The number π

Rajendra Bhatia

Indian Statistical Institute, DelhiIndia

December 20, 2012

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CORE MATHEMATICS

Nature of Mathematics

Reasoning and Proofs in Mathematics

History of Mathematics

Contributions of Indian Mathematicians.

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Circle of diameter D

Circumference C

Area A

Easy arguments show

C ∝ D C = πD

A ∝ D2 A = kD2

How do you know π and k are related?

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Shown by Archimedes ( ∼ 287-212 BCE)

Measurement of a circle ∼ 225 BCE

Theorem Let A be the area of a circle with circumference Cand radius r . Then

A = T

where T is the area of a right angled triangle with base C andheight r .

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Circumference Carea A area T

ThenA = T

(Reducing the more complicated problem to a simpler, morefamiliar, one.)

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Key ideas of proof

1. Method of exhaustion:

In any circle we can inscribe a regular polygon whosearea/circumference are arbitrarily close to those of thecircle.

Can also circumscribe a regular polygon with similarapproximation properties.

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Proof by contradiction (Reductio ad absurdum)

2. We want to prove two numbers a and b are equal.

Assume a > b and show it leads to a contradiction

Assume a < b and show it leads to a contradiction

So, we must have a = b.

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Step 1

Regular polygon with n sidesb = length of each sideA = area Q = perimeter

h = “apothem”.

Then A = 12bh.n, Q = bn

So A = 12hQ

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Step 2

We want to prove for the circle

A = T =12

r C (Statement of Archimedes’ Theorem)

Suppose A > T

Then p := A − T > 0 (1)

Inscribe a regular polygon of area A′ inside the circle, such that

A − A′ < p = A − T

Then − A′ < −T , i.e. T < A′. (2)

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By Step 1, A′ = 12hQ

where h is the apothem and Q the perimeter of the polygon.

But h < r and Q < C.

So A′ <12

r C =: T (3)

But we had T < A′ (See (2)).Contradiction. So the assumption A > T can’t be true.

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Step 3

Suppose A < T .

Then p := T − A > 0. (4)

Now choose a polygon, circumscribing the circle, having areaA′ such that

A′ − A < p = T − A

A′ < T . (5)

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By Step 1, A′ = 12hQ.

Now h = r , and Q > C.

So A′ >12

r C =: T (6)

Again (5) and (6) contradict each other.

So A < T can’t be true.

Conclusion: A = T .

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So

A = T =12

rC =12

r .2πr

i.e.A = πr2

where π is the constant in the formula

C = 2πr

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Archimedes estimated

31071

< π < 31070

i.e.3.140845 · · · < π < 3.142857 · · ·

which is accurate to two decimal places.

We now know

π = 3.1415926535879 . . . .

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Some other great formulas of Archimedes

Volume of sphere = 43πr3

Surface area of sphere = 4πr2

A cylinder circumscribed about a sphere hasVolume = 2πr3

Surface area = 6πr2

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Both the volume and the surface area of the cylinder are in ratio32 with the volume and surface area of the sphere.

Archimedes was so proud of this discovery that he requestedthat this should be put on his tomb.

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Archimedes as an experimenter

1 Eureka story

2 Levers, pulleys, catapults, lenses, mirrors, war machines

3 Archimedean screw

4 Measurements of the diameter of the earth by his pupils.

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Simple Ideas

1 The sun crosses all points on a meridian line (longitude) atthe same time.

2 At noon the sun is directly overhead at the equator. So astick will have no shadow. The further away you are fromthe equator the longer will be the shadow.

3 The measurements of these shadows can be used to inferthe angular separation between two points on the earth.From this the perimeter of the earth can be calculated.

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A,B two points on the same longitude.

length of sticks AP = ℓA, BQ = ℓB

length of shadows AL = sA, BM = sBsALA

= tan θA,sBLB

= tan θB.

If d(A,B) = distance between A and B, then

Perimeter of the earth =2π

θB − θAd(A,B)

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Ancient mathematicians thought of numbers as “lengths”.

The Greek mathematicians proved that√

2 is not a rationalnumber.

A number r is rational if it can be expressed as

r =pq

where p,q are integers.

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The proof that√

2 is irrational is easy, and beautiful.

(Exercise: Use the idea of the proof to investigate whether 3√

2is rational; for which n is

√n rational.)

π is irrational (proved by J. Lambert in 1761)

(Proof not as easy as for√

2)

π2 is irrational (A. Legendre, 1794)

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The number e is also irrational.

Surprisingly hard problems:

Is π + e rational?

Is πe rational?

