Making Waves in Vector Calculus

Transcription

How Cubic Equations
(and Not Quadratic)
Led to Complex Numbers
J. B. Thoo
Yuba College
2013 AMATYC Conference, Anaheim, California
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Are you sitting in the right room?
This talk is about the solution of the general cubic equation, and
how the solution of the general cubic equation led mathematicians
to study complex numbers seriously. This, in turn, encouraged
mathematicians to forge ahead in algebra to arrive at the modern
theory of groups and rings. Hence, the solution of the general cubic
equation marks a watershed in the history of algebra.
We shall spend a fair amount of time on Cardano’s solution of the
general cubic equation as it appears in his seminal work, Ars magna
(The Great Art).
Outline of the talk
Motivation: square roots of negative numbers
The quadratic formula: long known
The cubic formula: what is it?
Brief history of the solution of the cubic equation
Examples from Cardano’s Ars magna
Bombelli’s famous example
Brief history of complex numbers
The fundamental theorem of algebra
References
Girolamo Cardano, The Rules of Algebra (Ars Magna),
translated by T. Richard Witmer, Dover Publications, Inc.
(1968).
Roger Cooke, The History of Mathematics: A Brief Course,
second edition, Wiley Interscience (2005).
Victor J. Katz, A History of Mathematics: An Introduction,
third edition, Addison-Wesley (2009).
Amy Shell-Gellasch and J. B. Thoo, Intersecting Mathematics
and History: Topics in Elementary Mathematics from an
Historical Viewpoint, in preparation.
<http://ms.yccd.edu/cv-and-pubs.aspx#books>
Motivation: square roots of negative numbers
Myth
Many of today’s popular algebra textbooks introduce imaginary
numbers along these lines, saying1
The complex number system allows us to solve equations
such as x 2 + 1 = 0 that have no real number solutions.
The implication is that mathematicians became interested in
complex numbers because they were unable to solve certain
quadratic equations with only the real numbers.
1
Elayn Martin-Gay, Beginning & Intermediate Algebra, 5th edition, p. 634.
Myth buster
The truth is that, as late as the 16th century, when confronted with
an equation like
x 2 + 1 = 0 or x 2 + 2x + 2 = 0,
one could simply say that the equation has no solution, and that
would have been the end of it.
Square roots of negative numbers were considered to be “sophistic,”
“impossible,” “imaginary,” and “useless” objects. Indeed, imaginary
numbers were not understood as late as the 16th century beyond
their being a convenient contrivance.
The quadratic formula: long known
Quadratic problems
Ancient Egypt
Berlin Papyrus 6619: The area of a square of 100 is equal to that of two smaller
squares. The side of one is 2 4 the side of the other . Let me know the sides of
the two unknown squares [x 2 + ( 43 x)2 = 100].
Solved by the method of false position. In the solution, the square root of
1 2 16 is taken, yielding 1 4.
Ancient Mesopotamia
Cuneiform tablet AO 8862 (Louvre, Paris): I have multiplied the length and
width so as to make the area. Then I added to the area the amount by which
the length exceeds the width, obtaining 3,3. Then I added the length and width
together, obtaining 27. What are the length, width, and area [xy + x − y = 183;
x + y = 27]?
Often in two unknown quantities; often solved by “completing the square.”
The quadratic formula: rhetorical
Arabic: Abu Ja’far Muh.ammad ibn-Mūsā al-Khwārizmı̄
(ca. 780–850)
Al-kitāb al-muhtas.ar fı̄ h.isāb al-jabr wa’l muqābala
(The Condensed Book on the Calculation of al-Jabr and
al-Muqabala)
al-Khwārizmı̄
(a > 0, b > 0, c > 0)
Squares equal to roots
ax 2 = bx
Squares equal to numbers
ax 2 = c
Squares and roots equal to numbers
ax 2 + bx = c
Squares and numbers equal to roots
ax 2 + c = bx
Roots and numbers equal to squares
bx + c = ax 2
One square, and ten roots of the same, amount to
thirty-nine dirhems. . . .
One square, and ten roots of the same, amount to
thirty-nine dirhems. . . .
The solution is this: you halve the number of roots, which
in the present instance yields five. This you multiply by
itself; the product is twenty-five. Add this to thirty-nine;
the sum is sixty-four. Now take the root of which, which
is eight, and subtract from it half the number of roots,
which is five; the remainder is three. This is the root of
the square which you sought for; the square itself is nine.
x 2 + bx = c:
b→
b
2
→
b2
4
→c+
b2
4
→
q
c+
b2
4
→
q
c+
b2
4
−
b
2
=x
S
A
Al-Khwārizmı̄ provided two
geometric demonstrations of
his solution. This one is most
widely known, and is often
used to introduce the method
of completing the square.
G
B
25
D
H
S
b
2
x
A
x 2 + bx = c
b
2x
x2
x
x2 + 2 ·
B
b2
4
b
b2
b2
x+
=c+
2
4
4
2
b
b2
x+
=c+
2
4
x2 + 2 ·
b
2x
b
2
H
b
x =c
2
[A] square and twenty-one in numbers are equal to ten
roots of the same. . . .
Solution: Halve the number of the roots; the half is five.
Multiply this by itself; the product is twenty-five.
Subtract from this the twenty-one which are connected
with the square; the remainder is four. Extract its root; it
is two. Subtract this from the half of the roots, which is
five; the remainder is three. This is the root of the square
which you required, and the square is nine. Or you may
add the root to the half of the roots; the sum is seven;
this is the root of the square which you sought for, and
the square itself is forty-nine.
[A] square and twenty-one in numbers are equal to ten
roots of the same. . . .
Solution: Halve the number of the roots; the half is five.
Multiply this by itself; the product is twenty-five.
Subtract from this the twenty-one which are connected
with the square; the remainder is four. Extract its root; it
is two. Subtract this from the half of the roots, which is
five; the remainder is three. This is the root of the square
which you required, and the square is nine. Or you may
add the root to the half of the roots; the sum is seven;
this is the root of the square which you sought for, and
the square itself is forty-nine.
x 2 + c = bx:
b→
b
2
→
b2
4
→
b2
4
−c →
q
b2
4
−c →
b
2
−
q
b2
4
−c
or
b
2
q
+
c−
b2
4
=x
India
Brahmagupta (598–670): tells how to solve quadratic
equations by completing the square
Take the absolute number from the side opposite to that from
which the square and simple unkown are to be subtracted. To
the absolute number multiplied by four times the [coefficient
of the] square, add the square of the [coefficient of the] middle
term; the square root of the same, less the [coefficient of the]
middle term, being divided by twice the [coefficient of the]
square is the [value of the] middle term.
India
(Handled negative numbers without fanfare.)
ax 2 + bx = c
India
ax 2 + bx = c
c
India
ax 2 + bx = c
c → 4ac
India
ax 2 + bx = c
c → 4ac → 4ac + b 2
India
ax 2 + bx = c
c → 4ac → 4ac + b 2 →
p
4ac + b 2
India
ax 2 + bx = c
c → 4ac → 4ac + b 2 →
p
p
4ac + b 2 → 4ac + b 2 − b
India
ax 2 + bx = c
√
p
p
c → 4ac → 4ac + b → 4ac + b 2 → 4ac + b 2 − b →
2
4ac + b 2 − b
=x
2a
Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)
131. Śríd’hara’s rule on this point: Multiply both sides of the
equation by a number equal to four times the [coefficient] of the
square, and add to them a number equal to the square of the
original [coefficient] of the unknown quantity. (Then extract the
root.)
ax 2 + bx = c
root.)
ax 2 + bx = c
4a2 x 2 + 4abx = 4ac
root.)
ax 2 + bx = c
4a2 x 2 + 4abx = 4ac
2 2
4a x + 4abx + b 2 = 4ac + b 2
root.)
ax 2 + bx = c
4a2 x 2 + 4abx = 4ac
2 2
4a x + 4abx + b 2 = 4ac + b 2
(2ax + b)2 = 4ac + b 2
root.)
ax 2 + bx = c
4a2 x 2 + 4abx = 4ac
2 2
4a x + 4abx + b 2 = 4ac + b 2
(2ax + b)2 = 4ac + b 2
p
2ax + b = 4ac + b 2
√
4ac + b 2 − b
x=
2a
The square-root of half the number of a swarm of bees is
gone to a shrub of jasmine; and so are eight-ninths of the
whole swarm; a female is buzzing to one remaining male,
that is humming within a lotus, in which he is confined,
having been allured to it by its fragrance at night. Say,
lovely woman, the number of bees.
q
1
2S
+ 98 S + 2 = S
→ x+
16 2
9 x
+ 2 = 2x 2
q
1
2S
+ 98 S + 2 = S
→ x+
16 2
9 x
+ 2 = 2x 2
2x 2 − 9x = 18
q
1
2S
+ 98 S + 2 = S
→ x+
16 2
9 x
+ 2 = 2x 2
2x 2 − 9x = 18
ax 2 + bx = c
The quadratic formula: modern
x 2 + bx + c = 0
b
x =− ±
2
D = b 2 − 4c :
† We
√
b 2 − 4c
2
two real roots† if D > 0, one if D = 0, no real root if D < 0
will always mean distinct real roots.
The cubic formula: what is it?
Cubic problems
Ancient Mesopotamia
Cuneiform tablets found containing the sum of the square and the cube of an
integer for many different integers. This suggests they may have solved
problems that reduced to x 3 + x 2 = a, and perhaps even problems that reduced
to ay 3 + by 2 + cy = d.
China: Xugu Suanjing
(Continuation of Ancient Mathematics; 7th century)
compute the length of a leg of a right triangle given that the product of the
1
other leg and the hypotenuse is 1337 20
and the difference between the
1
64
1
hypotenuse and the leg is 1 10
[x 3 + 14 x 2 + 50
x − 8938513 125
= 0].
Generally used numerical methods (“Horner’s method”).
The cubic formula: modern
General cubic:
x 3 + bx 2 + cx + d = 0
General cubic:
x 3 + bx 2 + cx + d = 0
x =y−
b
3
General cubic:
x 3 + bx 2 + cx + d = 0
x =y−
b
3
Reduced cubic:
y 3 + py + q = 0,
where
p=c−
b2
3
and q =
2b 3
27
−
bc
3
+ d.
Let p and q be any real numbers. Then the reduced or depressed
cubic equation
x 3 + px + q = 0
has solutions x1 , x2 , and x3 given by
x1 = r + s,
x2 = r ω2 + sω3 ,
x3 = r ω3 + sω2 ,
where
ω2 = − 21 +
√
3
2 i,
ω3 = − 12 −
√
3
2 i.
cubic equation
x 3 + px + q = 0
x1 = r + s,
x2 = r ω2 + sω3 ,
x3 = r ω3 + sω2 ,
where
r
r=
3
− q2
q
+
r
q2
4
+
ω2 = − 21 +
p3
27 ,
√
3
2 i,
s=
3
− q2
−
ω3 = − 12 −
q
√
q2
4
3
2 i.
