[math-fun] That one-ninthish sextic
Our sextic solver fervently assumes (I think with Piezas's blessing) that the solvable ones all factor into either cubics with quadratic surd coefficients, or vice versa. In[343]:= Factor[27*x^6 + 3087*x - 343, Extension -> Sqrt[21]] Out[343]= -(1/12) (-147 - 35 Sqrt[21] + (-21 + 21 Sqrt[21]) x + 6 Sqrt[21] x^2 - 18 x^3) * (147 - 35 Sqrt[21] + (21 + 21 Sqrt[21]) x + 6 Sqrt[21] x^2 + 18 x^3) is the former. The interesting real root is Out[218]= -Sqrt[(7/3)]/3 - 1/3 Sqrt[7/3] (1/2 (-70 + 9 Sqrt[21] + 9 Sqrt[85 - 12 Sqrt[21]]))^(1/3) + (Sqrt[7] (1 + 3 Sqrt[21]))/(3 2^(2/3) Sqrt[3] (-70 + 9 Sqrt[21] + 9 Sqrt[85 - 12 Sqrt[21]])^(1/3)) In[219]:= Factor[27*x^6 + 3087*x - 343 - 3^-9] Out[219]= ((-1 + 9 x) (6751270 + 9 x + 81 x^2 + 729 x^3 + 6561 x^4 + 59049 x^5))/19683 In[220]:= ContinuedFraction[%%, 33] Out[220]= {0, 9, 750141, 1, 1, 4, 50009, 4, 3, 4, 6, 1, 389, 1, 5, 7, 7, 2, 5, 1, 1, 1, 4, 64, 2, 1, 1, 1, 1, 1, 1, 2, 7} NDE>Are you suggesting there's some significance to the large convergents? Well, it can't hold a candle to Brillhart's In[228]:= ContinuedFraction[(15 Sqrt[3]-Sqrt[163])^(1/3)/Sqrt[3]+(15 Sqrt[3]+Sqrt[163])^(1/3)/Sqrt[3],239] Out[228]= {3,3,7,4,2,30,1,8,3,1,1,1,9,2,2,1,3,22986,2,1,32,8,2,1,8,55,1,5,2,28,1,5,1,1501790,1,2,1,7,6,1,1,5,2,1,6,2,2,1,2,1,1,3,1,3,1,2,4,3,1,35657,1,17,2,15,1,1,2,1,1,5,3,2,1,1,7,2,1,7,1,3,25,49405,1,1,3,1,1,4,1,2,15,1,2,83,1,162,2,1,1,1,2,2,1,53460,1,6,4,3,4,13,5,15,6,1,4,1,4,1,1,2,1,16467250,1,3,1,7,2,6,1,95,20,1,2,1,6,1,1,8,1,48120,1,2,17,2,1,2,1,4,2,3,1,2,23,3,2,1,1,1,2,1,27,325927,1,60,1,87,1,2,1,5,1,1,1,2,2,2,2,2,17,4,9,9,1,7,11,1,2,9,1,14,4,6,1,22,11,1,1,1,1,4,1,3,2,1,2,1,1,2,4,2,1,5,1,8,2,2,5,1,2,1,1,1,1,1,3,1,2,6,10,1,3,1,3,2,2,1,1,2,1} (A002937) But I don't recall the phenomenon in sextics. (Is H. M. Stark, ‘An explanation of some exotic continued fractions found by Brillhart, in A. O. L. Atkin and B. J. Birch eds., Computers in Number Theory, Academic Press, London and New York, 1971 available online? Hmm, √163.) NDE>No septics here. This family of octics factors over quadratic extensions, so it's "just" solving a quartic over Q(sqrt(d)). Yow, it Gaussian-factors! In[338]:= Factor[4 x^8 + 8 x + 13, Extension -> I] Out[338]= ((3 - 2 I) + (4 + 4 I) x - (4 - 2 I) x^2 - 4 I x^3 + 2 x^4) * ((3 + 2 I) + (4 - 4 I) x - (4 + 2 I) x^2 + 4 I x^3 + 2 x^4) --rwg
Bill Gosper <billgosper@gmail.com> writes:
Our sextic solver fervently assumes (I think with Piezas's blessing) that the solvable ones all factor into either cubics with quadratic surd coefficients, or vice versa.
That's right, at least if "cubic surd" means the solution of an arbitrary cubic, not just expressions rational in n^(1/3).
NDE>Are you suggesting there's some significance to the large convergents?
Well, it can't hold a candle to Brillhart's In[228]:= ContinuedFraction[(15 Sqrt[3]-Sqrt[163])^(1/3)/Sqrt[3]+(15 Sqrt[3]+Sqrt[163])^(1/3)/Sqrt[3],239] [...] Hmm, \sqrt{163}.)
