The Third Book: Descartes’ theory of equations. I
16th March 2015
The choice of the appropriate equations
What is simplicity?
The compasses of Descartes’ youth surely describe geometrical
curves.
But their use to insert proportional means must be rejected.
It is possible to make use of curves of lower degree.
The simplicity to be requested is not that of the geometric
construction;
It is the algebraic one.
The fundamental Theorem of Algebra
Descartes’ Rule of Signs
The reduction of equations of fourth degree when the
problem is plane
“Au lieu de”
x4 + px2 + qx + r = 0,
(1)
“il faut escrire”
y 6 + 2py 4 + (p2 − 4r)y 2 − q 2 = 0.
If the value of y 2 can be found by using acceptable tools, “. . . au
lieu de (1) il faut escrire ces deux autres”:
x2 − yx +
y2
q
p
−
+ = 0,
2
2y 2
x2 + yx +
y2
q
p
+
+ = 0.
2
2y 2
Descartes’ idea?
x4 + px2 + qx + r = (x2 + yx + u)(x2 − yx + v).
u + v − y 2 = q,
(v − u)y = q,
uv = r.
We have the system of two equations

 v + u = q + y2,
q
v − u = .
y
The solution is:
u=
y2
q
p
−
+ ,
2
2y 2
v=
y2
q
p
+
+ .
2
2y 2
Since uv = r, we get what Descartes affirms.
Debeaune’s reconstruction
The two factors that Descartes proposes for an equation of the
form
x4 + px2 + qx + r = 0
are given by
x2 +
|
q
1 2 p
y +
±
− xy .
2
2
2y
{z
} |
{z
}
A
B
Hence
A2 − B 2 = x4 + px2 + qx + r.
The equality (2) is true if an only if y is a root of Descartes’
Resolvent.
(2)
An intriguing example
The equation (!)
1 2
1
5
4
2
z +
a − c z 2 − (a3 + ac2 )z + a4 − a2 c2 = 0
2
16
4
(3)
has the associated resolvent
y 6 + (a2 − 2c2 )y 4 − (a4 − c4 )y 2 − a2 (a2 + c2 )2 = 0
√
which has the root y = a2 + c2 . Hence the roots of (3) are given
by solving the second degree equations
p
a2 + c2 z +
p
z 2 + a2 + c2 z +
z2 −
3 2
a −
4
3 2
a +
4
1 p 2
a a + c2 = 0,
2
1 p 2
a a + c2 = 0.
2
(4)
After this premise Descartes writes
The problem he chooses has a well known solution. But Descartes
is not interested in its solution, but in the method by which it can
be obtained.
What he offers is a general strategy useful to analyse all the fourth
degree problems which can be solved by elementary tools.
The choice: the Problem of the Square
The square AD and the line BN are given.
It is required to extend the side AC up to E, so that EF , which
extended passes through B, is equal to BN .
Descartes assumes BD = a, BN = F E = c.
The equation which is associated to the problem in a
natural way
a(a − x)
.
x
CF 2 + CE 2 = c2 ⇒ x4 − 2ax3 + (2a2 − c2 )x2 − 2a3 x + a4 = 0.
DF = x, CF = a − x, CE =
The removal of the coefficient of x3 , by setting x = z + a2 , gives (!)
1 2
5
1
z4 +
a − c2 z 2 − (a3 + ac2 )z + a4 − a2 c2 = 0
2
16
4
We have seen√in the preceding pages that the associated resolvent
has the root a2 + c2 , and that the roots of the equation
5 4
z 4 + 21 a2 − c2 z 2 − (a3 + ac2 )z + 16
a − 14 a2 c2 = 0
are to be found between the roots given by
p
a2 + c2 z +
p
z 2 + a2 + c2 z +
z2 −
3 2
a −
4
3 2
a +
4
1 p 2
a a + c2 = 0,
2
1 p 2
a a + c2 = 0.
2
(5)
Only the first equation has positive roots, and remembering that
a
z + = x and that we must have 0 < x < a, the only solution is
2
given by
r
r
1
1 2 1 2
1 2 1 2 1 √ 2
x= a+
a + c −
c − a + a a + c2 .
2
4
4
4
2
2
Papus’solution
A Lemma (proved in algebraic terms)
Pappus’Lemma: CD2 + F E 2 = DG2
DF = x, CF = a − x, CE =
a(a−x)
, F E2
x
= (a − x)2 1 +
BD : DF = BE 0 : EE 0 , a : x = BE 0 : a ⇒ BE 0 =
2
a2
x2
.
a2
x.
E 0 G = x, DG = ax − a + x
2
2
2
(a − x)2 1 + xa2 + a2 = ax − a + x , as it is easily checked.
Descartes’ suggestion for the choice of indeterminates
The choice of DG
DE 0 = x, E 0 G = y, DG = z.
z = x + y,
(a + x)y = a2 .
ac
BF : F E = BD : DE 0 ⇒ BF = .
x
ac
2
+ c + a2 = (a + x)2 .
x
(6)
(7)
(8)
The simplest equation
The elimination of x and y between (6), (7), (8) give
az
,
a+z
(2a + z) z
,
y=
a+z
a (a2 + c2 − z 2 ) = 0.
x=−
The resulting equation in the variable z is indeed very simple, but
Descartes is right. It is not obtained in a simple way.
Lagrange
Take a careful look at the original equation!
The original equation: x4 − 2ax3 + (2a2 − c2 )x2 − 2a3 x + a4 = 0.
An obvious variable substitution: x ← ax:
2a2 − c2 2
x4 − 2x3 +
x − 2x + 1 = 0.
a2
1
2a2 − c2
1
2
2
+
= 0.
x x + 2 −2 x+
x
x
a2
x+
z 2 − 2 − 2z +
1
= z,
x
x2 +
2a2 − c2
=0
a2
1
= z 2 − 2.
x2
∧
x2 − z x + 1 = 0.