University of Southern California Dissertations and Theses

A development of some of the basic properties of the Riemann Zeta function

(USC Thesis Other)

A development of some of the basic properties of the Riemann Zeta function

PDF

Download a page range

Download transcript

Copy asset link

Request this asset

Transcript (if available)

Content A DEVELOPMENT OF SOME OF THE BASIC PROPERTIES OF THE RIEMANN ZETA FUNCTION by Branden Neese A Dissertation Presented to the FACULTY OF THE GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree MASTER OF ARTS (MATHEMATICS) December 2008 Copyright 2008 Branden Neese Dedication This work is dedicated to my loving friend Ko Hong Fu. ii Acknowledgements Many thanks to my advisor, Dr. Sheldon Kamienny, for our meetings and his support. I would also like to thank Dr. Igor Kukavica and Dr. Nicolai Haydn for their time. Thank you to Koo Tak-Lun, whose advice helped to make this paper possible. Special thanks to Scott Baumann. iii Table of Contents Dedication ii Acknowledgements iii Abstract v Chapter 1 Introduction 1 1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 The Riemann Zeta Function . . . . . . . . . . . . . . . . . . . . . . 13 1.4 A Functional Equation and the Riemann Hypothesis . . . . . . . . . . 19 Chapter 2 The Prime Number Theorem 28 References 39 iv Abstract I have adapted a development of the basic properties of the Riemann Zeta function. Chapter one includes the Euler product, a functional equation, and an analytic contin- uation. Chapter two shows the equivalence of the Prime Number Theorem with the fact that the Zeta function has no zeros with real part equal to one. v Chapter 1 Introduction 1.1 Background 1 Definition. A sequencefa n g 1 n=1 of complex numbers is said to CONVERGE if there exists a complex numbera such that 0<2R)9N2N ,which depends on, such thatN <n2N)ja n aj<. In this case we say the limit asn goes to infinity of a n EXISTS [9]. We write lim n!1 a n = a, ora n ! a asn!1. Iffa n g 1 n=1 is a sequence of real numbers such thatM 2 N)9N 2 N, which depends onM, such thatN n2 N) a n > M we write a n !1 as n!1 even though this sequence does not converge [9]. 2 Definition (Big O, Little o). Letfa n g 1 n=1 andfb n g 1 n=1 be sequences of complex numbers. If there existsN2N and 0<K2R such thatN <n2N) an bn <K, we write a n =O(b n ): 1 If lim n!1 an bn = 0 we write a n =o(b n ) [12]: 3 Definition. A series P 1 n=1 a n of complex numbers is said to CONVERGE if the sequencefS N = P N n=1 a n g 1 N=1 converges. Otherwise, the series is said to DIVERGE. A series P 1 n=1 a n of complex numbers is said to CONVERGE ABSOLUTELY if P 1 n=1 ja n j converges [9]. 1 Result (p-test). P 1 n=1 n p converges ifp> 1 and diverges ifp 1 [9]. 2 Result (Geometric Series). For 0<< 1, the geometric series P 1 n=0 n converges to (1) 1 [9]. 4 Definition. A sequenceff n g 1 n=1 of functions from a set E intoC is said to CON- VERGE UNIFORMLY onE to a functionf :E!C if 0<2R)9N2N ,which depends on, such that [N <n2N;x2E])jf n (x)f(x)j< [9]. A series P 1 n=1 f n , where n 2 N ) f n is a function from E into C is said to CONVERGE UNIFORMLY on E if there exists a function f : E ! C such that the sequencefS N = P N n=1 f n g 1 N=1 of partial sums converges uniformly onE tof [9]. 5 Definition. A REGION is defined to be a nonempty connected open subset ofC [1]. I will denote the regionfs2C : Re(s)> 1g by . The following criterion, due to Weierstrass, is useful for showing that a series con- verges uniformly. 3 Result. Supposeff n g 1 n=1 is a sequence of complex-valued functions on a set E. Suppose further thatn2 N)9 0 < M n 2 R such thatx2 E)jf n (x)j M n . Then P 1 n=1 M n converges) P 1 n=1 f n converges uniformly onE [9]. 2 4 Result. Furthermore, if a series with analytic termsf(z) = P 1 n=1 f n (z) converges uniformly on every compact subset of a region, then the sumf(z) is analytic in that region, and the series can be differentiated term by term [1]. 6 Definition. An infinite product of complex numbers Q 1 n=1 a n is said to CONVERGE if the sequence of partial productsfp N = Q N n=1 a n g 1 N=1 converges to a number 06= a2C [7]. We have the following criterion for the convergence of a product. 5 Result. The infinite product Q 1 n=1 (1 +a n ) withn2 N) 1 +a n 6= 0 converges () P 1 n=1 log(1 +a n ) converges. In this case 1 Y n=1 (1 +a n ) =e P 1 n=1 log(1+an) [7]: 7 Definition. If, in an infinite product, only a finite number of factors are zero, and if the product over the nonzero factors converges, the original product is said to CON- VERGE TO ZERO [12]. 8 Definition. A product Q 1 n=1 (a n ) where either lim N!1 f Q N n=1 a n g 1 N=1 = 0 or lim N!1 f Q N n=1 a n g 1 N=1 fails to exist is said to be DIVERGENT [7]. 1 Remark. Q 1 n=1 a n is divergent ifa n = 0 for infinitely manyn. 9 Definition. A product Q 1 n=1 (1 + a n ) is said to be ABSOLUTELY CONVERGENT provided Q 1 n=1 (1 +ja n j) is convergent [7]. 6 Result. For any complex sequencefa n g 1 n=1 tending to zero, the product Q 1 n=1 (1 + a n ) converges absolutely () P 1 n=1 ja n j converges [7]. 3 10 Definition. LetA C. Letff n g 1 n=1 be a sequence of complex-valued functions such that the domain-set of eachf n containsA. Then Q 1 n=1 f n is said to be POINT- WISE CONVERGENT IN A if for each fixedz inA, the product Q 1 n=1 f n (z) converges (possibly to zero) [7]. 2 Remark. In this case we have a functionf : A! C given byz7! Q 1 n=1 f n (z). Forz2 A,f(z) = 0 () 9n2N such thatf n (z) = 0. In this case there are only finitely many suchn by remark 1 [7]. 11 Definition. An infinite product of functions Q 1 n=1 g n is said to CONVERGE UNI- FORMLY ON A SET A if (i) Q 1 n=1 g n is pointwise convergent inA (ii)9N2N such that [N <n2N ,z2A])g n (z)6= 0 andfg N+1 g N+n g 1 n=1 converges uniformly on A to Q 1 n=N+1 g n [7]. 7 Result. Let E be a compact subset ofC and letfg n g 1 n=1 be a sequence of continuous functions from E intoC such that P 1 n=1 g n (z) converges absolutely for z2 E and P 1 n=1 g n converges uniformly onE. Then 1 Y n=1 (1 +g n (z)) converges uniformly and absolutely onE [5]. 8 Result. LetG be a region andfg n g 1 n=1 a sequence of analytic functions onG such that nog n is identically zero. If 1 X n=1 [g n (z) 1] 4 converges absolutely and uniformly on compact subsets of G, then 1 Y n=1 g n (z) converges to an analytic functiong onG. Ifa is a zero ofg, then it is a zero of only finitely manyg n . The multiplicity of the zero ofg ata is the sum of the multiplicities of the zeros of theg n s ata [5] . 12 Definition. LetG be a region. Leta2R and suppose thatf : [a;1)G!C is a continuous function. Define the integralF (z) = R 1 a f(t;z)dt to be UNIFORMLY CONVERGENT ON COMPACT SUBSETS of G if lim b!1 R b a f(t;z)dt exists uniformly for z in any compact subset ofG [5]. 9 Result. Suppose that the integral F (z) = R 1 a f(t;z)dt converges uniformly on compact subsets ofG and that for eacht2 (a;1); f(t;) :G!C is analytic. Then F is analytic andF 0 (z) = R 1 a d dz f(t;z)dt [5]. The following is De la Vall´ ee Poussin’s test for uniformity of convergence of an infinite integral. 10 Result. R 1 a f(x;)dx converges uniformly with regard to in a domain of values of ifjf(x;)j<(x), where is independent of and R 1 a (x)dx converges [12]. 13 Definition. Let a 2 R. Suppose a < b 2 R ) f is Riemann-integrable on [a;b]. Then R 1 a f(x)dx is defined to be lim b!1 R b a f(x)dx if this limit exists (and is finite) [9]. The following is the integral test for convergence of series. 11 Result. Assume thatf(x) 0 and thatf decreases monotonically on [1;1). Then R 1 1 f(x)dx converges if and only if P 1 n=1 f(n) converges [9]. 5 12 Result (Dominated Convergence Theorem). Letff n g be a sequence inL 1 such that (i) f n !f a:e., and (ii) There exists a nonnegativeg2L 1 such thatjf n jga:e: for alln. Thenf2L 1 and R f n = lim n!1 R f n [6]. The following result will be used. 13 Result. Suppose thatff j g 1 j=1 is a sequence inL 1 such that P 1 j=1 R jf j j<1. Then P 1 j=1 f j convergesa:e: to a function inL 1 , and RP f j = PR f j [6]. 14 Definition (Fourier Transform). Forf2L 1 (R), the FOURIER TRANSFORM off is b f2BC(R) given by b f(m) = Z R f(y)e 2imy dy [6]: 14 Result (Poisson Summation Formula). Suppose thatf : R! R is an integrable continuous function of bounded total variation. Suppose that there exists a second functiong : R! R + so thatg is even, decreasing onR + , and integrable. Suppose also thatt2R)jf(t)jg(t). Then X n2Z f(t +n) = lim M!1 X fm2Z:jmjMg b f(m)e 2imt : If the right hand side is absolutely convergent, we have X n2Z f(t +n) = X m2Z b f(m)e 2imt : Takingt = 0 gives X n2Z f(n) = X m2Z b f(m) [8]: (1.1) 6 1.2 The Gamma Function 1 Lemma (Euler’s Constant). lim m!1 P m n=1 1 n log(m) exists. Proof. Forn2N, put u n = 1 n Z 1 