“Deriving The Bring-Jerrard Quintic Using A Quadratic Transformation”
Titus Piezas III
Abstract: It can be shown that by passing through the Brioschi quintic form, a quadratic transformation can suffice to transform the general quintic to the Bring-Jerrard form. This is in contrast to the quartic transformation found by Erland Bring and independently by George Jerrard, and the cubic one recently found by this author. A new one-parameter quintic form (z-5)(z2+15)2 + p = 0 which the general quintic can be reduced to in radicals will also be discussed.
Dedicated to Shelli Manuel-Tomacruz, for remembering.
Contents:
I. Introduction
II. The Brioschi Quintic
III. The Quadratic Transformation
IV. Other One-Parameter Forms
V. Beyond the Quintic
VI. Conclusion
Given a version of the principal quintic,
x5 + 5rx2 + 5sx + t = 0
we want to know for what kth degree transformation can one derive a Bring-Jerrard quintic form,
y5 + J1y + J2 = 0
such that,
a) the Ji are derived in radicals from the r,s,t.
b) the roots xi are related to yi by the kth degree transformation for some minimum k.
To recall, a Tschirnhausen transformation of the equation f(x) = 0 could be of form y = x+r, y = x2+ax+b, y = 1/x, or in general y = g(x)/h(x) where g and h are polynomials such that h(x) vanishes for no root of f(x) = 0. (Dickson, p. 210)
It has been known since Tschirnhaus (1683) that a quadratic transformation y = g(x) can eliminate two terms from the general nth degree equation, or in the case of the general quintic, reduced it in radicals to principal quintic form. He also thought (Adamchik, Jeffrey, p.92) that a transformation y = g(x) where g was a cubic polynomial would suffice to eliminate, in radicals, three terms. However, as pointed out by Leibniz and others (Tignol), this entails that the Ji would be determined by roots of sextic equations which, in general, would not be solvable in radicals.
The first transformation known that satisfies the two conditions was found by Bring (1786) and later by Jerrard (c.1836) and was of form y = g(x) for k = 4. Recently, the author found that a cubic Tschirnhausen transformation could indeed derive in radicals the Bring-Jerrard form but it had to be the rational kind y = g(x)/h(x). (And is only applicable to the quintic and sextic.) In this paper, we will discuss that even a quadratic transformation of form g(x,y) = h(x,y) will do and this in fact is just a composition of two Tschirnhausen transformations. Since it is highly doubtful that a linear relationship can exist between the xi and yi, then k = 2 might be the minimum degree possible.
The Brioschi quintic form is important to this alternative way to derive the Bring-Jerrard form. As has already been pointed out, one of the uses of the Tschirnhausen transformation is that it enables one to find a formula to solve the general quintic, though one has to go beyond radicals. Felix Klein (1849-1925) found a hypergeometric formula for the principal quintic, though he also had to make use of Brioschi quintic. This form, named after Francesco Brioschi (1824-1897), is given by,
w5 -10cw3 + 45c2w - c2 = 0
which has the nice discriminant D = 55c8(-1+1728c)2. In addition to the Bring and the Euler quintics this is the third one-parameter form to which the general quintic can be reduced to in radicals. This can be done by using the rational quadratic Tschirnhausen transformation,
x = (aw+b)/(c-1w2-3)
By eliminating w between the two using resultants we get a new quintic,
f0x5 + f3x2 + f4x + f5 = 0
where the fi explicitly are,
f0 = -1+1728c
f3 = 5c(8b3 + ab2 + a3c + 72a2bc)
f4 = 5c(-b4 + a3bc + 18a2b2c + 27a4c2)
f5 = c(b5 - 10a2b3c + a5c2 + 45a4bc2)
which, after dividing by the leading coefficient f0, is already in principal quintic form,
x5 + 5rx2 + 5sx + t = 0
To transform the principal form to Brioschi form, given r,s,t the objective is to equate coefficients to have a system of three equations in three unknowns a,b,c. For simplicity, let V = (-1+1728c)/c, hence,
Vr = 8b3 + ab2 + a3c + 72a2bc (eq.1)
Vs = -b4 + a3bc + 18a2b2c + 27a4c2 (eq.2)
Vt = b5 - 10a2b3c + a5c2 + 45a4bc2 (eq.3)
By resolving this system to a single equation, the amazing thing is that the final resultant always has a quadratic factor! With modern computer algebra systems one in fact can do this mechanically and end up with a 24th degree equation which factors as a quadratic and a 22nd degree equation (not the expected 3*4*5 = 60th degree as it has trivial factors). Dickson, by a series of ingenious algebraic manipulations involving polynomial invariants of the icosahedral group, managed to find explicitly this quadratic factor (in the variable b) without elevation of degree. (The interested reader is referred to pp. 244-245 of Modern Algebraic Theories.) This is given by,
(r4-s3+rst)b2 + (-11r3s-2s2t+rt2)b + (64r2s2-27r3t-st2) = 0
Incidentally, the discriminant D of this quadratic,
D = (-11r3s-2s2t+rt2)2 - 4(r4-s3+rst)(64r<2s2-27r3t-st2)
disregarding the factor r2, is the same as the discriminant of x5 + 5rx2 + 5sx + t = 0 which is given by 55D, a result similar for the quadratic encountered in deriving the Bring-Jerrard form. To find a,c, the quadratics for them unfortunately are rather long and complicated expressions in terms of r,s,t. For convenience, they are usually defined also in terms of b. First define the variable V as,
r2(-bs2+brt-st) V = (b2r-3bs-3t)3
Then a and c are,
a = (-8b3r-72b2s-72bt+Vr2)/(b2r+bs+t)
c = 1/(1728-V)
(Duke, Tóth, p.10) and we have all the unknowns a,b,c! Example, given,
x5+5x2+10x+2 = 0
so r =1, s = 2, t = 2. Hence,
a = (16/27)(5071-179Ö871)
b = (1/3)(-17+Ö871)
c = (1/1875929344)(615193+20894Ö871)
and the roots xi of x5+5x2+10x+2 = 0 are related to the roots wi of the Brioschi quintic w5 -10cw3 + 45c2w - c2 = 0, by x = (aw+b)/(c-1w2-3).
As one can see, even with small integral values for r,s,t, the numbers defining c are quite large. This can be explained since, interestingly, if the quadratics for a and c were given explicitly, they would be expressions of form,
(r4-s3+rst)3a2 + Poly1(r,s,t) a + Poly2(r,s,t) = 0
Poly3(r,s,t)2c2 + Poly4(r,s,t) c + (r4-s3+rst)5 = 0
Hence, the leading coefficient of the quadratic in a is a cube, for c is a square, plus the constant term of the latter is a perfect fifth power! Considering that the Brioschi quintic is expressed in terms of c and this form enables one to solve the general quintic in terms of hypergeometric functions, it is quite reminiscent of solving solvable quintics (non-binomial) in terms of radicals since it can be shown that its quartic Lagrange resolvent also has a constant term that is a perfect fifth power. For example, given the solvable,
y5+10y3+15y2-140y+2897 = 0
then its solution in radicals is given by the quartic with a fifth-power constant term,
z4+597z3+26919z2-2614747z-195 = 0
such that y1 = z11/5 + z21/5 + z31/5 + z41/5 where the zi are the quartic roots.
To get to the Bring-Jerrard form we simply do a second transformation on the Brioschi quintic. This is given by y = (w+d)/(c-1w2-3). Again eliminating w, we get,
y5 + g3y2 + g4y + g5 = 0
where,
g3 = 5c(8d3 + d2 + 72cd + c)/(-1+1728c)
g4 = 5c(-d4 + 18cd2 + cd + 27c2)/(-1+1728c)
g5 = c(d5 - 10cd3 + 45c2d + c2)/(-1+1728c)
Since c is already known, one can just set g3 = 0 and solve for d to get the Bring-Jerrard quintic (or g4 = 0 for the Euler-Jerrard one). The principal quintic (in x) and the Bring-Jerrard (in y) are then connected by the “bridge” which is the Brioschi quintic (in w). By eliminating w between the two transformations,
x = (aw+b)/(c-1w2-3), y = (w+d)/(c-1w2-3)
we then get a quadratic relation between the roots xi and yi.
Theorem 1.
The general quintic can be transformed in radicals to Bring-Jerrard form using only quadratic transformations.
Given the principal quintic,
x5 + 5rx2 + 5sx + t = 0
one can derive a particular Bring-Jerrard form,
y5 + g4y + g5 = 0
such that only a quadratic transformation (or mapping) exists between the roots xi and yi and which is given by,
(-3c+d2)x2 + 2(3ac-bd)xy - c(b-ad)(x-ay) + (b2-3a2c)y2 = 0
where a,b,c,d and g4, g5 are radicals in terms of r,s,t. Explicitly, these are,
a = (-8b3r-72b2s-72bt+Vr2)/(b2r+bs+t)
c = 1/(1728-V)
g4 = 5c(-d4 + 18cd2 + cd + 27c2)/(-1+1728c)
g5 = c(d5 - 10cd3 + 45c2d + c2)/(-1+1728c)
where V is,
r2(-bs2+brt-st) V = (b2r-3bs-3t)3
and b,d is any root of,
(r4-s3+rst)b2 + (-11r3s-2s2t+rt2)b + (64r2s2-27r3t-st2) = 0
8d3 + d2 + 72cd + c = 0
Example. Using the same principal quintic,
x5+5x2+10x+2 = 0
where a,b,c have been solved for earlier what remains is to solve the cubic in d and choose its real root d = -0.02306458…, giving the Bring-Jerrrard form (with approximate coefficients),
y5 + (6.0977x10-8)y + (2.8128x10-10) = 0
which has the sole real root y1 = -0.004579… Using this value for y in the quadratic relation and solving for x,
x1 = -0.225328…, x2 = 0.604459…
with the first as the correct root of the principal quintic.
It was mentioned that there are three one-parameter forms that the general quintic can be reduced to in radicals, not counting equivalent forms that can be obtained by a multiplicative inverse. However, one in fact can find a distinct fourth. These four are given by,
w5 + w + B1 = 0
w5 + w2 + E1 = 0
w5 - 10cw3 + 45c2w - c2 = 0
w5 - 5gw3 + 10g2w - g2 = 0
with the last one already tantalizingly close to the solvable de Moivre’s quintic,
x5 - 5mx3 + 5m2x - m2 = 0
The fourth was found by analogy to the Brioschi quintic. Defining,
w5 + pcw3 + qc2w - c2 = 0
and the transformation y = (aw+b)/(c-1w2+d) with unknowns p,q and a,b,c,d, we get,
h0y5+h1y4+h2y3+h3y2+h4y+h5 = 0
with h1 = -c(5d2-3dp+q)(a+bd2-bdp+bq). Setting h1 = 0 and using the first factor, then,
q = -5d2+3dp
While the expression for h2 is not given, by substituting this into h2 = 0, we get,
h2 = c(10d-3p)(ab-2b2d2+2a2cd3+b2dp-a2cd2p) = 0
Solving for p using the first factor yields the Brioschi form while the second factor yields the new form. Sparing the reader the rest of the algebra, in summary, given,
w5 - 5gw3 + 10g2w - g2 = 0
with discriminant D = 55g8(-1+32g)(-1+36g) and the transformation,
y = (aw+b)/(g-1w2-1)
where g = -(ab+3b2)/(3a2) gives a quintic already in Bring-Jerrard form,
y5 - (5/27)(a+3b)b3 y + (1/27)(a+3b)(a+4b)b3 = 0
Equating coefficients with a generic Bring-Jerrard y5 + sy + t = 0 and solving for the unknown a,b involves only a quartic and in the variable b is given by,
5b4s+25b3t-27s2 = 0
As a last point, the last two forms can also be put into a more aesthetic form. Using the transformations w = 1/(z2+20) and w = 4/(z2+15) on the Brioschi and the fourth, respectively, yields,
c(z-5)(z2+20)2 + 1 = 0
g(z-5)(z2+15)2 + 32 = 0
Simple scaling on the variables c and g implies the general quintic can be reduced, in radicals, to any of the beautiful and simple forms,
(z-5)(z2+20)2 + p1 = 0
(z-5)(z2+15)2 + p2 = 0
A. Sextics
By extrapolating Jerrard’s quartic transformation to the nth degree equation, the sextic, after scaling variables, can be reduced to a form depending only on two parameters,
x6 + x2 + a1x + a2 = 0
