Skip to article frontmatterSkip to article content
Site not loading correctly?

This may be due to an incorrect BASE_URL configuration. See the MyST Documentation for reference.

Quintic Example

In this section, we’ll work through the procedure for computing the roots of a quintic in the case of a specific example with known roots, three real and two complex, of the polynomial given by

P(x):=(x+3)(x+1)(x−2)(x2+x+9)=x5+3 x4+6 x3+7 x2−51 x−54 .\begin{aligned} P(x) &:= (x+3)(x+1)(x-2)(x^2+x+9) \\ &= x^5 + 3\,x^4+6\,x^3+7\,x^2-51\,x-54 \,. \end{aligned}

Since the roots are known, we can run the polynomial coefficients through our algorithm and verify that the appropriate roots are found.

The first step is to perform a Tschirnhaus transformation, in order to get the coefficients of the corresponding principal quintic. Taking

y=x2+A x+By = x^2+A\,x+B

with A=−6A = -6, B=−3B=-3, we find

Q(y)=y5−11065 y2−88260 y+529056Q(y) = y^5 -11065\,y^2-88260\,y+529056

and so α=−11065\alpha = -11065, β=−17652\beta = -17652, and γ=529056\gamma = 529056.

Loading...
Loading...
Loading...
{alpha: -2213, beta: -17652, gamma: 529056}

Now that the coefficients of the principal quintic have been identified, we can proceed by solving for the values appearing in the expression

y=yr:=Wr(z0,1)f(z0,1)H(z0,1)(λ±+μ± tr(z0,1)f2(z0,1)T(z0,1)) .y = y_r := \frac{W_r(z_0,1)f(z_0,1)}{H(z_0,1)}\left(\lambda_\pm + \mu_\pm\,\frac{t_r(z_0,1)f^2(z_0,1)}{T(z_0,1)}\right) \,.

First of all, λ±\lambda_\pm is found by solving the quadratic equation

(α4−β3+αβγ)λ2+(αγ2−2 β2γ−11 α3β)λ+64 α2β2−27 α3γ−βγ2=0 .\left(\alpha^4-\beta^3+\alpha\beta\gamma\right)\lambda^2 + \left(\alpha\gamma^2-2\,\beta^2\gamma-11\,\alpha^3\beta\right)\lambda + 64\,\alpha^2\beta^2-27\,\alpha^3\gamma-\beta\gamma^2 =0 \,.

Solving for λ\lambda, given the values of (α,β,γ)(\alpha,\beta,\gamma) above, we find

λ=λ±=5551876412706182368882091±44703099031507i182368882091 .\lambda = \lambda_\pm = \frac{5551876412706}{182368882091} \pm \frac{4470309903150 \sqrt{7} i}{182368882091} \,.
Loading...
Loading...

From here, the corresponding values of II are

I=I±=(αλ±2−3 βλ±−3γ)3123α2(αγλ±−β2λ±−βγ)=1543339300632027612848218519779240304755275995712080000018338381126299237208601401250567585364768902047827121041±24776318349174328330534650476278740558986157268948800007i201722192389291609294615413756243439012457922526098331451 .\begin{aligned} I &= I_\pm = \frac{(\alpha\lambda_\pm^2-3\,\beta\lambda_\pm-3\gamma)^3}{12^3\alpha^2(\alpha\gamma\lambda_\pm-\beta^2\lambda_\pm-\beta\gamma)} \\ &= \frac{15433393006320276128482185197792403047552759957120800000}{18338381126299237208601401250567585364768902047827121041} \pm \frac{2477631834917432833053465047627874055898615726894880000 \sqrt{7} i}{201722192389291609294615413756243439012457922526098331451} \,. \end{aligned}
Loading...
Loading...

We also need a value for μ\mu, which we take to be

μ=μ±:=±−βλ±−γαB(z0)=110651913473185152697048763063406110447004453169057563354985863622888089636787850718980739066465358072503539144355288353260839478659255944041+19659954258155915921195319123883671250432162439436156927964376850169728007i36787850718980739066465358072503539144355288353260839478659255944041 .\begin{align*} \mu &= \mu_\pm := \pm\sqrt{\frac{-\beta\lambda_\pm-\gamma}{\alpha B(z_0)}}\\ &=\sqrt{\frac{1106519134731851526970487630634061104470044531690575633549858636228880896}{36787850718980739066465358072503539144355288353260839478659255944041} + \frac{1965995425815591592119531912388367125043216243943615692796437685016972800 \sqrt{7} i}{36787850718980739066465358072503539144355288353260839478659255944041}} \,. \end{align*}

(Recall that B(z0)=1+1/(1728 I(z0))B(z_0) = 1+1/(1728\,I(z_0)).)

Loading...
Loading...

Now, we do have to compute z0z_0 using the icosahedral equation below. (This is the only part of the computation that can’t be done using radicals.) In doing so, we recall that the inverse is given by a ratio of hypergeometric functions:

s(x)=−(1728 x)1/5 2F1(b,1+b−c;1+b−a;1/x)2F1(a,1+a−c;1+a−b;1/x) .s(x) = - (1728\,x)^{1/5}\,\frac{{}_2F_1(b,1+b-c;1+b-a;1/x)}{{}_2F_1(a, 1+a-c;1+a-b;1/x)} \,.
Loading...

With z0:=s(I0)z_0 := s(I_0) in tow, we can now compute the roots yry_r of the principal quintic:

yr=Wr(z0,1)f(z0,1)H(z0,1)(λ±+μ± tr(z0,1)f2(z0,1)T(z0,1)) .y_r = \frac{W_r(z_0,1)f(z_0,1)}{H(z_0,1)} \left(\lambda_\pm + \mu_\pm\,\frac{t_r(z_0,1)f^2(z_0,1)}{T(z_0,1)}\right) \,.
60
[-8.49999999999997 + 20.7062792408486*I, -8.50000000000002 - 20.7062792408486*I, -11.0 + 5.5169898597468e-15*I, 24.0 - 2.28072769004051e-14*I, 4.0 + 5.97253095504374e-16*I]
[4.00000000000001 + 3.2816089255705e-15*I, -8.49999999999997 + 20.7062792408486*I, -8.50000000000002 - 20.7062792408486*I, -11.0 + 2.99115116356163e-15*I, 24.0 - 2.01198649754653e-14*I]
[-8.49999999999997 + 20.7062792408486*I, -11.0 + 1.62564325065576e-14*I, -8.5 - 20.7062792408486*I, 3.99999999999999 - 1.54896305200672e-14*I, 24.0 - 1.91632733986813e-16*I]
[24.0 - 2.95001653313753e-14*I, 4.00000000000001 - 2.21564845463706e-15*I, -8.49999999999998 + 20.7062792408487*I, -8.50000000000001 - 20.7062792408486*I, -11.0 + 2.10476167744611e-15*I]
[4.00000000000001 - 9.60273798412264e-15*I, -8.50000000000001 - 20.7062792408486*I, -8.49999999999997 + 20.7062792408486*I, 24.0 - 2.03325854417069e-14*I, -11.0 + 2.07939134660995e-15*I]
120
[(1, 2, 3, 4, 5), (1, 2, 3, 5, 4), (1, 2, 4, 3, 5), (1, 2, 4, 5, 3), (1, 2, 5, 3, 4), (1, 2, 5, 4, 3)]
0
[-8.49999999999997 + 20.7062792408486*I, -8.50000000000002 - 20.7062792408486*I, -11.0 + 5.5169898597468e-15*I, 24.0 - 2.28072769004051e-14*I, 4.0 + 5.97253095504374e-16*I]
[[-0.499999999999996 - 2.9580398915498*I, 6.5 + 2.9580398915498*I], [-0.499999999999996 + 2.9580398915498*I, 6.5 - 2.9580398915498*I], [2.0 - 2.7584949298734e-15*I, 4.0 + 2.7584949298734e-15*I], [-3.0 + 1.90060640836709e-15*I, 9.0 - 1.90060640836709e-15*I], [-1.0 - 7.46566369380468e-17*I, 7.0 + 7.46566369380468e-17*I]]
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...