March 2013 - mathfun@mailman.xmission.com

[math-fun] Wolfram Alpha has disabled very large numbers
by Bill Gosper 19 Mar '13

19 Mar '13

I wonder if this explains a recent frustration. While puzzling out the condition for arithmetic bursts in A129935, I lost faith in 9.0.1 and 8.0.4 numerics and signed up for an Alpha Pro trial. But I couldn't perform the Floor[...] != Ceiling[...] test on numbers as small as 10^10000. --rwg Sometime in the 2-3 weeks since I last mentioned the "large factorials" problem, Wolfram Alpha has "fixed" it... by disabling all input that gives an answer higher than about 10^10^9. This is true for inputs like "10^10^10^34", which involve no factorials at all. For example, "2^3250000000" works: http://www.wolframalpha.com/input/?i=2%5E3250000000 but "2^3300000000" gives an error: http://www.wolframalpha.com/input/?i=2%5E3300000000 Of course, large factorials give an error, beyond about 101000000! (the factorial of 101 million), which is about 10^10^8.88... On Thu, Feb 28, 2013 at 12:29 PM, Robert Munafo <mrob27(a)gmail.com> wrote: Thanks to each of you for the help. Meeussen Wouter had the great idea of actually figuring out what Wolfram's "wrong answer" was. In fact it's easy to divide one answer by the other and see the ratio is 20000, which makes it clear what happened. That means that the internals of Wolfram Alpha are actually computing those two results separately, rather than just computing one and then converting it into the other representation. Quite odd. I had other saved results from Wolfram Alpha for larger factorials, and I can see now that their wrong answers for 30000! and 100000! fit the same pattern: they're computing Gamma[N] when they should be computing Gamma[N+1]. There's another bug for factorials of even larger numbers. In the case of (10^10)! they give only the "power of 10 representation", and their answer is too big rather than being too small, and by a far greater amount. They give 10^10^10^1.0413926851582..., but the correct answer (by GP/PARI and my own implementation) is 10^10^10^1.0406306991294... In this case it's easy to see that Wolfram's answer is exactly 10^10^11. They are approximating n! as n^n, which "isn't even close" to the truth. Using two terms of the factorial form of the Stirling series [1] and expressing everything as a power of 10: (10^10)! ~ 10^(0.5*ln(2*pi)/ln(10) + 5 + 10^11 - 10^10/ln(10) + 1/(12*ln(10)*10^10)) (where "ln" is natural logarithm). You can see the dominant term 10^11 in there, but WolframAlpha is ignoring all the rest, which makes a big difference. They are doing the same thing for larger values, for example if you put in (10^20)!, their answer is equivalent to (10^20)^(10^20) = 10^(20*10^20) = 10^10^(21+log(2)/log(10)) = 10^10^10^1.328400603937241. I have reported this to Wolfram as a separate problem. This started when I was investigating the difference between my computed value of (10^100)! (the factorial of a googol) and the value on someone else's webpage. Googol factorial has been discussed at least as early as 1996; see [2]. - Robert [1] http://mathworld.wolfram.com/StirlingsSeries.html [2] http://mathforum.org/library/drmath/view/57903.html

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[math-fun] Screwy one-liner Mma "derivation"
by Bill Gosper 19 Mar '13

19 Mar '13

of the binomial coefficient formula: In[60]:= FindSequenceFunction[Factor[FindSequenceFunction[#,k]&/@NestList[Accumulate,Table[1,{9}],7]],n] Out[60]= Pochhammer[k, -1 + n]/Pochhammer[1, -1 + n] In[61]:= TableForm[Table[%, {n, 5}, {k, 5}]] Out[61]= 1 1 1 1 1 1 2 3 4 5 1 3 6 10 15 1 4 10 20 35 1 5 15 35 70 --rwg

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[math-fun] double digit inflation
by Bill Gosper 19 Mar '13

19 Mar '13

In[52]:= Floor[Pi*10^20/11] Out[52]= 28559933214452665804 --rwg Decimal. Feh.

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[math-fun] Wolfram Alpha has disabled very large numbers
by Robert Munafo 19 Mar '13