x6 + x3 + a1x + a2 = 0
The general method utilizes what Klein calls an “accessory irrationality”, a radical whose usual function allows the elimination of the xn-1 and xn-2 terms. In 1861, P. Joubert derived what is called the Joubert sextic form,
x6 + a1x4 + a2x2 + a3x + a3 = 0
which can be attained “…without any accessory irrationalities”!1 Another form, the Maschke sextic form named after Heinrich Maschke (1853-1908), can be transformed into the Joubert form using a cubic Tschirnhausen transformation. (See http://www.math.rutgers.edu/courses/535/535-f02/cubics.pdf). Finally, the general 6th degree equation can also be reduced to the Valentiner sextic form (after Siegfried Valentiner(?)), which can be solved in terms of generalized hypergeometric functions, in a manner analogous to Klein’s solution of the general quintic by reducing it to the Brioschi form.
B. Septics
Likewise, the general septic2 (unfortunate name!) can be reduced to a form depending now on three parameters,
x7 + x3 + a1x2 + a2x + a3 = 0
x7 + x4 + a1x2 + a2x + a3 = 0
(x2 + ½)2 x3 + p1x2 + p2x + p3 = 0
with the last also found by Jerrard. One of Hilbert’s famous problems (the 13th) asks if the general seventh degree equation could be reduced even further and be expressed “…by means of functions of only two arguments”. (D. Joyce). The question was finally settled in 1957 by Andrei Kolmogorov and Vladimir Arnold.
There is a different kind of transformation, still in the radicals, for the septic (and above). Jerrard proposed that the general nth degree equation can be reduced to,
xn + a1xn-2 + ka12 xn-4 + a2xn-5 + … + an-3 = 0
for some arbitrary k, with the hope that as applied to n = 5, it would transform it to the solvable de Moivre quintic. However, Hamilton (p.3-4) stated that the method works only for n = 7 and higher. So by choosing k = ¼ and scaling the variable x, it then reduces the nth degree equation, for n > 6, to the form,
(x2 + ½)2 xn-4 + p1xn-5 +…+ pn-4 = 0
We can conclude this paper by asking some questions:
For [1], duplicates resulting from elementary transformations can be weeded out simply by looking at the discriminant. For the four, these are given by,
D1 = 44 + 55B1, D2 = 2233 + 55E1
D3 = 55c8(-1+123c)2, D4 = 55g8(-1+25g)(-1+62g)
hence those of additional reduced forms should be distinct. Furthermore, while we now know of transformations of degree k = 2,3,4 between the principal and Bring-Jerrard quintics, it seems unlikely there will be one for k = 1. As was pointed out by other authors, inverting these transformations involve getting spurious numerical solutions to the principal quintic, the one for k = 4 naturally generating the most, with the correct solutions ascertained only by testing. Minimum k would imply less extraneous solutions.
For [2] and [3], in a paper prior to this one, it was mentioned that the rational cubic transformation to eliminate three terms in radicals from the general nth degree equation would suffice for n = 5,6 but no higher, since the coefficients of these transformations were dependent on n, and for n > 6 already involve higher than quartic roots. Is k = 3 the least degree then of a transformation to eliminate three terms from the sextic? Similarly, is it k = 4 for the septic?
--End--
Footnotes:
© Titus Piezas III
April. 10, 2006
titus_piezasIII@yahoo.com (Pls. remove “III”)
www.oocities.org/titus_piezas/
References: