The number π. Rajendra Bhatia. December 20, Indian Statistical Institute, Delhi India 1 / 78

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1 1 / 78 The number π Rajendra Bhatia Indian Statistical Institute, Delhi India December 20, 2012

2 2 / 78 CORE MATHEMATICS Nature of Mathematics Reasoning and Proofs in Mathematics History of Mathematics Contributions of Indian Mathematicians.

3 Circle of diameter D Circumference C Area A Easy arguments show C D C = πd A D 2 A = kd 2 How do you know π and k are related? 3 / 78

4 4 / 78 Shown by Archimedes ( Measurement of a circle 225 BCE BCE) Theorem Let A be the area of a circle with circumference C and radius r. Then A = T where T is the area of a right angled triangle with base C and height r.

5 5 / 78 Circumference C area A area T Then A = T (Reducing the more complicated problem to a simpler, more familiar, one.)

6 6 / 78 Key ideas of proof 1. Method of exhaustion: In any circle we can inscribe a regular polygon whose area/circumference are arbitrarily close to those of the circle. Can also circumscribe a regular polygon with similar approximation properties.

7 7 / 78 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.

8 8 / 78 Step 1 Regular polygon with n sides b = length of each side A = area Q = perimeter h = apothem. Then A = 1 2 bh.n, Q = bn So A = 1 2 hq

9 9 / 78 Step 2 We want to prove for the circle A = T = 1 2 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)

10 10 / 78 By Step 1, A = 1 2 hq where h is the apothem and Q the perimeter of the polygon. But h < r and Q < C. So A < 1 2 r C =: T (3) But we had T < A (See (2)). Contradiction. So the assumption A > T can t be true.

11 11 / 78 Step 3 Suppose A < T. Then p := T A > 0. (4) Now choose a polygon, circumscribing the circle, having area A such that A A < p = T A A < T. (5)

12 12 / 78 By Step 1, A = 1 2 hq. Now h = r, and Q > C. So A > 1 2 r C =: T (6) Again (5) and (6) contradict each other. So A < T can t be true. Conclusion: A = T.

13 13 / 78 So A = T = 1 2 rc = 1 2 r.2πr i.e. A = πr 2 where π is the constant in the formula C = 2πr

14 14 / 78 Archimedes estimated i.e < π < < π < which is accurate to two decimal places. We now know π =

15 15 / 78 Some other great formulas of Archimedes Volume of sphere = 4 3 πr 3 Surface area of sphere = 4πr 2 A cylinder circumscribed about a sphere has Volume = 2πr 3 Surface area = 6πr 2

16 16 / 78 Both the volume and the surface area of the cylinder are in ratio with the volume and surface area of the sphere. 3 2 Archimedes was so proud of this discovery that he requested that this should be put on his tomb.

17 17 / 78 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.

18 18 / 78 Simple Ideas 1 The sun crosses all points on a meridian line (longitude) at the same time. 2 At noon the sun is directly overhead at the equator. So a stick will have no shadow. The further away you are from the equator the longer will be the shadow. 3 The measurements of these shadows can be used to infer the angular separation between two points on the earth. From this the perimeter of the earth can be calculated.

19 19 / 78

20 20 / 78

21 21 / 78 A, B two points on the same longitude. length of sticks AP = l A, BQ = l B length of shadows AL = s A, BM = s B s A s L A = tanθ A, B L B = tanθ B. If d(a, B) = distance between A and B, then Perimeter of the earth = 2π θ B θ A d(a, B)

22 22 / 78 Ancient mathematicians thought of numbers as lengths. The Greek mathematicians proved that 2 is not a rational number. A number r is rational if it can be expressed as where p, q are integers. r = p q

23 23 / 78 The proof that 2 is irrational is easy, and beautiful. (Exercise: Use the idea of the proof to investigate whether 3 2 is rational; for which n is n rational.) π is irrational (proved by J. Lambert in 1761) (Proof not as easy as for 2) π 2 is irrational (A. Legendre, 1794)

24 24 / 78 The number e is also irrational. Surprisingly hard problems: Is π + e rational? Is π e rational?

25 25 / 78 If a number x is a solution of a (nonconstant) polynomial equation x n + a 1 x n 1 + +a n = 0 with rational coefficients a 1,..., a n, then we say x is an algebraic number. All rational numbers are algebraic. Some irrational numbers are algebraic, e.g. 2 is a solution of the equation x 2 2 = 0.

26 26 / 78 π is not an algebraic number. It is a transcendental number. (Lindemann, Another proof by D. Hilbert) Consequences: π cannot be expressed as a combination of rational numbers and roots (e.g. it is not anything like )

27 27 / 78 It is impossible to square the circle. i.e. it is impossible to construct, using a compass and ruler alone, a square whose area is equal to the area of a given circle. (Theorem in algebra: All constructible numbers are algebraic) See: Courant and Robbins, What is Mathematics?

28 28 / 78 π and infinite series Does π have anything to do with natural numbers N = {1, 2, 3,...}. Surprisingly: a lot.

29 29 / 78 Infinite series a 1 + a 2 + a 3 + +a n +. Obviously diverges. The series converges. The sum is 1. Less obvious diverges n n +.

30 30 / 78 Madhava ( ) Nilakantha ( 1500) Tantrasangraha Madhava Series Gregory-Leibnitz Series ( 1670) Choose z = ( 1)n 2n+1 + = π 4. arctan z = z z3 3 + z5 5 z7 7 +

31 31 / 78

32 32 / 78 Translation: The diameter multiplied by four and divided by unity [is found and saved]. Again the products of the diameter and four are divided by the odd numbers like three, five, etc. and the results are subtracted and added in order [to the earlier result saved]. Take half of the succeeding even number as the multiplier at whichever [odd] number the division process is stopped, because of boredom. The square of that [even number] added to unity is the divisor. Their ratio has to be multiplied by the product of the diameter and four as earlier. The result obtained has to be added if the earlier term [in the series] has been subtracted and subtracted if the earlier term has been added.

33 33 / 78 The resulting circumference is very accurate; in fact more accurate than the one which may be obtained by continuing the division process [with large number of terms in the series]. { C = 4d 1 1 } p 1 1 p 1 p+1 + +( 1) 2 +( 1) 2 3 p 2 /((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 34 / 78 The first infinite product (F. Viète, 1593) 2 2 π = Another famous formula (J. Wallis, 1655) Intrigued? see R. Bhatia, Fourier Series π 2 =

35 35 / 78 A calculation that made L. Euler famous (1735) n=1 1 n 2 = π2 6. This was called the Basel Problem, posed by J. Bernoulli. This is one of the most famous formulas in mathematics.

36 36 / 78 Euler s first proof Relation between roots and coefficients of a polynomial: If α 1,α 2,...,α m are the roots of then a 0 + a 1 x + a 2 x 2 + +a m x m = 0, 1 α i = a 1 a 0. The function cos x has Taylor expansion cos x = 1 x 2! + x 2 It has zeros at (2n+1)2 π 2 4, n = 0, 1, 2,... 4!.

37 37 / 78 Leap of faith: Think of the infinite series as a polynomial of infinite degree and assume the same relation holds between roots and coefficients. This gives n=0 From this it is easy to see that 1 (2n+1) 2 = π2 8. n=1 1 n 2 = π2 6.

38 38 / 78 Euler was aware that there is a gap in this argument and found other proofs. Discovery of infinite product expansions: sin x x = (1 x 2 n=1 Several proofs are known now. n 2 π 2 ).

39 39 / 78 π can appear at unexpected places. An example: Z = {..., 2, 1, 0, 1, 2,...} integers Z 2 = {(m, n) : m, n Z} lattice" all points in the plane with integral coordinates. Question: How many points of the lattice are visible from the origin?

40 40 / 78

41 41 / 78 Observe: The point (m, n) is visible if and only if m and n are coprime. So our question is: Among all points of Z 2 what is the proportion of (m, n) that have m, n coprime. Many questions of this kind are studied in number theory: e.g. how many numbers are square-free?

42 42 / 78 Let n = p m 1 1 pm 2 2 pm k k (prime factoring) If m 1 = m 2 = = m k = 1, we say n is square-free. e.g. 70 = is square-free. 90 = 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 of its being square-free.

43 43 / 78 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 1 k. So, if p 1, p 2,... is the sequence of primes, then( the probability ) that a random number is not a multiple of pj 2 is 1 1. pj 2

44 44 / 78 So the probability that none of the pj 2 is a factor is ( ) 1 1 pj 2. j (The probability of two independent events happening simultaneously is the product of their individual probabilities). How to find the infinite product given above?

45 45 / 78 Riemann Zeta Function ζ(s) = n=1 1 n s (s > 1). This function is of great importance in the study of prime numbers. Theorem (Euler) ζ(s) = n= p s n where p 1, p 2,... is the sequence of primes.

46 46 / 78 A quick idea of the proof. Recall the geometric series, for 0 < x < 1 So 1 1 x = 1+x + x 2 + = x m. m=0 1 1 p s 1 = 1+p s 1 + p 2s 1 + p 3s 1 + = 1+ n s n:n=p m 1 (sum over all n such that n is a power of the prime p 1 )

47 47 / 78 = ( 1 1 p s 1 ( 1+p s = 1+ )( 1 1 p s p 2s 1 + n:n=p 1 m 1 p2 m 2 n s ) )( ) 1+p s 2 + p 2s 2 + (sum over all n which have only p 1, p 2 as prime factors). Similarly r 1 1 p s = 1+ n s (7) k k=1 n:n=p 1 m 1 pr mr (sum over all n which have only p 1, p 2,..., p r as prime factors)

48 As r, the sequence p 1, p 2,... runs over all primes and the n which have p 1, p 2,... as prime factors runs over all numbers n 2. So we get from (7) 1 k=1 1 1 p s k = n=1 1 n s = ζ(s). 48 / 78

49 49 / 78 So: the probability of a random positive integer being square-free is ( ) 1 1 pj 2 = 1 ζ(2) = 6 π j=1 Exercise: Show that the probability of two natural numbers m, n, picked at random, being coprime is also 6 π 2. So the proportion of lattice points visible from the origin is 6 π 2. See Hardy & Wright, An Introduction to the Theory of Numbers.

50 50 / 78 Elements of beauty in mathematics: Surprise Elegance Depth Brevity Power Connections between different things.

51 51 / 78 I had started reading mathematics when I was about thirteen or so. I had accidentally opened a book and come across Leibnitz s series for π/4 = 1 1/3+1/5 1/7+ involving reciprocals of the odd integers with alternating signs. Till then, school mathematics had always bored me but this seemed such a very strange and beautiful relationship that I determined I would read that book in order to find out how this formula came about. -Atle Selberg

52 52 / 78 Some amazing facts. ζ(4) = π4 π6, ζ(6) = ,... Euler found many such sums. A good way now is via Fourier series. Generally: ζ(2k) = π 2k (a rational number). Very little is known about the values ζ(2k + 1). R. Apéry (1978) showed ζ(3) is an irrational number. Proved in 2000: atleast one of ζ(5),ζ(7),...,ζ(21) is irrational.

53 53 / 78 π occurs often in statistics and probability. One obvious reason is that the area under the bell-shaped curve is e x2 dx = π.

54 54 / 78 A surprising occurrence is in the famous Buffon Needle Problem On a plane ruled by parallel lines distance d apart, a needle of length l d is thrown at random. What is the probability that the needle intersects one of the lines?