19 Mar '13

Sometime in the 2-3 weeks since I last mentioned the "large factorials" problem, Wolfram Alpha has "fixed" it... by disabling all input that gives an answer higher than about 10^10^9. This is true for inputs like "10^10^10^34", which involve no factorials at all. For example, "2^3250000000" works: http://www.wolframalpha.com/input/?i=2%5E3250000000 but "2^3300000000" gives an error: http://www.wolframalpha.com/input/?i=2%5E3300000000 Of course, large factorials give an error, beyond about 101000000! (the factorial of 101 million), which is about 10^10^8.88... On Thu, Feb 28, 2013 at 12:29 PM, Robert Munafo <mrob27(a)gmail.com> wrote: > Thanks to each of you for the help. > > Meeussen Wouter had the great idea of actually figuring out what Wolfram's > "wrong answer" was. In fact it's easy to divide one answer by the other and > see the ratio is 20000, which makes it clear what happened. That means that > the internals of Wolfram Alpha are actually computing those two results > separately, rather than just computing one and then converting it into the > other representation. Quite odd. > > I had other saved results from Wolfram Alpha for larger factorials, and I > can see now that their wrong answers for 30000! and 100000! fit the same > pattern: they're computing Gamma[N] when they should be computing > Gamma[N+1]. > > There's another bug for factorials of even larger numbers. In the case of > (10^10)! they give only the "power of 10 representation", and their answer > is too big rather than being too small, and by a far greater amount. They > give 10^10^10^1.0413926851582..., but the correct answer (by GP/PARI and my > own implementation) is 10^10^10^1.0406306991294... > > In this case it's easy to see that Wolfram's answer is exactly 10^10^11. > They are approximating n! as n^n, which "isn't even close" to the truth. > Using two terms of the factorial form of the Stirling series [1] and > expressing everything as a power of 10: > > (10^10)! ~ 10^(0.5*ln(2*pi)/ln(10) + 5 + 10^11 - 10^10/ln(10) + > 1/(12*ln(10)*10^10)) > > (where "ln" is natural logarithm). You can see the dominant term 10^11 in > there, but WolframAlpha is ignoring all the rest, which makes a big > difference. They are doing the same thing for larger values, for example if > you put in (10^20)!, their answer is equivalent to (10^20)^(10^20) = > 10^(20*10^20) = 10^10^(21+log(2)/log(10)) = 10^10^10^1.328400603937241. I > have reported this to Wolfram as a separate problem. > > This started when I was investigating the difference between my computed > value of (10^100)! (the factorial of a googol) and the value on someone > else's webpage. Googol factorial has been discussed at least as early as > 1996; see [2]. > > - Robert > > [1] http://mathworld.wolfram.com/StirlingsSeries.html > > [2] http://mathforum.org/library/drmath/view/57903.html > -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] Twenty-dollar question
by Henry Baker 18 Mar '13

18 Mar '13

Correct, although even if the chains/wires are supporting weight, their shapes for the free portions should still be _sections_ of catenaries. Thus, even the parabolic shape on a typical (Golden Gate-style) suspension bridge is a piecewise approximation from bits of catenary shapes. I hadn't realized it until I did some searching a few years ago, but the people most interested in catenaries on a day-to-day basis are people on ships who have to worry about the lengths and stresses on anchor chains. At 01:03 PM 3/18/2013, Michael Kleber wrote: >Are you talking about the chains holding up the light? There's a vertical >chain as well, so the "mathematically correct" shape depends on the >relative tensions of the chains. As long as the bill matches reality, >catenary vs. parabola is irrelevant. > >http://0.tqn.com/d/dc/1/0/C/J/whitehouse2.jpg > >--Michael

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Re: [math-fun] Twenty-dollar question
by Henry Baker 18 Mar '13

18 Mar '13

I googled "white house chandelier" & got a closeup. It looks like the chains/wires are for the purpose of keeping this outdoor chandelier from blowing in the wind (& possibly destroying itself). So the chains/wires aren't just hanging loose, and hence are neither parabolae or catenariae. But the catenary (~ e^x) function is an appropriate metaphor for the spending encouraged by every occupant of this building. At 12:03 PM 3/18/2013, James Propp wrote: >Is the hanging chain on the back of the $20 bill mathematically correct? It >looks more like a parabola than a catenary to me. > >Jim Propp

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[math-fun] Twenty-dollar question
by James Propp 18 Mar '13

18 Mar '13

Is the hanging chain on the back of the $20 bill mathematically correct? It looks more like a parabola than a catenary to me. Jim Propp

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Re: [math-fun] A129935
by Bill Gosper 17 Mar '13

