November 2013 - mathfun@mailman.xmission.com

[math-fun] Eccentricity of hyperbolic path of deflected light
by Mike Stay 30 Nov '13

30 Nov '13

I'm trying to follow the derivation in http://arxiv.org/pdf/physics/0508030v4.pdf, but I have no idea where equation 5 comes from: ------ For a light particle with mass m, total energy E and angular momentum L (with respect to the Sun's center) and where the Sun has mass M, the eccentricity epsilon is epsilon = sqrt(1 + (2 E L^2)/(G^2 m^3 M^2)) (5) ------ I could probably get some of that by dimensional analysis, but certainly not all of it. How would one derive this? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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[math-fun] Group theory question
by Eugene Salamin 30 Nov '13

30 Nov '13

Is the following a theorem? Let G be a group with center Z. If G/Z is abelian, then G is abelian. -- Gene

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[math-fun] GCD[1,π]
by Bill Gosper 30 Nov '13

30 Nov '13

On 11/29/2013 3:43 PM, Bill Gosper wrote: Hi Richard, had you intended to exclude math-fun? RJFateman>No, just not sure I had "write permission"... feel free to forward. Anyway, Mathematica has the separate function PolynomialGCD. Macsyma's gcd (and originally, factor) were semantically overloaded. I can't imagine why Integrate would take the numerator and denominator of ComplexInfinity, nor any harm that could follow taking them to be 1 and 0. This I agree with, except that the cost associated with Numerator[] will be the time for user-code testing whether the argument is ComplexInfinity. I have at various times proposed that the CL rational numbers or those in Maxima be extended to include 1/0 and -1/0 and even 0/0. I may have suggested it to Wolfram people. In some ways it makes the rational arithmetic programs shorter .. just do the arithmetic and it (sometimes) comes out. e.g. 1/0 * 1/0 = 1/0. 1/2 + 1/0 = (1*0+1*2)/(2*0) = 2/0. reduce to lowest terms given gcd(2,0) is 2, gives 1/0 I think that you only have to check if the answer comes out 0/0, in which case, see if you can you do better. e.g. 1/0+1/0 comes out 0/0 so you go back and do some checking to come up with 1/0. 1/0 + -1/0 comes out 0/0, and that is probably what you want. rwg>Here's an actual piece of Lisp code doing just that: ;Closest rational to x with numerator ≤ nhi (defun bstrat (x nhi &optional (c 0) (d 1)) (multiple-value-bind (f y) (floor x) (cond ((or (> f (* 2 nhi)) (and (= f (* 2 nhi)) (>= (* c y) d))) #.(SI::make-canonical-rational 1 0)) ((>= f nhi) nhi) ((= 0 (setf r (round (* nhi y) x))) f) ((+ f (/ (bstrat (/ y) (floor (- nhi r) f) (+ d (* f c)) c))))))) >gwr rjf>But I haven't thought about this for a while. Returning to your theme.. I think it does rely on gcd(0,n) coming out n for integer n. :) RJF PS feel free to forward this too. rwg>Holy cow, I totally forgot about nouuo. On Fri, Nov 29, 2013 at 2:52 PM, Richard Fateman <fateman(a)berkeley.edu>wrote: > An alternative view is that Pi (etc), is a symbol, and also a polynomial > in one variable, i.e. 1*Pi^1+0*Pi^0. > > Under those circumstances, GCD[1,Pi] would be 1. > > Given that the reals are not a unique factorization domain, the GCD is > unspecified > mathematically, so programmatically you can do something like analytically > continue GCD any old way that works when you wander onto some UFD, and > nothing expect that otherwise works will be broken. At least that seemed > to > be the Macsyma philosophy. > > Rationals are not a UFD either, but there is some consensus here. > > Both Mathematica and Macsyma/ Maxima say that gcd(1/6,1/9) is 1/18. > > Introducing Sqrt[2] is a somewhat different matter, which I'd rather not > write a dissertation about, here. > > I agree that Mma is accommodating at least superficially. However, > redefining > Numerator etc may have consequences in runtime. At least assuming that > the internal routines that call Numerator now use the redefined version, > what > happens to, say, Integrate[]? > > was it (setq nouuo t) in maclisp?? > > RJF >

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Re: [math-fun] GCD[1,π]
by Bill Gosper 30 Nov '13

30 Nov '13

On 2013-11-29 15:42, Dan Asimov wrote: > Hmm, this apparently presumes that the input fractions are reduced. > > 'Cause if say 2 divides both a and b, and we input (a/2)/(b/2) instead of a/b, > this would output (a/2 + c + e)/(b/2 + d + f) -- not the same thing. > > --Dan > > > On 2013-11-29, at 11:44 AM, Bill Gosper wrote: > >> In[904]:= mediant[a/b, c/d, e/f] >> >> Out[904]= (a + c + e)/(b + d + f) Indeed. Maybe this is why Mathematica doesn't provide Mediant. In fact, when the kids and I roll our own, we usually include optional weight arguments by which to UNreduce the input fractions. --rwg

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[math-fun] Heavy ice D2O sinks in light water H2O
by Eugene Salamin 29 Nov '13

29 Nov '13

A while ago we had a discussion about whether heavy ice floats or sinks in light water. It sinks. I found a picture and tried to post it, but it exceeded the 40 KB limit for math-fun postings. It's on page 6 of this document. http://mragheb.com/NPRE%20402%20ME%20405%20Nuclear%20Power%20Engineering/Is… Antinuclear people should don their tinfoil hat before viewing. -- Gene

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[math-fun] (no subject)
by Bill Gosper 28 Nov '13

28 Nov '13

Bill Gosper wrote: If your developers are leery of GCD[1,π]→0, I think I can muster authoritative corroborators, if not corroborative literature. Motivation: The gcd of two real quantities is the largest quantity that goes into each a whole number of times. [me:] I'm unconvinced. In this context (multiplicative rather than additive) 0 is *larger* than everything else, not *smaller*, no? [Bill:] Well, we *are* looking for the "greatest".-) Do you propose a different answer, or deny there is one? >I deny there is one, or at least that there is any number >that has a good claim. Please name one. >>> And 0 emphatically doesn't "go into" anything "a whole >>> number of times". >> Touché. New wording: >> GCD(a,b):= the limit of the Euclidean process of iteratively >> subtracting the smaller from the larger. >The *limit* of doing this for commensurable quantities is zero. >The gcd is the last thing you get immediately before 0. n[755]:= NestList[ Simplify[{Max[#] - Min[#], Min[#]}] &, {GoldenRatio - 1/2, 1/√5}, 6] Out[755]= {{-(1/2) + GoldenRatio, 1/√5}, {3/(2 √5), 1/√5}, {1/(2 √5), 1/√5}, {1/(2 √5), 1/(2 √5)}, {0, 1/(2 √5)}, {1/(2 √5), 0}, {1/(2 √5), 0}} So GCD[GoldenRatio - 1/2, 1/√5] is 1/(2√5) In[758]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {69, 105}, 16] Out[758]= {{69, 105}, {36, 69}, {33, 36}, {3, 33}, {30, 3}, {27, 3}, {24, 3}, {21, 3}, {18, 3}, {15, 3}, {12, 3}, {9, 3}, {6, 3}, {3, 3}, {0, 3}, {3, 0}, {3, 0}} So GCD[69,105] is 3. In[761]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {1, π}, 3 + 7 + 15 + 1] Out[761]= {{1, π}, {-1 + π, 1}, {-2 + π, 1}, {-3 + π, 1}, {4 - π, -3 + π}, {7 - 2 π, -3 + π}, {10 - 3 π, -3 + π}, {13 - 4 π, -3 + π}, {16 - 5 π, -3 + π}, {19 - 6 π, -3 + π}, {22 - 7 π, -3 + π}, {-25 + 8 π, 22 - 7 π}, {-47 + 15 π, 22 - 7 π}, {-69 + 22 π, 22 - 7 π}, {-91 + 29 π, 22 - 7 π}, {-113 + 36 π, 22 - 7 π}, {-135 + 43 π, 22 - 7 π}, {-157 + 50 π, 22 - 7 π}, {-179 + 57 π, 22 - 7 π}, {-201 + 64 π, 22 - 7 π}, {-223 + 71 π, 22 - 7 π}, {-245 + 78 π, 22 - 7 π}, {-267 + 85 π, 22 - 7 π}, {-289 + 92 π, 22 - 7 π}, {-311 + 99 π, 22 - 7 π}, {-333 + 106 π, 22 - 7 π}, {355 - 113 π, -333 + 106 π}} In[762]:= N[%] Out[762]= {{1., 3.14159}, {2.14159, 1.}, {1.14159, 1.}, {0.141593, 1.}, {0.858407, 0.141593}, {0.716815, 0.141593}, {0.575222, 0.141593}, {0.433629, 0.141593}, {0.292037, 0.141593}, {0.150444, 0.141593}, {0.00885142, 0.141593}, {0.132741, 0.00885142}, {0.12389, 0.00885142}, {0.115038, 0.00885142}, {0.106187, 0.00885142}, {0.0973355, 0.00885142}, {0.0884841, 0.00885142}, {0.0796327, 0.00885142}, {0.0707813, 0.00885142}, {0.0619298, 0.00885142}, {0.0530784, 0.00885142}, {0.044227, 0.00885142}, {0.0353756, 0.00885142}, {0.0265241, 0.00885142}, {0.0176727, 0.00885142}, {0.00882128, 0.00885142}, {0.0000301444, 0.00882128}} So GCD[1, π] is going to 0. Gareth>I agree that gcd(1,pi) can be thought of as a certain sort of >limit of real numbers that tend to 0 -- but I don't think the >usual sort of limit is the right one, because the metric it >implicitly invokes isn't the right one. -- > g What's the right one? The smallest square you can make by dividing a rectangle into squares by the greedy algorithm is GCD(length,width). What could be more natural? --rwg

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[math-fun] GCD(1,π)
by Eugene Salamin 28 Nov '13

28 Nov '13

Consider GCD ring theoretically. If a and b are elements of a ring R (which really should be an integral domain), GCD(a,b) is the smallest ideal in R containing a and b, and is the set of elements of the form ax + by as x and y range over R. An ideal containing 1 can only be the entire ring. If the ring is a field, then the only ideals are {0} and the entire field. GCD(a,b) depends on the containing ring R. If we take R to be the real numbers, then GCD(1,π) is R, which is also the ideal (a) for any nonzero a. So one can let GCD(1,π) be any nonzero real number. -- Gene

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Re: [math-fun] Fwd: [CASE:423927] GCD suggestions
by Bill Gosper 28 Nov '13

28 Nov '13

GarethM> On 27/11/2013 21:30, Bill Gosper wrote: If your developers are leery of GCD[1,π]→0, I think I can muster authoritative corroborators, if not corroborative literature. Motivation: The gcd of two real quantities is the largest quantity that goes into each a whole number of times. I'm unconvinced. In this context (multiplicative rather than additive) 0 is *larger* than everything else, not *smaller*, no? Well, we *are* looking for the "greatest".-) Do you propose a different answer, or deny there is one? And 0 emphatically doesn't "go into" anything "a whole number of times". -- g Touché. New wording: GCD(a,b):= the limit of the Euclidean process of iteratively subtracting the smaller from the larger. --rwg

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Re: [math-fun] "greatest", "least"
by James Propp 28 Nov '13

