March 2013 - mathfun@mailman.xmission.com

Re: [math-fun] speed-of-light-may-not-be-constant?
by Adam P. Goucher 28 Mar '13

28 Mar '13

The speed of light is actually constant. Sincerely, Adam P. Goucher http://cp4space.wordpress.com

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Re: [math-fun] Volume-preserving vector field paradox: Spoiler
by Andy Latto 27 Mar '13

27 Mar '13

On Tue, Mar 26, 2013 at 3:46 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > This question is directed to everyone whose initials are not E.S. or V.E. > > Assume any vector fields or surface mentioned is at least C^2. Or even real analytic. > The symbol == means "equals for all values of the domain". Since V is discontinuous at the origin, we'll have to work in some domain that doesn't include the origin. For example, we could use the region epsilon < r < 1. > > Under Divergenceless Field[1] at MathWorld, the first sentence asserts that > > (*) div(V) == 0 implies that there exists a vector field W such that V == curl(W). > > But their entry for Curl Theorem[2] states that the flux of a vector field of the form curl(W) > through a surface S is equal to the line integral of W around its boundary bd(S). Well, if we're working in 3 dimensions, it's a surface integral, not a line integral. You mean > > In particular, this implies: > (**) The flux of curl(W) through a closed surface must be 0, for any vector field W. > > Now consider the vector field V given by V(x,y,z) = (x,y,z) / (x^2 + y^2 + z^2)^(3/2), (x,y,z) unequal to (0,0,0). > > It's easy to check that div(V) == 0. > > By (*) there exists a W such that V == curl(W). > > But it's also easy to check that the flux of V through the unit sphere x^2 + y^2 + z^2 = 1 is 4pi. But the unit sphere is only one of the two components of the boundary of our region. No matter how small we make the other component of the boundary, the small sphere around the origin, the flux of V through that other component wil be -4pi. The divergence measures the "sourceness" of the vector field at a point, but V has a "point source" at the origin. Andy Latto But > This contradicts (**). > > PUZZLE: Explain how this paradox is possible. > > --Dan > _____________________________________________________________ > [1] < http://mathworld.wolfram.com/DivergencelessField.html > > [2] < http://mathworld.wolfram.com/CurlTheorem.html > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Andy.Latto(a)pobox.com

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[math-fun] Volume-preserving vector field paradox
by Dan Asimov 27 Mar '13

27 Mar '13

This question is directed to everyone whose initials are not E.S. or V.E. Assume any vector fields or surface mentioned is at least C^2. Or even real analytic. The symbol == means "equals for all values of the domain". Under Divergenceless Field[1] at MathWorld, the first sentence asserts that (*) div(V) == 0 implies that there exists a vector field W such that V == curl(W). But their entry for Curl Theorem[2] states that the flux of a vector field of the form curl(W) through a surface S is equal to the line integral of W around its boundary bd(S). In particular, this implies: (**) The flux of curl(W) through a closed surface must be 0, for any vector field W. Now consider the vector field V given by V(x,y,z) = (x,y,z) / (x^2 + y^2 + z^2)^(3/2), (x,y,z) unequal to (0,0,0). It's easy to check that div(V) == 0. By (*) there exists a W such that V == curl(W). But it's also easy to check that the flux of V through the unit sphere x^2 + y^2 + z^2 = 1 is 4pi. This contradicts (**). PUZZLE: Explain how this paradox is possible. --Dan _____________________________________________________________ [1] < http://mathworld.wolfram.com/DivergencelessField.html > [2] < http://mathworld.wolfram.com/CurlTheorem.html >

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[math-fun] "A superdiagonal, ~ again." -- Paul Erdos. (anagram)
by ed pegg 27 Mar '13

27 Mar '13

Paul would be 100 today. "A superdiagonal, ~ again." -- Paul Erdos. I'm putting words into Paul's mouth, but the diagonal argument is a very common proof method, and I've used it many times. I'm sure Paul did an order of magnitude more proofs than I did. Also... he gave a problem to Alex Soifer, and I solved it with a superdiagonal argument. Anyways, does anyone thing Paul might have said this at some point? Ed Pegg Jr

