December 2016 - mathfun@mailman.xmission.com

[math-fun] Julian has constructed a spacefill dense with sextuple points
by Bill Gosper 26 Jan '17

26 Jan '17

using a 25-way conditional. ClearAll[piecewiserecursivefractal]; piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])); (I cannot praise this function enough.) In[363]:= trifil25[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, #1 == 0}, {{Ceiling[25*#1]}, Inequality[0, Less, #1, LessEqual, 1]}}, {}] & , {25*#1 + 0 & , -25*#1 + 2 & , 25*#1 - 2 & , 25*#1 - 3 & , 25*#1 - 4 & , 25*#1 - 5 & , 25*#1 - 6 & , 25*#1 - 7 & , 25*#1 - 8 & , -25*#1 + 10 & , -25*#1 + 11 & , 25*#1 - 11 & , -25*#1 + 13 & , -25*#1 + 14 & , 25*#1 - 14 & , -25*#1 + 16 & , 25*#1 - 16 & , -25*#1 + 18 & , 25*#1 - 18 & , 25*#1 - 19 & , -25*#1 + 21 & , -25*#1 + 22 & , 25*#1 - 22 & , -25*#1 + 24 & , 25*#1 - 24 & }, {#1/5 & , (1/10)*(3 + I*Sqrt[3] + (-1 - I*Sqrt[3])* #1) & , (1/10)*(3 + I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (2 + #1)/5 & , (1/10)*(6 + #1 + I*Sqrt[3]*#1) & , (1/10)*(7 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/5)*(3 + I*Sqrt[3] - #1) & , (1/10)*(4 + 2*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/10)*(3 + I*Sqrt[3] - 2*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-2 + #1) & , (1/5)*(2 + I*Sqrt[3] - #1) & , (1/5)*(2 + I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-4 + #1) & , (1/5)*(2 + 2*I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(5 + 3*I*Sqrt[3] + 2*#1) & , (1/10)*(7 + 3*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/5)*(4 + I*Sqrt[3] - #1) & , (1/10)*(9 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/10)*(9 + I*Sqrt[3] - 2*#1) & , (1/10)*(8 + I*(I + Sqrt[3])*#1) & , (4 + #1)/5 & }][[1]] E.g., In[361]:= trifil25[2/25 + 2/75] // ComplexExpand Out[361]= {1/2 + (I Sqrt[3])/10} In[367]:= Graphics[{Hue@#, Line[ReIm@{trifil25@#, trifil25[# + 5^-5]}]} & /@ Range[0, 1 - 5^-5, 5^-5]] gosper.org/trifil25.png (Those little white hexagons are sextuple points.) The inverse fractal, constructed more or less automatically: In[380]:= untrifil25[t_] := piecewiserecursivefractal[t, Identity, If[0 <= Re[#1] <= 1 && 0 <= Im[#1] <= Sqrt[3]/2, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}, {}] &, {ComplexExpand[5 #1] &, ComplexExpand[1/4 (6 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-4 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 + 5 #1] &, ComplexExpand[1/4 (-6 + 6 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (4 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[3 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (10 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (3 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[2 + I Sqrt[3] - 5 #1] &, ComplexExpand[-2 - I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (4 - 8 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-10 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-2 + 6 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (16 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 - 2 I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (-14 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (-3 I Sqrt[3] + 5 (-1 + 2 #1))] &, ComplexExpand[1/4 (16 - 4 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[4 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (6 + 10 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (9 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[-4 + 5 #1] &}, {#1/25 &, (2 - #1)/25 &, (2 + #1)/ 25 &, (3 + #1)/25 &, (4 + #1)/25 &, (5 + #1)/25 &, (6 + #1)/ 25 &, (7 + #1)/25 &, (8 + #1)/25 &, (10 - #1)/25 &, (11 - #1)/ 25 &, (11 + #1)/25 &, (13 - #1)/25 &, (14 - #1)/25 &, (14 + #1)/ 25 &, (16 - #1)/25 &, (16 + #1)/25 &, (18 - #1)/25 &, (18 + #1)/ 25 &, (19 + #1)/25 &, (21 - #1)/25 &, (22 - #1)/25 &, (22 + #1)/ 25 &, (24 - #1)/25 &, (24 + #1)/25 &}] Here it finds the preimages of the simplest sextuple point: In[381]:= untrifil25[1/2 + I Sqrt[3]/10] Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} --rwg

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

30 Dec '16

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} > --rwg

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[math-fun] My coda to Legendre's conjecture
by David Wilson 30 Dec '16

30 Dec '16

Legendre's conjecture is that there is at least one prime between any two consecutive positive square numbers. I further conjecture that for any n > 1, there are two consecutive squares with exactly n primes between.

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Re: [math-fun] Two little sextics,
by Bill Gosper 29 Dec '16

29 Dec '16

On Wed, Dec 28, 2016 at 5:52 PM, <rcs(a)xmission.com> wrote: > For the 2nd, the subst x->iy converts it to symmetric, and then > y+1/y -> z gets it down to a cubic. I don't see anything for the 1st. > Rich > > Yeah, I just wrote up y+1/y -> z for a dangerous youngster named Gabe. gosper.org/gabesolve.nb . It's a Mathematica notebook, but there's a free reader. http://library.wolfram.com/infocenter/TechNotes/426/ Or at least there was. The link now redirects to their Computable Document player. Tell me if this won't view the .nb, and I'll write a cdf. --rwg Nice roots, nice coefficients: In[414]:= Factor@Decompose[MinimalPolynomial[1/2 (-1 - 5^(1/3) + 5^(1/6) Sqrt[2 + 5^(1/3) + 5^(2/3)]), x], x] Out[414]= {-1 + 3 x + 3 x^5 + x^6} Nice roots, nasty coefficients: Out[1020]= (-3 I + Sqrt[3 + 4 2^(1/3) 3^(2/3)])^6 In[1021]:= Factor@Decompose[MinimalPolynomial[%, x], x] Out[1021]= {570630428688384000000 + 4891824455002619904 x + 161093791317491712 x^2 + 12153384861696 x^3 + 984379392 x^4 - 17280 x^5 + x^6} So the "parenthesand" has degree 36. --rwg > --- > > Quoting Bill Gosper <billgosper(a)gmail.com>: > >> 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} >> --rwg >> > > >

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[math-fun] Two little sextics,
by Bill Gosper 29 Dec '16

29 Dec '16

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} --rwg

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[math-fun] the 4th rep-4 dragon
by Bill Gosper 27 Dec '16

27 Dec '16

I'm sure it's in Jörg's menagerie, but Julian had to lead me by the nose. http://gosper.org/4thdragon.png What a strange, camouflaged creature. --rwg

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[math-fun] Capabilities of Mathematica "Home Edition"
by Dan Asimov 26 Dec '16

26 Dec '16

Mathematica has recently advertised a discount on their Home Edition. About $250 while the sale lasts. Their website claims the Home Edition has "All of the power of Mathematica for personal use" . But it doesn't say specifically that there are no differences — or if there are differences, what those differences are. This *might* be just a pricing difference, the way some academic journals charge institutional libraries like $1500 / year but a mere $300 / year for an individual. But if that's the case, why wouldn't they just say so instead of giving it a whole nother name? Anyone know the answer to whether the Home Edition differs from the full Mathematica, and if so how? —Dan

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Re: [math-fun] Protect your address bits!
by Henry Baker 20 Dec '16

20 Dec '16

Luckily, my house number doesn't require 0 or 1 in hex. (If I use Common Lisp hex notation "xABC", not C hex notation "0xABC".) So, if you are denied the digits 0,1, what would be the best way to display your house number? At 10:17 AM 12/19/2016, Thane Plambeck wrote: >http://billypenn.com/2016/12/13/binary-bandits-heart-grows-3-sizes-83-1s-0s…

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