28 Nov '13

While we're discussing such matters, perhaps we should consider whether the "lowest common denominator" of two or more fractions has been properly defined. Most math teachers will tell you that the lowest common denominator of the fractions 5/12 and 11/18 is 36. But what about -36, which is lower? Or -72, which is lower still? Clearly the lowest common denominator of any set of fractions is minus infinity. And don't tell me that -72 is "higher" than -36, citing Joe Silverman or some other deluded arithmetician; any Alaskan meteorologist can tell you that -72 is much lower than -36. Jim Propp On Thursday, November 28, 2013, Bill Gosper <billgosper(a)gmail.com> wrote: > Bill Gosper wrote: > > If your developers are leery of GCD[1,π]→0, I think I can muster > authoritative corroborators, if not corroborative literature. > Motivation: The gcd of two real quantities is the largest quantity > that goes into each a whole number of times. > > > [me:] > > I'm unconvinced. In this context (multiplicative rather than > additive) 0 is *larger* than everything else, not *smaller*, > no? > > > [Bill:] > > Well, we *are* looking for the "greatest".-) Do you propose a different > answer, or deny there is one? > > >>I deny there is one, or at least that there is any number >>that has a good claim. > > Please name one. > >>>> And 0 emphatically doesn't "go into" anything "a whole >>>> number of times". > >>> Touché. New wording: >>> GCD(a,b):= the limit of the Euclidean process of iteratively >>> subtracting the smaller from the larger. > >>The *limit* of doing this for commensurable quantities is zero. >>The gcd is the last thing you get immediately before 0. > > n[755]:= NestList[ > Simplify[{Max[#] - Min[#], Min[#]}] &, {GoldenRatio - 1/2, 1/√5}, 6] > > Out[755]= {{-(1/2) + GoldenRatio, 1/√5}, {3/(2 √5), 1/√5}, {1/(2 √5), 1/√5}, > {1/(2 √5), 1/(2 √5)}, {0, 1/(2 √5)}, {1/(2 √5), 0}, {1/(2 √5), 0}} > > So GCD[GoldenRatio - 1/2, 1/√5] is 1/(2√5) > > In[758]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {69, 105}, 16] > > Out[758]= {{69, 105}, {36, 69}, {33, 36}, {3, 33}, {30, 3}, {27, > 3}, {24, 3}, {21, 3}, {18, 3}, {15, 3}, {12, 3}, {9, 3}, {6, 3}, {3, > 3}, {0, 3}, {3, 0}, {3, 0}} > > So GCD[69,105] is 3. > > > In[761]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {1, π}, > 3 + 7 + 15 + 1] > > Out[761]= {{1, π}, {-1 + π, 1}, {-2 + π, 1}, {-3 + π, 1}, {4 - π, -3 + π}, > {7 - 2 π, -3 + π}, {10 - 3 π, -3 + π}, {13 - 4 π, -3 + π}, > {16 - 5 π, -3 + π}, {19 - 6 π, -3 + π}, {22 - 7 π, -3 + π}, > {-25 + 8 π, 22 - 7 π}, {-47 + 15 π, 22 - 7 π}, {-69 + 22 π, 22 - 7 π}, > {-91 + 29 π, 22 - 7 π}, {-113 + 36 π, 22 - 7 π}, {-135 + 43 π, 22 - 7 π}, > {-157 + 50 π, 22 - 7 π}, {-179 + 57 π, 22 - 7 π}, {-201 + 64 π, 22 - 7 π}, > {-223 + 71 π, 22 - 7 π}, {-245 + 78 π, 22 - 7 π}, {-267 + 85 π, 22 - 7 π}, > {-289 + 92 π, 22 - 7 π}, {-311 + 99 π, 22 - 7 π}, {-333 + 106 π, 22 - 7 π}, > {355 - 113 π, -333 + 106 π}} > > In[762]:= N[%] > > Out[762]= {{1., 3.14159}, {2.14159, 1.}, {1.14159, 1.}, {0.141593, > 1.}, {0.858407, 0.141593}, {0.716815, 0.141593}, {0.575222, > 0.141593}, {0.433629, 0.141593}, {0.292037, 0.141593}, {0.150444, > 0.141593}, {0.00885142, 0.141593}, {0.132741, 0.00885142}, {0.12389, > 0.00885142}, {0.115038, 0.00885142}, {0.106187, > 0.00885142}, {0.0973355, 0.00885142}, {0.0884841, > 0.00885142}, {0.0796327, 0.00885142}, {0.0707813, > 0.00885142}, {0.0619298, 0.00885142}, {0.0530784, > 0.00885142}, {0.044227, 0.00885142}, {0.0353756, > 0.00885142}, {0.0265241, 0.00885142}, {0.0176727, > 0.00885142}, {0.00882128, 0.00885142}, {0.0000301444, 0.00882128}} > > So GCD[1, π] is going to 0. > > > Gareth>I agree that gcd(1,pi) can be thought of as a certain sort of >>limit of real numbers that tend to 0 -- but I don't think the >>usual sort of limit is the right one, because the metric it >>implicitly invokes isn't the right one. > -- >> > g > > What's the right one? > > The smallest square you can make by dividing a rectangle into > > squares by the greedy algorithm is GCD(length,width). What could > > be more natural? > > --rwg > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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Re: [math-fun] "greatest", "least"
by Adam P. Goucher 28 Nov '13

