The Wallis Product, a Connection to Pi and Probability, and Maybe the Gamma Function Perhaps?

Transcription

1 The Wallis Product, a Connection to Pi and Probability, and Maybe the Gamma Function Perhaps? Aba Mbirika Assistant Professor of Mathematics May 8th, 2017 A Math Presentation for the Riemann Seminar Class

2 Outline 1 Why pi? 2 A brief history of π 3 Coin toss and drunken walk problem 4 A simple geometric proof 5 Poop

3 Questions I may or may not address (in no particular order) Why is π so freaking cool?

4 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite?

5 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite? What does π have to do with combinatorics?

6 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite? What does π have to do with combinatorics? Coin flips?

7 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite? What does π have to do with combinatorics? Coin flips? Drunken walks?

8 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite? What does π have to do with combinatorics? Coin flips? Drunken walks? Are Ralph Waldo Emerson and Henry David Thoreau transcendentalists? And is π transcendental?

9 Questions I may or may not address (in no particular order) Why is π so freaking cool? How many digits of π can you recite? What does π have to do with combinatorics? Coin flips? Drunken walks? Are Ralph Waldo Emerson and Henry David Thoreau transcendentalists? And is π transcendental? Does this talk have ANYTHING at all to do with the gamma function and more precisely the value Γ ( ) 3 2 being 1 2 factorial?

10 Introducing my co-speaker, Sophie

11 Venn diagram of the Real Numbers

12 Outline 1 Why pi? 2 A brief history of π 3 Coin toss and drunken walk problem 4 A simple geometric proof 5 Poop

13 Sophie suggests

14 The 16th and 17th Centuries François Vièta (1579, France)

15 The 16th and 17th Centuries François Vièta (1579, France) π =

16 The 16th and 17th Centuries François Vièta (1579, France) π = 2 2 John Wallis (1650, England) π 2 = = 2 n=1 2n 2n (2n 1)(2n + 1)

17 The 16th and 17th Centuries François Vièta (1579, France) π = 2 2 John Wallis (1650, England) π 2 = = 2 n=1 2n 2n (2n 1)(2n + 1) Equation above comes up BIG TIME later in the talk!

18 The 16th and 17th Centuries François Vièta (1579, France) π = 2 2 John Wallis (1650, England) π 2 = = 2 n=1 2n 2n (2n 1)(2n + 1) Equation above comes up BIG TIME later in the talk! Lord Brouncker (1650, England) 4 π =

19 James Gregory (1668, Scotland)

20 James Gregory (1668, Scotland) π 4 =

21 James Gregory (1668, Scotland) π 4 = Does anyone see a Calc II way to prove this?

22 James Gregory (1668, Scotland) π 4 = Does anyone see a Calc II way to prove this? Maybe use the Taylor series expansion for arctan(x) and a judicious choice for x.

23 James Gregory (1668, Scotland) π 4 = Does anyone see a Calc II way to prove this? Maybe use the Taylor series expansion for arctan(x) and a judicious choice for x. Abraham Sharp (1699, England) ( π 1 6 = )

24 The 18th Century Georges-Louis Leclerc, Comte de Buffon (1760, France) A plane is ruled with parallel lines 1 inch apart. A needle of length 1 inch is dropped randomly on the plane. What is the probability that it will be lying across one of the lines?

25 The 18th Century Georges-Louis Leclerc, Comte de Buffon (1760, France) A plane is ruled with parallel lines 1 inch apart. A needle of length 1 inch is dropped randomly on the plane. What is the probability that it will be lying across one of the lines? 2 π 63.66%

26 The 18th Century Georges-Louis Leclerc, Comte de Buffon (1760, France) A plane is ruled with parallel lines 1 inch apart. A needle of length 1 inch is dropped randomly on the plane. What is the probability that it will be lying across one of the lines? 2 π 63.66% Johann Lambert (1761, Germany) proves that π is irrational. In this same paper, he conjectures that π is transcendental. Remember this for a few slides from now when I speak of Lindemann.

27 Leonhard Euler (1748, Switzerland) publishes Introductio in analysin infinitorum. Math historians say that it was this publication that catapulted the symbol π into popular use.

