Collinearity

In algebra, a Bring radical or ultraradical of a complex number a is a root of the polynomial

$x^5+x+a.$

(The root is chosen so the radical of a real number is real, and the radical is a differentiable function of a in the complex plane, with a branch cut along the negative real line below −1. See the "Bring radicals" section below.)

George Jerrard showed that some quintic equations can be solved using radicals and Bring radicals, which had been introduced by Erland Bring. They can be used to obtain closed-form solutions of quintic equations.

Normal forms

The quintic equation is rather difficult to obtain solutions for directly, with five independent coefficients in its most general form:

$x^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0=0$

The various methods for solving the quintic that have been developed generally attempt to simplify the quintic using Tschirnhaus transformations to reduce the number of independent coefficients.

Principal quintic form

The general quintic may be reduced into what is known as the principal quintic form, with the quartic and cubic terms removed:

$x^5+c_2{x}^2+c_{1}x+c_0=0$

If the roots of a general quintic and a principal quintic are related by a quadratic Tschirnhaus transformation:

$y_k=x_k^2+\alpha x_k+\beta$

the coefficients α and β may be determined by using the resultant, or by means of the power-sum formulae of the quintic roots. This leads to a system of equations in α and β consisting of a quadratic and a linear equation, and either of the two sets of solutions may be used to obtain the corresponding three coefficients of the principal quintic form.

This form is used by Felix Klein's solution to the quintic.

Bring–Jerrard normal form

It is possible to simplify the quintic still further and eliminate the quadratic term, producing the Bring–Jerrard normal form:

$x^5+d_{1}x+d_0=0$

Using the power-sum formulae again with a cubic transformation as Tschirnhaus tried does not work, since the resulting system of equations results in a sixth-degree equation.

In 1796 Bring found a way around this by using a quartic Tschirnhaus transformation to relate the roots of a principal quintic to those of a Bring–Jerrard quintic:

$z_k=x_k^4+\alpha x_k^3+\beta x_k^2+\gamma x_k+\delta$

The extra parameter this fourth-order transformation provides allowed Bring to decrease the degrees of the other parameters. This leads to a system of five equations in six unknowns, which then requires the solution of a cubic and a quadratic equation. This method was also discovered by Jerrard in 1852, but it is likely that he was unaware of Bring's previous work in this area. The full transformation may readily be accomplished using a computer algebra package such as Mathematica or Maple. As might be expected from the complexity of these transformations, the resulting expressions can be enormous, particularly when compared to the solutions in radicals for lower degree equations, taking many megabytes of storage for a general quintic with symbolic coefficients.

Regarded as an algebraic function, the solutions to

$x^5+d_{1}x+d_0=0$

involve two variables, d₁ and d₀, however the reduction is actually to an algebraic function of one variable, very much analogous to a solution in radicals, since we may further reduce the Bring–Jerrard form. If we for instance set

$z=\frac{x}{\sqrt[4]{-\frac{d_1}{5}}}$

then we reduce the equation to the form

$z^5-5z-4t=0$

which involves z as an algebraic function of a single variable t, where $t=-(d_0/4)(-d_1/5{)}^{-5/4}$. A similar transformation suffices to reduce the equation to

$y^5-y+a=0$

which is the form required by the Hermite-Kronecker-Brioschi method, Glasser's method, and the Cockle-Harley method of differential resolvents described below.

Brioschi normal form

There is another one-parameter normal form for the quintic equation, known as Brioschi normal form:

$x^5-10C{x}^3+45C^{2}x-C^2=0$

which can be derived by using the following rational Tschirnhaus transformation

$z_k=\frac{\lambda +\mu y_k}{\frac{y_k^2}{C}-3}$

to relate the roots of a principal quintic to a Brioschi quintic. The values of the parameters $\lambda$ and $\mu$ may be derived by using polyhedral functions on the Riemann sphere, and is related to the partition of an object of icosahedral symmetry into five objects of tetrahedral symmetry.

It is to be noted that this Tschirnhaus transformation is rather simpler than the difficult one used to transform a principal quintic into Bring–Jerrard form. This normal form is used by the Doyle–McMullen iteration method and the Kiepert method.

