A short history of equations –
The search for x
Roger House
Scientific Buzz Café
French Garden
Sebastopol, CA
2012 January 19
Copyright © 2012 Roger House
That's Daniel Osmer's title for my talk.
I think it's probably better than my original
title.
But I can't quite let the original title go:
“x is almost always 11”
A short history of equations
Roger House
Scientific Buzz Café
French Garden
Sebastopol, CA
2012 January 19
Copyright © 2012 Roger House
The Secret of Algebra
“I'll never get this second problem.”
4
The Secret of Algebra
“Just put down 'eleven,' Franklin, and
don't worry about it. That's what I did.”
5
The Secret of Algebra
“x is almost always eleven, and y is
almost always nine ...”
6
The Secret of Algebra
“One thing I've learned about algebra,
don't take it too seriously ...”
7
The subject of today's talk
A brief, selective history of solving
polynomial equations in one unknown over
the field of complex numbers.
Now, forget I said that.
That's intended for the three people in the
audience who will accuse me of
misrepresentation if I don't make precisely
clear at the outset what kind of equations,
what kind of history, etc.
8
Equations first, then history
For the rest of you, please resist the urge to
walk out before the talk is over.
We'll begin by looking at a few simple
equations, and then we'll look at the
history of solving them.
If at any point you begin to feel uneasy, just
mutter quietly under your breath,
“x is almost always eleven”
9
Travel is broadening
You're a tourist in Europe for the first time.
It's a bit chilly; you notice the bank
thermometer says -3 degrees.
You didn't think it was that chilly.
You don't feel so well; the doctor says your
temperature is 38.
You thought you had a fever, but that sounds
like a severe chill.
10
It's not the 3rd degree ...
What's going on?
Europeans (and most of the world except
the USA and Belize) use Celsius, not
Fahrenheit, to measure temperature.
In Celsius, water freezes at 0 degrees, and
boils at 100 degrees.
To convert from Celsius to Fahrenheit:
F = (9/5)*C + 32
11
Let's try the formula
C=
0 F = (9/5)*0
+ 32 = 32
C = 100 F = (9/5)*100 + 32 = 212
C=
-3 F = (9/5)*(-3) + 32 = 26.6
C = 38 F = (9/5)*38 + 32 = 100.4
12
A picture is worth a 1000 ...
13
All the points are in a line
14
Linear Equations
We've seen four points all on the same line,
and, in fact, given any real number C, if we
compute F by
F = (9/5)*C + 32
then the point (C, F) will fall on the green
line.
This kind of equation is called a linear
equation.
Here is the general form of a linear15equation:
Another linear equation
For our Celsius to Fahrenheit equation,
a = 9/5 and b = 32
But we can pick any a (≠ 0) and any b and
get another linear equation, e.g.,
a = 5/9 and b = -160/9
gives us this equation:
f(x) = (5/9)*x – 160/9
Here's a graph of this new equation:
16
What's this? Make a wild guess
17
C = (5/9)(F - 32)
18
See both equations at once
19
Let's get to the root of it
Given a function like
f(x) = ax + b
one is often asked to
–
–
–
–
Solve it.
Find its roots.
Find its zeros.
Etc.
These all mean the same thing: Find some
x such that f(x) = 0.
20
Solving a linear equation
For a linear equation, we want to find x such
that
ax + b = 0
This is pretty easy:
ax = -b
x = -b/a
(Remember, a ≠ 0, so we can divide by a.)
21
Let's try it
For the Fahrenheit to Celsius conversion,
f(x) = (5/9)*x – 160/9
so
a = 5/9 and b = -160/9
so the solution of
(5/9)*x – 160/9 = 0
is
x = -b/a = -(-160/9) / (5/9) = (160/9)*(9/5) =
22
32
Where is the x-axis intersected?
23
Now solve the inverse function
For the Celsius to Fahrenheit conversion,
f(x) = (9/5)*x + 32
so
a = 9/5 and b = 32
so the solution of
(9/5)*x + 32 = 0
is
x = -b/a = -32/(9/5) = -32*5/9 ≈ -17.78
24
Where is the x-axis intersected?
25
Summary of linear equations
A linear equation has this form:
ax + b = 0
where a ≠ 0.
A linear equation has exactly one root,
x = -b/a
which is the point where the graph of the
equation intersects the x-axis.
26
A Physics Problem
Toss a ball straight up in the air.
Release it at a height of 4 ft above the
ground.
