Demonstration of Riemann Hypothesis

Size: px

Start display at page:

Download "Demonstration of Riemann Hypothesis"

Lorin Manning
5 years ago
Views:

1 Demontration of Riemann Hypothei Diego arin June 2, 204 Abtract We define an infinite ummation which i proportional to the revere of Riemann Zeta function ζ(). Then we demontrate that uch function can have ingularitie only for Re = /n with n N \ 0. Finally, uing the functional equation, we reduce thee poibilitie to the only Re = /2. dmarin.math@gmail.com

2 Content Target 3 2 Strategy 4 3 Application of the Euler-acLaurin formula 5 4 Calculating the limiting function 3 2

3 Target Riemann hypothei, propoed by Bernhard Riemann in the 859, i a conjecture that regard an apparently imple function of complex variable. Such function, called Riemann Zeta Function, i defined for Re > via the following ummation ζ() = n For every integer n, an unique decompoition a product of (power of) prime number exit. In thi way ζ() = + p=2 j=0 n= p j = + p=2 p where we have ued the ummation rule for the geometric erie j= w j = w for w <. Riemann Zeta function can t have zero in the convergence area, becaue no term in the product can be equal to zero. Neverthele an holomorphic extenion of ζ() can be defined over the entire complex plane C, with the exception of =. Such extenion ha infinite zero correponding to all negative even integer; that i, ζ() = 0 when i one of 2, 4, 6,.... Thee are the o-called trivial zero. Again for the holomorphic extenion, for we can prove the functional equation ζ() = 2 π in ( π ) Γ( )ζ( ) 2 The firt proof wa given by Bernhard Riemann in the 0-page paper Ueber die Anzahl der Primzahlen unter einer gegebenen Gre (uual Englih tranlation: On the Number of Prime Le Than a Given agnitude) publihed in the November 859 edition by onatberichte der Kniglich Preuichen Akademie der Wienchaften zu Berlin. 3

4 Leaving out the zero of in ( ) π 2 which aren t pole of Γ( ), i.e. for = 2n, n N, any other zero 0 mut have a mate zero 0 = 0. Becaue there are no zero for Re >, functional equation implie that there are no zero alo for Re < 0 (except for = 2n). Other work have excluded alo the preence of zero for Re = 0 and Re =. A conequence, all the non trivial zero of ζ() tay in the critical trip 0 < Re <. Riemann hypothei conjecture that all the non-trivial zero have real part equal to. Thi i what we aim to demontrate in the following. 2 2 Strategy We tart from the definition of ζ() a an infinite product of term, one term for every prime number p, running from 2 to + : ζ() = + p=2 p () The product converge for Re >. From here we fix = x + iy with C and x, y R. So the convergence condition i x >. Out of convergence area, ζ() i defined via holomorphic extenion. The derivative of a ingle term in the product () give d d p = ( p ) (log p 2 p ) = p log p p Hence the derivative ζ () of ζ() reult ζ () = d d ζ() = + p=2 p log p p p =2 and then ζ + () ζ() = log p p 4 p=2 (2)

5 For briefne we et C() = ζ (). The um (2) converge for x >. Otherwie we ζ() define C() a the holomorphic extenion of (2), recognizing it by the label H.e. : C() = H.e. n= log p n p n The uniquene of the holomorphic extenion enure that C() = ζ () ζ() x. At thi point we rely upon the following evidence: alo for Any ingularity of C() correpond to a zero of ζ() and/or to a ingularity of ζ (). We can exclude that a zero of ζ () hide a zero of ζ(): in fact, for any holomorphic function f(), if a point 0 exit which i a zero both for f() and f (), we have urely f( 0 ) f ( 0 ) = lim 0 k= f k( 0 ) k k= kf = lim k( 0 ) k 0 fˆk( 0 )ˆk ˆkfˆk( 0 )ˆk = lim 0 0 ˆk = 0 where ˆk i the minor k for which a Taylor coefficient f k i 0. Hence the zero of ζ() are ingularitie for C() alo when they are zero of both ζ() and ζ (). For thi reaon we ll proceed with finding an enemble of point which include all the ingularitie of C() and o, among them, all the zero of ζ(). 3 Application of the Euler-acLaurin formula We move from prime to integer number: C() = n= log p n p n (3) 5

6 where p n i the n-th prime number. For 0, no partial um C () = n= log p n p n with < + ha ingularitie. Hence, any function C () = n= log p n p n ha the ame ingularitie of C() for 0. Thi permit u to work with C () in place of C(), in uch a way to exploit the freedom in chooing. Conider now the Euler-acLaurin theorem at leading order: N f(l, ) l= N F N () = f(w, )dw F N () R + N dw d f(w, ) dw for ome f(w, a) analytic in w R and holomorphic in C except at mot for iolated point, provided that in uch point we have w / N. Conider now lim N F N ()! = F (). If F () exit (i finite) in a open et x > A (with A ome real number) except at mot for iolated point, then the Dominated Convergence Theorem enure (for x > A) that N lim f(l, ) N l= N f(w, )dw F () (4) Now it i poible that the um and the integral in the left ide converge only for x > B with B > A. For A < x B we can eparate the holomorphic extenion of 6

7 the ummation (H.e.) from it divergent piece (D.p.): f(l, ) = H.e. l= We can do the ame for the integral: f(w, )dw = H.e. Inerting them in (4) we obtain: f(l, ) + D.p. l= H.e. f(l, ) + D.p. f(l, ) H.e. l= l= f(l, ) l= f(w, )dw + D.p. f(w, )dw D.p. f(w, )dw Being F () finite (except for iolated point), it ha to be true and o for A < x B. D.p. f(l, ) = D.p. l= H.e. f(l, ) H.e. l= f(w, )dw f(w, )dw F () f(w, )dw F () Hence we can apply the Euler-aclaurin formula not only for a comparion between um and integral, but alo between their holomorphic extenion, at leat until F () i finite. Our cae i n= log p n p n log p(t) dt p(t) F () Here p(t) i any analytic function which poe analytic invere t(p) and atifie 7

8 p(n) = p n. oreover dt d log p(t) dt p(t) [ dt dp(t) dt p(t)(p(t) ) p(t) log p(t) (p(t) ) 2 ] (5) The right ide converge for x > 0 with the exception of iolated point. The function t(p n ) return the cardinality n of the prime number p n. Thi i equivalent to return how many prime number exit that are le than or equal to p n, i.e. t(p) = π(p), where π(p) i the prime-counting function. A holomorphic definition wa given by Rieel, Edward and Derbyhire 2 : For p + we have π(p) = π (p) = lim T + lim T + 2+iT 2πi 2 it p log ζ()d 2+iT p log ζ()d 2πi 2 it π(p) π (p) = dπ(p) dp p log p ( ) log p log p log p Before going any further, we prove that (the holomorphic extenion of) the integral in, H.e. p dp dt(p) dp log p p = H.e. p dp π (p) log p p, (6) 2 Rieel, H. The Riemann Prime Number Formula Prime Number and Computer ethod for Factorization, 2nd ed. Boton, A: Birkhuer, pp , 994; Edward, H.. Riemann Zeta Function. New York: Dover, 200; Derbyhire, J. Prime Obeion: Bernhard Riemann and the Greatet Unolved Problem in athematic. New York: Penguin,

9 9

10 ha no ingularitie. We introduce firt a redefinition of the ame integral which permit to calculate it alo for x. Thi i dp π 2i in(π) Cε log p ( p) (p), (7) p which i equivalent to (6) for x >, o giving the relative holomorphic extenion. The Cauchy principle enure that integral maintain the ame value for every choice of ε (until the path doen t touch any pole of the integrated function); for demontration we find ueful to work in the limit ε 0 +. The contribute at i null for x >. Thi i unique zone of interet to check the equivalence between (6) and (7); o, at thi aim, we can forget about it. log p ( p) lim p + π (p) = 0 for x > p Having aid thi, we proceed by calculating eparately the contribute to the integral along Γ, Γ 3 and Γ 2 : Γ [ ] = lim ε 0 + p = e iπ dp π (p) p = e iπ p dt π (t + iε) log(t + iε) e log(t+iε)+iπ e log(t+iε) log p p p dp π (p) log p p Γ 3 [ ] = lim ε 0 + p + = e iπ dp π (p) p = e iπ p dt π log(t iε) iπ log(t iε) e (t iε) e log(t iε) log p p p dp π (p) log p p 0

11 For the integral in Γ 2, we poe p = εe iθ + p, dp = iεe iθ dθ, Γ 2 [ ] = lim 3π 4 ε 0 = lim ε 0 iε 3π 4 3π 4 dθ π (εe iθ + p ) log(εe iθ + p )(εe iθ + p ) e iπ iεe iθ (εe iθ + p ) 3π 4 dθ π (p ) log(p )(p ) e iπ e iθ (p ) = lim iε π (p ) log(p )e iπ ε 0 p = lim ε π (p ) log(p )e iπ [ ε 0 p = lim ε 0 iε 3π 4 3π 4 dθ e iθ e i 3π 4 e i 3π 4 2 π (p ) log(p )e iπ p = 0 ] In the end: dp π 2i in(π) Cε log p ( p) (p) = p = = = [ [ ] + 2i in(π) Γ 2i in(π) ] [ ] + [ ] Γ 2 Γ 3 ( e iπ e iπ) dp π (p) log p p 2i in(π) 2i in(π) p dp π (p) log p p p + p dp π (p) log p p CVD. At thi tage we can doubtle affirm that (7) i the holomorphic extenion of (6). Uing the Cauchy integral theorem, the integral in (7) can be tranformed into a um over the minimal circuitation around all the pole. Thee it at p = exp ( ) 2πik, k Z. A ingle term i the following: (k) iε in(π) [ ] = lim ε 0 + 2i 2π 0 e iθ dθ π (e 2πik + εe iθ ) 2πik log(e + εe iθ ) (e 2πik + εe iθ ) eiπ = ε = lim ε π 0 e iθ dθ π (e 2πik + εe iθ ) 2πik log( + εeiθ ) + 2πk e 2πik 2πik iθ ( + εe ) eiπ =

