Download Lesson 3 - approximatingrealrootsofpolynomials

is
√ a negative quantity, then it took until the 16th century AD before
d was even admitted for study, and “imaginary” numbers remained
mysterious until the 18th century.
Cubic equations
For many centuries, from the time of the Babylonians to the sixteenth
century A.D., mathematicians sought a method to find a root of the
cubic polynomial. The problem had particular import because of its
close relationship with the classical geometric problem of trisecting an
angle by ruler and compass. For example, to trisect the angle of 60
degrees is equivalent to constructing the cosine of 20 degrees, and since
cos 3θ = 4 cos3 θ − 3 cos θ and cos 60◦ = 1/2, cos 20◦ is a root of the
cubic polynomial 4x3 − 3x − 1/2.
The Greek geometers tried to construct the cube root of a positive
quantity by straightedge and compass methods, but that turned out to
be impossible.
Some ancient Greek mathematicians, and Arabic mathematicians of
the ninth and tenth centuries, including Omar Khayyam (the famous
poet), discovered that a root of a cubic could be obtained geometrically
as the intersection of two conics. For example, the equation x3 +px = q
(with p, q > 0) can be rewritten as x3 + a2 x = a2 b; then the positive
real solution x of the equation is then the positive value of x that solves
the two equations x2 = ya, y 2 = x(b − x), equations of a parabola and a
circle, respectively. (Of course, a parabola cannot be constructed with
straightedge and compass.) [c.f. Berggren (1986), p. 126ff.]
The first general algebraic method for finding a root of a cubic x3 +
px − q was discovered by the Italian mathematician del Ferro sometime
prior to his death in 1526, and later by Tartaglia in 1535. The method
was first published by Cardano in his Ars Magna in 1545.
To solve the equation
x3 + px = q,
(1)
set x = u + v, and substitute, to get
u3 + 3u2 v + 3uv 2 + v 3 + p(u + v) = q,
or
6
(u3 + v 3 ) + 3uv(u + v) = q − p(u + v).
FINDING ROOTS
This can be solved if we can find u, v satisfying
u3 + v 3 = q and 3uv = −p.
Cubing the second equation yields
(2)
u3 + v 3 = q
u3 v 3 = −(p3 /27).
These equations are Babylon normal equations for u3 and v 3 . The
solutions are
�
q 1
4p3
3
u = +
q2 +
2 2
27
and
�
q 1
4p3
3
v = −
q2 +
.
2 2
27
Taking the real cube roots of u3 and v 3 to find u and v, we obtain
x = u + v, a positive solution to (1). Notice that for p, q > 0, (1) has
a unique real root, and that root is positive, for the function f (x) =
x3 +px−q has f (0) = −q < 0 and the derivative f � (x) = 3x2 +p > 0 for
all x, so f (x) > 0 for all x sufficiently large and the graph of y = f (x)
Cubing the second equation yields
(2)
u3 + v 3 = q
u3 v 3 = −(p3 /27).
These equations are Babylon normal equations for u3 and v 3 . The
solutions are
�
q
1
4p3
u3 = +
q2 +
2 2
27
and
�
q 1
4p3
3
v = −
q2 +
.
2 2
27
Taking the real cube roots of u3 and v 3 to find u and v, we obtain
x = u + v, a positive solution to (1). Notice that for p, q > 0, (1) has
a unique real root, and that root is positive, for the function f (x) =
x3 +px−q has f (0) = −q < 0 and the derivative f � (x) = 3x2 +p > 0 for
all x, so f (x) > 0 for all x sufficiently large and the graph of y = f (x)
crosses the positive x-axis exactly once.
Example 2. (from Cardano’s Ars Magna.) Consider
x3 + 6x = 20.
Set x = u + v, to get
u3 + v 3 + 3uv(u + v) + 6(u + v) = 20.
Set u3 + v 3 = 20, 3uv = −6, and solve for u3 and v 3 to get
√
√
u3 = 10 + 108, v 3 = 10 − 108.
To get the obvious solution x = 2, we note that
√
√
√
10 + 108 = 10 + 6 3 = (1 + 3)3 ,
√
(the last√equality is not so obvious!) so u = 1 + 3, and similarly,
v = 1 − 3, so x = u + v = 2.
Example 3. Consider the equation
(3)
x3 + 3x = 14.
Setting x = u + v, as above, we obtain
then
u3 + v 3 = 14, uv = −1,
u3 = 7 +
√
50
√
v = 7 − 50.
