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[math-fun] "A physicist's proof of the Basel problem"
by Dan Asimov 05 Jan '21

05 Jan '21

I thought this paper had an interesting (fully rigorous!) take on Euler's (not fully rigorous) proof that 1 + 1/2^2 + ... + 1/n^2 + ... = π^2/6: <https://arxiv.org/abs/1908.07518>. —Dan

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[math-fun] about some films made on a Tektronik equipment in 1976
by Simon Plouffe 05 Jan '21

05 Jan '21

Hello mr Gosper, I think you will find this of interest, https://vimeo.com/98285334 these are films made in or around 1976 with a tektronik screen, with dragon curves, triangles and such. They were made by a certain Eric Cornwell, and some other Cornwell in the same family I presume. also here : https://vimeo.com/channels/bkcfilms and here : https://vimeo.com/user543333/videos Best regards, Simon Plouffe ps : happy new year.

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[math-fun] Aware-like seq
by Éric Angelini 03 Jan '21

03 Jan '21

Hello Math-Fun (we need help to check, compute and submit this to the OEIS in good mathematical English — as we fear Dan's furrowed brow ;-D) This sequence is based on the Aware strategic game. We empty a “pit” and sow to the right its content – one seed per pit. When this is done, we continue from there, empty the next pit and sow again one seed per pit to the right. S0 = 1, 109, 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … To start, we empty the first pit (1) and sow its unique seed to the right; 109 becomes 110 and S0 becomes S1 (the vertical stroke shows where we are in the process): S1 = _, 110, | 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … We now empty the first “untouched” pit (2) and sow its content in the same way. S1 becomes: S2 = _, 110, _, 9, 28, | 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … We proceed like this, step by step, ad infinitum: S3 = _, 110, _, 9, 28, _, 27, 32, 6, | 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S4 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, | 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S5 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, | 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S6 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, _, 750, 60, 97, 49, 59, 96, 48, 58, 95, | 10, … If we now compare the end result Se to the starting sequence S0, we see that… S0 = 1, 109, 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10 … Se = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, _, 750, 60, 97, 49, 59, 96, 48, 58, 95, _, … … S0 and Se share the same digit succession. S0 should be the lexicographically earliest sequence of distinct positive integers with this property. This is not too tricky, I guess – still, I hope I didn't leave too many mistake above (and that this is not old hat). Best, É.

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[math-fun] Transforming integer D into integer E (exponentiation)
by Éric Angelini 03 Jan '21

03 Jan '21

Hello Math-Fun, Say D has 5 digits: D = abcde Starting from the right, pick D’s smallest digit (say it is b) Then E = acde^b Repeat. Examples: 23 --> 3^2 = 9 = stop 34 --> 4^3 = 64 --> 6^4 = 1296 --> 296^1 = 296 --> 96^2 = 9216 --> 926^1 = 926 --> 96^2 = 9216 = loop 35 --> 5^3 = 125 --> 25^1 = 25 --> 5^2 = 25 = loop 3252 --> 325^2 = 105625 --> 15625^0 = 1 = stop Which integers loop? Which integers stop? Best, É. (sorry if this has been explored before)

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[math-fun] Follow up on C.A. for Lotus Fields
by Brad Klee 02 Jan '21

02 Jan '21

https://groups.google.com/g/alt.religion.buddhism.nichiren/c/-2v4wQvVd6g The short of it is: One point goes to Wolfram & Co. for Code 746, one of the more memorable results of New Kind Of Science: https://demonstrations.wolfram.com/AlmostCircularCellularAutomaton/ If a stolon network grows at a constant radial rate, then Code 746 is okay, but from other Japanese research we know that congregations involve multiple individuals: > Little is known about how many clones exist in a population > and how large an area does a single clone occupy. That is an interesting question. Is anyone in the United States working on this at all ? ? ? ? Happy New Year, --Brad

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[math-fun] Beltrami, Hadamard & Hamilton
by Henry Baker 01 Jan '21