I don't know whether the article is online, but I remember the result, and yes, it certainly has to do with sqrt(163) yoga. Even if it "doesn't hold a candle" to Brillhart's cubic, it might point the way to other sextics whose roots have more remarkable rational approximations.
NDE> No septics here. This family of octics factors over quadratic NDE> extensions, so it's "just" solving a quartic over Q(sqrt(d)).
Yow, it Gaussian-factors!
Right, in this case d=-1. But you'd still have to go through the solution of a quartic by radicals to fully write out a root this way. Likewise for the mutant trinomial x^8 + 4*x^5 + 8 -- and indeed this one also "Gaussian-factors" (I exhibited it as A^2+B^2), though it's unique up to scaling even without specifying the quadratic extension over which it factors into two quartics. The curves parametrizing octic trinomials with Galois group contained in (2^3).7 (the ax+b group over the 8-element field), or even ((2^3).7).3 (extending by the field automorphism), are much more complicated, and I wouldn't be surprised if there are no non-degenerate rational points. NDE P.S. Are all of those e-mails being seen on math-fun too? I'm not a subcriber and keep getting auto-replies "Your mail to 'math-fun' with the subject <...> Is being held until the list moderator can review it for approval. The reason it is being held: Post by non-member to a members-only list".
On Mon, Sep 2, 2013 at 2:51 PM, Noam Elkies <elkies@math.harvard.edu> wrote:
Bill Gosper <billgosper@gmail.com> writes:
Our sextic solver fervently assumes (I think with Piezas's blessing) that the solvable ones all factor into either cubics with quadratic surd coefficients, or vice versa.
That's right, at least if "cubic surd" means the solution of an arbitrary cubic, not just expressions rational in n^(1/3).
NDE>Are you suggesting there's some significance to the large convergents?
Well, it can't hold a candle to Brillhart's In[228]:= ContinuedFraction[(15 Sqrt[3]-Sqrt[163])^(1/3)/Sqrt[3]+(15 Sqrt[3]+Sqrt[163])^(1/3)/Sqrt[3],239] [...] Hmm, \sqrt{163}.)
I don't know whether the article is online, but I remember the result, and yes, it certainly has to do with sqrt(163) yoga. Even if it "doesn't hold a candle" to Brillhart's cubic, it might point the way to other sextics whose roots have more remarkable rational approximations.
Oops, Piezas (https://sites.google.com/site/tpiezas/001) gives the sextic 1 - 10 x^3 - 640320 x^5 + 5 x^6 with a root (also of the quadratic 264840 + 87 10^( 2/3) ((282094543 - 21 Sqrt[489])^( 1/3) + (282094543 + 21 Sqrt[489])^(1/3)) - 15 (256128 + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] - 54578363 Sqrt[163]))^(1/3) + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] + 54578363 Sqrt[163]))^(1/3)) x + 90 x^2) with CF = {128064, 8200194048, 62470, 4, 10, 6513259, 1, 2, 28, 1, 1, 2, 2, 4, 2, 1, 3, 9, 1, 329, 2, 2, 3, 1, 1, 1, 5, 2, 2, 2,...} (489=3*163) Are there more spectacular e^(π algebraic)?
NDE> No septics here. This family of octics factors over quadratic NDE> extensions, so it's "just" solving a quartic over Q(sqrt(d)).
Yow, it Gaussian-factors!
Right, in this case d=-1. But you'd still have to go through the solution of a quartic by radicals to fully write out a root this way. Likewise for the mutant trinomial x^8 + 4*x^5 + 8 -- and indeed this one also "Gaussian-factors" (I exhibited it as A^2+B^2), though it's unique up to scaling even without specifying the quadratic extension over which it factors into two quartics.
The curves parametrizing octic trinomials with Galois group contained in (2^3).7 (the ax+b group over the 8-element field), or even ((2^3).7).3 (extending by the field automorphism), are much more complicated, and I wouldn't be surprised if there are no non-degenerate rational points.
NDE
P.S. Are all of those e-mails being seen on math-fun too? I'm not a subcriber and keep getting auto-replies "Your mail to 'math-fun' with the subject <...> Is being held until the list moderator can review it for approval. The reason it is being held: Post by non-member to a members-only list".
They eventually came out in a bunch. People will think you're manic. --rwg
Bill Gosper <billgosper@gmail.com> wrote:
On Mon, Sep 2, 2013 at 2:51 PM, Noam Elkies <elkies@math.harvard.edu> wrote:
Bill Gosper <billgosper@gmail.com> writes:
Our sextic solver fervently assumes (I think with Piezas's blessing) that the solvable ones all factor into either cubics with quadratic surd coefficients, or vice versa.
That's right, at least if "cubic surd" means the solution of an arbitrary cubic, not just expressions rational in n^(1/3).
NDE>Are you suggesting there's some significance to the large convergents?
Well, it can't hold a candle to Brillhart's In[228]:= ContinuedFraction[(15 Sqrt[3]-Sqrt[163])^(1/3)/Sqrt[3]+(15 Sqrt[3]+Sqrt[163])^(1/3)/Sqrt[3],239] [...] Hmm, \sqrt{163}.)
I don't know whether the article is online, but I remember the result, and yes, it certainly has to do with sqrt(163) yoga. Even if it "doesn't hold a candle" to Brillhart's cubic, it might point the way to other sextics whose roots have more remarkable rational approximations.
Oops, Piezas (https://sites.google.com/site/tpiezas/001) gives the sextic 1 - 10 x^3 - 640320 x^5 + 5 x^6 with a root (also of the quadratic 264840 + 87 10^( 2/3) ((282094543 - 21 Sqrt[489])^( 1/3) + (282094543 + 21 Sqrt[489])^(1/3)) - 15 (256128 + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] - 54578363 Sqrt[163]))^(1/3) + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] + 54578363 Sqrt[163]))^(1/3)) x + 90 x^2)
with CF = {128064, 8200194048, 62470, 4, 10, 6513259, 1, 2, 28, 1, 1, 2, 2, 4, 2, 1, 3, 9, 1, 329, 2, 2, 3, 1, 1, 1, 5, 2, 2, 2,...} (489=3*163) Are there more spectacular e^(?? algebraic)?
NDE> No septics here. This family of octics factors over quadratic NDE> extensions, so it's "just" solving a quartic over Q(sqrt(d)).
Yow, it Gaussian-factors!
Right, in this case d=-1. But you'd still have to go through the solution of a quartic by radicals to fully write out a root this way. Likewise for the mutant trinomial x^8 + 4*x^5 + 8 -- and indeed this one also "Gaussian-factors" (I exhibited it as A^2+B^2), though it's unique up to scaling even without specifying the quadratic extension over which it factors into two quartics.
The curves parametrizing octic trinomials with Galois group contained in (2^3).7 (the ax+b group over the 8-element field), or even ((2^3).7).3 (extending by the field automorphism), are much more complicated, and I wouldn't be surprised if there are no non-degenerate rational points.
NDE
P.S. Are all of those e-mails being seen on math-fun too? I'm not a subcriber and keep getting auto-replies "Your mail to 'math-fun' with the subject <...> Is being held until the list moderator can review it for approval. The reason it is being held: Post by non-member to a members-only list".
They eventually came out in a bunch. People will think you're manic. --rwg
-- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe@googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out. The cubic is x^3-8x-10. The only papers on the work were written by other people, not by me. There are two papers: the first by Robert Churchhouse and Muir, (1969) Continued fractions, algebraic numbers, and modular invariants, J.I.M A.5, 518-328. The main paper was written by Harold Stark and appears in the Proc. of Computers in Number Theory of the Science Research Council Atlas Symposium No.2 held at Oxford from 18-23 August, 1969 Ed. Atkin and Birch 1971. Academic Press, London and New York
The greatest picture ever taken at a mathematics meeting is the photo of the attendees of that meeting. See online, Meeting at Oxford at 1969. It is also a marvelous collection of photos of computational number theorists. Clear and there is a full listing of the attendees put together by John Cosgrave. You will note the picture was taken when we were all hustled out of the sherry party outside because Marshall Hall and John Cosgrave are still holding their glasses.
Bill Gosper <billgosper@gmail.com> writes:
[...] Are there more spectacular e^(\pi algebraic)?
Do you mean "more spectacular than \sqrt{163}" or "other spectacular examples"? For the former, none that I know. For the latter, yes, all with the algebraic number being sqrt(d) for some rational d. The further integral examples d=67 and (barely) 43 are well-known; since the cube root of j is also a modular function (and some other things work out nicely), you can also use 163/9, 67/3, and (barely) 43/3. [I've suggested that a CM mile should be exactly exp(Pi*sqrt(67)/3) feet long. :-)] There's also the known example of d=58, and here it also 58/4 and 58/16 that yield near-integers. There are also some even less well-known examples like exp(Pi*sqrt(89/3)) and its cube root 300.000155555... Some of these also have small powers that are nearly integral but not as close, e.g. exp(Pi*sqrt(163)*4/3) = 168107956062137200957439.99999965975... NDE
participants (3)
-
Bill Gosper -
elkies＠math.harvard.edu -
jdb＠math.arizona.edu