17 Mar '13

NJAS>Bill, It seems that you are saying that A129935 should be corrected, but I don't quite follow what should be changed. This comes to me as a message from the cloud, and I will do whatever you say. What should be changed? Neil Neil, I'm not sure yet. The PARI code is wrong since it misses confirmed values among the first 100, so there's not much point in adding equivalent Mathematica code with the same bug. We need an algorithm, perhaps from Max Alekseyev and Robert Gerbicz, and presumably based more adaptively on the continued fraction of 2/ln 2. But even if we find one, it may be impossible at this time to give a correct program, because of the Mma 9.0.1 bug, and a new discrepancy Neil B. seems to have found between Mathematica 8.04 and PARI! We may need to add warnings that, as of March 2013, the programs fail for certain large n due to implementation bugs. Or if Maple or Maxima works, we can just punt the code for the flaky CAS(es). --Bill On Wed, Mar 13, 2013 at 9:18 PM, Bill Gosper <billgosper(a)gmail.com> wrote: On 2013-02-27, at 3:05 PM, Bill Gosper wrote: Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil [B.] empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) MRob>This reminds me of something Neil Sloane mentions in his OEIS talks, namely that the function f(n) = floor(2n/log(2)) has the same value as the function g(n) = ceiling(2/(2^(1/n)-1)) (Sloane's sequence A78608) for all integer n from 1 to [7]77451915729367, but differs at n=[7]77451915729368 (and at some other larger values, see oeis.org/A129935) -------------------------- rwg>Neil B. points out that Phil's PARI code misses eleven entries in http://oeis.org/A129935/b129935.txt, and suggests 13*cand (or larger) instead of 2*cand in default(realprecision, 500); c=contfrac(log(2)/2); for(n=2, #c, cand=contfracpnqn(vecextract(c, 2^n-1))[1, 1]; forstep(m=cand, 2*cand, cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), print(m)))) 13, [or] even 20, is in turn insufficient for longer tables. 20 gives 638215591788186030688409<<5153>>7190350518178849343531561 as A(1520), but 26 gives it as A(1527). Is there any finite alternative to 26 that works indefinitely? If not, what is a correct algorithm here? This may relate to the startling interludes of slow, arithmetic growth at, e.g., A(56)-A(68). Here is equivalent Mma code: $MaxExtraPrecision = 999999999; Select[Flatten[ Transpose[Outer[Times, Range[26], Denominator@Convergents[2/Log[2], 9999]]]], Floor[2*#/Log[2]] != Ceiling[2/(2^(1/#) - 1)] &]; This should really say $MaxExtraPrecision = ∞ but that tickles a numerics bug in the current version (9.0.1). --rwg

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[math-fun] N-body
by Warren D Smith 16 Mar '13

16 Mar '13

It is perhaps worth noting that I proved that Newtonian N-body evolution is undecidable. To be more precise, if I give you the initial velocities, masses, positions of N point masses (for some finite N) on some input tapes (the tapes are infinite so can store real numbers, most-significant digits first) and I provide two concentric spheres, and two concentric time-intervals, and I ask you: "will (A) any of the bodies enter the inner sphere during the inner time interval, or (B) will every body stay outside of both spheres, throughout the outer time interval?" [Note: I have phrased this in this way so that there is a 3rd physically possible intermediate option (C), and if it occurs the decider is allowed to provide any answer.] then Theorem: no Turing machine program can answer this question. (But the bodies themselves would answer it.) However, if Newton's laws are replaced by certain more-realistic laws (better approximating special and general relativity) then the problem becomes decidable. Also, I proved in one of my finest results that a quantum version of the problem is decidable, which is quite odd (quantum mechanics computationally easier than classical mechanics). Thus, believers in computability could have told the physicists that Newton's laws had to be wrong (in a superior alternate history). Another result of mine is that the laws (Navier Stokes) of hydrodynamics for water in a rigid container, also lead to undecidable problems, i.e. "will more than a liter of water end up in this sphere during the next hour?" with same three-way phrasing game as for Newton, is undecidable, OR certain other nasty things happen such as Navier Stokes solution nonexistence or nonuniqueness, or such as Navier Stokes solutions vastly contradicting experiment in certain simple situations. In any of these 3 permitted alternatives, computational fluid dynamics is basically killed -- there will never be an always-rigorous Navier-Stokes solver, and indeed the Navier-Stokes equations are physically wrong and we can devise innocent-looking situations where atomic simulations would simulate reality but Navier-Stokes could not...

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Re: [math-fun] Physicists Discover 13 New Solutions To Three-Body Problem
by Bill Gosper 16 Mar '13

16 Mar '13

CMoore>The figure-8 orbit is _slightly_ stable, in the sense that small perturbations of the initial positions and momenta, or small changes in the masses away from them being identical, result in smooth oscillations, phase changes, or rotations rather than escape. The same seems to be true of the "criss-cross" orbit, which was discovered by Henon; see some numerical evidence of this in http://arxiv.org/pdf/math/0511219.pdf . Still, you would have to be incredibly lucky to be in the basin of attraction of either orbit. Cris On Mar 9, 2013, at 9:57 PM, Henry Baker wrote: Are there any 3-body configurations that show anything like this type of stability, other than a hierarchy like Sun, Earth, Moon ? Cristopher Moore Professor, Santa Fe Institute ------ Since there's a (small) continuum of possible perturbations, is there a continuum of possible oscillation periods, mostly incommensurable with the main orbits? More generally, are there concrete examples of "immortal", aperiodic three body systems with comparable masses that never eject one? --rwg

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