Series representation

A Taylor series for Bring radicals, as well as a representation in terms of hypergeometric functions can be derived as follows. The equation $x^5+x+a=0$ can be rewritten as $x^5+x=-a$; by setting $f(x)=x^5+x$, the desired solution is $x=f^{-1}(-a)$.

The series for $f^{-1}$ can then be obtained by reversion of the Taylor series for $f(x)$ (which is simply $x+x^5$), giving:

$f^{-1}(a)=\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{5k}{k})\frac{(-1{)}^k{a}^{4k+1}}{4k+1}=a-a^5+5a^9-35a^{13}+...$

where the absolute values of the coefficients are sequence A002294 in the OEIS. The series confirms that $f^{-1}(a)$ is odd. This gives

$BR(a)=f^{-1}(-a)=-f^{-1}(a)=-a+a^5-5a^9+35a^{13}+...$

The series converges for $\left|a\right|<4/(5\cdot \sqrt[4]{5})\approx 0.53499$ and can be analytically continued in the complex plane. The above result can be written in hypergeometric form as:

$BR(a)=-a{}_4{F}_3(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};-{\left(\frac{5a}{4}\right)}^4)$

Compare with the hypergeometric functions that arise in Glasser's derivation and the method of differential resolvents below.

Solution of the general quintic

We now may express the roots of any polynomial

$x^5+px+q$

in terms of the Bring radical as

$\sqrt[4]{-\frac{p}{5}}\mathrm{BR}(-\frac{\sqrt[4]{\frac{5^5}{-p^5}}q}{4})$

and its four conjugates. We have a reduction to the Bring–Jerrard form in terms of solvable polynomial equations, and we used transformations involving polynomial expressions in the roots only up to the fourth degree, which means inverting the transformation may be done by finding the roots of a polynomial solvable in radicals. This procedure produces extraneous solutions, but when we have found the correct ones by numerical means we can also write down the roots of the quintic in terms of square roots, cube roots, and the Bring radical, which is therefore an algebraic solution in terms of algebraic functions of a single variable — an algebraic solution of the general quintic.

Other characterizations

Many other characterizations of the Bring radical have been developed, the first of which is in terms of elliptic modular functions by Charles Hermite in 1858, and further methods later developed by other mathematicians.

The Hermite–Kronecker–Brioschi characterization

In 1858, Charles Hermite published the first known solution to the general quintic equation in terms of elliptic transcendents, and at around the same time Francesco Brioschi and Leopold Kronecker came upon equivalent solutions. Hermite arrived at this solution by generalizing the well-known solution to the cubic equation in terms of trigonometric functions and finds the solution to a quintic in Bring–Jerrard form:

$x^5-x+a=0$

into which any quintic equation may be reduced by means of Tschirnhaus transformations as has been shown. He observed that elliptic functions had an analogous role to play in the solution of the Bring–Jerrard quintic as the trigonometric functions had for the cubic. If $K$ and $K^{\prime }$ are the periods of an elliptic integral of the first kind:

$K=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{1-k^2{\sin }^2\phi }}$

$K^{\prime }=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{1-k{\text{'}}^2{\sin }^2\phi }}$

the elliptic nome is given by:

$q=e^{-\frac{\pi K^{\prime }}{K}}$

and

$k^2+k{\text{'}}^2=1$

With

$q=e^{-\frac{\pi K^{\prime }}{K}}=e^{\mathrm{i}\pi \tau }$

define the two elliptic modular functions:

$\sqrt[4]{k}=\phi (\tau )=\sqrt{\frac{{\vartheta }_{10}(0;\tau )}{{\vartheta }_{00}(0;\tau )}}$

$\sqrt[4]{k^{\prime }}=\psi (\tau )=\sqrt{\frac{{\vartheta }_{01}(0;\tau )}{{\vartheta }_{00}(0;\tau )}}$

where ${\vartheta }_{00}(0;\tau )$ and similar are Jacobi theta functions.

