Approaching the Riemann   Hypothesis with Mathematica

Lev Aizenberg
Department of Mathematics and Computer Science,
Bar-Ilan University
Ramat-Gan 52900, Israel
e-mail: aizenbrg@bimacs.cs.biu.ac.il

Victor Adamchik
Wolfram Research, Inc.
100 Trade Center Dr.
Champaign IL 61820, USA
e-mail: victor@wolfram.com

Vadim E. Levit 
Department of Computer Systems
Center for Technological Education
Tel-Aviv University
52 Golomb St., P.O.Box 305
Holon 58102, Israel
e-mail: levitv@barley.cteh.ac.il

Abstract

The method of the analytic continuation of the function given on a part of the boundary into the whole domain is applied to the Riemann hypothesis on the Zeta function zeros. Comprehensive numerical experiments have been performed with the Mathematica Version 3.0. The computational approach confirmed the Riemann hypothesis.

Introduction

The famous Riemann zeta function ζ(z) is defined by

ζ(z) = (∑_(n = 1)^∞) (1/(n^( z)))

for ℜ𝔢(z) > 1, and by

ζ(1 - z)=(2^(1 - z)) (π^(-z)) cos(((π z)/2)) Γ(z) ζ(z)

for ℜ𝔢(z) < 1. The Riemann zeta function is meromorphic everywhere in the complex z-plane except for a simple pole at z = 1. In 1859 Riemann studying the problem of the distribution of prime numbers came up with the following

Conjecture

All complex zeros of ζ(z) lie on the critical line ℜ𝔢(z) = (1/2)

Though the great deal of efforts have been performed and many outstanding mathematicians have worked on it, the hypothesis neither been proved nor disproved. However, the latest numerical computations of complex zeros of the Riemann zeta function (see [4],[5] and [6]) allow us believe the hypothesis to be true.

In this article we shortly describe a new method of approaching the Riemann hypothesis. The idea is based on the solution of the Carleman boundary value problem, which were successfuly solved by L. Aizenberg (see [1], [2] and [3]). It turns out that the Riemann hypothesis can be envisaged as the Carleman problem. Armed with the new mathematical technique we performed here the computational approach confirmed the Riemann hypothesis. It is worth to mention that in comparison with the previous numerical verifications of the Riemann hypothesis using supercomputer tools, we arrived at our conclusion with only a regular Pentium computer at hand.

Underlying Mathematics

Let K = {z : |z| < 1} be a unit disk, G be a chord, connecting two points of ∂K and D be a segment between G and the upper part of ∂K:

\(\(Show[
      Graphics[\ 

        		{Circle[{0, 0}, 1], PointSize[0.015], \ Point[{0, 0}], \ 
			
          Text["0", {0, \(-0.15\)}, 
            TextStyle\  -> {FontSlant -> "Italic", FontSize -> 14}], 

          		\ \ Point[{1, 0}], PointSize[0.008], 
          Text["1", {1.05, \(-0.15\)}, 
            TextStyle\  -> {FontSlant -> "Italic", FontSize -> 14}], 

          			Text["K", {\(-0.4\), 0}, \ 
            TextStyle\  -> \ {FontSlant -> "Italic", FontSize -> 14}\ ], 
          
		\ \ Thickness[0.008], \ Line[{{\(-0.8\), 0.6}, {0.8, 0.6}}], 
          
		\ \ Thickness[0.004], \ Dashing[{0.01, 0.01}], 
		\ \ 
          Line[{{\(-1.3\), 0}, {\(-0.4\), 0.6}}], \ 
          Text["G", {\(-1.3\), \(-0.15\)}, \ 
            TextStyle\  -> \ {FontSlant -> "Italic", FontSize -> 14}\ ], 
          
			Text["D", {0, 0.75}, \ 
            TextStyle\  -> \ {FontSlant -> "Italic", FontSize -> 14}\ ], 
          
			Text["∂K", {0.55, \(-1\)}, \ 
            TextStyle\  -> \ {FontSlant -> "Italic", FontSize -> 14}\ ]

          		}]]
\)\)

Suppose that 0 ∉ \(D\&_\), and f ∈ C(G) ⋂ (L^1)(G). The Carleman problem is formulated as it follows: whether there exists an analytic continuation of function f given on a boundary G into domain D. In has been proved (see [Aiz,1995]) that the condition

(OverBar[(lim_(n->∞))]) ((\[LeftBracketingBar](Subscript[a, n])\[RightBracketingBar])^(1/n)) = 1,

where

(Subscript[a, n])=(∫_G)((f(ξ))/(ξ^(n + 1)))\[DifferentialD]ξ

is neccesary and sufficient for the analytic continuation of the function f from the chord G into the domain D. Moreover, such analytic continuation does not exist, if and only if

(OverBar[(lim_(n->∞))]) ((\[LeftBracketingBar](Subscript[a, n])\[RightBracketingBar])^(1/n)) > 1

