"Exhaustive" search (81 cases) sez there are no solvable irreducible quintics with coeffs restricted to {-1,0,1}. After x^5-2 (modulo root negation and reciprocation), the smallest, in some sense, seems to be -1 - 2 x - x^2 + x^3 + x^4 + x^5, whose five solutions are -1/5 +w*u_00+w^2*u_01+w^3*u_10+w^4*u_11, where w = e^(2 pi i n/5), n an integer mod 5, and u_jk := j j k (-1) 1371 - 145 (-1) (sqrt 5) + 65 (-1) sqrt(470 (1 - ------)) sqrt 5 R(-----------------------------------------------------------) 12500 where R is the *real* fifth root of a real: R(#):=#/(#^4)^(1/5). If you use the default complex 5th root, the negative ones will need a correction factor of e^(4 i pi/5). Beware: This is the kind of 5th root they mean in the Wikipedia article<http://en.wikipedia.org/wiki/Quintic_function> and in the examples in the (glitchy) Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf> . (InputForm in case Gmail murdered the above: (1371 - 145*(-1)^j*Sqrt[5] + 65*(-1)^k*Sqrt[470]*Sqrt[1 - (-1)^j/Sqrt[5]])/12500 ). The u_jk^5 are the (real) roots of the quartic 1 + 325793750 u + 4331113281250 u^2 - 41839599609375 u^3 + 95367431640625 u^4 (Recall that the quintic only had 2s and 1s.) So a convenient way to exhibit the solution to an irreducible quintic ax^5+bx^4+... is the constant -b/(5a) + one of these quartics. Recall that quartic solutions in general have quadratic surds inside cubic surds inside quartic surds, so for the quintic we appear doomed to suffer quadratic surds inside cubic surds inside ... inside quintic surds. But no! For some magic Galwological reason, these quartics seem always to solve down to (real!) quadratics in quadratics, at worst. Sometimes, as with -3 + 3 x + 3 x^2 + 3 x^3 + 3 x^4 + x^5, the quartic even factors: (59049 - 12543750 u + 9765625 u^2) (-7776 + 1050000 u + 9765625 u^2). But usually it doesn't even decompose, so your quartic solver needs to be smart. --rwg Is there a name for these nice quartics? They're not usually biquadratic. How about "nice"?-) PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>.
On 8/6/11, Bill Gosper <billgosper@gmail.com> wrote:
"Exhaustive" search (81 cases) sez there are no solvable irreducible quintics with coeffs restricted to {-1,0,1}. After x^5-2 (modulo root negation and reciprocation), the smallest, in some sense, seems to be -1 - 2 x - x^2 + x^3 + x^4 + x^5, whose five solutions are -1/5 +w*u_00+w^2*u_01+w^3*u_10+w^4*u_11, where w = e^(2 pi i n/5), n an integer mod 5, and u_jk :=
j j k (-1) 1371 - 145 (-1) (sqrt 5) + 65 (-1) sqrt(470 (1 - ------)) sqrt 5 R(-----------------------------------------------------------) 12500
where R is the *real* fifth root of a real: R(#):=#/(#^4)^(1/5). If you use the default complex 5th root, the negative ones will need a correction factor of e^(4 i pi/5). Beware: This is the kind of 5th root they mean in the Wikipedia article<http://en.wikipedia.org/wiki/Quintic_function> and in the examples in the (glitchy) Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf> .
(InputForm in case Gmail murdered the above: (1371 - 145*(-1)^j*Sqrt[5] + 65*(-1)^k*Sqrt[470]*Sqrt[1 - (-1)^j/Sqrt[5]])/12500 ).
The u_jk^5 are the (real) roots of the quartic
1 + 325793750 u + 4331113281250 u^2 - 41839599609375 u^3 + 95367431640625 u^4
(Recall that the quintic only had 2s and 1s.) So a convenient way to exhibit the solution to an irreducible quintic ax^5+bx^4+... is the constant -b/(5a) + one of these quartics. Recall that quartic solutions in general have quadratic surds inside cubic surds inside quartic surds, so for the quintic we appear doomed to suffer quadratic surds inside cubic surds inside ... inside quintic surds. But no! For some magic Galwological reason, these quartics seem always to solve down to (real!) quadratics in quadratics, at worst. Sometimes, as with -3 + 3 x + 3 x^2 + 3 x^3 + 3 x^4 + x^5, the quartic even factors: (59049 - 12543750 u + 9765625 u^2) (-7776 + 1050000 u + 9765625 u^2). But usually it doesn't even decompose, so your quartic solver needs to be smart. --rwg Is there a name for these nice quartics? They're not usually biquadratic. How about "nice"?-) PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Apologies for previous empty post (dodgy mouse disease) ... On 8/6/11, Bill Gosper <billgosper@gmail.com> wrote:
PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>.
