January 2014 - mathfun@mailman.xmission.com

Re: [math-fun] Flow of time
by Adam P. Goucher 17 Jan '14

17 Jan '14

> OK, here's this topic again; I'll bite: I think Albert Einstein pretty much > nailed the distinction between physical time and the perception of time: > > "When a man sits with a pretty girl for an hour, it seems like a minute. > But let him sit on a hot stove for a minute and it's longer than any hour. > That's relativity." Conclusion: There exists a unique positive real, epsilon, such that if a man sits on a hot stove with (1 + epsilon) pretty girls, time passes at its usual rate. Sincerely, Adam P. Goucher http://cp4space.wordpress.com

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[math-fun] Forbidden topics
by Bill Gosper 17 Jan '14

17 Jan '14

="Dan Asimov" <dasimov(a)earthlink.net> The rule as I understand it is to abstain from posts about politics or climate change. MLB>If I ask a goofy self-referential question about whether this post, which may or may not be about climate change and/or politics--is it permitted?<MLB Boko Haram reportedly has a website promising Sharia punishment for anyone caught using the Internet. --Bill Hey, Rich didn't say anything precluding Religion!

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[math-fun] Climate change - an actual economic estimate (claimed in press) for a refreshing change
by Warren D Smith 16 Jan '14

16 Jan '14

"Climate Protection May Cost 4% of World GDP by 2030: UN" By Alex Morales 16 January 2014, BloombergBusinessweek http://www.businessweek.com/news/2014-01-16/climate-protection-may-cut-worl… it is about a UN report which isn't officially out yet. -- Warren D. Smith http://RangeVoting.org

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[math-fun] Sullivan spooky knot untangler
by Warren D Smith 16 Jan '14

16 Jan '14

The E(K) energy functional is badly behaved for "summed" knots. I.e. tie two (or more) small trefoil knots in some string, then glue the two ends of the sting together. Now the self-energies of the little trefoils will be scale-invariant, but if you tighten them thus obtaining more rope to make the whole large loop of string get greater diameter, then you (energetically) win. Hence, the knots will tighten forever to reduce total energy, and there will be no nice energy minimum. But there is a nice energy min for "prime" knots (this is a published theorem).

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[math-fun] DanA's puzzle sum
by Bill Gosper 16 Jan '14

16 Jan '14

DanA> Okay, this has gone on entirely too long! Shall I explain the sum (1/2) + (1/3)(1-1/2) + (1/5)(1-1/2)(1-1/3) + (1/7)(1-1/2)(1-1/3)(1-1/5) + ... ? Or give a hint? --Dan -------------- On second thought, NO! --rwg

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[math-fun] Limit puzzle
by Bill Gosper 16 Jan '14

16 Jan '14

DWilson> For integer n >= 2, let x(n) be the smallest positive solution x of (ni + x)^n + (ni - x)^n = 0 where i is the imaginary unit. What is lim n->inf x(n)? -------- NeilB just got π/2, but refuses to post. --rwg

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[math-fun] two short papers about quantum gravity by WDS
by Warren D Smith 15 Jan '14

15 Jan '14

https://dl.dropboxusercontent.com/u/3507527/MinGravMass.html claims that two equal masses m will be unable to gravitationally bind if m<10^7 amu. https://dl.dropboxusercontent.com/u/3507527/DistNoise.html solves a problem about D-dimensional generalization of 1-dimensional "random walk" concept. Both papers should be very simple, short, find a rather surprising result, and which is comparatively (for a quantum gravity paper, anyhow) near to what is feasibly experimentally testable. (In case you were wondering, these 2 papers do NOT in any way attempt to provide "theories of everything," or grandiose quantum gravity theories; all the ideas used are ancient.) Comments? Questions? Revelations I'm an idiot?