At a speed of 12 ft/sec (≈ 8.2 mph).
Describe qualitatively what happens:
–
–
–
–
The speed of the ball steadily decreases.
Until it actually comes to a stop.
Then it begins falling downwards.
Until it hits the ground.
27
Take some wild guesses
How long will it take before the ball hits the
ground?
How long will it take to reach its maximum
height?
How high will the ball go?
28
We can do better than guessing
Given the time t in seconds after tossing the
ball, here is a formula which tells us the
height h of the ball at that time:
h = -16t2 +12t + 4
Quick check: When t = 0, what is h?
Where does the 4 come from?
Where does the 12 come from?
Where does the -16 come from?
29
Let's try some numbers
Let t range from 0 to 2 by half seconds:
t = 0 h = -16*02 +12*0 + 4
= 4
t = 0.5 h = -16*0.52 +12*0.5 + 4 = 6
t = 1 h = -16*12 +12*1 + 4
= 0
t = 1.5 h = -16*1.52 +12*1.5 + 4 = -14
t = 2 h = -16*22 +12*2 + 4
= -36
How to interpret these heights?
30
We need a picture ...
31
Fill in the blanks
32
Some things to note
The ball goes straight up and comes straight
back down, so ...
The graph is NOT the path of the ball.
The graph shows how the height of the ball
varies with time.
What does it mean when the height is zero?
What does it mean when the height is
negative?
33
Staying above ground
34
Now we have some answers
How long will it take before the ball hits the
ground? 1 second
How long will it take to reach its maximum
height? 0.375 second (= 3/8 second)
How high will the ball go? 6.25 ft (= 6 ft 3 in)
35
Quadratic equations
We are now dealing with a new type of
equation which is not a linear equation.
It's a quadratic equation, which has this
general form:
ax2 + bx + c = 0
where a ≠ 0.
Note the extra term ax2 containing the
square of x (quadratus means “square” in
Latin).
36
Solving a quadratic equation
As for a linear equation, there is a fairly easy
way to solve a quadratic equation.
The solution of
ax2 + bx + c = 0
is
− b± √
b − 4ac
x=
2a
2
This may look a bit scary, but watch:
37
When is the height zero?
Let's solve our favorite quadratic:
-16t2 +12t + 4 = 0
In this case,
a = -16, b = 12, c = 4,
so
12 − 4(− 16)(4)
− b± √
b − 4 a c − 12± √
x=
=
2a
2(− 16)
2
2
38
Grind through some arithmetic
When the dust settles, we have
12± 20
x=
32
so
x = (12 + 20) / 32 = 32 / 32 = 1
and
x = (12 – 20) / 32 = -8/32 = -1/4 = -0.25
39
Can this be right?
Plug 1 into -16t2 +12t + 4 = 0:
-16*12 +12*1 + 4 = -16 + 12 + 4 =0
Now try -0.25:
-16*(-0.25)2 +12*(-0.25) + 4 = -1 – 3 + 4 = 0
So, both 1 and -0.25 are solutions to the
equation.
How can this be? Let's look at a graph:
40
f(x) = -16x2 + 12x + 4
41
Para what?
Now we are looking at our favorite quadratic
equation as a mathematical object.
The physical scenario is no longer of
interest.
The graph of the equation is a shape called
a parabola.
The graph of every quadratic is a parabola.
Here are some examples:
42
f(x) = x2 - 6*x + 5
43
f(x) = x2 - 6*x + 9
44
f(x) = x2 - 6*x + 13
45
The parabola misses the x-axis
Let's find the roots of
x2 - 6*x + 13 = 0
Since
a = 1, b = -6, c = 13,
we have
(− 6) − 4(1)(13)
− b± √
b − 4 a c − (− 6)± √
x=
=
2a
2(1)
2
2
46
Can you image the roots?
Doing a bit of arithmetic:
6± √
− 16 6± 4 √
−1
x=
=
= 3± 2 √
−1
2
2
so,
and
x= 3+ 2 √
−1
x= 3− 2 √
−1
What is the square root of -1?
47
Summary of quadratic equations
A quadratic equation has this form:
ax2 + bx + c = 0
where a ≠ 0.
A quadratic equation has exactly two roots,
− b± √
b − 4ac
x=
2a
2
48
Linear equations go way back
The Egyptian Rhind Papyrus from 1650 BC
(but copied from a document from 18491801 BC - so it's 3800 years old) posed
problems like this:
“A quantity and its 1/7 added become 19.