0 t t +n dt = 1 n t log(t +n)j 1 0 Z 1 0 log(n +t)dt = 1 n log(1 +n) y log(y) y=n+1 y=n Z n+1 n dy = 1 n log(1 +n) (n + 1) log(n + 1)n log(n) (n + 1) +n = 1 n log(n + 1) (n + 1) log(n + 1) +n log(n) + 1 = log(n + 1) + log(n) + 1 n = 1 n log( n + 1 n ): Thenu n is positive and less than R 1 0 1 n 2 dt = 1 n 2 , so that1> P 1 n=1 1 n 2 > P 1 n=1 u n by the p-test withp = 2. Note: m X n=1 u n = m X n=1 log(n + 1) + log(n) + 1 n = m X n=1 ( 1 n ) log(m + 1): 7 Then lim x!1+ log(x) = 0) lim m!1 " m X n=1 1 n log(m) # = lim m!1 " m X n=1 u n + log(m + 1) log(m) # = lim m!1 " m X n=1 u n + log( m + 1 m ) # = 1 X n=1 u n ; so that this last limit exists. It is called Euler’s constant and is 0:5772157. It is denoted [12]. 2 Lemma. Forw2C withjwj 1, je w 1jjwje Proof. Letw2C withjwj 1, je w 1j = 1 X n=0 w n n! 1 = 1 X n=1 w n n! jwj 1 X n=1 w n1 n! =jwj 1 + jwj 2! + jwj 2 3! + jwj 3 4! +::: jwj 1 + 1 1! + 1 2! + 1 3! +::: =jwje: (1.2) 8 1 Proposition. z7!ze z 1 Y n=1 h (1 + z n )e z n i (1.3) is entire; it has a simple zero at each of the non-positive integers and no other zeros. Proof. Let E be a compact subset ofC. 9N 2 N such that z2 E )jzj < 1 2 N. log(z) is analytic in Cf(1; 0]g. Its power series expansion about z 0 = 1 is log(z) = P 1 n=1 (1) n1 (z1) n n . Now [N <n2N;z2C withjzj< 1 2 N]) log(1 + z n ) z n = 1 2 z 2 n 2 + 1 3 z 3 n 3 1 4 z 4 n 4 +::: jzj 2 n 2 n 1 +j z n j +j z 2 n 2 j +::: o 1 4 N 2 n 2 n 1 + 1 2 + 1 2 2 +::: o 1 2 N 2 n 2 : (1.4) Forn2N, put f n (z) =e log(1+ z n ) z n = (1 + z n )e z n : Then lemma 2 combined with (1:4) gives that [N <n2N;z2C withjzj< 1 2 N]) jf n (z) 1j 1 2 N 2 n 2 e: Thus by the p-test combined with results 3 and 7, 1 Y n=N+1 h (1 + z n )e z n i 9 converges absolutely and uniformly onE. Finally, sinceE was arbitrary, result 8 gives that z7!ze z 1 Y n=1 h (1 + z n )e z n i is entire; it has a simple zero at each of the non-positive integers and no other zeros [12]. 15 Definition (Gamma Function). We now have the gamma function (z) = e z z 1 Y n=1 h (1 + z n ) 1 e z n i : By proposition 1, it is meromorphic in the whole complex plane with simple poles at the non-positive integers and no zeros [12]. 3 Lemma. z2Cf0;1;2;3;:::g) lim n!1 n z n! z(z + 1)::: (z +n) =(z) (1.5) Proof. Forz2Cf0;1;2;3;:::g, (z) = 1 z e z limn!1(logn P n k=1 1 k ) lim n!1 n Y k=1 h (1 + z k ) 1 e z k i = lim n!1 n 1 z e z(logn P n k=1 1 k ) n Y k=1 k z +k e z k o = lim n!1 n z n! z(z + 1)::: (z +n) [7]: 10 3 Remark. Then forz2Cf0;1;2;3;:::g, (z + 1) = lim n!1 n!n z+1 (z + 1)(z + 2)::: (z +n)(z +n + 1) = lim n!1 nz z +n + 1 n!n z z(z + 1)::: (z +n) =z(z): Immediately (1) = lim n!1 n!n (n+1)! = lim n!1 n n+1 = 1. So forn2N, the functional equation for gives (n) = (n 1)! [7]. 4 Lemma. Forz2Cwith Re(z)> 0, lim n!1 Z n 0 t z1 1 t n n dt = (z) (1.6) Proof. Making the substitutiont =ns gives 11 Z n 0 t z1 1 t n n dt = Z 1 0 (ns) z1 1 ns n n nds =n z Z 1 0 s z1 (1s) n ds =n z h f(1s) n s z z g 1 0 Z 1 0 s z z n(1s) n1 (1)ds i =n z n z Z 1 0 s z (1s) n1 ds =n z n z h (1s) n1 s z+1 z + 1 1 0 Z 1 0 s z+1 z + 1 (n 1)(1s) n2 (1)ds i =n z n(n 1) z(z + 1) Z 1 0 s z+1 (1s) n2 ds =::: =n z n(n 1)::: (1) z(z + 1)::: (z + (n 1)) Z 1 0 s z+n1 ds = n z n! z(z + 1)::: (z +n 1) s z+n z +n 1 0 = n z n! z(z + 1)::: (z +n) : Then lim n!1 Z n 0 t z1 1 t n n dt = (z) [12]: 2 Proposition. Forz2Cwith Re(z)> 0, Z 1 0 t z1 e t dt = (z) (1.7) Proof. Letz2 C with Re(z) > 0. Let [0;n] be the characteristic function of [0;n]. Fort2 (0;1), letf n (t) = [0;n] t z1 (1 t n ) n . Thenff n g is a sequence inL 1 (0;1). Putf(t) =t z1 e t . Thenf n !f a:e: on (0;1). (1 t n ) n [0;n] !e t asn!1 also 12 implies there existsN2N such thatN <n2N;t2 (0;1) =) [0;n] (1 t n ) n e t 2 . Then N < n 2 N;t 2 (0;1) =) [0;n] (1 t n ) n t z1 e t 2 t z1 2 L 1 (0;1). Then the DCT implies lim n!1 Z 1 0 [0;n] t z1 1 t n n dt = Z 1 0 lim n!1 [0;n]t z1 1 t n n dt = Z 1 0 t z1 e t dt Then lemma 4 gives the desired result. 1.3 The Riemann Zeta Function 16 Definition. The Riemann Zeta function : !C is given bys7! P 1 n=1 n s . 3 Proposition. s7! P 1 n=1 n s is well defined and analytic on . Proof. Let E be a compact subset of . Then9 1 < 0 2 R such that Efs2 C : Re(s) 0 g. Now, [n 2 N; ;t 2 R] ) jn (+it) j = je (+it) logn j = e logn p cos 2 (t logn) + sin 2 (t logn) = n . For n2 N, define f n : ! C by s7! n s . Then s2 E )jf n (s)j n 0 . Then by the Weierstrass criterion P 1 n=1 f n converges uniformly on E. Since E was an arbitrary compact subset of , this series converges uniformly on every compact subset of . z7!e z entire) [n2N)f n analytic on ]. Thus by result 4 the Riemann Zeta function : !C given bys7! P 1 n=1 n s is defined and analytic on . 4 Proposition (Euler Product). Fors2 , (s) = 1 Y n=1 (1p s n ) 1 (1.8) 13 wherefp n g 1 n=1 is an enumeration of the primes. Proof. Letfp n g 1 n=1 be an enumeration of the primes in increasing order. Let 1< 0 2 R. Then 0 2 R) n n 0 , and P 1 n=1 p 0 n < P 1 n=1 n 0 <1 by the p-test. Then by result 7 Q 1 n=1 (1p s n ) converges uniformly and absolutely on fs2C :Re(s) 0 g, and result 8 gives 1 Y n=1 (1p s n ) defines an analytic function on . Fors2 , (s)(1 2 s ) = 1 X n=1 n s 1 X n=1 (2n) s = X m s ; wherem ranges over all positive integers not divisible by 2. Completely analogously, (s)(1p s 1 )::: (1p s n ) = P m s , where m ranges over all positive integers coprime to each ofp 1 ;:::;p n . The first term in this sum is 1, and the next term isp s n+1 . lim n!1 1 X m=n+1 p s m = 0 since P 1 m=1 p s m converges. Thus lim N!1 (s) N Y n=1 (1p s n ) = 1; and (s) = 1 Y n=1 (1p s n ) 1 [1]: 14 17 Definition (The von Mangoldt function). :N!R is defined by (n) = 8 > > < > > : log(p) ifn =p l for some primep,l2N; 0 otherwise. 5 Lemma. Fors2 , 0 (s) (s) = X n2N (n)n s : (1.9) Proof. By the Euler Product,s2 )(s)6= 0 so that log((s)) = 1 X n=1 log(1e s log(pn) ); so that 0 (s) (s) = 1 X n=1 log(p n )p s n (1p s n ) 1 = 1 X n=1 [log(p n )p s n 1 X m=0 p sm n ] = X n2N;m2N log(p n )p sm n : Then 0 (s) (s) = X n2N (n)n s [8]: (1.10) 5 Proposition. There exists an analytic continuation of to a meromorphic function onfs2C :Re(s)> 0g. In this region, has a simple pole of residue 1 ats = 1. Proof. Forx2R, denote by [x] the integer that satisfiesx 1< [x]x. Lets2 . Then 15 s Z N+1 1 [x]x 1s dx =s N X n=1 n Z n+1 n x 1s dx = N X n=1 n(n s (n + 1) s ) =1(1 s 2 s ) + 2(2 s 3 s ) + 3(3 s 4 s ) + +N(N s (N + 1) s )) =( N X n=1 n s )N(N + 1) s : AndN(N + 1) s ! 0 asN!1, so that fors2 , (s) =s Z 1 1 [x]x s1 dx =s Z 1 1 ([x]x)x 1s dx +s Z 1 1 x s dx = Z 1 1 ([x]x)x 1s dx + x s+1 s + 1 j 1 1 =s Z 1 1 ([x]x)x 1s dx + s s 1 : (1.11) Forx2 [1;1);j([x]x)j< 1: LetE be a compact subset offs2C : Re(s)> 0g. Then9 0 < 0 2 R such thats2 E) Re(s) 0 . Then [x2 [1;1);s2 E]) j([x]x)x 1s j x 1 0 . R 1 1 x 1 0 dx converges by the integral test combined with the p-test withp =1 0 , so that by results 9 and 10 R 1 1 ([x]x)x 1s dx defines a holomorphic function onfs 2 C : Re(s) > 0g. Thus, (1.11) gives an analytic continuation of to this region, which has a simple pole of residue 1 ats = 1 [10]. 6 Proposition. has no zeros with real part equal to one. 16 Proof. Let 0<< 1. We have (1) 1 = P 1 n=0 n . Whereby log(1) = Z (1) 1 d = Z 1 X n=0 n d = 1 X n=1 n n : Fors2 ,(s) = Q 1 n=1 (1p s n ) 1 ) log(s) = log 1 Y n=1 (1p s n ) 1 ! = 1 X n=1 log(1p s n ) 1 = 1 X n=1 1 X m=1 p sm n m ! : Now, for a complex number!, log(!) is defined to be logj!j+iArg!. Fix 06=t2R. Let 0<2R: Then logj( +it)j =Re log(( +it)) =Re ( X n;m2N m 1 p (+it)m n ) = X n;m2N m 1 p m n cos(tm logp n ) = X n;m2N m 1 p m n cos(tm logp n ) [11]: 17 We get logj 3 () 4 (+it)(+2it)j = X n;m2N m 1 p m n 3+4 cos(tm logp n )+cos(2tm logp n ) : (1.12) For x 2 R, we have cos(2x) = 2 cos 2 (x) 1, so 3 + 4 cos(x) + cos(2x) = 2(1 + cos(x)) 2 0. Then logj 3 () 4 ( +it)( + 2it)j 0: Also, forx> 0, [log(x) 0)x 1], so that j 3 () 4 ( +it)( + 2it)j 1: And j()( 1)j 3 ( +it) 1 4 j( + 2it)j 1 1 : Now if (1 + it) were zero, the LHS would tend to a finite limit, namelyj 0 (1 + it)jj(1 + 2it)j 4 , as ! 1+. This is because has a simple pole of residue 1 at s = 1, so that lim !1+ j( 1)()j = 1. Also, l’hopital’s rule, with the assumption that (1 +it) = 0, gives lim !1+ j (+it) 1 j 4 =j 0 (1 +it)j 4 . Finally, we know from the analytic continuation that is holomorphic in a neighborhood of 1 + 2it. Yet lim !1+ 1 1 =1. So (1 +it)6= 0, and has no zeros with real part equal to one [4]. 