**“A New Way To
Derive The Bring-Jerrard Quintic In Radicals”**

** **

**Titus Piezas III**

*Abstract*: We
derive, *in radicals*, the Bring-Jerrard quintic using a *cubic*
Tschirnhausen transformation instead of the usual quartic Tschirnhausen
transformation which was essentially the method employed by Erland Bring
(1736-1798) and George Jerrard (1804-1863).
Certain limitations of the new method as applied to higher degrees will
also be discussed.

* *

*Contents*:

I. Introduction

II. The Principal Quintic

III. The Bring-Jerrard Quintic via a quartic Tschirnhausen transformation

IV. The Bring-Jerrard Quintic via a cubic Tschirnhausen transformation

V. Conclusion: Beyond three terms and eliminating *four*
terms

Among the
applications of the *Tschirnhausen transformation*, its more notable use
is that it enables one to find a formula for the general quintic, though one
has to go beyond radicals and use other functions, such as hypergeometric ones
as was first done by Felix Klein (1849-1925) or elliptic functions as Charles
Hermite (1822-1901) did. This
transformation was named after Count Ehrenfried von Tschirnhaus** **(1651-1708)
who in a short four-page paper “*On a method for removing all intermediate
terms from a given equation*” proposed a method to solve the general *n*th
degree equation.

It starts from the simple observation that given, say, a general cubic equation,

x^{3}+ax^{2}+bx+c
= 0

by a change of variable y = x+r for some indeterminate *r*,
a new cubic is formed,

y^{3}
+(-3r+a)y^{2}+(3r^{2}-2ar+b)y+(-r^{3}+ar^{2}-br+c)
= 0

One can then eliminate any of the intermediate terms (the *y*^{2}
or *y* term) by equating the coefficient to zero and solving for *r*
involving an equation less than a cubic.
This can obviously be applied to any *n*th degree equation to
eliminate any of its intermediate terms using an equation of degree less than *n*. Tschirnhaus’ insight was to allow more
general substitutions,

y = x^{m}
+ r_{m-1}x^{m-1 }+…r_{1}x + r_{0}

where *m* intermediate terms can be eliminated *simultaneously*,
as the *m* parameters *r*_{k} enable us to fulfill *m*
conditions. The procedure will be
illustrated in the next two sections when we derive the principal and
Bring-Jerrard quintic forms. However,
this can be broadened even more by allowing *fractional* transformations. In its most general sense, the Tschirnhausen
transformation then of a polynomial equation *f*(x) = 0 is of the form *y*
= *g*(x)/*h*(x) where *g* and *h* are polynomials and *h*(x)
does not vanish at a root of *f*(x) = 0, (Weisstein). We will make use of such a fractional form
later.

The principal quintic lacks two terms,

y^{5}
+ P_{1}y^{2} + P_{2}y + P_{3} = 0

and the importance of this form is that it is solvable in terms of hypergeometric functions and a related icosahedral equation as was first demonstrated by Klein. Given the general quintic,

x^{5}+px^{4}+qx^{3}+rx^{2}+sx+t
= 0

and the quadratic Tschirnhausen transformation,

y = x^{2}+ax+b

the variable *x* can be eliminated between the two
using resultants to form a new quintic,

y^{5}+c_{1}y^{4}+c_{2}y^{3}+c_{3}y^{2}+c_{4}y+c_{5
}= 0

where the *c _{k}* are polynomials in terms of
the coefficients

Resultant[*poly1*, *poly2*,
*var*]

where *var* is the variable to be eliminated. (And
similarly also for other computer algebra systems.) Explicitly c_{1} and c_{2} are then,

c_{1} = -5b+ap-p^{2}+2q

c_{2} = 10b^{2}+4b(p^{2}-2q)+a^{2}q+q^{2}-2pr+a(-4bp-pq+3r)+2s

The principal quintic form can then be acquired by setting *c _{1}=c_{2}=0*
and solving for the unknowns