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[math-fun] speed of light may not be constant?
by Warren D Smith 26 Mar '13

26 Mar '13

http://blogs.voanews.com/science-world/2013/03/26/speed-of-light-may-not-be… described latest science fad claim that light's speed may fluctuate in space and time. Does this contradict the observed fact that light from halfway across the universe, hitting the left side of your telescope mirror, is still perfectly in phase with light hitting right side of your telescope mirror, allowing interferometry?

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[math-fun] speed-of-light-may-not-be-constant?
by Warren D Smith 26 Mar '13

26 Mar '13

http://blogs.voanews.com/science-world/2013/03/26/speed-of-light-may-not-be… Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Baker GCD question
by Henry Baker 26 Mar '13

26 Mar '13

I just found this very cool GCD algorithm by Stehle & Zimmerman; it is built on a *** backwards *** division algorithm (!), more reminiscent of something you might want to do with 2-adic ('backwards') integers. Divide (ahi,alo) by b to get q, r. But instead of getting (ahi,alo) - q*b to get r<b, we want q', r', s.t. (ahi,alo) - q'*b = (r',0), where r'<b. (Here, '*' & '-' have their usual meaning.) I.e., instead of q killing off the _high order_ digit ahi, we us q' to kill off the _low order_ digit alo !! http://perso.ens-lyon.fr/damien.stehle/downloads/recbinary.pdf Stehle, Damien & Zimmerman, Paul. "A Binary Recursive GCD Algorithm". ~2004. So when calculating gcd(m,n) we can use a gcd on the _low order_ parts of m & n to kill off these low order bits, and then work on the high order part. Henry Baker At 04:58 PM 3/25/2013, Warren D Smith wrote: >> From: Henry Baker <hbaker1(a)pipeline.com> >> I'm looking for the _simplest_ binary _recursive_ GCD algorithm that has >> "pretty good" performance. >> >> I.e., a GCD algorithm that takes two length 2m integers and reduces this to >> a number of computations on length m integers. > >--Let the two input numbers be X and Y with X<Y. >X=(Xhi,Xlo) and Y=(Yhi,Ylo) for the top and bottom halves, >same lengths for all four. >We assume Xhi>0 otherwise you can immediately replace Y by Y mod X to >reduce to half precision. > >Algorithm idea: >Use extended GCD to find a,b and g so that a*Xhi+b*Yhi=g=gcd(Xhi,Yhi). >Replace Y by a*X+b*Y. >Replace X by X mod Y. >Replace Y by Y mod X. >swap(X,Y). >Now continue on. > >Seems to me, each iteration will reduce Yhi to about its square root (or less) >thus multiplying the bitlength by 0.75 (or less). > >I've cheated a bit by using extended GCD as a subroutine inside plain GCD. >However, you can add extra bells & whistles to the plain gcd algorithm to keep >track of what happens to the coeffs, thus converting it to an extended >GCD algorithm...

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[math-fun] By RIES equation-finder has new capabilities
by Robert Munafo 26 Mar '13