01 Jan '21

No, not *that* (politician) Hamilton! [Although the following derivation requires only undergraduate level linear algebra, I've never seen this derivation before; I've Google'd and ScholarGoogle'd, but nothing of this sort shows up.] SVD's Solve Linear Equations in Quaternions The linear mapping captured by a 4x4 matrix M of real numbers can be computed by a 4-term linear equation in quaternions: M(Z) = AZB + CZD + EZF + GZH But how can we constructively compute A,B,C,D,E,F,G,H from this 4x4 matrix M? The most significant step is Beltrami's: compute the SVD of M: M = USV' where U,V are unitary, and S is a non-negative diagonal matrix. Actually, we want to tweak our SVD slightly by making |U|=|V|=+1, i.e., U,V are *special unitary*, and allow S to have negative numbers. The next step is to factor both U & V' into left and right quaternions [Van Elfrinkhof & 'Isoclinic decomposition' -- Wikipedia]: If Z is interpreted as a column vector, then UZ is the column vector rotated by U. If Z is interpreted as a quaternion, then we can find quaternions Ul & Ur, such that the quaternion (Ul)Z(Ur) is the same quaternion as (UZ), interpreted as a quaternion. Similarly, we can find Vl & Vr, such that (Vl)Z(Vr) is the same as (V'Z). So, once we have tweaked our SVD to utilize only special unitary matrices, the implementation of the pre and post rotations by means of quaternions is straight-forward. So now comes the harder part: how to implement the equivalent of a diagonal matrix using quaternions? The answer is Hadamard's 4x4 'H2' matrix: [1 1 1 1] [1 -1 1 -1] H2 = (1/2) [1 1 -1 -1] [1 -1 -1 1] If the diagonal entries of S are [s1,s2,s3,s4], then [s1 s2 s3 s4] H2 = [a b c d] and our linear quaternion expression implementing (SZ) is: f(Z) = (a*Z - c*iZi - b*jZj -d*kZk)/2 Here i,j,k are Hamilton's basic quaternions: i^2 = j^2 = k^2 = -1 = ijk. Some examples. Extracting the 'scalar part' of Z: s1=1,s2=s3=s4=0, so [a b c d] = [1 -1 -1 -1]/2 s(Z) = (Z - iZi - jZj - kZk)/4 Extracting the 'vector part' of Z: s1=0,s2=s3=s4=1, so [a b c d] = [3 1 1 1]/2 V(Z) = (3Z + iZi + jZj + kZk)/4 Computing the 'conjugate' of Z: s1=1,s2=-1,s3=-1,s4=-1, so [a b c d] = [-1 -1 -1 -1] K(Z) = -(Z + iZi + jZj + kZk)/2 But here's the really cool part: so long as diagonal matrices have no zero diagonal elements, then they are trivially invertible with diagonal elements: [1/s1 1/s2 1/s3 1/s4] So we can trivially compute the quaternion representation for (1/M) by using 1/s1, 1/s2, 1/s3, 1/s4 instead of s1, s2, s3, s4. But this means we can find an explicit representation for the inverse of the quaternion linear function: f(Z) = (a*Z - c*iZi - b*jZj -d*kZk)/2 If g(f(Z))=Z=f(g(Z)), then we have g(Z) = (e*Z - f*iZi - g*jZj - h*kZk)*2, where e = (a^3 - a(b^2+c^2+d^2) + 2bcd)/P f = (c^3 - c(a^2+b^2+d^2) + 2abd)/P g = (b^3 - b(a^2+c^2+d^2) + 2acd)/P h = (d^3 - d(a^2+b^2+c^2) + 2abc)/P and P = (a+b+c+d)(a-b+c-d)(a+b-c-d)(a-b-c+d) -------------------------------------------- To summarize, given an arbitrary 4x4 matrix M of real numbers, we can produce a quaternion linear expression with 4 terms which represents the same mapping, thus: MZ ~ M(Z) = AZB+CZD+EZF+GZH M(Z) = (Vl) f((Ul)Z(Ur)) (Vr) = (Vl)[a*(Ul)Z(Ur) - c*i(Ul)Z(Ur)i - b*j(Ul)Z(Ur)j - d*k(Ul)Z(Ur)k](Vr)/2 = [a*(VlUl)Z(UrVr) - c*(VliUl)Z(UriVr) - b*(VljUl)Z(UrjVr) - d*(VlkUl)Z(UrkVr)]/2 i.e., A = a*(Vl)(Ul)/2 B = (Ur)(Vr) C = -c*(Vl)i(Ul)/2 D = (Ur)i(Vr) E = -b*(Vl)j(Ul)/2 F = (Ur)j(Vr) G = -d*(Vl)k(Ul)/2 H = (Ur)k(Vr) If we can explicitly represent the inverse of f(Z), then we can trivially solve equations f(Z)=R by computing Z0 = f^-1(R).

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Re: [math-fun] Life turns 50
by Brad Klee 30 Dec '20

30 Dec '20

> > Does a lotus seed have an internal timer > that just goes off after a century? Or is it > waiting on an external signal, like a water > pulse? The seeds are 10-20 per pod, about the size of acorns, sometimes larger. They are protected by a strong waxy exterior, which needs to be scored open somehow. If not, seeds can be stored and shipped over seas. Assume growth needs mixture of acceptable chemistry and thermodynamics. there are videos on YouTube, tutorials elsewhere. > Now, as far as exponential growth after > the seed is triggered, no can do. > GoL has a ceiling of quadratic growth. This is a huge problem, because exponential population growth is common in nature—as we saw with “insect pandemics” earlier this year. Say each seed pod has 15 seeds, expect five positives and 5^n growth until saturation is reached. Just don’t mention Visistacaritra, or “the Later days of the Law”. Cheers, —Brad

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[math-fun] Life turns 50
by Bernie Cosell 30 Dec '20

30 Dec '20

via the NYtimes: https://www.nytimes.com/2020/12/28/science/math-conway-game-of-life.html /b\ Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

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[math-fun] Analytic continuation
by Dan Asimov 29 Dec '20

