 7, 11, 13, ,,,) = 2.3130367364335829..., (http://oeis.org/A064442) and f_n+1 = 1/(f_n - Floor(f_n)) the sequence of Floor(f_n) will be the sequence of primes. Or, as in the clip, starting with the average of http://oeis.org/A053669 = 2.920050977316134712..., and f_n+1 = Floor(f_n) * (f_n - Floor(f_n) + 1) will do the same. Or the Blazys continued fraction, see http://oeis.org/A233588 The question is, which constant/function pair encodes primes in the most compact way. More precisely, given a denominator for a rational approximation, which function will correctly produce the most primes starting from the approximation of the corresponding constant? For example, with the IEEE double (64 bit floating point) approximations of the constants, the continued fraction starts diverging at 29, the Blazys continued fraction - at 47, the 2.92... constant - at 59. I was able to come up with the function f_n+1 = Floor(f) + 1/(f_n-Floor(f_n)); which is the delta-coded continued fraction of the sequence of primes, gives rise to f_1 = 2.7101020234300968374... and given its IEEE double approximation, starts diverging at 67. Improvements are welcome. Thanks, Leo

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[math-fun] Connecting simple closed curves of the same length
by Dan Asimov 10 Dec '20

10 Dec '20

Let C be a rectifiable simple closed curve* (RSCC) in the plane having length L(C) = 2π. Question: --------- Does there always exist a continuous family {C_t} of RSCC such that C_0 = C and C_1 is the unit circle in R^2, such that L(C_t) = 2π for all t in [0,1] ??? —Dan _____ * If C ⊂ R^2 is a RSCC, then C = A([0,1]) for some continuous function A : [0,1] —> R^2 where A(s) = A(t) for s ≠ t if and only if {s,t} = {0,1}, and such that L(C) = sup{∑ ‖A(x_(j+1)) - A(x_j)‖} < oo where the supremum is taken over all 0 = x_0 < x_1 < ... < x_n = 1. The length of C is then defined as L(C).

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[math-fun] Life at the Echigo House in the Shimmachi Quarter
by Brad Klee 10 Dec '20

10 Dec '20

<< Umegawa (to proprietress): Kiyo, that blockhead of a country bumpkin has been bothering me all day and my head is splitting. Hasn't Chubei showed up yet? I sneaked away from my customer, hoping to see you, my only connection with Chubei. Narrator: She slides open the /shoji/ door as she enters, even as she will slide it tomorrow at dawning. Kiyo: I'm glad you've come. There's a crowd of girls upstairs relaxing. They're drinking and playing /ken/ to pass the time until their customers call. Why don't you join them in a game of /ken/ and have a cup of sake? It'll cheer you up. Some of your friends are there. >> Major Plays of Chikamatsu, *Meido no hikyaku, Act 2, p. 173* Tr. Donald Keene (1922-2019, RIP). According to a footnote from Keene: /ken/ is "A game of chinese origin. Each player holds out none to five fingers; the one who guesses the total held out by both wins." See also: https://en.wiktionary.org/wiki/%E6%8B%B3 https://en.wikipedia.org/wiki/Sansukumi-ken https://en.wikipedia.org/wiki/Ken%27ichi In case anyone is curious or wanted to hear, I successfully defended my PhD dissertation on Nov. 20, and now I am expecting a permalink from proquest. Advance review copies of the dissertation can be made available on request to: bradklee(a)gmail.com . Until then, it looks like I'm locked in a combat situation with Julian Stephenson and / or one or more of his psychotic, possibly schizoidal minions "Chas", about whether or not google groups can help us decide if Nichiren Daishonin is the One True Shakyamuni Lotus Preacher, see also: http://ptlslzb87.blogspot.com/ . Hopefully the situation will improve by mid January or thereafter. Thanks again for your help and contributions earlier in time, --Brad

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[math-fun] sin(x/3)
by Bill Gosper 08 Dec '20

08 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 03 Dec '20

03 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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