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Re: [math-fun] Shanks weirdness, DanA's puzzle sum
by Dan Asimov 15 Jan '14

15 Jan '14

My solution(s): [Quoted spoiler space deleted. Unquoted spoiler space follows.] It looks as if Neil, approximately, and Gareth solved the basic problem, and Gareth found essentially the generalization I had in mind. Congratulations! ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... The first version of this series is the probability that a random positive integer has a prime factor: (*) 1/2 + 1/3 (1-1/2) + 1/5 (1-1/2)(1-1/3) + 1/7 (1-1/2)(1-1/3)(1-1/5) + ... = 1 . (Or the limit of the probability that a random positive integer <= N has a prime factor, as N -> oo.) A bit more generally: If a(n), b(n) are real for n in Z+, with 0 < a(n), b(n) < 1 and a_(n)+b(n) = 1, then because oo S = Product (a_n + b_n) = 1, n=1 it follows by expanding that S = a(1) + a(2)*b(1) + a(3)*b(2)*b(1) + a(4)*b(3)*b(2)*b(1) + ... = 1 Setting a(n) = 1/Prime[n] gives (*). --Dan --------------------------------------------------------- Gareth wrote: ----- (Dan asked: What's sum{p} of 1/p prod{q<p} of 1-1/q ? ... Write A = 1/2 + (1-1/2) 1/3 + (1-1/2) (1-1/3) 1/5 + ... (the sum we are interested in), and then write B = (1-1/2) + (1-1/2) (1-1/3) + (1-1/2) (1-1/3) (1-1/5) + ... (a rather obvious complement thereto). Then, adding termwise, A+B = 1 + (1-1/2) + (1-1/2) (1-1/3) + ... which is obviously just 1+B. Hence A = 1. Clearly we have used nothing whatever about the primes here; the theorem in question is that for any sequence a1,a2,... (probably with some technical conditions) we have sum{i} ai prod{j<i} (1-aj) = 1. This cries out for a probabilistic interpretation, and lo there is one: consider doing something that succeeds with probability a1, and if that fails doing something that succeeds with probability a2, etc.; then, provided the a don't tend to 0 too fast, we are guaranteed to succeed eventually, and Dan's sum is precisely the probability of this, split up according to the stage at which we succeed. -- g ----- and Neil B. wrote: ----- [SPOILER] ... 1! I actually have two solutions for this, one somewhat more general (though not very rigorous) and one which is probably Dan's intended solution. The first one rearranges the sum into: (1/2)+(1-1/2)(1/3+(1-1/3)(1/5+(1-1/5)( ... ))) which can be seen as the composition of an infinite number of functions A_n[z]=1/Prime[n]+(1-1/Prime[n])*z (Bill points out this is essentially a matrix product, and that I'm just obscuring things by writing it out like this.) And now the risky bit: looking at the tail end of the infinite sum, we'll have something like A_1[A_2[ ... A_inf[A_inf[A_inf[ ... ]]]...]] Of course the only fixed point of A, and thus the fixed point of A_inf (whatever that is) is 1, and then the nested expression simplifies to 1. This can also be generalized- just replace 1/Prime[n] with any real, non-one sequence in (0,2) which converges to a non-negative number, and the sum will still come out to 1. (Caution: some of these constraints are only a few minutes old!) Second solution: Expand the entire sum (parentheses added for clarity): (1/2) + (1/3 - (1/3)*(1/2)) + (1/5 - (1/5)*(1/2)-(1/5)*(1/3) + (1/5)*(1/2)*(1/3)) +... and it's easy to see that this looks like an application of the inclusion-exclusion principle, except with multiplication instead of intersection, and asymptotic density instead of cardinality. One can carry the analogy further and claim this is the asymptotic density of the union of <composites divisible by 2>, <composites divisible by 3>, and so forth- in other words, the asymptotic density of the set of composites. Since PrimePi[x]<x/(Log[x]-4) (for x>55) and is thus less than linear, the sum must be equal to 1, although it will converge very, /very/ slowly. This could also be combined with the first solution for some set-theoretical* results which I'm almost certain have been discovered before. --Neil *number-theoretical? measure-theoretical? ----- On Wed, Jan 15, 2014 at 10:46 AM, Dan Asimov <dasimov(a)earthlink.net> wrote: > Okay, this has gone on entirely too long! > > Shall I explain the sum > > (1/2) + (1/3)(1-1/2) + (1/5)(1-1/2)(1-1/3) + (1/7)(1-1/2)(1-1/3)(1-1/5) > + ... > > ? > > Or give a hint? > > --Dan > > > On 2014-01-10, at 2:21 PM, Bill Gosper wrote: > >> Trying for several digits of DanA's > >> Out[197]= Sum[Product[1 - 1/Prime[k], {k, 1, -1 + n}]/Prime[n], {n, 1, > Infinity}] > >> (No need to special-case n=1 -- Mma does the right thing on >> empty products. >> ... > >> A tall pole in the tent! > > Bill! This is a family list! > >> ... >> Is it time to give up and beg Dan for his answer?