+
p3
27 ,
cubic equation
x 3 + px + q = 0
x1 = r + s,
x2 = r ω2 + sω3 ,
x3 = r ω3 + sω2 ,
where
r
r=
3
− q2
q
+
r
q2
4
+
ω2 = − 21 +
∆=
q2
4
+
p3
27
p3
27 ,
√
3
2 i,
s=
3
− q2
−
ω3 = − 12 −
: three real roots if ∆ < 0
q
q2
4
+
p3
27 ,
√
3
2 i.
(irreducible case),
two real roots if ∆ = 0, one if ∆ > 0
Brief history of the solution of
the cubic equation
The principal players were the Italian mathematicians
Luca Pacioli (1445–1514)
Scipione del Ferro (1465–1526)
Niccolò Tartaglia (1500–1557)
Girolamo Cardano (1501–1576)
– Ludovico Ferrari (1522–1565)
Raphael Bombelli (1526–1572)
Luca Pacioli (1445–1514)
Summa de arithmetica geometria
proportioni et proportionalita
(1494)
In the Summa, we find the
abbreviations co. (cosa) for x
(cosa means “thing”), ce. (censo)
for x 2 , and cu. (cubo) for x 3
Declared at the end that solving
the cubic equation would be as
impossible as solving the ancient
problem of squaring the circle
Scipione del Ferro (1465–1526)
Within twenty years of the Summa, he found a method for
solving the “cube and first power equal to the number,”
x 3 + px = q,
where p and q are any positive real numbers.
Revealed solution to his pupils Annibale della Nave
(1500–1558) and Antonio Maria Fiore (first half of the
sixteenth century) at his death
Niccolò Tartaglia (1500–1557)
Born Niccolò Fontana
In 1535 burst onto cubic scene with
announcement he discovered solution
of “cube and square equal to the
number,”
x 3 + px 2 = q
Tartaglia’s announcement prompted del Ferro’s pupil Fiore,
who believed Tartaglia to be bluffing, to challenge Tartaglia to
a public problem-solving contest
Each contestant proposed 30 problems to the other and had
50 days to solve as many of the problems as he could
All of the 30 problems that Fiore had proposed to Tartaglia
were of the form x 3 + px = q, which, after having worked
doggedly, Tartaglia finally discovered how to solve just before
the end of the 50 days
Fiore could not solve any of the problems proposed by
Tartaglia, most of which led to equations of the form
x 3 + px 2 = q
Fiore was humiliated; Tartaglia’s renown grew
Girolamo Cardano (1501–1576)
Also commonly called Cardan
Artis magnae, sive de regulis
algebraicis (The Great Art, or the
Rules of Algebra; 1545), commonly
referred to as the Ars magna
Ars magna gives solution of the cubic
and the quartic, the latter by his pupil
Ludovico Ferrari (1522–1565)
In Ars magna, Cardano gives credit to
del Ferro and Tartaglia for the solution
of “cube and first power equal to a
constant,” and to Ferrari for the
solution of the quartic
Tartaglia-Cardano “cubic dispute”
After Tartaglia beat Fiore in a public problem-solving contest,
Cardano immediately set out to obtain the secret of the cubic
solution from Tartaglia
Cardano offered to publish Tartaglia’s solution in a work,
giving full credit to Tartaglia; however, Tartaglia refused to
divulge his secret, saying that he would publish it in his own
work at the right time
Tartaglia had a change of heart, perhaps in the hope of
securing a favor from Cardano, and provided Cardano with his
solution of the cubic; but this was only after Cardano “swore a
most solemn oath, by the Sacred Gospels and his word as a
gentleman, never to publish the method, and he pledged by his
Christian faith to put it down in cipher, so that it would be
unintelligible to anyone after his death”
It seems that after pledging to Tartaglia that he would never
divulge the secret of Tartaglia’s method, Cardano and his
young pupil Ferrari, who had been a servant in Cardano’s
household, discovered that del Ferro had originally solved the
“cube and first power equal to the number”
After having confirmed this by inspecting the papers of the
late del Ferro, Cardano apparently no longer felt bound to his
oath of secrecy to Tartaglia because del Ferro, and not
Tartaglia, was the first discoverer
Nevertheless, Cardano gives credit to Tartaglia in the Ars
magna; and, nevertheless, Tartaglia was furious that Cardano
had broken his oath. This was followed by bitter exchanges
between Tartaglia and Cardano, with Ferrari being Cardano’s
champion
Examples from
Cardano’s Ars magna
In Ars magna (The Great Art) we find the solution of the general
cubic equation and the general quartic equation published for the
first time.
“For I had been deceived by the
words of Luca Paccioli, who
denied that any more general
rule could be discovered than
his own. Notwithstanding the
many things which I had
already discovered, as is well
known, I had despaired and had
not attempted to look any
further. Then, however, having
received Tartaglia’s solution
and seeking for the proof of it, I
came to understand that there
were a great many other things
that could also be had.”
p, q, a, b, c > 0
XI Cube and first power equal to the number
x 3 + px = q
XII Cube equal to the first power and number
x 3 = px + q
XIII Cube and number equal to the first power
x 3 + q = px
XIV Cube equal to the square and number
x 3 = bx 2 + d
XV Cube and square equal to the number
x 3 + bx 2 = d
x 3 + d = bx 2
XVI Cube and number equal to the square
XVII Cube, square, and first power equal to the number
x3
+ bx 2 + cx = d
XVIII Cube and first power equal to the square and number
x 3 + cx = bx 2 + d
XIX Cube and square equal to the first power and number
x 3 + bx 2 = cx + d
XX Cube equal to the square, first power, and number
x 3 = bx 2 + cx + d
XXI Cube and number equal to the square and first power
x 3 + d = bx 2 + cx
XXII Cube, first power, and number equal to the square
x 3 + cx + d = bx 2
XXIII Cube, square, and number equal to the first power
x 2 + bx 2 + d = cx
In treating the different cases of the cubic in Chapters XI to XXIII
of the Ars magna, with the exception of Chapter XVI, Cardano first
provides a proof of the method (“demonstratio”), then he states the
rule (“regula”) for solving the particular case, and follows the rule
with one or more examples (“exemplum”); in Chapter XVI, Cardano
states the rule before providing a proof of the method. His proofs
are in the Greek geometrical tradition. To ease the reading in the
following examples, we write, for instance, x 3 + 6x = 20 where
Cardano writes “a cube and 6 things equal 20” (“cubus & 6
positiones, æquantur 20”), and
q
q
3 √
3 √
108 + 10 −
108 − 10
instead of Cardano’s
“R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10”.
Ch XI, “On the Cube and First Power Equal to the Number”
Rule
Cube one-third the coefficient of x; add
to it the square of one-half the constant
of the equation; and take the square root
of the whole. You will duplicate this, and
to one of the two you add one-half the
number you have already squared and
from the other you subtract one-half the
same. You will then have a binomium and
its apotome. Then, subtracting the cube
root of the apotome from the cube root
of the binomium the remainder [or] that
which is left is the value of x.
For example,
cubus & 6 positiones, æquantur 20
x 3 + 6x = 20.
Cube 2, one-third of 6, making 8; square
10, one-half the constant; 100 results.
Add 100 and 8, making
108, the square
√
root of which is 108. This you will
duplicate: to one add 10, one-half the
constant, and from the other subtract the
same. Thus you will obtain the
√ binomium
√
108 + 10 and its apotome 108 − 10.
Take the cube roots of these. Subtract
[the cube root of the] apotome from that
of the binomium and you will have the
value of x:
q
q
3 √
3 √
108
+
10
−
108 − 10
R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10
For example,
cubus & 6 positiones, æquantur 20
x 3 + 6x = 20.
Cube 2, one-third of 6, making 8; square
10, one-half the constant; 100 results.
Add 100 and 8, making
108, the square
√
root of which is 108. This you will
duplicate: to one add 10, one-half the
constant, and from the other subtract the
same. Thus you will obtain the
√ binomium
√
108 + 10 and its apotome 108 − 10.
Take the cube roots of these. Subtract
[the cube root of the] apotome from that
of the binomium and you will have the
value of x:
q
q
3 √
3 √
108
+
10
−
108 − 10
R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10
Direct substitution shows 2 is a solution (the only real solution).
x 3 + 6x = 20
=⇒
x 3 + 6x − 20 = 0
p = 6, q = −20
∆=
q2
4
+
x1 =
p3
27
= 108 > 0
q
3 √
108 + 10 −
x2 = −1 + 3i,
=⇒
q
3 √
one real root
108 − 10 = 2,
x3 = −1 − 3i
The roots Cardano missed:
x2 =
− 12
q
q
√
√
3
3
10 + 108 + 10 − 108
√
+i
x3 =
− 12
3
2
q
q
√
√
3
3
10 + 108 − 10 − 108
q
q
√
√
3
3
10 + 108 + 10 − 108
√
−i
3
2
q
q
√
√
3
3
10 + 108 −
10 − 108
x2 =
− 12
q
q
√
√
3
3
10 + 108 + 10 − 108
√
+i
x3 =
− 12
3
2
q
q
√
√
3
3
10 + 108 − 10 − 108 = −1 + 3i
q
q
√
√
3
3
10 + 108 + 10 − 108
√
−i
3
2
q
q
√
√
3
3
10 + 108 −
10 − 108 = −1 − 3i
Ch XII, “On the Cube Equal to the First Power and Number”
The rule, therefore, is: When the cube of one-third the coefficient of x is not
greater than the square of one-half the constant of the equation, subtract the
former from the latter and add the square root of the remainder to one-half the
constant of the equation and, again, subtract it from the same half, and you
will have, as was said, a binomium and its apotome, the sum of the cube roots
of which constitutes the value of x.
For example,
x 3 = 6x + 40.
Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract
this from 400, the square of 20, one-half the constant, making 392; the square
√
root of this added to 20 makes 20 + 392, and subtracted from 20 makes
p
p
√
√
√
3
3
20 − 392; the sum of the cube roots of these, 20 + 392 + 20 − 392,
is the value of x.
For example,
x 3 = 6x + 40.
Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract
this from 400, the square of 20, one-half the constant, making 392; the square
√
root of this added to 20 makes 20 + 392, and subtracted from 20 makes
p
p
√
√
√
3
3
20 − 392; the sum of the cube roots of these, 20 + 392 + 20 − 392,
is the value of x.
Exercise: Show that
p
p
√
√
3
3
20 + 392 + 20 − 392 = 4.
x 3 = 6x + 40
=⇒
x 3 − 6x − 40 = 0
p = −6, q = −40
∆=
q2
4
+
p3
27
= 392 > 0
=⇒
one real root
q
q
√
√
3
3
x1 = 20 + 392 + 20 − 392 = 4,
√
x2 = −2 + i 6,
√
x3 = −2 − i 6
x2 =
− 21
q
q
√
√
3
3
20 + 392 + 20 − 392
√
+i
x3 =
− 12
3
2
q
q
√
√
√
3
3
20 + 392 − 20 − 392 = −2 + i 6
q
q
√
√
3
3
20 + 392 + 20 − 392
√
−i
3
2
q
q
√
√
√
3
3
20 + 392 − 20 − 392 = −2 − i 6
Ch XIII, “On the Cube and Number Equal to the First Power”
The rule, therefore, is: When the cube and the constant are equal to the first
power, find the solution for the cube equal to the same number of y ’s and the
same constant; take three times the square of one-half of this and subtract it
from the coefficient of the first power; and the square root of the remainder
added to or subtracted from one-half the solution for the cube equal to y plus
the constant gives the solution for the cube and constant equal to x.
The rule, therefore, is: When the cube and the constant are equal to the first
power, find the solution for the cube equal to the same number of y ’s and the
same constant; take three times the square of one-half of this and subtract it
from the coefficient of the first power; and the square root of the remainder
added to or subtracted from one-half the solution for the cube equal to y plus
the constant gives the solution for the cube and constant equal to x.
Example:
x 3 + 3 = 8x.
Solving
y 3 = 8y + 3
according to the preceding rule, I obtain 3. The square of one-half of this is 2 14 ,
which multiplied by 3 is 6 34 . Subtracting this from 8, the coefficient of x, leaves
1 41 , the square root of which added to or subtracted from 1 12 , which is one-half
the solution for the cube equal to the first powerq
and constant, gives both
q
solutions which were being sought. One is 1 12 + 1 14 , the other 1 12 − 1 14 .
x 3 + 3 = 8x
=⇒
x 3 − 8x + 3 = 0
p = −8, q = 3
∆=
q2
4
+
p3
27
= − 1805
108 < 0
=⇒
1 21
q
+ 1 14 ,
x1 =
x2 = −3,
x3 = 1 12 −
three real roots
q
1 14
The root Cardano missed:
x2 =
− 12
r
3
− 23
+i
q
r
1805
108
+
3
− 23
−i
q
1805
108
r
r
q
q
3
3
3
3
1805
3
1805
+i 2
− 2 + i 108 − − 2 − i 108 = −3
√
Cardano’s rules in modern notation
Ch XI, “On the Cube and First Power Equal to the Number”
rq
rq
3
3
3
q 2
p 3
q
q 2
x + px = q : x =
+ 3 +2−
+
2
2
p 3
3
−
r
r
q q
3 q
3 q
3
q 2
p 3
x = px + q : x =
+
+
−
+
2
2
3
2
−
p 3
3
q 2
2
q
x 3 + q = px → y 3 = py + q : x = y2 ± p − 3
y 2
2
q
2
Compare to the cubic formula: For any real numbers p and q,
x 3 + px + q = 0 :
x1 = r + s,
x2 = r ω2 + sω3 ,
x3 = r ω3 + sω2 ,
where
r
r=
3
− q2
q
+
r
q2
4
+
ω2 = − 12 +
p3
27 ,
√
3
2 i,
s=
3
− q2
−
ω3 = − 12 −
q
√
q2
4
3
2 i.
+
p3
27 ,
Cubics not in reduced form
Cardano tells how to transform general a cubic equation,
x 3 = bx 2 + d,
x 3 + bx 2 + cx = d,
&c.,
into a reduced cubic equation,
x 3 + px = q,
x 3 = px + q,
or x 3 + q = px,
essentially by making the change of variable x = y − b3 .
Ch XV, “On the Cube and Square Equal to the Number”
Cube one-third the coefficient of x 2 , multiply the result by 2, and take the
difference between this and the constant of the equation. [The result is the
constant of a new equation.] Then multiply the square of one-third the
coefficient of x 2 by 3 and you will have [the number of] y ’s which are equal to
the cube and the constant, if twice the cube is greater than the constant of the
equation, or [the number of] y ’s which equal the cube, if the difference
between these two numbers is zero. Having derived the solution [for this
equation], subtract one-third the coefficient of x 2 from it and the remainder is
the value of x.
Ch XV, “On the Cube and Square Equal to the Number”
Cube one-third the coefficient of x 2 , multiply the result by 2, and take the
difference between this and the constant of the equation. [The result is the
constant of a new equation.] Then multiply the square of one-third the
coefficient of x 2 by 3 and you will have [the number of] y ’s which are equal to
the cube and the constant, if twice the cube is greater than the constant of the
equation, or [the number of] y ’s which equal the cube, if the difference
between these two numbers is zero. Having derived the solution [for this
equation], subtract one-third the coefficient of x 2 from it and the remainder is
the value of x.
For example,
x 3 + 6x 2 = 100.
The cube of 2 is 8 which, being multiplied by 2, is 16. Subtract this from 100
and you will have
y 3 = 84 + 12y .
[. . . ]
Ch XVII, “On the Cube, Square, and First Power Equal to the Number”
Cube one-third the coefficient of the square (which we show by this sign, Tp q̃d
[Tertia pars numeri quadratorum]) and add it to the constant. Then multiply
the coefficient of the square by one-third itself, and the difference between this
product and the coefficient of x is the number of y ’s to be added to the cube if
the product is less than the given coefficient of x or to be added to the
constant if the product is greater than the given coefficient of x.
[. . . ]
Ch XVII, “On the Cube, Square, and First Power Equal to the Number”
Cube one-third the coefficient of the square (which we show by this sign, Tp q̃d
[Tertia pars numeri quadratorum]) and add it to the constant. Then multiply
the coefficient of the square by one-third itself, and the difference between this
product and the coefficient of x is the number of y ’s to be added to the cube if
the product is less than the given coefficient of x or to be added to the
constant if the product is greater than the given coefficient of x.
[. . . ]
Example . . .
x + 13 x 2 + 19 x 3 = 19.
Turn these into whole numbers and you will have
x 3 + 3x 2 + 9x = 171,
for all are multiplied by 9.. . . Therefore
y 3 + 6y = 178
[. . . ]
Problem III
An oracle ordered a prince to build a sacred building whose space should be
400 cubits, the length being six cubits more than the width, and the width
three cubits more than the height. These quantities are to be found.
Let the altitude be x; the width be x + 3; and the length be x + 9. Multiplied
in turn, you will have
x 3 + 12x 2 + 27x = 400.