18 1.4 A Functional Equation and the Riemann Hypothe- sis 6 Lemma. For 0<x2R, x 1 2 X k2Z e xk 2 = X m2Z e m 2 x : (1.13) Proof. Let 0<x2R. Fork2R, putf(k) =e xk 2 . Now, d dm b f(m) = d dm Z R f(y)e 2iym dy = Z R f(y)(2iy)e 2iym dy = k7!f(k)(2ik) b(m) = k7! i x d dk f(k) b(m) = i x Z R f 0 (y)e 2imy dy = i x f(y)e 2imy y=1 y=1 Z R f(y)(2im)e 2imy dy = i x (2im) b f(m) = 2m x b f(m): Then d dm e m 2 x b f(m) = 2m x e m 2 x b f(m) +e m 2 x 2m x b f(m) = 0: 19 Thene m 2 x b f(m) is a constant, so that e m 2 x b f(m) =e 0 2 x b f(0) = Z 1 1 e xy 2 dy = Z 1 1 Z 1 1 e x(y 2 +t 2 ) dydt 1 2 = Z 2 0 Z 1 0 e xr 2 rdrd 1 2 = Z 2 0 1 2x d 1 2 =x 1 2 : Thus b f(m) =x 1 2 e m 2 x , and X m2Z j b f(m)j = X m2Z x 1 2 e m 2 x <1: Then the Poisson Summation Formula gives X k2Z f(k) = X m2Z b f(m): Finally, x 1 2 X k2Z e xk 2 = X m2Z e m 2 x [6]: (1.14) 7 Lemma. Forx > 0, Let!(x) = P 1 n=1 e n 2 x . Then R 1 1 (x s 2 1 +x s 2 1 2 )!(x)dx defines an analytic function inC. 20 Proof. Forx2 [1;1), !(x) = 1 X n=1 e n 2 x < 1 X n=1 e nx = 1 + 1 X n=0 e nx = 1e x 1e x + 1 1e x e x (1e ) 1 [12]: LetE be a compact subset ofC.9 0 ; 1 2R such thats2 E) 0 Re(s) 1 . Then [x2 [1;1);s2E]) (x s 2 1 +x s 2 1 2 )!(x) x 1 2 1 +x 0 2 1 2 e x (1e ) 1 : There exist constantsC 0 ;C 1 2R such that x>C 0 )e x <x 0 2 1 ; x>C 1 )e x <x 1 2 1 2 : 21 Z 1 1 (x 1 2 1 +x 0 2 1 2 )e x (1e ) 1 dx = Z 1 1 (x 1 2 1 )e x (1e ) 1 dx + Z 1 1 (x 0 2 1 2 )e x (1e ) 1 dx < Z C 1 1 (x 1 2 1 )e x (1e ) 1 dx + Z 1 C 1 (x 1 2 1 )x 1 2 1 2 (1e ) 1 dx + Z C 0 1 (x 0 2 1 2 )e x (1e ) 1 dx + Z 1 C 0 (x 0 2 1 2 )x 0 2 1 (1e ) 1 dx = Z C 1 1 (x 1 2 1 )e x (1e ) 1 dx + Z 1 C 1 x 3 2 (1e ) 1 dx + Z C 0 1 (x 0 2 1 2 )e x (1e ) 1 dx + Z 1 C 0 (x 3 2 )(1e ) 1 dx <1: Then by results 9 and 10, R 1 1 (x s 2 1 +x s 2 1 2 )!(x)dx defines an analytic function inC. 7 Proposition. ForRe(s)> 1,(s) s 2 ( s 2 ) = 1 s(s1) + R 1 1 (x s 2 1 +x s 2 1 2 )!(x)dx. 22 Proof. For =Re(s)> 0, the gamma function (s) = R 1 0 e t t s1 dt converges and defines an analytic function. Making the substitutiont =n 2 x gives s 2 = Z 1 0 e t t s 2 1 dt = Z 1 0 e n 2 x (n 2 x) s 2 1 n 2 dx =(n 2 ) s 2 Z 1 0 x s 2 1 e n 2 x dx; whereby n s s 2 s 2 = Z 1 0 x s 2 1 e n 2 x dx: The functionx7!x 2 e n 2 x 2 defined on (0;1) has derivative x7! 2 x 2 1 e n 2 x 2 +x 2 e n 2 x 2 n 2 2 ; so that it is maximized when 0 = 2 x 2 1 e n 2 x 2 +x 2 e n 2 x 2 n 2 2 () 0 = 2 +x n 2 2 () x = n 2 : 23 Then 1 X n=1 Z 1 0 x 2 1 e n 2 x dx = 1 X n=1 x 2 2 e n 2 x 1 0 Z 1 0 x 2 2 (n 2 )e n 2 x dx = 1 X n=1 n 2 2 Z 1 0 e n 2 x 2 x 2 e n 2 x 2 dx < 1 X n=1 n 2 2 Max n e n 2 x 2 x 2 :x2 (0;1) o Z 1 0 e n 2 x 2 dx = 1 X n=1 n 2 2 (e n 2 x 2 x 2 ) x=(n 2 ) 1 2 n 2 = 1 X n=1 n 2 2 e 2 n 2 2 2 n 2 = 1 X n=1 4 e 2 2 2 n <1 by the p-test when> 1. Let 1< Re(s). Then result 13) (s) s 2 s 2 = Z 1 0 x s 2 1 1 X n=1 e n 2 x dx = Z y=1 y=0 y s 2 1 !(y)dy + Z x=1 x=1 x s 2 1 !(x)dx: Making the substitutiony = 1 x , we have Z y=1 y=0 y s 2 1 !(y)dy = Z 1 x=1 x s 2 +1 !(x 1 )(x 2 )dx = Z 1 1 x s 2 1 !(x 1 )dx; 24 so that (s) s 2 ( s 2 ) = Z 1 1 x s 2 1 !(x 1 )dx + Z 1 1 x s 2 1 !(x)dx: Put 1 + 2!(x) = P n=1 n=1 e n 2 x =(x). We have by lemma 6 that !(x 1 ) = 1 2 [(x 1 ) 1] = 1 2 [x 1 2 (x) 1] = 1 2 [x 1 2 (1 + 2!(x)) 1]; so that Z 1 1 x s 2 1 !(x 1 )dx = Z 1 1 1 2 x s 2 1 dx + Z 1 1 1 2 x s 2 1 2 dx + Z 1 1 x s 2 1 2 !(x)dx = 1 s + 1 s 1 + Z 1 1 x s 2 1 2 !(x)dx: Then (s) s 2 s 2 = 1 s(s 1) + Z 1 1 (x s 2 1 +x s 2 1 2 )!(x)dx [12]: (1.15) By lemma 7, the right hand side of (1.15) defines a meromorphic function inC, and this provides an analytic continuation of to a meromorphic function inC. 1 Theorem (Functional equation). (s) s 2 s 2 =(1s) (1s) 2 1s 2 : (1.16) 25 Proof. 1 s(s1) + R 1 1 (x s 2 1 +x s 2 1 2 )!(x)dx is invariant unders7! (1s), so that (s) s 2 s 2 =(1s) (1s) 2 1s 2 : (1.17) Since has no zeros in Re(s) 1, this functional equation gives that the only zeros that has in Re(s) 0 are at the negative even integers, where ( s 2 )has simple poles. These are called the trivial zeros. This functional equation also gives that the nontrivial zeros of lie symmetrically about the line Re(s) = 1 2 . The Riemann Hypothesis is the statement that all nontrivial zeros of have real part equal to 1 2 . Put (s) = 1 2 s ( s 2 ) ( 1s 2 ) . Then (1s) = (s)(s) [8]: Lemma 7 and 1.15 give that s 2 s(1s)( s 2 )(s) is entire. Then, sinces s 2 has no zeros, only has the pole ats = 1, which has already been discussed. 18 Definition. ForT > 0 and> 1 2 ,N(;T ) is the number of zeros of( +it) such that >, and 0t<T [8]. The Riemann hypothesis can be formulated as asserting thatT > 0 and > 1 2 ) N(;T ) = 0. The Lindelof Hypothesis states that for any > 0, ( 1 2 +it) = O(t ) as t! 1 [3]. 