Quintic Equation

A general quintic cannot be solved algebraically in terms of finite additions, multiplications, and root extractions, as rigorously demonstrated by Abel and Galois.

Euler reduced the general quintic to

$$x^5-10qx^2-p=0.$$ (1)

A quintic also can be algebraically reduced to Principal Quintic Form

$$x^5 + a_2 x^2 + a_1 x + a_0 = 0.$$ (2)

By solving a quartic, a quintic can be algebraically reduced to the Bring Quintic Form

$$x^5-x-a=0,$$ (3)

as was first done by Jerrard.

Consider the quintic

$$\prod_{j=0}^4 [x-(\omega^ju_1+\omega^{4j}u_2)]=0,$$ (4)

where $\omega=e^{2\pi i/5}$ and $u_1$ and $u_2$ are Complex Numbers. This is called de Moivre's Quintic. Generalize it to

$$\prod_{j=0}^4 [x-(\omega^ju_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4)]=0.$$ (5)

Expanding,

$(\omega^j u_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4)^5$
$-5U(\omega^ju_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4)^4$
$-5V(\omega^ju_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4)^2$
$+5W(\omega^ju_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4)$
$+[5(X-Y)-Z]=0,\quad$	(6)

where

$$U = u_1u_4+u2u_3$$ (7)

$$V = u_1{u_2}^2+u_2{u_4}^2+u_3{u_1}^2+u_4{u_3}^2$$ (8)

$$W = {u_1}^2{u_4}^2+{u_2}^2{u_3}^2-{u_1}^3u_2-{u_2}^3u_4-{u_3}^3u_1-{u_4}^3u_3-u_1u_2u_3u_4$$ (9)

$$X = {u_1}^3u_3u_4+{u_2}^3u_1u_3+{u_3}^3u_2u_4+{u_4}^3u_1u_2$$ (10)

$$Y = u_1{u_3}^2{u_4}^2+u_2{u_1}^2{u_3}^2+u_3{u_2}^2{u_4}^2+u_4{u_1}^2{u_2}^2$$ (11)

$$Z = {u_1}^5+{u_2}^5+{u_3}^5+{u_4}^5.$$ (12)

The $u_i$s satisfy

$\quad u_1u_4+u_2u_3=0$	(13)
$\quad u_1{u_2}^2+u_2{u_4}^2+u_3{u_1}^2+u_4{u_3}^2=0$	(14)
$\quad {u_1}^2{u_4}^2+{u_2}^2{u_3}^2-{u_1}^3u_2-{u_2}^3u_4-{u_3}^3u_1-{u_4}^3u_3-u_1u_2u_3u_4={\textstyle{1\over 5}} a$	(15)
$\quad 5[({u_1}^3u_3u_4+{u_2}^3u_1u_3+{u_3}^3u_3u_4+{u_4}^3u_1u_2)$
$\quad \phantom{=}-(u_1{u_3}^2{u_4}^2+u_2{u_1}^2{u_3}^2+u_3{u_2}^2{u_4}^2+u_4{u_1}^2{u_2}^2)]$
$\quad \phantom{=} -({u_1}^5+{u_2}^5+{u_3}^5+{u_4}^5)=b.$	(16)

Spearman and Williams (1994) show that an irreducible quintic

$$x^5+ax+b=0$$ (17)

with Rational Coefficients is solvable by radicals Iff there exist rational numbers $\epsilon=\pm 1$, $c\geq 0$, and $e\not=0$ such that

$$a = {5e^4(3-4\epsilon c)\over c^2+1}$$ (18)

$$b = {-4e^5(11\epsilon+2c)\over c^2+1}.$$ (19)

The Roots are then

$$x_j=e(\omega^ju_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4),$$ (20)

where

$$u_1 = \left({{v_1}^2v_3\over D^2}\right)^{1/5}$$ (21)

$$u_2 = \left({{v_3}^2v_4\over D^2}\right)^{1/5}$$ (22)

$$u_3 = \left({{v_2}^2v_1\over D^2}\right)^{1/5}$$ (23)

$$u_4 = \left({{v_4}^2v_2\over D^2}\right)^{1/5}$$ (24)

$$v_1 = \sqrt{D}+\sqrt{D-\epsilon\sqrt{D}}$$ (25)

$$v_2 = -\sqrt{D}-\sqrt{D+\epsilon\sqrt{D}}$$ (26)

$$v_3 = -\sqrt{D}+\sqrt{D+\epsilon\sqrt{D}}$$ (27)

$$v_4 = \sqrt{D}-\sqrt{D-\epsilon\sqrt{D}}$$ (28)

$$D = c^2+1.$$ (29)

The general quintic can be solved in terms of Theta Functions, as was first done by Hermite in 1858. Kronecker subsequently obtained the same solution more simply, and Brioshi also derived the equation. To do so, reduce the general quintic

$$a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$$ (30)

into Bring Quintic Form

$$x^5 - x + \rho = 0.$$ (31)

