September 2015 - mathfun@mailman.xmission.com

Re: [math-fun] something linguistic
by Warren D Smith 18 Sep '15

18 Sep '15

who, woo

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Re: [math-fun] something linguistic
by Warren D Smith 18 Sep '15

18 Sep '15

My mother recently was phoned by Hillary Clinton Inc, asking for money. Mom told them she could not support HCI because Hillary was a "hawk." The guy on the phone had never heard of the word "hawk" and interpreted it as the word "whore." They bitched at each other for quite a while. Eventually, she spelled the word letter by letter, whereupon the HCI guy could not comprehend this.

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[math-fun] Asimov's group-metric puzzle
by Warren D Smith 18 Sep '15

18 Sep '15

Er yes, Tits group T is not a sporadic according to the usual accounting, I just was following the ATLAS web pages like a monkey and those guys are a gazillion times more expert than I am. Anyhow, if my search thru the ATLAS to find out which simple groups have outer auts, is not to be largely a waste of time, then we need next to answer: when simple groups are glued together to make more complex groups, what happens to the count of their outer automorphisms? If you understood that and understood simples, then you'd understand everything about finite groups (at least as far as the question of their outer auts was concerned, which may or may not in turn provide answer to Asimov). At least if you believe in the Group Classification Theorem [which seems largely unverifiable by those of us who want to do anything else in their lives]. There might be no simple and complete answer to this follow-up question, though -- but probably at least partial answers can be got. Indeed, I've never even seen any simple answer to the question: "how, exactly, are simple groups glued together to build all other finite groups?" These gluings can sometimes be very messy and hard to be confident you've really found them all.

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[math-fun] Essay on .999...
by James Propp 18 Sep '15

18 Sep '15

The final version of my essay "The One About .999..." is available at http://mathenchant.wordpress.com . I especially like the new graphic contributed by Li-Mei Lim. I'd welcome comments in the Comments section; the blog has been too quiet. And surely the pedagogical aspects of the topic are subject to legitimate controversy. Do you think I'm wrong about the best way to deal with convergence confusion and decimal dissidence? Jim

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[math-fun] I just ported my ancient Macsyma Euler-Maclaurin fcn to Mma.
by Bill Gosper 17 Sep '15