2 Theorem. The Lindelof hypothesis is equivalent to the statement that for every> 1 2 one has N(;T + 1)N(;T ) =o(logT )asT!1 [3] 26 Then the truth of the Riemann Hypothesis implies that of the Lindelof Hypothesis [8]. 19 Definition (Dirichlet character). A function : Z! C is called a DIRICHLET CHARACTER if (i)9k2N such thatn2Z)(n) =(n +k) (ii) Forn2Z, gcd(n;k) = 1)(n)6= 0; gcd(n;k)> 1)(n) = 0 (iii) m;n2Z)(mn) =(m)(n) [3]. 20 Definition (Dirichlet L-function). Given a Dirichlet character, the corresponding DIRICHLET L-FUNCTION, defined onfs2C : Re(s)> 1g, is given by L(;s) = 1 X n=1 (n) n s [3]: L(;s) can be analytically continued to a meromorphic function defined onC. The Generalized Riemann Hypothesis is the statement that given a Dirichlet character, [s2C; 0 Re(s) 1; L(;s) = 0]) Re(s) = 1 2 [3]: In 1997 it was proved that assuming the Generalized Riemann Hypothesis, every odd number n > 5 can be expressed as a sum of three prime numbers. This was a strengthening of a 1922 result concerning Goldbach’s conjecture [3]. 27 Chapter 2 The Prime Number Theorem For 1 x2 R, (x) denotes the number of primesp satisfyingp x. The Prime Number theorem is the statement that lim x!1 (x) log(x) x = 1 [10]: (2.1) 21 Definition. For 0<x2R, define (x) = X nx (n); where is the von Mangoldt function defined earlier [10]. 8 Proposition. lim x!1 (x) x = 1) the prime number theroem. Proof. [2x2R;p prime])9!m2N, which depends onx andp, such that p m x<p m+1 : 28 Thenm logx logp , so thatm = h logx logp i is the number of powers ofp that do not exceed x. Then (x) = X p prime;px logx logp logp X p prime;px logx =(x) logx [2]: (2.2) Also, 1<y<x) (x)(y) = X p primey<px 1 X p primey<px logp logy (x) logy : Then(x)<y + (x) logy . Takingy = x log 2 (x) gives (x)< x log 2 (x) + (x) log( x log 2 (x) ) ; so that by (2.2) (x) x (x) logx x < 1 logx + (x) logx x log x log 2 x : 29 Then lim x!1 (x) x = 1) lim x!1 " 1 logx + (x) logx x log( x log 2 x ) # = lim x!1 logx log( x log 2 x ) = lim x!1 x 1 log 2 x x ( log 2 x x2logx x log 4 x ) = lim x!1 x 1 logx2 x logx = lim x!1 logx logx 2 =1: Then lim x!1 (x) x = 1) the prime number theroem [10]. 8 Lemma (Abel’s Summation Formula). Let 0 1 2 3 ::: be a sequence of real numbers such that lim n!1 n =1. Letfa n g 1 n=1 be a sequence of complex numbers. LetA(x) = P nx a n , and'(x) a complex-valued function defined forx 0. Then P k n=1 a n '( n ) = A( k )'( k ) P k1 n=1 A( n )f'( n+1 )'( n )g. If' has a continuous derivative in (0;1) andx 1 , then P nx a n '( n ) = A(x)'(x) R x 1 A(t)' 0 (t)dt. If, in addition, A(x)'(x)! 0 as x!1, then P 1 n=1 a n '( n ) = R 1 1 A(t)' 0 (t)dt provided that either side is convergent. Proof. DefineA( 0 ) = 0. Then k X n=1 a n '( n ) = k X n=1 h A( n )A( n1 ) i '( n ) =A( k )'( k ) k1 X n=1 A( n ) h '( n+1 )'( n ) i : 30 Now assume' has a continuous derivative. Letx 1 , and letk be the largest integer such that k x. Then X nx a n '( n ) = k X n=1 a n '( n ) =A( k )'( k ) k1 X n=1 A( n ) '( n+1 )'( n ) =A(x)'(x) Z x k A(t)' 0 (t)dt k1 X n=1 A( n ) Z n+1 n ' 0 (t)dt =A(x)'(x) Z x k A(t)' 0 (t)dt Z k 1 A(t)' 0 (t)dt =A(x)'(x) Z x 1 A(t)' 0 (t)dt: WhenA(x)'(x)! 0 asx!1, we have P 1 n=1 a n '( n ) = R 1 1 A(t)' 0 (t)dt [4]. 9 Proposition. Fors2 , Z 1 0 (e x )e xs dx = 0 (s) s(s) ; Proof. Let s2 . In Abel’s summation formula, taking n = n, a n = (n), and '(x) =x s givesA(x) = (x) so that 1 X n=1 (n) n s = Z 1 1 (s)x s1 (x)dx By 2.2, (x)x s (x) log(x)x s <x log(x)x s =x 1s logx ! 0 asx!1: 31 Then by lemma 5 0 (s) (s) = 1 X n=1 (n) n s = Z 1 1 (s)x s1 (x)dx = Z 1 1 (x) x s+1 dx = Z 1 0 (e x )e xs dx [4]: 9 Lemma. Iff(s) has a pole of orderk ats = , then the quotient f 0 (s) f(s) has a first order pole ats = with residuek. Proof. By hypothesis we havef(s) = g(s) (s) k whereg is analytic in a neighborhood of andg()6= 0. Then fors in a neighborhood of, we have f 0 (s) = g 0 (s)(s) k g(s)k(s) k1 (s) 2k = g 0 (s) (s) k kg(s) (s) k+1 = g(s) (s) k k s + g 0 (s) g(s) : So that f 0 (s) f(s) = k s + g 0 (s) g(s) [2]: 15 Result ( Wiener-Ikehara Theorem ). LetA(x) be a non-negative, non-decreasing function ofx, defined for 0x<1. Fors = +it, let the integral R 1 0 A(x)e xs dx converge for > 1 to the functionf(s). Letf(s) be analytic for 1, except for a simple pole ats = 1 with residue 1. Then lim x!1 e x A(x) = 1 [4]. 32 10 Proposition. lim x!1 (x) x = 1: Proof. We know that has a simple pole of residue one ats = 1. Then by lemma 9 s7! 0 (s) (s) has a simple pole ats = 1 of residue1. Finallys7! 