The Wallis Product, a Connection to Pi and Probability, and ......Why pi?A brief history of ˇ Coin...

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

The Wallis Product, a Connection to Pi andProbability, and Maybe the Gamma Function

Perhaps?

Aba MbirikaAssistant Professor of Mathematics

May 8th, 2017A Math Presentation for the Riemann Seminar Class


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 Thoreautranscendentalists? And is π transcendental?

Does this talk have ANYTHING at all to do with the gammafunction and more precisely the value Γ

(32

)being 1

2 factorial?


Introducing my co-speaker, Sophie


Venn diagram of the Real Numbers


Sophie suggests


The 16th and 17th Centuries

Francois Vieta (1579, France)

2

π=

√2

2·√

2 +√

2

2·

√2 +

√2 +√

2

2· · ·

John Wallis (1650, England)

π

2=

2 · 21 · 3

· 4 · 43 · 5

· 6 · 65 · 7

· 8 · 87 · 9· · · =

∞∏n=1

2n · 2n(2n− 1)(2n+ 1)

Equation above comes up BIG TIME later in the talk!Lord Brouncker (1650, England)

4

π= 1 +

12

2 +32

2 +52

2 +. . .


James Gregory (1668, Scotland)

π

4= 1− 1

3+

1

5− 1

7· · ·

Does anyone see a Calc II way to prove this? Maybe use theTaylor series expansion for arctan(x) and a judicious choicefor x.

Abraham Sharp (1699, England)

π

6=

√1

3

(1− 1

3× 3+

1

32 × 5− 1

33 × 7+

1

34 × 9+ · · ·

)


The 18th Century

Georges-Louis Leclerc, Comte de Buffon (1760, France)A plane is ruled with parallel lines 1 inch apart. A needle oflength 1 inch is dropped randomly on the plane. What is theprobability 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 ofLindemann.


Leonhard Euler (1748, Switzerland) publishes Introductio inanalysin infinitorum. Math historians say that it was thispublication that catapulted the symbol π into popular use.QUESTION: What is the coolest of all of Euler’s formula?ANSWER: eiπ + 1 = 0


How cool do these folks think this equation is?


Can this guy wash his shirt anymore?


The 19th Century to the present

William Rutherford (1841, England)

π

4= 4 arctan

(1

5

)− arctan

(1

70

)+ arctan

(1

99

)He uses this to estimate π to 208 places (152 correct). Buteven 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 istranscendental.

Srinivasa Ramanujan (1913, India) gives many remarkableapproximations of π:(

92 + 192

22

) 14

= 3.1415926525826461252 . . .355113

(1− .0003

3533

)= 3.1415926535897943 . . ..


Sophie wonders


Recall from a previous slide, John Wallis (1655) arrived at hiscelebrated formula

π

2=

2 · 21 · 3

· 4 · 43 · 5

· 6 · 65 · 7

· 8 · 87 · 9· · · =

∞∏n=1

2n · 2n(2n− 1)(2n+ 1)

Many textbook proofs of this formula rely the family {In} ofdefinite integral

In :=

∫ π2

0(sinx)ndx

by repeated partial integration. (Hint: Let u = sinn−1 x anddv = sinx dx. And derive the recursion formula In = n−1

n In−2with initial values I0 = π

2 and I1 = 1.)

This is cool problem for Calculus II. I highly suggest youwrite this down and give it a go later if you want.


Overview for the rest of the talk

Introduce two equivalent combinatorial problems

Coin flips problem, andDrunken 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 perhapsthe gamma function or more precisely the value of Γ

(32

)which is 1

2 factorial.


Flipping Coins

Suppose we flip a fair coin 2n times. What are the chances thatwe get an equal amount of heads and tails?

Label the head and tail sides of the coin as H and Trespectively.

We can view the 2n coin flips as sequences of H’s and T ’s.


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 acoin 2 times.


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

HHHH THHHHHHT THHTHHTH THTHHHTT THTTHTHH TTHHHTHT TTHTHTTH TTTHHTTT TTTT

QUESTION: What is the probability that you get an equalnumber of heads and tails when you flip a coin 4 times?

ANSWER: Well this is silly question since I have highlightedcertain sequences above in bold red!


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 possiblesequences!

Suppose we flip a fair coin 2n times. What are the chances thatwe get an equal amount of heads and tails?

We can then rephrase the original question:

Out of the 22n possible sequences, how many sequencescontain exactly n H’s and n T ’s?


Example

Consider the case for n = 5. Then the number of ways of gettingexactly five H’s is the number of ways of choosing 5 of the 10sequence slots in which to place them. Hence there are(

10

5

)=

10!

5! 5!= 252

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

1024 = .24609375.

This number is close to 1√5π

which is approximately .252313.

For n = 20, the probability is .125371 while 1√20π

is approximately.126157.

QUESTION: What is your conjecture????


Goal:

As n gets large, the probabilityof getting exactly n heads andn tails out of 2n coin flipsapproaches the number 1√

nπ.


Drunken Walks

The coin flip problem can be recast in the following drunken walkmodel:


Drunken Walks

Here is the drunken walk model:

Consider a walk where we go forward each time, however ateach step forward we veer 45 degrees left or right.

Let H be denoted by a step forward that veers left and T tobe one that veers right.

If after 2n steps we return to the original horizontal startinglevel, then we must have taken exactly n left-steps and nright-steps.


Definition

Let Mn be the number of cases of 2n-walks that return to theoriginal horizontal starting level. And let Nn be the number ofcases of 2n-walks that never revisit the original horizontal startinglevel after initially leaving it.


Drunken Walks

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

Figure: M2 walks


Now consider the number of drunken walks of four steps that neverreturn to the original horizontal starting level. It is no coincidencethat this also yields six possibilities!

Figure: N2 walks


Challenge: Prove Mn = Nn.

Figure: M2 walks

Figure: N2 walks

Proof Strategy: Describe a bijection between the 6 above andthe 6 below to prove Mn = Nn when n = 2, and then generalizeto all n values.


Two facts:

The probability of getting exactly n heads and n tails in 2n

coin flips isMn

22n.

Furthermore, we can prove the following remarkable identity:

22n = MnN0 +Mn−1N1 + · · ·+M1Nn−1 +M0Nn.

The next slide (and some added comments from me) hopefullyprovides a picture of why this second fact makes sense.


Remark

Consider an arbitrary walk. In any such walk there will be a portionof the walk of the form Mi for some i ≤ n, followed by a walk thatis 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 ofthe form M2N1. We give two such walks in the figure.


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

22n = MnN0 +Mn−1N1 + · · ·+M1Nn−1 +M0Nn

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

26 = M3N0 +M2N1 +M1N2 +M0N3

= 20 · 1 + 6 · 2 + 2 · 6 + 1 · 20.


Definition

Let an denote the quotient

an =Mn

22n=

(2nn

)22n

.

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

n an sn

0 1 0

1 12 1

2 38

32

3 516

158

4 35128

3516

For example, for n = 3 we have

s3 = a0 + a1 + a2 = 1 +1

2+

3

8=

15

8.


More “homework” for you? Yay!

ELFS: Use the following three equalities:

Mn = Nn,

22n = MnN0 +Mn−1N1 + · · ·+M1Nn−1 +M0Nn, and

an =Mn

22n

to show the remarkable (and useful, you’ll soon see) identity holds

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


Sophie’s Choice


Oh yeah, this talk is about π.

As alluded to earlier, the probability of getting exactly n heads andn tails in 2n coin flips involves π. We will show the probability Mn

22n

(that is, an) asymptotically approaches the value 1√nπ

:

limn→∞

an =1√nπ

QUESTION: How hard will this proof be?

ANSWER: The proof uses nothing more than the Pythagoreantheorem, the area formula for a circle, and some elementaryalgebra.


The heart of the proof that an −→ 1√nπ

is a clever geometric

insight by Johan Wastlund using a dreadfully dull looking picturein the black and white journal Amer. Math. Monthly, Dec. 2007.

Question: Why do I call that picture dull?


ANSWER: With a little splash of color, it can provide moreinsight into the simplicity of this geometric proof. Plus, it’s prettierto look at.


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).


Since we showed the products of the formana0 + 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.


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

P1 has area 1

P2 has area 2

......

Pn has area n


Summary

As we add more rectangles, the outermost rectangles appear moreand more to sit exactly on the circumference of the quarter circle.That is, the value 1

4πs2n is approximately n as n gets large. This

follows since

n ≈ 1

4π(sn)2 =

1

4π(2nan)2 = πn2a2n.

Hence a2n ≈ 1nπ and thus an approaches 1√

πnas n gets large.

MY PLEA TO THE AUDIENCE: The rest of the slides prettymuch spell out the fun “work” most of which I leave as exercises(full of hints) in my write-up called On a Coin Flip Problemavailable at: http://people.uwec.edu/mbirika/CoinFlipProblem.pdf


Example

HEY!! Before we read the words and mathy stuff below, letsremind ourself of the rainbow (next slide). Then we’ll come backto this slide.

Observations about the Rainbow Slide

The rectangles arranged in this fashion begin to resemble aquarter circle with radius s5 = a0 + a1 + · · ·+ a4.

Hence the area 14π(s5)

2 is approximately equal to 5.

Precisely, since s5 = 315128 the value 1

4π(s5)2 equals 4.75654.

I’m going to click back and forth on these two slides untilEVERYONE in the room feels “Yah, I get it!”


We have the relation n− 1 < 14πs

2n < n+ 1. The following figure

motivates our reasoning.


Theorem

The following relation n− 1 < 14πs

2n < n+ 1 holds.

Proof by picture: The regionP3 (red, brown, yellow) hasarea 3. The third 1

4 -circlecontains P3 in its interior andhas radius s4 and thus an area14π(s4)

2. So 3 < 14π(s4)

2.The region P5 (all 5 colors)has area 5. This region con-tains the third 1

4 -circle. Weconclude 3 < 1

4π(s4)2 < 5.


Challenge: Prove these facts.

Fact

For n ≥ 1, each an can be written as the product 12345678 · · ·

2n−12n .

Sketch: For the an-case rewrite 12 ·

34 · · ·

2n−12n as

1222 ·

3444 · · ·

2n−12n ·

2n2n and show this equals

Mn

22n.

Fact

For n > 1, each sn can be written as the product 32547698 · · ·

2n−12n−2 .

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

We will use the two facts above to prove the Wallis

product equalsπ

2as desired.


We denote the Wallis product (remember this from long ago?) by

W =2 · 21 · 3

· 4 · 43 · 5

· 6 · 65 · 7

· 8 · 87 · 9· · · =

∞∏n=1

2n · 2n(2n− 1)(2n+ 1)

Let s0 = 0, s1 = 1, and

sn =3

2· 5

4· · · 2n− 1

2n− 2.

Then we have the following equalities: (Time-permitting, I can don = 3 and n = 4 examples to illustrate.)

2n− 1

s2n=

22 · 42 · · · (2n− 2)2

1 · 32 · · · (2n− 3)2 · (2n− 1)(n is even)

2n

s2n=

22 · 42 · · · (2n)2

1 · 32 · · · (2n− 3)2 · (2n− 1)2(n is odd)

and the relations2n− 1

s2n< W <

2n

s2n.


Since n− 1 < 14πs

2n < n+ 1 (from 3 slides ago), it follows that

4(n− 1)

π< s2n <

4(n+ 1)

π,

and thus

π

4

(2n− 1

n+ 1

)<

2n− 1

s2n< W <

2n

s2n︸︷︷︸from previous slide

<π

2

(n

n− 1

).

Taking the limit limn→∞ we obtain

π

2≤W ≤ π

2.

This proves Wallis’ formula. YAY!


Back to flipping coins and drunken walks:

Recall that

an = Mn22n

sn = a0 + · · ·+ an−1

sn = 2nan

Since2n− 1

s2n< W <

2n

s2n,

we have2n− 1

(2n · an)2< W <

2n

(2n · an)2.


From2n− 1

(2n · an)2< W <

2n

(2n · an)2

it follows that2n− 1

4n2 ·W< a2n <

2n

4n2 ·W,

and

2n− 1

4n2 · π2(

nn−1

) < a2n <2n

4n2 · π4(2n−1n+1

) .2n− 1

2n· 1

n · π· 1(

nn−1

) < a2n <1

n · π· 2(

2n−1n+1

) .So taking the limit as n→∞ we obtain

1

n · π≤ a2n ≤

1

n · π.


Sophie requests a cat nap


Summary and The END?

We used coin flipping, drunken walks, and rainbows to prove:

limn→∞ an = 1√π·n , and

the fact that the Wallis product W equals π/2. That is,

W =2

1· 2

3· 4

3· 4

5· 6

5· 6

7· · · = π

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 do you smell poop in this talk?

The speaker NEVERmentioned the gamma

function!!!

Let’s rectify this situation


What is the gamma function?

Recall from Emily Gullerud and michael vaughn’s talk, they defined thegamma function Γ(x) as the analytic continuation of the factorialfunction as follows:

Γ(x) =

∫ ∞0

e−t tx−1 dt.

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.


Theorem

Γ(12

)=√π.

Proof.

By the definition Γ(x) =∫∞0 e−t tx−1 dt, we need to compute

Γ

(1

2

)=

∫ ∞0

e−t t12−1 dt.

Let u =√t and so dt = 2udu = 2

√t du, yielding

Γ

(1

2

)= 2

∫ ∞0

e−u2

du =

∫ ∞−∞

e−u2

du.

This integral is well-known to equal√π. The result follows.


A proof of Wallis’ product formula using Γ(

12

)Theorem (Euler’s Reflection Formula)

For z ∈ C\Z, the following holds:

Γ(z) Γ(1− z) =π

sinπz.

An as a corollary of the above, the following holds:

(i) Γ(z) = limN→∞

N z ·N !

z(z + 1) · · · (z +N).

(ii)(Γ(12)

)2= 2 lim

N→∞

(2 · 2 · 4 · 4 · 6 · 6 · · · (2N)(2N)(2N + 2)

1 · 3 · 3 · 5 · 5 · · · (2N + 1)(2N + 1)

).

Putting z = 12 in (i) and in Euler’s reflection formula and

noting Γ(12

)=√π, we easily deduce Wallis’ product formula!


Last slide (fo’ real)

We find the value of 12!

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

then Γ(32

)equals 1

2 !.

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

12! =

√π2

The Wallis Product, a Connection to Pi and Probability, and ......Why pi?A brief history of ˇ Coin...

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