Bill Gosper <billgosper@gmail.com> wrote:
Wouter> Bill,
I tried it, but no go:
On Tue, Aug 23, 2011 at 6:10 AM, Bill Gosper <billgosper@gmail.com> wrote:
[...]
should work as was advertised, but alas, I've found a screw case: Squint[9 - 84 x - 80 x^2 + 1024 x^3 - 256 x^4 - 1024 x^5] which should give radicals for Cos[2*Pi/41]*Cos[4*Pi/41] - Sin[5*Pi/82]*Cos[5*Pi/41], gives nonsense which is only a correct root for %@1 . I may have to abandon condensing all five roots into an f[Range[5]] and just return all five in a list.
Like so:
Squint[q_] := Block[{rts = #[[1, 2]] & /@ NSolve[q == 0, WorkingPrecision -> 69], boa, foos, S3 = Permutations[Range[3]]}, foos = InverseFourier[ rts[[#]]/Sqrt[5]]^5 & /@ (Join[{}, #] & /@ (Permutations[Range[5]])); foos = (Together[#[[1, 2]]]^(1/5) &) /@ Solve[0 ==Rationalize[ Expand[Times @@ (# - Rest[foos[[Ordering[ Denominator[Rationalize[Plus @@ #, 9.^-69]] & /@ foos, 1][[1]]]]])]]]; boa = Rationalize[Plus @@ rts/5]; boa + foos.# & /@Select[fiveto4th, MemberQ[Chop[#.foos + boa - rts], 0] &]]
In[112]:= (fiveto4th = Exp[2*I*\[Pi]*Flatten[Outer[List, Range[0, 4], Range[0, 4], Range[0, 4], Range[0, 4]], 3]/5]) // Short
Out[112]{{1,1,1,1},<<623>>,{E^(-((2 I Pi)/5)),E^(-(<<1>>/5)), E^(-(<<1>>/5)),E^(-((2<<1>><<1>>Pi)/5))}}
In[116]:= DeleteCases[ Flatten[Outer[ Equal, {Cos[((8*Pi)/41)]*Cos[((10*Pi)/41)] + Sin[((7*Pi)/82)]*Sin[((19*Pi)/82)], Sin[((3*Pi)/82)]*Cos[((7*Pi)/41)] + Sin[Pi/82]*Sin[((9*Pi)/82)], Sin[Pi/82]*Cos[((10*Pi)/41)] + Sin[((9*Pi)/82)]*Cos[((8*Pi)/41)], -Sin[((9*Pi)/82)]* Cos[((9*Pi)/41)] - Sin[Pi/82]*Cos[Pi/41], -Sin[((5*Pi)/82)]*Cos[((9*Pi)/41)] - Cos[Pi/41]*Cos[((2*Pi)/41)]}, Squint[9 - 84 x - 80 x^2 + 1024 x^3 - 256 x^4 - 1024 x^5]]], False]
Out[116]= {Cos[(8 \[Pi])/41] Cos[(10 \[Pi])/41] + Sin[(7 \[Pi])/82] Sin[(19 \[Pi])/82] == -(1/ 20) + (41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^( 1/5)/(20 2^( 2/5)) + (41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5)/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^(-((2 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^((2 I \[Pi])/5))/( 20 2^(2/5)), Cos[(7 \[Pi])/41] Sin[(3 \[Pi])/82] + Sin[\[Pi]/82] Sin[(9 \[Pi])/82] == -(1/ 20) + (41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^( 1/5)/(20 2^( 2/5)) + (41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5)/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^(-((4 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^((4 I \[Pi])/5))/( 20 2^(2/5)), Cos[(10 \[Pi])/41] Sin[\[Pi]/82] + Cos[(8 \[Pi])/41] Sin[(9 \[Pi])/82] == -(1/ 20) + ((41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^(-((2 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^(-((2 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^((2 I \[Pi])/5))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^((2 I \[Pi])/5))/( 20 2^(2/5)), -Cos[\[Pi]/41] Sin[\[Pi]/82] - Cos[(9 \[Pi])/41] Sin[(9 \[Pi])/82] == -(1/ 20) + ((41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^(-((2 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^((2 I \[Pi])/5))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^(-((4 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^((4 I \[Pi])/5))/( 20 2^(2/5)), -Cos[\[Pi]/41] Cos[(2 \[Pi])/41] - Cos[(9 \[Pi])/41] Sin[(5 \[Pi])/82] == -(1/ 20) + ((41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^(-((4 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^(-((4 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^((4 I \[Pi])/5))/( 20 2^(2/5)) + ((41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5) E^((4 I \[Pi])/5))/( 20 2^(2/5))}
Apologies. And thanks for trying it!
(Anyone else?)
--rwg
Dear Bill, I was wondering what Mathemaica's take and research on quintics relates to what you're doing? Best, John