THE ZETA FUNCTION AND THE RIEMANN HYPOTHESIS. Contents 1. History 1

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1 THE ZETA FUNCTION AND THE RIEMANN HYPOTHESIS VIKTOR MOROS Abstract. The zeta fuctio has bee studied for ceturies but mathematicias are still learig about it. I this paper, I will discuss some of the zeta fuctio s properties ad itroduce the Riema hypothesis, a importat ope questio. Cotets. History. Defiitio ad basic properties 3. Euler product 4. Evaluatig zeta at particular poits 5. Aalytic cotiuatio 4 6. Riema hypothesis 5 Ackowledgmets 5 Refereces 5. History The zeta fuctio, usually referred to as the Riema zeta fuctio today, has bee studied i may differet forms for ceturies. The harmoic series, ζ(, was prove to be diverget as far back as the 4th cetury []. I the 8th cetury, the Swiss mathematicia Leohard Euler foud a closed form epressio for the sum of the reciprocals of the squared itegers i.e. ζ(. He also geeralized this result ad foud a closed form epressio for ζ( for N []. I the 9th cetury, the Germa mathematicia Berhard Riema cosidered ζ as a comple fuctio. He published his work i the 859 paper O the Number of Primes Less Tha a Give Magitude, which is oe of the most ifluetial works of moder mathematics [5]. I this paper, he cojectured that all o-trivial zeros of ζ have real part, a cojecture that has become kow as the Riema hypothesis ad whose proof is perhaps the most sought after i all of mathematics. The Riema hypothesis remais uprove ad its resolutio, either i the egative or affirmative, would have eormous cosequeces across may areas of mathematics ad sciece [3].. Defiitio ad basic properties Defiitio (Zeta fuctio. ζ( = = = = I particular, ζ( is the harmoic series ad ζ( is the subject of the Basel problem. Before cosiderig the domai of ζ, ote that if is i N ad a, b are i

2 VIKTOR MOROS R, ad a < b, the a < b. Therefore, for ay i i N, the i-th term i is greater tha the i-th term i = = = ζ(a a = ζ(b, so if both sums coverge, the former b is greater tha the latter. Therefore, i R, ζ is decreasig wherever it is defied. Propositio (Domai of ζ. The domai of ζ, cosidered as a comple fuctio, is {s C Re(s > }. Proof. First, we cosider ζ as a real-valued fuctio. We use the itegral test for covergece of ifiite sums, which tells us that = ζ( coverges if ad oly if d eists. But d eists if ad oly if > so the domai of ζ is { R > }. Now, we broade our attetio from R to C. Cosider s, a arbitrary elemet of C. The = = =. As s e s log e Re(s log Re(s show previously, coverges if ad oly if Re(s > so the domai of ζ, Re(s = = cosidered ow as a comple fuctio, is {s C Re(s > }. Euler gave this idetity: 3. Euler product ζ(s = where p k is the k-th prime umber. ( Proof. = ( p p k= k k k= =0 ( = + p = + i= + p p i p i,j= p i p j ( + p i,j,k= k= + p p s k p 3 p + = + i p j p k = ζ( Usig this idetity, we prove the ifiitude of the prime umbers. Proof. Note that ζ( is the harmoic series. So ζ( = = k= p k but we kow that the harmoic series diverges, so this product must also diverge. But for a product of positive umbers to diverge, the product must have a ifiite umber of terms, so we coclude that there are a ifiite umber of prime umbers. While the Euler product is beautiful ad surprisig, it is difficult to evaluate a ifiite product over all prime umbers, so mathematicias typically study ζ usig the summatio defiitio ad related tools, ad use this to lear about the Euler product formula. 4. Evaluatig zeta at particular poits Now we fid some values of ζ. Numerical evaluatio with a computer would be straightforward but fidig eact values is more ivolved. First we itroduce the Weierstrass factorizatio theorem, which will be useful for evaluatig ζ.

3 THE ZETA FUNCTION AND THE RIEMANN HYPOTHESIS 3 Theorem (Weierstrass Factorizatio Theorem. Let f be a etire fuctio ad let {a } be the o-zero zeros of f repeated accordig to multiplicity; suppose f has a zero at z = 0 of order m 0. The there is a etire fuctio g ad a sequece of itegers {p } such that ( z f(z = z m e g(z E p a [4, p70] Eample. Si is etire ad has a zero at z = 0 of order so we ca apply the previous theorem. Doig so gives (4. si(πz = πz ( z = = [4, p75] Now we evaluate ζ( by usig two differet formulas for si(. Propositio. ζ( = π 6. Proof. By the Taylor series of si, si( = = =0 ( ( +! = 3! + 4 5!... ad by equatio (4. si( = ( π = ( ( π 4π... By epadig the product ad equatig coefficiets for the term, we see 6 = π Therefore, ζ( = π 6. Cotiuig with this approach by equatig coefficiets of higher degree terms, as Euler did, gives = ζ( = B π (! for N [5] where B deotes the -th Beroulli umber, defied by B = k=0 k + k ( k ( r r r [9]. No closed form epressio is kow for the odd positive iteger values of ζ, but some approimate values are give i the followig table. r=0

4 4 VIKTOR MOROS Table. Some values of ζ ζ( π 6 = π 4 90 = π = Aalytic cotiuatio Defiitio (Aalytic cotiuatio. Suppose f is a aalytic fuctio o U C ad g is a aalytic fuctio o V C where V U ad V ad U are ope. Suppose further that g(z = f(z for all z U. The g is a aalytic cotiuatio of f. Ituitively, a aalytic cotiuatio of a fuctio has a larger domai tha the origial fuctio but agrees with the origial fuctio o the origial domai. A powerful theorem of comple aalysis states that if a aalytic cotiuatio eists, it is uique [6, p59]. As discussed earlier, the domai of the zeta fuctio as defied above is {s C Re(s > }. Now we will fid the aalytic cotiuatio to {s C Re(s > 0, s }. Defiitio (Alteratig zeta fuctio. Z(s = = ( + s = s + 3 s... Propositio (Epadig the domai of ζ to Re(s > 0, s. aalytic cotiuatio of ζ to {s C Re(s > 0 s }. Z(s s is a Proof. If Re(s >, ζ(s = Z(s+ ( + s 4 + s = Z(s+ ζ(s s. Solvig for s ζ gives ζ(s = Z(s [7]. But Z(s is aalytic ad coverges for {s C Re(s > 0} s so this ew formula for ζ is a aalytic cotiuatio to {s C s }. Usig further techiques of comple aalysis, ζ has bee aalytically cotiued to C \ {} ad the followig fuctioal equatio has bee prove [6, p373]. ( πs ζ(s = s π s si Γ( sζ( s From this fuctioal equatio, it ca be see that ζ( = 0 for all N because si vaishes ad all other factors eist. The zeros at egative eve itegers are called trivial zeros [6, p375]. The aalytic cotiuatio of ζ gives surprisig results such as ζ(0 = ad ζ( =. ζ(, by the origial defiitio of ζ, would be = = but this sum diverges, of course. is ot = = i the domai of ζ as origially defied but it is i the domai of the aalytic cotiuatio so evaluatig ζ( is ot the same as evaluatig the sum.

5 THE ZETA FUNCTION AND THE RIEMANN HYPOTHESIS 5 6. Riema hypothesis The Riema Hypothesis is perhaps the greatest ope questio i mathematics. The claim is simple: all o-trivial zeros of ζ have real part of [6, p375]. This hypothesis, if foud to be true, would have may powerful cosequeces, especially with regards to the distributio of prime umbers. Also, may theorems have bee prove with the assumptio that the Riema hypothesis is true, so a proof of the hypothesis would validate the proofs of these theorems. For more iformatio o the cosequeces of the Riema hypothesis, see [3] ad [8]. Ackowledgmets. It is a pleasure to thak Professor May ad my metors, Yu Cheg ad Presto Wake, for their iput ad guidace. Refereces [] Weisstei, Eric. Harmoic Series. MathWorld. Web. 4 August 04. [] Weisstei, Eric. Riema Zeta Fuctio zeta(. MathWorld. Web. 4 August 04. [3] Goodma, Le ad Weisstei, Eric. Riema Hypothesis. MathWorld. Web. 4 August 04. [4] Coway, Joh. Fuctios of Oe Comple Variable. Prit. [5] Sodow, Joatha ad Weisstei, Eric. Riema Zeta Fuctio. MathWorld. Web. 5 August 04. [6] Gameli, Theodore. Comple Aalysis. Prit. [7] Riema zeta fuctio. Art of Problem Solvig. Web. 5 August 04. [8] Cosequeces of the geeralized Riema hypothesis. Wikipedia. Web. 5 August 04. [9] Weisstei, Eric. Beroulli Number. MathWorld. Web. 9 September 04.