3
FINDING ROOTS
7
If we choose for u and v the real cube roots of u3 and v 3 , then uv is a
real number whose cube is u3 v 3 = −1, hence uv = −1. Then
√
√
x = u + v = (7 + 50)1/3 + (7 − 50)1/3
is the desired solution to (3).
Del Ferro solved one particular case of the cubic, namely the case (1),
x + px = q. The sixteenth century Italians, not recognizing negative
numbers, had to consider three different cases of the cubic:
3
x3 + px = q,
x3 + q = px,
and
x3 + px = q. The sixteenth century Italians, not recognizing negative
numbers, had to consider three different cases of the cubic:
x3 + px = q,
x3 + q = px,
and
x3 = px + q,
where in each case, p and q are > 0. (The case x3 + px + q = 0, p, q > 0,
did not arise because that equation has a unique real solution which
is negative, so, from their point of view, had no solution of interest.)
Each of the three cases involved slightly different methods of solution.
Cardano, in the Ars Magna (1545) was the first to publish solutions to
all three cases.
The general cubic equation has the form
t3 + at2 + bt + c = 0.
Cardano showed how to reduce this general equation to one of the
three cases above: eliminate the t2 term by making an appropriate
substitution, namely, x = t + a/3; then f (t) is transformed into a
polynomial (a “depressed” cubic) of the form
p(x) = x3 + px + q
for some p and q. Which case this represents depends on the signs of
p and q.
Example 4. Let g(t) = t3 − 6t2 − 4t + 9. We set x = t − 2. Then,
substituting t = x + 2 into g(t) gives
(x + 2)3 − 6(x + 2)2 − 4(x + 2) + 9 = x3 − 16x − 15 = f (x).
If x = a is a root of f (x), then t = a + 2 is a root of g(t).
8
FINDING ROOTS
The “casus irreducibilis”. One of the cases solved by Cardano was
particularly mysterious, namely the case where the cubic polynomial
has three real roots. For those polynomials, the solution of Cardano
involves imaginary numbers.
Example 5. We illustrate this situation with the equation
x3 = 7x + 6,
which has roots x = −1, −2 and 3 (Cardano would have sought only
the root x = 3.) We set x = u + v in this case, to get
(u3 + v 3 ) + 3uv(u + v) = 7(u + v) + 6,
which is solved if we can solve
u3 + v 3 = 6, 3uv = 7.
We set u3 = 3 + z, v 3 = 3 − z, then clearly u3 + v 3 = 6, while
343/27 = u3 v 3 = 9 − z 2 ,
hence
2
x3 = 7x + 6,
which has roots x = −1, −2 and 3 (Cardano would have sought only
the root x = 3.) We set x = u + v in this case, to get
(u3 + v 3 ) + 3uv(u + v) = 7(u + v) + 6,
which is solved if we can solve
u3 + v 3 = 6, 3uv = 7.
We set u3 = 3 + z, v 3 = 3 − z, then clearly u3 + v 3 = 6, while
hence
so
343/27 = u3 v 3 = 9 − z 2 ,
z 2 = 9 − 343/27 = −100/27,
√
10 −3
z=±
.
9
Thus
√
√
10 −3 3
10 −3
3
u =3+
,v = 3 −
.
9
9
If we happen to observe that
√
√
9 + −3 3 3
9 + −3 3
3
u =[
] ,v = [
],
6
6
then we can set
√
√
9 + −3
9 − −3
u=
,v =
6
6
and
√
√
x = u + v = (9 + −3)/6 + (9 − −3)/6] = 3.
Now let
√
−1 + −3
ω=
,
2
a cube root of unity in C. Then u3 is also the cube of
√
√
9 + −3
−3 + 2 −3
(
)ω =
6
3
and of
√
√
9 + −3 2 −3 − 5 −3
(
)ω =
.
6
6
FINDING
ROOTS
So we can let
9
√
√
−3 + 2 −3 −3 − 2 −3
x=
+
= −2.
3
3
or
√
√
−3 + 5 −3 −3 − 5 −3
x=
+
= −1.
6
6
Thus x = −1, −2 and 3 are the solutions of the equation
x3 − 7x − 6 = (x − 3)(x + 1)(x + 2).
In this example, Cardano’s method finds the three real roots of the
polynomial x3 − 7x − 6, but expresses all three as the sums of complex
numbers.
Cardano’s solution of the cubic in this case is the first situation in
the history of mathematics in which complex numbers appeared in an
essential way in the solution of a “real” problem.
We can show that for an equation such as x3 = 7x + 6, Cardano’s
method must always express the real roots as differences of non-real
complex numbers. To do so, we first obtain a criterion for a cubic to
x3 − 7x − 6 = (x − 3)(x + 1)(x + 2).