*Example*:

Given the non-solvable quintic x^{5}-x^{4}-x^{3}-x^{2}-x-1
= 0 with discriminant *D* = 2^{4}(599), the characteristic
polynomial of the so-called *pentanacci numbers*, we use the resultant to
eliminate *x* between it and the Tschirnhausen transformation y = x^{2}+ax+b. Collecting the variable *y*,

Collect[Resultant[x^{5}-x^{4}-x^{3}-x^{2}-x-1,y-(x^{2}+ax+b),x],y]

we get the new quintic y^{5}+c_{1}y^{4}+c_{2}y^{3}+c_{3}y^{2}+c_{4}y+c_{5
}= 0 where,

c_{1} = -3-a-5b

c_{2} = -3-4a-a^{2}+12b+4ab+10b^{2}

Solving c_{1 }= c_{2 }= 0 for this
particular example conveniently involves only rational numbers and we find that
*a = -3, b = 0*. So, using the
transformation y = x^{2}-3x, the pentanacci equation has the principal
quintic form,

y^{5}+2y^{2}+47y+122
= 0

The Bring-Jerrard quintic on the other hand is important to Hermite’s solution of the general quintic in terms of elliptic functions. It was named after Erland Bring (1736-1798) and George Jerrard (1804-1863) who worked independently of each other. This form lacks three terms,

z^{5}
+ J_{1}z + J_{2} = 0

and simple scaling can reduce it even further to the *Bring
quintic form*,

z^{5}
+ z + B_{1} = 0

To eliminate *three* terms
from the general quintic, it is reasonable to assume using a cubic
Tschirnhausen transformation,

y = x^{3}+ax^{2}+bx+c

However, as was seen, the coefficients *c*_{k}
would in general involve the *k*th powers of the unknowns *a,b,c* and
by resolving this system of three equations one ends up with a sextic. This was noticed by the
philosopher-mathematician Gottfried Liebniz (1646-1716), (Tignol), as a major
difficulty of Tschirnhaus’ method.

Around
1786, Bring (and later c. 1836, Jerrard in Hamilton’s report) found a way
around the problem by a method equivalent to using a *quartic*
transformation, with the extra parameter used to prevent elevation of the
degree of the final equation. In
Adamchik’s and Jeffrey’s paper “*Polynomial Transformations of Tschirnhaus,
Bring, and Jerrard*” [1] they give a very elegant derivation of this form
and we’ll reproduce this here with a small departure in the last step. Given the principal quintic form,

y^{5} +ry^{2}+sy+t
= 0

and the quartic Tschirnhausen transformation

z = y^{4}+ay^{3}+by^{2}+cy+d

eliminating *y* between the two again using resultants,
we get the quintic,

z^{5}+c_{1}z^{4}+c_{2}z^{3}+c_{3}z^{2}+c_{4}z+c_{5}
= 0

where the *c _{k}* are in the coefficients

c_{1} = -5d+3ar+4s

c_{2} = 10d^{2}-12adr+3a^{2}r^{2}-3br^{2}+2b^{2}s-16ds+5ars+6s^{2}+5abt-4rt+c(3br+4as+5t)

Let *c _{1} = 0* and solving for d,

d = (3ar+4s)/5

The next step is the ingenious algebraic trick of
“freeing-up” the variable *c* for other tasks *by eliminating it in c _{2}*. Obviously this can be done by letting,

3br+4as+5t = 0

and solving for *b*,

b = -(4as+5t)/(3r)

Setting c_{2} = 0 and substituting into it these two
expressions for *b,d*, we get a quadratic solely in the variable *a*
and the coefficients *r,s,t* given by,

a^{2}(-27r^{4}+160s^{3}-300rst)+a(-27r^{3}s+400s^{2}t-375rt^{2})+(-18r^{2}s^{2}+45r^{3}t+250st^{2})
= 0 (eq.1)

and we find the first unknown! The discriminant *D* of this quadratic,

*D* = (-27r^{3}s+400s^{2}t-375rt^{2})^{2}-4(-27r^{4}+160s^{3}-300rst)(-18r^{2}s^{2}+45r^{3}t+250st^{2})

is in fact the discriminant of the principal quintic, up to
the factor 45r^{2}. Curiously,
note that the analogous process applied to the “principal *sextic*” will
yield a quadratic in the variable *a* whose discriminant is *not* the
discriminant of the principal sextic (a result, in general, for the principal *n*th-ic
for *n* > 5), hence making the quintic rather special.