Then define

$$k \equiv \tan\left[{{\textstyle{1\over 4}}\sin^{-1}\left({16\over 25 \sqrt{5}\,\rho^2}\right)}\right]$$ (32)

$$s \equiv \left\{\begin{array}{ll} -\mathop{\rm sgn}\nolimits (\Im[\rho]) &... ...{\rm sgn}\nolimits (\Re[\rho]) & \mbox{for $\Re[\rho]\not=0$}\end{array}\right.$$ (33)

$$b = {s(k^2)^{1/8}\over 2\cdot 5^{3/4} \sqrt{k(1 - k^2)}}$$ (34)

$$q = q(k^2)=e^{i\pi K'(k^2)/K(k^2)},$$ (35)

where $k$ is the Modulus, $m\equiv k^2$ is the Parameter, and $q$ is the Nome. Solving

$$q(m)=e^{i\pi K'(m)/K(m)}$$ (36)

for $m$ gives the inverse parameter

$$m(q)={{\vartheta_2}^4(q)\over{\vartheta_3}^4(q)}.$$ (37)

The Roots are then given by

$x_1 = (-1)^{3/4}b \{[m(e^{-2\pi i/5}q^{1/5})]^{1/8}+ i [m(e^{2\pi i/5}q^{1/5})]^{1/8}\}$
$\hskip10pt \times\{[m(e^{-4\pi i/5}q^{1/5})]^{1/8} + [m(e^{4\pi i/5}q^{1/5})]^{1/8}\}$
$\hskip10pt \times \{[m(q^{1/5})]^{1/8} + q^{5/8}(q^5)^{-1/8} [m(q^5)]^{1/8}\}$	(38)
$x_2 = b\{-[m(q^{1/5})]^{1/8} + e^{3 \pi i/4} [m(e^{2\pi i/5} q^{1/5})]^{1/8}\}$
$\hskip10pt \times\{e^{-3\pi i/4}[m(e^{-2\pi i/5}q^{1/5})]^{1/8} + i [m(e^{4\pi i/5} q^{1/5})]^{1/8}\}$
$\hskip10pt \times \{i [m(e^{-4\pi i/5}q^{1/5})]^{1/8} + q^{5/8}(q^5)^{-1/8}[m(q^5)]^{1/8}\}$	(39)
$x_3 = b\{e^{-3\pi i/4}[m(e^{-2 \pi i/5}q^{1/5})]^{1/8}- i [m(e^{-4\pi i/5}q^{1/5})]^{1/8})\}$	(40)
$\hskip10pt \times \{-[m(q^{1/5})]^{1/8}- i [m(e^{4\pi i/5} q^{1/5})]^{1/8}\}$
$\hskip10pt \times \{e^{3 \pi i/4}[m(e^{2\pi i/5} q^{1/5})]^{1/8}+q^{5/8}(q^5)^{-1/8}[m(q^5)]^{1/8}\}$	(41)
$x_4= b\{[m(q^{1/5})]^{1/8} - i [m(e^{-4\pi i/5}q^{1/5})]^{1/8})\}$
$\hskip10pt \times \{-e^{3 \pi i/4}[m(e^{2\pi i/5} q^{1/5})]^{1/8} - i [m(e^{4\pi i/5} q^{1/5})]^{1/8}\}$
	(42)
$x_5= b\{[m(q^{1/5})]^{1/8} - e^{-3\pi i/4}[m(e^{-2\pi i/5}q^{1/5})]^{1/8}\}$
$\hskip10pt \times \{-e^{3 \pi i/4}[m(e^{2\pi i/5} q^{1/5})]^{1/8} + i [m(e^{-4\pi i/5}q^{1/5})]^{1/8}\}$
$\hskip10pt \times \{(-i [m(e^{4\pi i/5} q^{1/5})]^{1/8} + q^{5/8}(q^5)^{-1/8}[m(q^5)]^{1/8}\}.$
	(43)

Felix Klein used a Tschirnhausen Transformation to reduce the general quintic to the form

$$z^5+5az^2+5bz+c=0.$$ (44)

He then solved the related Icosahedral Equation

$I(z, 1, Z)=z^5(-1+11z^5+z^{10})^5\quad$
$-[1+z^{30}-10005(z^{10}+z^{20})+522(-z^5+z^{25})]^2Z=0,$
	(45)

where $Z$ is a function of radicals of $a$, $b$, and $c$. The solution of this equation can be given in terms of Hypergeometric Functions as

$${Z^{-1/60}{}_2F_1(-{\textstyle{1\over 60}}, {\textstyle{29\o... ...}}, {\textstyle{41\over 60}}, {\textstyle{6\over 5}}, 1728Z)}.$$ (46)

Another possible approach uses a series expansion, which gives one root (the first one in the list below) of

$$t^5-t-\rho.$$ (47)