In this example, Cardano’s method finds the three real roots of the
polynomial x3 − 7x − 6, but expresses all three as the sums of complex
numbers.
Cardano’s solution of the cubic in this case is the first situation in
the history of mathematics in which complex numbers appeared in an
essential way in the solution of a “real” problem.
We can show that for an equation such as x3 = 7x + 6, Cardano’s
method must always express the real roots as differences of non-real
complex numbers. To do so, we first obtain a criterion for a cubic to
have three real roots.
Proposition 1. Let f (x) = x3 − px + q have distinct complex roots.
Then f (x) has three real roots if and only if 27q 2 < 4p3 .
Proof. The derivative f � (x) = 3x2 − p. If p < 0, then 27q 2 > 4p3 , and
also f � (x) > 0 for all x, so f (x) has exactly one real
√ roots. If p >√0,
write p = s2 with s > 0. Then f � (x) √
= 0 for x = s/ 3 and
√ x = −s/ 3,
and f has three real roots iff f (−s/ 3) > 0 and f (s/ 3) < 0. Now
−s
−s3
s
2s3
f ( √ ) = √ + s2 √ + q = √ + q > 0
3
3 3
3
3 3
iff
q>
−2s3
√ ,
3 3
and similarly, f ( √s3 ) < 0 iff
2s3
q< √ .
3 3
Thus f has three real roots iff
10
2s3
|q| < | √ |,
FINDING 3ROOTS
3
which is equivalent to
q2 <
4s6
4p3
=
,
27
27
the desired inequality.
�
Proposition 2. Let f (x) = x3 − px + q have distinct complex roots.
Then f (x) has three real roots iff Cardano’s method gives roots that are
sums of non-real complex numbers.
Proof. In Cardano’s method we set x = u + v, then set
u3 + v 3 = −q
3uv = p;
we introduce z so that
q
q
u3 = − + z, v 3 = − − z
2
2
from which it follows that
q2
p3
z2 =
− .
4
27
Cardano’s method will yield u and v non-real complex numbers if and
only if z 2 < 0. As shown above, f (x) has three real roots if and only if
4p3
3uv = p;
we introduce z so that
q
q
u3 = − + z, v 3 = − − z
2
2
from which it follows that
q2
p3
z2 =
− .
4
27
Cardano’s method will yield u and v non-real complex numbers if and
only if z 2 < 0. As shown above, f (x) has three real roots if and only if
q2 <
4p3
,
27
�
exactly the condition that z 2 < 0.
Vieta’s method. Later in the 16th century, Vieta (1593) discovered
a way to use trigonometry to find the three real roots of the polynomial
f (x) = x3 − px − q
when 27q 2 < 4p3 . The method uses the trigonometric identity
4 cos3 θ = 3 cos θ + cos 3θ.
(∗)
Multiply (*) by 2m3 for any real number m to get
8m3 cos3 θ = 6m3 cos θ + 2m3 cos 3θ.
If we set x = 2m cos θ, then this equation becomes
x3 − 3m2 x − 2m3 cos 3θ = 0.
If we can solve the equations
p = 3m2 , q = 2m3 cos 3θ
for m and 3θ in terms of p and q, then f (x) will have the solutions
x = 2m cos θ, x = 2m cos(θ + 2π/3) and x = 2m cos(θ − 2π/3).
FINDING ROOTS
Now the equation q = 2m3 cos 3θ is solvable iff −1 ≤
equivalently,
q 2 ≤ 4m6 .
When p = 3m2 , this becomes
11
q
2m3
≤ 1, or
p3
).
27
But f (x) has three real roots iff 27q 2 < 4p3 . Thus whenever f (x) has
three real roots we can solve for 3θ and find the three roots of f (x) by
Vieta’s trigonometric method.
q 2 ≤ 4(
Quartic equations. Once having learned how to solve a cubic, it
was only a short time before Cardano’s student Ferrari (born 1522)
discovered how to solve a quartic, sometime before 1541. Here is how
it is done.
Given the polynomial equation
y 4 + ay 3 + by 2 + cy + d = 0,
we first make the substitution y = z − a/4 to get a new equation in z
in which the coefficient of z 3 is 0. Thus we reduce to an equation of
the form
z 4 + pz 2 + qz + r = 0,
a “depressed” quartic. Now isolate the term z 4 and put the other terms
on the right side, then add to both sides t2 z 2 + t4 /4, to get
z 4 + t2 z 2 + t4 /4 = t2 z 2 + t4 /4 − pz 2 − qz − r.
The left side is a perfect square, namely (z 2 + t2 /2)2 , and we can solve