Hypothesis Riemann part II

HYPOTHESIS RIEMANN

PART II

Μ

antzakouras Nikos

University of Athens

–

Greece

Distribution and finding intervals contained of the zeros

of the Riemann Zeta function.

Abstract.. One of the most celebrated problems of mathematics is the Riemann hypothesis which states that all the non trivial zeros of the Zeta-function lie on the critical line s = 1/2.

all the non trivial zeros lie in the critical strip, the number of such zeros with ordinate less than T is proportional to T log T, most zeros concentrate along the critical line s = 1/2, there exists an infinity of zeros on the critical line and moreover, more than two fifth of the zeros are on the critical line.Finding the roots by Lagrange method demarcating the intervals whose are not trivial zeros of the Zeta-function.

INTRODUCTION

1.1Forms of the Riemann ζfunction..

The Riemann ζ function has two types of zeros: even negative integers see the func- tional equation set..

usually refered to as the trivial zeros, and non-trivial complex zeros.

It is proved that any non-trivial zero lies in the open strip

{

s



C

:

0



Re(

s

)



1

}

called the critical strip.Riemann hypothesis: All the complex zeros of the function ζ lie in the line

}

2

/

1

)

Re(

:

{

s



C

s



called the critical line.

#1.1 From Riemann Siegel Formula

An important part of evaluation of ζ is along the critical line. The Riemann Siegel method involves estimates of the Rieman Siegel Z-function, which is denned as

where _ is the Riemann-Siegel _ function:

Both Z(t) and θ(t) can be estimated relatively fast, which yields an efficient algorithm to calculate on the critical line. Riemann-Siegel ζ function has an asymptotic expansion

which is not convergent, but the terms decrease very rapidly for t at all large.

where Φj are entire functions which may be expressed in terms of derivatives of

The error term Rn( ) is bounded above by O

(



(2n3)/4

)

. Typically, mathematicians

have chosen n = 2, and use some conservative bound for their calculation. For example, Brent used |R2( )| < 3

(



7/4

)

for > 2000.

#1.2 Asymptotic equation satisfied by the n-th zero.

Let us start with the completed Riemann zeta function defined by

where s = 1/2 + it. In quantum statistical physics, this function is the free energy of a gas of mass less bosonic particles in d spatial dimensions when s = d + 1 up to the overall power of the temperature

1 

d

T

.Under a \modular" transformation that exchanges one spatial coordinate with Euclidean time, if one analytically continues d, physical arguments shows that it must have the symmetry..

This is the fundamental, and amazing, functional equation satisfied by the _ζ-function, which was proven by Riemann using only complex analysis. For several different ways of proving (2) see [10].

Now consider Stirling's approximation

which is valid for large t. Under this condition we also have

Therefore, using the polar representation

and the above expansions, we can write

Where

The above approximation is very accurate. For t as low as 100, it evaluates

correctly to one part in 106. Above, we are assuming t > 0. The results for t < 0 follows trivially from the relation

(1)

Now let ρ = + it be a Riemann zero. Then Αrgζ(ρ) can be defined by the limit (2)

For reasons that are explained below, it is important that 0 < δ<= 1. This limit in general is not zero. For instance, for the first Riemann zero given by

ρ=1/2 + i 14:1347, we have arg ζ (ρ) = 0.157873919880941213041945. On the critical line s = 1/2 + it, if t does not correspond to the imaginary part of a zero, the well-known function

Is defined by continuous variation along the straight lines starting from 2, then up to 2+it and finally to 1/2 + it, where argζ(2) = 0. From (1) we have

which implies that

And θ( ,-t) = -θ( , t). Denoting

we then have

From (2) we also have

therefore

Let us now approach a zero ρ =1/2+it through the δ 0+ limit. From (1) it follows that ζ(s) and χ(s) have the same zeros on the critical strip, so it is enough to consider the zeros of χ(s). From (2) we see that if ρ is a zero so is 1-ρ. Then we clearly have

where we have de_ned

The second equality in (8) follows from A = A0. Then, in the limit _ δ 0+, a zero corresponds to A = 0, B = 0 or both. They can simultaneously be zero since they are not independent.If B = 0 then A = 0, since A |ζ|. However, the converse is not necessarily true. In order to be more rigorous, one should consider the limits _ ! 0+ separately in A verses B; below we will consider taking the limit in B Firrst

