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[math-fun] patterns in numbers : vertical view
by Simon Plouffe 07 Jan '17

07 Jan '17

Hello everyone, I made one experiment with n*n*phi, phi being the golden ratio. I use a precision of let's say 1000 digits and n = 1 to 16000000. If we look at the square grid made of the numbers (without the decimal point), there are no clear patterns in the lines of digits BUT there is a definite pattern in vertical. for each k (from 1 to 1000) I constructed 1 image made out of the first 16 000 000 digits , making one image of 4000 x 4000. This is the result : http://plouffe.fr/phi_time_n_n%20vertical_base_10/index.html These digits are in base 10, so I used a colourization with blue tones : light blue = 0 , darker blue = 9. The thing is : there is a pattern for every k. Very definite pattern. I used that color because in gray scale it does not comes out clearly enough, the colourization. if we take another constant like pi, e, gamma or even a large rational, the same patterns appear, not surprising. Now, the good question is : ok there is a pattern , then what, what is the difference in those patterns. If n*n*n is used instead of n*n, the same but it does scramble after a few digits. Any explanation for such patterns ? best regards, Simon Plouffe

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[math-fun] 250+ digits of pi on NFL football jerseys suspended on a clothesline on Alcatraz Island
by Thane Plambeck 06 Jan '17

06 Jan '17

I took this video yesterday afternoon https://www.youtube.com/watch?v=HGXTxpkqR9A&feature=youtu.be -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] Fwd: The moving sofa problem
by Brent Meeker 05 Jan '17

05 Jan '17

More fun than moving it upstairs. Brent -------- Forwarded Message -------- https://www.math.ucdavis.edu/~romik/movingsofa/ <https://www.math.ucdavis.edu/%7Eromik/movingsofa/> Christine

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[math-fun] Fwd: The moving sofa problem
by Bill Gosper 04 Jan '17

04 Jan '17

Wow, I thought Gerver once claimed a proof. I'd like to see the analytic expressions for the 15 curved pieces. And maybe the whole area. Too bad it's not π/2 + 2/π (the expected value of a random 3D rotation). --rwg Fred's 3D remark is not just another idle case of Let's add a dimension. Who makes floor-to-ceiling cylindrical furniture? Or does Ikea have a branch in Flatland? Date: 2017-01-04 11:50 From: Brent Meeker <meekerdb(a)verizon.net> More fun than moving it upstairs. Brent -------- Forwarded Message -------- https://www.math.ucdavis.edu/~romik/movingsofa/ < https://www.math.ucdavis.edu/%7Eromik/movingsofa/> Christine

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[math-fun] Mercury's locked rotation & orbit
by Henry Baker 03 Jan '17

03 Jan '17

When I was in high school, I was taught that Mercury's orbit and rotation were locked 1:1. I recently discovered (to my chagrin) that NASA's knowledge had improved, but mine had not; Mercury actually rotates 1.5x faster that it's orbital period -- i.e., two full elliptical orbits = 3 full rotations. Here's a link to a .gif animation of Mercury's rotation & orbit. https://commons.wikimedia.org/wiki/File:Orbital_resonance_of_Mercury.gif *I'm not 100% convinced that this particular gif animation is correct w.r.t. timing; it doesn't seem to follow Kepler's Second (Equal Area) Law, whereby equal areas are swept out in equal times, and thus Mercury should actually be moving faster when closer to the Sun.* https://en.wikipedia.org/wiki/Mercury_%28planet%29 Astronomers call Mercury's orbit/rotation a "resonance", but I don't understand why. The whole point of a 1:1 orbit/rotation locking is to eliminate the tidal effects on the planet, but any other ratio still produces tides. I'm not sure what effect is causing a "lock" to the 3:2 ratio; perhaps it is related to the excessive eccentricity of Mercury's orbit? In any case, when I was in high school, we talked about the "dark side of Mercury", still thinking that it had a 1:1 ratio -- like that of the Earth's Moon. In such a case, the "dark side" should be quite cold -- particularly since Mercury has essentially no atmosphere to transfer energy from the "bright side" to the "dark side". I would imagine that the temperature of the "dark side" in such a case would be determined by an equilibrium between the temperature of deep space and the *mean* temperature of Mercury (averaged over the entire body), because in that case, Mercury would be *conducting* heat from the bright side to the dark side solely by conducting the heat through the interior of the planet's body. However, there is actually no "dark side" of Mercury, which is a pity, because it might be an interesting place for various scientific measurements. I do hope that no science fiction stories were harmed in the making of the discovery that Mercury was not in 1:1 resonance!