All five roots can be derived using differential equations (Cockle 1860, Harley 1862). Let

$$F_1(\rho) = F_2(\rho)$$ (48)

$$F_2(\rho) = {}_4F_3({\textstyle{1\over 5}}, {\textstyle{2\over 5}}, {\textsty... ...extstyle{3\over 4}}, {\textstyle{5\over 4}}; {\textstyle{3125\over 256}}\rho^4)$$ (49)

$$F_3(\rho) = {}_4F_3({\textstyle{9\over 20}}, {\textstyle{13\over 20}}, {\text... ...textstyle{5\over 4}},{\textstyle{3\over 2}}; {\textstyle{3125\over 256}}\rho^4)$$ (50)

$$F_4(\rho) = {}_4F_3({\textstyle{7\over 10}}, {\textstyle{9\over 10}}, {\texts... ...xtstyle{3\over 2}}, {\textstyle{7\over 4}}; {\textstyle{3125\over 256}}\rho^4),$$ (51)

then the Roots are

$$t_1 = -\rho\,{}_4F_3({\textstyle{1\over 5}}, {\textstyle{2\over 5}}, {\... ...xtstyle{3\over 4}}, {\textstyle{5\over 4}}; {\textstyle{3125\over 256}} \rho^4)$$ (52)

$$t_2 = -F_1(\rho)+{\textstyle{1\over 4}}\rho F_2(\rho)+{\textstyle{5\over 32}}\rho^2 F_3(\rho)+{\textstyle{5\over 32}}\rho^3 F_4(\rho)$$ (53)

$$t_3 = -F_1(\rho)+{\textstyle{1\over 4}}\rho F_2(\rho)-{\textstyle{5\over 32}}\rho^2 F_3(\rho)+{\textstyle{5\over 32}}\rho^3 F_4(\rho)$$ (54)

$$t_4 = -i F_1(\rho)+{\textstyle{1\over 4}}\rho F_2(\rho)-{\textstyle{5\over 32}}i \rho^2 F_3(\rho)-{\textstyle{5\over 32}}\rho^3 F_4(\rho)$$ (55)

$$t_5 = i F_1(\rho)+{\textstyle{1\over 4}}\rho F_2(\rho)+{\textstyle{5\over 32}}i \rho^2 F_3(\rho)-{\textstyle{5\over 32}}\rho^3 F_4(\rho).$$ (56)

This technique gives closed form solutions in terms of Hypergeometric Functions in one variable for any Polynomial equation which can be written in the form

$$x^p+bx^q+c.$$ (57)

Cadenhad, Young, and Runge showed in 1885 that all irreducible solvable quintics with Coefficients of $x^4$, $x^3$, and $x^2$ missing have the following form

$$x^5+{5\mu^4(4\nu+3)\over\nu^2+1} x+{4\mu^5(2\nu+1)(4\nu+3)\over \nu^2+1}=0,$$ (58)

where $\mu$ and $\nu$ are Rational.

See also Bring Quintic Form, Bring-Jerrard Quintic Form, Cubic Equation, de Moivre's Quintic, Principal Quintic Form, Quadratic Equation, Quartic Equation, Sextic Equation

References

Birkhoff, G. and Mac Lane, S. A Survey of Modern Algebra, 3rd ed. New York: Macmillan, pp. 418-421, 1965.

Chowla, S. ``On Quintic Equations Soluble by Radicals.'' Math. Student 13, 84, 1945.

Cockle, J. ``Sketch of a Theory of Transcendental Roots.'' Phil. Mag. 20, 145-148, 1860.

Cockle, J. `` On Transcendental and Algebraic Solution--Supplemental Paper.'' Phil. Mag. 13, 135-139, 1862.

Davis, H. T. Introduction to Nonlinear Differential and Integral Equations. New York: Dover, p. 172, 1960.

Dummit, D. S. ``Solving Solvable Quintics.'' Math. Comput. 57, 387-401, 1991.

Glashan, J. C. ``Notes on the Quintic.'' Amer. J. Math. 8, 178-179, 1885.

Harley, R. ``On the Solution of the Transcendental Solution of Algebraic Equations.'' Quart. J. Pure Appl. Math. 5, 337-361, 1862.

Hermite, C. ``Sulla risoluzione delle equazioni del quinto grado.'' Annali di math. pura ed appl. 1, 256-259, 1858.

King, R. B. Beyond the Quartic Equation. Boston, MA: Birkhäuser, 1996.

King, R. B. and Cranfield, E. R. ``An Algorithm for Calculating the Roots of a General Quintic Equation from Its Coefficients.'' J. Math. Phys. 32, 823-825, 1991.

Rosen, M. I. ``Niels Hendrik Abel and Equations of the Fifth Degree.'' Amer. Math. Monthly 102, 495-505, 1995.

Shurman, J. Geometry of the Quintic. New York: Wiley, 1997.

Spearman, B. K. and Williams, K. S. ``Characterization of Solvable Quintics $x^5+ax+b$.'' Amer. Math. Monthly 101, 986-992, 1994.

Young, G. P. ``Solution of Solvable Irreducible Quintic Equations, Without the Aid of a Resolvent Sextic.'' Amer. J. Math. 7, 170-177, 1885.