28 Nov '13

The greatest common divisor is *not* the *largest* element d such that d|a and d|b. (That is meaningless in unordered Euclidean domains.) Instead, it is the unique element (up to multiplication by units) d such that d|a and d|b, and for any c such that c|a and c|b, we also have c|d. Likewise for lowest common multiple. So, in the integers, gcd(12,18) is 6 or -6, which are considered to be the same thing for this purpose (since they are associates). Similarly, in the Gaussians, gcd(5, 2+i) = 2+i, 2i-1, -2-i or -2i+1, which are associates (differ by multiplication by a unit). Sincerely, Adam P. Goucher > ----- Original Message ----- > From: James Propp > Sent: 11/28/13 02:49 PM > To: math-fun > Subject: Re: [math-fun] "greatest", "least" > > While we're discussing such matters, perhaps we should consider whether the > "lowest common denominator" of two or more fractions has been properly > defined. > > Most math teachers will tell you that the lowest common denominator of the > fractions 5/12 and 11/18 is 36. But what about -36, which is lower? Or > -72, which is lower still? > > Clearly the lowest common denominator of any set of fractions is minus > infinity. > > And don't tell me that -72 is "higher" than -36, citing Joe Silverman or > some other deluded arithmetician; any Alaskan meteorologist can tell you > that -72 is much lower than -36. > > Jim Propp > > On Thursday, November 28, 2013, Bill Gosper <billgosper(a)gmail.com> wrote: > > Bill Gosper wrote: > > > > If your developers are leery of GCD[1,π]→0, I think I can muster > > authoritative corroborators, if not corroborative literature. > > Motivation: The gcd of two real quantities is the largest quantity > > that goes into each a whole number of times. > > > > > > [me:] > > > > I'm unconvinced. In this context (multiplicative rather than > > additive) 0 is *larger* than everything else, not *smaller*, > > no? > > > > > > [Bill:] > > > > Well, we *are* looking for the "greatest".-) Do you propose a different > > answer, or deny there is one? > > > > > >>I deny there is one, or at least that there is any number > >>that has a good claim. > > > > Please name one. > > > >>>> And 0 emphatically doesn't "go into" anything "a whole > >>>> number of times". > > > >>> Touché. New wording: > >>> GCD(a,b):= the limit of the Euclidean process of iteratively > >>> subtracting the smaller from the larger. > > > >>The *limit* of doing this for commensurable quantities is zero. > >>The gcd is the last thing you get immediately before 0. > > > > n[755]:= NestList[ > > Simplify[{Max[#] - Min[#], Min[#]}] &, {GoldenRatio - 1/2, 1/√5}, 6] > > > > Out[755]= {{-(1/2) + GoldenRatio, 1/√5}, {3/(2 √5), 1/√5}, {1/(2 √5), > 1/√5}, > > {1/(2 √5), 1/(2 √5)}, {0, 1/(2 √5)}, {1/(2 √5), 0}, {1/(2 √5), 0}} > > > > So GCD[GoldenRatio - 1/2, 1/√5] is 1/(2√5) > > > > In[758]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {69, 105}, 16] > > > > Out[758]= {{69, 105}, {36, 69}, {33, 36}, {3, 33}, {30, 3}, {27, > > 3}, {24, 3}, {21, 3}, {18, 3}, {15, 3}, {12, 3}, {9, 3}, {6, 3}, {3, > > 3}, {0, 3}, {3, 0}, {3, 0}} > > > > So GCD[69,105] is 3. > > > > > > In[761]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {1, π}, > > 3 + 7 + 15 + 1] > > > > Out[761]= {{1, π}, {-1 + π, 1}, {-2 + π, 1}, {-3 + π, 1}, {4 - π, -3 + π}, > > {7 - 2 π, -3 + π}, {10 - 3 π, -3 + π}, {13 - 4 π, -3 + π}, > > {16 - 5 π, -3 + π}, {19 - 6 π, -3 + π}, {22 - 7 π, -3 + π}, > > {-25 + 8 π, 22 - 7 π}, {-47 + 15 π, 22 - 7 π}, {-69 + 22 π, 22 - 7 π}, > > {-91 + 29 π, 22 - 7 π}, {-113 + 36 π, 22 - 7 π}, {-135 + 43 π, 22 - 7 π}, > > {-157 + 50 π, 22 - 7 π}, {-179 + 57 π, 22 - 7 π}, {-201 + 64 π, 22 - 7 > π}, > > {-223 + 71 π, 22 - 7 π}, {-245 + 78 π, 22 - 7 π}, {-267 + 85 π, 22 - 7 > π}, > > {-289 + 92 π, 22 - 7 π}, {-311 + 99 π, 22 - 7 π}, {-333 + 106 π, 22 - 7 > π}, > > {355 - 113 π, -333 + 106 π}} > > > > In[762]:= N[%] > > > > Out[762]= {{1., 3.14159}, {2.14159, 1.}, {1.14159, 1.}, {0.141593, > > 1.}, {0.858407, 0.141593}, {0.716815, 0.141593}, {0.575222, > > 0.141593}, {0.433629, 0.141593}, {0.292037, 0.141593}, {0.150444, > > 0.141593}, {0.00885142, 0.141593}, {0.132741, 0.00885142}, {0.12389, > > 0.00885142}, {0.115038, 0.00885142}, {0.106187, > > 0.00885142}, {0.0973355, 0.00885142}, {0.0884841, > > 0.00885142}, {0.0796327, 0.00885142}, {0.0707813, > > 0.00885142}, {0.0619298, 0.00885142}, {0.0530784, > > 0.00885142}, {0.044227, 0.00885142}, {0.0353756, > > 0.00885142}, {0.0265241, 0.00885142}, {0.0176727, > > 0.00885142}, {0.00882128, 0.00885142}, {0.0000301444, 0.00882128}} > > > > So GCD[1, π] is going to 0. > > > > > > Gareth>I agree that gcd(1,pi) can be thought of as a certain sort of > >>limit of real numbers that tend to 0 -- but I don't think the > >>usual sort of limit is the right one, because the metric it > >>implicitly invokes isn't the right one. > > -- > >> > > g > > > > What's the right one? > > > > The smallest square you can make by dividing a rectangle into > > > > squares by the greedy algorithm is GCD(length,width). What could > > > > be more natural? > > > > --rwg > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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