With *a,b,d* now determined,
the role of the variable *c* appears.
In [1], the authors suggested using the *power sums* of the
coefficients to eliminate the z^{2} term (as well as for the z^{4}
and z^{3}). However, since as
was pointed out the coefficients *c _{k}* (easily given by

c_{3} = v_{0}c^{3
}+ v_{1}c^{2 }+ v_{2}c + v_{3}

c_{4} = w_{0}c^{4
}+ w_{1}c^{3 }+ w_{2}c^{2 }+ w_{3}c
+ w_{4}

(where both the *v _{i}* and

x^{5} + J_{1}x + J_{2}
= 0

as was first done by Bring, or set c_{4} = 0, solve
the quartic in *c* to get the *Euler-Jerrard quintic*,

z^{5} + E_{1}z^{2}
+ E_{2} = 0

as was first achieved by Euler (Weisstein, “*Quintic
Equation*”). In general, this kind
of Tschirnhausen transformation can simultaneously eliminate in radicals the x^{n-1},
x^{n-2}, and x^{n-3} (*or* x^{n-4}) terms of the
general *n*th degree equation for *n* > 3, though it is pointless
to apply it to the case *n* = 4 as one ends up solving the same quartic
(Hamilton, p.10). This is *not* to
say though that the quartic cannot be transformed into binomial form in
radicals in a non-trivial manner. One
can use a cubic transformation to reduce it so. While indeed it would involve a
final equation that is a sextic, it is quite easy to show that this would be
solvable in radicals. In general, we
can prove that irreducible solvable equations of the *n*th degree can be
reduced to *binomial* form in radicals using a Tschirnhausen
transformation of *n*-1 degree.
See “*Solving Solvable Quintics Using One Fifth Root Extraction*”
by this author.

Using the principal form given earlier,

y^{5}+2y^{2}+47y+122
= 0

and the quartic Tschirnhausen transformation z = y^{4}+ay^{3}+by^{2}+cy+d,
by eliminating *y* between the two we get,

z^{5}+c_{1}z^{4}+c_{2}z^{3}+c_{3}z^{2}+c_{4}z+c_{5}
= 0

where,

c_{1} = 188+6a-5d

c_{2} = 2(6139+235a+6a^{2}-6b+305ab+47b^{2})-8(94+3a)d+10d^{2}+2(305+94a+3b)c

Following the procedure outlined earlier, we find,

a = (-1509782+3243Ö2995)/411589

b = (5461621-101614Ö2995)/411589

d = 2(34160020+9729Ö2995)/2057945

Note that the discriminant of the pentanacci equation is 2^{4}(599)
and that 5*599 = 2995. These values are
enough to set c_{1}=c_{2}=0.
To set c_{3} = 0, we need to solve a rather complicated cubic in
*c*, which for this case is a one-real root cubic. Approximately this is *c* =
1010.29006103… and with all *a,b,c,d* known the Bring-Jerrard quintic
form, with approximate coefficients, is then,

z^{5 }+ (4.840918x10^{13})z
+ (1.258842x10^{17}) = 0

with the unique real root z_{1} = -1976.819519…
Reversing the transformation,

z_{1} = y^{4}+ay^{3}+by^{2}+cy+d

by solving this quartic, three of the roots will be extraneous but one,

y_{1} = -2.032892…

is precisely the real root of the given principal quintic y^{5}+2y^{2}+47y+122
= 0.

