Lecture 3. January 9, 2018

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1 Lecture 3 January 9, 208 Some complex analyi Although you might have never taken a complex analyi coure, you perhap till know what a complex number i. It i a number of the form z = x + iy, where x and y are real and i =. The exponential function e z can be defined in term of it erie e z = + z + z zn + = z n. () n 0 More preciely, one can check, uing the ratio tet or the root tet that the erie appearing in the right hand ide converge abolutely for all complex number z. Thu, thi erie define a function which we denote by e z. When z = x i real, we recognize in the right hand ide of formula () the Taylor expanion of the uual exponential function e x, o it coincide with it. When z = iy, where y i real, we have e iy = n 0 (iy) n = k 0 (iy) 2k (2k)! + (iy) 2k+ (2k + )! k 0 = k 0 ( ) k y2k (2k)! + i ( ) k y 2k+ (2k + )! k 0 (2) and in the right hand ide of the lat formula (2) we recognize the familiar Taylor expanion of co y and in y. Thu, we get e iy = co y + i in y.

2 The exponential function on the real number ha the important property that e x+y = e x e y. The ame i true for the complex exponential function ince e z +z 2 = n 0 = n 0 = (z + z 2 ) n = n 0 n k=0 z u z k z n k 2 k! (n k)! = u 0 z v 2 ( n v 0 k=0 z u u! = e z e z 2, u! v! u 0 v 0 ( ) ) n z k z2 n k k where in the above calculation we ued the binomial formula, the change in the order of ummation u = k, v = n k (for all n 0 and 0 k n whoe invere i k = u, n = u + v), a well a the fact that we can rearrange the order of the term anyway we want ince the erie we are working with i abolutely convergent. In particular, z v 2 v! e x+iy = e x e iy = e x (co y + i in y). If a > 0 i any real number, then we can ue the fact that a = e log a and thu define a z = e (log a)z. If z = x + iy, then a z = e (log a)(x+iy) = e (log a)x e i(log a)y = a x (co(a log y) + i in(a log y)). In particular, a z = a x. 2 The Riemann Zeta Function We are now ready to define the Riemann zeta function. We will ue Riemann notation where C i a complex number written a = σ + it, where σ and t are real number. Definition. For C with σ >, we define ζ() = n 0 n. 2

3 The erie n n converge abolutely for σ >. Indeed, given N N, we have N N N n = n σ, and the erie n n= n= ζ(σ) = n i convergent for all σ >. Furthermore, we have the Euler product repreentation ( ) p = n. (3) To ee the above formula, note that n σ n= n z = + z + z2 + + z n + = i valid for all complex number z < (to prove it, note that if we top the um at N we get zn+ z which tend to z when N tend to infinity becaue z < ). Thu, n=0 z n ( ) p = ( + p + ) p 2 +, and if we expand the above product, then a typical term of it i p α p α k = k (p α, pα k k ) and now the concluion that the formula (3) hold follow from the Fundamental Theorem of Arithmetic. Note that formula (3) allow u to give another proof that there are infinitely many prime. Indeed, aume that there are only finitely many. Then ( ) p i bounded a we let tend to from above on the real line. However, n n i divergent, which i a contradiction. Thi argument goe back to Euler. 3

4 3 The Riemann Zeta Function when σ > 0 In the preceding ection, we defined ζ() for all σ >. In thi ection, we will extend thi definition in a nice way (here, by nice we mean continuou, differentiable, analytic, etc.) to a function defined for all C with σ > 0 except for =. Theorem. The following formula ζ() = (x x ) x + dx (4) i valid for the Riemann Zeta function whenever i a complex number with σ >. Proof. Let x be any poitive real number. Ue Abel ummation formula with a n = and f(t) = t. In thi cae, A(x) = n x a n = x and we get that n = x x x + u du. u+ Letting x, we get n x u ζ() = 0 + u + = (u (u u ) u + du du = u (u u u + du ( u ) = u= (u u ) u= u + du = x x x + dx, which i what we wanted to prove. Note that the improper integral x x x + dx converge abolutely for all σ > 0. Thu, we may jut adopt formula (4) a the definition of ζ(), and then we ee that the Riemann Zeta function can be defined for all C, with σ > 0. In thi domain, the function ζ() i very nice. It i continuou and in fact even differentiable everywhere in 4

5 the domain C with σ > 0. There are way to extend it to all the complex number in uch a way that it remain continuou and differentiable, but we hall not need thi. Being alo a function of a complex variable which i differentiable everywhere for all C with σ > 0 with, it i in fact analytic in thi domain. If you do have never een thi word, we will explain it in one of the future lecture. You might have heard of the Riemann Hypothei. Here it i: Conjecture. If ζ() = 0 for ome C with σ > 0 and, then σ = /2. It i eay to ee that ζ() = ζ(). In particular, if i real o i ζ(). Thu, the complex zero of ζ() with σ > 0 come in pair coniting of a zero and it conjugate. Thu, it uffice to look at thoe one lying in the part of the complex plane for which t 0. The Riemann Hypothei Conjecture ay that all zero of the Riemann Zeta function with σ > 0 have σ = /2. It ha been checked to be true for the firt (i.e., thoe with mallet t), 500, 000, 000 zero. It doen t look like it will be hard to prove doe it? Well, many tried and failed. If you prove it, not only do you become famou, but you alo cah in the prize of $,000,000 offered by the Clay Mathematical Intitute for a proof thi conjecture (check out the web ite the Clay Mathematical Intitute). You get no money if you find a counterexample. For u, the two mot important propertie are the following. The firt one concern the zero of ζ(). Theorem 2. The function ζ() ha no zero with σ. The econd one concern the derivative of ζ(), a a function of a complex variable. Definition 2. Let f be a function defined everywhere in ome open dik D(z; δ) = {w : w z < δ} centered at z of radiu δ. Then f ha a derivative in z if f(z + h) f(z) lim h 0 h exit. In thi cae, it i denoted f (z). If f ha a derivative everywhere in D(z, δ), then f i called holomorphic in D(z; δ). 5

6 Example. We how here that the derivative of e z exit and i itelf. For thi it uffice to how that for all fixed z, we have Note that e z+h e z h e z+h e z lim h 0 h = e z. e z = e z ( e h h Since z i fixed, it uffice to prove that Aume h <. Then, o e h h e h h h n 2 ). e h lim = 0. (5) h 0 h = h 2! + h2 3! + + hn +, h n 2 which clearly implie limit (5). < h = e h, h <, (6) n 0 All rule that you know about derivative apply here: product rule, um rule, chain rule, etc. We hall not mention them. We now prove that ζ() ha a derivative for all complex number = σ + it C, with σ > 0. Propoition. The function ζ() ha a derivative for all complex number C, with σ > 0. We hall not prove thi. 6

Riemann s Functional Equation is Not Valid and its Implication on the Riemann Hypothesis. Armando M. Evangelista Jr.

Riemann s Functional Equation is Not Valid and its Implication on the Riemann Hypothesis. Armando M. Evangelista Jr. Riemann Functional Equation i Not Valid and it Implication on the Riemann Hypothei By Armando M. Evangelita Jr. On November 4, 28 ABSTRACT Riemann functional equation wa formulated by Riemann that uppoedly