28 Leonhard Euler (1748, Switzerland) publishes Introductio in analysin infinitorum. Math historians say that it was this publication that catapulted the symbol π into popular use. QUESTION: What is the coolest of all of Euler s formula?

29 Leonhard Euler (1748, Switzerland) publishes Introductio in analysin infinitorum. Math historians say that it was this publication that catapulted the symbol π into popular use. QUESTION: What is the coolest of all of Euler s formula? ANSWER: e iπ + 1 = 0

30 How cool do these folks think this equation is?

31 How cool do these folks think this equation is?

32 How cool do these folks think this equation is?

33 Can this guy wash his shirt anymore?

34 The 19th Century to the present William Rutherford (1841, England)

35 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99

36 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99 He uses this to estimate π to 208 places (152 correct). But even 152 decimal places broke the world record at the time. The modern quest for EVER-more digits begins.

37 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99 He uses this to estimate π to 208 places (152 correct). But even 152 decimal places broke the world record at the time. The modern quest for EVER-more digits begins. Ferdinand Lindemann (1882, Germany) proves that π is transcendental.

38 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99 He uses this to estimate π to 208 places (152 correct). But even 152 decimal places broke the world record at the time. The modern quest for EVER-more digits begins. Ferdinand Lindemann (1882, Germany) proves that π is transcendental. Srinivasa Ramanujan (1913, India) gives many remarkable approximations of π:

39 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99 He uses this to estimate π to 208 places (152 correct). But even 152 decimal places broke the world record at the time. The modern quest for EVER-more digits begins. Ferdinand Lindemann (1882, Germany) proves that π is transcendental. Srinivasa Ramanujan (1913, India) gives many remarkable approximations of π: ( ) 1 4 =

40 The 19th Century to the present William Rutherford (1841, England) ( ) π 1 4 = 4 arctan arctan 5 ( 1 70 ) ( ) 1 + arctan 99 He uses this to estimate π to 208 places (152 correct). But even 152 decimal places broke the world record at the time. The modern quest for EVER-more digits begins. Ferdinand Lindemann (1882, Germany) proves that π is transcendental. Srinivasa Ramanujan (1913, India) gives many remarkable approximations of π: ( ) 1 4 ( = ) =

41 Sophie wonders

42 Outline 1 Why pi? 2 A brief history of π 3 Coin toss and drunken walk problem 4 A simple geometric proof 5 Poop

43 Recall from a previous slide, John Wallis (1655) arrived at his celebrated formula π 2 = = n=1 2n 2n (2n 1)(2n + 1)

44 Recall from a previous slide, John Wallis (1655) arrived at his celebrated formula π 2 = = n=1 2n 2n (2n 1)(2n + 1) Many textbook proofs of this formula rely the family {I n } of definite integral I n := π 2 0 (sin x) n dx by repeated partial integration. (Hint: Let u = sin n 1 x and dv = sin x dx. And derive the recursion formula I n = n 1 n with initial values I 0 = π 2 and I 1 = 1.) I n 2

45 Recall from a previous slide, John Wallis (1655) arrived at his celebrated formula π 2 = = n=1 2n 2n (2n 1)(2n + 1) Many textbook proofs of this formula rely the family {I n } of definite integral I n := π 2 0 (sin x) n dx by repeated partial integration. (Hint: Let u = sin n 1 x and dv = sin x dx. And derive the recursion formula I n = n 1 n with initial values I 0 = π 2 and I 1 = 1.) I n 2 This is cool problem for Calculus II. I highly suggest you write this down and give it a go later if you want.

46 Overview for the rest of the talk Introduce two equivalent combinatorial problems

47 Overview for the rest of the talk Introduce two equivalent combinatorial problems Coin flips problem, and Drunken walks problem.

48 Overview for the rest of the talk Introduce two equivalent combinatorial problems Coin flips problem, and Drunken walks problem. Give a geometric proof of Wallis product formula.

49 Overview for the rest of the talk Introduce two equivalent combinatorial problems Coin flips problem, and Drunken walks problem. Give a geometric proof of Wallis product formula. Use this to derive the solution to the combinatorial problems.

50 Overview for the rest of the talk Introduce two equivalent combinatorial problems Coin flips problem, and Drunken walks problem. Give a geometric proof of Wallis product formula. Use this to derive the solution to the combinatorial problems. And of course, we will connect ALL of this to π and perhaps the gamma function or more precisely the value of Γ ( ) 3 2 which is 1 2 factorial.

51 Flipping Coins Suppose we flip a fair coin 2n times. What are the chances that we get an equal amount of heads and tails? Label the head and tail sides of the coin as H and T respectively.

52 Flipping Coins Suppose we flip a fair coin 2n times. What are the chances that we get an equal amount of heads and tails? Label the head and tail sides of the coin as H and T respectively. We can view the 2n coin flips as sequences of H s and T s.

53 The sequences for n = 1 (i.e., flipping a coin 2 1 = 2 times) HH TT HT TH

54 The sequences for n = 1 (i.e., flipping a coin 2 1 = 2 times) HH TT HT TH So there are exactly 2 ways to get exactly one head and one tail. So there is a.5 probability that this event occurs when we flip a coin 2 times.

55 The sequences for n = 2 (i.e., flipping a coin 2 2 = 4 times): HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT

56 The sequences for n = 2 (i.e., flipping a coin 2 2 = 4 times): HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT QUESTION: What is the probability that you get an equal number of heads and tails when you flip a coin 4 times?

57 The sequences for n = 2 (i.e., flipping a coin 2 2 = 4 times): HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT QUESTION: What is the probability that you get an equal number of heads and tails when you flip a coin 4 times? ANSWER: Well this is silly question since I have highlighted certain sequences above in bold red!

58 Flipping Coins QUESTION: How many sequences for a given n are there (i.e., flipping a coin 2 n = 2n times)? Why?

59 Flipping Coins QUESTION: How many sequences for a given n are there (i.e., flipping a coin 2 n = 2n times)? Why? ANSWER: You re absolutely right! There are exactly 2 2n possible sequences!

60 Flipping Coins QUESTION: How many sequences for a given n are there (i.e., flipping a coin 2 n = 2n times)? Why? ANSWER: You re absolutely right! There are exactly 2 2n possible sequences! Suppose we flip a fair coin 2n times. What are the chances that we get an equal amount of heads and tails? We can then rephrase the original question:

61 Flipping Coins QUESTION: How many sequences for a given n are there (i.e., flipping a coin 2 n = 2n times)? Why? ANSWER: You re absolutely right! There are exactly 2 2n possible sequences! Suppose we flip a fair coin 2n times. What are the chances that we get an equal amount of heads and tails? We can then rephrase the original question: Out of the 2 2n possible sequences, how many sequences contain exactly n H s and n T s?

62 Example Consider the case for n = 5. Then the number of ways of getting exactly five H s is the number of ways of choosing 5 of the 10 sequence slots in which to place them. Hence there are ( ) 10 5 = 10! 5! 5! = 252 ways. Now since there are exactly 2 10 = 1024 possible sequences of H s and T s, the probability of getting exactly five H s and five T s is =

63 Example Consider the case for n = 5. Then the number of ways of getting exactly five H s is the number of ways of choosing 5 of the 10 sequence slots in which to place them. Hence there are ( ) 10 5 = 10! 5! 5! = 252 ways. Now since there are exactly 2 10 = 1024 possible sequences of H s and T s, the probability of getting exactly five H s and five T s is = This number is close to 1 5π which is approximately

64 Example Consider the case for n = 5. Then the number of ways of getting exactly five H s is the number of ways of choosing 5 of the 10 sequence slots in which to place them. Hence there are ( ) 10 5 = 10! 5! 5! = 252 ways. Now since there are exactly 2 10 = 1024 possible sequences of H s and T s, the probability of getting exactly five H s and five T s is = This number is close to 1 5π which is approximately For n = 20, the probability is while 1 20π is approximately

65 Example Consider the case for n = 5. Then the number of ways of getting exactly five H s is the number of ways of choosing 5 of the 10 sequence slots in which to place them. Hence there are ( ) 10 5 = 10! 5! 5! = 252 ways. Now since there are exactly 2 10 = 1024 possible sequences of H s and T s, the probability of getting exactly five H s and five T s is = This number is close to 1 5π which is approximately For n = 20, the probability is while 1 20π is approximately QUESTION: What is your conjecture????

66 Goal: As n gets large, the probability of getting exactly n heads and n tails out of 2n coin flips approaches the number 1 nπ.

67 Drunken Walks The coin flip problem can be recast in the following drunken walk model:

68 Drunken Walks Here is the drunken walk model:

69 Drunken Walks Here is the drunken walk model: Consider a walk where we go forward each time, however at each step forward we veer 45 degrees left or right.

70 Drunken Walks Here is the drunken walk model: Consider a walk where we go forward each time, however at each step forward we veer 45 degrees left or right. Let H be denoted by a step forward that veers left and T to be one that veers right.

71 Drunken Walks Here is the drunken walk model: Consider a walk where we go forward each time, however at each step forward we veer 45 degrees left or right. Let H be denoted by a step forward that veers left and T to be one that veers right. If after 2n steps we return to the original horizontal starting level, then we must have taken exactly n left-steps and n right-steps.

72 Definition Let M n be the number of cases of 2n-walks that return to the original horizontal starting level. And let N n be the number of cases of 2n-walks that never revisit the original horizontal starting level after initially leaving it.

73 Drunken Walks Consider the number of drunken walks of four steps that return to the original horizontal starting level. There are six possibilities. Figure: M 2 walks

74 Now consider the number of drunken walks of four steps that never return to the original horizontal starting level. It is no coincidence that this also yields six possibilities! Figure: N 2 walks

75 Challenge: Prove M n = N n. Figure: M 2 walks Figure: N 2 walks

76 Proof Strategy: Describe a bijection between the 6 above and the 6 below to prove M n = N n when n = 2, and then generalize to all n values. Challenge: Prove M n = N n. Figure: M 2 walks Figure: N 2 walks

77 Two facts: The probability of getting exactly n heads and n tails in 2n coin flips is M n 2 2n.

78 Two facts: The probability of getting exactly n heads and n tails in 2n coin flips is M n 2 2n. Furthermore, we can prove the following remarkable identity: 2 2n = M n N 0 + M n 1 N M 1 N n 1 + M 0 N n. The next slide (and some added comments from me) hopefully provides a picture of why this second fact makes sense.

79 Remark Consider an arbitrary walk. In any such walk there will be a portion of the walk of the form M i for some i n, followed by a walk that is of the form N j for j = n i. Figure: Two of the 12 possible M 2 N 1 walks Since M 2 = 6 and N 1 = 2, there are exactly 12 drunken walks of the form M 2 N 1. We give two such walks in the figure.

80 We list the first few M n and N n values in the table below. n M n N n Let s verify that our favorite identity from two slides back 2 2n = M n N 0 + M n 1 N M 1 N n 1 + M 0 N n holds for small n values, like n = 3 for instance.

81 We list the first few M n and N n values in the table below. n M n N n Let s verify that our favorite identity from two slides back 2 2n = M n N 0 + M n 1 N M 1 N n 1 + M 0 N n holds for small n values, like n = 3 for instance. 2 6 = M 3 N 0 + M 2 N 1 + M 1 N 2 + M 0 N 3 =

82 Definition Let a n denote the quotient a n = M n 2 2n = ( 2n n ) 2 2n. Define s n to be the sum a 0 + a a n 1 and s 0 = 0. The following table gives the first five a n and s n values. n a n s n

83 Definition Let a n denote the quotient a n = M n 2 2n = ( 2n n ) 2 2n. Define s n to be the sum a 0 + a a n 1 and s 0 = 0. The following table gives the first five a n and s n values. For example, for n = 3 we have n a n s n s 3 = a 0 + a 1 + a 2 = = 15 8.