**IV. The Bring-Jerrard Quintic via a Cubic Tschirnhausen
Transformation**

It turns
out we *can* use a cubic Tschirnhausen transformation, *though it has to
be of the fractional sort*. Again,
given the principal quintic,

y^{5} +ry^{2}+sy+t
= 0

and the cubic Tschirnhausen transformation

z = (y^{3}+ay^{2}+by+c)/(y+d)

we eliminate the variable *y* still using resultants to
get the quintic,

c_{0}z^{5}+c_{1}z^{4}+c_{2}z^{3}+c_{3}z^{2}+c_{4}z+c_{5}
= 0

where the *c _{k}* are slightly messier expressions
in the coefficients

c_{1} = -ad(3d^{2}r-4ds+5t)+b(3d^{2}r-4ds+5t)-c(5d^{4}-2dr+s)+d^{2}(3d^{2}r-4ds+5t)

Set c_{1} = 0 and among the linear variables, solve
for *a*. (Solving *b* or *c* needs more work later.) Substitute this to c_{2} = 0 and we
get the equation,

*P(c,d)*-b(5c-3r)(3d^{2}r-4ds+5t)^{2}
= 0

where *P(c,d)* is a polynomial in *r,s,t* and *c,d*
and which is complicated to write down.
The more important point is that obviously the above equation is
susceptible to the same algebraic trick used earlier, namely “freeing-up” a
variable, this time *b* by letting,

5c-3r = 0

or,

c = 3r/5.

Substituting this into *P(c,d)* = 0 (which makes c_{2}
= 0), we get the quadratic solely in the unknown *d*,

d^{2}(-27r^{4}+160s^{3}-300rst)+d(27r^{3}s-400s^{2}t+375rt^{2})+(-18r^{2}s^{2}+45r^{3}t+250st^{2})
= 0 (eq.2)

which if one notices is, up to sign, essentially the same
equation as (eq.1)! Solving for *d*,
and substituting *a,c,d* into c_{3} = 0, one has to solve a cubic
in *b*. Alternatively, if the
Euler-Jerrard quintic form is desired, into c_{4} = 0 and solve a
quartic in *b*. These known values of *b,c,d* will then define the
numerical value of *a*.

*Example*:
Using the same principal quintic for comparison,

y^{5}+2y^{2}+47y+122
= 0

but this time the cubic Tschirnhausen transformation z = (y^{3}+ay^{2}+by+c)/(y+d),
by eliminating *y* between the two we get,

c_{0}z^{5}+c_{1}z^{4}+c_{2}z^{3}+c_{3}z^{2}+c_{4}z+c_{5}
= 0

where the explicit expressions are still a bit messy. Using the procedure and starting with the known result that c = 3r/5, we find that,

c = 6/5

ad(305-94d+3d^{2}) =
b(305-94d+3d^{2})-(1/5)(141-12d-1525d^{2}+470d^{3})

d = (1509782+3243Ö2995)/411589

Substituting these values for *a,c,d* into c_{3}
= 0, we have to solve a cubic in *b*, which this time has three real
roots. Arbitrarily choosing one, *b*
= -435.63831050…, which together with *d* should define the value of *a*. The Bring-Jerrard form is then,

(3.73359x10^{6})z^{5}
+ (1.49365x10^{16})z – (6.17569x10^{18}) = 0

(one can just divide by the leading coefficient) which has
the unique real root z_{1} = 234.85708481…. such that by solving the
cubic

z_{1}(y+d) = (y^{3}+ay^{2}+by+c)

which has three real roots, with two extraneous but one,

y_{1} = -2.032892…

is again the real root of the given principal quintic y^{5}+2y^{2}+47y+122
= 0.

As was pointed out in [1], by
inverting the Tschirnhausen transformation one has to deal with extraneous
solutions. To quote, “…*It is
interesting to note that if one used Tschirnhaus’ cubic transformation to solve
a quintic (using something other than radicals), then one would obtain 15
solution candidates. By using a quartic
transformation, Bring and Jerrard simplified the intermediate expressions at
the price of now generating 20 solution candidates.*” (p.93)

It turns out that one in fact *can*
use a cubic Tschirnhausen transformation, though of the fractional sort, and
the extra variable in the denominator is enough to enable us to derive the
Bring-Jerrard quintic form still in the radicals. On the downside, while the quartic transformation suggested by
Bring-Jerrard can be non-trivially applied to equations of degree *n* = 5
and above, this fractional cubic transformation can be non-trivial in radicals *only
for n = 5 and 6*. For *n* = 6,
one can use the same steps to have a final equation in the unknown *d*,
but now it is a quartic! For *n* =
7 and above, it steadily gets higher hence giving a limit to the applicability
of the method.

**V. Conclusion: Beyond three terms and eliminating four
terms**

While
Jerrard is noted for the transformation that can eliminate three terms from the
general quintic and higher degrees, he in fact proposed *four* kinds of
transformations in radicals. In
Hamilton’s report [3], these are described, namely:

1)
Eliminating x^{n-1}, x^{n-2}, x^{n-3}.

2)
Eliminating x^{n-1}, x^{n-2}, x^{n-4}.

3)
Eliminating x^{n-1}, x^{n-3} and setting c_{2}^{2}
= *m*c_{4} for an arbitrary *m*.

4)
Eliminating x^{n-1}, x^{n-2}, x^{n-3},
x^{n-4}.

The first two are familiar and we
know how to attain them. The third, a
variant of which is to eliminate x^{n-1}, x^{n-3}, x^{n-5},
would reduce solving sextics to solving a quintic. Finally, with the fourth, Jerrard hoped it could solve the general
quintic by reducing it to binomial form.
Unfortunately, in Hamilton’s analysis while these ingenious
transformations of an *n*th degree equation were indeed valid, each had an
effective lower limit, being *n* = 5 for (1) and (2), *n* = 7 for
(3), and *n* = 10 for (4). It then
implies that starting with the *decic*, as much as four terms can be
eliminated in radicals from the general *n*th degree equation!

With the
advent of computer algebra systems, it might be feasible and interesting to
find the explicit Tschirnhausen transformation for the fourth Jerrard
transformation and present it in a relatively concise manner. For those interested, the paper is: “*Inquiry
into the validity of a method recently proposed by George B., Jerrard, Esq.,
for transforming and resolving equations of elevated degrees*”, http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Jerrard/

--End--

© Titus Piezas III

Mar. 18, 2006

titus_piezasIII@yahoo.com (Pls. remove “III”)

www.oocities.org/titus_piezas/

References:

- Adamchik, V., and Jeffrey, D., “Polynomial Transformations of Tschirnhaus, Bring, and Jerrard”, ACM SIGSAM Bulletin, Vol 37, No. 3, Sept 2003, http://www.apmaths.uwo,ca/~djeffrey/Offprints/Adamchik.pdf
- Doyle, P., and McMullen, C., “Solving The Quintic By Iteration”, http://abel.math.harvard.edu/_ctm/papers/home/text/papers/icos/icos.pdf
- Hamilton, W.R., “Inquiry into the validity of a method recently proposed by George B., Jerrard, Esq., for transforming and resolving equations of elevated degrees”, http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Jerrard/
- MacTutor History of Mathematics Archive, http://www-gap.dcs.st-and.ac.uk/~history/index.html
- Piezas, T., “Solving Solvable Quintics Using One Fifth Root Extraction”, http://www.oocities.org/titus_piezas/quintics.html.
- Tignol, J., Galois’ Theory of Algebraic Equations, World Scientifc, 2001
- Weisstein, E., “Quintic Equation”, CRC Concise Encyclopedia of Mathematics, Chapman & Hall/CRC, 1999, or at http://mathworld.wolfram.com.