I noticed the same thing (briefly) using Adobe Reader 7 --- file length is a remarkably generous 4.6Mbyte, apparently after scanning at high resolution, which probably probably accounts for the conniptions. WFL
It seems you still need to screw with 5th roots of unity, as I did in this rather terse quintic solver based on reconstructing the quartic from the approximate inverse Fourier transformation of the solution vector of the quintic: Squint[q_] := Block[{rts = #[[1, 2]] & /@ NSolve[q == 0, WorkingPrecision -> 69],foos, S3 = Permutations[Range[3]]}, foos = InverseFourier[rts[[#]]/Sqrt[5]]^5 & /@ (Join[{1, 2}, #] & /@ (2 + S3)); foos = (Sign[#[[1, 2]]]*Abs[#[[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]; Evaluate[boa + foos.Select[Exp[2*I*\[Pi]*Join[{1}, #] & /@ (1 + S3)/5], MemberQ[Chop[#.foos + boa - rts], 0] &][[1]]^#] &] E.g., In[498]:= Squint[x^5+3*(x^4+x^3+x^2+x-1)] Out[498]= -(3/5)+1/5 3^(2/5) (223+70 Sqrt[10])^(1/5) (E^(-((2 I \[Pi])/5)))^#1+1/5 3^(2/5) (223-70 Sqrt[10])^(1/5) (E^((2 I \[Pi])/5))^#1+1/5 2^(2/5) (3 (-14+5 Sqrt[10]))^(1/5) (E^(-((4 I \[Pi])/5)))^#1-1/5 2^(2/5) (3 (14+5 Sqrt[10]))^(1/5) (E^((4 I \[Pi])/5))^#1& In[499]:= Union[MinimalPolynomial/@%/@Range[5]] Out[499]= {-3+3 #1+3 #1^2+3 #1^3+3 #1^4+#1^5&} In[512]:= Squint[52+39 y+2 y^5] Out[512]= -(((13/2 (-5+10 Sqrt[10]-Sqrt[1/5 (5125-7197/Sqrt[10])]))^(1/5) (E^(-((2 I \[Pi])/5)))^#1)/5^(3/5))-((13/2 (-5+10 Sqrt[10]+Sqrt[1/5 (5125-7197/Sqrt[10])]))^(1/5) (E^((2 I \[Pi])/5))^#1)/5^(3/5)+(E^(-((4 I \[Pi])/5)))^#1/(5^(2/5) (2/(13+26 Sqrt[10]+25 Sqrt[6929/625+1216293/(78125 Sqrt[10])]))^(1/5))-(E^((4 I \[Pi])/5))^#1/(5^(2/5) (2/(-13-26 Sqrt[10]+25 Sqrt[6929/625+1216293/(78125 Sqrt[10])]))^(1/5))& In[515]:= 52+39 y+2 y^5/.y->#&/@N[%512/@Range[5],69] Out[515]= {0.*10^-67+0.*10^-67 I,0.*10^-66+0.*10^-66 I,0.*10^-66+0.*10^-66 I,0.*10^-67+0.*10^-67 I,0.*10^-68} In[511]:= Squint[x^5-5*x+12] Out[511]= (-1+2/Sqrt[5]+3/5 Sqrt[5-11/Sqrt[5]])^(1/5) (E^(-((2 I \[Pi])/5)))^#1-(1-2/Sqrt[5]+3/5 Sqrt[5-11/Sqrt[5]])^(1/5) (E^((2 I \[Pi])/5))^#1-(E^(-((4 I \[Pi])/5)))^#1/(2/(2+4/Sqrt[5]+Sqrt[36/5+396/(25 Sqrt[5])]))^(1/5)-(E^((4 I \[Pi])/5))^#1/(2/(2+4/Sqrt[5]-Sqrt[36/5+396/(25 Sqrt[5])]))^(1/5)& In[514]:= MinimalPolynomial/@%511/@Range[4,5] During evaluation of In[514]:= MinimalPolynomial::nalg: -(1-2/Sqrt[5]+3/5 Sqrt[5+Times[<<2>>]])^(1/5) E^(-((2 I \[Pi])/5))+(-1+2/Sqrt[5]+3/5 Sqrt[5+Times[<<2>>]])^(1/5) E^((2 I \[Pi])/5)-E^(-((4 I \[Pi])/5))/(2/(2+<<1>><<1>>Power[<<1>>]))^(1/5)-E^((4 I \[Pi])/5)/(2/(2+<<1>>+Sqrt[Plus[<<2>>]]))^(1/5) is not an explicit algebraic number. >> Out[514]= {MinimalPolynomial[-(1-2/Sqrt[5]+3/5 Sqrt[5-11/Sqrt[5]])^(1/5) E^(-((2 I \[Pi])/5))+(-1+2/Sqrt[5]+3/5 Sqrt[5-11/Sqrt[5]])^(1/5) E^((2 I \[Pi])/5)-E^(-((4 I \[Pi])/5))/(2/(2+4/Sqrt[5]-Sqrt[36/5+396/(25 Sqrt[5])]))^(1/5)-E^((4 I \[Pi])/5)/(2/(2+4/Sqrt[5]+Sqrt[36/5+396/(25 Sqrt[5])]))^(1/5)],12-5 #1+#1^5&} (Mma bizarreness.) In releasable code, the builtin WorkingPrecision of 69 digits clearly needs to scale with the input coefficients. It might be possible to avoid approximate arithmetic entirely if only InverseFourier didn't impose it. --rwg On Sat, Aug 6, 2011 at 3:11 PM, Bill Gosper <billgosper@gmail.com> wrote:
"Exhaustive" search (81 cases) sez there are no solvable irreducible quintics with coeffs restricted to {-1,0,1}. After x^5-2 (modulo root negation and reciprocation), the smallest, in some sense, seems to be -1 - 2 x - x^2 + x^3 + x^4 + x^5, whose five solutions are -1/5 +w*u_00+w^2*u_01+w^3*u_10+w^4*u_11, where w = e^(2 pi i n/5), n an integer mod 5, and u_jk :=
j j k (-1) 1371 - 145 (-1) (sqrt 5) + 65 (-1) sqrt(470 (1 - ------)) sqrt 5 R(-----------------------------------------------------------) 12500
where R is the *real* fifth root of a real: R(#):=#/(#^4)^(1/5). If you use the default complex 5th root, the negative ones will need a correction factor of e^(4 i pi/5). Beware: This is the kind of 5th root they mean in the Wikipedia article<http://en.wikipedia.org/wiki/Quintic_function> and in the examples in the (glitchy) Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf> .
(InputForm in case Gmail murdered the above: (1371 - 145*(-1)^j*Sqrt[5] + 65*(-1)^k*Sqrt[470]*Sqrt[1 - (-1)^j/Sqrt[5]])/12500 ).
The u_jk^5 are the (real) roots of the quartic
1 + 325793750 u + 4331113281250 u^2 - 41839599609375 u^3 + 95367431640625 u^4
(Recall that the quintic only had 2s and 1s.) So a convenient way to exhibit the solution to an irreducible quintic ax^5+bx^4+... is the constant -b/(5a) + one of these quartics. Recall that quartic solutions in general have quadratic surds inside cubic surds inside quartic surds, so for the quintic we appear doomed to suffer quadratic surds inside cubic surds inside ... inside quintic surds. But no! For some magic Galwological reason, these quartics seem always to solve down to (real!) quadratics in quadratics, at worst. Sometimes, as with -3 + 3 x + 3 x^2 + 3 x^3 + 3 x^4 + x^5, the quartic even factors: (59049 - 12543750 u + 9765625 u^2) (-7776 + 1050000 u + 9765625 u^2). But usually it doesn't even decompose, so your quartic solver needs to be smart. --rwg Is there a name for these nice quartics? They're not usually biquadratic. How about "nice"?-) PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>.
On Sat, Aug 6, 2011 at 3:11 PM, Bill Gosper <billgosper@gmail.com>wrote:[...] Recall that quartic solutions in general have quadratic surds inside cubic
surds inside quartic surds,
No, Julian points out that In[642]:= Position[Solve[x^4 + c*x^2 + d*x + e == 0, x][[1]], _^(1/4)|_^(-1/4)] Out[642]= {} I.e. Mathematica does it with square roots of square roots of cube roots of square roots. However in Macsyma, reveal(solve(x^4+c*x^2+d*x+e,x)[1],7) Sum(4) sqrt(------) sqrt(- Quotient) Expt x = - ---------------- - ------------ 1/4 6 6 Quotient --rwg
From the header line "quintic trivia", the resolvent quartic of the solvable quintic needs only square roots of square roots (no cube roots).
I believe it is the same with the next Fermat degree, namely the 17-deg. It's resolvent 16-deg needs only square roots of square roots of square roots, etc. Here's a "natural" solvable 17-th deg with small coefficients: -1 + x + x^2 + x^3 - x^5 - 13 x^6 - 7 x^7 - 11 x^8 - 4 x^9 - x^10 - 7 x^11 - 23 x^12 - 31 x^13 - 42 x^14 - 24 x^15 - 6 x^16 + x^17 = 0 It's unique real root is exactly given by (in Mathematica) as x = zeta_48 DedekindEta[tau]/(Sqrt[2]DedekindEta[2tau]) = 9.1630942... with the root of unity zeta_48 = exp(2Pi I/48), tau = (1+Sqrt[-d])/2, and d = 383. This d has class number h(-d) = 17. I was able to find it's 16-deg resolvent, but I don't have my notes here with me. :-( - Tito On Sun, Sep 4, 2011 at 4:34 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Sat, Aug 6, 2011 at 3:11 PM, Bill Gosper <billgosper@gmail.com>wrote:[...]
Recall that quartic solutions in general have quadratic surds inside cubic
surds inside quartic surds,
No, Julian points out that In[642]:= Position[Solve[x^4 + c*x^2 + d*x + e == 0, x][[1]], _^(1/4)|_^(-1/4)]
Out[642]= {}
I.e. Mathematica does it with square roots of square roots of cube roots of square roots. However in Macsyma, reveal(solve(x^4+c*x^2+d*x+e,x)[1],7) Sum(4) sqrt(------) sqrt(- Quotient) Expt x = - ---------------- - ------------ 1/4 6 6 Quotient --rwg
participants (3)
-
Bill Gosper -
Fred lunnon -
Tito Piezas