0 (s) s(s) has a simple pole of residue 1 ats = 1. Forx 0, (e x ) 0, and is non-decreasing. TakeA(x) = (e x ) in result 15. we have from proposition 9 that fors2 , Z 1 0 (e x )e xs dx converges to 0 (s) s(s) ; which has a simple pole of residue 1 ats = 1. Due to the fact that has no zeros with real part equal to one, 0 (s) s(s) is analytic infs2C : Re(s) 1g except for the pole at s = 1. We thus have lim x!1 e x (e x ) = 1; and lim x!1 (x) x = 1 [4]: 22 Definition. Define# : (0;1)!R by x7! X p prime;px log(p) [2]: 11 Proposition. lim x!1 #(x) x = 1) lim x!1 (x) x = 1. 33 Proof. We have (x) = X nx (n) = 1 X m=1 X p prime;p m x (p m ) = 1 X m=1 X p prime;px 1 m log(p): If 2 m >x, then there are no primes withpx 1 m . 2 m >x ()x 1 m < 2 () 1 m < log 2 logx ()m> logx log 2 = log 2 (x): Then (x) = X mlog 2 (x) X p prime;px 1 m log(p): Which gives (x) = X mlog 2 (x) #(x 1 m ) =#(x) + X 2mlog 2 (x) #(x 1 m ): So 0 (x)#(x) = X 2mlog 2 (x) #(x 1 m ): #(x) X p prime;px log(x)x log(x): 34 So 0 (x)#(x) X 2mlog 2 (x) x 1 m log(x 1 m ) log 2 (x) p x log( p x) = logx log 2 p x 2 logx = p x log 2 (x) 2 log 2 : Finally, 0 (x) x #(x) x log 2 (x) 2 log(2) p x : lim x!1 log 2 (x) p x2 log 2 = lim x!1 " 2 logx 1 x 1 2 x 1 2 2 log 2 # = lim x!1 2 logx p x log 2 = lim x!1 2 1 x 1 2 x 1 2 log 2 = lim x!1 4 p x log 2 =0: So if lim x!1 #(x) x = 1, then lim x!1 (x) x = 1 [2]. 10 Lemma. Forx> 2, #(x) x = (x) log(x) x 1 x Z x 2 (t) t dt: Proof. In Abel’s summation formula, taking n =n; 35 a n = 8 > > < > > : 1; if n is prime; 0 otherwise. '(x) = logx, givesA(x) = P nx a n =(x). Then #(x) = X nx a n logn = X nx a n '( n ) =A(x)'(x) Z x 1 A(t)' 0 (t)dt =(x) log(x) Z x 1 (t) t dt =(x) log(x) Z x 2 (t) t dt since(x) is zero forx< 2. Then #(x) x = (x) log(x) x 1 x Z x 2 (t) t dt [4]: 12 Proposition. The prime number theorem ) lim x!1 (x) x = 1. Proof. Assume the Prime Number Theorem. Then lemma 10 gives lim x!1 1 x Z x 2 (t) t dt =0 ) lim x!1 #(x) x = 1: 36 The Prime Number Theorem also implies that lim x!1 (x) x = lim x!1 1 logx : That is,9 constants 0<C2R and 0<M2R such that M <x2R) (x) x <C 1 logx : Then M <x2R) 1 x Z x 2 (t) t dt < 1 x Z M 2 (t) t dt + Z x M C logt dt < 1 x Z M 2 (t) t dt + 1 x C " Z p x 2 dt logt + Z x p x dt logt # 1 x Z M 2 (t) t dt + 1 x C p x log 2 + x p x log p x ! 0 as x!1: Then by lemma 11, lim x!1 (x) x = 1 [2]. 13 Proposition. The prime number theorem implies that has no zeros with real part equal to one. Proof. :fs2C : Re(s)> 0g!C given by s7! 0 (s) s(s) 1 s 1 = Z 1 1 (x)x x s+1 dx 37 has a removable pole ats = 1 and is otherwise analytic except for simple poles at the zeros of. By propostion 12, the prime number theorem implies that (x) =x +b(x) such that lim x!1 b(x) x = 0: Let 0<2R. Then9 0<M2R, which depends on, such that M <x2R)j (x)xj<x: For Re(s) => 1, j(s)j< Z M 1 j (x)xj x 2 dx + Z 1 M x dx <K + Z 1 1 x dx =K + 1 ; since R M 1 j (x)xj x 2 dx is a constant. Then ( 1)j(s)j< ( 1)K +: (2.3) Fix 06=t2R. Then lim !1+ ( 1)j( +it)j = 0. If(1 +it) were equal to zero, would have a simple pole at 1 +it, and this limit would be its residue and not equal to zero [4]. 38 References [1] L. Ahlfors Complex Analysis, McGraw-Hill Book Company, New York (1979). [2] T. Apostol Introduction to Analytic Number Theory, Springer-Verlag, New York (1976). [3] P. Borwein, S. Choi, B. Rooney, A. Weirathmueller The Riemann Hypothesis A Resource for the Afficionado and Virtuoso Alike, Springer, New York (2008). [4] K. Chandrasekharan Introduction to Analytic Number Theory, Springer-Verlag, New York (1968). [5] J. Conway Functions of One Complex Variable, Springer-Verlag, New York (1978). [6] G. Folland Real Analysis: Modern Techniques and Their Applications, John Wiley and Sons, Inc., New York (1999). [7] B. Palka An Introduction to Complex Function Theory, Springer-Verlag, New York (1991). [8] S. Patterson An Introduction to the Theory of the Riemann Zeta-Function, Cam- bridge University Press, Cambridge (1988). [9] W. Rudin Functional Analysis, McGraw-Hill Book Company, New York (1991). [10] W. Rudin Principles of Mathematical Analysis, McGraw-Hill Book Company, New York (1976). [11] J. Stopple A Primer of Analytic Number Theory From Pythagoras to Riemann, Cambridge University Press, Cambridge (2003). [12] E. Whittaker, and G. Watson A Course of Modern Analysis , The Macmillan Company, New York (1947). 39

