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

Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Aba Mbirika Assistant Professor of Mathematics

May 8th, 2017 A Math Presentation for the Riemann Seminar Class Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 3 1 function and more precisely the value Γ 2 being 2 factorial? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Introducing my co-speaker, Sophie Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Venn diagram of the Real Numbers Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Sophie suggests Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop The 16th and 17th Centuries

Fran¸coisVi`eta(1579, France) √ p √ q p √ 2 2 2 + 2 2 + 2 + 2 = · · ··· π 2 2 2

John Wallis (1650, England) ∞ π 2 · 2 4 · 4 6 · 6 8 · 8 Y 2n · 2n = · · · ··· = 2 1 · 3 3 · 5 5 · 7 7 · 9 (2n − 1)(2n + 1) n=1 Equation above comes up BIG TIME later in the talk! Lord Brouncker (1650, England) 4 12 = 1 + π 32 2 + 52 2 + . 2 + .. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

James Gregory (1668, Scotland) π 1 1 1 = 1 − + − ··· 4 3 5 7 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 r1 1 1 1 1 = 1 − + − + + ··· 6 3 3 × 3 32 × 5 33 × 7 34 × 9 Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 ≈ 63.66% π

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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

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: eiπ + 1 = 0 Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop How cool do these folks think this equation is? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop How cool do these folks think this equation is? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop How cool do these folks think this equation is? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Can this guy wash his shirt anymore? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop The 19th Century to the present

William Rutherford (1841, England)

π 1 1 1 = 4 arctan − arctan + arctan 4 5 70 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 2 192 4 9 + 22 =3 .1415926525826461252 ... 355 .0003 113 1 − 3533 =3 .1415926535897943 .... Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Sophie wonders Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Recall from a previous slide, John Wallis (1655) arrived at his celebrated formula ∞ π 2 · 2 4 · 4 6 · 6 8 · 8 Y 2n · 2n = · · · ··· = 2 1 · 3 3 · 5 5 · 7 7 · 9 (2n − 1)(2n + 1) n=1

Many textbook proofs of this formula rely the family {In} of deﬁnite integral Z π 2 n In := (sin x) dx 0 by repeated partial integration. (Hint: Let u = sinn−1 x and n−1 dv = sin x dx. And derive the recursion formula In = n In−2 π with initial values I0 = 2 and I1 = 1.)

This is cool problem for Calculus II. I highly suggest you write this down and give it a go later if you want. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 3 the gamma function or more precisely the value of Γ 2 1 which is 2 factorial. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

The sequences for n = 2 (i.e., ﬂipping a coin 2 × 2 = 4 times):

HHHH THHH HHHT THHT HHTH THTH HHTT THTT HTHH TTHH HTHT TTHT HTTH TTTH HTTT 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! Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 22n 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 22n possible sequences, how many sequences contain exactly n H’s and n T ’s? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

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 10! = = 252 5 5! 5!

ways. Now since there are exactly 210 = 1024 possible sequences of H’s and T ’s, the probability of getting exactly ﬁve H’s and ﬁve 252 T ’s is 1024 = .24609375.

This number is close to √1 which is approximately .252313. 5π For n = 20, the probability is .125371 while √ 1 is approximately 20π .126157. QUESTION: What is your conjecture???? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Drunken Walks

The coin ﬂip problem can be recast in the following drunken walk model: Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Deﬁnition

Let Mn be the number of cases of 2n-walks that return to the original horizontal starting level. And let Nn be the number of cases of 2n-walks that never revisit the original horizontal starting level after initially leaving it. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Drunken Walks

Consider the number of drunken walks of four steps that return to the original horizontal starting level. There are six possibilities.

Figure: M2 walks Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

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: N2 walks Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Challenge: Prove Mn = Nn.

Figure: M2 walks

Figure: N2 walks

Proof Strategy: Describe a bijection between the 6 above and the 6 below to prove Mn = Nn when n = 2, and then generalize to all n values. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Two facts:

The probability of getting exactly n heads and n tails in 2n M coin ﬂips is n . 22n Furthermore, we can prove the following remarkable identity:

2n 2 = MnN0 + Mn−1N1 + ··· + M1Nn−1 + M0Nn.

The next slide (and some added comments from me) hopefully provides a picture of why this second fact makes sense. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Remark Consider an arbitrary walk. In any such walk there will be a portion of the walk of the form Mi for some i ≤ n, followed by a walk that is of the form Nj for j = n − i.

Figure: Two of the 12 possible M2N1 walks

Since M2 = 6 and N1 = 2, there are exactly 12 drunken walks of the form M2N1. We give two such walks in the ﬁgure. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

We list the ﬁrst few Mn and Nn values in the table below.

n Mn Nn 0 1 1 1 2 2 2 6 6 3 20 20 4 70 70 5 252 252 Let’s verify that our favorite identity from two slides back

2n 2 = MnN0 + Mn−1N1 + ··· + M1Nn−1 + M0Nn

holds for small n values, like n = 3 for instance.

6 2 = M3N0 + M2N1 + M1N2 + M0N3 = 20 · 1 + 6 · 2 + 2 · 6 + 1 · 20. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Deﬁnition

Let an denote the quotient

M 2n a = n = n . n 22n 22n

Define sn to be the sum a0 + a1 + ··· + an−1 and s0 = 0. The following table gives the first five an and sn values.

n an sn 0 1 0 1 1 2 1 3 3 2 8 2 5 15 3 16 8 35 35 4 128 16 For example, for n = 3 we have 1 3 15 s = a + a + a = 1 + + = . 3 0 1 2 2 8 8 Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop More “homework” for you? Yay!

ELFS: Use the following three equalities:

Mn = Nn, 2n 2 = MnN0 + Mn−1N1 + ··· + M1Nn−1 + M0Nn, and M a = n n 22n to show the remarkable (and useful, you’ll soon see) identity holds

1 = ana0 + an−1a1 + ··· + a1an−1 + a0an. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Sophie’s Choice Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Oh yeah, this talk is about π.

As alluded to earlier, the probability of getting exactly n heads and Mn n tails in 2n coin ﬂips involves π. We will show the probability 22n √1 (that is, an) asymptotically approaches the value nπ : 1 lim an = √ 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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

√1 The heart of the proof that an −→ nπ is a clever geometric insight by Johan W¨astlundusing a dreadfully dull looking picture in the black and white journal Amer. Math. Monthly, Dec. 2007.

Question: Why do I call that picture dull? Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Recall: We used drunken walks to prove that

1 = ana0 + an−1a1 + ··· + a1an−1 + a0an.

By the term “we”, I meant you. Recall ELFS (a few slides ago). Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Since we showed the products of the form ana0 + an−1a1 + ··· + a1an−1 + a0an each equal one, we have:

Red rectangle area = a0a0 = 1

Orange rectangle areas = a1a0 + a0a1 = 1

Yellow rectangle areas = a2a0 + a1a1 + a0a2 = 1

Green rectangle areas = a3a0 + a2a1 + a1a2 + a0a3 = 1

Blue rectangle areas = a4a0 + a3a1 + a2a2 + a1a3 + a0a4 = 1. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Let Pn denote the region of rectangles labeled aiaj such that i + j < n. Then,

P1 has area 1

P2 has area 2 . . . .

Pn has area n Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Summary

As we add more rectangles, the outermost rectangles appear more and more to sit exactly on the circumference of the quarter circle. 1 2 That is, the value 4 πsn is approximately n as n gets large. This follows since 1 1 n ≈ π(s )2 = π(2na )2 = πn2a2 . 4 n 4 n n

2 1 √1 Hence an ≈ nπ and thus an approaches 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: http://people.uwec.edu/mbirika/CoinFlipProblem.pdf Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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 s5 = a0 + a1 + ··· + a4. 1 2 Hence the area 4 π(s5) is approximately equal to 5. 315 1 2 Precisely, since s5 = 128 the value 4 π(s5) equals 4.75654.

I’m going to click back and forth on these two slides until EVERYONE in the room feels “Yah, I get it!” Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

1 2 We have the relation n − 1 < 4 πsn < n + 1. The following ﬁgure motivates our reasoning. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Theorem 1 2 The following relation n − 1 < 4 πsn < n + 1 holds.

Proof by picture: The region P3 (red, brown, yellow) has 1 area 3. The third 4 -circle contains P3 in its interior and has radius s4 and thus an area 1 2 1 2 4 π(s4) . So 3 < 4 π(s4) . The region P5 (all 5 colors) has area 5. This region con- 1 tains the third 4 -circle. We 1 2 conclude 3 < 4 π(s4) < 5. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Challenge: Prove these facts. Fact 1 3 5 7 2n−1 For n ≥ 1, each an can be written as the product 2 4 6 8 ··· 2n .

1 3 2n−1 Sketch: For the an-case rewrite 2 · 4 ··· 2n as M 1 2 · 3 4 ··· 2n−1 · 2n and show this equals n . 2 2 4 4 2n 2n 22n Fact 3 5 7 9 2n−1 For n > 1, each sn can be written as the product 2 4 6 8 ··· 2n−2 .

Putting these two facts together, we see sn = 2n an.

We will use the two facts above to prove the Wallis π product equals as desired. 2 Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

We denote the Wallis product (remember this from long ago?) by ∞ 2 · 2 4 · 4 6 · 6 8 · 8 Y 2n · 2n W = · · · ··· = 1 · 3 3 · 5 5 · 7 7 · 9 (2n − 1)(2n + 1) n=1

Let s0 = 0, s1 = 1, and 3 5 2n − 1 s = · ··· . n 2 4 2n − 2 Then we have the following equalities: (Time-permitting, I can do n = 3 and n = 4 examples to illustrate.) 2n − 1 22 · 42 ··· (2n − 2)2 2 = 2 2 (n is even) sn 1 · 3 ··· (2n − 3) · (2n − 1) 2n 22 · 42 ··· (2n)2 2 = 2 2 2 (n is odd) sn 1 · 3 ··· (2n − 3) · (2n − 1) and the relations 2n − 1 2n 2 < W < 2 . sn sn Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

1 2 Since n − 1 < 4 πsn < n + 1 (from 3 slides ago), it follows that 4(n − 1) 4(n + 1) < s2 < , π n π and thus π 2n − 1 2n − 1 2n π n < 2 < W < 2 < . 4 n + 1 sn sn 2 n − 1 | {z } from previous slide

Taking the limit limn→∞ we obtain π π ≤ W ≤ . 2 2 This proves Wallis’ formula. YAY! Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Back to ﬂipping coins and drunken walks:

Recall that

Mn an = 22n sn = a0 + ··· + an−1

sn = 2n an Since 2n − 1 2n 2 < W < 2 , sn sn we have 2n − 1 2n 2 < W < 2 . (2n · an) (2n · an) Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

From 2n − 1 2n 2 < W < 2 (2n · an) (2n · an) it follows that 2n − 1 2n < a2 < , 4n2 · W n 4n2 · W and

2n − 1 2 2n < an < . 2 π n 2 π 2n−1 4n · 2 n−1 4n · 4 n+1

2n − 1 1 1 2 1 2 · · < an < · . 2n n · π n n · π 2n−1 n−1 n+1 So taking the limit as n → ∞ we obtain 1 1 ≤ a2 ≤ . n · π n n · π Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Sophie requests a cat nap Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Summary and The END?

We used coin ﬂipping, drunken walks, and rainbows to prove: √1 limn→∞ an = π·n , and the fact that the Wallis product W equals π/2. That is, 2 2 4 4 6 6 π W = · · · · · ··· = . 1 3 3 5 5 7 2

For references and more fun “homework” problems (full of hints), see my write-up called On a Coin Flip Problem available at: http://people.uwec.edu/mbirika/CoinFlipProblem.pdf. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Why do you smell poop in this talk?

The speaker NEVER mentioned the gamma function!!!

Let’s rectify this situation Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 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: Z ∞ Γ(x) = e−t tx−1 dt. 0

Integrating by parts yields the relation

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. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop

Theorem 1 √ Γ 2 = π.

Proof. R ∞ −t x−1 By the deﬁnition Γ(x) = 0 e t dt, we need to compute Z ∞ 1 −t 1 −1 Γ = e t 2 dt. 2 0 √ √ Let u = t and so dt = 2u du = 2 t du, yielding

∞ ∞ 1 Z 2 Z 2 Γ = 2 e−u du = e−u du. 2 0 −∞ √ This integral is well-known to equal π. The result follows. Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop 1 A proof of Wallis’ product formula using Γ 2

Theorem (Euler’s Reﬂection Formula) For z ∈ C\Z, the following holds: π Γ(z) Γ(1 − z) = . 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) 2 · 2 · 4 · 4 · 6 · 6 ··· (2N)(2N)(2N + 2) (ii) Γ( 1 )2 = 2 lim . 2 N→∞ 1 · 3 · 3 · 5 · 5 ··· (2N + 1)(2N + 1)

1 Putting z = 2 in (i) and in Euler’s reﬂection formula and 1 √ noting Γ 2 = π, we easily deduce Wallis’ product formula! Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop Last slide (fo’ real)

1 We ﬁnd the value of 2!

If you suspend belief that Γ(n + 1) = n! can work for rationals,

3 1 then Γ 2 equals 2 !.

But Γ(x + 1) = x Γ(x). So what can we “conclude”? √ 1 π 2! = 2