What is the quantity?”
Today we would write this as
x + x/7 = 19
49
A modern solution
Transforming to standard form:
(8/7)x - 19 = 0
Since
a = 8/7 and b = -19
we know the solution is
x = -b/a = -(-19)/(8/7) = (7/8)19 = 16.625
Of course the Egyptians did not have the
nice notation we use today, nor the
50
method x = -b/a.
An Egyptian solution
The Egyptians used a method called false
position.
Looking at
x + x/7 = 19
make a 'wild' guess that x = 7, so
x + x/7 = 7 + 7/7 = 7 + 1 = 8
(Do you see why 7 is a convenient guess?)
51
How'd they do that?
Clearly 7 is not correct because we want x +
x/7 to be 19, not 8.
But, if we multiply 8 by 19/8, then we get 19,
so let's multiply our guess of 7 by 19/8 to
get (19/8)7 = (7/8)19 = 16.625.
Voilà! That's precisely the answer we got
with fancy modern methods.
3800 years ago Egyptians knew a thing or
two
–
52
Remember the pyramids? the Sphinx?
False position has a long history
Known in these times and places:
–
–
–
–
–
–
1850 BC
200 BC
3rd century
10th century
1200
1550
Egyptians
Chinese
Hindus
Arabs
Fibonacci
Robert Record
Even employed in late 19th century
53
Quadratic equations are old
Babylonian clay tablets from 2000 BC
contained problems like this:
–
“I have added the area and two-thirds of
the side of my square and it is 0;35.
What is the side of my square?”
What's “0;35”? It's the fraction 35/60. So
the problem is asking for the solution of
x2 + (2/3) x = 35/60
54
x2 + (2/3) x = 35/60
Here is the solution given on the clay tablet:
“You take 1, the coefficient [of x]. Two-thirds
of 1 is 0;40. Half of this, 0:20, you multiply
by 0;20 and it [the result] you add to 0;35
and [the result] 0;41,40 has 0;50 as its
square root. The 0;20, which you have
multiplied by itself, you subtract from 0;50,
and 0;30 is [the side of] the square.”
55
x2 + bx – c = 0
In effect, the solution uses our modern
solution with a = 1 and c negated:
√
2
b
b
x= ( ) + c−
2
2
The Babylonians did not have this form nor
notation. They solved example after
example, and at the end wrote:
“Such is the procedure.”
56
al-Khwārizmī (c.780 – c. 850)
About 830, the Persian al-Khwārizmī wrote
an important book: Al-Kitāb al-mukhtaṣar
fī ḥisāb al-jabr wa-l-muqābala
The Compendious Book on Calculation by
Completion and Balancing
From al-jabr (= 'completion' or subtracting
the same value from both sides of an
equation) we get our word algebra.
57
Avoid negative numbers
In this book various forms of quadratic
equations were treated:
x2 + ax = b
x2 + b = ax
x2 = ax + b
To us, these are essentially the same, but
they are carefully arranged so all
coefficients are postive.
58
Step by step
al-Khwārizmī's book was extremely
important to Europe, not so much for
solving quadratic equations,
but because it introduced Arabic numerals
and the place-value decimal system.
Also, we got another important word from alKhwārizmī: algorithm is a corruption of his
name.
59
François Viète (1540 – 1603)
A French mathematician who introduced
notation for algebra:
–
–
knowns represented by consonants
unknowns represented by vowels
Thus, instead of always working with a
particular problem stated in words, a
general problem could be expressed:
dee + be + c = 0
60
René Descartes (1596 – 1650)
Within 50 years Viète's notation was
replaced by Descartes's:
–
–
–
knowns represented by letters from the
beginning of the alphabet (a, b, c, ...)
unknowns represented by letters from the
end of the alphabet (x, y, z, …)
superscripts for powers (x3 instead of
xxx)
We are still using Descartes notation:
ax3 + bx2 + cx +d = 0 61
Review and preview
type
linear
quadratic
form
ax + b = 0
ax2 + bx + c = 0
no. roots
1
2
What's next?