If n is a prime number, we can define two values u and v as follows:

$v=\phi (n\tau )$

and

$u=\phi (\tau )$

The parameters $u$ and $v$ are linked by an equation of degree n + 1 known as the modular equation, whose n + 1 roots are given by:

$ϵ\phi (n\tau )$

and

$\phi \left(\frac{\tau +16m}{n}\right)$

where ε is 1 or −1 depending on whether 2 is a quadratic residue with respect to n or not, and m is an integer modulo n. For n = 5, we have the modular equation of the sixth degree:

$u^6-v^6+5u^2{v}^2(u^2-v^2)+4uv(1-u^4{v}^4)=0$

with six roots as shown above.

The modular equation of the sixth degree may be related to the Bring–Jerrard quintic by the following function of the six roots of the modular equation:

$\mathrm{\Phi }(\tau )=[\phi (5\tau )+\phi \left(\frac{\tau }{5}\right)][\phi \left(\frac{\tau +16}{5}\right)-\phi \left(\frac{\tau +64}{5}\right)][\phi \left(\frac{\tau +32}{5}\right)-\phi \left(\frac{\tau +48}{5}\right)]$

The five quantities $\mathrm{\Phi }(\tau )$, $\mathrm{\Phi }(\tau +16)$, $\mathrm{\Phi }(\tau +32)$, $\mathrm{\Phi }(\tau +48)$, $\mathrm{\Phi }(\tau +64)$ are the roots of a quintic equation with coefficients rational in $\phi (\tau )$:

${\mathrm{\Phi }}^5-2000{\phi }^4(\tau ){\psi }^{16}(\tau )\mathrm{\Phi }-1600\sqrt{5}{\phi }^3(\tau ){\psi }^{16}(\tau )[1+{\phi }^8(\tau )]=0$

which may be readily converted into the Bring–Jerrard form by the substitution:

$\mathrm{\Phi }=2\sqrt[4]{125}\phi (\tau ){\psi }^4(\tau )x$

leading to the Bring–Jerrard quintic:

$x^5-x+a=0$

where

$a=\frac{2[1+{\phi }^8(\tau )]}{\sqrt[4]{5^5}{\phi }^2(\tau ){\psi }^4(\tau )}$

The Hermite–Kronecker–Brioschi method then amounts to finding a value for τ that corresponds to the value of a, and then using that value of τ to obtain the roots of the corresponding modular equation. To do this, let

$A=\frac{a\sqrt[4]{5^5}}{2}$

and calculate the required elliptic modulus $k$ by solving the quartic equation:

$k^4+A^2{k}^3+2k^2-A^{2}k+1=0$

The roots of this equation are:

$k=\tan \frac{\alpha }{4},\tan \frac{\alpha +2\pi }{4},\tan \frac{\pi -\alpha }{4},\tan \frac{3\pi -\alpha }{4}$

where $\sin \alpha =\frac{4}{A^2}$ (note that some important references erroneously give it as $\sin \alpha =\frac{1}{4A^2}$ ). Any of these roots may be used as the elliptic modulus for the purposes of the method. The value of $\tau$ may be easily obtained from the elliptic modulus $k$ by the relations given above. The roots of the Bring–Jerrard quintic are then given by:

$x_i=\frac{\mathrm{\Phi }(\tau +16i)}{2\sqrt[4]{125}\phi (\tau ){\psi }^4(\tau )}$

for $i=0,\dots ,4$.

It may be seen that this process uses a generalization of the nth root, which may be expressed as:

$\sqrt[n]{x}=\exp \left(\frac{1}{n}\ln x\right)$

or more to the point, as

$\sqrt[n]{x}=\exp \left(\frac{1}{n}\underset{1}{\overset{x}{\int }}\frac{dt}{t}\right).$

The Hermite–Kronecker–Brioschi method essentially replaces the exponential by an elliptic modular function, and the integral $\underset{1}{\overset{x}{\int }}\frac{dt}{t}$ by an elliptic integral. Kronecker thought that this generalization was a special case of a still more general theorem, which would be applicable to equations of arbitrarily high degree. This theorem, known as Thomae's formula, was fully expressed by Hiroshi Umemura in 1984, who used Siegel modular forms in place of the exponential/elliptic modular function, and the integral by a hyperelliptic integral.

Glasser's derivation

This derivation due to M. L. Glasser generalizes the series method presented earlier in this article to find a solution to any trinomial equation of the form:

$x^N-x+t=0$

In particular, the quintic equation can be reduced to this form by the use of Tschirnhaus transformations as shown above. Let $x={\zeta }^{-\frac{1}{N-1}}$, the general form becomes:

$\zeta =e^{2\pi i}+t\varphi (\zeta )$

where

$\varphi (\zeta )={\zeta }^{\frac{N}{N-1}}$

A formula due to Lagrange states that for any analytic function $f$, in the neighborhood of a root of the transformed general equation in terms of $\zeta$, above may be expressed as an infinite series:

$f(\zeta )=f(e^{2\pi \mathrm{i}})+\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n!}\frac{d^{n-1}}{d{a}^{n-1}}[f^{\prime }(a)|\varphi (a){|}^n{]}_{a=e^{2\pi \mathrm{i}}}$

If we let $f(\zeta )={\zeta }^{-\frac{1}{N-1}}$ in this formula, we can come up with the root:

$x_k=e^{-\frac{2k\pi \mathrm{i}}{N-1}}-\frac{t}{N-1}\sum _{n=0}^{\mathrm{\infty }}\frac{(t{e}^{\frac{2k\pi \mathrm{i}}{N-1}}{)}^n}{\mathrm{\Gamma }(n+2)}\cdot \frac{\mathrm{\Gamma }(\frac{Nn}{N-1}+1)}{\mathrm{\Gamma }(\frac{n}{N-1}+1)}$

$k=1,2,3,\dots ,N-1$

By the use of the Gauss multiplication theorem the infinite series above may be broken up into a finite series of hypergeometric functions:

${\psi }_n(q)={\left(\frac{e^{\frac{2n\pi \mathrm{i}}{N-1}}t}{N-1}\right)}^q{N}^{\frac{qN}{N-1}}\frac{\prod _{k=0}^{N-1}\mathrm{\Gamma }(\frac{q}{N-1}+\frac{1+k}{N})}{\mathrm{\Gamma }(\frac{q}{N-1}+1)\prod _{k=0}^{N-2}\mathrm{\Gamma }\left(\frac{q+k+2}{N-1}\right)}=={\left(\frac{t{e}^{\frac{2n\pi \mathrm{i}}{N-1}}}{N-1}\right)}^q{N}^{\frac{qN}{N-1}}\prod _{k==2}^N\frac{\mathrm{\Gamma }(\frac{q}{N-1}+\frac{k-1}{N})}{\mathrm{\Gamma }\left(\frac{q+k}{N-1}\right)}$

$x_n=e^{-\frac{2n\pi \mathrm{i}}{N-1}}-\frac{t}{(N-1{)}^2}\sqrt{\frac{N}{2\pi (N-1)}}\sum _{q=0}^{N-2}{\psi }_n(q{)}_{(N+1)}{F}_N\left[\begin{array}{c}\frac{qN+N-1}{N(N-1)},\dots ,\frac{q+N-1}{N-1},1;\\ \frac{q+2}{N-1},\dots ,\frac{q+N}{N-1},\frac{q+N-1}{N-1};\\ {\left(\frac{t{e}^{\frac{2n\pi \mathrm{i}}{N-1}}}{N-1}\right)}^{N-1}{N}^N\end{array}\right],n=1,2,3,\dots ,N-1$

$x_N=\sum _{m=1}^{N-1}\frac{t}{(N-1{)}^2}\sqrt{\frac{N}{2\pi (N-1)}}\sum _{q=0}^{N-2}{\psi }_m(q{)}_{(N+1)}{F}_N\left[\begin{array}{c}\frac{qN+N-1}{N(N-1)},\dots ,\frac{q+N-1}{N-1},1;\\ \frac{q+2}{N-1},\dots ,\frac{q+N}{N-1},\frac{q+N-1}{N-1};\\ {\left(\frac{t{e}^{\frac{2m\pi \mathrm{i}}{N-1}}}{N-1}\right)}^{N-1}{N}^N\end{array}\right]$

and the trinomial of the form,has roots

${}_{a{x}^N+b{x}^2+c=0,N\equiv 1(\mathrm{mod}2)}$

${}_{x_N=-\frac{a}{2b}\sqrt{{\left(\frac{c}{b}\right)}^{N-1}}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}\frac{N+1}{2N},\frac{N+3}{2N},\cdots ,\frac{N-2}{N},\frac{N-1}{N},\frac{N+1}{N},\frac{N+2}{N},\cdots ,\frac{3N-3}{2N},\frac{3N-1}{2N};\\ \frac{N+1}{2N-4},\frac{N+3}{2N-4},\cdots ,\frac{N-4}{N-2},\frac{N-3}{N-2},\frac{N-1}{N-2},\frac{N}{N-2},\cdots ,\frac{3N-5}{2N-4},\frac{3}{2};\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right]+\sqrt{\frac{c}{b}}\mathrm{i}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}\frac{1}{2N},\frac{3}{2N},\cdots ,\frac{N-4}{2N},\frac{N-2}{2N},\frac{N+2}{2N},\frac{N+4}{2N},\cdots ,\frac{2N-3}{2N},\frac{2N-1}{2N};\\ \frac{3}{2N-4},\frac{5}{2N-4},\cdots ,\frac{2N-3}{2N-4};\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right]}$

${}_{x_{N-1}=-\frac{a}{2b}\sqrt{{\left(\frac{c}{b}\right)}^{N-1}}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}\frac{N+1}{2N},\frac{N+3}{2N},\cdots ,\frac{N-2}{N},\frac{N-1}{N},\frac{N+1}{N},\frac{N+2}{N},\cdots ,\frac{3N-3}{2N},\frac{3N-1}{2N};\\ \frac{N+1}{2N-4},\frac{N+3}{2N-4},\cdots ,\frac{N-4}{N-2},\frac{N-3}{N-2},\frac{N-1}{N-2},\frac{N}{N-2},\cdots ,\frac{3N-5}{2N-4},\frac{3}{2};\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right]-\sqrt{\frac{c}{b}}\mathrm{i}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}\frac{1}{2N},\frac{3}{2N},\cdots ,\frac{N-4}{2N},\frac{N-2}{2N},\frac{N+2}{2N},\frac{N+4}{2N},\cdots ,\frac{2N-3}{2N},\frac{2N-1}{2N};\\ \frac{3}{2N-4},\frac{5}{2N-4},\cdots ,\frac{2N-3}{2N-4};\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right]}$

${}_{x_n=-e^{\frac{2n\pi \mathrm{i}}{N-2}}\sqrt[N-2]{\frac{b}{a}}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}-\frac{1}{N(N-2)},-\frac{1}{N(N-2)}+\frac{1}{N},-\frac{1}{N(N-2)}+\frac{2}{N},\cdots ,-\frac{1}{N(N-2)}+\frac{1}{N},\frac{N-5}{2N},-\frac{1}{N(N-2)}+\frac{N-3}{2N},-\frac{1}{N(N-2)}+\frac{N+1}{2N},-\frac{1}{N(N-2)}+\frac{N+3}{2N},\cdots ,-\frac{1}{N(N-2)}+\frac{N-1}{N},;\\ \frac{1}{N-2},\frac{2}{N-2},\cdots ,\frac{2N-5}{2N-4},;\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right]+\sqrt[N-2]{\frac{b}{a}}\sum _{q=1}^{N-3}\frac{\mathrm{\Gamma }(\frac{2q-1}{N-2}+q)}{\mathrm{\Gamma }(\frac{2q-1}{N-2}+1)}\cdot {(-\frac{c}{b}\sqrt[N-2]{\frac{a^2}{b^2}})}^q\cdot \frac{e^{\frac{2n(1-2q)}{N-2}\pi \mathrm{i}}}{q!}{}_{N-1}{F}_{N-2}\left[\begin{array}{c}\frac{Nq-1}{N(N-2)},\frac{Nq-1}{N(N-2)}+\frac{1}{N},\frac{Nq-1}{N(N-2)}+\frac{2}{N},\cdots ,\frac{Nq-1}{N(N-2)}+\frac{N-3}{2N},\frac{Nq-1}{N(N-2)}+\frac{N+1}{2N},\cdots ,\frac{Nq-1}{N(N-2)}+\frac{N-1}{N};\\ \frac{q+1}{N-2},\frac{q+2}{N-2},\cdots ,\frac{N-4}{N-2},\frac{N-3}{N-2},\frac{N-1}{N-2},\frac{N}{N-2},\cdots ,\frac{q+N-2}{N-2},\frac{2q+2N-5}{2N-4};\\ -\frac{a^2{c}^{N-2}}{4b^N{(N-2)}^{N-2}}\end{array}\right],n=1,2,\cdots ,N-2}$

