Open Science Repository Mathematics

doi: dx.doi.org/10.7392/Mathematics.70081941


Riemann Hypothesis


Jarmo Kiukkonen

University of Eastern Finland


Abstract

With connection of Riemann zeta-function and Euler product I got infinitive series, which converges to 0. John-Derbyshire's book Prime Obsession has Riemann-invariant equation, which I counted to be true from value by mathematical induction using derivates to infinite series.

Keywords: Riemann hypothesis, mathematics.



Citation: Kiukkonen, J. (2013). Riemann Hypothesis . Open Science Repository Mathematics, Online(open-access), e70081941. doi:10.7392/Mathematics.70081941

Received: February 12, 2013

Published: March 14, 2013

Copyright: © 2013  Kiukkonen, J. Creative Commons Attribution 3.0 Unported License.

Contact: research@open-science-repository.com



Full text


In John Derbyshire's book Prime Obsession is for Riemann hypothesis equivalent inequation:
 
            M( s ) = O(s^(½ + ε)), ε>0,
 
where  M ( s ) is defined in the following way:
 
            1.         M( 1 ) = 1,
            2.         M(s)   = u( 1 ) + u( 2 ) + u( 3 ) + ... + u( s ),
                        s = 2, 3, 4, ... ,
           
where  u( s ) is defined in the following way: 
 
            1.        u ( 1 ) = 1,
            2.        u ( s ) = -1, when  s is product of different primes so that factors are odd amount,
            3.        u ( s ) = +1, when  s  is product of  different primes so that factors are even amount,
            4.        u ( s ) = 0, for others.
 
Function  u  is Möbius-function and  M  cumulative Möbius-function.
 
I use Riemann zeta-function:
 
            zeta(t) = 1/(1^t) + 1/(2^t) + 1/(3^t) + and so on.
           
 
If I take Euler product and Riemann zeta-function, and suppose  t # 1:
 
            1/(zeta(t))  = big phi k=1 to infinite (1-1/(p(v)^t))
            here  p(v) = v. prime,
            big phi = infinite product.
 
And now multiply right side step by step, I get:
 
            1/(zeta(t) = (1-1/(p(1)^t))*(1-1/(p(2)^t)*(1-1/(p(3)^t)* … .
 
            Because  zeta(t) à + infinite, when  t à 1+  and
            zeta(t) à - infinite, when  t à  1-,
            1/(zeta(t)) à 0, when  t à 1.
                       
            1/zeta(t) = (1-(1/p(1)^t+1/p(2)^t+1/p(3)^t+…)
            + (1/((p(1)^t)*(p(2)^t)) + 1/((p(1)^t)*(p(3)^t))+1/((p(1)^t)*(p(4)^t))…
            +  1/((p(2)^t)*(p(3)^t)) + 1/((p(2)^t)*((p(4)^t)) + 1/((p(2)^t)*(p(5)^t)) + … ) …
            =  1 – ( sigma 0 < i 1/((p(i)^t) + ( sigma 0< i < j 1/((p(i)^t)*(p(j)^t)) -
            sigma 0< i < j< k 1/((p(i)^t)*(p(j)^t)*(p(k)^t)) + …
            = sigma n=1 to infinite  u(n)/(n^t) = 1/zeta(t) à 0, when  t à 1,
            here  sigma = infinite sum.
 
Above are all values n, because if n belongs to set “others” in denotion of  Möbius-map value u(n) = 0 then.
 
The signs  +  or  -  are also valid because  (-1)^n=  -1  or  +1, depending whether is value n odd or even.
 
From this follows:
 
            sigma n=1 to infinite. | u(n)/n | =  0,
            here  sigma = infinite sum.
 
Because   n==1/k, is for  k, integer
 
            | sigma n=1 to k  u(n)/k | <= | sigma n=1 to k  u(n)/n | à 0,
            when  k à infinite.
 
            | sigma  n=1 to k  u(n)/k | = | M(k)/k | à 0, when  k à
            infinite,
            here sigma = finite sum.
 
Now I have straight formulas to calculate value
 
            M ( k ).
 
