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

29 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 29 Dec '16

29 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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[math-fun] science museum has new math gallery (London)
by James Buddenhagen 19 Dec '16

19 Dec '16

see here: http://artdaily.com/news/92261/Maths-explored-in-new-Science-Museum-gallery…

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[math-fun] Variants of the Etch-a-Sketch
by James Propp 19 Dec '16

19 Dec '16

Has anyone made variants of the Etch-a-Sketch featuring knobs that allow other sorts of control (e.g., drawing a circular arc with a prescribed center)? Jim Propp

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Re: [math-fun] The difficulty of integration vs. the ease of differentiation
by Henry Baker 19 Dec '16

19 Dec '16

I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors. At 02:27 PM 12/16/2016, Fred Lunnon wrote: >Numerically speaking, integration is in contrast easier than differentiation ... WFL > >On 12/16/16, Henry Baker <hbaker1(a)pipeline.com> wrote: >> 2. Some closed form integrals may require polynomial root-finding (or root >> expressing), so you may not like the look of the resulting expression. >> E.g., you might find floating point approximations to certain numbers >> instead of a "perfectly precise" answer.

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