A short history of equations – The search for x

Roger HouseScientific Buzz Café

French GardenSebastopol, CA2012 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 HouseScientific Buzz Café

French GardenSebastopol, CA2012 January 19

Copyright © 2012 Roger House

4

The Secret of Algebra

“I'll never get this second problem.”

5

The Secret of Algebra

“Just put down 'eleven,' Franklin, and don't worry about it. That's what I did.”

6

The Secret of Algebra

“x is almost always eleven, and y isalmost always nine ...”

7

The Secret of Algebra

“One thing I've learned about algebra,don't take it too seriously ...”

8

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.

9

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”

10

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.

11

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

12

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

13

A picture is worth a 1000 ...

14

All the points are in a line

15

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 + 32then 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 linear equation:

f(x) = ax + b,where a and b are real numbers, a ≠ 0.

(a and b can be complex numbers, but nevermind.)

16

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 – 160/9Here's a graph of this new equation:

17

What's this? Make a wild guess

18

C = (5/9)(F - 32)

19

See both equations at once

20

Let's get to the root of it

Given a function like

f(x) = ax + bone 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.

21

Solving a linear equation

For a linear equation, we want to find x such that

ax + b = 0This is pretty easy:

ax = -b

x = -b/a(Remember, a ≠ 0, so we can divide by a.)

22

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) = 32

23

Where is the x-axis intersected?

24

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

25

Where is the x-axis intersected?

26

Summary of linear equations

A linear equation has this form:

ax + b = 0where a ≠ 0.

A linear equation has exactly one root,

x = -b/awhich is the point where the graph of the

equation intersects the x-axis.

27

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.

28

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?

29

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 + 4Quick 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?

30

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?

31

We need a picture ...

32

Fill in the blanks

33

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?

34

Staying above ground

35

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)

36

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 = 0where a ≠ 0.

Note the extra term ax2 containing the square of x (quadratus means “square” in Latin).

37

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 = 0is

This may look a bit scary, but watch:

x=−b±√b2−4a c2a

38

When is the height zero?

Let's solve our favorite quadratic:

-16t2 +12t + 4 = 0In this case,

a = -16, b = 12, c = 4,

so

x=−b±√b2−4a c2a

=−12±√122−4(−16)(4)

2(−16)

39

Grind through some arithmetic

When the dust settles, we have

so

x = (12 + 20) / 32 = 32 / 32 = 1

and

x = (12 – 20) / 32 = -8/32 = -1/4 = -0.25

x= 12±2032

40

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:

41

f(x) = -16x2 + 12x + 4

42

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:

43

f(x) = x2 - 6*x + 5

44

f(x) = x2 - 6*x + 9

45

f(x) = x2 - 6*x + 13

46

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

x=−b±√b2−4a c2a

=−(−6)±√(−6)2−4(1)(13)

2(1)

47

Can you image the roots?

Doing a bit of arithmetic:

so,

and

What is the square root of -1?

x= 6±√−162

= 6±4√−12

=3±2√−1

x=3+ 2√−1

x=3−2√−1

48

Summary of quadratic equations

A quadratic equation has this form:

ax2 + bx + c = 0where a ≠ 0.

A quadratic equation has exactly two roots,

x=−b±√b2−4a c2a

49

Linear equations go way back

The Egyptian Rhind Papyrus from 1650 BC (but copied from a document from 1849-1801 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

50

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 method x = -b/a.

51

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?)

52

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– Remember the pyramids? the Sphinx?

53

False position has a long history

Known in these times and places:– 1850 BC Egyptians– 200 BC Chinese– 3rd century Hindus– 10th century Arabs– 1200 Fibonacci– 1550 Robert Record

Even employed in late 19th century

54

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

55

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.”

56

x2 + bx – c = 0

In effect, the solution uses our modern solution with a = 1 and c negated:

The Babylonians did not have this form nor notation. They solved example after example, and at the end wrote:

“Such is the procedure.”

x=√( b2)2

+ c− b2

57

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.

58

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.

59

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 al-Khwārizmī: algorithm is a corruption of his name.

60

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

61

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

62

Review and preview

type form no. roots

linear ax + b = 0 1

quadratic ax2 + bx + c = 0 2

What's next?

cubic ax3 + bx2 + cx + d = 0 3

63

f(x) = x3 + 3x2 - 6x - 8

64

f(x) = x3 + 2x2 + 10x - 20

65

f(x) = x3 + 3x2 - 4

66

f(x) = x3 - 3x2 + 3x - 1

67

All four cubics together

68

Solving a cubic equation

First compute some auxiliary values:

D=2b3−9abc+ 27a2d

E= 3√ 12 [D+ √D2−4(b2−3ac)3]

F= 3√ 12 [D−√D2−4(b2−3ac)3]

69

The three solutions are ...

x2=− b3 a

+ 1+ i√36 a

E+ 1−i √36a

F

x1=− b3a

− 13aE− 1

3aF

x3=− b3 a

+ 1−i √36 a

E+ 1+ i√36 a

F

70

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.

71

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”

72

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?

73

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.

Both went to work on their problem sets.

74

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.

75

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.

76

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.

77

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.

78

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).

79

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.

80

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, there's similarities, but I kept my word.”

81

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.

82

Review and preview - again

type form no. roots

linear ax + b = 0 1

quadratic ax2 + bx + c = 0 2

cubic ax3 + bx2 + cx + d = 0 3

What's next?

quartic ax4 + bx3 + cx2 + dx + e = 0 4

83

f(x) = x4 - x3 - 7x2 + x + 6

84

f(x) = x4 - 3x3 - 7x2 + 15x + 18

85

f(x) = x4 - 7x3 + 9x2 + 27x - 54

86

f(x) = x4 - 12x3 + 54x2 - 108x +81

87

f(x) = x4 - 2x3 - 11x2 + 12x + 36

88

f(x) = x4 - x3 - 5x2 - x - 6

89

f(x) = x4 + 5x2 + 4

90

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.

91

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 + e = 0 4

What's next?

quintic ax5 + bx4 + cx3 + dx2 + ex + f = 0 5

92

f(x) = x5 - 3x4 - 5x3 + 15x2 + 4x - 12

93

f(x) = x5 - x4 + 5x3 - 5x2 + 4x - 4

94

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.

95

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:

96

Quadratic formula is algebraic

Note that the only operations are

addition, subtraction, multiplication, division, and square root,

applied to the coefficients a, b, and c.

x=−b±√b2−4a c2a

97

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.

98

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.

99

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.

100

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.

101

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.

102

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.

103

Corollary to Patty's Theorem

“One thing I've learned about algebra, don't take

it too seriously ...”