            M ( 1 ) = 1,
            M ( 2 ) = u ( 1 ) + u ( 2 ) = 1 - 1 = 0,
           
            M ( 3 ) = ∑ u( i ) i=1 to 3 = 0 - 1 = -1,
 
            M ( 4 ) = ∑ u( i )  i=1 to 4 = -1 + u ( 2^2 ) = -1,
 
            M ( 5 ) = ∑ u ( i ) i=1 to 5 = -1 -1 = - 2,
 
            M ( 6 ) = ∑ u ( i ) i=1 to 6 = -2+u(2*3) = -2+1 = -1,
 
            M ( 7 ) = ∑ u ( i ) i=1 to 7 = -1+ u(7) = -1-1 = -2,
 
            M ( 8 ) = ∑ u ( i ) i=1 to 8 = -2 + u(2^3) = -2 + 0 = -2,
 
            M ( 9 ) = ∑ u ( i ) i=1 to 9 = -2 + u(3^2) = -2 + 0 = -2,
 
            M ( 10 ) = ∑ u ( i ) i=1 to 10 = -2 + u(2*5) = -2+1 = -1,
 
            M ( 11 ) = ∑ u ( i ) i=1to 11 = -1 + u(11) = -1-1 = -2,
 
            M ( 12 ) = ∑ u ( i ) i=1 to 12 = -2+u(2*2*3) = -2+0 = -2,
 
            M ( 13 ) = ∑ u ( i ) i=1 to 13 = -2+ú(13) = -2-1= -3,
 
            M ( 14 ) = ∑ u ( i ) i=1 to 14 = -3+u(2*7) = -3+1= -2,
 
            M ( 15 ) = ∑ u ( i ) i=1 to 15 = -2+u(3*5) = -2+1= -1,
 
            M ( 16 ) = ∑ u ( i ) i=1 to 16 = -1+u(2^4) = -1+0= -1…
 
John Derbyshire’s equation is same as
 
 
            | M(k) | <= k^(½+epsilon)
            epsilon > 0.
            k  positive integer,
            when integer  k  increases over known  K > 0.
 
The demanded inequation is eqivalent with
 
            | M(k) |/k <=  1 /k^(½+epsilon),
            k = positive integer
 
This I will prove by mathematical induction.
 
 
Induction step  I):
 
 
For value  k=4, 4>=4^(½+epsilon),  0  < epsilon <  ½:
 
            | M(4)/4 | <= 1 / 4^(½+epsilon)
 
            ¼  <=   1/(4^(½+epsilon)
 
Multiplying this by value  4  I get
 
            1 = | M(4) | <= 4^(½+epsilon).
 
 
Induction step II):
 
Supposition:
 
            | M(k)/k | <= k^(-½+epsilon).
 
Insist:
 
            | M(k+1)/(k+1) | <= (k+1)^(-½+epsilon).
 
 
With this mathematical induction I can  soIve John  Derbyshire's inequation compIeteIy
 
            | M( k ) |= O(k^(½ + ε)), ε>0,
 
where   O  is  the  symbol that function on right-hand side Ieaves between interval  
 
            ](k^(½+epsiIon)), + (k^(½+epsilon)[
 
.Let  ½ > epsilon > 0.
 
I analyse formulas
 
            3) | (M(k+1)/(k+1) | <= (k+1)^(-½+epsilon).
            k = 4, 5, 6, … ,
 
when it is true
 
            | M(k)/k | <= k^(-½+epsilon).
 
 
Possibilities there are   u(k+1) =  -1  or  1  or  0  gives valid for formulas
 
            | M(k)/(k+1) + 1/(k+1) |  or
            | M(k)/(k+1) – 1/(k+1) |  or
            | M(k)/(k+1) + 0 | .
 
And these gives by triangle formula
 
            | M(k)/(k+1) | + 1/(k+1)  or
            | M(k)/(k+1) | + 0.
 
Trivially
 
            | M(k)/(k+1) |. <= | M(k)/(k+1) | + 1/(k+1)
 
 
I must prove
 
            4) | M(k)/(k+1) | <= | M(k)/(k+1) | + 1/(k+1) <=
            (k+1)^(-½+epsilon).
 
 
and the proof is done by mathematical induction.
 
Because  k < k+1, it is true  1/k > 1/(k+1), so that demand  4)  configures into form
 
            5) | M(k)/(k+1) | + 1/(m+1) <= (m+1)^(-½+epsilon),
            0 < epsilon < ½.
 
The supposition for mathematical induction is:
 
I put  k = x,  k+1 = x+1  and analyse derivatives
 
            a) D(x^(-½+epsilon)+(x+1)^(-1))   and
            b) D((x+1)^(-½+epsilon)).
 
Derivate  a)
 
            = (-½+epsilon)x^(-3/2+epsilon)-((x+1)^(-2))
 
Derivate  b):
 
            = (-½+epsilon)(x+1)^(-3/2+epsilon).
 
With collage mathematics you can count that function with derivative a) is decreasing by higher speed than map with derivative b), because all terms are negative and even the one term = negative root of derivate a) (=-2) is greater by self-value than the whole term of derivate b) = smaller by self-value, negative root (=-3/2+epsilon), 0 < epsilon < ½, if only functions are continued.
 
So that soon or later
 
            x^(-½+epsilon)+(x+1)^(-1)
            <= (x+1)^(-½+epsilon).
 
It is true
            | M(k)/(k) | <= (k)^(-½+epsilon),
m is  4,
 
function  x^(-½+epsilon)  is continued and so is map  x^(-½+epsilon)+(x+1)^(-1), if only epsilon is solid.
 
Next I applicate the derivates  a)  and   b).
 
            k = x.
 
            | M(k+1)/(k+1) | <=
            | M(k)/(k+1) | + 1/(k+1) <=
            | M(k)/(k) | + 1/(k+1) <=
            (k)^(-½+epsilon) + 1/(k+1) =
            x^(-½+epsilon) + (x+1)^(-1) <=
            (x+1)^(-½+epsilon) =
            ( k+1)^(-½+epsilon),
 
beginning from  5, 6, 7, and so on.
 
Multiplying first and last term of above inequation chain by value  k+1  I get the finishing conclusion below.
 
            | M(k) | = O(k^(½+epsiIon)), ½ > epsiIon > 0.


References

1. Derbyshire, J. (2004). Prime obsession. The Mathematical Intelligencer26(1), 55-59. FIND ONLINE

2. Weisstein, Eric W. "Möbius Function." From MathWorld--A Wolfram Web Resource. FIND ONLINE

3. Average Value Continued: φ and μ. Dep. of Math. Univ. of Georgia. FIND ONLINE

4. Sutantyo, Daniel Armand. "Elementary and Analytic Methods in Number Theory." PhD diss., Macquarie University, 2007. FIND ONLINE

Connections

1. Goodman, Len and Weisstein, Eric W. "Riemann Hypothesis." From MathWorld--A Wolfram Web Resource. FIND ONLINE

2. Katz, N. M., & Messing, W. (1974). Some consequences of the Riemann hypothesis for varieties over finite fields. Inventiones mathematicae23(1), 73-77. FIND ONLINE

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5. Balazard, M. (2012). Elementary remarks on Möbius’ function. Proceedings of the Steklov Institute of Mathematics276(1), 33-39. FIND ONLINE

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Cite this paper

APA

Kiukkonen, J. (2013). Riemann Hypothesis. Open Science Repository Mathematics, Online(open-access), e70081941. doi:10.7392/Mathematics.70081941

MLA

Kiukkonen, Jarmo. “Riemann Hypothesis.” Open Science Repository Mathematics Online.open-access (2013): e70081941.

Chicago

Kiukkonen, Jarmo. “Riemann Hypothesis.” Open Science Repository Mathematics Online, no. open-access (2013): e70081941. http://www.open-science-repository.com/riemann-hypothesis.html.

Harvard

Kiukkonen, J., 2013. Riemann Hypothesis. Open Science Repository Mathematics, Online(open-access), p.e70081941. Available at: http://www.open-science-repository.com/riemann-hypothesis.html.

Science

1. J. Kiukkonen, Riemann Hypothesis, Open Science Repository Mathematics Online, e70081941 (2013).

Nature

1. Kiukkonen, J. Riemann Hypothesis. Open Science Repository Mathematics Online, e70081941 (2013).


doi

Research registered in the DOI resolution system as: 10.7392/Mathematics.70081941.


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