Bring radical

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In algebra, a Bring radical or ultraradical is a real zero of the polynomial

$x^5+x+a, \,$

George Jerrard (1804-1863) showed that some quintic equations can be solved using radicals and Bring radicals, which had been introduced by Erland Bring (1736-1798).

1 Bring-Jerrard normal form
2 Bring radicals
3 Solution of the general quintic
4 Other Characterizations
- 4.1 Glasser's derivation
5 See also
6 External links

[edit] Bring-Jerrard normal form

$x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0\,$

then if

$y = x^4+b_1x^3+b_2x^2+b_3x+b_4\,$

we may obtain a polynomial of degree five in y, a Tschirnhaus transformation, for instance using the resultant to eliminate x. We might then seek particular values of the coefficients b_i which make the coefficients for the polynomial for y of the form

$y^5 + py + q\,$

This reduction, discovered by Bring and rediscovered by Jerrard, is called Bring-Jerrard normal form. A direct attack on the reduction to Bring-Jerrard normal form does not work; the trick is to do it in stages, using more than one Tschirnhaus transformation, in which case modern computer algebra systems make the computations relatively easy.

First, substituting x - a₁/5 in place of x removes the trace (degree four) term. We then may employ an idea due to Tschirnhaus to eliminate the x³ term also, by setting y = x² + px + q and solving for p and q so as to eliminate the x⁴ and x³ terms both, we find that setting q = 2c/5 and

$p = {\sqrt{5c(3c^2-10d)} \over 5c}\,$

eliminates both the third and fourth degree terms from

$x^5 + cx^3 + dx^2 + ex + f\,$

We now may successfully set

$y = x^4+b_1x^3+b_2x^2+b_3x+b_4\,$

$x^5 + dx^2+ex+f\,$

and eliminate the degree two term also, in a way which does not require the solution of any equation above degree three. This requires taking square roots for the values of b₁, b₂ and b₄, and finding the root of a cubic for b₃.

The general form is easy enough to compute using a computer algebra package such as Maple or Mathematica, but is messy enough that it seems advisable to simply explain the method, which can then be applied in any particular case. However, it should be noted that what is entailed is a solution to the general quintic. In any particular case, one may set up the system of three equations, and then solve for the coefficients b_i. One of the solutions so obtained will be as described, involving the roots of no polynomial higher than the third degree; taking the resultant with the coefficients so computed reduces the equation to Bring-Jerrard normal form. The roots of the original equation are now expressible in terms of the roots of the transformed equation.

Regarded as an algebraic function, the solutions to

$x^5+ux+v = 0\,$

involves two variables, u and v, 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

$z = {x \over (-u/5)^{1/5}}\,$

then we reduce the equation to the form

$x^5 - 5x - 4t = 0\,$

which involves x' as an algebraic function of a single variable t.

[edit] Bring radicals

Bring radicals can be used to obtain closed form solutions of quintic equations.

As a function of the complex variable t, the roots x of

$x^5 - 5x - 4t = 0\,$

have branch points where the discriminant 800000(t⁴ - 1) is zero, which means at 1, -1, i and -i. Monodromy around any of the branch points exchanges two of the roots, leaving the rest fixed. For real values of t greater than or equal to -1, the largest real root is a function of t increasing monotonically from 1; we may call this function the Bring radical, BR(t). By taking a branch cut along the real axis from minus infinity to -1, we may extend the Bring radical to the entire complex plane, setting the value along the branch cut to be that obtained by analytically continuing around the upper half-plane.

More explicitly, let $a_0 = 3, a_1 = {1\over100}, a_2 = -{27\over400000}, a_3 = {549/800000000}$ , with subsequent a_i defined by the recurrence relationship

$a_{n+4} = -{\frac {185193}{5278000}}\,{\frac {2\,n+5}{n+4}}a_{n+3}$

$-{\frac {9747}{ 52780000}}\,{\frac {10\,{n}^{2}+40\,n+39}{ \left( n+4 \right) \left( n+3 \right) }}a_{n+2}$

$-{\frac {57}{52780000}}\,{\frac { \left( 2\,n+3 \right) \left( 10\,{n}^{2}+30\,n+17 \right) }{ \left( n+4 \right) \left( n+3 \right) \left( n+2 \right) }}a_{n+1}$

$-{\frac {1}{6597500000}}\,{\frac { \left( 5\,n+11 \right) \left( 5\,n+7 \right) \left( 5\,n+3 \right) \left( 5\,n-1 \right) }{ \left( n+4 \right) \left( n+3 \right) \left( n+2 \right) \left( n+1 \right) }}a_n.$ For complex values of t such that |t - 57| < 58, we then have

$\operatorname{BR}(t) = \sum_{n=0}^\infty a_n (t-57)^n,\,$

which then can be analytically continued in the manner described.

