对黎曼猜想的证明

+
β2 4
⎤ ⎥ ⎥⎦
⎡⎢⎢⎣⎛⎜⎝
n
+
1 4
−
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
(17)
then
Γ′ Γ
⎛1− s ⎜⎝ 2
⎞ ⎟⎠
+
Γ′ ⎛ 1− s Γ ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ Γ
⎛ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
0
⇔
α
=
0.
(18)
Combining (15) and (18), we obtain
i.e., Proof 2:
so,
α
=
0
⇔
A(s)
A(s
)
=
A
⎛ ⎜⎝
1 2
+ iβ
⎞ ⎟⎠
A ⎛⎜⎝
1 2
− iβ
⎞ ⎟⎠
=1,
(13)
Re ( s )
=
1 2
⇔
A(s)
A(s
)
=1.
⎛ ⎜⎝
s
=
1 2
+α
+ iβ
⎞ ⎟⎠
∨
⎛ ⎜⎝
s
=
1 2
+α
− iβ
⎞ ⎟⎠
⇒
A(s)
A(s
)
=πα
Γ
⎛ ⎜⎝
Γ
∞
ζ (s) = ∑n−s .
(1)
n=1
Recall that this function has the equation
ζ (s) = A(s)ζ (1− s) ,
(2)
where
A(
s
)
=
π
s−
Γ 1 2
⎛ ⎜⎝
1
− 2
s
⎞ ⎟⎠
.
(3)
Γ
⎛ ⎜⎝
s 2
⎞ ⎟⎠
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called trivial zero points;
z ζ ( s) has infinite complex zero points ρ = σ + iβ in region 0 < σ < 1. These zero
points are called non-trivial zero points. Furthermore, ρ,1− ρ, ρ ,1− ρ are also
2
A(s)
A (1 −
s)
≡1⇔
A ⎛⎜⎝
1 2
+α
+
iβ
⎞ ⎟⎠
A
⎛ ⎜⎝
1 2
−α
− iβ
⎞ ⎟⎠
≡1.
(10)
Since
α
≠
0
⇔
A
⎛ ⎜⎝
1 2
−α
−
iβ
⎞ ⎟⎠
≠
A
⎛ ⎜⎝
1 2
+α
− iβ
⎞ ⎟⎠
,
(11)
i.e.,
α ≠ 0 ⇔ A(s) A(s ) ≠1,
(12)
then
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( ) ( ) 0 < σ
<1 ∧
t
→∞
⇒
( ) A s
～
π
σ
−
1 2
t
1 2 −σ
,
(25)
then
Re(s) ≠
1 2
⇒
A(s)
≠ 1.
(26)
Combining
Re(s) =
1 2
⇒
A(s) ≡ 1, we obtain
Re(s) =
1 2
⇔
A(s) ≡ 1.
(27)
Lemma 4 is true.
ζ (s)
have
real
part
equal
to
1 2
.
It
is
the
well-known Riemann Hypothesis (RH). Even it was collected into Hilbert’s 23 famous mathematic problems. Although some converging insights into RH have been provided [2−9] , this mysterious and challenging conjecture has not be completely proved or disproved yet up to now. .The purpose of this paper is to prove RH by a new approach.
non-trivial zero points of ζ ( s) ;
z
ζ (ρ)
exists infinite non-trivial zero points lying on line
Re ( s )
=
1 2
.
Riemann further conjectured that all non-trivial zeros of
Re(s) =
1 2
⇔
A(s) A(s
)
=1
(9)
Proof 1: Let
s = 1 +α + iβ 2
and
s = 1 +α − iβ 2
(where
α
<
1 2
is any of real constants,
β ≠ 0 is real variable), such that Re ( s) = 1 + α , Im ( s) ≠ 0 , then
s
−
1 n
⎞ ⎟⎠ ,
(7)
where γ is Euler constant, then
∑ ∑ Γ′
Γ
(s1 ) −
Γ′ Γ
(
s2
)
=
1 s2
+
∞ n=1
n
1 + s2
−
1 s1
−
∞ n=1
n
1 + s1
∑ =
∞⎛ ⎜
n=0 ⎝
n
1 + s2
−
n
1 + s1
⎞ ⎟ ⎠
(8)
Hence, Lemma 1 is established. Lemma 2:
0
<
Re
(
s)
<
1
⇒
ζ
(n)
(s)
=
n! 2π i
∫C
(
ζ (z) ) z − s n+1
dz
,
(22)
and
∫ ∫ 0 < Re s < 1 ⇒ ζ (n) (1− s) = n!
2π i
C
ζ (1− z) ( z − s)n+1
dz
=
n! 2π i
ζ (z) C ( z − s)n+1 dz .
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3. Proof of RH
Let ρ = σ + iβ ( 0 < σ < 1) is anyone of non-trivial zero points of (n) order of ζ ( s) , then
the functions ζ ( s) and ζ (1− s) ) are both equal to zero at point ρ = σ + it . Namely,
+
Γ′ ⎛ 1− s Γ ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
0 .（15）
Since
∑ Γ′ ⎛ 1− s
Γ ⎜⎝ 2
⎞ ⎟⎠
+
Γ′ Γ
⎛1− s ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
−α
∞ n=o
notations (e.g., ⇒ , ⇔ ， ∀ ， ∧ , ∨ ). We also define a new notation Φ as the set of total
non-trivial zeros of ζ ( s) .
Riemann Hypothesis: All non-trivial zero points of
z ζ ( s) has an analytic continuation to the whole complex plane except for a simple pole at
s =1;
z ζ ( s) has zero points at s = −2, −4," , in half plane Re ( s) < 0 . These zero points are
In case of a simplified and convenient description in following sections, we need some
derivative notations (e.g., f ′( s) , f (n) ( s) ) and mathematic logic propositional connectives
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A Proof of Riemann Hypothesis
Jinzhu Han Baotou North Hospital, Baotao, Inner Mongolia, P. R. China, 014010
zzhhan@bnu.edu.cn
Abstract
2. Lemmas
Lemma 1:
∑ Re ( s )
>
0
⇒
Γ′ Γ
(s1 )
−
Γ′ Γ
( s2
)
=
∞⎛ ⎜
n=0 ⎝
n
1 + s2
−
n
1 + s1
⎞ ⎟ ⎠
,
(6)
where
Γ′ Γ
(
s
)
=
Γ′ ( s ) Γ(s)
.
Proof: Since
∑ −
Γ′ (s)
Γ
=
1 s
+γ
+
∞ n=1
⎛ ⎜⎝
1 n+
⎛ ⎜⎝
1 4 1 4
− +
α 2 α 2
− +
iβ 2 iβ 2
⎞ ⎟⎠ ⎞ ⎟⎠
⋅
Γ
⎛ ⎜⎝
Γ
⎛ ⎜⎝
1 4 1 4
− +
α 2 α 2
+ −
iβ 2 iβ 2
⎞ ⎟⎠ ⎞ ⎟⎠
,.
(14)
A(s) A(s ) = 1⇒ ( A(s) A(s ))′ = 0
⇒
Γ′ ⎛ 1− Γ ⎜⎝ 2
s
⎞ ⎟⎠
In this paper, a proof of the Riemann Hypothesis is given. We first obtain the ratio
ζ (s) ζ (1− s)
limited value of the Riemann zeta functions
to
at non-trivial
⎡⎛ ⎢⎢⎣⎜⎝
n
+
1 4
+
⎛ ⎜⎝
n
+
1 4
⎞2 ⎟⎠
−
α2 4
−
β2 4
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
⎡⎛ ⎢⎢⎣⎜⎝
n
+
ห้องสมุดไป่ตู้
1 4
−
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
,
(16)
and
∑∞
⎛ ⎜⎝
n
+
1 4
⎞2 ⎟⎠
−
α2 4
−
β2 4
≠0,
n=o
⎡⎢⎢⎣⎛⎜⎝
n
+
1 4
+
α 2
⎞2 ⎟⎠
Riemann Hypothesis.
ζ (s)
Γ (s)
Keywords: Riemann Hypothesis, zeta function
, gamma function
AMS: 11M26
1. Introduction
In 1859, Riemann [1] defined the zeta function
Re ( s )
=
1 2
⇔
A(s) A(s ) =1.
Lemma 2 is true.
Lemma 3:
0 < Re ( s) < 1 ⇒ ζ (n) (s) = ζ (n) (1− s)
(21)
Proof: In region 0 < Re ( s) < 1, if we select any suitable simple closed contour C, which is
(23)
Consequently, 0 < Re ( s) < 1 ⇒ ζ (n) ( s) = ζ (n) (1− s) .
Lemma 3 is established.
Lemma 4:
Re(s) =
1 2
⇔
A(s) ≡ 1.
(24)
Proof: Let s = σ + it , some mathematicians [10] had proved that
A(s)
A(s
)
=1⇒
Re ( s )
=
1 2
.
(19)
Note that
Re ( s )
=
1 2
⇒
A(s)
A(s
)
=
A
⎛ ⎜⎝
1 2
+ iβ
⎞ ⎟⎠
A
⎛ ⎜⎝
1 2
− iβ
⎞ ⎟⎠
=1.
(20)
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Combining (19) and (20), we also have
zero points by L’Hospital Rule. And then we prove that all non-trivial zero points of
ζ (s)
1
have real part 2 . Thus, we provide evidence for the correctness of the
ζ (s)
have
real
part
equal
to
1 2
.
It
can
be
expressed as
∀ρ = σ + iβ ∈ Φ ⇒
σ
=
1 2
(5)
Before we give a proof of RH, we require the following Lemmas.
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symmetry for real axis and line
Re ( s )
=
1 2
,
then
the
two
functions,
ζ (s)
and
ζ (1− s) , have
completely identical boundary values in total. According to Cauchy derivative formula, we have
Γ ( s) here is the Euler gamma function
∏ 1
Γ(s)
=
s
∞ n=1
⎛⎜⎝1 +
s n
⎞ ⎟⎠
⎛⎜⎝1 +
1 n
⎞−s ⎟⎠
.
(4)
Moreover, this zeta function ζ ( s) has the following properties: