April 2016 - mathfun@mailman.xmission.com

[math-fun] Fwd: cubes passing thru holes in cubes
by Bill Gosper 07 Apr '16

07 Apr '16

(Per Greg) Warren, what does your "naive look" predict for the degree of the univariate polynomial describing the cube-in-tesseract, which winds up reducing to 8? --rwg -------- Original Message -------- Subject: [math-fun] cubes passing thru holes in cubes Date: 2016-04-07 12:06 From: Warren D Smith <warren.wds(a)gmail.com> To: math-fun(a)mailman.xmission.com Reply-To: math-fun <math-fun(a)mailman.xmission.com> Gosper: The side of the largest square inscribable in the unit cube is 1.06066017177982... = 3/?2. Thus a unit cube can pass though a hole in another unit cube. The side of the largest cube inscribable in the unit tesseract can be found numerically: 1.00743475688427937609825359524... https://oeis.org/A243309 http://mathforum.org/kb/message.jspa?messageID=4637111 which is readily detected by integer lattice reduction to be a bi-quartic: In[1]:= RootApproximant[1.00743475688427937609825359524] Out[1]= Root[16 + 16 #1^2 - 7 #1^4 - 28 #1^6 + 4 #1^8 &, 3] (and thus is expressible in radicals). At G4G12, Greg Huber unveiled a mystery: Using the same trusted numerical methods (Huber & Ligocki, unpublished), he finds the largest tesseract inscribable in a 5-cube has side 1.0008394468593497886019289217565945828768033618266262653807572778501302485680777652768153338725025731471415139085497252961195484534254852511063402098761485174829601977310949912597155616952089698595643, yet this constant has so far resisted every effort (RootApproximant, FindIntegerNullVector, LatticeReduce, ...) to find its polynomial. --WDS: The largest tesseract inscribable in a 5-cube has side that certainly is an algebraic number. This also should be true of any local but nonglobal max (if there are any) and in any fixed number of dimensions. This should be obvious. So if you cannot find the algebraic, it is because his calculation was wrong, or since it has too high degree to find with that number of digits. I do not know how he calculated it, but a naive look by me at how to calculate it involves a set of simultaneous quadratic equations in 21 unknowns, plus a further maximization. If that were solved "symbolically" via methods of resultants, etc to reduce it to a univariate polynomial equation, then -- at least if you did so naively -- the degree of said polynomial would be very large indeed, probably at least a million. (There's various fancy assed resultants techniques out there, I do not know which is the best.) So it is not surprising at all that you are unable to determine the polynomial from a mere 200 or so digits. Here is perhaps a more interesting question. What is the asymptotic behavior, for large D, of the biggest (D-1)-hypercube inscribable in a unit D-hypercube? Your numbers for D=3,4,5 make it looks like answer behaves something like 1+5/9^D. However, why should behavior be exponential? Presumably it is since we perturb the Kth coordinate by some amount exponentially decreasing with K... something like that.

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[math-fun] cubes passing thru holes in cubes
by Warren D Smith 07 Apr '16

07 Apr '16

Gosper: The side of the largest square inscribable in the unit cube is 1.06066017177982... = 3/?2. Thus a unit cube can pass though a hole in another unit cube. The side of the largest cube inscribable in the unit tesseract can be found numerically: 1.00743475688427937609825359524... https://oeis.org/A243309 http://mathforum.org/kb/message.jspa?messageID=4637111 which is readily detected by integer lattice reduction to be a bi-quartic: In[1]:= RootApproximant[1.00743475688427937609825359524] Out[1]= Root[16 + 16 #1^2 - 7 #1^4 - 28 #1^6 + 4 #1^8 &, 3] (and thus is expressible in radicals). At G4G12, Greg Huber unveiled a mystery: Using the same trusted numerical methods (Huber & Ligocki, unpublished), he finds the largest tesseract inscribable in a 5-cube has side 1.0008394468593497886019289217565945828768033618266262653807572778501302485680777652768153338725025731471415139085497252961195484534254852511063402098761485174829601977310949912597155616952089698595643, yet this constant has so far resisted every effort (RootApproximant, FindIntegerNullVector, LatticeReduce, ...) to find its polynomial. --WDS: The largest tesseract inscribable in a 5-cube has side that certainly is an algebraic number. This also should be true of any local but nonglobal max (if there are any) and in any fixed number of dimensions. This should be obvious. So if you cannot find the algebraic, it is because his calculation was wrong, or since it has too high degree to find with that number of digits. I do not know how he calculated it, but a naive look by me at how to calculate it involves a set of simultaneous quadratic equations in 21 unknowns, plus a further maximization. If that were solved "symbolically" via methods of resultants, etc to reduce it to a univariate polynomial equation, then -- at least if you did so naively -- the degree of said polynomial would be very large indeed, probably at least a million. (There's various fancy assed resultants techniques out there, I do not know which is the best.) So it is not surprising at all that you are unable to determine the polynomial from a mere 200 or so digits. Here is perhaps a more interesting question. What is the asymptotic behavior, for large D, of the biggest (D-1)-hypercube inscribable in a unit D-hypercube? Your numbers for D=3,4,5 make it looks like answer behaves something like 1+5/9^D. However, why should behavior be exponential? Presumably it is since we perturb the Kth coordinate by some amount exponentially decreasing with K... something like that. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Fwd: Rupert's algebraic mystery
by Bill Gosper 07 Apr '16

