Riemann Hypothesis Elementary Discussion

Transcription

1 Progress in Applied Mathematics Vol. 6, No., 03, pp. [74 80] DOI: /j.pam ISSN 95-5X [Print] ISSN [Online] Riemann Hypothesis Elementary Discussion LIU Dan [a],* and LIU Jingfu [b] [a] Department of Mathematics, Neijiang Normal University, Neijiang, Sichuan, China. [b] Department of Mathematics, Sichuan Normal University, Chengdu, Sichuan, China. * Corresponding author. Address: Department of Mathematics, Neijiang Normal University, Neijiang 64003, China; zxc @qq.com Received: March, 03/ Accepted: June 4, 03/ Published: July 3, 03 Abstract: Areas of prime number theorem is proposed in this paper, and the area of prime number theorem. The basic theorem of prime number distribution are obtained. To prove the Rienann conjecture. Key words: Prime number; Area; Guess; Theorem Liu, D., & Liu, J. (03). Riemann Hypothesis Elementary Discussion. Progress in Applied Mathematics, 6 (), Available from article/view/j.pam DOI: /j.pam INTRODUCTION In 859, a German mathematician Riemann proposed: Riemann zeta function [ 3]: ζ(s) = From Equation (), we can get: n s, Re(s) >, () n= ]ζ(s) = Γ( s)(π) s sin(πs/)ζ( s), () Surrounded by () of zero point is referred to as: zero. Riemann hypothesis: Riemann zeta function all nontrivial zero point is in the critical line. 74

2 Liu, D., & Liu, J./Progress in Applied Mathematics, 6 (), 03. RIEMANN PRIME DISTRIBUTION FORMULA Set π(x) is not greater than x number of prime Numbers, then: π(x) = n µ(n) n J(x/n ), (3) J(x) = Li(x) Im ρ>0 [Li(x ρ ) + Li(x) ρ dt ] + x t(t ln, (4) ) ln t Here Equation (3) is Riemann prime distribution formula. The main conclusions of the Riemann was obtained in 859. The function µ(x) is called: M obius function, which is defined as follows [5]:, n =, µ(n) = ( ) k, n = p, p, p 3, p k, 0, The rest. The p k is not the same Prime number. Here Equation (4) refers to: the step function. Li(x ρ ) + Li(x ρ ) involves the Riemann content distribution of zero. Obviously, Riemann prime distribution formula of calculation is very complicated. In 90, Swedish mathematician von Koch proved that if Riemann content was established, then [5,6]: π(x) = Li(x) + O(x / ln x), (5) Li(x) = x ln u du Here Equation (5) is a prime number theorem. If Riemann hypothesis was established, we can also get the prime number theorem: π(x) = Li(x) + O(x /+ε ), In turn: if the prime number theorem Equation (5) was established, then the Riemann hypothesis was established. In fact for the x limited was set up. So as long as can prove sufficiently large x is set up, then Equation (5) is set up. 3. THEOREM OF PRIME NUMBER DISTRIBUTION SE- RIES Set a large number x, parameter lambda λ >, prime number p, get: x/ n= π(x) s(x), (6) n p n+ p, λ = n + n, Here Equation (6) is referred to: Theorem of prime number distribution series. x/ is an integer. For example, set x = 0, the following can easily obtained by 75

3 Riemann Hypothesis Elementary Discussion Equation (6): s(0) = 4 n= ln n+ n n p n+ so s(0) = , actual π(0) = 4. = ln + ln 3 + ln(3/) 5 + ln(4/3) 7, Proof. Set n < p < n +, if there is a prime number in the interval (n, n + ), it must be n +, which means and then λ = n + n By Equation (6), we can get: p p = n +, p >, x/ n= = n + + n + = p + p, = x/ ln + = ln + = ln + n= n p n+ p n<p<n+ <p<(x/ )+ <p<x p, p p Therefore, the following equation is obtained: ln + p, λ = p + p. (7) Set x > y, form Equation (7), <p<x y<p<x p, λ = p + p, If x is a large number, and y = [x / /], then: ( = ln + ) = p p, = π(x) π(y), Obviously, The theorem (6) was proved. y<p<x π(x) = s(x) s(y) + π(y), π(x) s(x). 76

4 Liu, D., & Liu, J./Progress in Applied Mathematics, 6 (), 03 For example, x π(x) s(x) INTERVAL PRIME NUMBER THEOREM Set x > y, ignore the reminder term and table as an integer, we can get the following from Equation (6): π(x) π(y) = s(x) s(y), x/ n=y/ n p n+ p, λ = n + n, (8) Here Equation (8) is: interval prime number theorem. Where x/ and y/ are integers. Proof. The same as that prove Theorem (6). By Equation (8), we can get: π(x) = s(x) s(y) + π(y). For example, set y = [x / /], by Equation (8), the following can be obtained: x π(x) s(x) s(y) + π(y) Ignore the remainder term and table as an integer, π(x) = s(x) s(y) + π(y). 5. TRANSFORM THEOREM OF PRIME NUMBER DIS- TRIBUTION From Equation (6), we have Substitute Equation (9) into Equation (8), we have π(y) > s(y), (9) π(x) = s(x) s(y) + π(y) > s(x) π(y) + π(y), and π(x) > s(x) π(y), (0) 77

5 Riemann Hypothesis Elementary Discussion and From Equation (8), we have π(x) = s(x) s(y) + π(y) < s(x) + π(y), () Combining Equation (0) and Equation (): s(x) π(y) < π(x) < s(x) + π(x), () We can prove a new prime number theorem by Equation (). In 874, mathematician Mertens proved [4]: p x p = ln ln x + A + O ( ), (3) ln x Here Equation (3) is called: Mertens theory. Where A is a constant. Set ln x, by Equation (3) [7,8]: n p n+ p x x/ p = ln x/ n=y/ n=y/ n p n+ p = ln ln x + A, ln(n + ) ln(n) = x/ n=y/ ( ln + ) x/ = ln(n) n=y/ ( ln + ), ln(n) ln(n), x/ p = ln(n), λ = n + n, n=y/ Here Equation (4) is called interval prime number theorem. 6. PROOF A PRIME NUMBER THEOREM Substitute Equation (4) into Equation (8), we have x/ n= x/ n=y/ ln(n) = x/ ln(n) = n= x n= ln(n), y/ ln(n) n= ln(n), If x is a large number and y = x /, then the following can be obtained from Equation (5) and Equation (): (4) (5) s(x) π(x / ) < π(x) < s(x) + π(x / ), (x ), x ln(n), (6) n= Here Equation (6) is a new prime number theorem. 78

6 Liu, D., & Liu, J./Progress in Applied Mathematics, 6 (), PROVE RIEMANN HYPOTHESIS In integral sense, Equation (5) and Equation (6) are the same, therefore Li(x) = x Li(x) = s(x), ln u du = x n= ln(n), For example: x π(x) Li(x) s(x) Substitute Li(x) = s(x) into Equation (6), we have Li(x) π(x / ) < π(x) < Li(x) + π(x / ), (x ). (7) Here Equation (7) Shows that Riemann hypothesis is established when x tends to infinity. By Mertens theore (3), if of the small x, can is π(x / ), then all the x, can is π(x / ). For example: set π(x) = Li(x) π(x / )c(x), then: x π(x) Li(x) c(x) Clearly all the x, can is π(x) > Li(x) π(x / ). And from Equation (7), table as an integer, we have: Li(x) π(x / ) π(x) Li(x) + π(x / ), π(x) = Li(x) + O(x /+ɛ ), x. So the theorem (5) was established. And Riemann hypothesis was established. 8. DISCUSSION This is the focus of the prove Theorem of prime number distribution series. So Proving a and Riemann hypothesi The prime number theorem of equivalence. It is proved that: Riemann zeta function all nontrivial zero point is in the critical line. 79

7 Riemann Hypothesis Elementary Discussion In addition, Theorem of prime number distribution series can There are different forms. Set Positive number s, has: From (8), we have: π(x) = s(x), x/ n= x + y y p x π(x) = Li(x) + O(x /s ), x c, y = x /s, p, (8) where c is an arbitrary large number. This problem is beyond over This subject, should not be discussed in detail. REFERENCES [] Manin, Yu. I., et al. (006). Introduction to modern number theory. Beijing: Science Press. [] Hua, Loo-Keng. (979). An introduction to number theory (In Chinese). Beijing: Science Press. [3] Neukirch, Jurgen. (007). Algebraic number theory. Beijing: Science Press. [4] Hua, Loo-Keng, & Wang, Yuan. (963). Numerical integration and its application. Beijing: Science Press. [5] Pan, Cheng-Dong, & Pan, Cheng-Biao. (988). The elementary proof of prime number theorem (In Chinese). Shanghai: Shanghai Science and Technology Press. [6] Lu, Changhai. (004). The Riemann hypothesis (In Chinese). Beijing: Tsinghua University Press. [7] Liu, Dan. (005). Prime symmetry. Neijiang Technology, (3), 6. [8] Liu, Dan (03). Elementary discussion of the distribution of prime numbers. Progress in Applied Mathematics, 5 (),