chan Euler products

� THE SINE PRODUC T FORMULA AND THE GAMMA FUNCTION ERICA CHAN DECEMBER 12, 2006 Abstra ct. The function sin x is very importan t in math ematics and has many application s. In addition to its series expansion, it can also be written as an infinite product. The infinite product of sin x can be used to prove certain values of ζ(s), such as ζ(2) and ζ(4). The gamma function is related directly to the sin x function and can be used to prove the infinite product expansion. Also used are Weierstrass’ product formula and Legend re’s relati on. 1. Introduc tion There are a few special function s in mathemat ics that have particular significa nce and many applicat ions. The gamma functio n is one of those functions. The gamma function can be defined as ∞ Γ(x) = e−ttx−1dt. 0 We can also get the formula (1) Γ(x +1) = xΓ(x) by repla cing x with x + 1 and integra ting by parts. In addition, since Γ(1) = 1, using Equatio n (1), by inductio n, we can relate the gamma function to the factoria l formula (2) Γ(n)=(n − 1)!. The gamma functio n has the properties that it is log convex and mono­ tonic, which will be used in a later proof. Anot her important function in mathematics is the sine function. The trigonometric functio n sin x can be writt en as an infinite series 3 5 7x x xsin x = x − 3! + 5! − 7! + ... Date : December 12, 2006. 1 2 ERICA CHAN The functio n sin x can also be written as an infinite product expa nsion. The gamma functio n is directly related to the sine function. To deriv e the infinite product expa nsion of the sine functio n, the Weierstr ass product formula, Legendre relation, and the gamma functio n are all used. The sine product formula is important in mathemat ics because it has many applicat ions, including the proofs of other problems. One such applica tion is the calculat ion of the values of ζ(2) and ζ(4), where �∞1 1 1 1 ζ(s)= =1+ + + + ... ns 2s 3s 4s n=1 is the Riemann zeta function. Section 2 deriv es Weierstra ss’ product formula and Euler’s constan t. Section 3 introduces Stirling ’s formulas, Gauss’ multiplicatio n formula, and the Legendre relatio n. In Section 4, the sine product formula is produced from the gamma function. Finally , Section 5 discuss the applica tions of the sine product formula, including the calcula tion of ζ(2) and ζ(4). 2. Weierstrass’ product formula Weierstr ass deriv ed a formula which, when applied to the gamma function, can be used to prove the sine product formula. To find Weier­ strass’ product formula, we first begin with a theorem. Theorem 2.1. The function Γ(x) is equal to the limit as n goes to infinity of nxn! (3) Γ(x) = lim . n→∞ x(x +1) (x + n) ··· Proof. Begin with a difference quotient express ed as the inequa lity logΓ(−1+ n) − logΓ(n) log Γ(x + n) − logΓ(n) log Γ(1 + n) − logΓ(n) . (−1+ n) − n ≤ (x + n) − n ≤ (1 + n) − n This inequa lity is true because of the gamma function s prop erties that it is monot onically increa sing and log convex. Substitut e for Γ(n) using Equat ion (2) and simplify to get log(n − 1) ≤ logΓ(x + n) − log(n − 1)! ≤ log n. x � � 3 THE SINE PRODUCT FORMULA AND THE GAMMA FUNCTION Rearrang e the terms of the inequality and since the logarithm is a mono tonica lly increasing function, we get (n − 1)x(n − 1)! ≤ Γ(x + n) ≤ n x(n − 1)!. Using Equa tion (1), we know that Γ(x + n)=(x + n − 1)(x + n − 2) (x + 1)xΓ(x). Substitut ing it into the inequality gives us ··· (n − 1)x(n − 1)! nxn! x + n. (x + n − 1) ≤ Γ(x) ≤x(x + 1) x(x +1) (x + n) n ··· ··· If we repla ce n with n + 1 on the left hand side and then rearrange the inequalit y, we get n nxn! Γ(x) x + n ≤ x(x +1) (x + n) ≤ Γ(x). ··· Taking the limit as n →∞ gives us nxn! Γ(x) = lim . n→∞ x(x +1) (x + n) ··· Equat ion (3) was deriv ed by Gauss. From there, Weierstrass was able to deriv e another form of the same equat ion x(log n−1/1−1/2−...−1/n) 1 ex/1 ex/2 ex/n Γ(x) = e . x 1+ x/1 · 1+ x/2 ··· 1+ x/n The limit of 11 1 lim(+ + + n − log n) n→∞ 12 ··· exists, is equal to C, and is often called Eule r’s constant. So we can rewrite Weierstra ss’ product formula as � x/i1 ∞e(4) Γ(x)= e−Cx . x 1+ x/ii=1 3. Multiplication Formula There are three formulas, Γ(x) = √ 2πxx−1/2 e−x+µ(x), where ∞1 1 θ µ(x) = (x + n + )log(1+ ) − 1= , and 2 x + n 12x n=0 e−n+θ/12n n!= √ 2πnn+1/2 , 4 ERICA CHAN which are known as Stirling ’s formulas. Where θ is a number inde­ pendent of other values and where 0 ≤ θ ≤ 1. Stirling ’s formulas are approxima tions of Γ(x) for large values of x and the accuracy increases as x increases. Gauss discovered a formula, which expresse s Γ(x) as a product of its facto rs. Gauss’ multiplicat ion formula is (2π)(p−1)/2 x x +1 x + p − 1 (5) Γ(x) =Γ( )Γ( ) Γ( ), px−1/2 p p ··· p where p is a positive integer. There is the special case disco vered by Legendre, where p = 2, which is called Legendr e’s relatio n. Legendre’s relation states x x +1 √π (6) Γ( )Γ( )= Γ(x). 2 2 2x−1 The deriv ation and proof of these formulas can be found at [1]. They are based on finding an approxima tion for Γ(x) in terms of an estimate for n!. 4. The Sine and and Gamma Func tions To deriv e the sine product formula, we first find a relationship be­ tween the sine and gamma functions. We define a function φ(x) and find that φ(x +1) = φ(x). Theorem 4.1. Define the function φ(x), for noninte gral x, to be (7) φ(x) = Γ(x)Γ( 1 − x)sin πx, then φ(x +1) = φ(x). Proof. If we use Equatio n (1) and substit ute −x + 1 for x, then we get that (8) Γ(−x +1) = −xΓ(−x). Finding φ(x + 1), we get φ(x +1) = Γ(x + 1)Γ(−x)sin(π(x + 1)). Note that sin(π x+π)= − sin πx. Use Equa tion (1), rearrange Equat ion (8), and substitu te them in to get φ(x +1) = xΓ(x)Γ(−x + 1) (− sin πx) = Γ(x)Γ(−x + 1) sin πx. −x This is equal to the original equatio n for φ(x), so φ(x+1) = φ(x). � � � � � � ��� ��� ���� ���� � 5 THE SINE PRODUCT FORMULA AND THE GAMMA FUNCTION The Legendre relation (Equa tion (6)) can be written as x x +1 Γ( )Γ( )= b2−xΓ(x),2 2 where b =2√π is a constant. Replacing x with 1 − x gives x Γ(1 − x )Γ(1 − )= b2x−1Γ(1 − x). 2 2 From there, we get that φ( x )φ( x +1 ) = Γ( x )Γ(1 − x )sin πx Γ( x + 1 )Γ(1 − x )cos πx . 2 2 2 2 2 2 2 2 Simplifying the above, we find x x +1 b2 b2 φ()φ( )= Γ(x)Γ( 1 − x)sin πx = φ(x)= cφ(x)2 2 4 4 where c = b2 is a constant. 4 Using Equat ion (1) and the infinit e series expa nsion of sin x, we get that (πx)3 (πx)5 (πx)7Γ(1 + x)φ(x) = Γ(1 − x) πx −x + + ... − 3! 5! 7! π3x2 π5x4 π7x6 = Γ(1 + x)Γ(1 − x) π − 3! + 5! − 7! + ... . The right hand side of the equa tion equa ls π when x = 0. From there we see that φ(0) = π. Let g(x) be a periodic function that is equal the second derivative of log φ(x). It is periodic because log φ(x)= log(Γ(x)Γ( 1 − x)sin πx) is periodic and so the second deriv ative will also be periodic. Since g(x) is periodic, then it satisfies the equa tion 1 x x +1 (9) g(x)= (g( )+ g( )). 4 2 2 Since g(x) is continuous on the interval 0 ≤ x ≤ 1, it is bounded by a consta nt M, |g(x)|≤ M. Because g(x) is periodic, it is bounded by M for all x. From Equatio n (9), we get that ) MFrom this we see that g(x) can actua lly be bounded by 2 . We can continue to repeat this process until the bound of g(x) goes to 0. There­ fore g(x) = 0, which means that log φ(x) is a linear function, because g(x) = 0 is its second derivative. Since log φ(x) is periodic, this implies 1 x +1 M x g(x) g() g( + | |≤ .≤ 2 4 2 2 � � � 6 ERICA CHAN that it is a constant, which also implie s that φ(x) is consta nt. We know that φ(0) = π and therefor e φ(x) must equal π for all x. Rearranging Equat ion (7) and using the fact that φ(x)= π, π Γ(x)Γ( 1 − x)= . sin πx Using (1), the above equat ion can be rewritten as π sin πx = . −xΓ(x)Γ( −x) The Weierstr ass product formula allows us to replace the gamma func­ tion and rewrit e sin πx as an infinite product expansion � 2∞x(10) sin πx = πx 1 − i2 . i=1 5. Applications of the Sine Product Formula Applicatio ns of the sine product formula include the calculatio n of certain values of the Riemann zeta functio n. The proof that ζ(2) = π2/6 is often called Euler’s Theorem. Theorem 5.1. The sum of the reciprocal of the perfect squar es is π2/6: ∞1 π2 = . n2 6 1 Proof. Consider the function sin x = 0, which has an infinite number of roots ±π, ±2π, ±3π,.... Using the infinite series expa nsion of sin x and dividing sin x by x gives us the infinite series 2 4 6x x x(11) 1 − 3! + 5! − 7! + ... =0. Applying Equat ion (10) and dividing by πx, we get the infinite product expansio n sin x x2 x2 x2 (12) = (1 − )(1 − )(1 − )... x π2 4π2 9π2 When Equa tion (12) is expa nded, the coefficien t of x2 will be 1 1 1 + + + ... π2 4π2 9π2 Using Equatio n (11), we get 1 1 1 1 + + + ... = . π2 4π2 9π2 3! � � � � THE SINE PRODUCT FORMULA AND THE GAMMA FUNCTION 7 Multiplying both sides of the equatio n by π2 gives us 1 1 1 π2 + + + ... = . 12 22 32 6 Using a similar metho d, we can also calculate ζ(4), which is the sum of the reciprocals of numbers to the fourth power. Theorem 5.2. The value of ζ(4) is π4/90: ∞1 π4 = . n4 90 1 Proof. From the calculatio n of ζ(2), we know that �∞1 1 (13) = . i2π2 6 i=1 If we square Equatio n (13), we get that ��2�∞1 1 (14) = . i2π2 62 i=1 Expand ing the left hand side of Equatio n (14) we get ��2∞1 1 � 1 (15) i2π2 = i4π4 +2 i2j2π4 . i=1 i<j If we use the sine infinite product expansio n, Equat ion (12), we get that the coefficien t of x4 for sin xx is equa l to the sum of the product of i2x π2 2 . In other words, the coefficien t of x4 is � 1 . i2j2π4 i<j From Equatio n (11), we also know that the coefficien t of x4 must be 1 , so 120 � � � � � 8 ERICA CHAN � 1 1 (16) = . i2j2π4 120 i<j Therefore, substit uting Equat ion (16) into Equat ion (15) gives us ��2 �� ∞1 1 1 1 = = +2 . i2π2 36 i4π4 120 i=0 To calculat e ζ(4), simply solve for i41 π4 and multiply both sides by π4, to get ∞1 π4 = . i4 90 i=1 References [1] Artin, Emil. The Gamma Function. New York: Holt, Rineha rt and Winston , 1964. [2] Simmons, George F. “Calculu s with Anala ytic Geometry .” New York: McGra w Hill, Second Edition, 1996. [3] Weisstein, Eric W. ”Gamma Function.” From MathW orld–A Wolfram Web Re­ source. http://mathworld.wolfram.com /GammaFunction.html [4] Young, Robert M. Excursions in Calculus: An Interpl ay of the Continuous and Discre te. United States of Ame rica: The Math ematical Association of Ame rica, 1992.