Proof of the Riemann Hypothesis

Μantzakouras Nikos(nikmantza@gmail.com)

August 2015 - Athens

Abstract: The Riemann zeta function is one of the most Leonhard Euler important and fascinating functions in mathematics. Analyzing the matter of conjecture of Riemann divide our analysis in the zeta function and in the proof of conjecture, which has consequences on the distribution of prime numbers.

.Introduction #1.The Riemann Zeta Function

Let C denote the complex numbers. They form a two dimensional real vector space spanned by 1i and where i is a fixed square root of -1, that is, C = {x + iy : x,y R}

#2.Definition of Zeta. The Riemann zeta function is the function of a complex variable

It is conventional to write it + σ= s and it - σ=s where σ, t R. As the equality on the second line follows from unique prime factorization, we can say that the equation

is an analytic statement of unique prime factorization." It is known as the Euler product”. This gives us a first example of a connection between the zeta function and the primes.[3]

#3. Definition of complex function,Inverse functions.

Suppose A, B two subsets of the complex plane C. Each f tally the sets A and B f: A → B with

the restriction that each z∈A corresponds one and only one point w = f (z) ∈B, called

complex function with domain the set A and the range set B. The z called independent

variable and the dependent variable w.In cases where the above restriction does not apply,

ie where that each value of the variable z∈A match more than one value f (z), the f match is

called multi-valued function in contrast to the previous case where you use the term single-

valued function or simply function. If we set z = x + iy and w = u + iv, then the function w = f

(z) corresponds to each complex number z∈A with coordinates x and y two real numbers u

and v.In other words, the set A set of two real functions u = u (x, y) and v = v (x, y) of two

real variables x and y. So a complex function w = f (z) is equivalent to two real functions u =

u (x, y) and v = v(x, y) of two real variables.

Example.. The complex function: ixyyxiyxz 2)( 2222

It is equivalent to the actual functions: xyvyxu 2,22

If a complex function w = f (z) is one to one (1-1) and on (i.e. z1 ≠ z2⇒ f (z1) ≠ f (z2) or

f (z1) = f (z2) ⇒ z1 = z2 and f (A) = B), then the correlation that corresponds to each w∈V

that z∈A such that f (z) = w, defines a new function, called reverse the function f (z) and is

denoted by 1f ,that is )(wz 1f .So if we have Zeta(z1)=Zeta(z2) ⇒ z1 = z2 in the domain

of set A, if occurring earlier.In this case Zeta[z] is 1-1 which is acceptable.

#4.Extensions of holomorphic functions

One of the main properties of holomorphic functions is uniqueness in the sense that if two

holomorphic functions f and g defined in a domain G are equal on a sequence Gzn ,

GzGzLim nn

0 i.e., )()( nn zgzf for n = 1, 2..,, then f = g in G, see Fig.1.

Fig..1

Of course such a property is not true for functions in real calculus. In particular, if a

holomorphic function f is de_ned in a domain CG 1 and another holomorphic function g

is defined in a domain CG 2 with 021 GG and f = g on the intersection, then g is

determined uniquely by f; see Fig. 2.

Fig..2

f cannot be holomorphically extended beyond this disc. A simple example:

1,)(0

n

n zzzf

Obviously the series diverges for values 1z . However, the function f can be

holomorphically extended to the entire complex plane C except z = 1, by the formula

A natural question appears: whether the Riemann zeta function can be holomo-rphically extended beyond the half-plane Re s > 1? The answer is yes, which we show in two steps. The first step is easy, the second more dificult.

#5.Extension of ζ(s) from {Re s > 1} to {Re s > 0}

Let us calculate

We obtained another formula for ζ(s),

Fig..3

so the alternating series converges in a bigger half-plane (see Fig. 3) than the originally defined function ζ(s) in (3), but we have to remove s = 1 since the denominator

s 121 vanishes there. This rather easy extension of ζ(s) from s with Re s > 1 to s with Re s > 0 is already signi_cant as it allows us to formulate the Riemann Hypothesis about the zeros of ζ(s) in the critical strip. #6.Functional equations for the Riemann zeta function.

