A positive answer to the Riemann hypothesis:
A new result predicting the location of zeros
Zeraoulia Elhadj
Department of Mathematics, University of Tébessa, (12002), Algeria
E-mail: [email protected] and [email protected]
December 8, 2012
Abstract
In this paper, a positive answer to the Riemann hypothesis is given
by using a new result that predict the exact location of zeros of the
alternating zeta function on the critical strip.
Keywords: Riemann hypothesis, alternating zeta function, location of
zeros.
AMS 2010 Mathematics Subject Classification. Primary 11M26, Secondary 11A41, 11N05.
1
Introduction
The Riemann zeta function is the function of the complex variable s = α+iβ,
defined in the half-plane α > 1 by the absolutely convergent series
∞
X
1
ζ (s) =
ns
n=1
(1)
and in the whole complex plane C by analytic continuation. As shown by
Riemann, ζ (s) extends to C as a meromorphic function with only a simple
pole at s = 1 with residue 1. In [1] Riemann obtained an analytic formula
for the number of primes up to a preassigned limit in terms of the zeros of
1
the zeta function (1). This principal result implies that natural primes are
distributed as regularly as possible if the Riemann hypothesis is true.
Riemann Hypothesis. The nontrivial zeros of ζ (s) have real part equal
to α = 12 .
The Riemann hypothesis is probably the most important open problem
in pure mathematics today [2]. The unsolved Riemann hypothesis is part
of the Hilbert’s eighth problem, along with the Goldbach conjecture. It is
also one of the Clay mathematics institute millennium prize problems. This
hypothesis has been checked to be true for the first 1500000000 solutions.
However, a mathematical proof is not available since its formulation in 1859.
It is well known that the Riemann hypothesis is equivalent to the statement that all the zeros of the Dirichlet eta function (the alternating zeta
function)
∞
X
(−1)n−1
η (s) =
(2)
s
n
n=1
falling in the critical strip D = {s = α + iβ ∈ C with 0 < α < 1} lie on the
critical line α = 12 [3-4-8, pp. 49]. We have η (s) = (1 − 21−s ) ζ (s) and the
series η (s) given by (2) converges only for s = α + iβ ∈ C with α > 0. The
function η (s) is a non-constant analytic for all s = α + iβ ∈ C with α > 0.
2
Evaluating the possible values of α
The functional equation for η (s) restricted to the domain D is given by
(
η (s) = ϕ (s) η (1 − s)
¡ ¢
(3)
(1−2−1+s )
Γ (1 − s)
ϕ (s) = 2 (1−2s ) π s−1 sin πs
2
Let η (s) = x (s) + iy (s) , η (1 − s) = u (s) + iv (s) , where
⎧
∞
∞
X
X
⎪
(−1)n−1
(−1)n−1
⎪
x
(s)
=
cos
(β
ln
n)
,
y
(s)
=
−
sin (β ln n)
⎪
α
n
nα
⎨
n=1
∞
n=1
∞
X (−1)n−1
X (−1)n−1
⎪
⎪
⎪
u
(s)
=
cos
(β
ln
n)
,
v
(s)
=
sin (β ln n)
⎩
n1−α
n1−α
n=1
(4)
n=1
The function ϕ (s) can be written as ϕ (s) = ϕ1 (s) + iϕ2 (s) and therefore it
is given uniquely by the polar form ϕ (s) = ρ (s) exp (iθ (s)) , where, ρ (s) =
2
p
ϕ21 (s) + ϕ22 (s) 6= 0 and θ (s) = arg ϕ (s) ∈ R, the argument of ϕ (s) . Here,
we have ϕ1 (s) = ρ (s) cos θ (s) and ϕ2 (s) = ρ (s) sin θ (s) since such complex
numbers are entirely determined by their modulus and angle.
Let s ∈ D is be an arbitrary solution of η (s) = 0. The main result for
the proof of the Riemann hypothesis is given as follow:
Theorem 1 The complex number s ∈ D is a solution of η (s) = 0 with α =
if and only if θ (s) 6= 2kπ, k ∈ Z, x (s) = u (s) and y (s) = −v (s) .
1
2
Proof. We note that assumptions of Theorem 1 are valid only for single
s ∈ D since all zeros of non-constant analyatic functions are isolated. Hence,
in all what follow we deals only with a single root s.
By substituting the algebraic forms of η (s) and η (1 − s) into equation
(3), we obtain
⎧
x (s) = u (s) ϕ1 (s) − v (s) ϕ2 (s)
⎪
⎪
⎪
⎪
y (s) = u (s) ϕ2 (s) + v (s) ϕ1 (s)
⎪
⎪
⎪
⎪
⎨
µ
¶ or µ
¶
x (s)
u (s)
(5)
= A (s)
⎪
⎪
y
(s)
v
(s)
⎪
⎪
¶
µ
⎪
⎪
cos
θ
(s)
−
sin
θ
(s)
⎪
⎪
⎩ A (s) = ρ (s)
sin θ (s) cos θ (s)
and the inverse transformation is given by
(
y(s)
u (s) = ρx(s)
2 (s) ϕ1 (s) + ρ2 (s) ϕ2 (s)
y(s)
v (s) = − ρx(s)
2 (s) ϕ2 (s) + ρ2 (s) ϕ1 (s)
(6)
and w (s) = y(s)
, we obtain a rotation transformation
By setting z (s) = x(s)
ρ(s)
ρ(s)
of the form
⎧ µ
¶
¶
µ
u (s)
z (s)
⎪
⎪
= B (s)
⎨
v (s) ¶
w (s)
µ
(7)
cos
θ
(s)
−
sin
θ (s)
⎪
⎪
⎩ B (s) =
sin θ (s) cos θ (s)
The matrix B (s) in (7) is invertible for all s ∈ D since its determinant is 1.
It is known that a non trivial rotation must have a unique fixed point, its
rotocenter. The rotation in (5) is non trivial if ϕ (s) 6= 1 (here we assumed
that θ (s) 6= 2kπ, k ∈ Z in the second part of Theorem 1). The reason is
3
that the trivial rotation corresponding to the identity matrix, in which no
rotation takes place. The fixed point of the rotation in (7) must satisfies
(I2 − B (s)) (u (s) , v (s)) = 0 where I2 is the 2 × 2 unit matrix. The determinant of the matrix (I2 − B (s)) is −2 (cos θ (s) − 1) and it is not zero since
θ (s) 6= 2kπ, k ∈ Z, this means that s is a solution of η (s) = 0 by using (3).
For a single s ∈ D, the assumptions x (s) = u (s) and y (s) = −v (s) can
be reformulated by using (5) and (6) as follow:
½
x (s) = u (s) ϕ1 (s) − v (s) ϕ2 (s) = u (s) = x (s) ϕ1 (s) + y (s) ϕ2 (s)
y (s) = u (s) ϕ2 (s) + v (s) ϕ1 (s) = −v (s) = − (y (s) ϕ1 (s) − x (s) ϕ2 (s))
that is,
½
u (s) ϕ1 (s) − v (s) ϕ2 (s) = x (s) ϕ1 (s) + y (s) ϕ2 (s)
u (s) ϕ2 (s) + v (s) ϕ1 (s) = − (−x (s) ϕ2 (s) + y (s) ϕ1 (s))
by replacing the values of ϕ1 (s) = ρ (s) cos θ (s) , ϕ2 (s) = ρ (s) sin θ (s) , and
the values of x (s) , y (s) , u (s) and v (s) from (4), we obtain the following
equation
∞
X
¢
(−1)n−1 ¡
2α−1 2
ρ
(s)
exp (iβ ln n) = 0
(8)
1
−
n
nα
n=1
Since, we speak about a single and isolated point s, the only solution of (8)
is when
1
α=
(9)
2
for some β ∈ R. Indeed, for α = 12 , we have ρ (s) = 1. See [9] for more details.
In this case s = 12 + iβ is a root of equation (9).
Now, assume that s ∈ D is a solution of η (s) = 0 with α = 12 , then,
trivially, we have x (s) = u (s) = 0 and y (s) = −v (s) = 0. Thus we only
need to prove that θ (s) 6= 2kπ, k ∈ Z. We have
!µ
Ã
µ
¶
¶ µ
¶
√ iβ− 1 2iβ− 12 − 1
1
1
1
1
ϕ
cosh πβ + i sinh πβ Γ
+ iβ = 2π 2
− iβ
1
2
2
2
2
2iβ+ 2 − 1
(10)
4
and by direct calculations, we get
⎧
1
π iβ− 2 = √1π (cos (β ln π) + i sin (β ln π))
⎪
⎪
√
⎪
√
iβ− 1
⎪
cos(β ln 2)− 23 2
sin(β ln 2)
2 −1
2
3
⎪
√ − 1i 2 √
⎪
=
1
3
⎨
iβ+ 2
4
2
cos(β
ln
2)−
2
2
2
cos(β ln 2)−3
2
−1
4
¶
µ
1
1
exp
πβ
+exp
−
πβ
exp( 12 πβ )−exp(− 12 πβ )
(
)
(
)
1
1
2
2
⎪
cosh 2 πβ + i sinh 2 πβ =
+i
⎪
2
2
⎪
⎪
⎪
√
1
⎪
¡
¢
π exp(−i(2ϑ(β)+β ln 2π+arctan(tanh 2 πβ )))
⎩
√
Γ 12 − iβ =
cosh πβ
(11)
where ϑ (β) is the Riemann Siegel function [10].
On one hand, the last formula of equation (12) can be proved easily for
z = µ + it by using the following equations
p
¡ πz ¢
¡
¡
¡ πt ¢¢¢
⎧
√1
cos
cosh
(πt)
exp
−i
arctan
tanh
=
⎪
2
2
⎪
¢¢
¢2 ¯ ¡ 1
¢¯
¡ ¡
¡1
⎨
t
t ¯
t
¯
i ϑ (t) + 2 ln π
Γ 2 + i 2 = Γ¡ 4 + i¢2 exp √
⎪
Γ (z) Γ z + 12 = 21−2z πΓ (2z)
⎪
⎩
Γ (z) Γ (1 − z) = sinππz
and then we get the formula by setting z = 14 + i 2t and replacing t by −β.
Indeed, we have
1
π
Γ (z)
¡1
¢
Γ (2z) = 22z−1 π − 2
cos πz Γ 2 − z
For z =
1
4
+ i 2t , t ∈ R, we have
¡
¢
µ
¶
1
1
Γ 14 + i 2t
1
2− 2 +it π 2
¡ ¡
¢¢ ¡
¢
Γ
+ it =
2
cos π 14 + i 2t Γ 14 − i 2t
Hence, we have
¶
µ
1
2it π 2 π it exp (2iϑ (t))
1
+ it = √
Γ
¡
¡
¡ ¢¢¢
2
cosh πt exp −i arctan tanh πt
2
Thus, we can get
¡ ¡
¡
¡ ¢¢¢¢
µ
¶ √
π exp i 2ϑ (t) + t ln (2π) + arctan tanh πt
1
2
p
Γ
+ it =
2
cosh (πt)
For t = −β, we have
¡ ¡
¡
¡ ¢¢¢¢
¶ √
µ
π exp −i 2ϑ (β) + β ln (2π) + arctan tanh πβ
1
2
p
− iβ =
Γ
2
cosh (πβ)
5
¡
¡ ¢¢
because the function β → arctan tanh πβ
is odd and since the function
2
ϑ (β) is also odd as shown in [10]. We note that other proof of the last
equation of (12) uses the functional equation of the zeta function.
On the other hand, we can calculate the arguments of the quantities in
(12) as follow
³
´
⎧
iβ− 12
⎪
arg π
= β ln π
⎪
⎪
³ iβ− 1 ´
³
´
⎪
⎪
⎨
sin(β ln 2)
√
arg 2iβ+ 12 −1 = = − arctan 13 cos(β
ln 2)− 23 2
2 −1
2
(12)
³ − 1 πβ 1 πβ ´
¢
¡
⎪
1
1
e 2 −e 2
⎪
⎪
arg cosh 2 πβ + i sinh 2 πβ = φ = − arctan − 1 πβ 1 πβ
⎪
⎪
e 2 ¡ +e 2
¢¢
¡
¢¢
¡ ¡1
⎩
arg Γ 2 − iβ = ψ = − 2ϑ (β) + β ln 2π + arctan tanh 12 πβ
Since the complex argument of a product of two numbers is equal to the sum
of their arguments, then from (13) we have
¶
µ
1
+ iβ = 0 + β ln π + + φ + ψ
(13)
θ
2
Hence, we have
(
θ
¡1
2
¢
+ iβ³ = 2 (g (β) − ϑ´(β))
g (β) = − 12 β ln 2 − 12 arctan
sin(β ln 2)
1
√
3 cos(β ln 2)− 23 2
− arctan
³
1
1
1
1
e− 2 πβ −e 2 πβ
e− 2 πβ +e 2 πβ
´
(14)
Thus, from (15) we can get
µ
¶
1
1
+ iβ
(15)
ϑ (β) = g (β) − θ
2
2
¡1
¢
The only possible
value
of
equations
θ
+
iβ
= 2kπ is when k = 0. Indeed,
2
¡1
¢
assume that θ 2 + iβ = 2kπ. We know from [10] that ϑ (β) is an odd
function and g (β) is also an odd function by direct calculations. Hence,
from (16) we have ϑ (−β) =¡ −g (β)¢ − kπ = −g (β) + kπ, that is, k = 0.
Hence, we must prove that θ 12 + iβ 6= 0 if 12 + iβ is a root of η (s) = 0. We
know that the function ϑ (β) has
¡ 1 only ¢three roots 0 and ±17.8455995405...
1
Hence, the equation g ¡(β) − 2¢θ 2 + iβ = 0 has only the same three roots,
that is the equation θ 12 +
¡ iβ =¢2g (β) has only the same three roots. The
only possible case to get θ 12 + iβ = 0 is when β = 0 since we have g (0) = 0,
g (17.8455995405) = −4. 877 4 and g (−17.8455995405) = 4. 877 4. However,
6
the real s = 12 + i0 is not a root of η (s) = 0. Hence, we have proved that if
s ∈ D is a solution of η (s) = 0 and α = 12 , then θ (s) 6= 2kπ, k ∈ Z.
Theorem 1 proves that the Riemann hypothesis is true for the function
η (s) if and only if θ (s) 6= 2kπ, k ∈ Z, x (s) = u (s) and y (s) = −v (s) for
every root s of η (s) = 0. Hence, to prove that the Riemann hypothesis is true
for the function η (s), we need only a proof of the second case in Theorem
1 since we have an equivalence between two cases. Clearly, if s is a root of
η (s) = 0, then x (s) = u (s) = 0, y (s) = −v (s) = 0. Hence, we need only
to prove that if s is any root of η (s) = 0, then θ (s) 6= 2kπ, k ∈ Z. For
this purpose assume that s is any root of η (s) = 0 and assume by contraduction that θ (s) = 2kπ, k ∈ Z. From the relation η (s) = (1 − 21−s ) ζ (s) ,
we know that if β > 0, then the first zero of η (s) = 0 in D is computable
by¡ hand and its value
is s = 12 + 14.134725142i. For this root, we have
¢
ϕ 12 + 14.134725142i = 0.992 62 + 0.121 28i and θ (s) = 0.121 58. This is a
contraduction with the assumption θ (s) = 2kπ, k ∈ Z for any s a zero of
η (s) = 0 in D.
Finally, we can concludes that
Theorem 2 The Riemann hypothesis is true, i.e., all the nontrivial zeros of
ζ (s) have real part equal to α = 12 .
Proof. The fact that the Dirichlet eta function (2) have the same zeros as
the zeta function (1) in the critical strip D, implies that all nontrivial zeros
of ζ (s) have real part equal to α = 12 .
At this end, the hope that primes are distributed as regularly as possible
is now become a truth by this unique solution of the Riemann hypothesis.
Also, all propositions which are known to be equivalent to or true under the
Riemann hypothesis are now correct. Examples includes, growth of arithmetic functions, Lindelöf hypothesis and growth of the zeta function, large
prime gap conjecture...etc.
Acknowledgment
We would like the thank many colleagues for their valuable comments
improving the quality of this paper: Michel L. Lapidus (University of California), Andrew Odlyzko (University of Minnesota), Sergey. K. Sekatskii
(Ecole Polytechnique Fédérale de Lausanne), Marek Wolf (Cardinal Stefan
Wyszynski University, Poland), Chris King (University of Auckland, Australia), Jinhua Fei, Peter Braun and Craig Feinstein.
7
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8