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If a number x is a solution of a (nonconstant) polynomialequation

xn + a1xn−1 + · · ·+ an = 0

with rational coefficients a1, . . . ,an, then we say

x is an algebraic number.

All rational numbers are algebraic.

Some irrational numbers are algebraic, e.g.√

2 is a solution ofthe equation

x2 − 2 = 0.

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π is not an algebraic number.

It is a “transcendental number.”

(Lindemann, 1882. Another proof by D. Hilbert)

Consequences:π cannot be expressed as a combination of rational numbersand roots

(e.g. it is not anything like 3√

97 + 2√

6.)

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It is impossible to “square the circle”.

i.e. it is impossible to construct, using a compass and ruleralone, a square whose area is equal to the area of a givencircle.

(Theorem in algebra: All “constructible numbers” are algebraic)

See: Courant and Robbins, What is Mathematics?

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π and infinite series

Does π have anything to do with natural numbers

N = {1,2,3, . . .} .

Surprisingly: a lot.

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Infinite series

a1 + a2 + a3 + · · ·+ an + · · · .

Obviously1 + 1 + 1 + · · ·+ 1 + · · · .

“diverges”. The series

12+

14+ · · ·+ 1

2n + · · · .

“converges”. The sum is 1.Less obvious

1 +12+

13+ · · · + 1

n+ · · · .

diverges.

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Madhava (∼ 1350-1425)

Nilakantha (∼ 1500) Tantrasangraha

1 − 13+

15− 1

7+ · · ·+ (−1)n

2n + 1+ · · · = π

4.

Madhava SeriesGregory-Leibnitz Series (∼ 1670)

arctan z = z − z3

3+

z5

5− z7

7+ · · ·

Choose z = 1.

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Translation:

The diameter multiplied by four and divided by unity [is foundand saved]. Again the products of the diameter and four aredivided by the odd numbers like three, five, etc. and the resultsare subtracted and added in order [to the earlier result saved].Take half of the succeeding even number as the multiplier atwhichever [odd] number the division process is stopped,because of boredom. The square of that [even number] addedto unity is the divisor. Their ratio has to be multiplied by theproduct of the diameter and four as earlier. The result obtainedhas to be added if the earlier term [in the series] has beensubtracted and subtracted if the earlier term has been added.

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The resulting circumference is very accurate; in fact moreaccurate than the one which may be obtained by continuing thedivision process [with large number of terms in the series].

C = 4d{

1 − 13+ · · ·+ (−1)

p−12

1p+ (−1)

p−12

p + 12

/((p + 1)2 + 1)}

,

where p runs over odd positive integers.

Here not only is the series given but also the truncation error.This was then used to get series that converge more rapidly.For example,

π

4=

34+

133 − 3

− 153 − 5

+1

73 − 7− · · ·

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The first infinite product (F. Viète, 1593)

2π=

√2

2.

√

2 +√

22

.2 +

√2

2· · · .

Another famous formula (J. Wallis, 1655)

π

2=

21

23

43

45· · · .

Intrigued?

see R. Bhatia, Fourier Series

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A calculation that made L. Euler famous (1735)

∞∑

n=1

1n2 =

π2

6.

This was called the Basel Problem, posed by J. Bernoulli.

This is one of the most famous formulas in mathematics.

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Euler’s first “proof”Relation between roots and coefficients of a polynomial:If α1, α2, . . . , αm are the roots of

a0 + a1x + a2x2 + · · · + amxm = 0,

then∑ 1

αi=

−a1

a0.

The function cos√

x has Taylor expansion

cos√

x = 1 − x2!

+x2

4!− · · · .

It has zeros at (2n+1)2π2

4 , n = 0,1,2, . . . .

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Leap of faith: Think of the infinite series as a “polynomial ofinfinite degree” and assume the same relation holds betweenroots and coefficients. This gives

∞∑

n=0

1(2n + 1)2 =

π2

8.

From this it is easy to see that

∞∑

n=1

1n2 =

π2

6.

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Euler was aware that there is a gap in this argument and foundother proofs. Discovery of infinite product expansions:

sin xx

=

∞∏

n=1

(

1 − x2

n2π2

)

.

Several proofs are known now.

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π can appear at unexpected places.

An example:

Z = {. . . ,−2,−1,0,1,2, . . .} integers

Z2 = {(m,n) : m,n ∈ Z} “lattice"

all points in the plane with integral coordinates.

Question: How many points of the lattice are visible from theorigin?

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Observe: The point (m,n) is visible if and only if m and n arecoprime.

So our question is:Among all points of Z2 what is the proportion of (m,n) that havem,n coprime.