The second step, which provides a holomorphic extension for ζ(s) from {Re s > 0, 1s } το {Re s < 0}, see Fig. 4, was proved by Riemann in 1859.We do not give a proof here of the so-called functional equation, but the proof can be found, e.g. in the book by Titchmarsh . Alternatively one can first holomorphically extend ζ(s) step by step to half-planes {Re s > k,

1s }, where k is any negative integer. For details of this method, see for example the papers [5] and [6]. There are few versions of the functional equation; here we formulate two of them:

where

Fig..4

Before we give more information about the function ζ(s), we mention that each of the two before equations , give an extension of ζ(s) on the entire plane C except s = 1, as is illustrated in Fig. 4. The gamma function was already known to Euler. It generalizes the factorial n!, namely

Its basic properties are that ζ(z) is holomorphic on the entire plane C except for the points z = 0,-1,-2,... . At these points there are simple singularities, called poles, where we have the limits

From the de_nition of the gamma function it is not clear that it can be extended onto the entire plane except for z = 0,-1,-2,... , and that is non-vanishing. Fortunately there are other equivalent de_nitions of ζ(z) from which these properties follow more easily; see[8,9]. Namely we have

From the second formula for the gamma function we see that ζ(z) is holomorphic and nonvanishing for Re z > 0.

#7.The Riemann Hypothesis The Riemann Hypothesis is the most famous open problem in mathematics. Originally formulated by Riemann, David Hilbert then included the conjecture on his list of the most important problems during the Congress of Mathematicians in 1900, and recently the hypothesis found a place on the list of the Clay Institute's seven greatest unsolved problems in mathematics.It follows from the formula of ζ(s) as the product that the function does not

vanish for Re s > 1.Next, using the functional equation and the fact that 0Γ(z) for Re z >

0, we see that ζ(s) vanishes in the half-plane Re s < 0 only at the points where the function sine is zero, namely we obtain...

The above considerations do not tell us about the zeros of ζ(s) in the strip 0 <Re s < 1. Actually there are zeros in this strip and they are called nontrivial zeros. Calculation of some number of these nontrivial zeros shows that they are lying exactly on the line Re s = 1/2 , called the critical line; see Fig. 5. Now with the help of computers it is possible to calculate

Fig..5

an enormous number of zeros, currently at the level of 1013 (ten trillion). It is interesting to mention that before the computer era began roughly in the middle of the twentieth century, only about a thousand zeros were calculated. Of course all of these zeros are calculated with some (high) accuracy: they are lying on the critical line. However, there is no proof that really all nontrivial zeros lie on this line and this conjecture is called the Riemann Hypothesis. Riemann Hypothesis: All nontrivial zeros are on the line Re s = 1/2

Many great mathematicians have contributed to a better understanding of the Riemann Hypothesis. There is no room to even partially list them here. Only we mention four of them: Andr_e Weil (1906 - 1998), Atle Selberg (1917 - 2007), Enrico Bombieri (1940 ), and Alain Connes (1947 ). The last three received the Fields Medal (in 1950, 1974, and 1982, respectively), which is considered an equivalent to a Nobel Prize in mathematics. The Fields Medal is awarded only to scientists under the age of forty. If someone proves the Riemann Hypothesis and is relatively young, then they surely will receive this prize.

.Proof of Hypothesis Riemann

#8.The elementary theorems for the nontrivial zeros of ζ(z) = 0 .

All previous constitute general theoretical knowledge that helps to prove the Hypothesis. The essence but the proof is in 3 theorems that will develop and follow specifically in the Τheorem.1 which mostly bordered the upper and lower bound of the the nontrivial zeros roots in each case the equations of paragraph 6. Then in Theorem 2 is approached sufficiently generalized finding method roots of Lagrange, real root of ζ (z) = 0 ie special Re s = 1/2.Ιn the other theorem 3 and using imaging one to one (1-1) of Zeta Function, calculated on exactly the premise of Conjecture’s. So the ability to understand the evidence of the Hypothesis of Riemann and given step by step with the most simplest theoretical background. But we must accept the amazing effort they achieved two researchers with names Carles F. Pradas[1] also Kaida Shi [2] and arrived very close if not exactly with the achievement of proof.

Lemma 1. (Zeta-Riemann modulus in the critical strip). • The non-negative real-valued function |ζ(s)| : C → R is analytic in the critical strip. • Furthermore, on the critical line, namely when ℜ(s) = ½ one has: |ζ(s)| =|ζ(1 − s)|. • One has

. 1/2 = · 0 = |s) - ζ(1||f(s)|lim(0,0)y)(x,

• |ζ(s)| is zero in the critical strip, 0 < ℜ(s) < 1, if |ζ(1 − x − iy)| = 0, with 0 < x < 1 and y ∈ R.

Lemma 2. (Criterion to know whether a zero is on the critical line).

Let 000 iyxs be a zero of |ζ(s)| with 0 < ℜ(s0) < 1. Then 0s belongs to the critical line if

condition 1 |s)-ζ(1|

|ζ(s)|lim

0ss

is satisfied.

Proof. If s is a zero of |ζ(s)|, the results summarized in Lemma 1 in order to prove whether

0s belongs to the critical line it is enough to look to the limit 1 |s)-ζ(1|

|ζ(s)|lim

0ss

In fact

since |s)-ζ(1|/|ζ(s)| = |f(s)| and |f(s)| is a positive function in the critical strip,

it follows that when s0 is a zero of |ζ(s)|, one should have 0/0 = |)s-ζ(1|/|)ζ(s| 00 but

also |)f(s| |s)-ζ(1|/|)ζ(s| 00 .In other words one should have |)f(s| |s)-ζ(1|

|ζ(s)|lim 0

ss 0

.

On the other hands |f(s)| = 1 on the critical line, hence when condition 1 |s)-ζ(1|

|ζ(s)|lim

0ss

is

satisfied, the zero 0s belongs to the critical line.

Theorem 1.

For the non- trivial zeroes of the Riemann Zeta Function ζ(s) apply

i) Exist upper-lower bound of Re s of the Riemann Zeta Function ζ(s) and more

specifically the closed space

2,

2

2

Ln

Ln

Ln

Ln.

ii) The non-trivial zeroes Riemann Zeta function ζ(s) of upper-lower bound distribute

symmetrically on the straight line Re s=1/2.

iii) THE AVERAGE value of upper lower bound of Re s=1/2.

iv) If we accept the non- trivial zeroes of the Riemann Zeta Function ζ(s) as

kk its 0 and 101 kk its with kk ss 1 ,then

NnknLn

ktt kk ,,

)(

21

.

Proof ..

i) Here we formulate two of the functional equations…

We look at each one individually in order to identify and set of values that we want each

time ..

a. For the first equation and for real values with Re s>0 and if take logarithm of 2 parts of

equation then we have…

ikssCosLog

sLogLogssLogssCosss s

2)]()2/([

]2[]2[)](/)1([)()2/()2(2)(/)1(

but solving for s and if )()2/()( ssCossf and from Lemma 2 if

1)(/)1(lim0

ssss

or 0)](/)1(lim[0

ssLogss

we get…

]2[

2)]([

]2[

]2[

Log

iksfLog

Log

Logs

with 0

]2[

2)]([

Log

iksfLog

Finally because we need real s we will have Re s 3771.0]2[

]2[

Log

Log.This is the lower

bound which gives us, the first of Riemann Zeta Function of ζ(s).

b.For the second equation for real values with Re s<1 and if take logarithm of 2 parts of

equation then we have…

ikssSinLog

LogsLogssLogssSinss s

2)]1()2/([

]2[)1(]2[)]1(/)([)1()2/()2(2)1(/)( 1

but solving for s and if )1()2/()( ssSinsf and from Lemma 2

if 1)1(/)(lim0

ssss

or 0)]1(/)(lim[0

ssLogss

we get…

]2[

2)]([

]2[

][

Log

iksfLog

Log

Logs

with 0

]2[

2)]([

Log

iksfLog

In Follow because we need real s we will take Re s 62286.0]2[

][

Log

Log.This is the upper

bound which gives us, the second of Riemnn Zeta Function of ζ(s). So we see that exists the

lower and upper bound for Re s and is well defined.

ii) Assuming klowk its and kupper its with ]2[

]2[

Log

Loglow and

]2[

][

Log

Logupper

then apply low 1 and upper 2 in Next do the differences…

1228.02/12/1ReRe upperss

1228.02/1Re2/1Re lowkk ss

Then this suggests our absolute symmetry around from the middle value is 2/1Re s ,

generally as shown in Figure 6…

Fig. 6

iii) THE AVERAGE value of upper lower bound is Re s=1/2 because

2/1Re

s 2/1]2[

][]2[

Log

LogLog

iv)If we accept the non-trivial zeroes of the Riemann Zeta Function ζ(s) as 101 its and

202 its with 21 ss ,

suppose that the real coordinate 0 of each non-trivial zero of the Riemann Zeta, [2]

function ζ(s ) corresponds with two imaginative coordinates t1 and t2 , then, we have folloing equation group:

Taking the first equation minus the second equation, we obtain

where

=0

Enable above expression equals to zero, we must have

so, we obtain

Thus, theorem 1 has been proved.