Add the cube of 4, one-third the coefficient of x 2 , which is 64, to 400 and 464
results. Multiply 12, the coefficient of x 2 , by one-third itself, making 48, the
difference between which and 27 is 21, the number of y ’s which equal the cube
and constant. Therefore multiply 21 by 4, one-third the coefficient of x 2 , and
84 results; take the difference between this and 464, which is 380; add this to
y , since the first sum of numbers was greater than the second number
produced; and you will have
y 3 = 21y + 380.
[. . . ]
The quartic formula
Ch XXXIX, “On the Rule by Which We Find an Unknown Quantity in Several Stages”
Rule II
2. There is another rule, more noble than the preceding. It is Lodovico
Ferrari’s, who gave it to me on my request. Through it we have all the
solutions for equations of the fourth power, square, first power, and number, or
of the fourth power, cube, square, and number.. . .
[. . . ]
Reducing this to a rule: It is always sufficient to have b 3 plus one and a quarter
times the coefficient of x 2 [for the coefficient of b 2 ] plus such a number of b’s
as was the original constant of the equation. Thus, if we had
x 4 + 12x 2 + 36 = 6x 2 + 60x,
we will have
b 3 + 15b 2 + 36b = 450,
....
Given the general quartic equation
x 4 + bx 3 + cx 2 + dx + e = 0,
make the linear substitution x = y − b/4 to obtain the reduced
quartic equation
y 4 + py 2 + qy + r = 0
x 4 + bx 3 + cx 2 + dx + e = 0,
quartic equation
y 4 + py 2 + qy + r = 0
z 3 + 2pz 2 + (p 2 − 4r )z − q 2 = 0
x 4 + bx 3 + cx 2 + dx + e = 0,
quartic equation
y 4 + py 2 + qy + r = 0
that has solutions
√
√
√
√
√
√
y1 = 12 ( z1 + z2 + z3 ), y3 = − 12 ( z1 − z2 + z3 ),
√
√
√
√
√
√
y2 = 21 ( z1 − z2 − z3 ), y4 = − 12 ( z1 + z2 − z3 ),
where z1 , z2 , and z3 satisfy the cubic resolvant
z 3 + 2pz 2 + (p 2 − 4r )z − q 2 = 0
and z1 z2 z3 = q 2 > 0.
The quintic formula
There is not one using only radicals.
2
<http://library.wolfram.com/examples/quintic/main.html>
There is not one using only radicals.
That is, there is not one for the general quintic, but particular
quintics may be solved using only radicals. E.g., Wolfram2 gives the
example
x 5 + 20x + 32 = 0
that has for one solution
2
<http://library.wolfram.com/examples/quintic/main.html>
1
−
5
(r
5
√
2500 5 + 250
r
√
50 − 10 5 − 750
q
√
50 + 10 5
q
q
√
√
− 2500 5 − 750 50 − 10 5 − 250 50 + 10 5
r
q
q
√
√
√
5
+ 2500 5 + 750 50 − 10 5 + 250 50 + 10 5
r
)
q
q
√
√
√
5
+ 2500 5 − 250 50 − 10 5 + 750 50 + 10 5 .
5
√
q
First proposed proof by Italian Paolo Ruffini (1765–1822) in
1798; no one could understand it
Complete proof by Niels Henrik Abel (1802–1829) in
mid-1820s
Abel went on to attempt (i) to find all equations of any given
degree that can be solved by radicals; (ii) to decide whether a
given equation can be solved by radicals. Not successful before
his death
Picked up by Evariste Galois (1811–1832). This led to what is
now called Galois theory that describes exactly when a
polynomial equation is solvable by radicals
Complex numbers
Ch XXXVII, “On the Rule (Triple) for Postulating a Negative”
Imaginary numbers were not understood during Cardano’s day
beyond their being a convenient contrivance. When confronted
with an equation like
x 2 + 1 = 0 or x 2 + 2x + 2 = 0,
one could simply say that the equation has no solution.
Nevertheless, Cardano did address the appearance of the square
root of negative numbers in Chapter XXXVII of the Ars magna.
This was the first time that the square root of a negative number
was written down explicitly.
In “Rule II” of that chapter, Cardano considers a problem that is
equivalent to solving the equation x(10 − x) = 30 or
x(10 − x) = 40:
The second species of negative assumption involves the square root
of a negative. I will give an example: If it should be said, Divide 10
into two parts the product of which is 30 or 40, it is clear that this
case is impossible. Nevertheless, we will work thus: We divide 10
into two equal parts, making each 5. These we square, making 25.
Subtract 40, if you will, from the 25 thus produced . . . leaving a
remainder of −15, the square root of which added to or subtracted
from√5 gives parts the
√ product of which is 40. These will be
5 + −15 and 5 − −15.
y
(5, 25)
We can see from the graph of
x(10 − x) = y
to the left why the solutions of
x(10 − x) = 40
are complex:
√
√
5 + −15 and 5 − −15.
x
Cardano goes on to give a geometrical demonstration of the
solution, in which he says:
√
. . . you will have to imagine
−15 . . . and
√
√you will have that
√ which
you seek,√namely 5 + 25 − 40 and 5 − 25 − 40, or 5 + −15
and √
5 − −15. Putting
aside the mental tortures involved, multiply
√
5 + −15 by 5 − −15, making 25 − (−15) which is +15. Hence
this product is 40.. . .
Cardano goes on to give a geometrical demonstration of the
solution, in which he says:
√
. . . you will have to imagine
−15 . . . and
√
√you will have that
√ which
you seek,√namely 5 + 25 − 40 and 5 − 25 − 40, or 5 + −15
and √
5 − −15. Putting
aside the mental tortures involved, multiply
√
5 + −15 by 5 − −15, making 25 − (−15) which is +15. Hence
this product is 40.. . . So progresses arithmetic subtlety the end of
which, as is said, is as refined as it is useless.
