After the discovery of the general solutions
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Quite
lot of mathematicians came forward with different variations of the solution
of equations of third and fourth degree after the Ars Magna. Most known are: Gottfried
Wilhelm von Leibniz (1646--1716) seems to be the first to verify del
Ferros formulas and thereby giving an algebraic proof in contrary to the
earlier existing geometrical proofs. This was done by inserting the three
solutions x1,x2,x3 in the
expression (x-x1)(x-x2)(x-x3)
which is documented in a letter he sent to Christian Huygens
(1629--1695) in March 1673. Gauss (1777-1855)
showed in his thesis (1799) that there for every polynomial p
of degree > 1 exists a complex number (z) such
that p(z) = 0. (The Fundamental Theorem of Algebra) That (x-x1)(x-x2)(x-x3)
= 0 gives a cubic equation had much earlier been observed by
Harriot. Alexandre
Théophile Vandermonde (1735--1796) and Joseph-Louis Lagrange
(1646--1716) did independent of each other find a description of the solution
of the cubic equation. In 1591
Viète suggested the following solution for the cubic equation Let x
= k cos z . The equation then becomes k3cos3z
- 3pkcos z =2q. Note
that cos 3z =4cos3z - 3cos z, so if k
is chosen such that k2= 4p this
will give us 4cos3z - 3cos z =2q/pk which leads
to cos 3z = q/(p3/2) The
solutions finally becomes:
The
values could now be found using tables for trigonometric functions. This
solution evades the problem with a real solution being the sum of two complex
numbers, which is the case using del Ferros method. His method has this
problem especially in the case when q2 < p2.
Harriots solution:
this
gives us y6-2qy3=p3 which
is an equation of second degree in y3 from
which we can get y3= sqrt(p3+q2)+q. Vandermondes and Lagranges solution
Let x,y,z
be the solutions to a given cubic equation. Then every symmetrical polynom in
x,y,z can be written as a polynom in the coefficients of the cubic
equation. This was known by Newton at least for the cubic equation and notes
about that can be found in his papers dated to the years 1665--1666,
according to Whiteside. Then if a is a primitive
cubic root to 1, which means that a is a complex
number such that a3 = 1, a ¹ 1. Note that (a3 -1) = (a2 +a +1)( a- 1) Lagrange saw that the
expression t = x+ay +a2z would take six
different values depending on the order of the roots x,y,z.
Those six values are solutions to an equation of degree six: (x-t1)
(x-t2) (x-t3) (x-t4) (x-t5) (x-t6)=0,
and which coefficients can be shown symmetrical in x,y,z and hence
possible to express as polynom in the coefficients of the cubic equation.
Lagrange names this equation the resolvent and it is
possible to solve as it an equation of second degree in x3.
This can easily be shown by ordering t1... t6
as follows: t1=
x+ay +a2z, t2= at1, t3=
a2t1 t4=
x+az +a2y, t5= at4, t6=
a2t4 (x-t1)
(x-t2) (x-t3) = (x-t1) (x-at1) (x-a2t1) =x3
- t13 and the
resolvent then becomes: (x3
- t13)(x3 - t43)=
x4 - (t13+ t43)x3
- t13t43 = 0 When the
six solutions to the resolvent equation then are found the solutions to the
cubic equation are found by: x=1/3((x+y+z)
+t1 +t4) x=1/3((x+y+z)
+a2t1 +at4) x=1/3((x+y+z)
+at1 +a2t4) and the problem
will now be to identify t1 and t4
among the solutions to the resolvent equation. Vandermonde had a slightly
different way of approaching the problem and got the variables u = t13
and v = t43 but did not find any way
of choosing suitable cubic roots to insert in the formulas but suggested
instead that the three roots should be sorted out by inserting the candidates
to the solution in the original equation. But
Lagrange saw that if t now is a solution to the resolvent
equation, then it is possible to rearrange x,y,z so that t1=
x+ay +a2z. He then
observed that (x+ay+a2z)(x+a2y +az) is symmetric in x,y,z
so its value can be determined from the coefficients
of the cubic equation. Now it
is possible to get the roots by: x1=1/3((x+y+z)
+t + w/t) x2=1/3((x+y+z)
+a2t + w/a2t) x3=1/3((x+y+z)
+at + w/at) where t
can be any of the six roots of the resolvent equation. This can
be generalized to equations of fourth degree, which was done by both
Vandermonde and Lagrange. x1=1/4((x+y+z+r)+
t1 + t2 + w/t1 t2) x2=1/4((x+y+z+r)
- t1 + t2 - w/t1 t2) x3=1/4((x+y+z+r)
+ t1 - t2 - w/t1 t2) x4=1/4((x+y+z+r)
- t1 - t2 + w/t1 t2) In 1775 John
Landen published an entirely different approach to solution of a cubic
equation, namely by using calculus tools, differentiation, integration and
separation of variables. The q in
the equation x3 + px + q =0 he treated as a
function of x. Thus q
= -x3 - px = - x(x2+p), q(0)
= 0, q´(0)= - p ,q"(0) = 0 we can
then easily see that q" = -18q/(q´-2p), from which we
get that q"q´-2pq=-18q multiplying by q´
then gives q"(q´)2 -2pqq´= -18qq´ and
integrating both sides (q´)3 - 3p(q´)2 = -27(q)2
+c and by using the initial conditions, we will find that c
= - 4p3 . Thus we
have that (q´)3 - 3p(q´)2 = -27(q)2
- 4p3 that is (q´)2(q´-3p)=-27(q)2-4p3 By
substituting explicit formula for q´ in q´-3p
on the left side we get that (q´)2
(3x2 +4p)= -27(q)2 - 4p3 It can now be
observed that the above dx/(3x2
+4p)0,5 =dq/(27q2 +4p3)0,5 If p>0
the integration will yield 1/ from
which x = - If p<0
, the integration will yield , in the above formula Arcsin
in place of Arcsinh; x = - Since in
this formula, Arcsin is a multivalued function, by selecting
an appropriate branch, that is n in the formula belov, we
can find all the roots; x = + An analytical
solution, taken largely from Patel's 1980 Ph.D. thesis (Patel, 1980)
is presented here. A
general cubic equation can be written as: C1 x3
+ C1 x2 + C1 x + C1 = 0
(A-1) or, where: The
quadratic term in equation A-2 can be eliminated as follows: let, Now, The solution
to equation A-7 depends upon the sign of the discriminant: D may be zero,
greater than zero, or less than zero. The solution to equation A-7 for each
of these three cases is now shown. Case 1:
For D > 0, equation A-7 has one real root and two
imaginary roots. The real root is given by: where,
(A-10) (A-11) The two
imaginary roots are given by, (A-12) (A-13) Case 2:
For D = 0, there are three real roots and at least two are
equal. Since both A and B can be positive as well as negative, D
can become zero either by A and B
simultaneously becoming zero, or by and canceling each other. When
A and B are both zero, then there are three equal roots.
This is an inflection point and is the case when predicting the critical
point with a cubic equation of state. The three real roots are given by:
(A-14) (A-15) Case 3:
For D < 0, there are three, distinct, real roots which are given by the
following trigonometric functions: (A-16) where, i
= 1, 2, or 3, k = 0, 1, or 2, and, (A-17) In
equation A-17, is in degrees. The minus sign applies when B
> 0, and the plus sign applies when B < 0.
For all three cases, the corresponding three roots of equation A-2 are given
by A-4 which is rewritten as: Note,
while usually very convenient, there are cases when this analytical solution
will not work, or worse, predict incorrect roots. In such cases, an iterative
solution must be used.
For a
more complete treatment, this analytical solutions is presented in the book Theory
of Equations by J.V. Uspensky (Uspensky, 1948). |
+Arcsin(
