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- 1. Disprove of equality between Riemann zeta function and Euler product Chris De Corte chrisdecorte@yahoo.com KAIZY BVBA Beekveldstraat 22 9300 Aalst, Belgium November 18, 2014 1
- 2. The goal of this document is to share with the mathematical community the finding that the equality between the Riemann z`eta function and the Euler product does not seem to hold. 2
- 3. Contents 1 Key-Words 4 2 Introduction 4 3 De
- 4. nition and examples 4 3.1 Definition of finite zeta function . . . . . . . . . . . . . . . . . 4 3.2 example k=2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3.3 example k=4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 3.4 example k=8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 3.5 Generalized error formula . . . . . . . . . . . . . . . . . . . . 6 3.6 Intermediate Summary . . . . . . . . . . . . . . . . . . . . . . 6 4 Testing the formula's 6 4.1 Calculating absolute error . . . . . . . . . . . . . . . . . . . . 6 4.2 Calculating the relative error . . . . . . . . . . . . . . . . . . 7 4.3 Relationship between absolute and relative error . . . . . . . 8 4.4 Some other calculation results . . . . . . . . . . . . . . . . . . 8 4.5 Using the `eta function for 0 s 1 . . . . . . . . . . . . . . 9 5 Discussions 10 6 References 10 3
- 5. 1 Key-Words Riemann, Euler z`eta function, Euler product, Prime counting. 2 Introduction The following document originated out of our interest for primes. We remain surprised by the fact that so much attention goes to the investi-gation of the non trivial zero’s of the Riemann zeta function in the critical strip while we could not find an exact link with the distribution of the primes [1]. For this reason, we started to investigate the fundamentals all over again. The esteemed proof of the equality between the Riemann zeta function and the Euler product can be found in [2]. This document approaches this proof from a different angle. 3 De
- 6. nition and examples 3.1 De
- 7. nition of
- 8. nite zeta function We define the finite z`eta function as follows: k = Σk i=1 (1=i) (1) We will first redo the proofs found in [2] for a few examples of this limited zeta function and then discuss k →∞ once we have clarified our thoughts. 3.2 example k=2 2 = 1 + 1=2 (2) 1=22 = 1=2 + 1=2(1=2) (3) 4
- 9. Subtracting both equations, we get: (1 − 1=2)2 = 1 − 1=2(1=2) = 1 − errA(2) (4) We notice that when redoing the proof found in [2], an error term appears which for k = 2 equals to errA(2) = 0:25. We call this type of error the absolute error as we will later also introduce a relative error. 3.3 example k=4 4 = 1 + 1=2 + 1=3 + 1=4 (5) 1=24 = 1=2 + 1=4 + 1=2(1=3 + 1=4) (6) Subtracting equation (6) from (5), we get: (1 − 1=2)4 = 1 + 1=3 − 1=2(1=3 + 1=4) (7) Preparing to eliminate prime number 3 from the right side of the equation: 1=3(1 − 1=2)4 = 1=3 + 1=3:1=3 − 1=2:1=3(1=3 + 1=4) (8) Subtracting equation (8) from (7), we get: (1−1=3)(1−1=2)4 = 1−1=2(1=3+1=4)(1−1=3)−1=3:1=3 = 1−errA(4) (9) We notice that when redoing the proof for k = 4, a bigger error appears of errA(4) = 0:3055 3.4 example k=8 4 = 1 + 1=2 + 1=3 + 1=4 + 1=5 + 1=6 + 1=7 + 1=8 (10) Trying to eliminate all the fractions from the right side of the equation results in: (1 − 1=7)(1 − 1=5)(1 − 1=3)(1 − 1=2)8 = 1 − 1=2(1=5 + 1=6 + 1=7 + 1=8)(1 − 1=3)(1 − 1=5)(1 − 1=7) − 1=3(1=3 + 1=5 + 1=7)(1 − 1=5)(1 − 1=7) − 1=5(1=5 + 1=7)(1 − 1=7) = 1 − errA(8) (11) 5
- 10. We notice that when redoing the proof for k = 8, an even bigger error appears of errA(8) = 0:3584 3.5 Generalized error formula Based on formula (11), the error term for k → ∞ can be assumed to be of a size that can be represented as follows: errA(∞) ≈ Σ 8pi (1=pi) Σ 8pjpi (1=pj) Π 8pkpi (12) (1 − 1=pk) Note: we will not use this formula in the remainder of this document. 3.6 Intermediate Summary When using a limited zeta function and trying to eliminate the original fractions on the right side of the equation, we are actually creating extra fractions in stead and the frightening thing is that when increasing the k of our limited zeta function, the error factor only becomes bigger. We can actually recognize a pattern in the structure of the error factor which gives us no reason to believe that it will become zero for k → ∞ nor do we expect it to change signs in between. Moreover, this error factor is not of a size that we can neglect as it will seem. The general accepted proof for the equality [2] seems to neglect the impact of these infinite amount of extra fractions created. 4 Testing the formula's 4.1 Calculating absolute error Based on the above, we can now define the formula for the absolute error as follows: errA(k) = 1 − Π 8pik (1 − 1=pi) Σk i=1 (1=i) (13) 6
- 11. We calculated this formula for a set of primes in PARI/GP and the results are demonstrated in the table 1. For the large primes, we had to rely on data from [3] and extra calculations done in C++. Table 1: This table shows the absolute errors calculated for a list of values using formula (12) k errA 2 0.2500 11 0.3725 101 0.3809 1009 0.3938 10007 0.4041 100003 0.4106 500509 0.4139 999999937 0.4229 2147321881 0.4235 4.2 Calculating the relative error So far, we have been occupied with calculating what we called the absolute error. Maybe, it is more interesting to calculate the relative error. We define the relative error as follows: errR(k) = Π 8pik 1 (1

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