Exit Ars magna
Enter Bombelli’s famous example
Raphael Bombelli (1526–1572)
L’Algebra, first printed in 1569
First serious study of complex numbers
is motivated by solutions of the cubic
equation, not the quadratic
“I have found another kind of cubic
root of a polynomial which is very
different from the others. This [cubic
root] arises in the chapter dealing with
the equation of the kind x 3 = px + q,
when p 3 /27 > q 2 /4, as we will show
in that chapter.”
It was the combination of Cardano’s cubic formula and Bombelli’s
famous equation,
x 3 = 15x + 4,
that began a serious study of imaginary and complex numbers.
famous equation,
x 3 = 15x + 4,
The Cardano’s formula produced the solution
q
q
√
√
3
3
2 + −121 + 2 − −121
famous equation,
x 3 = 15x + 4,
The Cardano’s formula produced the solution
q
q
√
√
3
3
2 + −121 + 2 − −121
Direct substitution shows that 4 is a solution
Bombelli showed that, in fact,
q
q
√
√
3
3
2 + −121 + 2 − −121 = 4
Thus, somehow real numbers can be expressed using square roots
of negative numbers in a way that is not obvious.
Recall Cardano’s cubic formula
cubic equation
x 3 + px + q = 0
x1 = r + s,
x2 = r ω2 + sω3 ,
x3 = r ω3 + sω2 ,
where
r
r=
3
− q2
q
+
r
q2
4
+
ω2 = − 12 +
∆=
q2
4
+
p3
27
p3
27 ,
√
3
2 i,
s=
3
− q2 −
ω3 = − 12 −
: three real roots if ∆ < 0
q
q2
4
+
p3
27 ,
√
3
2 i.
(irreducible case),
two real roots if ∆ = 0, one if ∆ > 0
“This kind of square root has in its calculation [algorismo] different
operations than the others and has a different name. Since when
p 3 /27 > q 2 /4, the square root of their difference can be neither
positive nor negative, therefore I will call it ‘more than minus’ [‘plus
of minus’] when it should be added and ‘less than minus’ [‘minus of
minus’] when it should be subtracted.. . . I will first treat
multiplication, giving the law of plus and minus”
Plus times more than minus makes more than minus
(+)(+i) = +i
Minus times more than minus makes less than minus
(−)(+i) = −i
Plus times less than minus makes les than minus
(+)(−i) = −i
Minus times less than minus makes more than minus
(−)(−i) = +i
More than minus times more than minus makes minus
(+i)(+i) = −
More than minus times less than minus makes plus
(+i)(−i) = +
Less than minus times more than minus makes plus
(−i)(+i) = +
Less than minus times less than minus makes minus
(−i)(−i) = −
Bombelli’s famous equation: x 3 = 15x + 4
Ars magna:
L’Algebra:
x 3 = 6x + 40.
Raise 2, one-third the coefficient of x, to the
cube, which makes 8; subtract this from 400,
the square of 20, one-half the constant,
making 392; the square root of this added to
√
20 makes 20 + 392, and subtracted from 20
√
makes 20 − 392; the sum of the cube roots
p
p
√
√
3
3
of these, 20 + 392 + 20 − 392, is the
value of x.
q
x=
3
2+
√
−121 +
q
√
3
2 − −121
x 3 = 15x + 4
cubic formula produces the solution
q
q
√
√
3
3
x = 2 + −121 + 2 − −121
=
√
3
2 + 11i +
√
3
2 − 11i
direct substitution shows 4 is a solution
could it be that
√
3
2 + 11i +
√
3
2 − 11i = 4 ?
Bombelli’s genious was to presume that the two cube roots are,
themselves, complex conjugates, that is to say,
√
√
3
2 + 11i = a + bi and 3 2 − 11i = a − bi.
(There was no reason at that time to think that the cube root of a
complex number may be a complex number.) Then, by cubing the
relations and setting equal real and imaginary parts, he proceeded
to show that, in fact,
√
√
3
2 + 11i = 2 + i and 3 2 − 11i = 2 − i,
so that
√
3
2 + 11i +
√
3
2 − 11i = (2 + i) + (2 − i) = 4.
About all this, Bombelli says interestingly:
It was a wild thought in the judgement of many; and I
too for a long time was of the same opinion. The whole
matter seemed to rest on sophistry rather than on truth.
Yet I sought so long, until I actually proved this to be
case.
About all this, Bombelli says interestingly:
It was a wild thought in the judgement of many; and I
too for a long time was of the same opinion. The whole
matter seemed to rest on sophistry rather than on truth.
Yet I sought so long, until I actually proved this to be
case.
And now complex numbers could no longer be swept under the rug,
for the cubic formula expresses some real number solutions of cubic
equations using square roots of negative numbers: a real puzzler!
Thus, these “sophistic,” “impossible,” “imaginary,” and “useless”
objects no longer could be ignored.
Brief history of
complex numbers
Complex numbers: 17th century
Even as late as the seventeenth
century, prominent mathematicians like
René Descartes (1596–1650) referred to
positive roots of an equation as “true”
roots and negative roots as “false”
roots; Descartes called roots that
contained a square root of a negative
number “imaginary.”
By this time, finally, negative numbers were beginning to gain full
acceptance in Europe, even though they had long ago been
accepted in India and China, but it would be a while longer (not
until the nineteenth century) before imaginary or complex numbers
would enjoy an equivalent standing.
Leonhard Euler
√ (1707–1783) introduced
using i for −1, a notation that he
adopted late in his life. Even so, he did
not consider complex numbers to be
“true” numbers.
In his Elements of Algebra (1770), Euler wrote:
143. And, since all numbers which it is possible to conceive, are
either greater or less than 0, or are 0 itself, it is evident that we
cannot rank the square root of a negative number amongst possible
numbers, and we must therefore say that is an impossible quantity.
In this manner we are led to the idea of numbers, which from their
nature are impossible; and therefore they are usually called
imaginary quantities, because they exist merely in the imagination.
√
√
√
√
144. All such expressions, as −1, −2, −3, −4, &c. are
consequently impossible, or imaginary numbers, since they represent
roots of negative quantities; and of such numbers we may truly
assert that they are neither nothing, nor greater than nothing, nor
less than nothing; which necessarily constitutes them imaginary, or
impossible.
145. But notwithstanding this, these numbers present themselves to
the mind; they exists in our imagination, and we
√ still have a
sufficient idea of them; since we know that by −4 is meant a
number which, multiplied by itself, produces −4; for this reason
also, nothing prevents us from making use of these imaginary
numbers and employing them in calculation.
The idea to represent complex numbers in a plane occurred to
three persons from three different parts of Europe all about the
turn of the nineteenth century: Caspar Wessel (1745–1818), a
Norwegian surveyor and cartographer; Jean Robert Argand
(1768–1822), a French-Swiss bookkeeper; and Carl Friedrich Gauss
(1777–1855), arguably the greatest of all German mathematicians.
Both Wessel and Argand proposed using line segments or “vectors”
to represent complex numbers. Wessel presented his idea first to
the Royal Academy of Sciences of Denmark in 1797, and then in a
paper published in the Philosophical Transactions of the Academy
in 1799. Argand published his idea in a booklet titled Essai sur une
manière de représenter les quantités imaginarie dans les
constructions géométriques that was privately printed in 1806.
However, perhaps because neither Wessel nor Argand was well
known in mathematics, their ideas went largely overlooked.
It was not until Gauss proposed
representing a complex number as a point
in the plane that the plane geometric
interpretation took hold. Apparently, Gauss
had used his idea implicitly in his 1799
doctoral dissertation on the fundamental
theorem of algebra; it was not until 1831
before Gauss’s idea was described publicly
in a commentary on his paper Theoria
Residuorum Biquadraticorum.
The fundamental theorem of algebra
Today, the fundamental theorem of algebra may be stated thus:
Let a0 , a1 , . . . , an−1 be complex numbers. If n ≥ 1, then
the equation
x n + an−1 x n−1 + · · · + a2 x 2 + a1 x + a0 = 0
has at least one complex number solution.
Albert Girard (1595–1632)
The first explicit statement of the theorem, without a proof, was by
Albert Girard in his work Invention nouvelle en l’algèbre (A New
Discovery in Algebra, 1629):
Theorem. Every algebraic equation . . . admits of as
many solutions as the denomination of the highest
quantity indicates.. . .
By the “denomination of the highest quantity” is meant the degree
of the equation, so Girard asserts that a polynomial of degree n has
n solutions.
Katz* tells us that Girard realized that some solutions may be
repeated (multiplicity greater than one) and some may be imaginary
(“impossible”). According to Katz, “In answer to the anticipated
question of the value of these impossible solutions, Girard answered
that ‘they are good for three things: for the certainty of the general
rule, for being sure that there are no other solutions, and for its
utility.’ ” Katz indicates that Girard provides√the example √
x 4 + 3 = 4x that has solutions 1, 1, −1 + i 2, and −1 − i 2.
*Victor J. Katz, A History of Mathematics: An Introduction, 3rd ed., Addision-Wesley (2009).
René Descartes (1596–1650)
Some eight years later, Descartes, in his Third Book of La
géométrie, would enunciate essentially Girard’s theorem, writing:
Every equation can have as many distinct roots (values of
the unknown quantity) as the number of dimensions of
the unknown quantity in the equation.
Katz remarks that Descartes uses the phrase “can have” instead of
Girards “admits of” regarding the number of roots or solutions
because, unlike Girard, Descartes considers only distinct roots and,
at least earlier on, does not consider imaginary roots.
Carl Friedrich Gauss (1777–1855)
Gauss was the first to prove the fundamental theorem of algebra;
he did so in his doctoral thesis of 1799. Katz informs us that
“Gauss was so intrigued with the fundamental theorem—that every
polynomial p(x) with real coefficients has a real or complex
root—that he published four different proofs of it, in 1799, 1815,
1816, and 1848.
Each proof used in some form or other the geometric intepretation
of complex numbers, although in the first three proofs Gauss hid
this notion by considering the real and imaginary parts of the
numbers separately.. . . It was only in his final proof in 1848 that
Gauss believed mathematicians would be comfortable enough with
the geometric interpretation of complex numbers so that he could
use it explicitly. In fact, in that proof, similar to his first one, he
even permitted the coefficients of the polynomial to be complex.”
Ironically, Gauss himself did not consider a geometric interpretation
to be a sufficient justification for complex numbers.
References
Girolamo Cardano, The Rules of Algebra (Ars Magna),
translated by T. Richard Witmer, Dover Publications, Inc.
(1968).
Roger Cooke, The History of Mathematics: A Brief Course,
second edition, Wiley Interscience (2005).
Victor J. Katz, A History of Mathematics: An Introduction,
third edition, Addison-Wesley (2009).
Amy Shell-Gellasch and J. B. Thoo, Intersecting Mathematics
and History: Topics in Elementary Mathematics from an
Historical Viewpoint, in preparation.
<http://ms.yccd.edu/cv-and-pubs.aspx#books>

Making Waves in Vector Calculus

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