12 ε = lim ε π 0 e iθ dθ π (e 2πik ε = lim ε π By umming over all the pole: in(π) H.e. p 0 + εe iθ ) e iθ dθ π (e 2πik = lim ε π 2πik log( + εeiθ ) + 2πk 2πik iθ ( + εe ) 2πik + εe iθ εeiθ ) e iθ dθ π (e 2πik + 2πk 2πik iθ εe 2πk ) 2πik iθ 0 e = πk 2π 2 π (e 2πik )e 2πik +iπ = 2π2 k 2 dp π (p) log p p = eiπ = π 2 eiπ log ζ(2 + iw)dw = π 2 eiπ log ζ(2 + iw)dw Ue now the ummation rule k= k= = e iπ + k= k= k= ke ka = d e ka = d [ ] da da e a k= π (e 2πik 2π 2 k 2 2π 2 k 2 0 )e 2πik +iπ e iπ = e iπ = e iπ = dθ = π (e 2πik )e 2πki π (e 2πik )e 2πik 2 ke 2πik(+iw) 2 e 2πik 2 [ k = e 2πik(2+iw) 2 ] e 2πik(2+iw) 2 e a ( e a ) = 2 (e a/2 e a/2 ) 2 The lat term in the right i the correct value of ummation only for Re a < 0. Neverthele, when we climb over the line Re a = 0, it give the correponding holomorphic extenion. Hence we can ue the rule without care of convergence 2

13 criterium: = π 2 eiπ log ζ(2 + iw)dw (e πi(2+iw) 2 e πi(2+iw) 2 ) 2 For w + the integrand goe like π 2 eiπ log ζ(2 + iw) e 2πw 2 and o the integral converge at + (remember that Re = x > 0). Similarly, for w the integrand goe like π 2 eiπ log ζ(2 + iw) e 2πw 2 and o the integral converge at. Being know that ζ(2 + iw) ha neither zero nor pole for w R, we can ay that (the Holomorphic Extenion of) the integral in ha no ingularitie for C \ R. The inecapable concluion i that all ingularitie of C() in the critical trip emerge from the difference between it and the correponding integral, i.e. they are among the iolated point where F () = +. 4 Calculating the limiting function Let recover the reult (5): dt dp(t) dt dt dp(t) dt [ p(t)(p(t) ) p(t) log p(t) (p(t) ) 2 I t [ ] dt dp(t) dt ] ] [ I t ] [ where I t = if dp(t) dt 0 and I t = 0 otherwie. Ue now dt dp(t) dt = dp to achieve an 3

14 advantageou change of variable: p + p p p [ ] dp I t(p) dp [ I t(p) ] [ ] p [ ] dp I t(p) + dp [ I t(p) ] [ ] p [ ] dp p(p ) p log p (p ) 2 dp p p log p p(p ) 2 Now conider the following inequality: p 2 = (p )(p ) = p + p p + = p 2x 2p x co(y log p) + p 2x 2p x + = (p x ) 2 For x > 0 we have alo p p log p 2 = (p p log p)(p p log p) = p 2x 2p x co(y log p) + 2xp 2x log p + 2xp x log p co(y log p) 2yp x log p in(y log p) + 2 p 2x 2 log 2 p p 2x + 2p x + + 2xp 2x log p + 2xp x log p + +2 y p x log p + (x 2 + y 2 )p 2x 2 log 2 p 4

15 p 2x + 2p x + + 2xp 2x log p + 2xp x log p + +2 y p x log p + (x 2 + y 2 )p 2x 2 log 2 p + +2x y p 2x 2 log 2 p + 2 y p 2x log p = = (p x + + (x + y )p x log p) 2 Hence p dp px + + (x + y )p x log p p(p x ) 2 2 x( p x ) log( p x ) x (x+ y ) log p((px )Φ ( px,, ) +x) x x 2 p(p x ) p Φ(w, a, b) i the Lerch Trancendent 3 which goe at + in the firt variable a w ; in our cae a p x. A conequence, the contribute at + i null. Finally 2 log( p x + x(p x ) x ) + (x+ y ) log p((px )Φ( p x,, ) x +x) x 2 p (p x ) The Lerch Trancendent Φ(w, a, b) ha ingularitie (if a N\0) only for b N\0. In our cae, a = and o we have ingularitie for x = n with n N \ 0 Thi mean that the zero of ζ() in the trip 0 < x < have to atify x = n for ome n N \ 0. oreover the functional equation ζ() = 2 π in ( π ) Γ( )ζ( ) 2 reveal that if 0 i a non-trivial zero, then 0 i a zero too. Hence we earch 3 Φ(w, a, b) i uually defined a the holomorphic extenion of Φ(w, a, b) = + t a e bt Γ(w) 0 we dt t which work for {Re b > 0 Re a > 0 w < } {Re b > 0 Re a > w = }. 5

16 for two integer m, n uch that m = n, but the unique olution of thi equation i m = n = 2. non-trivial zero of zeta mut have real part equal to /2. CVD Conequently, all the 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