[math-fun] quintic solver safety recall, sin 1 (almost)
That terse solver sailed right through the examples in Quintic function - Wikipedia, the free encyclopedia<http://en.wikipedia.org/wiki/Quintic_function> until x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1, whose sol'ns are 2*Cos[2*k*π/11], 0<k<11. Watson-Williams's correct solution gives a quartic with four roots all of magnitude (5/Sqrt[11])^5, whereas the terse one finds four *different* roots of that magnitude, and then oviposits when it can't figure out the phase corrections. Fudged manually, the first solution gives Out[859]= Sin[7 Pi/22] == -1/10 + ( (-1)^( 2/5) (11 (-89 - 25 Sqrt[5] - 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) - (-1)^( 3/5) (11 (-89 - 25 Sqrt[5] + 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) + (-1)^(2/5) (11 (-89 + 25 Sqrt[5] - 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5) - (-1)^(3/5) (11 (-89 + 25 Sqrt[5] + 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5))/ (10 2^(2/5)) In[860]:= N[{%[[1]]], %[[2]], Sin[1]}, 22] Out[860]= {0.8412535328311811688618, 0.8412535328311811688618 + 0.*10^-23 I, 0.8414709848078965066525} Titus Piezas, a name I wish I'd learned much earlier, has an apparently much neater rigorous solution http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.2.3197&rep=rep1&typ... which he found by walking through a 22nd degree wall with computer algebra. It will be interesting to try on 2 cos(2k pi/11). His paper cites http://www.math.carleton.ca/~williams/papers/pdf/244.pdf but apparently slightly predates http://www.math.carleton.ca/~williams/papers/pdf/276.pdf He seems to have only one exceptional case, and not even that if you always take a limit where you might divide by 0. --rwg PS, I believe the image WFL is trying to attach is http://gosper.org/flowsnakefill_4.gif . What I'm looking for is a recursive construction of (one third of) http://gosper.org/hexflo.png (to arbitrary order) by clustering solid color tiles. PPS, there's a much higher boundary dimension variant of the France fractal, grouped three around three around one vs six around one. I'll try to unearth one.
[...]
PPS, there's a much higher boundary dimension variant of the France fractal, grouped three around three around one vs six around one. I'll try to unearth one.
http://gosper.org/fjord.png (Conjugate for base 2+omega.) An automatic, closed loop spacefill for all such base|digit systems is to "sausage" the natural tree induced by the recursion. (E.g. the boundary of http://www.tweedledum.com/rwg/7posies.bmp is a closed loop spacefill of "France".) --rwg
2011/8/13 Bill Gosper <billgosper@gmail.com>
That terse solver sailed right through the examples in Quintic function - Wikipedia, the free encyclopedia<http://en.wikipedia.org/wiki/Quintic_function> until x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1, whose sol'ns are 2*Cos[2*k*π/11], 0<k<11. Watson-Williams's correct solution gives a quartic with four roots all of magnitude (5/Sqrt[11])^5, whereas the terse one finds four *different* roots of that magnitude, and then oviposits when it can't figure out the phase corrections. Fudged manually, the first solution gives
Out[859]= Sin[7 Pi/22] == -1/10 + ( (-1)^( 2/5) (11 (-89 - 25 Sqrt[5] - 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) - (-1)^( 3/5) (11 (-89 - 25 Sqrt[5] + 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) + (-1)^(2/5) (11 (-89 + 25 Sqrt[5] - 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5) - (-1)^(3/5) (11 (-89 + 25 Sqrt[5] + 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5))/ (10 2^(2/5))
In[860]:= N[{%[[1]]], %[[2]], Sin[1]}, 22]
Out[860]= {0.8412535328311811688618, 0.8412535328311811688618 + 0.*10^-23 I, 0.8414709848078965066525} [...]
This is shameless, but seems to work: In[375]:= 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 = (fifth[Together[#[[1, 2]]]] &) /@ Solve[0 == Rationalize[ Expand[Times @@ (# - Rest[foos[[Ordering[ Denominator[Rationalize[Plus @@ #, 9.^-69]] & /@ foos, 1][[1]]]]])]]]; boa = Rationalize[Plus @@ rts/5]; Evaluate[boa + foos.Select[ Exp[2*I*\[Pi]*Join[{}, #] & /@ Permutations[Range[4]]/5], MemberQ[Chop[#.foos + boa - rts], 0] &][[1]]^#] &] In[371]:= fifth[z_] := If[Sign[z] == -1, -(-z)^(1/5), z^(1/5)] In[377]:= Squint[-3 - 5 x^2 + x^5] Out[377]= ((25 - 5 Sqrt[5] + Sqrt[30 (25 - 11 Sqrt[5])])^( 1/5) (E^(-((2 I \[Pi])/5)))^#1)/10^( 2/5) + ((25 - 5 Sqrt[5] - Sqrt[30 (25 - 11 Sqrt[5])])^( 1/5) (E^((2 I \[Pi])/5))^#1)/10^( 2/5) - ((-25 - 5 Sqrt[5] + Sqrt[30 (25 + 11 Sqrt[5])])^( 1/5) (E^(-((4 I \[Pi])/5)))^#1)/10^( 2/5) + ((25 + 5 Sqrt[5] + Sqrt[30 (25 + 11 Sqrt[5])])^( 1/5) (E^((4 I \[Pi])/5))^#1)/10^(2/5) & These triggy cases used to fail: Out[419]= Cos[1/31 2^(11 + 2 n) 3^(5 + n) \[Pi]] + Cos[1/31 2^(21 + 2 n) 3^(10 + n) \[Pi]] + Cos[1/31 2^(31 + 2 n) 3^(15 + n) \[Pi]] == -1/10 + ((31 I (409 I + 125 I Sqrt[5] + 5 Sqrt[10 (925 - 409 Sqrt[5])]))^(1/5) (E^(-((2 I \[Pi])/5)))^ n)/(10 2^( 2/5)) + ((-31 I (-409 I - 125 I Sqrt[5] + 5 Sqrt[10 (925 - 409 Sqrt[5])]))^(1/5) (E^((2 I \[Pi])/5))^n)/( 10 2^(2/5)) + ((31 (-409 + 125 Sqrt[5] - 5 I Sqrt[10 (925 + 409 Sqrt[5])]))^( 1/5) (E^(-((4 I \[Pi])/5)))^n)/(10 2^(2/5)) + ((31 (-409 + 125 Sqrt[5] + 5 I Sqrt[10 (925 + 409 Sqrt[5])]))^(1/5) (E^((4 I \[Pi])/5))^ n)/(10 2^(2/5)) In[420]:= N[Table[%, {n, 6}]] Out[420]= {True, True, True, True, True, True} In[422]:= TableForm@ Table[FindIntegerNullVector[(Cos[1/31 2^(11 + 2 n) 3^(5 + n) \[Pi]] + Cos[1/31 2^(21 + 2 n) 3^(10 + n) \[Pi]] + Cos[1/31 2^(31 + 2 n) 3^(15 + n) \[Pi]])^-Range[0, 5]], {n, 5}] Out[422] -32 -16 96 84 -2 -5 -32 -16 96 84 -2 -5 -32 -16 96 84 -2 -5 32 16 -96 -84 2 5 -32 -16 96 84 -2 -5 I.e., these five trig trinomials all satisfy In[425]:= Expand[FromDigits[%422[[4]], x]] Out[425]= 5 + 2 x - 84 x^2 - 96 x^3 + 16 x^4 + 32 x^5 In[426]:= Squint[%425] Out[426]= -1/10 + ((31 I (409 I + 125 I Sqrt[5] + 5 Sqrt[10 (925 - 409 Sqrt[5])]))^( 1/5) (E^(-((2 I \[Pi])/5)))^#1)/(10 2^(2/5)) + ((-31 I (-409 I - 125 I Sqrt[5] +5 Sqrt[10 (925 - 409 Sqrt[5])]))^( 1/5) (E^((2 I \[Pi])/5))^#1)/(10 2^(2/5)) + ((31 (-409 + 125 Sqrt[5] - 5 I Sqrt[10 (925 + 409 Sqrt[5])]))^( 1/5) (E^(-((4 I \[Pi])/5)))^#1)/(10 2^(2/5)) + ((31 (-409 + 125 Sqrt[5] + 5 I Sqrt[10 (925 + 409 Sqrt[5])]))^( 1/5) (E^((4 I \[Pi])/5))^#1)/(10 2^(2/5)) & Caution: Internal WorkingPrecision still doesn't scale with problem size, so sufficiently large coefficients will likely fail. --rwg (with the usual huge help from Julian) PS, I keep forgetting to mention that APGoucher's electrocuted geometer scenario is the default when using my aluminum laptop in my Gypsy-wired bedroom. And I'm not even completing a circuit-- just capacitance to the waterbed, I guess.
Bill, I tried it, but no go: Evaluation of Squint leads to an unsucessful evaluation of NSolve[q == 0,boa, WorkingPrecision -> 69] and "boa" is undefined at that point (* while inside the local variables definitions of Block *) This rather constricts my Mma 4.0 and it closes down. Does it really run ok on Mma8.0? Wouter. ----- Original Message ----- From: "Bill Gosper" <billgosper@gmail.com> To: <math-fun@mailman.xmission.com> Sent: Monday, August 22, 2011 11:45 PM Subject: Re: [math-fun] quintic solver safety recall, sin 1 (almost)
This is shameless, but seems to work: In[375]:= 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 = (fifth[Together[#[[1, 2]]]] &) /@ Solve[0 == Rationalize[ Expand[Times @@ (# - Rest[foos[[Ordering[ Denominator[Rationalize[Plus @@ #, 9.^-69]] & /@ foos, 1][[1]]]]])]]]; boa = Rationalize[Plus @@ rts/5]; Evaluate[boa + foos.Select[ Exp[2*I*\[Pi]*Join[{}, #] & /@ Permutations[Range[4]]/5], MemberQ[Chop[#.foos + boa - rts], 0] &][[1]]^#] &]
In[371]:= fifth[z_] := If[Sign[z] == -1, -(-z)^(1/5), z^(1/5)]
In[377]:= Squint[-3 - 5 x^2 + x^5]
Bill,
I tried it, but no go: Evaluation of Squint leads to an unsucessful evaluation of NSolve[q == 0,boa, WorkingPrecision -> 69] and "boa" is undefined at that point (* while inside the local variables definitions of Block *) This rather constricts my Mma 4.0 and it closes down. Does it really run ok on Mma8.0?
Wouter.
AARRGGHH!! I'm a careless idiot. This 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 = (fifth[Together[#[[1, 2]]]] &) /@ Solve[0 == Rationalize[ Expand[Times @@ (# - Rest[foos[[Ordering[ Denominator[Rationalize[Plus @@ #, 9.^-69]] & /@ foos, 1][[1]]]]])]]]; boa = Rationalize[Plus @@ rts/5]; Evaluate[boa + foos.Select[ Exp[2*I*\[Pi]*Join[{}, #] & /@ Permutations[Range[4]]/5], MemberQ[Chop[#.foos + boa - rts], 0] &][[1]]^#] &] fifth[z_] := If[Sign[z] == -1, -(-z)^(1/5), z^(1/5)] 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. Apologies. And thanks for trying it! --rwg
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
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
Bill Gosper <billgosper@gmail.com> wrote:
[...]
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)),[...]}
Note that Out[125]= Sin[a] Sin[b] + Sin[c] Sin[d] == Cos[a/2 - b/2 + c/2 - d/2] Cos[a/2 - b/2 - c/2 + d/2] - Cos[a/2 + b/2 - c/2 - d/2] Cos[a/2 + b/2 + c/2 + d/2] == Sin[a/2 + b/2 + c/2 - d/2] Sin[a/2 + b/2 - c/2 + d/2] - Sin[a/2 - b/2 - c/2 - d/2] Sin[a/2 - b/2 + c/2 + d/2] In[126]:= Simplify[%] Out[126]= True so there are three ways to write each of those five identities.
Apologies. And thanks for trying it!
(Anyone else?)
--rwg
Dear Bill, I was wondering what Mathema[t]ica's take and research on quintics relates to what you're doing? Best, John
Because being able to solve quintics intoxicates me with power! Seriously, they come up all the time, e.g. with D. Cantrell's disk packings, Corey's denesting adventures, recent offlist discussion with Ed Fredkin,... And they're intrinsically interesting. Rich points out that we can write a 1/5th angle formula sin a/5 = quinticsolution(trigs(a)) I've long been amazed and frustrated by the absence of quintic solvers from available CASes. Apparently, Mathematica still lacks GaloisGroup. I guess Mathematica's take is that anybody in his right mind uses Root. --rwg
On Thu, Aug 25, 2011 at 12:09 PM, Bill Gosper <billgosper@gmail.com> wrote:
Bill Gosper <billgosper@gmail.com> wrote:
[...]
Rich points out that we can write a 1/5th angle formula sin a/5 = quinticsolution(trigs(a))
Oh, foo, it's going to come down to complex 5th roots anyway: Out[271]= Cos[x/5] == 1/2 ((Cos[x] - I Sin[x])^(1/5) + (Cos[x] + I Sin[x])^(1/5)) In[272]:= FullSimplify[%271, \[Pi] > x > -\[Pi]] Out[272]= True [...]
--rwg
participants (3)
-
Bill Gosper -
jdb＠math.arizona.edu -
wouter meeussen