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A Geometric Proof of Riemann Hypothesis ∑ A Geometric Proof of Riemann Hypothesis Kaida Shi Department of Mathematics, Zhejiang Ocean University, Zhoushan City, Zip.316004, Zhejiang Province, China Abstract Beginning from the resolution of Riemann Zeta functionζ (s) , using the inner product formula of infinite-dimensional vectors in the complex space, the author proved the world’s problem Riemann hypothesis raised by German mathematician Riemannn in 1859. Keywords: complex space, solid of rotation, axis-cross section, bary-centric coordinate, inner product between two infinite-dimensional vectors in complex space. MR(2000): 11M 1 Introduction In 1859, German mathematician B. Riemann raised six very important hypothesizes aboutζ (s) in his thesis entitled Ueber die Anzahl der Primzahlen unter einer gegebenen Große [1] . Since his thesis was published for 30 years, five of the six hypothesizes have been solved satisfactorily. The remained one is the Super baffling problemRiemann hypothesis. That is the non-trivial zeroes of the Riemann 1 Zeta functionζ (s) are all lie on the straight line Re(s) = within the complex plane s. The well- 2 known Riemann Zeta function ζ ()s raised by Swiss mathematician Leonard Euler on 1730 to 1750. In 1749, he had proved the Riemann Zeta functionζ ()s satisfied another function equation. For Re(s) > 1, the series 1 1 1 1 +++++ 1sss2 3 LLn s is convergent, it can be defined asζ ()s . Although the definitional domain of the Riemann Zeta ∞ 1 function is , but German mathematician Hecke used the ζ (s) = ∑ s (s = σ + it) Re(s) > 1 n=1 n 1 analytic continuation method to continue this definitional domain to whole complex plane (except Re(s) = 1). In 1914, English mathematician G. H. Hardy declared that he had proved Riemann Zeta function 1 ζ ()s equation has infinite non-trivial zeroes lie on the straight line Re(s ) = . 2 In fact, about the Riemann Zeta functional equality ss1− −−s 1− s πζπ2 ΓΓ()()s = 2 ( )( ζ1− s ), (*) 2 2 because 1 1− 1 − 2 1 1 − 2 1− 1 1 π 2 Γ( 2 )ζ ( ) = π 2 Γ( 2 )ζ (1− ), 2 2 2 2 1 1 therefore, we have discovered that the solution of above equality is s = , namely Re(s) = . 2 2 1 That is to say no matter how the value t takes, σ always equals to . This explains that the 2 definitional domain of the Riemann Zeta function ζ (s) can be continued to at least 1 Re(s) = . 2 2 The formal resolution of Riemann Zeta function ζ ()s Considering the geometric meaning of the Riemann Zeta function 1 1 1 1 , ζ (s) = s + s + s +L+ s +L 1 2 3 n we will have the following figure: 2 w C(1) A1 1 A 2 1 s 1 B1 B2 A 3 2 s 1 B3 3 s O D1(1) D2(2) D3(3) … z Fig. 1 In above figure, the areas of the rectangles ADOCADDB11,,, 2 211 ADDB 3322L are respectively: 1 1 1 , s , s , s , L 1 2 3 therefore, the geometric meaning of the Riemann Zeta functionζ ()s is the sum of the areas of a series of rectangles within the complex space s. Using the inner product formula between two infinite-dimensional vectors, the Riemann Zeta functionζ ()s equation 1 1 1 1 ζ ()s =+++++=0 ()s = σ + it 1sss2 3 LLn s can be resolved as 1 1 1 1 1 1 1 1 ( σ , σ , σ ,L, σ ,L) ⋅ ( it , it , it ,L, it ,L) = 0. 1 2 3 n 1 2 3 n So, we should have (Please refer to the Appendix in the end) 1 1 1 1 1 1 1 1 rr +++++⋅⋅++++cos(NN , )= 0 , (1) 1222σσσ2 3 LLn 2 σ 1222it2 it3 itLLn 2 it 12 but in the complex space, if the inner product between two vectors equals to zero, then these two vectors 3 r r π are perpendicular, namely, (N , N ) = , therefore 1 2 2 r r cos(NN12 , )= 0 . So long as the values of two radical expressions in the left side of (1) are finite (real or imaginary), then we think the expression (1) is tenable. Because Riemann had proved that when Re(s) > 1, the equation ζ (s) = 0 has no solutions; when Re(s) < 0 , the Riemann Zeta function ζ ()s has only the trivial zeroes such as s = −2, −4, −6, L , therefore, the non-trivial zeroes of the Riemann Zeta function ζ ()s will lie on the banded domain 0 ≤ σ < 1. This explains that German mathematician Hecke used the analytic continuation method to continue the definitional domain of the Riemann Zeta function ζ ()s is tenable. 3 The relationship between the volume of the rotation solid and the area of its axis-cross section within the complex space 1 From the expression (1), we can know that when Re(s) = σ = , the series (harmonic series) 2 r r within the first radical expression of left side is divergent. Because cos(NN12 , )= 0 , therefore the situation of the complex series within the second radical expression of left side will change unable to research. Hence, we must transform the Riemann Zeta functionζ ()s equation. For this aim, let’s derive the relationship between the volume of the rotation solid and the area of its axis-cross section within the complex space. We call the cross section which pass through the axis z and intersects the rotation solid as axis- cross section. According to the barycentric formula of the complex plane lamina: 4 b ∫ zf[() z− gzdz ()] ξ = a , b [()fz− gzdz ()] ∫a b 1 22 2 [()()]fz− gzdz ∫a η = b . [()fz− gzdz ()] ∫a we have b π [()()]fz22− gzdz ∫a 2πη = b . (**) [()fz− gzdz ()] ∫a The numerator of the fraction of right side of the formula (**) is the volume of the rotation solid, and the denominator is the area of the axis-cross section. Theη of left side of the formula (**) is the longitudinal coordinate of the barycenter of axis-cross section. Its geometric explanation is: taking the axis-cross section around the axis z to rotate the angle of 2π , we will obtain the volume of the rotation solid. 4 The proof of divergence of two concerned serieses ∞ ∞ Now, let’s prove the divergence of the series ∑ cos(4tn ln ) and the series ∑sin(4t ln n) . n=1 n=1 y ytx= cos(4 ln ) ()t = 2 1 0 1 2 3 4 5 6 7 8 x -1 Fig. 2 ∞ Obviously, from Fig.2, we can find that the area of c ⋅ ∑ cos(4t ln n) is contained by the area n=1 5 +∞ of cos(4t ln x)dx . ∫ 1 Suppose that c is a very small positive number, then the sum of the areas of a series of rectangles (which take cos(4123tnn ln )(= , , ,L ) as high and take c as width) is: ∞ c ⋅ ∑ cos(4t ln n) = c ⋅ cos(4t ln1) + c ⋅cos(4t ln 2) + c ⋅cos(4t ln 3) +L. n=1 Because +∞ 1 w cos(4t ln x)dx = lim x[cos(4t ln x) + 4t sin(4t ln x)]1 ∫ 1 w→+∞ 2 t→∞ 1+16t 1 = lim [w(cos(4t ln w) + 4t sin(4t ln w)) −1] w→+∞ 2 t→∞ 1+16t w 1 4t 1 = lim cos(4t ln w) + sin(4t ln w) − lim w→+∞ 2 2 2 t→∞ 2 t→∞ 1+16t 1+16t 1+16t 1+16t 1 4t Suppose that = sinδ and = cosδ , then above equals to 1+16t 2 1+16t 2 w 1 lim sin(δ + 4t ln w) − lim w→+∞ 2 t→∞ 2 t→∞ 1+16t 1+16t d = lim sin(δ + 4t ln w) w→+∞ 1 t→∞ +16 t 2 d = lim sin(δ + 4t ln w). 4 w→+∞ t→∞ Because the relationship between w and t is linear, therefore, we denote w = dt . Obviously, d d d − ≤ lim sin(δ + 4t ln w) ≤ , 4 4 w→+∞ 4 t→∞ therefore 6 d ∞ d − ≤ c ⋅ ∑cos(4t ln n) ≤ , 4 n=1 4 namely, d ∞ d − ≤ ∑cos(4t ln n) ≤ . 4c n=1 4c ∞ Thus, although the series ∑ cos(4t ln n) is divergent, but it is oscillatory and bounded. n=1 ∞ Similarly, although the series ∑sin(4t ln n) is divergent, but it is also oscillatory and bounded. n=1 ∞ In a word, when t = 0 , the series ∑ cos(4t ln n) is divergent (goes to infinite); when t ≠ 0 , n=1 it is oscillatory and bounded. ∞ Thus, the divergence (oscillatory and bounded) of ∑ cos(4t ln n) has been proved. n=1 Similarly, we can prove the divergence (oscillatory and bounded) of the series ∞ ∑sin(4t ln n) (t ≠ 0) . n=1 . 5 Two theorems and their corollaries Suppose that 1−s 1 − 1− s ξπζ()sss=− (1 )2 Γ ( ) (1− s ), 2 2 then, we have ξ(s) = ζ (s) . According to the following: 7 [2] Theorem 1 The function ξ()s is the one class integer function, it has infinite zero points ρ n , ∞ ∞ −1 −−1 ε ， 01≤≤Re(ρ n ) , the series ∑ ρ n is divergent. But for ∀ ε > 0 the series ∑ ρ n is n=1 n=1 convergent. The zeroes of the functionξ()s are non-trivial zeroes of the Riemann Zeta function. We can obtain: [2] Corollary 1 The formula s ∞ s ξ()se=−ABs+ ∏1 e ρn n=1 ρ n is tenable. Corollary 2[2] The non-trivial zeroes of the Riemann Zeta function ζ ()s distribute 1 symmetrically on the straight line Re s = and Ims = 0 . 2 i t σ 0 + it2 σ1 + it0 • • •σ 2 + it0 1 o Re s = 2 σ σ 0 + it1 • Fig.
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