The Basel Problem The Basel Problem Abstract Because of its relation to the distribution of prime numbers, the Riemann zeta function ζ (s) is one of the most important functions in mathematics. The zeta function is defined by the following formula for any complex ∞ s number s with the real component greater than 1. ζ (s) = ∑n=1 1/n .Taking s=2, we see that ζ(2) is equal to the sum of the squares of reciprocals of all positive integers. This leads to the famous problem by Basel in mathematical analysis with important relevance to number theory, solved by Leonhard Euler in 1734. In this paper we discuss some of the notable proofs given by mathematicians to the basal problem. Most of the theorems are very well known whereas some can be found as proofs of problems present in textbooks. Keywords: Zeta Function, Pi, Basal Problem. 1. Introduction The Basel problem was first introduced in 1644 by Pietro Mengoli, Italian mathematician and clergyman (1626–1686) who is known for his Known (nowadays) for work in infinite series. The Problem remained open for 90 years, until by solving it, Euler gave his first proof in 1734. The Basel problem asks for the sum of the series: 1 1 1 1 ∑∞ = + + + ⋯ (1) n=1 n2 12 22 32 휋2 It was Euler who found the exact sum to be . In this paper we discuss some of the notable proofs given by 6 mathematicians to the basal problem. Most of the theorems are very well known whereas some can be found as proofs of problems present in textbooks. The R.H.S. of (1) can be written as 1 1 1 1 1 3 1 ∑∞ = ∑∞ − ∑∞ = (1 − ) ∑∞ = ∑∞ (2) 푛=0 (2푛+1)2 n=1 n2 n=1 (2n)2 4 n=1 n2 4 n=1 n2 Proof 1: Let us first take a proof that is due to Calabi, Beukers and Kock1 and also mentioned in Robin Chapman14. They first noticed the following equality: 1 1 1 푑푥푑푦 ∑∞ = ∫ ∫ (3) 푘=0 (2푘+1)2 0 0 1−푥2푦2 2 2 −1 1−푦 −1 1−푥 After that they made the substitution 푢 = 푡푎푛 푥√ 푎푛푑 푣 = 푡푎푛 푦√ . We now get the transformed 1−푥2 1−푦2 sin 푢 sin 푣 variables after substitution as 푥 = 푎푛푑 푦 = . Now notice the following important equation which cos 푣 cos 푢 connects 휁(2) and the left hand side of the equation (3). 3 ∞ 1 ∞ 1 ∞ 1 휁(2) = ∑ − ∑ = ∑ (4) 4 푛=1 푛2 푚=1 2푚2 푘=0 (2푘+1)2 The Right hand side of the equation (3) can be rewritten after the substitution as Sourangshu Ghosh Page 1 The Basel Problem 1 1 푑푥푑푦 푑푢푑푣 휋 ∫ ∫ =∫ ∫ 퐽 ℎ푒푟푒 퐴 = {(푢, 푣): 푢 > 0, 푣 > 0, 푢 + 푣 < } (5) 0 0 1−푥2푦2 퐴 1−푢2푣2 2 Since the 퐽 the Jacobian matrix will be 2 휕(푥,푦) 푐표푠 푢/ 푐표푠 푣 푠푖푛 푢 푠푖푛 푣/ 푐표푠 푣 퐽 = = det | | = 1 − 푢2푣2 (6) 휕(푢,푣) 푠푖푛 푢 푠푖푛 푣/ 푐표푠2푢 푐표푠 푣/ 푐표푠 푢 1 휋 휋 휋2 휋2 Now as the area of 퐴 = ( ) ∗ ( ) ∗ ( ) = . We get from equation (1),(2) and (3) that 휁(2) = . 2 2 2 8 6 Proof 2: We now derive a proof that comes from a note by Boo Rim Choe2 in the American Mathematical Monthly in 1987 and also mentioned in Robin Chapman14. We start by taking a note of the power series expansion of the inverse sine function. 2푛+1 −1 ∞ 1.3…(2푛−1) 푥 푠푖푛 (푥) = ∑ 푓표푟 |푥| ≤ 1 (7) 푛=0 2.4…(2푛) 2푛+1 Now let us put 푡 = 푠푖푛−1(푥). After the substitution we get 2푛+1 ∞ 1.3…(2푛−1) sin (푡) 푡 = ∑ 푓표푟 |푡| ≤ 휋/2 (8) 푛=0 2.4…(2푛) 2푛+1 2푛+1 To proceed further we first need to find the integration from 0 푡표 휋/2 of sin (푡) or 퐼2푛+1 = 휋/2 2푛+1 ∫0 (sin (푥)) 푑푥, we need to use the recursion formula which can be expressed as: 휋/2 2푛−1 퐼2푛+1 = 퐼2푛−1 − ∫0 [(sin (푥)) cos (푥)] 푐표푠푥. 푑푥 (9) Now using integration by parts now we can derive that 2푛+1 퐼 = 퐼 (10a) 2푛 2푛+1 2푛−1 Therefore by recursion we can write that 2푛 2푛.2(푛−1) 2.4…(2푛) 퐼 = 퐼 = 퐼 = (10b) 2푛+1 2푛+1 2푛−1 (2푛+1)(2푛−1) 2푛−3 3.5…(2푛+1) We therefore get that 휋/2 2.4…(2푛) 퐼 = ∫ (sin (푥))2푛+1 푑푥 = (11) 2푛+1 0 3.5…(2푛+1) Now if we use equation (11) and (8) we get 휋2 휋⁄ 1 =∫ 2 푡. 푑푡 = ∑∞ (12) 8 0 푛=0 (2푛+1)2 휋2 The right hand side of the equality (12) is (3/4)휁(2). Therefore we get that 휁(2) = . 6 Proof 3: We now take a proof from an article in the Mathematical Intelligencer by Apostol3 in 1983 and also mentioned in Robin Chapman14. We first start noting the following equation: Sourangshu Ghosh Page 2 The Basel Problem 1 1 1 = ∫ ∫ 푥푗−1푦푗−1푑푥푑푦 (13) 푗2 0 0 It is also to note that we get by the monotone convergence theorem 1 1 1 1 1 푑푥푑푦 ∑∞ = ∫ ∫ (∑∞ (푥푦)(푛−1)). 푑푥푑푦 = ∫ ∫ (14) 푘=1 푘2 0 0 푘=1 0 0 1−푥푦 Let us now introduce change of variables by changing the variables 푢 = (푥 + 푦)/2) 푎푛푑 푣 = (푥 − 푦)/2)). After the substitution of the variables we get the relationship between changed variables and actual variables as 푥 = 푢 – 푣 and = 푢 + 푣 . Now notice that the Left hand side of the equation (14) is 휁(2). We therefore can rewrite equation (14) in terms of the changed variables as: 푑푢푑푣 휁(2) = 2 ∬ (15) 푆 1−푢2+푣2 Here S is the square with vertices (0, 0), (1/2, −1/2), (1, 0) and (1/2, 1/2). Now because of the symmetry present (symmetrical about the line y=0). We can find the value by only doubling the integration of the upper 1 1 1 half of the square which can be itself be divided into two triangles with coordinates ((0, 0), ( , ) , ( , 0)) and 2 2 2 1 1 1 (( , ) , ( , 0) , (1,0)) respectively. We can therefore express 휁(2) as 2 2 2 1 푢 푑푢푑푣 1 1−푢 푑푢푑푣 ( ) 2 1 휁 2 = 4 ∫0 ∫0 2 2 + 4 ∫ ∫0 2 2 1−푢 +푣 2 1−푢 +푣 1 1 푢 1 1 1−푢 = 4 ∫2 푡푎푛−1 ( ) . 푑푢 + 4 ∫ 푡푎푛−1 ( ) . 푑푢 (16) 0 √1−푢2 √1−푢2 1/2 √1−푢2 √1−푢2 푢 Now we notice that 푡푎푛−1 ( ) = 푠푖푛−1(푢) . Let us this identity by again changing the variables by √1−푢2 1−푢 substituting 휃 = 푡푎푛−1( ). After substitution we get √1−푢2 2 2 2 푡푎푛 휃 = (1 − 푢)/(1 + 푢) and 푠푒푐 휃 = . (17) 1+푢 1 휋 1 It follows that 푢 = 2푐표푠2휃 − 1 = 푐표푠2휃 and therefore we get that 휃 = ( ) 푐표푠−1푢 = − 푠푖푛−1푢. We can 2 4 2 hence rewrite equation (16) in terms of the changed variables as 1 −1 −1 2 2 푠푖푛 푢 1 1 휋 푠푖푛 푢 −1 2 1/2 −1 −1 2 1 휋 휋 휁(2) = 4 2 푑푢 + 4 1 ( − ) . 푑푢 = [2(푠푖푛 푢) ] + [휋푠푖푛 푢 − (푠푖푛 푢) ] = + − ∫0 2 ∫ 2 0 1/2 √1−푢 2 √1−푢 4 2 18 2 휋2 휋2 휋2 휋2 − + = . (18) 4 6 36 6 Proof 4: We now present some textbook proofs, found in many books on Fourier analysis to find the value of 휁(2). In this proof we are going to discuss next we will use the 퐿2 completeness of the trigonometric functions. Suppose that 푯 is a Hilbert space with inner product. Let (푒푛) be an ortho-normal basis of H such that 〈푒푚, 푒푛〉 = 1 푖푓 푚 = 푛 0 푖푓 푚 ≠ 푛 (19) Then we know that by Parseval's identity, we can write for every 푔 ∈ 푯, Sourangshu Ghosh Page 3 The Basel Problem 2 2 〈푔, 푔〉 = ||푔|| = ∑푛|〈푔, 푒푛〉| (20) Let the Hilbert Space 푯 be 퐿2[0,1]. Let us now apply this theorem 푔(푥) = 푥. The Left hand side of the 1 1 1 equation (20) will be 〈푔, 푔〉 = and 〈푔, 푒 〉 = and 〈푔, 푒 〉 = for 푛 ≠ 0. Therefore we can re express 3 0 2 푛 2휋푖푛 equation (20) as: 1 1 1 = + ∑ (21) 3 4 풏∈풁,풏≠ퟎ 4휋2푛2 Which therefore gives us 휁(2) = 휋2/6. Proof 5: Let us now see another similar way of proving the identity by fourier transform. We first take a function f which is continuous in [ 0, 1 ] and 푓(0) = 푓(1). Let us first explore the criteria for point-wise convergence of a periodic function f. The criteria for point-wise convergence of a periodic function f are as follows: • Fourier series converges uniformly if 푓 satisfies a Holder condition. • Fourier series converges everywhere if 푓 is of bounded variation. • Fourier series converges uniformly if 푓 is continuous and its Fourier coefficients are absolutely summable. Then the Fourier series of 푓 converges to 푓 pointwise. As our function holds true in both the circumstances we can say that the Fourier series of f converges to f point wise. Let us apply this to the function with these properties let’s say 푓(푥) = 푥(1 − 푥) which then gives us the following equation: 1 ∞ cos(2휋푛푥) 푥(1 − 푥) = − ∑ (22) 6 푛=1 휋2푛2 If we put x=0 in the LHS of the equation (22) we get (2) = 휋2/6.