Theorem 2.

For real part of non-trivial zeroes Riemann Zeta function ζ(s), the upper-lower bound converge in line Re s=1/2, then solving the 2 equations of the Riemann zeta functions. Proof.. Using the Generalized theorem of Lagrange (GRLE),[7] to solve an equation attempt to bring the equation in such form, as to give us the limit of the roots, which converge on one number. Taking advantage of the method, we develop resolve in 2 parts to show the convergence as a number, using the two equations of the Riemann zeta functions .

a)Solve of first equation…

For real values with Re s>0 and if take logarithm of 2 parts of equation[7] then we have…

ikssSinLog

LogsLogssLogssSinss s

2)]1()2/([

]2[)1(]2[)]1(/)([)1()2/()2(2)1(/)( 1

the correlation theory we ill have after using Lemma 2 ie.. 0)]1(/)(lim[0

ssLogss

and

relations 3 groups fields, and therefore for our case we get..

ysp )(1 this means that sypyf )()( 1

1 ,but with an initial value

]2[

2

Log

yiks

With yLog[π] and total form

With q is the count of repetitions of the Sum(Σ). We very simply with a language

mathematica to calculate with data q=25,and we get the program….

In():Clear[k,q,y] k := 1; q := 25; t := (Log[π]); s = N[(2 I π k + y)/(Log[2] + Log[π])] +

w 1

q

1w

Gamma w 1

D 1 Log 2

Log Sin 22 I k y

Log 2 LogLog Gamma 1

2 I k y

Log 2 Log

w

,

y, w 1

FQ = N[s /. y -> t, 30]; N[1 - 2^(FQ)*(π)^(FQ - 1)*Sin[π*FQ/2]*Gamma[1 - FQ]]

Out(): 0.500006 +3.43623 I -3.71908*10

-6-6.00526*10

-6 I

we see a good approximation of upper Re s=0.500006 of the order 10-6

Schematically illustrate get fig 7 with program mathematica…

Clear [z]; f=Log[1-z]/Log[z]-2*(2*π)^(-z)*Cos[π*z/2]*Gamma[z];ContourPlot[{Re[f/.z->x+I y]==0,Im[f/.z->x+I y]==0},{x,-15,15},{y,-10,10},ContourStyle->{Red,Blue}]

15 10 5 0 5 10 15

10

5

0

5

10

Fig 7

b)Solve of second equation…

For real values with Re s<1 and if take logarithm of 2 parts of equation then we have…

ikssCosLog

sLogLogssLogssCosss s

2)]()2/([

]2[]2[)](/)1([)()2/()2(2)(/)1(

the correlation theory we ill have after using Lemma 2 ie.. 0)](/)1(lim[0

ssLogss

and

relations 3 groups fields, and therefore for our case we get..

ysp )(1 this means that sypyf )()( 1

1 ,but with an initial value

]2[

2

Log

yiks

With yLog[2] and total form

With q is the count of repetitions of the Sum(Σ). Similarly forming the program a language

mathematica to calculate with data q=25,we have….

In():k := 1; q := 25; t := -Log[2]; S =-(2 I π k + y)/Log[2π] +

w 1

q

1w

Gamma w 1

yw 1

1 Log 2 Log Cos 2

2 I k y

Log 2 LogLog Gamma

2 I k y

Log 2 Log

w

FQ = N[der /. y -> t, 30] N[1 - 2*(2*π)^(-FQ)*Cos[π*FQ/2]*Gamma[FQ]] Out(): 0.499994 -3.43623 I, -3.71908*10

-6-6.00526*10

-6 I

a good approximation of lower Re s=0.499994 of the order 10-6

Schematically illustrate get fig 8 with program mathematica…

Clear [z];f=Log[z]/Log[1-z]-2^z*(π)^(z-1)*Sin[π*z/2]*Gamma[1-z];ContourPlot[{Re[f/.z->x+I y]==0,

Im[f/.z->x+I y]==0},{x,-15,15},{y,-10,10},ContourStyle->{Red,Blue}]

15 10 5 0 5 10 15

10

5

0

5

10

Fig.8

With this analysis as we see the lower and upper limit converge in value Re s=1/2, what we

want results which proves the theorem2.