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[math-fun] Orbital period determined by major axis
by Henry Baker 02 Jan '17

02 Jan '17

What is the shortest/simplest/most elegant proof that the orbital period of the 2-body problem depends only on the length of the major axis, and NOT on the eccentricity of the ellipse? I'm looking for a Feynmann style proof; perhaps Feynmann himself came up with an elegant proof? I've read "Feynmann's lost lecture", in which Feynmann reproduces Maxwell's elegant proof that Kepler's Laws are equivalent to the inverse square law and elliptical orbits, but I don't recall that Feynmann addressed this particular question by itself. In any case, this fact means that one can have simultaneous bodies orbiting with the same period in the same plane, which should be able to produce some pretty animations. For example, one might color a circular orbit with the colors from a color wheel, and then deform that circle into an ellipse of the same major axis.

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Re: [math-fun] Two little sextics,
by Dan Asimov 01 Jan '17

01 Jan '17

Yes, Happy New Year to Warut, Zak, and everyone else on math-fun! —Dan Zak Seidov wrote: >Happy New Year 2017 >and Happy all next years >to all math-fun subscribers. > >>Суббота, 31 декабря 2016, 20:38 +03:00 от Warut Roonguthai <warut822(a)gmail.com>: >> >>Happy New Year 2017 to all math-fun subscribers. >>Warut

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[math-fun] Two little sextics,
by Bill Gosper 31 Dec '16

31 Dec '16

Date: 2016-12-31 09:38 From: Warut Roonguthai <warut822(a)gmail.com> >I guess not. So does ANYBODY (Save http://community.wolfram.com/groups/-/m/t/988309)? >Happy New Year 2017 to all math-fun subscribers. >Warut Well, emperor PENGUINS, at least, have no clothes. But as I say in the Wolfram post, Galois Theory seems not to be doing the job. Though Julian says it's the only hope for proving that Corey's kludge actually works. I guess the establishment line would be a) Galois theory is hard (dare I say simply hard?), and b) Denesting is a kludgy problem. --rwg On Sat, Dec 31, 2016 at 5:37 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Whoa. Can PARI (or GAP) reliably denest radicals? > http://community.wolfram.com/groups/-/m/t/988309 > Maple, Mathematica, and (presumably) Maxima cannot. --rwg > > Date: 2016-12-30 02:19 > From: Warut Roonguthai <warut822(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > The Galois group of x^6-3*x+3 told me that it had a quadratic subfield (but > no cubic subfield). The form of the exact value of x told me which > quadratic field. (Otherwise, some more guessing would be needed, or I had > to solve it the hard way.) I computed the Galois group and factored over > Q(sqrt(-3)) with PARI. > > Warut > > On Fri, Dec 30, 2016 at 3:57 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > > > Date: 2016-12-29 22:20 > > From: Warut Roonguthai <warut822(a)gmail.com> > > To: math-fun <math-fun(a)mailman.xmission.com> > > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > > > 4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2 > > tells that x^6-3*x+3 can be factored into two cubic over Q(sqrt(-3)). > > > > BTW, 822 is my lucky number but doesn't seem to work very well so far. > > Hahaha > > Warut > > > > How could it not be at least doubly informative? > > > > DAWK! I plead senility. Ok, how did you find those squares? I see how > to > > go the other way--i.e. from a factorization over √n to f = p² - n q². > > E.g., if f = the duodecic for x2 in > > http://mathworld.wolfram.com/SquareDissection.html , > > guessing Factor[f,Extension->√{-1,2,3}] immediately cracked it with n = > 2. > > Then > > 4 - 32 x2 + 80 x2^2 - 32 x2^3 - 168 x2^4 + 288 x2^5 - 164 x2^6 - > > 16 x2^7 + 60 x2^8 - 16 x2^9 - 4 x2^10 + x2^12 == > > -32 (-1 + x2)^4 (-1 + 2 x2)^2 + > > (6 - 24 x2 + 28 x2^2 - 8 x2^3 - 2 x2^4 + x2^6)^2 > > > > Is this what you did? --rwg > > Sadly, Noam Elkies tells me the Q(√2) sextic factors don't solve in > radicals. > > > > On Fri, Dec 30, 2016 at 11:49 AM, Bill Gosper <billgosper(a)gmail.com> > > wrote: > > > > > Two extremes: > > > In[466]:= Factor@Decompose[%462,x] > > > Out[466]= {-4+24 x+12 x^2+4 x^3-6 x^4+x^6} > > > > > > In[467]:= PolySolve6@@%//FullSimplify > > > > > > Out[467]= (*All 6 solutions*) > > > {{x->(-2)^(1/3)-Sqrt[2]},{x->(-2)^(1/3)+Sqrt[2]},{x->-2^(1/3) > > > ((-1)^(2/3)+2^(1/6))}, > > > {x->Sqrt[2]+Root[2+#1^3&,2]},{x->-2^(1/3) (1+2^(1/6))},{x->2^(1/3) > > > (-1+2^(1/6))}} > > > > > > Whereas Simplify can barely reduce just this one solution (a mere unit > !-) > > > Out[436]= x->Root[1+3 #1+4 #1^2+2 #1^3+#1^6&,6] > > > In[469]:= Simplify@%435[[6,1]] > > > Out[469]= x->1/2 (1/3+(1-I Sqrt[3])/(2^(2/3) (27-3 > Sqrt[69])^(1/3))+((1+I > > > Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))+1/6 I (I+Sqrt[3]) > (1/2(11+3 > > > Sqrt[69]))^(1/3)+(5 (1+I Sqrt[3]))/(3 2^(2/3) (11+3 > Sqrt[69])^(1/3))-((1-I > [... chop ...] > > > to display on a page. (FullSimplify just spits back the sextic.) > [... chop ...]

