# Demonstration of Riemann Hypo Riemann hypothesis, proposed by Bernhard Riemann in the 1859, is a...

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Demonstration of Riemann Hypothesis

Diego Marin ∗

June 12, 2014

Abstract

We define an infinite summation which is proportional to the reverse of Riemann Zeta

function ζ(s). Then we demonstrate that such function can have singularities only for

Re s = 1/n with n ∈ N \ 0. Finally, using the functional equation, we reduce these possibilities to the only Re s = 1/2.

∗dmarin.math@gmail.com

1

Contents

1 Target 3

2 Strategy 4

3 Application of the Euler-MacLaurin formula 5

4 Calculating the limiting function 13

2

1 Target

Riemann hypothesis, proposed by Bernhard Riemann in the 1859, is a conjecture

that regards an apparently simple function of complex variable s. Such function,

called Riemann Zeta Function, is defined for Re s > 1 via the following summation

ζ(s) = +∞∑ n=1

n−s

For every integer n, an unique decomposition as product of (powers of) prime

numbers exists. In this way

ζ(s) = +∞∏ p=2

+∞∑ j=0

p−sj = +∞∏ p=2

1

1− p−s

where we have used the summation rule for the geometric series

+∞∑ j=1

wj = 1

1− w

for |w| < 1. Riemann Zeta function can’t have zeros in the convergence area, because no term in the product can be equal to zero. Nevertheless an holomorphic

extension of ζ(s) can be defined over the entire complex plane C, with the exception

of s = 1. Such extension has infinite zeros corresponding to all negative even

integers; that is, ζ(s) = 0 when s is one of −2,−4,−6, . . .. These are the so-called “trivial zeros”.

Again for the holomorphic extension, for s 6= 1 we can prove the functional equation1

ζ(s) = 2sπs−1 sin (πs

2

) Γ(1− s)ζ(1− s)

1The first proof was given by Bernhard Riemann in the 10-page paper Ueber die Anzahl der Primzahlen unter einer gegebenen Grsse (usual English translation: On the Number of Primes Less Than a Given Magnitude) published in the November 1859 edition by Monatsberichte der Kniglich Preuischen Akademie der Wissenschaften zu Berlin.

3

Leaving out the zeros of sin ( πs 2

) which aren’t poles of Γ(1− s), i.e. for s = −2n,

n ∈ N, any other zero s0 must have a “mate” zero s′0 = 1− s0. Because there are no zeros for Re s > 1, functional equation implies that there are no zeros also for

Re s < 0 (except for s = −2n). Other works have excluded also the presence of zeros for Re s = 0 and Re s = 1. As consequence, all the non trivial zeros of ζ(s)

stay in the “critical strip” 0 < Re s < 1.

Riemann hypothesis conjectures that all the non-trivial zeros have real part

equal to 1 2 . This is what we aim to demonstrate in the following.

2 Strategy

We start from the definition of ζ(s) as an infinite product of terms, one term for

every prime number p, running from 2 to +∞:

ζ(s) = +∞∏ p=2

1

1− p−s (1)

The product converges for Re s > 1. From here we fix s = x+ iy with s ∈ C and x, y ∈ R. So the convergence condition is x > 1. Out of convergence area, ζ(s) is defined via holomorphic extension.

The derivative of a single term in the product (1) gives

d

ds

1

1− p−s = − 1

(1− p−s)2 (log p p−s) = − 1

1− p−s · log p ps − 1

Hence the derivative ζ ′(s) of ζ(s) results

ζ ′(s) = d

ds ζ(s) = −

+∞∏ p=2

1

1− p−s ·

+∞∑ p′=2

log p′

p′s − 1

and then ζ ′(s)

ζ(s) = −

+∞∑ p=2

log p

ps − 1 (2)

4

For briefness we set C(s) = ζ ′(s) ζ(s)

. The sum (2) converges for x > 1. Otherwise we

define C(s) as the holomorphic extension of (2), recognizing it by the label “H.e.”:

C(s) = −H.e. ∞∑ n=1

log pn psn − 1

The uniqueness of the holomorphic extension ensures that C(s) = ζ ′(s) ζ(s)

also for

x ≤ 1. At this point we rely upon the following evidence:

Any singularity of C(s) corresponds to a zero of ζ(s) and/or to a

singularity of ζ ′(s).

We can exclude that a zero of ζ ′(s) hides a zero of ζ(s): in fact, for any holomorphic

function f(s), if a point s0 exists which is a zero both for f(s) and f ′(s), we have

surely

f(s0)

f ′(s0) = lim

s→s0

∑∞ k=1 fk(s− s0)k∑∞

k=1 kfk(s− s0)k−1 = lim

s→s0

fk̂(s− s0)k̂

k̂fk̂(s− s0)k̂−1 = lim

s→s0

s− s0 k̂

= 0

where k̂ is the minor k for which a Taylor coefficient fk is 6= 0. Hence the zeros of ζ(s) are singularities for C(s) also when they are zeros of both ζ(s) and ζ ′(s).

For this reason we’ll proceed with finding an ensemble of points which includes

all the singularities of C(s) and so, among them, all the zeros of ζ(s).

3 Application of the Euler-MacLaurin formula

We move from prime to integer numbers:

C(s) = − ∞∑ n=1

log pn psn − 1

(3)

5

where pn is the n-th prime number. For s 6= 0, no partial sum

CM1 (s) = − M∑ n=1

log pn psn − 1

with M < +∞ has singularities. Hence, any function

CM(s) = − +∞∑ n=M

log pn psn − 1

has the same singularities of C(s) for s 6= 0. This permit us to work with CM(s) in place of C(s), in such a way to exploit the freedom in choosing M . Consider

now the Euler-MacLaurin theorem at leading order:∣∣∣∣∣ N∑ l=M

f(l, s)− ∫ N M

f(w, s)dw

∣∣∣∣∣ ≤ FN(s) ∈ R+

FN(s) =

∫ N M

dw

∣∣∣∣ ddwf(w, s) ∣∣∣∣

for some f(w, a) analytic in w ∈ R and holomorphic in s ∈ C except at most for isolated points, provided that in such points we have w /∈ N.

Consider now limN→∞ F N(s)

! = F (s). If F (s) exists (is finite) in a open set

x > A (with A some real number) except at most for isolated points, then the

Dominated Convergence Theorem ensures (for x > A) that

lim N→∞

∣∣∣∣∣ N∑ l=M

f(l, s)− ∫ N M

f(w, s)dw

∣∣∣∣∣ ≤ F (s) (4) Now it is possible that the sum and the integral in the left side converge only for

x > B with B > A. For A < x ≤ B we can separate the holomorphic extension of

6

the summation (H.e.) from its divergent piece (D.p.):

∞∑ l=M

f(l, s) = H.e. ∞∑ l=M

f(l, s) +D.p. ∞∑ l=M

f(l, s)

We can do the same for the integral:∫ ∞ M

f(w, s)dw = H.e.

∫ ∞ M

f(w, s)dw +D.p.

∫ ∞ M

f(w, s)dw

Inserting them in (4) we obtain:∣∣∣∣∣H.e. ∞∑ l=M

f(l, s) +D.p. ∞∑ l=M

f(l, s)−H.e. ∫ ∞ M

f(w, s)dw −D.p. ∫ ∞ M

f(w, s)dw

∣∣∣∣∣ ≤ F (s) Being F (s) finite (except for isolated points), it has to be true

D.p. ∞∑ l=M

f(l, s) = D.p.

∫ ∞ M

f(w, s)dw

and so ∣∣∣∣∣H.e. ∞∑ l=M

f(l, s)−H.e. ∫ ∞ M

f(w, s)dw

∣∣∣∣∣ ≤ F (s) for A < x ≤ B. Hence we can apply the Euler-Maclaurin formula not only for a comparison between sum and integral, but also between their holomorphic

extensions, at least until F (s) is finite. Our case is∣∣∣∣∣ ∞∑

n=M

log pn psn − 1

− ∫ ∞ M

dt log p(t)

p(t)s − 1

∣∣∣∣∣ ≤ F (s) Here p(t) is any analytic function which posses analytic inverse t(p) and satisfies

7

p(n) = pn. Moreover

| ∑ − ∫ | ≤

∫ ∞ M

dt

∣∣∣∣ ddt log p(t)p(t)s − 1 ∣∣∣∣

≤ ∫ ∞ M

dt

∣∣∣∣dp(t)dt [

1

p(t)(p(t)s − 1) − sp(t)

s−1 log p(t)

(p(t)s − 1)2

]∣∣∣∣ (5) The right side converges for x > 0 with the exceptions of isolated points.

The function t(pn) returns the cardinality n of the prime number pn. This is

equivalent to return how many prime numbers exist that are less than or equal to

pn, i.e. t(p) = π(p), where π(p) is the prime-counting function. A holomorphic

definition was given by Riesel, Edwards and Derbyshire2:

π(p) = lim T→+∞

1

2πi

∫ 2+iT 2−iT

ps

s log ζ(s)ds

π′(p) = lim T→+∞

1

2πi

∫ 2+iT 2−iT

ps−1 log ζ(s)ds

For p→ +∞ we have

π(p) ∼ p log p

π′(p) = dπ(p)

dp ∼ 1

log p

( 1− 1

log p

) ∼ 1

log p

Before going any further, we prove that (the holomorphic extension of) the integral

in | ∑ − ∫ |,

H.e.

∫ +∞ pM

dp dt(p)

dp

log p

ps − 1 = H.e.

∫ +∞ pM

dp π′(p) log p

ps − 1 , (6)

2Riesel, H. “The Riemann Prime Number Formula” Prime Numbers and Computer Methods for Factorization, 2nd ed. Boston, MA: Birkhuser, pp. 50-52, 1994; Edwards, H. M. Riemann’s Zeta Function. New York: Dover, 2001; Derbyshire, J. Prime Obsess

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