Theorem 3.

The Riemann Hypothesis states that the nontrivial zeros of ζ (s) have real part equal to 1/2.

Proof…

Constant Hypothesis ζ (s)=0. In this case we use the 2 equations of the Riemann zeta functions, so if they apply what they represent the ζ (s) and ζ (1-s) to equality. BEFORE We developed the method make three assumptions:

i) a)For Cz

1. { 0Re00 zaa z }

2. { 001 zaa z } which It refers to the inherent function similar in two Riemann zeta functions as

0)2( z or 0)2( 1 z and it seems they do not have roots in C-Z because 0)2( .

b)For all ZCz true the form zSin

zGammazGamma

)1()( .

From this form resulting that Gamma(z)=0 or Gamma(1-z)=0 don’t has roots in C-Z. c)Sin(π/2z)=0 or Cos(π/2z)=0 More specifically if z=x+yi then 1. 0y)/2] Sinh[( x)/2]Cos[( I x)/2]Sin[( y)/2] Cosh[( 0)2/( zSin

Has solution

of the generalized solution seems that the pairs (x, y) will always arise integers which makes impossible the case is 0 <x <1, therefore it has not roots the equation Sin(π/2z)=0 2. 0y)/2] Sinh[( x)/2]Cos[( I x)/2]Sin[( y)/2] Cosh[( 0)2/( zCos

Has solution

as we can see again of the generalized solution seems that the pairs (x, y) will always arise integers which makes impossible the case is 0 <x <1, therefore it has not roots the equation Cos(π/2z)=0. With these three cases exclude the case one of these three equations to zero for z=x+yi And be valid concurrently 0<x<1, because x in Z.

ii) Therefore analyze the two equations of the Riemann zeta functions and try to find common solutions… 1.For the first equation and for real values with Re s>0 apply..

)()()1()()2/()2(2)(/)1( ssfsssCosss s

,where )()2/()2(2)( ssCossf s but this means that occur two cases…

a. )()1( ss ,s complex number.

With this assumption implies that 0)(0))(1)(( ssfs .In Theorem 1, iv we

showed that the function ζ(s) is 1-1 and if ζ(xo+yi)=ζ(x’o+y’i) then y=y’ and xo=xo’ ,but also

appy that ζ(xo+yi)=ζ(xo-yi)=0. The form )()1( ss means if s=xo+yi that (1-xo)=xo

namely xo=1/2 because in this case will be verified

0)2/1()2/1()1( yiyiyixo and like that verified the definition of any

complex equation. Therefore if s=xo+iy then xo=1/2 to verify the equation 0)( s .

b. )()1( ss ,s complex number.

The case for this to be verified should 0)(0)()()1()( sfssfss .But

this case is not possible because, as we have shown in Section i, the individual functions of f (s) is not zero for s complex number. 2.For the second equation and for real values with Re s<1 apply..

)1()()()1()2/()2(2)1(/)( 1 ssfsssSinss s ,where

)1()2/()2(2)( 1 ssSinsf s but this means also that occur two cases…

a. )()1( ss ,s complex number.

This case is equivalent to 1.a, and therefore if s=xo+iy then xo=1/2 to verify the equation

0)( s .

b. )()1( ss ,s complex number.

Similarly, the case is equivalent to 1.b and therefore can not be happening, as proven.

References

[1]. Carles F. Pradas Azimut S. I.Barcelona [2]. A Geometric Proof of Riemann Hypothesis,Kaida Shi [3]. An Exploration of the Riemann Zeta Function and its

Applic.to the Theory of Prime Number Distribution, Elan Segarra

[4].R. Dwilewicz and J. Min,The Hurwitz zeta function as a convergent series. Rocky Mountain J. Math. 36 (2006),

[5].G. Everest, C. Rottger and T. Ward, The continuing story of zeta.The Math. Intelligencer 31 (2009) [6].T.M. Apostol, Introduction to Analytic Number Theory.

Springer -Verlag, New York - Heidelberg - Berlin 1976. [7].Solve Polynomials and Transcetental equations use Mathematica, Mantzakouras Nikos 2014

[8].RobertE, Green and StevenG,Krantz,Fuction Theory of One Complex variables [9].Elias M.Stein and Rami Shakarchi; Complex Analysis Prinseton 2007 [10].L.Ahlfors, Complex An. 3rd Edition, McGraw Hill, NY 1979: