Maybe we need an Electronic Journal of Unoriginal Research with a quick turn-around time. If Wikipedia rejects part of a submitted article as being Original Research, just publish it in EJUR and resubmit the Wikipedia article with the EJUR citation included. :-) Jim Propp On Mon, Aug 26, 2013 at 7:10 AM, Bill Gosper <billgosper@gmail.com> wrote:
According to http://www.math.harvard.edu/~elkies/trinomial.html, modulo canonicalization, there are only five solvable, irreducible quintics with quadratic middle terms. Only one, x^5 - 100 x^2 - 1000, has roots that are the sums of fifth roots of monomials. Its real root is 2 2^(1/5) + 2 2^(2/5) - 2^(3/5) + 2^(4/5). Raising this to consecutive powers as before, In[199]:= Table[FullSimplify[Expand[%^k]], {k, 12}]
Out[199]= {Root[-1000 - 100 #1^2 + #1^5 &, 1], 10 (2^(1/5) + 2^(3/5)), 10 (6 - 2 2^(1/5) + 4 2^(2/5) + 2 2^(3/5) + 2^(4/5)), 100 2^(1/5) (2 + 2^(1/5) + 2 2^(3/5)), 1000 (1 + 2^(1/5) + 2^(3/5)), 1000 (6 + 6 2^(2/5) + 2^(3/5) + 2 2^(4/5)), 10000 2^(1/5) (3 + 2^(1/5) + 2^(2/5) + 2 2^(3/5)), 10000 (16 + 8 2^(1/5) + 4 2^(2/5) + 12 2^(3/5) + 2^(4/5)), 100000 (6 + 2 2^(1/5) + 7 2^(2/5) + 2^(3/5) + 4 2^(4/5)), 1000000 (1 + 4 2^(1/5) + 2^(2/5) + 2 2^(3/5) + 2 2^(4/5)), 1000000 (22 + 8 2^(1/5) + 10 2^(2/5) + 13 2^(3/5) + 3 2^(4/5)), 10000000 (6 + 5 2^(1/5) + 8 2^(2/5) + 2 2^(3/5) + 6 2^(4/5))}
The square gives this familiar-looking denesting that I can't find in my notes: In[193]:= DenestRadicals3[Sqrt[1 + 4^(1/5)]] {f[Sqrt[1+2^(2/5)]]}
problem Sqrt[1+2^(2/5)]
outer root: 2 extra root: 1
multiplier count: 2 multipliers: {1,2^(3/5)}
multiplier: 1 minpoly degree: 10
adding new multipliers from: 1
multiplier: 2^(3/5) minpoly degree: 10
adding new multipliers from: 2^(3/5)
multiplier count: 3 from: {1,10}
multiplier: 2 minpoly degree: 10
multiplier: 5 minpoly degree: 5
PossibleZeroQ::ztest1: Unable to decide whether numeric quantity -(1/Sqrt[5])+2^(1/5)/Sqrt[5]+2^(3/5)/Sqrt[5]+2^(4/5)/Sqrt[5]-Sqrt[1+2^(2/5)] is equal to zero. Assuming it is. >>
success: {5,5,{{5,1}}}
history: {{0,\[Infinity]},{1,10},{5,5}}
(-1+2^(1/5)+2^(3/5)+2^(4/5))/Sqrt[5]
Out[193]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/Sqrt[5]
(Heuristic denester by Corey Ziegler Hunts.)
The fifth power gives a trinomial^(1/5) -> quadrinomial denesting
In[206]:= DenestRadicals3[(1 + 2^(1/5) + 8^(1/5))^(1/5)]
{f[(1+2^(1/5)+2^(3/5))^(1/5)]}
problem (1+2^(1/5)+2^(3/5))^(1/5)
outer root: 5 extra root: 1
multiplier count: 3 multipliers: {1,2^(4/5),2^(2/5)}
multiplier: 1 minpoly degree: 25
adding new multipliers from: 1
multiplier: 2^(4/5) minpoly degree: 25
adding new multipliers from: 2^(4/5)
multiplier: 2^(2/5) minpoly degree: 25
adding new multipliers from: 2^(2/5)
multiplier count: 124 from: {1,25}
multiplier: 2 minpoly degree: 25
multiplier: 4 minpoly degree: 25
multiplier: 5 minpoly degree: 25
multiplier: 8 minpoly degree: 25
multiplier: 10 minpoly degree: 25
multiplier: 16 minpoly degree: 25
multiplier: 20 minpoly degree: 25
multiplier: 25 minpoly degree: 25
multiplier: 29 minpoly degree: 25
multiplier: 40 minpoly degree: 25
multiplier: 50 minpoly degree: 25
multiplier: 58 minpoly degree: 25
multiplier: 80 minpoly degree: 25
multiplier: 100 minpoly degree: 25
multiplier: 116 minpoly degree: 25
multiplier: 125 minpoly degree: 5
success: {125,5,{{5,3}}}
history: {{0,\[Infinity]},{1,25},{125,5}}
(-1+2^(1/5)+2^(3/5)+2^(4/5))/5^(3/5)
Out[206]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/125^(1/5)
This I would nominate to replace the "more complicated" yet meaningless, undenestable trinomial^(1/3) in http://en.wikipedia.org/wiki/Nested_radical . But that would smack of the ever contemptible Original Research. Also missing from that article is a remark to the effect that denestings with outer radicand of the form a+b^(k/n) are quite rare when n>2. Citation needed. --rwg
http://www.math.harvard.edu/~elkies/trinomial.html gives a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2
but my sextic solver fails for u≠0. Can anybody propose a correction?
I found a djvu of B.H. Matzat: *Konstruktive Galoistheorie*, LNM 1284 (1987), where Elkies claims he got the sextic, but instead Matzat has the two families x^6 + (200000 (5 + u) (8 + u)^2 (-5 + 6 x))/((-120 + u) u^5) and x^6 - (84375 (-16 + v) v^3 (-5 + 6 x))/(4 - 14 v + v^2)
Maybe one of these is Elkies's under some tricky change of variable?
But woe, my sextic solver fails on these as well! Perhaps it's just broken. But it works on the cases I can find which are substantiated with actual solutions, leaving me to wonder if *both* Elkies and Matzat failed to computer-check their drafts. --rwg
On Fri, Aug 23, 2013 at 11:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
Early in the sequence of solvable quintic trinomials comes In[105]:= Squint[5*x^5 + 6*x + 6];
whose only real root is
In[107]:= FullSimplify /@ %105[[4]]
Out[107]= -(1/5) 2^(3/5) 3^(1/5) - 1/5 2^(1/5) 3^(2/5) - 1/5 2^(4/5) 3^(3/5) + 1/5 2^(2/5) 3^(4/5)
Raising this to consecutive powers, we find the third gives a binomial:
In[108]:= Table[FullSimplify[%^k], {k, 3}]
Out[108]= {Root[6 + 6 #1 + 5 #1^5 &, 1], Root[-36 + 36 #1 + 60 #1^3 + 25 #1^5 &, 1], 1/5 2^(3/5) 3^(1/5) (-3 + 2^(1/5) 3^(2/5))}
thus rediscovering one of my favorite denestings: (-2^(1/5) + 3^(3/5))^(1/3) == (2^(2/5) + 3^(1/5) + 2^(3/5) 3^(2/5) - 2^(1/5) 3^(3/5))/5^(2/3) --rwg
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