Riemann Hypothesis

We will approach the Riemann hypothesis in the following way.  Consider the strip K = { z : 0 < ℜ𝔢(z) < 1} and the function(1/(ζ(z))) defined on the interval (0,1). Our goal is to analytically continue the function(1/(ζ(z))) from the given interval into the whole strip K. Taking into account that all complex zeros of the zeta function lie in the strip K and they are symmetrical so that if ρ is a zero then \(TraditionalForm\`ρ\&_\), 1-ρ and 1-\(TraditionalForm\`ρ\&_\) are also zeros, it is quite sufficient to consider the half strip Ω = { z : 0 < ℜ𝔢(z) < (1/2)} and find out whether (1/(ζ(z))) can be analytically continued into Ω. If that is a case then it would mean that (1/(ζ(z)))does not have any singular points in Ω, or, other words, that there are no zeros of the zeta function in Ω, and thus in K. It would follow then that all zeros are on a critical line ℜ𝔢(z) = (1/2).

To make a use of the necessary and sufficient conditions decribed in the previous section, we have to map the strip Ω into the sector of the unit disk. This could be done by the function tan(z), which maps the strip Ξ = { z : -(π/4) < ℜ𝔢(z) < (π/4) } into the unit disk.  In the same way the function  tan(z + i δ), where δ>0 is an arbitrary constant maps the strip Ξ into the sector of the unit disk, so that the interval (- (π/4), (π/4)) maps onto the chord. Performing necessary calculations we arrive at the following

Proposition

The Riemann hypothesis on zeros of the Zeta function is valid if and only if

(OverBar[(lim_(n->∞))]) ((\[LeftBracketingBar](Subscript[a, n])\[RightBracketingBar])^(1/n)) =(OverBar[(lim_(n->∞))]) ((\[LeftBracketingBar](Subscript[b, n])\[RightBracketingBar])^(1/n)) = 1,

where

(Subscript[a, n])=(∫_0^((1/2)))(((((i ((e^(-2 δ)) - 1)-2 (e^(-δ)) cos(2  π x) )/(1 + (e^(-2 δ)) - 2 (e^(-δ)) sin(2 π x)))))^n)(1/(ζ(x)))\[DifferentialD]x

(Subscript[b, n])=(∫_((1/2))^1)(((((i ((e^(-2 δ)) - 1)+2 (e^(-δ)) cos(2  π x) )/(1 + (e^(-2 δ)) + 2 (e^(-δ)) sin(2 π x)))))^n)(1/(ζ(x)))\[DifferentialD]x

and δ > 0 is an arbitrary constant.

Numerical Tests

We shall begin to test the Riemann hypothesis with the following simple numeric routines

a(δ_, n_):=((\[LeftBracketingBar]NIntegrate[(((((I 
                              ((E^(-2 δ)) - 1)
                                 - 2 (E^(-δ)) 
                              cos(2 π x))/(1 + 
                            (E^(-2 δ)) - 
                            2 (E^(-δ)) 
                              sin(2 π x)))))^n)(1/(ζ(x))),{x, 0, (1/2)}]\[RightBracketingBar])^(1/n))

b(δ_, n_):=((\[LeftBracketingBar]NIntegrate[(((((I 
                              ((E^(-2 δ)) - 1)
                                 + 2 (E^(-δ)) 
                              cos(2 π x))/(1 + 
                            (E^(-2 δ)) + 
                            2 (E^(-δ)) 
                              sin(2 π x)))))^n)(1/(ζ(x))),{x, (1/2), 1}]\[RightBracketingBar])^(1/n))

Let us plot a graphic of the behaviour of these sequences with respect to n. Since δ is an arbitrary positive parameter we can set it to any number, for example, to 2. Then changing n from 10 to 70, we have

Plot[{a[2,  n],  b[2,  n],  1},  {n,  10,  70}]

The plot is in favor of the Riemann hypothesis. It shows that both sequences monotonically approach one. However, that does not guarantee that sequences would behave in the similar way for bigger values of n. Before to perform further numerical tests we need to investigate the behaviour of the integrands . Let

\(TraditionalForm
    \`h(δ_, n_, x_) := 
      \((\(I\ \((E\^\(\(-2\)\ δ\) - 1)\) - 
              2\ E\^\(-δ\)\ 
                \(cos(2\ π\ x)\)\)\/\(E\^\(\(-2\)\ 
                  δ\) - 
              2\ E\^\(-δ\)\ \(sin(2\ π\ x)\) + 1\))
          \)\^n\)

Then setting δ=2 we plot two graphics of the integrand, one with n = 100:

Plot[intgrd[2, 100, x] // Re,  {x,  0.18,  0.32}];

and second. with n = 400:

Plot[h[2, 400,  x] // Re,  {x,  0.18,  0.32}];

We see than the bigger n than the higher oscillation of the integrand. Now  let us set δ to 10 and n to 400

Plot[h[10, 400,  x] // Re,  {x,  0,  1/2}] l

It is quite obvious to guess that we can eliminate the high oscillation of the integrand by increasing the arbitrary parameter δ. However it could be easily proven by considering two limiting cases of the integrand, one when δ->0

(lim_(δ -> 0)) h(δ, n, x)

and second, when δ->∞

(lim_(δ -> ∞)) h(δ, n, x)

Let us set δ to 50 and continue the  investigation of the behaviour of sequences a(δ, n) and b(δ, n) . For n running from (10^4) to (10^5) we have

Plot[{a[50,  n],  b[50,  n],  1},  {n,  10^4,  10^5}];

In the next picture  n runs from (10^10) to (10^11).  For the better visualization we show how those sequences are close to one:

Plot[{1 - a[50,  n],  1 - b[50,  n]},  {n,  10^10,  10^11}];

Again the numerical experiments are in favor of the Riemann hypothesis. For further n we shall use the following code

\(\(a[δ_, \ n_, \ prec_]\  := 
	
      Module[\ \ {exp1\  = \ N[Exp[\(-2\)*δ], \ prec], \ 
          nn\  = \ SetPrecision[1/n, \ prec], 

          		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
          exp2\  = \ N[2\ Exp[\(-δ\)], \ prec], \ 
          pi\  = \ N[2\ Pi, \ prec], \ tmp\ }, 
	
        tmp\  = \((
                \((I*\((\(-1\)\  + \ exp1)\)\  - \ exp2*Cos[pi*z]\ )\)/

                  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
                  \((1\  + \ exp1\  - \ exp2*Sin[pi*z])\))\)^n/Zeta[z]; 
	
        Abs[NIntegrate[tmp, \ {z, \ 0, \ 1/2}, 
						\ 
              WorkingPrecision\  -> \ prec, \ PrecisionGoal -> \ prec - 5, 

              	\ \ \ \ \ \ \ \ \ \ \ \ AccuracyGoal\  -> \ prec - 5]]^nn]

    \)\)

\(b[δ_, \ n_, \ prec_]\  := 
	
      Module[\ {exp1\  = \ N[Exp[\(-2\)*δ], \ prec], \ \ 
          nn\  = \ SetPrecision[1/n, \ prec], 

          		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
          exp2\  = \ N[2\ Exp[\(-δ\)], \ prec], \ 
          pi\  = \ N[2\ Pi, \ prec], \ tmp\ }, 
	
        tmp\  = \ 
          \((\((I*\((\(-1\)\  + \ exp1)\)\  + \ exp2*Cos[pi*z]\ )\)/

                  \ \ \ \ \ \ \ \ \ \((1\  + \ exp1\  + \ exp2*Sin[pi*z])\))
                \)^n/Zeta[z]; 
	
        Abs[\ NIntegrate[tmp, \ {z, \ 1/2, \ 1}, 
	\ \ \ \ \ \ \ \ \ \ \ \ 
              WorkingPrecision\  -> \ prec, \ PrecisionGoal -> \ prec - 5, 

              	\ \ \ \ \ \ \ \ \ \ \ \ AccuracyGoal\  -> \ prec - 5]\ ]^nn]
        \)

Then for n running from (10^12) to (10^20) we have :

Table[1 - a[50,  10^(12  +  n),  17 + n],  {n,  0,  8}] // 
      TableForm

Table[1 - b[50,  10^(12  +  n),  17 + n],  {n,  0,  8}] // 
      TableForm

Conclusion

Though our findings are far from being conclusive, the high level of computations presented in the paper results in real support of the Riemann Hypothesis. It is important to mention that all calculations were done on a regular Pentium machine.  We are sure that was possible just because our novel and inventive theoretical approach to the problem is much simpler and more suitable for numerical testing of the Riemann Hypothesis.

References

1. L.A. Aizenberg and A.M. Kutmanov, On the Possibility of Holomorphic Extension into a Domain of Functions Defined on a Connected Piece of its Boundary, Math. USSR Sbornik, 2, 72 (1992), 467-483.

2. L. Aizenberg, Carleman's Formulas in Complex Analysis, Kluwer Acad. Publ., Dozdrecht, 1993.

3. L. Aizenberg, Carleman's Formulas and Conditions of Analytic Extendability, Banach Center Publ., 31(1995), 37-34.

4. J. van de Lune, H.J.J. te Riele, and D.T. Winter, On the Zeroes of the Riemann Zeta Function in the Critical Strip, IV, Math. Comp., 46 (1986), 667-681.

5. A.M. Odlyzko,   The  (10^20) zero of the Riemann Zeta Function and its Neighbors, preprint, 1989.

6. R. Varga, Theoretical and Computational Aspects of the Riemann Hypothesis, chapter 3 in the book: Scientific Computations on Mathematical Problems and Conjectures, Society for Industrial and Applied Mathematics, Philadelhia, Pensylvania, 1990.