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 s = σ + it and s = σ - it 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 singlevalued 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. 2 2 2 2 Example.. The complex function: z ( x iy) x y 2ixy
It is equivalent to the actual functions: u x 2 y 2 , v 2 xy 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 f 1 ,that is z f 1 (w) .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 z n G ,
Lim z n G z0 G i.e., f ( z n ) g ( z n ) for n = 1, 2..,, then f = g in G, see Fig.1. n
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 G1 C and another holomorphic function g is defined in a domain G2 C with G1 G2 0 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:
f ( z) z n , z 1 n 0
Obviously the series diverges for values z 1 . 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 1 21s 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, s 1 } το {Re s < 0}, see Fig. 4, was proved by Riemann in 1859.We do not give a proof here of the socalled 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, s 1 }, 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 Γ(z) 0 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
lim
(x,y )(0,0)
| f(s)|| ζ(1- s) | = 0 · = 1/2 .
• |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 s0 x0 iy0 be a zero of |ζ(s)| with 0 < ℜ(s0) < 1. Then s 0 belongs to the critical line if
| ζ(s) | 1 is satisfied. ss 0 | ζ(1 - s) |
condition lim
Proof. If s is a zero of |ζ(s)|, the results summarized in Lemma 1 in order to prove whether
| ζ(s) | 1 In fact ss 0 | ζ(1 - s) |
s 0 belongs to the critical line it is enough to look to the limit lim
since | f(s)| = | ζ(s) | / | ζ(1- 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 | ζ(s 0 ) | / | ζ(1 - s 0 ) | = 0/0 but
| ζ(s) | | f(s 0 ) | . ss 0 | ζ(1 - s) | | ζ(s) | 1 is On the other hands |f(s)| = 1 on the critical line, hence when condition lim ss 0 | ζ(1 - s) | satisfied, the zero s 0 belongs to the critical line. also | ζ(s 0 ) | / | ζ(1 - s) | | f(s 0 ) | .In other words one should have lim
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
Ln 2
Ln
, specifically the closed space . Ln 2 Ln 2 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
s k 0 itk and s k 1 0 it k 1 with s k 1 s k ,then
t k 1 t k
2k , k, n N . Ln (n)
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…
(1 s) / (s) 2(2 ) s Cos(s / 2)(s) Log[ (1 s) / (s)] Log[2] sLog[2 ] Log[Cos(s / 2)(s)] 2ki but solving for s and if f ( s) Cos(s / 2)(s) and from Lemma 2 if
lim (1 s) / (s) 1 or Log[ lim (1 s) / (s)] 0 we get…
s s0
s s0
s
Log [ f ( s )] 2ki Log [2] Log [ f ( s)] 2ki 0 with Log [2 ] Log [2 ] Log [2 ]
Finally because we need real s we will have Re s
Log [2] 0.3771 .This is the lower Log [2 ]
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…
(s) / (1 s) 2(2 ) s 1 Sin(s / 2)(1 s) Log[ (s) / (1 s)] Log[2] (s 1) Log[2 ] Log[Sin(s / 2)(1 s)] 2ki but solving for s and if f ( s) Sin(s / 2)(1 s) and from Lemma 2 if lim ( s) / (1 s) 1 or Log[ lim ( s) / (1 s)] 0 we get… s s0
s
s s0
Log [ f ( s )] 2ki Log [ ] Log [ f ( s)] 2ki 0 with Log [2 ] Log [2 ] Log [2 ]
In Follow because we need real s we will take Re s
Log [ ] 0.62286 .This is the upper Log [2 ]
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 s k low it k and s upper it k with low
Log [2] Log [ ] and upper Log [2 ] Log [2 ]
then apply 1 low and 2 upper in Next do the differences…
Re s Re s 1 / 2 upper 1 / 2 0.1228
Re s k 1 / 2 Re s k 1 / 2 low 0.1228 Then this suggests our absolute symmetry around from the middle value is Re s 1 / 2 , generally as shown in Figure 6…
Fig. 6 iii) THE AVERAGE value of upper lower bound is Re s=1/2 because
Re s 1 / 2
Log [2] Log [ ] 1/ 2 Log [2 ]
iv)If we accept the non-trivial zeroes of the Riemann Zeta Function ζ(s) as s 1 0 it1 and
s 2 0 it 2 with s1 s2 , 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…
(s) / (1 s) 2(2 ) s 1 Sin(s / 2)(1 s) Log[ (s) / (1 s)] Log[2] (s 1) Log[2 ] Log[Sin(s / 2)(1 s)] 2ki the correlation theory we ill have after using Lemma 2 ie.. Log[ lim ( s) / (1 s)] 0 and s s0
relations 3 groups fields, and therefore for our case we get..
p1 ( s) y this means that f ( y ) p11 ( y ) s ,but with an initial value
s
2ik y With yLog[π] and total form Log [2 ]
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])] + q
1
w
Gamma w
1
w 1
D
1 Log 2 Log Sin y, w
2
2I
k
Log 2
y
Log
Log Gamma 1
2I Log 2
k
y
Log
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 -6 -6 -3.71908*10 -6.00526*10 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}] 10
5
0
5
10 15
10
5
0
5
10
15
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…
(1 s) / (s) 2(2 ) s Cos(s / 2)(s) Log[ (1 s) / (s)] Log[2] sLog[2 ] Log[Cos(s / 2)(s)] 2ki
the correlation theory we ill have after using Lemma 2 ie.. Log[ lim (1 s) / ( s)] 0 and s s0
relations 3 groups fields, and therefore for our case we get..
p1 ( s) y this means that f ( y ) p11 ( y ) s ,but with an initial value s
2ik y With yLog[2] and total form Log [2 ]
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] + q
1 w
Gamma w
1
w 1 w 1 y
2I 1
Log 2
Log Cos
2
Log 2
k
y
Log
2I Log Gamma
Log 2
k
y
Log
FQ = N[der /. y -> t, 30] N[1 - 2*(2*π)^(-FQ)*Cos[π*FQ/2]*Gamma[FQ]] -6
-6
Out(): 0.499994 -3.43623 I, -3.71908*10 -6.00526*10 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}] 10
5
0
5
10 15
10
5
0
Fig.8
5
10
15
w
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 z C 1. { a z 0 a 0 Re z 0 } 2. { a z 1 a 0 z 0 } which It refers to the inherent function similar in two Riemann zeta functions as (2 ) z 0 or (2 ) z 1 0 and it seems they do not have roots in C-Z because (2 ) 0 . b)For all z C Z true the form Gamma( z ) Gamma(1 z )
. Sinz
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. Sin( / 2 z ) 0 Cosh[( y)/2]Sin[( x)/2] I Cos[( x)/2]Sinh[( y)/2] 0 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. Cos( / 2 z ) 0 Cosh[( y)/2]Sin[( x)/2] I Cos[( x)/2]Sinh[( y)/2] 0 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 s) / ( s) 2(2 ) s Cos(s / 2)( s) (1 s) f ( s) ( s)
,where f ( s) 2(2 ) s Cos(s / 2)( s) but this means that occur two cases… a. (1 s) (s) ,s complex number. With this assumption implies that ( s)(1 f ( s)) 0 ( s) 0 .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 s) (s) means if s=xo+yi that (1-xo)=xo namely xo=1/2 because in this case will be verified (1 xo yi) (1/ 2 yi) (1/ 2 yi) 0 and like that verified the definition of any complex equation. Therefore if s=xo+iy then xo=1/2 to verify the equation (s) 0 . b. (1 s) (s) ,s complex number. The case for this to be verified should (s) (1 s) f (s) (s) 0 f (s) 0 .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.. ( s) / (1 s) 2(2 ) s 1 Sin(s / 2)(1 s) ( s) f ( s) (1 s) ,where f ( s) 2(2 ) s 1 Sin(s / 2)(1 s) but this means also that occur two cases… a. (1 s) (s) ,s complex number. This case is equivalent to 1.a, and therefore if s=xo+iy then xo=1/2 to verify the equation ( s) 0 . b. (1 s) (s) ,s complex number. Similarly, the case is equivalent to 1.b and therefore can not be happening, as proven.
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