A root of the equation can thus be expressed as the sum of at most N − 1 hypergeometric functions. Applying this method to the reduced Bring–Jerrard quintic, define the following functions:

$\begin{array}{rl}{F}_1(t)& ={}_4{F}_3(\frac{-1}{20},\frac{3}{20},\frac{7}{20},\frac{11}{20};\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{3125t^4}{256})\\ F_2(t)& ={}_4{F}_3(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{3125t^4}{256})\\ F_3(t)& ={}_4{F}_3(\frac{9}{20},\frac{13}{20},\frac{17}{20},\frac{21}{20};\frac{3}{4},\frac{5}{4},\frac{3}{2};\frac{3125t^4}{256})\\ F_4(t)& ={}_4{F}_3(\frac{7}{10},\frac{9}{10},\frac{11}{10},\frac{13}{10};\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3125t^4}{256})\end{array}$

which are the hypergeometric functions that appear in the series formula above. The roots of the quintic are thus:

$\begin{array}{rcr}{x}_1& =& -t{F}_2(t)\\ x_2& =& -F_1(t)& +& \frac{1}{4}t{F}_2(t)& +& \frac{5}{32}{t}^2{F}_3(t)& +& \frac{5}{32}{t}^3{F}_4(t)\\ x_3& =& F_1(t)& +& \frac{1}{4}t{F}_2(t)& +& \frac{5}{32}{t}^2{F}_3(t)& +& \frac{5}{32}{t}^3{F}_4(t)\\ x_4& =& -\mathrm{i}{F}_1(t)& +& \frac{1}{4}t{F}_2(t)& -& \frac{5}{32}\mathrm{i}{t}^2{F}_3(t)& -& \frac{5}{32}{t}^3{F}_4(t)\\ x_5& =& \mathrm{i}{F}_1(t)& +& \frac{1}{4}t{F}_2(t)& -& \frac{5}{32}\mathrm{i}{t}^2{F}_3(t)& -& \frac{5}{32}{t}^3{F}_4(t)\end{array}$

This is essentially the same result as that obtained by the method of differential resolvents developed by James Cockle and Robert Harley in 1860.

The method of differential resolvents

James Cockle and Robert Harley developed a method for solving the quintic by means of differential equations. They consider the roots as being functions of the coefficients, and calculate a differential resolvent based on these equations. The Bring–Jerrard quintic is expressed as a function:

$f(x)=x^5-x+a$

and a function $\varphi (a)$ is to be determined such that:

$f[\varphi (a)]=0$

The function $\varphi$ must also satisfy the following four differential equations:

$\begin{array}{r}\frac{df[\varphi (a)]}{da}=0\\ \frac{d^{2}f[\varphi (a)]}{d{a}^2}=0\\ \frac{d^{3}f[\varphi (a)]}{d{a}^3}=0\\ \frac{d^{4}f[\varphi (a)]}{d{a}^4}=0\end{array}$

Expanding these and combining them together yields the differential resolvent:

$\frac{(256-3125a^4)}{1155}\frac{d^4\varphi }{d{a}^4}-\frac{6250a^3}{231}\frac{d^3\varphi }{d{a}^3}-\frac{4875a^2}{77}\frac{d^2\varphi }{d{a}^2}-\frac{2125a}{77}\frac{d\varphi }{da}+\varphi =0$

The solution of the differential resolvent, being a fourth order ordinary differential equation, depends on four constants of integration, which should be chosen so as to satisfy the original quintic. This is a Fuchsian ordinary differential equation of hypergeometric type, whose solution turns out to be identical to the series of hypergeometric functions that arose in Glasser's derivation above.

