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# Quintic equation

In mathematics, a quintic equation is a polynomial equation in which the greatest exponent on the independent variable is five. For example:

$x^5-4x^4+2x^3-3x+7=0\,$

Finding the zeroes of a polynomial — values of x which satisfy such an equation — given its coefficients was long a prominent mathematical problem. The linear and quadratic cases fell fairly quickly; after a while cubic and quartic succumbed. But if there was some pattern to the formulæ no one could see it, and the quintic was proving to be extremely stubborn.

Eventually, Paolo Ruffini and Niels Abel were able to prove that there is no single finite expression of +, -, ×, ÷, and radicals that can produce them from the coefficients for all quintics. This is sometimes, mistakenly, taken to mean that there is no algebraic solution to the general quintic, which is false. Such a solution is given below. It should be noted however that numerical methods such as Newton's method give excellent results if all we require are numerical values for the roots, and that various transcendental functions such as the theta function or the Dedekind eta function can be used to give closed expressions.

The honour of proving the quartic formula to be the last of its kind, i.e. that there was no solution in radicals to the general sextic, septic, octic, formula, and so on, fell to Evariste Galois, who had an ingenious insight which reduced the issue to a question of group theory.

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## Bring-Jerrard normal form

If

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

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, by for instance using the resultant to eliminate x. We might then seek particular values of the coefficients bi which make the coefficients for the polynomial for y of the form

$y^5 + px + q\,$

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.

We first note that substituting x5 - 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

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

in

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

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 bi. 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.

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 ai 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

r4 = - r0 - r1 - r2 - r3

for the fifth root.

## Algebraic 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.