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[math-fun] Two little sextics,
by Bill Gosper 31 Dec '16

31 Dec '16

Whoa. Can PARI (or GAP) reliably denest radicals? http://community.wolfram.com/groups/-/m/t/988309 Maple, Mathematica, and (presumably) Maxima cannot. --rwg Date: 2016-12-30 02:19 From: Warut Roonguthai <warut822(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> The Galois group of x^6-3*x+3 told me that it had a quadratic subfield (but no cubic subfield). The form of the exact value of x told me which quadratic field. (Otherwise, some more guessing would be needed, or I had to solve it the hard way.) I computed the Galois group and factored over Q(sqrt(-3)) with PARI. Warut On Fri, Dec 30, 2016 at 3:57 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > Date: 2016-12-29 22:20 > From: Warut Roonguthai <warut822(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > 4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2 > tells that x^6-3*x+3 can be factored into two cubic over Q(sqrt(-3)). > > BTW, 822 is my lucky number but doesn't seem to work very well so far. > Hahaha > Warut > > How could it not be at least doubly informative? > > DAWK! I plead senility. Ok, how did you find those squares? I see how to > go the other way--i.e. from a factorization over √n to f = p² - n q². > E.g., if f = the duodecic for x2 in > http://mathworld.wolfram.com/SquareDissection.html , > guessing Factor[f,Extension->√{-1,2,3}] immediately cracked it with n = 2. > Then > 4 - 32 x2 + 80 x2^2 - 32 x2^3 - 168 x2^4 + 288 x2^5 - 164 x2^6 - > 16 x2^7 + 60 x2^8 - 16 x2^9 - 4 x2^10 + x2^12 == > -32 (-1 + x2)^4 (-1 + 2 x2)^2 + > (6 - 24 x2 + 28 x2^2 - 8 x2^3 - 2 x2^4 + x2^6)^2 > > Is this what you did? --rwg > Sadly, Noam Elkies tells me the Q(√2) sextic factors don't solve in radicals. > > On Fri, Dec 30, 2016 at 11:49 AM, Bill Gosper <billgosper(a)gmail.com> > wrote: > > > Two extremes: > > In[466]:= Factor@Decompose[%462,x] > > Out[466]= {-4+24 x+12 x^2+4 x^3-6 x^4+x^6} > > > > In[467]:= PolySolve6@@%//FullSimplify > > > > Out[467]= (*All 6 solutions*) > > {{x->(-2)^(1/3)-Sqrt[2]},{x->(-2)^(1/3)+Sqrt[2]},{x->-2^(1/3) > > ((-1)^(2/3)+2^(1/6))}, > > {x->Sqrt[2]+Root[2+#1^3&,2]},{x->-2^(1/3) (1+2^(1/6))},{x->2^(1/3) > > (-1+2^(1/6))}} > > > > Whereas Simplify can barely reduce just this one solution (a mere unit !-) > > Out[436]= x->Root[1+3 #1+4 #1^2+2 #1^3+#1^6&,6] > > In[469]:= Simplify@%435[[6,1]] > > Out[469]= x->1/2 (1/3+(1-I Sqrt[3])/(2^(2/3) (27-3 Sqrt[69])^(1/3))+((1+I > > Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))+1/6 I (I+Sqrt[3]) (1/2(11+3 > > Sqrt[69]))^(1/3)+(5 (1+I Sqrt[3]))/(3 2^(2/3) (11+3 Sqrt[69])^(1/3))-((1-I [... chop ...] > > to display on a page. (FullSimplify just spits back the sextic.) [... chop ...]

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[math-fun] Two little sextics,
by Bill Gosper 30 Dec '16

30 Dec '16

Date: 2016-12-29 22:20 From: Warut Roonguthai <warut822(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> 4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2 tells that x^6-3*x+3 can be factored into two cubic over Q(sqrt(-3)). BTW, 822 is my lucky number but doesn't seem to work very well so far. Hahaha Warut How could it not be at least doubly informative? DAWK! I plead senility. Ok, how did you find those squares? I see how to go the other way--i.e. from a factorization over √n to f = p² - n q². E.g., if f = the duodecic for x2 in http://mathworld.wolfram.com/SquareDissection.html , guessing Factor[f,Extension->√{-1,2,3}] immediately cracked it with n = 2. Then 4 - 32 x2 + 80 x2^2 - 32 x2^3 - 168 x2^4 + 288 x2^5 - 164 x2^6 - 16 x2^7 + 60 x2^8 - 16 x2^9 - 4 x2^10 + x2^12 == -32 (-1 + x2)^4 (-1 + 2 x2)^2 + (6 - 24 x2 + 28 x2^2 - 8 x2^3 - 2 x2^4 + x2^6)^2 Is this what you did? --rwg Sadly, Noam Elkies tells me the Q(√2) sextic factors don't solve in radicals. On Fri, Dec 30, 2016 at 11:49 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Two extremes: > In[466]:= Factor@Decompose[%462,x] > Out[466]= {-4+24 x+12 x^2+4 x^3-6 x^4+x^6} > > In[467]:= PolySolve6@@%//FullSimplify > > Out[467]= (*All 6 solutions*) > {{x->(-2)^(1/3)-Sqrt[2]},{x->(-2)^(1/3)+Sqrt[2]},{x->-2^(1/3) > ((-1)^(2/3)+2^(1/6))}, > {x->Sqrt[2]+Root[2+#1^3&,2]},{x->-2^(1/3) (1+2^(1/6))},{x->2^(1/3) > (-1+2^(1/6))}} > > Whereas Simplify can barely reduce just this one solution (a mere unit !-) > Out[436]= x->Root[1+3 #1+4 #1^2+2 #1^3+#1^6&,6] > In[469]:= Simplify@%435[[6,1]] > Out[469]= x->1/2 (1/3+(1-I Sqrt[3])/(2^(2/3) (27-3 Sqrt[69])^(1/3))+((1+I > Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))+1/6 I (I+Sqrt[3]) (1/2 (11+3 > Sqrt[69]))^(1/3)+(5 (1+I Sqrt[3]))/(3 2^(2/3) (11+3 Sqrt[69])^(1/3))-((1-I > Sqrt[3])/(2^(2/3) (27-3 Sqrt[69])^(1/3))+((1+I Sqrt[3]) (1/2 > (9-Sqrt[69]))^(1/3))/(2 3^(2/3)))^2-1/6 ((1-I Sqrt[3])/(2^(2/3) (27-3 > Sqrt[69])^(1/3))+((1+I Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))) > (4+10 (1+I Sqrt[3]) (2/(11+3 Sqrt[69]))^(1/3)+I 2^(2/3) (I+Sqrt[3]) (11+3 > Sqrt[69])^(1/3))+1/144 (-4 I+10 (-I+Sqrt[3]) (2/(11+3 > Sqrt[69]))^(1/3)+2^(2/3) (I+Sqrt[3]) (11+3 Sqrt[69])^(1/3))^2+(1/(-54+8 > Sqrt[69]))(\[Sqrt](1/6 (14568 2^(2/3) (11+3 Sqrt[69])^(1/3)-14568 I 2^(2/3) > Sqrt[3] (11+3 Sqrt[69])^(1/3)+5058 I 2^(2/3) Sqrt[23] (11+3 > Sqrt[69])^(1/3)-1686 2^(2/3) Sqrt[69] (11+3 Sqrt[69])^(1/3)+18879 2^(1/3) > (11+3 Sqrt[69])^(2/3)+18879 I 2^(1/3) Sqrt[3] (11+3 Sqrt[69])^(2/3)-6849 I > 2^(1/3) Sqrt[23] (11+3 Sqrt[69])^(2/3)-2283 2^(1/3) Sqrt[69] (11+3 > Sqrt[69])^(2/3)-4496 3^(5/6) Sqrt[23] (-27+4 Sqrt[69])^(1/3)-2224 3^(1/6) > Sqrt[23] (-27+4 Sqrt[69])^(2/3)+38848 (-81+12 Sqrt[69])^(1/3)+5504 (-81+12 > Sqrt[69])^(2/3)+(27-3 Sqrt[69])^(2/3) (2^(1/3) (12043+4283 I Sqrt[23])+4 > (498-263 I Sqrt[23]) (11+3 Sqrt[69])^(1/3))-3^(1/6) (9-Sqrt[69])^(2/3) > (2^(1/3) (36129 I+4283 Sqrt[23])+4 (-1494 I+263 Sqrt[23]) (11+3 > Sqrt[69])^(1/3))+2 (27-3 Sqrt[69])^(1/3) (2^(2/3) (2044-339 I > Sqrt[23])+(-5396-1926 I Sqrt[23]) (11+3 Sqrt[69])^(2/3))+2 3^(5/6) > (9-Sqrt[69])^(1/3) (2^(2/3) (2044 I-113 Sqrt[23])+(5396 I+642 Sqrt[23]) > (11+3 Sqrt[69])^(2/3)))))) > > to display on a page. (FullSimplify just spits back the sextic.) > > Date: 2016-12-29 09:56 > From: Warut Roonguthai <warut822(a)gmail.com> > > For the first one, 4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2. > Warut > > So there are no real roots. But does this tell me what the roots actually > are? > (E.g., In[996].) > --rwg > And are there ≥822 GMail customers named Warut? > > On Thu, Dec 29, 2016 at 5:55 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > > > irreducible, indecomposable, nonreciprocal, solvable: > > In[996]:= Factor@Decompose[MinimalPolynomial[I/Sqrt[3]-(I (1/2 (9 I-2 > > Sqrt[3]+3 3^(1/4) Sqrt[-4 I-3 > > Sqrt[3]]))^(1/3))/3^(2/3)+(I+Sqrt[3])/(2^(2/3) (27 I-6 Sqrt[3]+9 3^(1/4) > > Sqrt[-4 I-3 Sqrt[3]])^(1/3)),x],x] > > Out[996]= {3-3 x+x^6} > > > > In[997]:= Factor@Decompose[MinimalPolynomial[1/12 (2-2^(2/3) (-29+3 > > Sqrt[321])^(1/3)+2^(2/3) (29+3 Sqrt[321])^(1/3)-2^(5/6) Sqrt[3 (7 > > 2^(1/3)+(83-3 Sqrt[321])^(1/3)+(83+3 Sqrt[321])^(1/3))]),x],x] > > Out[997]= {-1-x-x^5+x^6}

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