26 Mar '13

Over the past couple months I've been making my RIES program [1] a bit more useful. This is partly in response to people on this mailing list, which is why I'm sharing it here. For those new to RIES, it is a bit like the Inverse Symbolic Calculator, but optimized to report solutions of lowest Kolmogorov complexity [4]. If you want more specific, targeted representations, such as is provided by the various integer relation algorithms [5], you can use Mathematica or Wolfram Alpha. To give a simple example, when presented with the target number 2.506184, Wolfram Alpha reports [6] things like "e^(-4/e-e+pi) pi^(e-1)" and "csc(1/79 (23+3 pi))", while RIES finds x^x=10 and 2*(ln(8)+4)^(1/8). In addition to computing the value and the derivative with respect to x, RIES now "tags" all calculations throughout its recursive search. Tags include integer, rational, constructible [2], algebraic [3], "elementary" (by one definition that excludes the root of "x^x-7=0"), "transcendental" (meaning everything that RIES is capable of finding), and "unknown" (used for the variable x that stands for the user-specified target number). It plays it safe when in doubt, like assuming that anything raised to a non-integer rational power is irrational. But it tries a little bit, e.g. it knows that a constructible number raised to the power of 1/4 is also a constructible number. In addition, new options *--rational-exponents* and *--rational-trig-args*, and the existing *--trig-argument-scale* allow control over the discovery of implicit vs. explicit transcendental values. With these tags and the existing options to control which constants and functions are applied in a search, RIES now provides all the following: *-i* : the existing *--integer-subexpressions* option for integer-only searches *-r* or *--rational-subexpressions* : All reported equations must have rational roots *-c* or *--constructible-subexpressions* : All reported equations must have "constructible" (with straightedge and compass) roots [2]. *-a* or *--algebraic-subexpressions* : All reported equations must have algebraic roots (i.e. roots of a polynomial in one variable with rational coefficients) [3]. *-a* with *-Ox* : to limit it to algebraic roots that have a closed-form expression (with a reasonably low Kolmogorov complexity). *-Ox* : the existing option to allow roots to be any transcendental number expressible in closed form. *-l* or *--liouvillian-subexpressions* : A superset of -a allowing x=sqrt(2)^sqrt(2), but not allowing x^x=7. This is not a strict hierarchy: *-a* and *-Ox* are supersets of *-c*, but neither is a superset of the other, but *-l* is a superset of both. There is a table and more details at: http://mrob.com/pub/ries/index.html#classes The online search does not yet provide these options but that is planned. footnotes: [1] mrob.com/ries [2] http://en.wikipedia.org/wiki/Constructible_number [3] http://en.wikipedia.org/wiki/Algebraic_number [4] http://en.wikipedia.org/wiki/Kolmogorov_complexity [5] http://en.wikipedia.org/wiki/Integer_relation_algorithm [6] http://www.wolframalpha.com/input/?i=2.506184 -- 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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[math-fun] A cute number theory problem
by Victor Miller 26 Mar '13

26 Mar '13

I came across the following problem: Show that there are an infinite number of pairs (x,y) of positive integers such that 2^(x^2) + 2^(y^2) + 1 is a square. I would suspect that the answer that you'll come up with is the same as what I came up with, but I wondered if it's possible to characterize all (or all but a finite number) of the pairs which work. By this I mean to show that all (x,y) are in one of a finite set of specific sequences, with a possible finite set of exceptions. Victor

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[math-fun] Baker GCD question
by Warren D Smith 25 Mar '13

25 Mar '13

> From: Henry Baker <hbaker1(a)pipeline.com> > I'm looking for the _simplest_ binary _recursive_ GCD algorithm that has > "pretty good" performance. > > I.e., a GCD algorithm that takes two length 2m integers and reduces this to > a number of computations on length m integers. --Let the two input numbers be X and Y with X<Y. X=(Xhi,Xlo) and Y=(Yhi,Ylo) for the top and bottom halves, same lengths for all four. We assume Xhi>0 otherwise you can immediately replace Y by Y mod X to reduce to half precision. Algorithm idea: Use extended GCD to find a,b and g so that a*Xhi+b*Yhi=g=gcd(Xhi,Yhi). Replace Y by a*X+b*Y. Replace X by X mod Y. Replace Y by Y mod X. swap(X,Y). Now continue on. Seems to me, each iteration will reduce Yhi to about its square root (or less) thus multiplying the bitlength by 0.75 (or less). I've cheated a bit by using extended GCD as a subroutine inside plain GCD. However, you can add extra bells & whistles to the plain gcd algorithm to keep track of what happens to the coeffs, thus converting it to an extended GCD algorithm...

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