1. 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

2. 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:

3. “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

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 is
almost 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 ... F = (9/5)*C + 32 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

13. A picture is worth a

14. All the points are in a line

15. Linear Equations F = (9/5)*C + 32 f(x) = ax + b,
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 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 – 160/9
Here'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 + 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.

21. 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.)

22. x = -b/a = -(-160/9) / (5/9) = (160/9)*(9/5) = 32
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 ≈

25. Where is the x-axis intersected?

26. 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.

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 + 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?

30. Let's try some numbers Let t range from 0 to 2 by half seconds:
t = 0 ⇒ h = -16*02 +12* = 4
t = 0.5 ⇒ h = -16* * = 6
t = 1 ⇒ h = -16*12 +12* = 0
t = 1.5 ⇒ h = -16* * = -14
t = 2 ⇒ h = -16*22 +12* = -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? second (= 3/8 second)
How high will the ball go? ft (= 6 ft 3 in)

36. Quadratic equations ax2 + bx + c = 0
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).

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 = 0 is
This may look a bit scary, but watch:

38. When is the height zero? -16t2 +12t + 4 = 0
Let's solve our favorite quadratic:
-16t2 +12t + 4 = 0
In this case,
a = -16, b = 12, c = 4, so

39. Grind through some arithmetic
When the dust settles, we have so
x = ( ) / 32 = 32 / 32 = 1
and
x = (12 – 20) / 32 = -8/32 = -1/4 = -0.25

40. Can this be right? Plug 1 into -16t2 +12t + 4 = 0:
-16*12 +12*1 + 4 = =0
Now try -0.25:
-16*(-0.25)2 +12*(-0.25) + 4 = -1 – = 0
So, both 1 and 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

47. Can you image the roots? Doing a bit of arithmetic: so, and
What is the square root of -1?

48. Summary of quadratic equations
A quadratic equation has this form:
ax2 + bx + c = 0
where a ≠ 0.
A quadratic equation has exactly two roots,

49. Linear equations go way back
The Egyptian Rhind Papyrus from 1650 BC (but copied from a document from BC - so it's 3800 years old) posed problems like this:
“A quantity and its 1/7 added become What is the quantity?”
Today we would write this as
x + x/7 = 19

50. x = -b/a = -(-19)/(8/7) = (7/8)19 = 16.625
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 =
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 = = 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 =
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
Fibonacci
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. “Such is the procedure.”
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.”

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 =
What's next?
cubic ax3 + bx2 + cx + d =

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:

69. The three solutions are ...

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 ( )
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 ?) the secret of solving depressed cubics.
Fior immediately challenged Niccolo Fontana ( ), 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 ( ) 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 ( ) 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 =
quadratic ax2 + bx + c =
cubic ax3 + bx2 + cx + d =
What's next?
quartic ax4 + bx3 + cx2 + dx + e =

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 + 54x x +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 =
quadratic ax2 + bx + c =
cubic ax3 + bx2 + cx + d =
quartic ax4 + bx3 + cx2 + dx + e =
What's next?
quintic ax5 + bx4 + cx3 + dx2 + ex + f =

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. x = some algebraic expression
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.

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 ( ), 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 ( ) 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 –
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 ...”