The non-trivial behavior of A is mostly dictated by |ζ|. On the other hand there is much more structure in B since it contains the phases of χ(s) and χ(1 - s). It describes oscillations on the complex plane and involves t log t and _ itself. Thus let us consider B = 0. We will provide sample evidence that all zeros are characterized by this equation. The general solution of B = 0 is given by

which are a family of curves t( ). However, since χ(s) is analytic on the critical strip, we know that the zeros must be isolated points rather than curves, thus this general solution must be restricted. Let us choose the particular solution

(11)

On the critical line = ½ Пrom (7) аe СКve tСКt tСe _rst equation in (11) is already

satisfied. Then from the second equation in (11) we obtain

for n = 0,(+/-)1,(+/-)2,..., hence

Establishing the convention that zeros are labeled by positive integers,

where n = 1, 2,.... , we must replace n  n – 2. Therefore, the imaginary parts of these zeros satisfy the transcendental equation

#1.3 Explicit formula

We now show that it is possible to obtain an approximate solution to the previous transcendental equations with an explicit formula. In this approximation, there is indeed a unique solution to the equation for every n. Let us introduce the Lambert W-function,

The multi-valued W-function cannot be expressed in terms of other known elementary functions.If we restrict attention to real-valued W(x) there are two branches. Since we are interested only in positive real-valued solutions, we just need

the principal branch where W is single-valued.

Let us start with the zeros of the W-function, described by equation (13). Consider its leading order approximation, or equivalently its average since

Then we have the transcendental equation

Through the transformation

this equation can be written a

Comparing with (61) we thus we obtain

where n = 1,2,...

Using the asymptotic behavior W(x) ~ log x for large x, the n-th zero is approximately given by

as already known. The distance between consecutive ordinates is then approximately equal to which tends to zero when n

#2. Categories of complex roots and intervals contained of the Riemann zeta function .

From page 9,10 in my the first part of the proof of the Riemann , applicable according to the method of solving equations of Lagrange,

Theorem of [Lagrange Expansion].Let f(z) and g(z) be functions of z analytic on and inside a Мontour C surrounНinР г = К, КnН let e sКtisПв tСe inequКlitв еpР(г)е < ег − Ке Пor Кll г

on the perimeter of C. Then the equation in has precisely one root u = u(a) in the interior of C..[7]. Let f(z) be analytic on and insideC. Then

So, according the theorem of Lagrange we have 3 categories complex roots of the Riemann zeta functions (Eq.set)

#2.1 1st type roots of the Riemann zeta functions (2st Eq.set).

For the first category roots and by taking the logarithm of two sides of the equations, and thus we Рet…

we will have for total roots 3 groups fields(but i have interest for first group), and therefore for our case we will get, if we replace s=x:

which means that ,

but with an initial value for x is

and total form from theory Lagrange for the root is..

With k:=0,1,2,3,4,5…

……… …………..

……… …………..

But because the infinite sum approaching zero theoretically x get initial value

So we have in this case, in part, the consecutive intervals with k=n and k=n+1 for any n>=4 and for the imaginary roots.

#2.2 1st type roots of the Riemann zeta functions (1st Eq.set).

Same as in the first category roots

by taking the logarithm of two sides of the equations,

and

thus we get..

i k s s Cos Log sLog Log s s Log s s Cos s s s



















2 )] ( ) 2 / ( [ ] 2 [ ] 2 [ )] ( / ) 1 ( [ ) ( ) 2 / ( ) 2 ( 2 ) ( / ) 1 (             

and total form from theory Lagrange for the root is..

WitС k:=0,1,2,3,4,5… аe СКve..

……… ……….. ……… ………..