cubic
ax3 + bx2 + cx + d = 0
3
62
f(x) = x3 + 3x2 - 6x - 8
63
f(x) = x3 + 2x2 + 10x - 20
64
f(x) = x3 + 3x2 - 4
65
f(x) = x3 - 3x2 + 3x - 1
66
All four cubics together
67
Solving a cubic equation
First compute some auxiliary values:
3
2
D= 2 b − 9 abc+ 27 a d
√
√
1
2
2
3
E=
[ D+ √
D − 4(b − 3 ac) ]
2
3
1
2
2
3
F=
[ D− √
D − 4(b − 3 ac) ]
2
3
68
The three solutions are ...
b
1
1
x 1= −
−
E−
F
3a 3a
3a
b 1+ i √
3
1− i √
3
x 2= −
+
E+
F
3a
6a
6a
b 1− i √
3
1+ i √
3
x 3= −
+
E+
F
3a
6a
6a
69
Who figured this monster out?
Now that you've seen the solution to the
general cubic equation, you should never
again complain about the quadratic
formula as complicated or difficult.
The complete solution of the general cubic
equation was found about 500 years ago.
It was essentially an Italian affair, but first a
word about a Persian.
70
Omar Khayyám (1048-1131)
Some work was done on special cases of
cubic equations by the Babylonians,
Egyptians, Greeks, Chinese, Indians, …
In 1070 the Persian Omar Khayyám wrote
Treatise on Demonstrations of Problems of
Algebra, in which he solved a cubic by
intersecting a circle and a hyperbola.
He is perhaps better known for the Rubáiyát:
“A Jug of Wine, a Loaf of Bread--and a Cubic”
71
How low can an equation get?
Scipione del Ferro (1465 – 1526) of Bologna
solved the 'depressed' cubic:
x3 + cx + d = 0
'Depressed' means there is no x2 term (and
note that a = 1).
del Ferro didn't tell a soul about his
discovery until he was on his deathbed.
Why?
72
Dueling mathematicians
With his dying breath (almost), del Ferro told
his student Antonio Fior (c. 1506 - ?) the
secret of solving depressed cubics.
Fior immediately challenged Niccolo
Fontana (1499-1557), a well-known
mathematician, with 30 depressed cubic
equations to solve.
Fontana in turn sent 30 problems of various
sorts to Fior.
73
Both went to work on their problem sets.
Western civilization at work
Niccolo Fontana was popularly known by his
nickname, Tartaglia, which is Italian for
“stammerer.”
In 1512 the French invaded Fontana's town,
Brescia, and promptly massacred 45,000.
Fontana was wounded by a saber and
thereafter was never able to speak
properly.
Hence, he became known as Tartaglia.
74
All cubics? Not a good idea
Tartaglia was in trouble; he worked day and
night trying to solve the depressed cubics.
Finally, on February 13, 1535, he figured out
how to solve the general depresed cubic.
He then solved all 30 problems in two hours.
Fior, on the other hand, did a miserable job
on his problems; he faded into obscurity.
Tartaglia was the undisputed king of cubics.
75
I promise never to tell
When Gerolamo Cardano (1501-1576) of
Milan heard of Tartaglia's feat, he began a
concerted effort to convince Tartaglia to
show him the method.
Finally, on March 25, 1539, after Cardano
swore a solemn oath never to publish the
method, Tartaglia gave him the method,
encrypted in a cipher.
76
Solved at last
Cardano soon let his student/colleague
Ludovico Ferrari (1522-1565) in on the
secret (note that Cardano did not publish
anything).
The two of them made great progress.
Cardano figured out how to transform the
general cubic into a depressed cubic, so
then he could solve any cubic whatsoever.
77
If you only knew what I know
Cardano was dying to publish his results.
But it wasn't really satisfactory to write a
paper along the lines of
–
“If you only knew how to solve a
depressed cubic, then here's how you
could solve a general cubic”
Ferrari was also dying to publish a
significant result (which we'll talk about in
a minute).
78
del Ferro's secret out at last
Remember del Ferro who first solved the
depressed cubic and told nobody but Fior?
Well, del Ferro did write up his solution, but
never published it.
In 1543, Cardano and Ferrari stumbled on
del Ferro's notes.
In 1545 Cardano published his masterpiece,
Ars Magna, telling all.
79
I didn't publish your secret ...
The Ars Magna showed how to solve
–
–
–
the depressed cubic
the general cubic (by first depressing it)
another important equation (coming
soon)
Tartaglia was not happy.
Remember, Cardano swore a solemn oath.
Cardano said: “Hey, I didn't publish your
method; I published del Ferro's. Ok, ok,
80
there's similarities, but I kept my word.”