07 Apr '16

The side of the largest square inscribable in the unit cube is 1.06066017177982... = 3/√2. Thus a unit cube can pass though a hole in another unit cube. The side of the largest cube inscribable in the unit tesseract can be found numerically: 1.00743475688427937609825359524... https://oeis.org/A243309 http://mathforum.org/kb/message.jspa?messageID=4637111 which is readily detected by integer lattice reduction to be a bi-quartic: In[1]:= RootApproximant[1.00743475688427937609825359524] Out[1]= Root[16 + 16 #1^2 - 7 #1^4 - 28 #1^6 + 4 #1^8 &, 3] (and thus is expressible in radicals). At G4G12, Greg Huber unveiled a mystery: Using the same trusted numerical methods (Huber & Ligocki, unpublished), he finds the largest tesseract inscribable in a 5-cube has side 1.0008394468593497886019289217565945828768033618266262653807572778501302485680777652768153338725025731471415139085497252961195484534254852511063402098761485174829601977310949912597155616952089698595643, yet this constant has so far resisted every effort (RootApproximant, FindIntegerNullVector, LatticeReduce, ...) to find its polynomial. This is really strange. I'd expect even a flawed methodology to find an algebraic maximum, albeit an erroneous one. Has anyone a clue here? --rwg

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Re: [math-fun] Holding much water
by Bill Gosper 07 Apr '16

07 Apr '16

On 2016-04-06 23:04, Henry Baker wrote: > Aren't epicycles the earliest version of Fourier series? > Yes, http://gosper.org/gibbstriangle.gif --rwg > https://en.wikipedia.org/wiki/Deferent_and_epicycle > > Now if someone could show that Ptolemy invented the FFT... > > At 05:04 PM 4/6/2016, Marc LeBrun wrote: >>Or there's Gosper's observation that enough epicycles actually could have >>accounted for any observed orbit...

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Re: [math-fun] Holding much water
by Henry Baker 07 Apr '16

07 Apr '16

Methinks "atmosphere-strip theory" might better be called "hydrogen-strip theory", since Venus's atmosphere is a heck of a lot thicker than Earth's. I don't know how much hydrogen is bound up in the crust of Venus in various compounds, but I doubt that anyone else does either. There are other potential effects the Earth's Moon could have aside from the possible effects on the Earth's magnetic field. It could have helped to sweep up various light materials for the Earth during the early history of the Solar system. If Venus & the Earth are similarly sized & similarly located, and yet Venus has this incredibly thick atmosphere and the Earth doesn't, then I'm willing to listen to any theories in which the Moon helps to get rid of some of the Earth's primordial atmosphere -- e.g., carbon dioxide. So far, the hydrogen stripping theory kills off hydrogen on Venus, but how does the non-existence of that stripping on Earth kill off all the garbage in the Earth's early atmosphere? I know that some of the planet formation theories have the big planets -- Jupiter, Saturn, etc. -- "shepherding" some of the smaller planets into formation. I've seen some computer animations which show the inner planet eccentricities varying quite a bit during the last 4.5 billion years. Do any of the planet formation theories contemplate the Earth and Venus swapping orbits? Could there have been some sort of orbit-crossing event -- perhaps the one that created the Moon -- that could have dramatically shifted either the Earth's or Venus's orbital position? At 01:46 PM 4/5/2016, Warren D Smith wrote: >--Venus's hydrogen is gone. Hence no water, no ammonia, no chemistry involving H, no life. > > Their ref 4 hopefully provided evidence. If so, then HB's disputation >of atmosphere-strip theory, is busted by observational evidence. > >--diddlysquat. Venus is extremely dry and has about 10^(-8) the absolute amount of H on Earth.

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Re: [math-fun] Moon needed for earth's magnetic field to last?
by Henry Baker 06 Apr '16

06 Apr '16

Other than to align grazing cows in a N-S direction, why does "life" depend upon the Earth's magnetic field? http://news.nationalgeographic.com/news/2008/08/080825-magnetic-cows.html At 02:30 PM 4/4/2016, Warren D Smith wrote: >Denis Andrault, Julien Monteux, Michael Le Bars, Henri Samuel: >The deep Earth may not be cooling down, >Earth and Planetary Science Letters >Available online 30 March 2016 > >Claims without the moon, the Earth's core would have cooled and shut off >its magnetic field long ago. And we and most life would probably be in deep trouble. >Mercury, Venus, and Mars all lack magnetic fields (anyhow 100X below Earth's). >Which is presumably not a coincidence. But some Jupiter moons still have a field >exceeding any of the above 3 planets. Ditto. > >-- >Warren D. Smith >http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Database of math problems
by Keith F. Lynch 06 Apr '16

06 Apr '16

Veit Elser <ve10(a)cornell.edu> wrote: > There must be a better example. For this problem I would generate > a list of even perfect numbers and get lots of hits on that. > Somewhere in all that material there surely will be mention of > the odd perfect number problem. Fair enough. How about the problem of whether every even number greater than 2 is the sum of two primes? If I didn't know that was called the Goldbach conjecture, how would I find it? Well, okay, Googling "sum of two primes" finds it. Never mind. > I imagine that most interesting problems can be perturbed in some > way so that their solutions generate a searchable signature of large > integers. Meta-problem: Find a problem that doesn't generate anything searchable. Not sequences, not large integers, not unusual real numbers, not even unusual keywords. Bonus points if it's not obvious what, if any, existing branch of math it belongs to.

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Re: [math-fun] about the polygamma function
by Bill Gosper 05 Apr '16

05 Apr '16

On 2016-04-01 11:12, Fred Lunnon wrote: > Hogamous, higamous, > Plouffe is polygammous. > Higamous, hogamous, > Gosper monogammous. > > WFL > ! I believe Fred refers to gosper.org/facfun.pdf But I occasionally succumb to infidelity: gosper.org/filds.pdf --rwg > > > On 4/1/16, Simon Plouffe <simon.plouffe(a)gmail.com> wrote: >> Hello, >> >> the Psi function is one of my favorite function, >> it has so many aspects and formulas, very interesting, >> >> for example, you can express primes with it, >> >> 691 = Psi(11,1/2)/(Pi^12*2^8). >> >> or Euler numbers, (absolute value), I hate that - sign all over. ;-). >> >> E(2n) = >> -1/2*(Psi(k,1/8)-Psi(k,3/8)+Psi(k,5/8)-Psi(k,7/8))*Pi^(-k-1)*4^(-k). >> >> we can also use Psi(n,1/4) and Psi(n,3/4) for the same. >> >> another example is 67 = >> >> 1/64*(Psi(5,1/5)-Psi(5,2/5)-Psi(5,3/5)+Psi(5,4/5))*5^(1/2)/Pi^6 >> >> a very good question is : are there other primes like that ? >> >> I have such a formula for 17, 31, 41, 61, 67, ...this type >> of numbers include prime euler numbers and (with some arrangement with >> the denominator) >> prime bernoulli numbers as well. >> >> Another direction with that idea , a fact known to Ramanujan was >> that (for example) >> >> 24 * sum(n^673/(exp(2*Pi*n)-1),n=1..infinity) = 156344681616153723364... >> 465357092320036059151 >> a prime of 1077 digits, that number is in fact close to a Bernoulli >> number in disguise, B(674) >> we can express that infinite series with the polygamma function if we >> like instead. >> >> >> also a fact which I find quite interesting, we have here an expression >> for some primes which is made with a sum of irrational numbers. >> >> other representations are with binomial sums which are simplified if >> we use >> the polygamma function with a negative argument, >> , integer relations with known constants. >> >> you may look at this document for more formulas here : >> >> http://plouffe.fr/The%20many%20faces%20of%20the%20polygamma%20function.pdf >> >> one interesting finding is that I have a polygamma expression for >> Li1(1/2), Li2(1/2), Li3(1/2), >> which are <the same> in term of polygamma function, i could not find yet >> an expression >> for Li4(1/2), but I am working on it since the first 3 values are very >> similar. >> see the formulas on page 3. >> >> all formulas are with that same polygamma function, and yes : one last >> point, >> the one I use is the one implemented in Maple and not mathematica mainly >> because >> the mathematica program uses the wrong (in my opinion) generalization of >> polygamma with >> negative arguments, the correct one is the one of Espinosa and Moll. >> (see the references). >> >> bonne lecture, >> >> Simon plouffe >>

