Taylor 2

1. (First lecture, on Monday 2011 Jan 31,
was with no Keynote slides at all)

2. i
What is i ?

3. i
What is i ?

4. i
What is i ?

5. i
What is i ?

6. i
What is i ?

7. i
What is i ?

8. Define z ≡ cos θ + i sin θ

9. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ

10. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 

11. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 

12. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ }

13. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z

14. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz
So = i dθ
z

15. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz
So = i dθ Let’s integrate this:
z

16. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ

17. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0

18. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0

19. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iθ
z=e

20. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iθ
z=e = cosθ + i sin θ

21. Define z ≡ cos θ + i sin θ
Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iθ
z=e = cosθ + i sin θ
iθ
The Euler formula e = cosθ + i sin θ

22. Define z ≡ cos θ + i sin θ
eiπ = cos π + i sin π = −1 Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iθ
z=e = cosθ + i sin θ
iθ
The Euler formula e = cosθ + i sin θ

23. Define z ≡ cos θ + i sin θ
eiπ = cos π + i sin π = −1 Then dz = − sin θ dθ + i cos θ dθ
 1 
= i dθ − sin θ + cosθ 
 i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iπ iθ
e +1= 0 z=e = cosθ + i sin θ
iθ
The Euler formula e = cosθ + i sin θ

24. Define z ≡ cos θ + i sin θ
eiπ = cos π + i sin π = −1 Then dz = − sin θ dθ + i cos θ dθ
i
π
π π  1 
e = cos + i sin = i
2 = i dθ − sin θ + cosθ 
2 2  i 
 i 
= i dθ − 2 sin θ + cosθ 
 i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iπ iθ
e +1= 0 z=e = cosθ + i sin θ
iθ
The Euler formula e = cosθ + i sin θ

25. Define z ≡ cos θ + i sin θ
eiπ = cos π + i sin π = −1 Then dz = − sin θ dθ + i cos θ dθ
i
π
π π  1 
e = cos + i sin = i
2 = i dθ − sin θ + cosθ 
2 2  i 
π i π
 i2  π π i
−  i 
e  = cos + i sin = i = e 2 = 0.208
  = i dθ − 2 sin θ + cosθ 
 2 2  i 
= i dθ {i sin θ + cosθ } = i dθ z
dz dz
So
z
= i dθ Let’s integrate this: ∫ z = i ∫ dθ
ln z = i θ + ( ln z )θ = 0
When θ = 0, z = cos 0 + i sin 0 = 1 and ( ln z )θ = 0 = 0
iπ iθ
e +1= 0 z=e = cosθ + i sin θ
iθ
The Euler formula e = cosθ + i sin θ

26. 4 Square Questions
B A
C D
4

27. 4 Square Questions
B A
Look carefully to the
diagram
Now I will ask you 4
questions about
this square.
Are you
ready?
C D
4

28. 4 Square Questions
B A
C D
5

29. 4 Square Questions
B A
Q1
Q1
Divide the white
area in square A
into two equal
pieces.
Easy!!
Isn`t it?
C D
5

30. 4 Square Questions
B A
Q1
Q1
Divide the white
area in square A
into two equal
pieces.
C D
6

31. 4 Square Questions
B A
Q1
Q1
Divide the white
area in square A
into two equal
pieces.
Here is the answer!
C D
6

32. 4 Square Questions
B A
Q1
Q1
Divide the white
area in square A
into two equal
pieces.
C D
7

33. 4 Square Questions
B A
Q1
Q1
Divide the white
area in square A
into two equal
pieces.
Of course you solved
C D it!
:)
7

34. 4 Square Questions
B A
C D
8

35. 4 Square Questions
B A
Q2
Q2
Divide the white
area in square B
into three equal
pieces.
Come on it is not
C D soo difficult!
8

36. 4 Square Questions
B A
Q2
Q2
Divide the white
area in square B
into three equal
pieces.
C D
9

37. 4 Square Questions
B A
Q2
Q2
Divide the white
area in square B
into three equal
pieces.
Here is the answer!
C D
9

38. 4 Square Questions
B A
Q2
Q2
Divide the white
area in square B
into three equal
pieces.
C D
10

39. 4 Square Questions
B A
Q2
Q2
Divide the white
area in square B
into three equal
pieces.
You knew the answer
C D anyways or??
:))
10

40. 4 Square Questions
B A
C D
11

41. 4 Square Questions
B A
OK!!!
C D
11

42. 4 Square Questions
B A
C D
12

43. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
Very difficult??
That`s right!
C D
12

44. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
C D
13

45. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
You haven´t found
C D it yet???
13

46. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
C D
14

47. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
Come on! You can do
C D it!!
14

48. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
C D
15

49. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
Take your time.
Click If you want to see
C D the solution!
15

50. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
C D
16

51. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
Here is the solution!
C D
16

52. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
C D
17

53. 4 Square Questions
B A
Q3
Q3
Divide the white
area in square C
into four equal
pieces.
Could you solve it?
C D :)))
17

54. 4 Square Questions
B A
C D
18

55. 4 Square Questions
B A
Be ready
here comes
the last
Question!
C D
18

56. 4 Square Questions
B A
C D
19

57. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
World record is 7
seconds!!
C D
19

58. 4 Square Questions
B A
Divide the area D
into seven equal
pieces.
World record is 7
seconds!!
C D
20

59. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
World record is 7
seconds!!
C D
20

60. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
21

61. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
Time is up!
C D
21

62. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
22

63. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
Any idea??
C D
22

64. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
23

65. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
I can wait!!
Click when you are
C D bored!
23

66. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
24

67. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
Here is the answer!
C D
24

68. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
25

69. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
Was it really that
C D difficult?
25

70. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
C D
26

71. 4 Square Questions
B A
Q4
Q4
Divide the area D
into seven equal
pieces.
It was just to see how
our minds can be
conditioned!
:))))
C D
26

72. 4 Square Questions
HAVE A NICE DAY!!
27

74. Paul C. W. Davies: “I contrived to sit opposite her one
day, charged with the homework task of computing the
trajectory of a ball projected up an inclined plane. As I
was partway through several sheets of mathematics,
the ravishing Lindsay looked across at me with a mixture
of admiration and puzzlement. "What are you doing?" she
asked. When I explained, she seemed completely
mystified. "But how can you tell where a ball will go by
writing squiggles on paper?”

75. Paul C. W. Davies: “I contrived to sit opposite her one
day, charged with the homework task of computing the
trajectory of a ball projected up an inclined plane. As I
was partway through several sheets of mathematics,
the ravishing Lindsay looked across at me with a mixture
of admiration and puzzlement. "What are you doing?" she
asked. When I explained, she seemed completely
mystified."But how can you tell where a ball will go by
"But how can you tell where a ball
will go by writingsquiggles on paper?”
writing squiggles on paper?”

76. 1, 2, 3, many

77. 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12 ....

78. 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12 ....

79. There are 10 kinds
of people: those
that know binary,
and those that
don’t!

80. 1, 10, 11, 100, 101,
110, 111, 1000, ...

81. 0, 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 ....

82. 454
404

83. John Barrow’s book, Pi in the Sky
545

84. John Barrow’s book, Pi in the Sky
The Indian system of counting has been the most successful intellectual
innovation ever made on our planet. It has spread and been adopted
almost universally, for more extensively even than the letters of the
Phoenician alphabet which we now employ. It constitutes the nearest
thing we have to a universal language.
545

85. John Barrow’s book, Pi in the Sky
The Indian system of counting has been the most successful intellectual
innovation ever made on our planet. It has spread and been adopted
almost universally, for more extensively even than the letters of the
Phoenician alphabet which we now employ. It constitutes the nearest
thing we have to a universal language.
Invariably, the result of any contact between the Indian
scheme and any other system of counting was the adoption of
the Indian system...
545