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[math-fun] Limit problem
by David Wilson 15 Jan '14

15 Jan '14

For integer n >= 2, let x(n) be the smallest positive solution x of (ni + x)^n + (ni - x)^n = 0 where i is the imaginary unit. What is lim n->inf x(n)?

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[math-fun] Shanks weirdness, DanA's puzzle sum
by Bill Gosper 15 Jan '14

15 Jan '14

Trying for several digits of DanA's Out[197]= Sum[Product[1 - 1/Prime[k], {k, 1, -1 + n}]/Prime[n], {n, 1, Infinity}] (No need to special-case n=1 -- Mma does the right thing on empty products. And the *wrong* thing when they're of negative length!) In[198]:= shanks[L_List] := (#1*#3 - #2^2)/(#1 - 2*#2 + #3) &[Drop[L, -2], L[[2 ;; -2]], Drop[L, 2]] Take exponentially longer segments of the series: In[187]:= numMProd[ Table[MProd[ MatrixForm[{{1 - 1./Prime[i], 1/Prime[i]}, {0, 1}}], {i, 2^n + 1, 2^(n + 1)}], {n, 0, 11}]] Out[187]= {{{0.666667, 1/3}, {0, 1}}, {{0.685714, 0.314286}, {0., 1.}}, {{0.74823, 0.25177}, {0., 1.}}, {{0.795719, 0.204281}, {0., 1.}}, {{0.83033, 0.16967}, {0., 1.}}, {{0.856913, 0.143087}, {0., 1.}}, {{0.876387, 0.123613}, {0., 1.}}, {{0.891033, 0.108967}, {0., 1.}}, {{0.902836, 0.0971637}, {0., 1.}}, {{0.912323, 0.0876769}, {0., 1.}}, {{0.920145, 0.079855}, {0., 1.}}, {{0.926663, 0.0733374}, {0., 1.}}} Restore the first term, and accumulate: In[188]:= FoldList[Dot, {{1, 1}/2, {0, 1}}, %] Out[188]= {{{1/2, 1/2}, {0, 1}}, {{0.333333, 0.666667}, {0., 1.}}, {{0.228571, 0.771429}, {0., 1.}}, {{0.171024, 0.828976}, {0., 1.}}, {{0.136087, 0.863913}, {0., 1.}}, {{0.112997, 0.887003}, {0., 1.}}, {{0.0968287, 0.903171}, {0., 1.}}, {{0.0848595, 0.915141}, {0., 1.}}, {{0.0756126, 0.924387}, {0., 1.}}, {{0.0682658, 0.931734}, {0., 1.}}, {{0.0622804, 0.93772}, {0., 1.}}, {{0.057307, 0.942693}, {0., 1.}}, {{0.0531043, 0.946896}, {0., 1.}}} The partial sums are in the upper right [[1,2]]: In[189]:= shanks[#[[1, 2]] & /@ %] Out[189]= {0.948718, 0.899118, 0.917897, 0.932005, 0.94094, 0.949258, \ 0.955796, 0.96014, 0.964033, 0.967136, 0.969815} Second tier: In[190]:= shanks[%] Out[190]= {0.912739, 0.974621, 0.956372, 1.06137, 0.97982, 0.968741, \ 0.997558, 0.979348, 0.986716} Third (losing both ends each time.) In[191]:= shanks[%] Out[191]= {0.960528, 0.971919, 1.01547, 0.966999, 0.976743, 0.986399, \ 0.984594} Fourth. In[192]:= shanks[%] Out[192]= {0.956494, 0.99253, 0.975112, 2.0492, 0.984878} A tall pole in the tent! The immediate explanation is that 0.966999, 0.976743, 0.986399 is nearly in arithmetic progression, provoking division by nearly 0. In fact, the irregularity of the primes makes the process so noisy that the second differences change sign wantonly, and this whole computation is meaningless. But suppose we smooth it