(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. 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The differential resolvent of this quintic is a linear differential equation of order 4 of the form: :[font = input; preserveAspect] diffeqn = a1 t''''[rho] + a2 t'''[rho] + a3 t''[rho] + a4 t'[rho] + a5 t[rho] + a6 == 0 :[font = text; inactive; preserveAspect] The coefficients aj are polynomial functions in rho. We will find them explicitly in the next several steps. We differentiate the quintic equation eqn with respect to rho: ;[s] 3:0,0;147,1;150,2;171,-1; 3:1,11,8,Times,0,12,0,0,0;1,10,8,Courier,1,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = input; preserveAspect] deriv = Flatten[Table[ Solve[ D[eqn, {rho, k}], D[t[rho], {rho, k}]], {k, 1, 4}]] :[font = text; inactive; preserveAspect] We substitute these derivatives into the differential equation diffeqn: :[font = input; preserveAspect] algeqn = Simplify[diffeqn //. deriv] :[font = text; inactive; preserveAspect] We replace powers t[rho]^k with exponents k greater than 4 using the original quintic equation: :[font = input; preserveAspect] algeqn = Collect[Numerator[algeqn], t[rho] ] :[font = text; inactive; preserveAspect] This gives a new algebraic equation with respect to t[rho] of degree 4. We set the coefficients of the powers of t[rho] to zero: :[font = input; preserveAspect] algeqn = (# == 0)& /@ CoefficientList[algeqn, t[rho]] :[font = text; inactive; preserveAspect] This is a system of linear equations for the coefficients aj. Its solution is: :[font = input; preserveAspect] coeffs = Solve[algeqn, {a1,a2,a3,a4,a5,a6}] // Simplify :[font = text; inactive; preserveAspect] Hence the differential resolvent is: :[font = input; preserveAspect] diffeqn = First[diffeqn /. coeffs] :[font = text; inactive; preserveAspect] We solve this differntial equation: :[font = input; preserveAspect] < 1 :[font = input; preserveAspect] algeqn = eqn /. approximation :[font = text; inactive; preserveAspect] Now we collect like terms in rho and set the coefficients to zero. We solve the resulting system of four linear equations for the coefficients C[j]: :[font = input; preserveAspect] system = Map[# == 0 &, Take[CoefficientList[First[algeqn], rho], 4]] :[font = input; preserveAspect] coeffs = Solve[system, Table[C[j], {j, 4}]] :[font = text; inactive; preserveAspect] Finally we obtain the five roots of the given quintic: :[font = input; preserveAspect; endGroup] First[solution] /. coeffs :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Second example :[font = text; inactive; preserveAspect] Here we consider another example with an arbitray parameter z: :[font = input; preserveAspect] x^5 - 5 x^3 + 5 x - z = 0 :[font = text; inactive; preserveAspect] Surprisingly this equation is solvable by radicals for any values of z. We construct the differential resolvent associated with this equation as above: :[font = input; preserveAspect] 25 (4 - z^2) x''[z] - 25 z x'[z] + x[z] == 0 :[font = text; inactive; preserveAspect] Its general solution is: :[font = input; preserveAspect] DSolve[resolvent, x[z], z] :[font = text; inactive; preserveAspect] To determine the constants C[1] and C[2] we substitute this solution into the quintic and expand with respect to z at z = 0. Equating the first two terms of the expansion to zero, we obtain a system of two equations in C[1] and C[2]. Solving this system, we get the roots of the given quintic: :[font = input; preserveAspect; endGroup] {{x[z] -> 2 Sin[1/5 ArcSin[z/2]]}, {x[z] -> -((5 + 5^(1/2))/2)^(1/2) Cos[ArcSin[z/2]/5] + (-1 + 5^(1/2))/2 Sin[ArcSin[z/2]/5]}, {x[z] -> ((5 + 5^(1/2))/2)^(1/2) Cos[ArcSin[z/2]/5] + (-1 + 5^(1/2))/2 Sin[ArcSin[z/2]/5]}, {x[z] -> -((5 - 5^(1/2))/2)^(1/2) Cos[ArcSin[z/2]/5] - (1 + 5^(1/2))/2 Sin[ArcSin[z/2]/5]}, {x[z] -> ((5 - 5^(1/2))/2)^(1/2) Cos[ArcSin[z/2]/5] - (1 + 5^(1/2))/2 Sin[ArcSin[z/2]/5]} } :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Cubic and quartic :[font = text; inactive; preserveAspect] Cubic and quartic equations can be solved by the same method. Consider the cubic: :[font = input; preserveAspect] x^3 - x - z == 0 :[font = text; inactive; preserveAspect] Its solution is: :[font = input; preserveAspect] { {x[z] -> -z Hypergeometric2F1[2/3, 1/3, 3/2, 27 z^2/4]}, {x[z] -> Hypergeometric2F1[1/6, -1/6, 1/2, 27 z^2/4] + z/2 Hypergeometric2F1[2/3, 1/3, 3/2, 27 z^2/4]}, {x[z] -> -Hypergeometric2F1[1/6, -1/6, 1/2, 27 z^2/4] + z/2 Hypergeometric2F1[2/3, 1/3, 3/2, 27 z^2/4]} } :[font = text; inactive; preserveAspect] Consider the quartic: :[font = input; preserveAspect] x^4 - x - z == 0 :[font = text; inactive; preserveAspect] Its solution is: :[font = input; preserveAspect; endGroup; endGroup] { {x[z] -> -z HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -256 z^3/27]}, {x[z] -> HypergeometricPFQ[ {-1/12, 1/6, 5/12}, {1/3, 2/3}, -256 z^3/27] + z/3 HypergeometricPFQ[ {1/4, 1/2, 3/4}, {2/3, 4/3}, -256 z^3/27] - 2 z^2/9 HypergeometricPFQ[ {7/12, 5/6, 13/12}, {4/3, 5/3}, -256 z^3/27]}, {x[z] -> (-1)^(2/3) HypergeometricPFQ[ {-1/12, 1/6, 5/12}, {1/3, 2/3}, -256 z^3/27] + z/3 HypergeometricPFQ[ {1/4, 1/2, 3/4}, {2/3, 4/3}, -256 z^3/27] + 2 (-1)^(1/3) z^2/9 HypergeometricPFQ[ {7/12, 5/6, 13/12}, {4/3, 5/3}, -256 z^3/27]}, {x[z] -> -(-1)^(1/3) HypergeometricPFQ[ {-1/12, 1/6, 5/12}, {1/3, 2/3}, -256 z^3/27] + z/3 HypergeometricPFQ[ {1/4, 1/2, 3/4}, {2/3, 4/3}, -256 z^3/27] - 2 (-1)^(2/3) z^2/9 HypergeometricPFQ[ {7/12, 5/6, 13/12}, {4/3, 5/3}, -256 z^3/27]} } ^*)