17 Sep '15

(With exact remainder and optional phase parameter.) EulMac[fn,var lo,hi,trms,phase] Here are three terms of Stirling's formula: In[305]:= Assuming[n > 1, EulMac[Log[z], z, 1, n, 3]] Out[305]= 1 - n + n*Log[n] - Inactive[Sum][Log[z], {z, 2, n}] == (-(1/24))*Inactive[Sum][ ConditionalExpression[ (-1 + 3*¢*(1 + ¢)* (-1 + 10*¢*(1 + ¢)* (1 + 12*¢*(1 + ¢))) + 180*¢^3*(1 + ¢)^3* (1 + 2*¢)*(Log[¢] - Log[1 + ¢]))/(15*¢^3*(1 + ¢)^3), Re[¢] > 0 || Re[¢] < -1 || NotElement[¢, Reals]], {¢, 1, -1 + n}] + Inactive[Sum][((-1)^¢*BernoulliB[1 + ¢]* Derivative[¢][-Log[1 + #1] + Log[n + #1] & ][0])/ (1 + ¢)!, {¢, 0, 3}] The self-evidently superfluous ConditionalExpression surrounds the remainder summand. It is typically textually cumbersome but numerically small. The Sums are inactive to discourage Mathematica from struggling to "simplify" them. Activating gives Log[Pochhammer] instead of Log[Factoria] if you don't call n a positive integer. In[311]:= Assuming[n > 1 && Element[n,Integers], FullSimplify /@ Activate[Out[305]]]; In[314]:= Out[311] /. ConditionalExpression[ x_, __] :> FullSimplify[x]; In[315]:= Solve[%, Log[n!]] Out[315]= {{Log[n!] -> (1/(360*n^3))* (-1 + 30*n^2 + 331*n^3 - 360*n^4 + 180*n^3*Log[n] + 360*n^4*Log[n] + 15*n^3*Sum[(1/(15*¢^3* (1 + ¢)^3))*(-1 + 3*¢*(1 + ¢)*(-1 + 10*¢*(1 + ¢)* (1 + 12*¢*(1 + ¢))) + 180*¢^3*(1 + ¢)^3* (1 + 2*¢)*(Log[¢] - Log[1 + ¢])), {¢, 1, -1 + n}])}} In[318]:= Expand /@ %315[[1, 1]] Out[318]= Log[n!] -> 331/360 - 1/(360*n^3) + 1/(12*n) - n + Log[n]/2 + n*Log[n] + (1/24)* Sum[(1/(15*¢^3*(1 + ¢)^3))* (-1 + 3*¢*(1 + ¢)* (-1 + 10*¢*(1 + ¢)* (1 + 12*¢*(1 + ¢))) + 180*¢^3*(1 + ¢)^3* (1 + 2*¢)*(Log[¢] - Log[1 + ¢])), {¢, 1, -1 + n}] Note 331/360 vs ln (rt2 pi): In[44]:= N[{Log[Sqrt[2*Pi]], 331/360}] Out[44]= {0.918938533, 0.919444444} Testing on 8! In[319]:= %318 /. n -> 8; In[320]:= Expand /@ % Out[320]= Log[40320] -> Log[2] + Log[3] + Log[4] + Log[5] + Log[6] + Log[7] + Log[8] (because the remainder sum is exact). Punting it, In[321]:= %318 /. Sum[__] -> 0 Out[321]= Log[n!] -> 331/360 - 1/(360 n^3) + 1/(12 n) - n + Log[n]/2 + n Log[n] Testing with n=8, In[322]:= % /. n -> 8 Out[322]= Log[40320] -> -(1303169/184320) + (17 Log[8])/2 In[323]:= N[%] Out[323]= 10.6046029 -> 10.6051088 Repeating the analysis starting with Assuming[n > 1, EulMac[Log[z], z, 1, n, 5, 1/2]] (i.e., one more Bernoulli term and phase = 1/2) Out[46]= Log[n!] -> 628991/612360 - 31/(1260 (1 + 2 n)^5) + 7/(360 (1 + 2 n)^3) - 1/( 12 + 24 n) + Log[2] - n (1 + Log[2]) - (3 Log[3])/ 2 + (1/2 + n) Log[1 + 2 n] In[47]:= % /. n -> 8. Out[47]= 10.6046029 -> 10.6046256 with the ^-5 term contributing only In[48]:= %%[[2, 2]] /. {{n -> 8}, {n -> 8.}} Out[48]= {-31/1789019820, -1.73279243*10^-8} so most of the .00002 error resides in the (discarded) remainder term: Out[49]= -433349316889/62104545180 - 8 Log[2] - 5 Log[3]/2 - Log[4] - Log[5] - Log[6] - Log[7] - Log[8] + 17 Log[17]/2 In[50]:= N[%] Out[50]= 0.0000226652532 The function is EulMac[xp_, v_, lo_, hi_, deg_, y_: 0] := Integrate[xp /. v -> y + v, {v, lo, hi}] - Inactive[Sum][xp, {v, lo + 1, hi}] == (-1)^deg*Inactive[Sum][ Integrate[(deg + 1)*v^deg* D[xp /. v -> v*y + ¢ + 1, {v, deg + 1}] - BernoulliB[deg + 1, y + v]* D[xp /. v :> y + v + ¢, {v, deg + 1}], {v, 0, 1}], {¢, lo, hi - 1}]/(deg + 1)! + Inactive[Sum][(-1)^¢* Derivative[¢][Evaluate[(xp /. v -> # + hi) - (xp /. v -> # + lo)] &][y]* BernoulliB[¢ + 1, y]/(¢ + 1)!, {¢, 0, deg}] The motivation for porting was NSum's seeming inability to get so much as four digits of Sum 2 to oo of 1/Log[n!] - 1/Log[(2 n)!] - 1/Log[(-1 + 2 n)!] apparently due to NSum's internal Euler-Maclaurin failing to try Method -> "DoubleExponential" on the corresponding NIntegrate. I now believe the sum ~ 1.4424630245831935 . This is still not accurate enough for PSLQ or the like, and ISC humorously proposes 1/Log[2 + 1/(300 Khinchin Gamma[1/6])]. --rwg

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Re: [math-fun] Asimov's group-metric puzzle
by Warren D Smith 17 Sep '15