out, replacing Prime[n] with n*Log[n]. That's an interesting sum, too. Suppose we sum 10^6 terms directly, and Shanks the tail: numMProd[Table[ MProd[MatrixForm[{{1 - 1`33/(n*Log[n]), 1/(n*Log[n])}, {0, 1}}], {n, 2^k + 1000000, 999999 + 2^(k + 1)}], {k, 0, 11}]]; (Here, numMProd just performs the matrix products, to 33 places due to the 1`33.) Form the sequence of exponentially longer partial sums: FoldList[Dot, IdentityMatrix[2], %]; Run one Shanks: shanks[#[[1, 2]] & /@ %] {-7.23825846114180023820963784936630936956*10^-8, \ -7.2383081775534954986936726671956993755*10^-8, \ -7.238507042592270807744603740319705184*10^-8, \ -7.239302497883386891719990656592430349*10^-8, \ -7.24248428013694135359940241455532980*10^-8, \ -7.25521109787942075472466273513777568*10^-8, \ -7.3061158789240489622769357685310492*10^-8, \ -7.509715087677159821674775767630926*10^-8, \ -8.323952662885734390860079101633870*10^-8, \ -1.157962990187933870110393101592588*10^-7, \ -2.459217059666940024454702903981525*10^-7} How can these conceivably be <0?? Try the 2nd order Shanks: shanks2[#[[1, 2]] & /@ %198] {-7.2382418887852421433705793226704*10^-8, \ -7.238241887452633043864365057607*10^-8, \ -7.238241876791840693390633496226*10^-8, \ -7.238241791506788978747515379514*10^-8, \ -7.23824110924696740587213638954*10^-8, \ -7.23823565149782811295872404833*10^-8, \ -7.23819199477433435746658414992*10^-8, \ -7.2378428252583618557144316956*10^-8, \ -7.2350508161419247635566250253*10^-8} Still negative! The only glimmer of sanity comes from summing bursts of length 3^n, shanks[#[[1, 2]] & /@ %] {-7.23827917631153038724969102081745002169*10^-8, \ -7.2385774734354090466183845664296486922*10^-8, \ -7.241262106902162871015196793342498037*10^-8, \ -7.26542271070426051438560418211001551*10^-8, \ -7.4828385233749802945842107930570293*10^-8, \ -9.438781720404053555835539434832801*10^-8, \ -2.702074800500521053964558787965094*10^-7, \ -1.846816420328610065496655139531113*10^-6, \ -0.00001588394699143661453072982592933523, \ -0.0001383570856200631342075355905173333, \ -0.001152755525647733026707277401399129} and the using Shanks2: shanks2[#[[1, 2]] & /@ %%] {-7.23824188680716125090267195281569*10^-8, \ -7.2382418304287692662035666190882*10^-8, \ -7.238240308325929783457067074808*10^-8, \ -7.238199220757133558198054009671*10^-8, \ -7.237090600928513738314351741322*10^-8, \ -7.20721785747766461202604502914*10^-8, \ -6.40543875690906529608871726514*10^-8, 1.48716801605978891479361949759*10^-7, 5.62589299046691128953938459235*10^-6} I have never before seen this wrong sign behavior on a smooth sequence. Unnerving. --rwg Does anyone have a nice little Richardson accelerator to share? Is it time to give up and beg Dan for his answer?

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