The Wallis Product, a Connection to Pi and Probability, and ......Why pi?A brief history of ˇ Coin...
Transcript of 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
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 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?
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
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 +. . .
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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+ · · ·
)
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 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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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
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
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 . . ..
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 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.
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, 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.
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 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.
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 acoin 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 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!
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 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?
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 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????
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
Goal:
As n gets large, the probabilityof getting exactly n heads andn tails out of 2n coin flipsapproaches 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 walkmodel:
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 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.
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 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.
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 tothe 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 neverreturn to the original horizontal starting level. It is no coincidencethat 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 andthe 6 below to prove Mn = Nn when n = 2, and then generalizeto 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
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.
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 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.
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
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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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.
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,
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.
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 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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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?
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 moreinsight into the simplicity of this geometric proof. Plus, it’s prettierto 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 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.
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 thati+ 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 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
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, 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!”
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
We have the relation n− 1 < 14πs
2n < 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
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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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.
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
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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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!
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
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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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.
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:
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 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 NEVERmentioned 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 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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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.
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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!
Why pi? A brief history of π Coin toss and drunken walk problem A simple geometric proof Poop
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