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Solution of quintics based on differential equations
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Cockle (1860) and Harley (1862) developed a method for solving algebraic equations based on differential equations..
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First example
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We illustrate their approach with the quintic equation:
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eqn = t[rho]^5 - t[rho] - rho == 0;
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Each root is a function of the parameter rho. The differential resolvent of this quintic
is a linear differential equation of order 4 of the form:
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diffeqn = a1 t''''[rho] + a2 t'''[rho] + a3 t''[rho] +
a4 t'[rho] + a5 t[rho] + a6 == 0
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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;
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deriv = Flatten[Table[ Solve[ D[eqn, {rho, k}],
D[t[rho], {rho, k}]], {k, 1, 4}]]
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We substitute these derivatives into the differential equation diffeqn:
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algeqn = Simplify[diffeqn //. deriv]
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We replace powers t[rho]^k with exponents k greater than 4 using the original quintic equation:
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algeqn = Collect[Numerator[algeqn], t[rho] ]
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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:
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algeqn = (# == 0)& /@ CoefficientList[algeqn, t[rho]]
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This is a system of linear equations for the coefficients aj. Its solution is:
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coeffs = Solve[algeqn, {a1,a2,a3,a4,a5,a6}] // Simplify
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Hence the differential resolvent is:
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diffeqn = First[diffeqn /. coeffs]
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We solve this differntial equation:
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< 1
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algeqn = eqn /. approximation
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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]:
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system = Map[# == 0 &,
Take[CoefficientList[First[algeqn], rho], 4]]
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coeffs = Solve[system, Table[C[j], {j, 4}]]
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Finally we obtain the five roots of the given quintic:
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First[solution] /. coeffs
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Second example
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Here we consider another example with an arbitray parameter z:
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x^5 - 5 x^3 + 5 x - z = 0
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Surprisingly this equation is solvable by radicals for any values of z. We construct the differential resolvent associated with this equation as above:
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25 (4 - z^2) x''[z] - 25 z x'[z] + x[z] == 0
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Its general solution is:
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DSolve[resolvent, x[z], z]
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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:
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{{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]} }
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Cubic and quartic
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Cubic and quartic equations can be solved by the same method.
Consider the cubic:
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x^3 - x - z == 0
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Its solution is:
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{ {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]} }
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Consider the quartic:
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x^4 - x - z == 0
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Its solution is:
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{ {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]} }
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