Many questions of this kind are studied in number theory: e.g.how many numbers are “square-free”?

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Letn = pm1

1 pm22 · · · pmk

k (prime factoring)

If m1 = m2 = · · · = mk = 1, we say n is square-free.

e.g.70 = 2 × 5 × 7 is square-free.

90 = 2 × 32 × 5 is not square-free.

Q : What is the proportion of square-free numbers

i.e. If we pick up a number “at random” what is the probability ofits being square-free.

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Probability that a number is odd/even = 1/2.

Probability that a number is multiple of k = 1/k .

Probability that a number is not a multiple of k = 1 − 1k .

So, if p1,p2, . . . is the sequence of primes, then the probability

that a random number is not a multiple of p2j is

(

1 − 1p2

j

)

.

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So the probability that none of the p2j is a factor is

∏

j

(

1 − 1p2

j

)

.

(The probability of two independent events happeningsimultaneously is the product of their individual probabilities).

How to find the infinite product given above?

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Riemann Zeta Function

ζ(s) =∞∑

n=1

1ns (s > 1).

This function is of great importance in the study of primenumbers.

Theorem (Euler)

ζ(s) =∞∏

n=1

1

1 − 1ps

n

where p1,p2, . . . is the sequence of primes.

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A quick idea of the proof.

Recall the geometric series, for 0 < x < 1

11 − x

= 1 + x + x2 + · · · =∞∑

m=0

xm.

So

11 − p−s

1

= 1 + p−s1 + p−2s

1 + p−3s1 + · · ·

= 1 +∑

n:n=pm1

n−s

(sum over all n such that n is a power of the prime p1)

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(

11 − p−s

1

)(

11 − p−s

2

)

=(

1 + p−s1 + p−2s

1 + · · ·)(

1 + p−s2 + p−2s

2 + · · ·)

= 1 +∑

n:n=p1m1 p2

m2

n−s

(sum over all n which have only p1,p2 as prime factors).Similarly

r∏

k=1

11 − p−s

k

= 1 +∑

n:n=p1m1 ···pr mr

n−s (7)

(sum over all n which have only p1,p2, . . . ,pr as prime factors)

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As r → ∞, the sequence p1,p2, . . . runs over all primes and then which have p1,p2, . . . as prime factors runs over all numbersn ≥ 2. So we get from (7)

∞∏

k=1

1

1 − 1ps

k

=

∞∑

n=1

1ns = ζ(s).

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So: the probability of a random positive integer beingsquare-free is

∞∏

j=1

(

1 − 1p2

j

)

=1

ζ(2)=

6π2 ≈ 0.6079 . . .

Exercise: Show that the probability of two natural numbersm,n, picked at random, being coprime is also 6

π2 .

So the proportion of lattice points visible from the origin is 6π2 .

See Hardy & Wright, An Introduction to the Theory of Numbers.

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Elements of beauty in mathematics:

Surprise

Elegance

Depth

Brevity

Power

Connections between different things...

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I had started reading mathematics when I was about thirteen orso. I had accidentally opened a book and come across

Leibnitz’s series for π/4 = 1 − 1/3 + 1/5 − 1/7 + · · · involvingreciprocals of the odd integers with alternating signs. Till then,

school mathematics had always bored me but this seemedsuch a very strange and beautiful relationship that I determined

I would read that book in order to find out how this formulacame about.

-Atle Selberg

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Some amazing facts.

ζ(4) =π4

90, ζ(6) =

π6

945, . . .

Euler found many such sums. A good way now is via Fourierseries. Generally:

ζ(2k) = π2k · (a rational number).

Very little is known about the values ζ(2k + 1).

R. Apery (1978) showed ζ(3) is an irrational number.

Proved in 2000: atleast one of ζ(5), ζ(7), . . . , ζ(21) is irrational.

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π occurs often in statistics and probability.

One obvious reason is that the area under the bell-shapedcurve

is∫

∞

−∞

e−x2dx =

√π.

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A surprising occurrence is in the famous Buffon Needle

ProblemOn a plane ruled by parallel lines distance d apart, a needle oflength ℓ ≤ d is thrown at random. What is the probability thatthe needle intersects one of the lines?

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Note that the probability p satisfies

p ∝ ℓ and p ∝ 1d.

So

p = cℓ

d.

What is the constant c?

Answer

p = 2π

ℓd

Geometric probability, Integral geometry.

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Proof 1.