55 55 / 78 Note that the probability p satisfies So What is the constant c? Answer p l and p 1 d. p = c l d. p = 2 π Geometric probability, Integral geometry. l d

56 56 / 78 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 needle intersect line: for to a x < l 2 cos ϕ

57 57 / 78 Assumption: x, ϕ uniformly distributed, independent random variables, 0 x d 2, 0 ϕ π 2 The probability of the event x < l 2 cosϕ is the ratio area of the red region area of the blue rectangle

58 58 / 78 This is equal to π/2 0 l 2 cos ϕ dϕ π 2 d 2 = l/2 πd/4 = 2 π l d.

59 59 / 78 Proof 2 Assume, for convenience, d = 1. Instead of probability of the needle crossing a line, think of the expected number of crossings E(l) when a needle of length l is thrown. Suppose the needle is broken into two. What happens? E is a linear function of l; i.e., E(l 1 +l 2 ) = E(l 1 )+E(l 2 ) E(rl) = r E(l) for every r > 0

60 Suppose the needle is bent over double (like a hair pin). What happens? 60 / 78

61 61 / 78 E(l) depends only on l and is unchanged if the needle is bent into a shape like We could also bend it into a circle.

62 62 / 78 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 at distance 1. Each time it is thrown it either crosses one of the lines twice, or touches two of the lines once each. So, there are two crossings.

63 63 / 78 When l = π and d = 1 the expected number of crossings with each throw is 2. So for any l and d it is 2 l π 1 d = 2 π l d. (Proofs from THE BOOK.)

64 64 / 78 Two formulas due to S. Ramanujan (1914) n=0 n=0 ( ) 2n 3 42n+5 n 2 12n+4 = 1 π. (4n)! ( n) (n!) 4 (396) 4n = 1 π. (Connected with other problems in number theory.)

65 65 / 78 Return to the beginning: We said the ratio circumference diameter is 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.

66 66 / 78 Spherical geometry (positive curvature) Hyperbolic geometry (negative curvature)

67 67 / 78 Surface of a sphere S Through any two points passes a unique geodesic (a curve of minimal length) Exception: if two points are antipodal (like the North pole and the South pole) then there are infinitely many geodesics passing through them. The distance between any two points a, b is the distance along this geodesic. Call this d(a, b).

68 68 / 78

69 69 / 78 Circle with centre a and radius r is the set of all points x at distance r from a, i.e., C(a, r) = {x : d(a, x) = r}. For convenience, choose S to be the sphere with radius equal to 1. If a is the North pole, then for 0 < r < π the circles C(a, r) are the lines called parallels or latitudinal lines on the globe.

70 70 / 78 The circle C(a, π/2) is the equator (all points at distance π/2 from the North pole). In this case the quantity 2πr = 2π π 2 = π2. On the other hand the circumference of the equator (on the sphere of radius 1) is 2π. So in this case circumference = 2π < π 2 = 2π radius.

71 71 / 78 In this geometry the sum of angles of a triangle is bigger than 2π. The opposite happens in hyperbolic geometry.

72 72 / 78 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 x 3 3! + x 5 5! x 7 7! + cos x = 1 x 2 2! + x 4 4! x 6 6! + Both series converge for all real numbers x. So sin x, cos x are defined for all real numbers (not just angles of a triangle).

73 73 / 78 Relations like sin 2 x + cos 2 x = 1 link them to the circle and to triangles.

74 74 / 78 It can be shown that there exist (several) numbers t such that sin t = 0 Analyst s definition: The smallest positive number t 0 for which sin t 0 = 0 is called π. This defines π without any reference to circles. Of course the two notions are related, and are consistent.

75 75 / 78 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 was denounced as doing un-german mathematics and dismissed from his job.

76 76 / 78 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 the circumference, as the area of an equilateral triangle is to the square on one side. i.e. area of a circle circumference = 4?

77 Are there lessons for us in these two episodes? 77 / 78

78 78 / 78 In a lighter vein: A pneumonic for the decimal expansion of π. How I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics. π =

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