Abstract (if available)

Abstract I have adapted a development of the basic properties of the Riemann Zeta function. Chapter one includes the Euler product, a functional equation, and an analytic continuation. Chapter two shows the equivalence of the Prime Number Theorem with the fact that the Zeta function has no zeros with real part equal to one.

Linked assets

University of Southern California Dissertations and Theses

Conceptually similar

PDF

Zeta functions: from prime numbers to K-theory

PDF

Analyticity and Gevrey-class regularity for the Euler equations

PDF

The class number formula

PDF

Mach limits and free boundary problems in fluid dynamics

PDF

On prime power elements of GLd(q) acting irreducibly on large subspaces

PDF

A solvable version of the Baer-Suzuki theorem and generalizations

PDF

Well posedness and asymptotic analysis for the stochastic equations of geophysical fluid dynamics

PDF

(p,q)-Adic analysis and the Collatz Conjecture

PDF

From card shuffling to random walks on chambers of hyperplane arrangements

PDF

Asymptotic expansion for solutions of the Navier-Stokes equations with potential forces

PDF

Geometric properties of Anosov representations

PDF

Object adaptation in the lateral occipital complex: shape or semantics?

PDF

A fully discrete approach for estimating local volatility in a generalized Black-Scholes setting

PDF

Population modeling and Bayesian estimation for the deconvolution of blood alcohol concentration from transdermal alcohol biosensor data

PDF

Hyperbolic geometry and canonical triangulations in dimension three

PDF

Applications on symmetric and quasisymmetric functions

PDF

Design and synthesis of a series of methylenebisphosphonates: a nucleotide analogue toolkit to probe nucleic acid polymerase structure and function

Asset Metadata

Creator Neese, Branden (author)

Core Title A development of some of the basic properties of the Riemann Zeta function

School College of Letters, Arts and Sciences

Degree Master of Arts

Degree Program Mathematics

Publication Date 08/26/2008

Defense Date 08/14/2008

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag OAI-PMH Harvest,Riemann,Zeta

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Kamienny, Sheldon (committee chair), Haydn, Nicolai (committee member), Kukavica, Igor (committee member)

Creator Email brandenneese@gmail.com,neese@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-m1579

Unique identifier UC1305874

Identifier etd-Neese-2372 (filename),usctheses-m40 (legacy collection record id),usctheses-c127-105635 (legacy record id),usctheses-m1579 (legacy record id)

Legacy Identifier etd-Neese-2372.pdf

Dmrecord 105635

Document Type Thesis

Rights Neese, Branden

Type texts

Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Repository Name Libraries, University of Southern California

Repository Location Los Angeles, California

Repository Email cisadmin@lib.usc.edu