29 Dec '20

One of the most interesting ideas in complex analysis is called "analytic continuation. Suppose we start with a power series of form f(z) = ∑ a_n (z-c)^n where n = 0, 1, 2, ... and each a_n ∊ C and also c ∊ C. If this power series has a positive radius of convergence R, 0 < R ≤ oo, it defines the analytic function f(z) inside an open (circular) disk D_0 of radius R, centered at the complex number c. This can be viewed as shining a flashlight on the function to see how it's defined inside a certain disk. If R = oo, the function is defined for all z ∊ C and such an f(z) is called an "entire function". If R < oo, the power series won't converge for any z with |z - c| > R. (Its convergence for z exactly on the circle |z - c| = R can be quite complicated.) But if we try to use a power series with a different center, say c_1 ∊ D_0, we'll often find that the new disk of convergence, say D_1, extends beyond the boundary of D_0, thereby defining f(z) for some new values of z. Repeating this process, we can get a sequence of disks D_0, D_1, ..., D_k each overlapping the one before but extending beyond its boundary, defining f(z) on an expanding set. The sequence can be described just by its sequence of centers c = c_0, c_1, ..., c_k. It turns out that sometimes a new disk can overlap an old disk, however with values of f(z) that disagree for the same z. This happens notably for the function √(z-1) for a sequence of centers that goes around the point z = 1. Going once around, say picking centers lying on the circle C(t) = 1 + exp(it), 0 ≤ t ≤ 2π we find that the new value is the negative of the old value after traversing this circle once around. Indeed, going around a second time returns f(z) to its original values. We could take this kind of phenomenon to mean that f(z) = √(z-1) is a multi-valued function. Instead, Riemann's idea was to use this process to build a surface on which f(z) ends up being single-valued. The maximal surface on which this is possible turns out to be uniquely defined, and came to be called the "Riemann surface of f". In this overly simplified description, we've overlooked how some "defects" in the function and its Riemann surface can be "fixed". One is that, initially, a point z_1 where |f(z)| —> oo as z —> z_1 is thought of as a singularity of f, known as a "pole". The function f(z) = 1/z^3 for instance has a pole at z = 0. But we can add the point named "oo" to the complex plane, obtaining its one-point compactification C u {oo} sometimes called the "Riemann sphere" because, topologically, it is indeed homeomorphic to the sphere S^2 and we may think of it as just being S^2. (A convenient way to go back and forth between S^2 and C u {oo} is stereographic projection.) A second kind of "defect" is the branch point, such as f(z) = √(z-1) has at the point z = 1. This is any point for which going around it along an arbitrarily small circle and coming back to where you began results in a new value of the function. There are finite branch points like those of √(z-1), where enough circumnavigation will return the function to its original values. But there also exist branch points of infinite order, like that of f(z) = log(z), where no amount of going around the branch point gets you back to the function's original values. The latter kind can't be "fixed". But a finite branch point can be patched up so that the function's domain extends over the patch in an analytic way. Thus, every power series initially defined only inside its radius of convergence ends up defining a unique maximal domain of definition of the function, its "Riemann surface". A Riemann surface can be any smooth surface (that's Hausdorff and has a countable base to its topology). Every such has a well-defined universal covering space that's simply connected (i.e., every closed curve can be shrunk to a point). One of the greatest theorems in complex analysis is the Uniformization Theorem of Poincaré and Koebe, which proves that a simply connected Riemann surface can be just one of three possibilities, up to conformal equivalence: * the sphere S^2, * the complex plane C, or * the (open) unit disk D. Where "conformal equivalence between two Riemann surfaces" means there is a smooth, angle-preserving homeomorphism between them. This implies that *every* Riemann surface is the result of taking the quotient of one of these three simply connected surfaces by a discrete group acting on it, without fixed points and what's called "properly discontinuously". (Explained here if you're curious: <https://www.math.ucdavis.edu/~kapovich/EPR/prop-disc.pdf <https://www.math.ucdavis.edu/~kapovich/EPR/prop-disc.pdf>>.) —Dan

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[math-fun] Man/Math-hating at Netflix.com reaches new high/low?
by Brad Klee 29 Dec '20

29 Dec '20

Quoting from https://the-queens-gambit.fandom.com/ : *> Alice Harmon* is the biological mother to Beth Harmon <https://the-queens-gambit.fandom.com/wiki/Beth_Harmon> > and the divorced wife of Paul <https://the-queens-gambit.fandom.com/wiki/Paul>. She received her PhD in > mathematics from Cornell University, with the dissertation *> Monomial Representations and Symmetric Presentations*. > She lived with her daughter in a trailer in Wakefield, Kentucky, > until her vehicular suicide on 24th of July, 1957, leaving Beth > orphaned. Though not much is known about Paul, including whether or not he is actually the father of Beth? The title "Monomial Representations and Symmetric Presentations" sounds interesting, but snippets of Alice's dialogue in flashback suggest foul play on behalf of University Math Professors. Obviously Alice suffered from some sort of Trauma. Could Beth, in fact, be the illegitimate child of an over-enthusiastic Cornell dissertation advisor? Allegedly the TV-Adaptation was made from a book, but I am too busy reading hate mail and Bunraku plays to check and see if the source text has more details on Alice's conditions leading to suicide by vehicular accident. Be careful when driving, avoid inclement weather! --Brad

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