17 Sep '15

Message: 2 Date: Thu, 17 Sep 2015 12:03:46 -0700 From: Marc LeBrun <mlb(a)well.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: Re: [math-fun] Asimov's group-metric puzzle Message-ID: <D2205CA2.1C037%mlb(a)well.com> Content-Type: text/plain; charset="US-ASCII" [sorry for the previous spam] >> ="Warren D Smith" <warren.wds(a)gmail.com> >> Here is a list of the sporadic simple groups without Outer auts: >> M11, M23, M24, Co1, Co2, Co3, Th, Fi23, B, M, J1, Ru, J3, Ly >> (source: http://brauer.maths.qmul.ac.uk/Atlas/v3/spor/) >> Here is a list of the sporadic simple groups with outer auts: >> M12, M22, HS, J2, McL, Suz, He, HN, Fi22, Fi24', O'N, J3, T LeBrun: J3 appears in both lists, but J4 in neither? WDS: sorry. J3 has an outer aut, J4 does not.

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[math-fun] I wonder if this would wiggle the needle
by Bill Gosper 17 Sep '15

17 Sep '15

on the milliramanujanometer: Sum[n/(-1 + E^(2*Sqrt[11]*n*Pi)), {n, 1, Infinity}] == 1/24 - 1/(8*Sqrt[11]*Pi) - ((8 - (4*6^(1/3)*(-6127 + 63*Sqrt[33]))/ (8199 + 56809*Sqrt[33])^(2/3) + (99 + 205*Sqrt[33])/(6*(8199 + 56809*Sqrt[33]))^ (1/3))*Gamma[1/22]*Gamma[5/22]*Gamma[3/11]* Gamma[15/22])/(2112*2^(8/11)*Sqrt[11]*Pi^2* Gamma[7/22]*Gamma[10/11]) --rwg

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[math-fun] Asimov's group-metric puzzle
by Warren D Smith 17 Sep '15

17 Sep '15

> From: Dan Asimov <asimov(a)msri.org> > Let (G, *) be a finite group. > 1. Then (say by left multiplication) G acts as a group LG of permutations on > itself: > LG := {L_g | g in G} is a subgroup of S(G), where > L_g(x) := gx for all x in G and all g in G, > and S(G) is the group of all permutations of the set G. > 2. Definition: > A left-invariant metric D on G is one such that for every g in G, the > permutation > L_g: G -> G > is an isometry. > 3. There is always a left-invariant metric D on G. > For, pick any metric d on G, and average it over the action of LG... > D(x,y) := (1/|G|) Sum_{g in G} d(L_g(x), L_g(y)) --WDS COMMENT: this proof has a gap: you implicitly assumed such a metric d existed to start with. If we allow trivial metrics e.g. all distances are 1 except dist(x,x)=0, then ok, but it is not immediately obvious a nontrivial metric exists? Oh, I see: if a,b unequal then dist(a,b)=1+epsilon*random should work as "d". > 4. Question: > Which finite groups G have some left-invariant metric D whose full group of > isometries is precisely the left multiplications > Isom((G,D)) = {L_g | g in G} > ??? > 5. Example: > Let G be the alternating group A_3. Every left-invariant metric D makes G > into an equilateral triangle. But every equilateral triangle has a 6-element > isometry group, S_3. --WDS: The reason the A3 example fails is that A3 has an "outer automorphism" consisting of conjugation by an element of S3 that is not a member of A3. For any group, if g is a member, any conjugation map x --> g^(-1) x g is an automorphism. An automorphism not of this form, is "outer." So one might make the tentative conjecture, that the answer to Dan's question is "every group that has no outer automorphisms." Here is a list of the sporadic simple groups without Outer auts: M11, M23, M24, Co1, Co2, Co3, Th, Fi23, B, M, J1, Ru, J3, Ly (source: http://brauer.maths.qmul.ac.uk/Atlas/v3/spor/) Here is a list of nonsporadic simple groups without outer auts: S_a(2) apparently for all even a>=6, G2(5), E7(p), E8(p), E8(p) apparently for all primes p, Here is a list of nonsporadic simple groups with outer auts: A_n for each n>=3, L_a(b) apparently for all a>=2 and b>=2, S_a(b) apparently for all b>=3, U_a(b) apparently for all a,b, O_a(b), O+_a(b), O-_a(b) apparently for all a,b Here is a list of the sporadic simple groups with outer auts: M12, M22, HS, J2, McL, Suz, He, HN, Fi22, Fi24', O'N, J3, T As a positive example: There are some well known metrics on S_n, the group of n! permutations, including the number of inversions, and the euclidean and L1 distance metrics in n-space. In general: all we need to know is the distances(x,Id) where Id is the identity element. What if we consider the multiplicative group of nonsingular kXk matrices over some field F? Well, for Lie groups, see John Milnor: Curvatures of left invariant metrics on lie groups, Advances in Mathematics 21,3 (September 1976) 293-329 which begins with the claim "each n-dimensional Lie group possesses a (n-1)n/2-dimensional family of distinct left invariant metrics." If F is the complex numbers, using dist(Id,M) is some suitable function of the eigenvalues of M, such as dist=SUM |log(eigenvalue)|, perhaps works? But if F is (say) a finite field, then what? For any finite group G, the full set of left invariant metrics is describable as the solution of a linear program, i.e. is a polytope. One might ask whether and when the symmetries of this polytope are G. So... interesting puzzle, possibly with rich connections to a lot of stuff; sorry I have not solved it. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Doomsday argument
by Steve Witham 17 Sep '15

17 Sep '15

The point of the Doomsday argument as I understand it is that we should expect to be near the median of (human?) *births*. If so, then only about the same number of births as have already happened should be expected to happen in the future. This doesn't mean there will be no individuals living far into the future, just that there will only be a certain total number of lives in all of time (200 billion by the numbers below). So, rather than a continuing exponential growth of population across the asteroid belt, never mind across the galaxies, something drastically less profligate has to happen, that's the "doom". I am not smart enough to understand why this is not nonsense. For instance, does the outcome change with each change I make to the definition of my affinity group? E.g. conscious entities, original templates of consciousnesses, copies of consciousnesses, contiguous-over-spacetime graph structures of protoplasm (e.g. my family tree), waking days, etc? Some of the people who make the Doomsday argument also make the Simulation argument (we should expect that we are in a simulation), and I don't see how they can be reconciled or combined. Is what counts the number of different personalities, or the number of copies of people summed across all simulations, or what? Do these differences all cancel out, or what? (For the Akond of Swat.) --Steve p.s. I was going to ask whether it matters whether I later become a genetic reengineeree, cyborg, or upload. I got zero hits on google for "reengineeree," except from autospam "search for <what you searched for>" sites, so I claim the word for Spain. On 9/15/15 9:59 PM, math-fun-request(a)mailman.xmission.com wrote: > Date: Tue, 15 Sep 2015 14:12:03 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> >> From: Charles Greathouse <charles.greathouse(a)case.edu> >> The doomsday argument suggests considering oneself as a random sample from >> the total number of humans to be born. Estimating the number of humans born >> so far at 100 billion and the number of annual births at ~0.13 billion per >> year (assuming stabilization around 10 billion, if current trends continue) >> a 95% confidence interval suggests that humans will be around at least 20 >> more years but not more than 30,000 years. Adjust assumptions to suit. > --That argument yields very different conclusions if we assume that > future "humans" (which by the way might not be considered human) > will have much longer lifespans. > > For example I think cybernetic "life" organisms will become important > and their lifespans could be comparable to the lifespans of stars. > In that case, assuming you are near-median chronologically does NOT at all > imply that intelligent life will end in the next 100K years. >

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[math-fun] polynomial "pixerpolation"
by Warren D Smith 17 Sep '15

17 Sep '15

Interpolation: given N points (x1,y1), (x2,y2), ..., (xN,yN), find a polynomial P(x) of degree=N-1 so that yk=P(xk). Pixerpolation: There are N "one-dimensional pixels" which each are finite subintervals of the real x-axis, and the goal is to find a polynomial P(x) whose value AVERAGED over the Kth pixel, equals y_K. I'll assume these intervals are ADJACENT, i.e.they tile one big interval. Solution: Pixerpolation can be done in essentially the same time & space (as a function of N) as ordinary interpolation. Simply interpolate the INTEGRATED P-value at all the N+1 pixel endpoints, using a degree=N polynomial, then take its derivative. (Integral=interval length times average value...) What about if we drop the adjacent-pixels assumption? Then it is not so easy anymore. One could do it by solving an NxN linear system, e.g. by order N^3 time using Gauss's method. One also could do an interval-averaged analogue of Lagrange interpolation. That is, precompute polynomials of degree=N-1 which have average=1 on pixel K, but average=0 on (and exactly one root within) each of the other N-1 pixels. Anyhow, it clearly can be done in a polynomial number of operations, but it is not obvious how fast. One would hope for a method taking O(N * (logN)^2) operations, which would equal the time bounds known for ordinary interpolation via clever algorithm tricks. But that might be a vain hope.

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