86. John Barrow’s book, Pi in the Sky
The Indian system of counting has been the most successful intellectual
innovation ever made on our planet. It has spread and been adopted
almost universally, for more extensively even than the letters of the
Phoenician alphabet which we now employ. It constitutes the nearest
thing we have to a universal language.
Invariably, the result of any contact between the Indian
scheme and any other system of counting was the adoption of
the Indian system...
545
The Hebrew tradition imported the Indian ideas through the
traveling scholar Rabbi Ben Ezra (1092-1167)...He described the
Indian system in his influential Book of Number

87. John Barrow’s book, Pi in the Sky
The Indian system of counting has been the most successful intellectual
innovation ever made on our planet. It has spread and been adopted
almost universally, for more extensively even than the letters of the
Phoenician alphabet which we now employ. It constitutes the nearest
thing we have to a universal language.
Invariably, the result of any contact between the Indian
scheme and any other system of counting was the adoption of
the Indian system...
545
The Hebrew tradition imported the Indian ideas through the
traveling scholar Rabbi Ben Ezra (1092-1167)...He described the
Indian system in his influential Book of Number
Although Arabic writing reads from right to left, the Indian
convention of reading numbers from left to right has been
retained.

88. John Barrow’s book, Pi in the Sky
The Indian system of counting has been the most successful intellectual
innovation ever made on our planet. It has spread and been adopted
almost universally, for more extensively even than the letters of the
Phoenician alphabet which we now employ. It constitutes the nearest
thing we have to a universal language.
Invariably, the result of any contact between the Indian
scheme and any other system of counting was the adoption of
the Indian system...
Robert Browning’s poem, “Rabbi Ben Ezra”
545
Grow old along with me!
The best is yet to be,
The last of life, for which the first was made.
Although Arabic writing reads from right to left, the Indian
convention of reading numbers from left to right has been
retained.

89. -6, -5, -4, -3, -2, -1,
0, 1, 2, 3, 4, 5, 6

91. The Theorem of Pythagoras
c b
a

92. The Theorem of Pythagoras
c
c
c
c b
a

93. The Theorem of Pythagoras
b
b c
c
c
c b
b a

94. The Theorem of Pythagoras
a b
b c
c a
a c
c b
b a

95. The Theorem of Pythagoras
a b
b c
c a
area =
a c
c b
b a

96. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b)
a c
c b
b a

97. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
a c
c b
b a

98. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
area =
a c
c b
b a

99. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
2 1 
area = c + 4  ab 
2 
a c
c b
b a

100. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
2  1  = c 2 + 2ab
area = c + 4  ab 
2 
a c
c b
b a

101. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
2  1  = c 2 + 2ab
area = c + 4  ab 
2 
2 2 2
∴ a + 2ab + b = c + 2ab
a c
c b
b a

102. The Theorem of Pythagoras
a b
b c
c a
area = (a + b)(a + b) = a 2 + 2ab + b 2
2  1  = c 2 + 2ab
area = c + 4  ab 
2 
2 2 2
∴ a + 2ab + b = c + 2ab
a c 2 2 2
a +b = c
c b
b a

103. The Theorem of Pythagoras
z
s
z
y
c
O x

104. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y
c
O x

105. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2
c c =
O x

106. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x

107. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2
s =

108. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z

109. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z
2
s =

110. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z
2 2 2 2
s =x +y +z

111. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z
2 2 2 2
s =x +y +z

112. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z
2 2 2 2
Now write it in Eleven Dimensions: s =x +y +z
2
s =

113. The Theorem of Pythagoras
in Three Dimensions:
z
s
z
y 2 2 2
c c =x +y
O x 2 2 2
s =c +z
2 2 2 2
Now write it in Eleven Dimensions: s =x +y +z
2 2 2 2 2 2 2 2 2 2 2 2
s = a +b +e + f +g +h +m +n + p +q +r