A POSSIBLE EXPLANATION FOR THE VALIDITY
OF THE RIEMANN HYPOTHESIS
Kevin Duffy
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A POSSIBLE EXPLANATION FOR THE VALIDITY OF THE
RIEMANN HYPOTHESIS
KEVIN JAN DUFFY
Abstract. The renowned Riemann Hypothesis is considered from the
perspective of symmetry in the functional equation for the Riemann zeta
function developed by Bernhard Riemann. The validity of the Riemann
Hypothesis is shown by considering the complex conjugations of the leading terms of this equation.
1. Introduction
1
where s = α + βi with α, β real
ns
and n a positive integer, and where ζ(s) is convergent for α > 1. The Riemann
ζ(s) is defined as the series ζ(s) =
P∞
n=1
zeta function is an analytical continuation of this series by Bernhard Riemann
which is holomorphic for all s except at s = 1 [3]. Using a natural symmetry
in the result Riemann [3, 2] obtained the functional equation:
(1)
sπ
ζ(s) = ℵ(s)ζ(1 − s) = 2s π s−1 sin Γ(1 − s)ζ(1 − s)
2
R ∞ −z −s
with Γ(1 − s) = 0 e z dz.
The renowned Riemann Hypothesis (RH) is that non-trivial zeroes of the
Riemann zeta function ζ(s) all have real part on the ’critical line’ Re(s) = α =
1
2
[1]-[4].
My explanation for the validity of RH will focus on the complex conjugations
of this equation. Starting with ζ(s) = 0 and using the fact that ℵ(s) 6= 0 on
the critical strip 0 < α < 1 where all non-trivial zeroes are known to occur
[1]-[4] it is shown that ζ(s) = 0 ⇒ ℵ(s).ℵ(s) = 1 and why this is only possible
if α = 0.5.
1991 Mathematics Subject Classification. AMS Subject Classification 11S40.
Key words and phrases. Riemann zeta-function; functional equation.
1
2
KEVIN JAN DUFFY
2. Details
Here I will define the complex functions resulting from (1) as:
ζ(s) = u + vi, ζ(1 − s) = w + ti, ℵ(s) = x + yi
(2)
where u, v, w, t, x, y are all real. Thus,
(3)
(u + vi) = (x + yi)(w + ti)
resulting in
(4)
x=
uw + vt
vw − ut 2
u2 + v 2
,y = 2
, x + y2 = 2
>0
2
2
2
w +t
w +t
w + t2
Note in (4) x2 + y 2 6= 0 and exists on the critical strip 0 < α < 1 where
all non-trivial zeroes are known to occur [1]-[4] because: 2s 6= 0, π s−1 6= 0,
R ∞ −z −s
e z dz 6= 0, and sin sπ
2 6= 0 for 0 < α < 1, and all of these are holomor0
phic for the same range, ⇒ ℵ(s) 6= 0 and is holomorphic for 0 < α < 1. Thus,
using (1), ζ(s) = 0 ⇒ ζ(1 − s) = 0 ⇒ ζ(s) = ζ(1 − s) ⇒ u = w, v = t ⇒ x =
1, y = 0 ⇒ x2 + y 2 = 1 from (4).
Thus the RH is true if
ℵ(s).ℵ(s) = x2 + y 2 = 1 ⇐⇒ α = 0.5.
(5)
Theorem 1. ℵ(s).ℵ(s) = 1 ⇐⇒ α = 0.5 for 0 < α < 1 and β > 10.
Proof. First, for ℵ(0.5 + βi)ℵ(0.5 + βi) using (1):
(i) 2s π s−1 2s π s−1 = 20.5+βi π −0.5+βi 20.5−βi π −0.5−βi =
βiπ
βiπ
sπ
απ
απ
(ii) sin sπ
2 sin 2 = (sin 2 cos 2 +cos 2 sin 2 )(sin
= 0.5(cos2
βπi
2
− sin2
βπi
2 )
2
π
βiπ
απ
απ
2 cos 2 −cos 2
at α = 0.5
= 0.5 cos (βπi) = 0.5 sin (βπi + π2 )
(iii) Γ(1 − s) = Γ(1 − s) = Γ(s) at α = 0.5 and so Γ(1 − s)Γ(1 − s)
= Γ(s)Γ(1 − s) =
π
sin (βπi+ π
2)
by Euler’s reflection formula.
Thus, combining these equalities, ℵ(0.5 + βi).ℵ(0.5 + βi) = 1.
sin βiπ
2 )
A POSSIBLE EXPLANATION FOR THE VALIDITY OF THE RIEMANN HYPOTHESIS 3
Second, consider ℵ(α+βi).ℵ(α + βi) in general on the critical strip 0 < α < 1
where all non-trivial zeroes are known to occur. To simplify some of what
follows define functions f,g,h such that:
(i) f = 2s π s−1 2s π s−1 = 4α π 2α−2 and 0 < 4a π 2a−2 < 4 for 0 < α < 1.
βiπ
βiπ
βiπ
sπ
απ
απ
απ
(ii) g = sin sπ
2 sin 2 = (sin 2 cos 2 + cos 2 sin 2 )(sin 2 cos 2 −
βiπ
cos απ
2 sin 2 )
2 βiπ
απ
2 βiπ
2 απ
2 cos
2 − cos 2 sin
2 )
2 απ
2 βiπ
2 απ
' cos 2 (cos 2 + sin 2 ) for 0 < α < 1 and as for β large enough
2 βiπ
cos2 βiπ
2 ' − sin
2
= cos2 βiπ
which
is
constant on 0 < α < 1 for each β large enough.
2
sπ
sπ
Thus, sin 2 sin 2 ' constant for β large enough and 0 < α < 1.
= (sin2
Even β > 10 suffices which is less than the first known zero of the zeta
function (for the same reason this criteria will be used in what follows).
Also, the value increasingly converges as β increases.
(iii) h = Γ(1 − s)Γ(1 − s).
Now
∂|ℵ(s)|2
∂α
(i) f 0 =
0
(ii) h =
= f 0 gh + f g 0 h + f gh0 and using the partial derivatives:
∂(4α π 2α−2 )
= 2(2α+1) π (2α−2) (ln(π)
∂α
∂(Γ(1−s)Γ(1−s))
=
∂α
+ ln(2))
−(Ψ(1 − α − iβ) + Ψ(1 − α + iβ))Γ(1 − α + iβ)Γ(1 − α − iβ),
where Ψ(s) =
(iii) g 0 =
∂ sin
Γ0 (s)
Γ(s)
sπ
sπ
2 sin 2
∂α
= ( 12 )cos( π2 (α + iβ))πsin(( π2 (α − iβ)) + ( 21 )sin( π2 (α +
iβ))cos( π2 (α − iβ))π
= πcos( π2 α)cosh( π2 β)2 sin( π2 α) − πsin( π2 α)sinh( π2 β)2 cos( π2 α)
= cos( π2 α)sin( π2 α)π
and 0 < cos( π2 α)sin( π2 α)π < 1.6 for 0 < α < 1,
then for 0 < α < 1 and β > 10,
∂|ℵ(s)|2
∂α
0
= f gh( hh +
0
' f gh( hh +
f0
f )
g0
g
+
f0
f )
because cos2
0
< f gh( hh +
βiπ
2
1.6
g
+
f0
f )
0
' f gh( hh +
will be very large.
1.6
cos2 βiπ
2
+
f0
f )
4
KEVIN JAN DUFFY
So, taking the overall partial derivative and simplifying gives
(6)
∂|ℵ(s)|2
πβ
< −4α π (2α−2) cosh2
Γ(1 − α + iβ)Γ(1 − α − iβ)
∂α
2
(Ψ(1 − α − iβ) + Ψ(1 − α + iβ) − 2ln(π) − 2ln(2))
The <((Ψ(α + iβ)) > 0 if β > 5/3 and α ≥ 0 (see [5] for details). Thus,
<(Ψ(1 − α − iβ) + Ψ(1 − α + iβ)) > 0 on 0 < α < 1 and β > 5/3. Now
(Ψ(1 − α − iβ) + Ψ(1 − α + iβ) − 2ln(π) − 2ln(2)) > 0 for β > 10 on 0 < α < 1
because the first two terms are then at least > 0.3676, the approximate absolute
value of the last two terms.
The term 4α π (2α−2) cosh2
πβ
2
is a very large positive real number for all 0 <
α < 1 and β > 10.
The remaining term Γ(1 − α + iβ)Γ(1 − α − iβ) is by definition a positive real
number.
From these points, and considering (6), we have
∂|ℵ(s)|2
∂α
< 0 and so |ℵ(s)|2
is strictly monotonic in terms of α on 0 < α < 1 for each β > 10.
Now, as ℵ(α + βi).ℵ(α + βi) is strictly monotonic, this function can equal 1
at only a single unique value of α for each β > 10 on the critical strip 0 < α < 1.
It is already shown above that this occurs at α = 0.5.
Thus, ℵ(s).ℵ(s) = 1 =⇒ α = 0.5 for β > 10.
3. Summary
In this simple way, as ζ(s) = 0 ⇒ ℵ(s).ℵ(s) = 1 and ℵ(s).ℵ(s) = 1 ⇐⇒ α =
0.5, the Riemann Hypothesis is shown to be true. If these results are correct
then Riemann had already laid the foundation for the proof of his hypothesis
by establishing the functional equation in his original paper [3].
Acknowledgments
Obiora Collins is thanked for reading the manuscript many times, very fruitful discussions and valuable suggestions. Kian Duffy is thanked for pointing
out the requirement that ℵ(s) 6= 0.
A POSSIBLE EXPLANATION FOR THE VALIDITY OF THE RIEMANN HYPOTHESIS 5
References
[1] E. Bombieri, Problems of the Millenium the Riemann Hypothesis. Clay Mathematics Institute (2000).
http://www.claymath.org/sites/default/files/official problem description.pdf.
Accessed
January 2015.
[2] H.M. Edwards, Riemann’s Zeta Function, Dover Publications, New York (1974).
[3] B. Riemann, Ueber die Anzahl der Primzahlen unter einer gegebenen Grosse (On the
Number of Prime Numbers less than a Given Quantity). Translated by D. R. Wilkins,
(1998).
[4] P. Sarnak P, Problems of the Millenium: the Riemann Hypothesis. Clay Mathematics
Institute (2004). http://www.claymath.org/sites/default/files/sarnak rh 0.pdf. Accessed
January 2015.
[5] G.K. Srinavasan and P. Zvengrowski, On the horizontal monotonicity of |Γ(s)|, Can.
Math. Bull., 54 (2011), 538–543.
Institute of Systems Science, Durban University of Technology, Durban 4000,
South Africa and School of Mathematics, Statistics and Computer Science, University of Kwazulu-Natal, Durban 4000, South Africa.
E-mail address: kevind@dut.ac.za