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Re: [math-fun] Holding much water
by Warren D Smith 05 Apr '16

05 Apr '16

> HB: > I'm less worried about atmospheric percentages than I am about > absolute amounts. Venus's atmosphere is so thick that it drowns > a lot of percentages. > > For example, the last time I did the calculation, Venus has > approximately the *same* *absolute* amount of nitrogen in its > atmosphere as does the Earth. --actually about Venus = 4 * Earth. > I'm not worried about helium & neon, as I doubt that they are > important for life -- either for or against -- unless you > consider the neon lights of sleazy bars & hotels to be > important for nightlife. > > So that leaves us with hydrogen. --Venus's hydrogen is gone. Hence no water, no ammonia, no chemistry involving H, no life. The He and Ne are not important for life, they only are important as evidence for the atmosphere stripping theory. And if you counted Earth's oceans as part of its "atmosphere," then the comparison with Venus would be far more shocking about H. The fact Venus has more He than Earth could be considered as the only evidence AGAINST the stripping theory. So it needs to be explained somehow by proponents of that theory: 1. venus has more alpha emitting rocks? 2. venus gets recharged with He from solar wind? 3. earth's helium is trapped in rocks but baked out of venus rocks? Wikipedia: "Lighter gases, including water vapour, are continuously blown away [from Venus] by the solar wind through the induced magnetotail.[4] It is speculated that the atmosphere of Venus up to around 4 billion years ago was more like that of the Earth with liquid water on the surface. A runaway greenhouse effect may have been caused by the evaporation of the surface water and subsequent rise of the levels of other greenhouse gases.[8][9]" Their ref 4 hopefully provided evidence. If so, then HB's disputation of atmosphere-strip theory, is busted by observational evidence. > interesting to know the absolute amounts of hydrogen on Venus -- > including whatever hydrogen is bound up in other compounds -- > not necessarily water. --diddlysquat. Venus is extremely dry and has about 10^(-8) the absolute amount of H on Earth. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] about the polygamma function
by Simon Plouffe 05 Apr '16

05 Apr '16

Hello, the Psi function is one of my favorite function, it has so many aspects and formulas, very interesting, for example, you can express primes with it, 691 = Psi(11,1/2)/(Pi^12*2^8). or Euler numbers, (absolute value), I hate that - sign all over. ;-). E(2n) = -1/2*(Psi(k,1/8)-Psi(k,3/8)+Psi(k,5/8)-Psi(k,7/8))*Pi^(-k-1)*4^(-k). we can also use Psi(n,1/4) and Psi(n,3/4) for the same. another example is 67 = 1/64*(Psi(5,1/5)-Psi(5,2/5)-Psi(5,3/5)+Psi(5,4/5))*5^(1/2)/Pi^6 a very good question is : are there other primes like that ? I have such a formula for 17, 31, 41, 61, 67, ...this type of numbers include prime euler numbers and (with some arrangement with the denominator) prime bernoulli numbers as well. Another direction with that idea , a fact known to Ramanujan was that (for example) 24 * sum(n^673/(exp(2*Pi*n)-1),n=1..infinity) = 156344681616153723364... 465357092320036059151 a prime of 1077 digits, that number is in fact close to a Bernoulli number in disguise, B(674) we can express that infinite series with the polygamma function if we like instead. also a fact which I find quite interesting, we have here an expression for some primes which is made with a sum of irrational numbers. other representations are with binomial sums which are simplified if we use the polygamma function with a negative argument, , integer relations with known constants. you may look at this document for more formulas here : http://plouffe.fr/The%20many%20faces%20of%20the%20polygamma%20function.pdf one interesting finding is that I have a polygamma expression for Li1(1/2), Li2(1/2), Li3(1/2), which are <the same> in term of polygamma function, i could not find yet an expression for Li4(1/2), but I am working on it since the first 3 values are very similar. see the formulas on page 3. all formulas are with that same polygamma function, and yes : one last point, the one I use is the one implemented in Maple and not mathematica mainly because the mathematica program uses the wrong (in my opinion) generalization of polygamma with negative arguments, the correct one is the one of Espinosa and Moll. (see the references). bonne lecture, Simon plouffe

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