84 More homework for you? Yay! ELFS: Use the following three equalities: M n = N n, 2 2n = M n N 0 + M n 1 N M 1 N n 1 + M 0 N n, and a n = M n 2 2n to show the remarkable (and useful, you ll soon see) identity holds 1 = a n a 0 + a n 1 a a 1 a n 1 + a 0 a n.

85 Sophie s Choice

86 Outline 1 Why pi? 2 A brief history of π 3 Coin toss and drunken walk problem 4 A simple geometric proof 5 Poop

87 Oh yeah, this talk is about π. As alluded to earlier, the probability of getting exactly n heads and n tails in 2n coin flips involves π. We will show the probability Mn 2 1 2n (that is, a n ) asymptotically approaches the value nπ : lim n a n = 1 nπ

88 Oh yeah, this talk is about π. As alluded to earlier, the probability of getting exactly n heads and n tails in 2n coin flips involves π. We will show the probability Mn 2 1 2n (that is, a n ) asymptotically approaches the value nπ : lim a n = 1 n nπ QUESTION: How hard will this proof be?

89 Oh yeah, this talk is about π. As alluded to earlier, the probability of getting exactly n heads and n tails in 2n coin flips involves π. We will show the probability Mn 2 1 2n (that is, a n ) asymptotically approaches the value nπ : lim a n = 1 n nπ QUESTION: How hard will this proof be? ANSWER: The proof uses nothing more than the Pythagorean theorem, the area formula for a circle, and some elementary algebra.

90 The heart of the proof that a n 1 nπ is a clever geometric insight by Johan Wästlund using a dreadfully dull looking picture in the black and white journal Amer. Math. Monthly, Dec Question: Why do I call that picture dull?

91 ANSWER: With a little splash of color, it can provide more insight into the simplicity of this geometric proof. Plus, it s prettier to look at.

92 Recall: We used drunken walks to prove that 1 = a n a 0 + a n 1 a a 1 a n 1 + a 0 a n. By the term we, I meant you. Recall ELFS (a few slides ago).

93 Since we showed the products of the form a n a 0 + a n 1 a a 1 a n 1 + a 0 a n each equal one, we have: Red rectangle area = a 0 a 0 = 1 Orange rectangle areas = a 1 a 0 + a 0 a 1 = 1 Yellow rectangle areas = a 2 a 0 + a 1 a 1 + a 0 a 2 = 1 Green rectangle areas = a 3 a 0 + a 2 a 1 + a 1 a 2 + a 0 a 3 = 1 Blue rectangle areas = a 4 a 0 + a 3 a 1 + a 2 a 2 + a 1 a 3 + a 0 a 4 = 1.

94 Let P n denote the region of rectangles labeled a i a j such that i + j < n. Then, P 1 has area 1 P 2 has area 2. P n has area n.

95 Summary As we add more rectangles, the outermost rectangles appear more and more to sit exactly on the circumference of the quarter circle. That is, the value 1 4 πs2 n is approximately n as n gets large. This follows since n 1 4 π(s n) 2 = 1 4 π(2na n) 2 = πn 2 a 2 n. Hence a 2 n 1 nπ and thus a n approaches 1 πn as n gets large.

96 Summary As we add more rectangles, the outermost rectangles appear more and more to sit exactly on the circumference of the quarter circle. That is, the value 1 4 πs2 n is approximately n as n gets large. This follows since n 1 4 π(s n) 2 = 1 4 π(2na n) 2 = πn 2 a 2 n. Hence a 2 n 1 nπ and thus a n approaches 1 πn as n gets large. MY PLEA TO THE AUDIENCE: The rest of the slides pretty much spell out the fun work most of which I leave as exercises (full of hints) in my write-up called On a Coin Flip Problem available at:

97 Example HEY!! Before we read the words and mathy stuff below, lets remind ourself of the rainbow (next slide). Then we ll come back to this slide. Observations about the Rainbow Slide The rectangles arranged in this fashion begin to resemble a quarter circle with radius s 5 = a 0 + a a 4. Hence the area 1 4 π(s 5) 2 is approximately equal to 5. Precisely, since s 5 = the value 1 4 π(s 5) 2 equals I m going to click back and forth on these two slides until EVERYONE in the room feels Yah, I get it!

