PROOF OF THE RIEMANN HYPOTHESIS

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Transcript of PROOF OF THE RIEMANN HYPOTHESIS

  1. 1. Proof of the Riemann Hypothesis antzakouras 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+=s and it-=s where , 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 zA 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 zA 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 zA 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
  2. 2. 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 wV that zA such that f (z) = w, defines a new function, called reverse the function f (z) and is denoted by 1 f ,that is )(wz 1 f .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 n n 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.
  3. 3. #5.Extension of (s) from {Re s > 1} to {Re s > 0} Let us calculate We obtained another formula for (s), Fig..3 so the alternating series converges in a bigger half-plane (see Fig. 3) than the originally defined function (s) in (3), but we have to remove s = 1 since the denominator s 1 21 vanishes there. This rather easy extension of (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 (s) in the critical strip. #6.Functional equations for the Riemann zeta function. The second step, which provides a holomorphic extension for (s) from {Re s > 0, 1s } {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 (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:
  4. 4. where Fig..4 Before we give more information about the function (s), we mention that each of the two before equations , give an extension of (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 (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 (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 (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 (s) as the product that the function does not vanish for Re s > 1.Next, using the functional equation and the fact that 0(z) for Re z > 0, we see that (s) vanishes in the half-plane Re s < 0 only at the points where the function sine is zero, namely we obtain...
  5. 5. The above considerations do not tell us about the zeros of (s) in the strip 0
  6. 6. Lemma 1. (Zeta-Riemann modulus in the critical strip). The non-negative real-valued function |(s)| : C R is analytic in the critical strip. Furthermore, on the critical line, namely when (s) = one has: |(s)| =|(1 s)|. One has .1/2=0=|s)-(1||f(s)|lim (0,0)y)(x, |(s)| is zero in the critical strip, 0 < (s) < 1, if |(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 |(s)| with 0 < (s0) < 1. Then 0s belongs to the critical line if condition 1 |s)-(1| |(s)| lim 0ss is satisfied. Proof. If s is a zero of |(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)-(1| |(s)| lim 0ss In fact since |s)-(1|/|(s)|=|f(s)| and |f(s)| is a positive function in the critical strip, it follows that when s0 is a zero of |(s)|, one should have 0/0=|)s-(1|/|)(s| 00 but also |)f(s||s)-(1|/|)(s| 00 .In other words one should have |)f(s| |s)-(1| |(s)| lim 0 ss 0 . On the other hands |f(s)| = 1 on the critical line, hence when condition 1 |s)-(1| |(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 (s) apply i) Exist upper-lower bound of Re s of the Riemann Zeta Function (s) and more specifically the closed space 2 , 2 2 Ln Ln Ln Ln . ii) The non-trivial zeroes Riemann Zeta function (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 (s) as kk its 0 and 101 kk its with kk ss 1 ,then Nnk nLn k tt kk ,, )( 2 1 . Proof .. i) Here we formulate two of the functional equations
  7. 7. 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(lim 0 ss ss or 0)](/)1(lim[ 0 ssLog ss we get ]2[ 2)]([ ]2[ ]2[ Log iksfLog Log Log s 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 (s). b.For the second equation for real values with Re s0 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 ssLog ss 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 yik s With yLog[] and total form
  8. 10. With q is the count of repetitions of the Sum(). 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[]); s = N[(2 I k + y)/(Log[2] + Log[])] + w 1 q 1 w Gamma w 1 D 1 Log 2 Log Sin 2 2 I k y Log 2 Log Log Gamma 1 2 I k y Log 2 Log w , y, w 1 FQ = N[s /. y -> t, 30]; N[1 - 2^(FQ)*()^(FQ - 1)*Sin[*FQ/2]*Gamma[1 - FQ]] Out(): 0.500006 +3.43623 I -3.71908*10 -6