This method may also be generalized to equations of arbitrarily high degree, with differential resolvents which are partial differential equations, whose solutions involve hypergeometric functions of several variables. A general formula for differential resolvents of arbitrary univariate polynomials is given by Nahay's powersum formula.

Doyle–McMullen iteration

In 1989, Peter Doyle and Curt McMullen derived an iteration method that solves a quintic in Brioschi normal form:

$x^5-10C{x}^3+45C^{2}x-C^2=0.$

The iteration algorithm proceeds as follows:

1. Set $Z=1-1728C$ 2. Compute the rational function

$T_Z(w)=w-12\frac{g(Z,w)}{g^{\prime }(Z,w)}$

where $g(Z,w)$ is a polynomial function given below, and $g^{\prime }$ is the derivative of $g(Z,w)$ with respect to $w$

3. Iterate $T_Z[T_Z(w)]$ on a random starting guess until it converges. Call the limit point $w_1$ and let $w_2=T_Z(w_1)$. 4. Compute

${\mu }_i=\frac{100Z(Z-1)h(Z,w_i)}{g(Z,w_i)}$

where $h(Z,w)$ is a polynomial function given below. Do this for both $w_1$ and $w_2=T_Z(w_1)$.

5. Finally, compute

$x_i=\frac{(9+\sqrt{15}\mathrm{i}){\mu }_i+(9-\sqrt{15}\mathrm{i}){\mu }_{3-i}}{90}$

for i = 1, 2. These are two of the roots of the Brioschi quintic.

The two polynomial functions $g(Z,w)$ and $h(Z,w)$ are as follows:

$\begin{array}{rl}g(Z,w)& =91125Z^6\\ & +(-133650w^2+61560w-193536)Z^5\\ & +(-66825w^4+142560w^3+133056w^2-61140w+102400)Z^4\\ & +(5940w^6+4752w^5+63360w^4-140800w^3)Z^3\\ & +(-1485w^8+3168w^7-10560w^6)Z^2\\ & +(-66w^{10}+440w^9)Z\\ & +w^{12}\\ h(Z,w)=& (1215w-648)Z^4\\ & +(-540w^3-216w^2-1152w+640)Z^3\\ & +(378w^5-504w^4+960w^3)Z^2\\ & +(36w^7-168w^6)Z\\ & +w^9\end{array}$

This iteration method produces two roots of the quintic. The remaining three roots can be obtained by using synthetic division to divide the two roots out, producing a cubic equation. It is to be noted that due to the way the iteration is formulated, this method seems to always find two complex conjugate roots of the quintic even when all the quintic coefficients are real and the starting guess is real. This iteration method is derived by from the symmetries of the icosahedron and is closely related to the method Felix Klein describes in his book.

See also

• Theory of equations

Notes

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

References

• Felix Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree, trans. George Gavin Morrice, Trübner & Co., 1888. ISBN 0-486-49528-0.
• R. Bruce King, Beyond the Quartic Equation, Birkhäuser, 1996. ISBN 3-7643-3776-1
• Harold T. Davis, Introduction to Nonlinear Differential and Integral Equations, Dover, 1962, ISBN 0-486-60971-5, Chapter 6, especially Sections 20 and 21

External links

• 53 yrs old Fitter (Common ) Batterton from Carp, likes to spend some time kid advocate, property developers in singapore and handball. Completed a cruise liner experience that was comprised of passing by Gusuku Sites and Related Properties of the Kingdom of Ryukyu.
  
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  I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
  
  
  
  Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.