Both names live on
For years afterward, there were very nasty
words exchanged between Tartaglia on
one side and Cardano and Ferrari on the
other.
There was even another mathematical duel,
this time between Tartaglia and Ferrari.
Tartaglia left town before the duel ended.
But today the solution to the cubic is called
the Cardan-Tartaglia formula.
81
Review and preview - again
type
form
roots
linear
ax + b = 0
quadratic ax2 + bx + c = 0
cubic
ax3 + bx2 + cx + d = 0
no.
1
2
3
What's next?
quartic
ax4 + bx3 + cx2 + dx + 82
e=0
4
f(x) = x4 - x3 - 7x2 + x + 6
83
f(x) = x4 - 3x3 - 7x2 + 15x + 18
84
f(x) = x4 - 7x3 + 9x2 + 27x - 54
85
f(x) =
4
x
-
3
12x
+
+81
2
54x
- 108x
86
f(x) = x4 - 2x3 - 11x2 + 12x + 36
87
f(x) = x4 - x3 - 5x2 - x - 6
88
f(x) = x4 + 5x2 + 4
89
Quartics are easy (sort of)
Remember that Ferrari discovered
something important that was written up in
Ars Magna?
Ferrari solved the quartic!
He figured out how to reduce a quartic to a
cubic, use Cardano's method to solve the
cubic, and then get the quartic solution.
It gets real messy - see Wikipedia.
90
Review and preview – yet again
type
form
no.
roots
linear
ax + b = 0
1
quadratic ax2 + bx + c = 0
2
cubic
ax3 + bx2 + cx + d = 0
3
quartic
ax4 + bx3 + cx2 + dx + 91
e=0
4
f(x) =
5
x
-
4
3x
-
3
5x
+
2
15x
12
92
+ 4x -
f(x) = x5 - x4 + 5x3 - 5x2 + 4x - 4
93
What are we looking for?
So now we expect to hear about some bright
mathematician figuring out how to solve
the quintic.
But let's pause a moment and consider just
what it means to “solve the quintic.”
We haven't made it explicit until now, but we
have been looking for algebraic solutions,
sometimes called solutions by radicals.
94
How radical is that?
A solution by radicals of the quintic would
look like this:
x = some algebraic expression
in a, b, c, d, e, and f
where algebraic means only +, -, *, /, and nth
roots (square roots, cube roots, etc.)
appear in the expression.
Recall the quadratic formula:
95
Quadratic formula is algebraic
− b± √
b − 4ac
x=
2a
2
Note that the only operations are
addition, subtraction, multiplication, division,
and square root,
applied to the coefficients a, b, and c.
96
A lull in the action
For about 250 years, nobody could find an
algebraic solution for the quintic.
Not that lots of mathematicians were
spending all their time on the problem.
In the late 17th century calculus was born,
and lots of people focused on that.
And slowly, doubts began to form about the
whole issue of an algebraic solution.
97
Work not appreciated until later
In 1799 Paolo Ruffini (1765-1822), an
Italian, presented what he thought was a
proof that in general there is no algebraic
solution of the quintic and higher degree
equations.
He published three more proofs over the
years, but he was mostly ignored, and his
proofs weren't complete.
98
One Italian + One Norwegian
In 1824 the Norwegian Niels Henrik Abel
(1802-1829) published the definitive proof:
For the general equation of degree 5 or
greater, there is no solution by radicals.
This is now known as the Abel-Ruffini
theorem.
So, we've reached the end of the line.
99
Fundamental Theorem of
Algebra
It must be stressed that some equations of
degree 5 or greater can be solved by
radicals.
But in general this is not the case.
However, every equation of whatever
degree does have a root (possibly
complex rather than real).
In fact, an equation of degree n has n roots.
100
Disclaimer
If you ever have a real-world problem which
requires that you find a root of a quadratic,
cubic, or quartic equation,
don't ever, ever, ever, use the solutions by
radicals you've seen tonight.
As soon as you want to compute an actual
number as the root of an equation, you
need to use numerical analysis.
101
An example of Patty's Theorem
Say we are given this equation:
y = x2 - 11x + 9
Peppermint Patty's Theorem tells us that x is
almost always 11, so we plug in x = 11:
y = 112 – 11*11 + 9
y = 121 – 121 + 9
y=9
So we see that y is almost always 9.
102
Corollary to Patty's Theorem
“One thing I've learned
about algebra, don't take
it too seriously ...”
103