But because the infinite sum approaching zero theoretically x get initial value

So we have in this case, in part, the consecutive intervals with k=n and k=n+1 for any n>=4 and for the imaginary roots.

AnН toа МКses СКve to Пor Im(б) tСe relКtionsСip…

N

k

With

Log

k

x









,

,

,

]

2

[

2

)

Im(





#2.3 2st type roots of the Riemann zeta functions (1st Eq.set).

For the second category roots taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

))

k2

(2

i]

k1

2

[y]

ArcCos[Exp

(

)

(2/

)]

2

/

(

[

)

(

1

x



y



Log

Cos



x



x

in

























p

And

))

k2

(2

i]

k1

2

[y]

ArcCos[Exp

(

)

(2/

)]

2

/

(

[

)

(

1

x



y



Log

Cos



x



x

in



























p

and total form from theory Lagrange for the root is.. for 1st form

And for 2st form

] 2 [ ],

2 [ ,

, , ,

, k₁ k₂ N m Log x Log

With   





#2.4 2st type roots of the Riemann zeta functions (2st Eq.set).

For the second category roots taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

))

1)

k2

(2

i]

k1

2

[y]

ArcSin[Exp

(

)

(2/

)]

2

/

(

[

)

(

1

x



y



Log

Sin



x



x

in





























p

And

))

k2

(2

]

1

2

[y]

ArcSin[Exp

(

)

(2/

)]

2

/

(

[

)

(

1

x



y



Log

Sin



x



x











k







i









p

_in

which means that ,

] [ ],

2 [ ,

, , ,

, k₁ k₂ N m Log



x Log



With    

#2.5 3st type roots of the Riemann zeta functions (1st Eq.set).

In the third category roots we have specifically for Gamma function taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

b])

*

*

i

*

2

[Exp[x]

Gamma

)]

(

[

)

(

-1

1

x



y



Log

Gamma

x



x

in







p

And which means

that ,

and total form from theory Lagrange for the root is.. for 1st form …

]

2

[

],

2

[

,

,

,

,

b

N

m

Log

x

Log

With













#2.6 3st type roots of the Riemann zeta functions (2st Eq.set).

For the second category roots taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

b])

*

*

i

*

2

[Exp[x]

Gamma

-1

)]

1

(

[

)

(

-1

1

x



y



Log

Gamma



x



x

in







p

And which means that ,

]

[

],

2

[

,

,

,

,

b

N

m

Log



x

Log



With











PROGRAMMING

#3.The M Function – Bisection Method… The zeros of the Riemann Zeta function.

Knowing the time of the successive steps (k,k+1) of the relationship

]

2

[

2

)

Im(





Log

k

x







with

k



N

,we can calculate the roots on solving the equation



(

1

/

2



i



x

)



0

using the method Bisection .

3.1 Bisection Method.

Bisection is the division of a given curve, figure, or interval into two equal parts (halves).A simple bisection procedure for iteratively converging on a solution which is known to lie inside some interval

[

a

,

b

]

proceeds by evaluating the function in question at the midpoint of the original interval

2 b a

x  and testing to see in which of the subintervals

[

a

,

(

a



b

)

/

2

]

or

[(

a



b

)

/

2

],

b

]

the solution lies. The procedure is then repeated with the new interval as often as needed to locate the solution to the desired accuracy.

Leta_n and b_n be the endpoints at the

n

th iteration (with a₁a and b₁b) and let r_n be the

n

th approximate solution. Then the number of iterations required to obtain an error smaller than



is found by noting that

and that r_n is defined by

In order for the error to be smaller than



,

So

3.2 M-function of the Bisection Method..

We define functions

M

_d

,

M

_don an interval (α,b) according to the scheme:

I.

1

,

2

of

larger

Nearest

the

From

,

2

,

2

,

2

1 1 1



















    

k

b

a

M

M

M

d

m

d

d

M

M

M

m d d m d d d II.