98

99 We have the relation n 1 < 1 4 πs2 n < n + 1. The following figure motivates our reasoning.

100 Theorem The following relation n 1 < 1 4 πs2 n < n + 1 holds. Proof by picture: The region P 3 (red, brown, yellow) has area 3. The third 1 4 -circle contains P 3 in its interior and has radius s 4 and thus an area 1 4 π(s 4) 2. So 3 < 1 4 π(s 4) 2. The region P 5 (all 5 colors) has area 5. This region contains the third 1 4 -circle. We conclude 3 < 1 4 π(s 4) 2 < 5.

101 Challenge: Prove these facts. Fact For n 1, each a n can be written as the product 1 2 Sketch: For the a n -case rewrite n 1 2n as n 1 2n 2n 2n and show this equals M n 2 2n. Fact For n > 1, each s n can be written as the product 3 2 Putting these two facts together, we see s n = 2n a n n 1 2n. 8 2n 1 2n 2.

102 Challenge: Prove these facts. Fact For n 1, each a n can be written as the product 1 2 Sketch: For the a n -case rewrite n 1 2n as n 1 2n 2n 2n and show this equals M n 2 2n. Fact For n > 1, each s n can be written as the product 3 2 Putting these two facts together, we see s n = 2n a n n 1 2n. 8 2n 1 We will use the two facts above to prove the Wallis product equals π as desired. 2 2n 2.

103 We denote the Wallis product (remember this from long ago?) by W = = Let s 0 = 0, s 1 = 1, and n=1 s n = n 1 2n 2. 2n 2n (2n 1)(2n + 1) Then we have the following equalities: (Time-permitting, I can do n = 3 and n = 4 examples to illustrate.) 2n 1 s 2 n = 2n s 2 = n and the relations (2n 2) (2n 3) 2 (2n 1) (n is even) (2n) (2n 3) 2 (2n 1) 2 (n is odd) 2n 1 s 2 n < W < 2n s 2. n

104 Since n 1 < 1 4 πs2 n < n + 1 (from 3 slides ago), it follows that and thus π 4 ( ) 2n 1 n + 1 4(n 1) π < s 2 n < 4(n + 1), π < 2n 1 s 2 < W < 2n n s 2 n }{{} from previous slide < π 2 ( ) n. n 1

105 Since n 1 < 1 4 πs2 n < n + 1 (from 3 slides ago), it follows that and thus π 4 ( ) 2n 1 n + 1 4(n 1) π < s 2 n < 4(n + 1), π < 2n 1 s 2 < W < 2n n s 2 n }{{} from previous slide Taking the limit lim n we obtain π 2 W π 2. This proves Wallis formula. YAY! < π 2 ( ) n. n 1

106 Back to flipping coins and drunken walks: Recall that a n = Mn 2 2n s n = a a n 1 Since s n = 2n a n 2n 1 s 2 n < W < 2n s 2, n we have 2n 1 (2n a n ) 2 < W < 2n (2n a n ) 2.

107 From it follows that and 4n 2 π 2 2n 1 2n 2n 1 (2n a n ) 2 < W < 2n (2n a n ) 2 2n 1 4n 2 W < a2 n < 2n 1 ( n n 1 1 n π ) < a 2 n < 2n 4n 2 W, 4n 2 π 4 2n ( 2n 1 n+1 1 ( ) < a 2 n < 1 n n π n 1 So taking the limit as n we obtain 1 n π a2 n 1 n π. ). 2 ( 2n 1 n+1 ).

108 Sophie requests a cat nap

109 Summary and The END? We used coin flipping, drunken walks, and rainbows to prove: lim n a n = 1 π n, and the fact that the Wallis product W equals π/2. That is, W = = π 2. For references and more fun homework problems (full of hints), see my write-up called On a Coin Flip Problem available at:

110 Outline 1 Why pi? 2 A brief history of π 3 Coin toss and drunken walk problem 4 A simple geometric proof 5 Poop

111 Why do you smell poop in this talk?

112 Why do you smell poop in this talk? The speaker NEVER mentioned the gamma function!!!

113 Why do you smell poop in this talk? The speaker NEVER mentioned the gamma function!!! Let s rectify this situation

114 What is the gamma function? Recall from Emily Gullerud and michael vaughn s talk, they defined the gamma function Γ(x) as the analytic continuation of the factorial function as follows: Γ(x) = 0 e t t x 1 dt.

115 What is the gamma function? Recall from Emily Gullerud and michael vaughn s talk, they defined the gamma function Γ(x) as the analytic continuation of the factorial function as follows: Γ(x) = Integrating by parts yields the relation 0 e t t x 1 dt. Γ(x + 1) = x Γ(x).

116 What is the gamma function? Recall from Emily Gullerud and michael vaughn s talk, they defined the gamma function Γ(x) as the analytic continuation of the factorial function as follows: Γ(x) = Integrating by parts yields the relation 0 e t t x 1 dt. Γ(x + 1) = x Γ(x). And from the fact that Γ(1) = 1, we see that for n N we have Γ(n + 1) = n! thus generalizing the factorial function.

117 Theorem Γ ( 1 2) = π.

118 Theorem Γ ( 1 2) = π. Proof. By the definition Γ(x) = 0 e t t x 1 dt, we need to compute ( ) 1 Γ = e t t dt. 2 0

119 Theorem Γ ( 1 2) = π. Proof. By the definition Γ(x) = 0 e t t x 1 dt, we need to compute ( ) 1 Γ = e t t dt. 2 Let u = t and so dt = 2u du = 2 t du, yielding ( ) 1 Γ = 2 e u2 du = e u2 du

120 Theorem Γ ( 1 2) = π. Proof. By the definition Γ(x) = 0 e t t x 1 dt, we need to compute ( ) 1 Γ = e t t dt. 2 Let u = t and so dt = 2u du = 2 t du, yielding ( ) 1 Γ = 2 e u2 du = e u2 du This integral is well-known to equal π. The result follows.

121 A proof of Wallis product formula using Γ ( ) 1 2 Theorem (Euler s Reflection Formula) For z C\Z, the following holds: Γ(z) Γ(1 z) = π sin πz.

122 A proof of Wallis product formula using Γ ( ) 1 2 Theorem (Euler s Reflection Formula) For z C\Z, the following holds: Γ(z) Γ(1 z) = π sin πz. An as a corollary of the above, the following holds: N z N! (i) Γ(z) = lim N z(z + 1) (z + N). (ii) ( Γ( 1 2 )) 2 = 2 lim N ( (2N)(2N)(2N + 2) (2N + 1)(2N + 1) ).

123 A proof of Wallis product formula using Γ ( ) 1 2 Theorem (Euler s Reflection Formula) For z C\Z, the following holds: Γ(z) Γ(1 z) = π sin πz. An as a corollary of the above, the following holds: N z N! (i) Γ(z) = lim N z(z + 1) (z + N). (ii) ( Γ( 1 2 )) 2 = 2 lim N ( (2N)(2N)(2N + 2) (2N + 1)(2N + 1) Putting z = 1 2 in (i) and in Euler s reflection formula and noting Γ ( 1 2) = π, we easily deduce Wallis product formula! ).

124 Last slide (fo real) We find the value of 1 2!

125 Last slide (fo real) We find the value of 1 2! If you suspend belief that Γ(n + 1) = n! can work for rationals,

126 Last slide (fo real) We find the value of 1 2! If you suspend belief that Γ(n + 1) = n! can work for rationals, then Γ ( ) 3 2 equals 1 2!.

127 Last slide (fo real) We find the value of 1 2! If you suspend belief that Γ(n + 1) = n! can work for rationals, then Γ ( ) 3 2 equals 1 2!. But Γ(x + 1) = x Γ(x). So what can we conclude?

128 Last slide (fo real) We find the value of 1 2! If you suspend belief that Γ(n + 1) = n! can work for rationals, then Γ ( ) 3 2 equals 1 2!. But Γ(x + 1) = x Γ(x). So what can we conclude? 1 2! = π 2