O mid point of needle

x = OP = distance of O to the nearest line

ϕ acute angle between OP and needle

restrictions 0 ≤ x ≤ d/2, 0 ≤ ϕ ≤ π/2

Condition forneedle tointersect aline:

x <ℓ

2cos ϕ

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Assumption: x , ϕ uniformly distributed, independent randomvariables, 0 ≤ x ≤ d

2 , 0 ≤ ϕ ≤ π2

The probability of the event x < ℓ2 cosϕ is the ratio

area of the red regionarea of the blue rectangle

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This is equal to

∫ π/20

ℓ2 cos ϕ dϕ

π2

d2

=ℓ/2πd/4

=2π

ℓ

d.

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Proof 2

Assume, for convenience, d = 1.

Instead of probability of the needle crossing a line, think ofthe “expected number” of crossings E(ℓ) when a needle oflength ℓ is thrown.

Suppose the needle is broken into two. What happens?

E is a linear function of ℓ; i.e.,

E(ℓ1 + ℓ2) = E(ℓ1) + E(ℓ2)

E(rℓ) = r E(ℓ) for every r > 0

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Suppose the needle is bent over double (like a hair pin).What happens?

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E(ℓ) depends only on ℓ and is unchanged if the needle is bentinto a shape like

We could also bend it into a circle.

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Suppose ℓ = π, and we have a circular “needle” of length π(i.e., of diameter 1). It is thrown on a paper with parallel lines atdistance 1. Each time it is thrown

it either crosses one of the lines twice, or touches two of thelines once each. So, there are two “crossings”.

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When ℓ = π and d = 1 the expected number of crossings witheach throw is 2.

So for any ℓ and d it is

2 ℓ

π

1d

=2π

ℓ

d.

(Proofs from THE BOOK.)

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Two formulas due to S. Ramanujan (1914)

∞∑

n=0

(

2nn

)3 42n + 5212n+4 =

1π.

2√

29801

∞∑

n=0

(4n)! (1103 + 26390n)(n!)4(396)4n =

1π.

(Connected with other problems in number theory.)

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Return to the beginning:

We said the ratiocircumference

diameteris the same for every circle. This constant ratio is called π.

Caveat: This is true for plain/flat/Euclidean geometry.Not in curved/non-Euclidean geometry.

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Spherical geometry (positive curvature)

Hyperbolic geometry (negative curvature)

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Surface of a sphere S

Through “any” two points passes a unique geodesic (a curve ofminimal length)

Exception: if two points are antipodal (like the North pole andthe South pole) then there are infinitely many geodesicspassing through them.

The distance between any two points a,b is the distance alongthis geodesic. Call this d(a,b).

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Circle with centre a and radius r is the set of all points x atdistance r from a, i.e.,

C(a, r) = {x : d(a, x) = r}.

For convenience, choose S to be the sphere with radius equalto 1.

If a is the North pole, then for 0 < r < π the circles C(a, r) arethe lines called “parallels” or “latitudinal lines” on the globe.

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The circle C(a, π/2) is the equator (all points at distance π/2from the North pole).In this case the quantity

2πr = 2π · π2= π2.

On the other hand the circumference of the equator (on thesphere of radius 1) is 2π.So in this case

circumference = 2π < π2 = 2π · radius.

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In this geometry the sum of angles of a triangle is bigger than2π.

The opposite happens in hyperbolic geometry.

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π can be defined without any reference to the circle.

This can be done via trigonometric functions. These, in turn,can be defined without any reference to triangles.

sin x = x − x3

3!+

x5

5!− x7

7!+ · · ·

cos x = 1 − x2

2!+

x4

4!− x6

6!+ · · ·

Both series converge for all real numbers x . So sin x , cos x aredefined for all real numbers (not just angles of a triangle).

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Relations likesin2 x + cos2 x = 1

link them to the circle and to triangles.

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It can be shown that there exist (several) numbers t such that

sin t = 0

Analyst’s definition:

The smallest positive number t0 for which

sin t0 = 0

is called π.

This defines π without any reference to circles.

Of course the two notions are related, and are consistent.

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A somewhat tragic piece of history:

Edmund Landau gave this as the definition of π in his lectures,and in his textbook. For this, and for other things, he wasdenounced as doing “un-German” mathematics and dismissedfrom his job.

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A comic piece of history:

Bill No. 246, State of Indiana (U.S.A.), 1897“Be it enacted by the General Assembly of the State of Indiana:It has been found that the circular area is to the quadrant of thecircumference, as the area of an equilateral triangle is to thesquare on one side.”i.e.

area of a circle14circumference

=

√3

4?

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Are there lessons for us in these two episodes?

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In a lighter vein: A pneumonic for the decimal expansion of π.

How I need a drink, alcoholic of course, after the heavy lecturesinvolving quantum mechanics.

π = 3.14159265358979 . . .

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