1

,

2

of

smaller

Nearest

the

From

,

2

,

2

,

2

1 1 1



















    

k

b

a

M

M

M

d

m

d

d

M

M

M

m d d m d d d

for calculating the roots on solving the equation



(

1

/

2



i



x

)



0

take the limit r_S according to the sМСeme… N s M M r _n n n n

s   

    

 lim ,

lim with

M

_d and

M

_d, belong in interval

(

a

,

b

)

,

d



N

and also

)

]

2

[

)

1

(

2

,

]

2

[

2

(









Log

k

b

Log

k

a















,kN.k>=3 for the x of



(

1

/

2



i



r

_s

)



0

.

3.2.1 Programm in mathematica for Bisection method for the Zeros of



(

1

/

2



i



x

)



0

.

Using the Intervals

)

]

2

[

)

1

(

2

,

]

2

[

2

(









Log

k

b

Log

k

a















and successive steps we can compute all the roots of



(

1

/

2



i



x

)



0

. The most important is to calculate all the roots in each successive interval and therefore we will have the program for data {Integer k=4, Error approximate tol=10^-6 and Trials n=22}..

This value is the approximate root of the



(

1

/

2



i



x

)



0

by nearest error <10^-7.

In the event that within the interval we have two or more roots uncommon of course, divide similar successive intervals in the order of discovery of the first root either above or below. Such a case we have to with k=13, interval (47.8620664,51.280785) by 2 roots 48.0051088 and 49.7738324.

3.3 Explicit formula and the Zeros of



(

1

/

2



i



x

)



0

.

Consider its leading order approximation, or equivalently its average since

0

)

2

/

1

(









Arg



i

x

Then we have the transcendental equation

Through the transformation

this equation can be written a

Comparing with (61) we thus we obtain

3.3.1 Programm in mathematica for Newton method.. for the Zeros of



(

1

/

2



i



x

)



0

.

Using Newton method can reach the roots of the equation



(

1

/

2



i



x

)



0

at a very good initial value for Explicit formula.

3.4 Programm in mathematica of

Arg

(



(

1

/

2



i



x

))



m

for Explicit formula.

Consider its leading order approximation

Arg

(



(

1

/

2



i



x

))



m

,





/

2



m





/

2

. Then we have the system

Im(



(

1

/

2



i



x

))



Re(



(

1

/

2



i



x

))



tan(

m

)

with m=n-11/8 from before equation

Explicit formula.

4.Epilogue

In accordance with the foregoing we can calculate the imaginary part of the Zeta function of various methods, which are approximate approach with as much want. The method Bisection suggests that the best because it is based on interval defined, which fully defines the multitude of roots but also the position of prime numbers.

Reference

[1] B. RiemКnn, Ueber Нie AnгКСl Нer PrimгКСlen unter einer РeРebenen Gr¨osse, MonКtsberiМСte der Berliner Akademie. In Gesammelte Werke, Teubner, Leipzig (1982). See english translation in Д2], “τn tСe number oП primes less tСКn К Рiven mКРnituНe”

Д2] H. M. EНаКrНs, RiemКnn’s ГetК FunМtion, Dover PubliМКtions InМ., 1974

[3] P. Sarnak, Problems of the Millennium: The Riemann hypothesis, Clay Mathematics Institute (2004)

[4] E. Bombieri, Problems of the Millennium: The Riemann hypothesis, Clay Mathematics Institute (2000)

[5] J. B. Conrey, The Riemann Hypothesis, Notices of the AMS 50 (2003) 342

Д6] σ. Levinson, More tСКn one tСirН oП tСe гeros oП RiemКnn’s гetК-function are on = 1/2, Advances in Math. 13 (1974) 383–436

[7] J. B. Conrey, More than two fifths of the zeros of the Riemann zeta function are on the critical 47 line, J. reine angew. Math. 399 (1989) 1–26

[8] H. Bui, J. B. Conrey, M. Young, More than 41% of the zeros of the zeta function are on the critical line, Acta Arith. 150 (2011) 35–64

[9] S. Feng, Zeros of the Riemann zeta function on the critical line, arXiv:1003.0059 [math.NT] (2010) [10] E. C. Titchmarsh, The Theory of the Riemann Zeta-Function, Oxford University Press, 1988

[11] Balazard, . SAIAS, . " -Бё ф

." Вы авок. Ма ма ка 18 , 131-138, 2000.

Д12] Б , У. У. . , . . .Ма ма ч к Recreations оч к , 13 - . -: , . 75, 1987.

[13] Б , . " ч :.

"http://www.claymath.org/millennium/Riemann_Hypothesis/Official_Problem_Description.pdf

[14] S. J. Patterson, An introduction to the theory of the Riemann Zzta function, Cambridge University Press: Cambridge 1988.

[15] G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypoth`ese de Riemann, J. Math. Pures Appl. 63 (1984), 187–213.

] [ ],

2 [ ,

, , ,

, k₁ k₂ N m Log



x Log



With    

#2.5 3st type roots of the Riemann zeta functions (1st Eq.set).

In the third category roots we have specifically for Gamma function taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

b])

*

*

i

*

2

[Exp[x]

Gamma

)]

(

[

)

(

-1

1

x



y



Log

Gamma

x



x

in







p

And which means

that ,

and total form from theory Lagrange for the root is.. for 1st form …

]

2

[

],

2

[

,

,

,

,

b

N

m

Log

x

Log

With













#2.6 3st type roots of the Riemann zeta functions (2st Eq.set).

For the second category roots taking the logarithm of two sides of the equations, КnН tСus аe Рet…

and therefore for our case we will get, if we replace s=x:

b])

*

*

i

*

2

[Exp[x]

Gamma

-1

)]

1

(

[

)

(

-1

1

x



y



Log

Gamma



x



x

in







p

And which means that ,

]

[

],

2

[

,

,

,

,

b

N

m

Log



x

Log



With











PROGRAMMING

#3.The M Function – Bisection Method… The zeros of the Riemann Zeta function.

Knowing the time of the successive steps (k,k+1) of the relationship

]

2

[

2

)

Im(





Log

k

x







with

k



N

,we can calculate the roots on solving the equation



(

1

/

2



i



x

)



0

using the method Bisection . 3.1 Bisection Method.

Bisection is the division of a given curve, figure, or interval into two equal parts (halves).A simple bisection procedure for iteratively converging on a solution which is known to lie inside some interval

[

a

,

b

]

proceeds by evaluating the function in question at the midpoint of the original interval

2

b a

x  and testing to see in which of the subintervals

[

a

,

(

a



b

)

/

2

]

or

[(

a



b

)

/

2

],

b

]

the solution lies. The procedure is then repeated with the new interval as often as needed to locate the solution to the desired accuracy.

Leta_n and b_n be the endpoints at the

n

th iteration (with a₁a and b₁b) and let r_n be the

n

th approximate solution. Then the number of iterations required to obtain an error smaller than



is found by noting that

and that r_n is defined by

In order for the error to be smaller than



,

3.2 M-function of the Bisection Method..

We define functions

M

_d

,

M

_don an interval (α,b) according to the scheme:

I.

1

,

2

of

larger

Nearest

the

From

,

2

,

2

,

2

1 1 1



















    

k

b

a

M

M

M

d

m

d

d

M

M

M

m d d m d d d II.

1

,

2

of

smaller

Nearest

the

From

,

2

,

2

,

2

1 1 1



















    

k

b

a

M

M

M

d

m

d

d

M

M

M

m d d m d d d

for calculating the roots on solving the equation



(

1

/

2



i



x

)



0

take the limit r_S according to the sМСeme… N s M M r _n n n n

s   

    

 lim ,

lim with

M

_d and

M

_d, belong in interval

(

a

,

b

)

,

d



N

and also

)

]

2

[

)

1

(

2

,

]

2

[

2

(









Log

k

b

Log

k

a















,kN.k>=3 for the x of



(

1

/

2



i



r

_s

)



0

.

3.2.1 Programm in mathematica for Bisection method for the Zeros of



(

1

/

2



i



x

)



0

.