The roots of x⁵ - 5x - 4t = 0 can now be expressed in terms of the Bring radical as

$r_n = i^{-n} \operatorname{BR}(i^n t)$

for n from 0 through 3, and

r 4 = - r 0 - r 1 - r 2 - r 3

for the fifth root.

[edit] 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

$\left(-\frac{p}{4}\right)^\frac{1}{4}\operatorname{BR}\left(\frac{(-5/p)^\frac{5}{4} q}{4}\right)$

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.

[edit] Other Characterizations

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

[edit] Glasser's derivation

This derivation due to M. L. Glasser finds a solution to any trinomial equation of the form:

x N - x + t

In particular, the quintic equation can be reduced to this form by the use of Tschirnhaus transformations as shown above. Let $x = ζ - 1 / (N - 1)$ , the general form becomes:

ζ = e 2π i + t φ(ζ)

where

φ(ζ) = ζ 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 ζ above may be expressed as an infinite series:

$f(\zeta) = f(e^{2\pi i}) + \sum^\infty_{n=1} \frac{t^n}{n!}\frac{d^{n-1}}{da^{n-1}}[f'(a)|\phi(a)|^n]_{a = e^{2\pi i}}$

If we let $f (ζ) = ζ - 1 / (N - 1)$ in this formula, we can come up with the root:

$x_1 = \exp(-2\pi i/(N -1)) - \frac{t}{N-1}\sum^\infty_{n=0}\frac{(te^{2\pi i/(N-1)})^n}{\Gamma(n + 2)}\frac{\Gamma(\frac{Nn}{N-1} + 1)}{\Gamma(\frac{n}{N-1} + 1)}$

N-2 other roots may be found by replacing $exp( - 2π i / (N - 1))$ by the other (N-1)th roots of unity, and the last root by using any of the symmetric function relations between the roots of a polynomial (e.g. the sum of all the roots of any polynomial in the trinomial form above must be 1). By the use of the Gauss multiplication theorem the infinite series above may be broken up into a finite series of hypergeometric functions:

$\psi(q) = (\frac{\omega t}{N-1})^q n^{qN/(N-1)}\frac{\prod^{N-1}_{k=0}\Gamma(\frac{Nq/(N-1) + 1 + k}{N})}{\Gamma(\frac{q}{N-1} + 1)\prod^{N-2}_{k=0}\Gamma(\frac{q+k+2}{N-1})}$

$x_1 = \omega^{-1} - \frac{t}{(N-1)^2}\sqrt{\frac{N}{2\pi(N-1)}}\sum^{N-2}_{q=0}\psi(q)_{N+1}F_N \begin{bmatrix} \frac{qN/(N-1)+1}{N}, \ldots, \frac{qN/(N-1) + N}{N}, 1; \\ \frac{q+2}{N-1}, \ldots, \frac{q+N}{N-1}, \frac{q}{N-1}+1; \\ (\frac{t\omega}{N-1})^{N-1}N^N) \end{bmatrix}$

where $ω = exp(2π i / (N - 1))$ . 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:

$\begin{matrix} F_1(t) & = & F_2(t)\\ F_2(t) & = & \,_4F_3(1/5, & 2/5, & 3/5, & 4/5; & 1/2, & 3/4, & 5/4; & 3125t^4/256)\\ F_3(t) & = & \,_4F_3(9/20, & 13/20, & 17/20, & 21/20; & 3/4, & 5/4, & 3/2; & 3125t^4/256)\\ F_4(t) & = & \,_4F_3(7/10, & 9/10 , & 11/10, & 13/10; & 5/4, & 3/2, & 7/4; & 3125t^4/256) \end{matrix}$

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

$\begin{matrix} x_1 & = & -t^4F_1(t) \\ x_2 & = & -F_1(t) & + \frac{1}{4}tF_2(t) & + \frac{5}{32}t^2F_3(t) & + \frac{5}{32}t^3F_3(t)\\ x_3 & = & -F_1(t) & + \frac{1}{4}tF_2(t) & - \frac{5}{32}t^2F_3(t) & + \frac{5}{32}t^3F_3(t)\\ x_4 & = & -iF_1(t) & + \frac{1}{4}tF_2(t) & - \frac{5}{32}it^2F_3(t) & - \frac{5}{32}t^3F_3(t)\\ x_5 & = & iF_1(t) & + \frac{1}{4}tF_2(t) & + \frac{5}{32}it^2F_3(t) & - \frac{5}{32}t^3F_3(t)\\ \end{matrix}$