Using the Intervals

)

]

2

[

)

1

(

2

,

]

2

[

2

(









Log

k

b

Log

k

a















and successive steps we can compute all the roots of



(

1

/

2



i



x

)



0

. The most important is to calculate all the roots in each successive interval and therefore we will have the program for data {Integer k=4, Error approximate tol=10^-6 and Trials n=22}..

This value is the approximate root of the



(

1

/

2



i



x

)



0

by nearest error <10^-7.

In the event that within the interval we have two or more roots uncommon of course, divide similar successive intervals in the order of discovery of the first root either above or below. Such a case we have to with k=13, interval (47.8620664,51.280785) by 2 roots 48.0051088 and 49.7738324. 3.3 Explicit formula and the Zeros of



(

1

/

2



i



x

)



0

.

Consider its leading order approximation, or equivalently its average since

0

)

2

/

1

(









Arg



i

x

Then we have the transcendental equation

Through the transformation

this equation can be written a

Comparing with (61) we thus we obtain

3.3.1 Programm in mathematica for Newton method.. for the Zeros of



(

1

/

2



i



x

)



0

. Using Newton method can reach the roots of the equation



(

1

/

2



i



x

)



0

at a very good initial value for Explicit formula.

3.4 Programm in mathematica of

Arg

(



(

1

/

2



i



x

))



m

for Explicit formula.

Consider its leading order approximation

Arg

(



(

1

/

2



i



x

))



m

,





/

2



m





/

2

. Then we have the system

Im(



(

1

/

2



i



x

))



Re(



(

1

/

2



i



x

))



tan(

m

)

with m=n-11/8 from before equation

Explicit formula.

4.Epilogue

In accordance with the foregoing we can calculate the imaginary part of the Zeta function of various methods, which are approximate approach with as much want. The method Bisection suggests that the best because it is based on interval defined, which fully defines the multitude of roots but also the position of prime numbers.

Reference

[1] B. RiemКnn, Ueber Нie AnгКСl Нer PrimгКСlen unter einer РeРebenen Gr¨osse, MonКtsberiМСte der Berliner Akademie. In Gesammelte Werke, Teubner, Leipzig (1982). See english translation in Д2], “τn tСe number oП primes less tСКn К Рiven mКРnituНe”

Д2] H. M. EНаКrНs, RiemКnn’s ГetК FunМtion, Dover PubliМКtions InМ., 1974

[3] P. Sarnak, Problems of the Millennium: The Riemann hypothesis, Clay Mathematics Institute (2004)

[4] E. Bombieri, Problems of the Millennium: The Riemann hypothesis, Clay Mathematics Institute (2000)

[5] J. B. Conrey, The Riemann Hypothesis, Notices of the AMS 50 (2003) 342

Д6] σ. Levinson, More tСКn one tСirН oП tСe гeros oП RiemКnn’s гetК-function are on = 1/2, Advances in Math. 13 (1974) 383–436

[7] J. B. Conrey, More than two fifths of the zeros of the Riemann zeta function are on the critical 47 line, J. reine angew. Math. 399 (1989) 1–26

[8] H. Bui, J. B. Conrey, M. Young, More than 41% of the zeros of the zeta function are on the critical line, Acta Arith. 150 (2011) 35–64

[9] S. Feng, Zeros of the Riemann zeta function on the critical line, arXiv:1003.0059 [math.NT] (2010) [10] E. C. Titchmarsh, The Theory of the Riemann Zeta-Function, Oxford University Press, 1988 [11] Balazard, . SAIAS, . " -Бё ф

."Вы авок.Ма ма ка18 , 131-138, 2000.

Д12] Б , У. У. . , . . .Ма ма ч к Recreations оч к , 13 - . -: , . 75, 1987.

[13] Б , . " ч :.

"http://www.claymath.org/millennium/Riemann_Hypothesis/Official_Problem_Description.pdf

[14] S. J. Patterson, An introduction to the theory of the Riemann Zzta function, Cambridge University Press: Cambridge 1988.

[15] G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypoth`ese de Riemann, J. Math. Pures Appl. 63 (1984), 187–213.