A geometrical summation method for the Riemann zêta function
AA GEOMETRICAL SUMMATION METHOD FOR THE R IEMANNZÊTA FUNCTION
A P
REPRINT
Ulysse REGLADE
Student at Mines Paristech [email protected]
March 27, 2019 A BSTRACT
Disclaimer : I am no professional mathematician. Though, it seems to me that the intuition presentedin this preprint has not yet been explored by the mathematical community. This intuition seems veryconsistent and makes predictions, these predictions can be verified numerically. Because this articleis a preliminary work, some parts of the proof are not complete. They will be explicitly indicated .In this paper, we introduce a geometrical summation method that makes the original Riemann seriesconverge over the critical strip. This method gives an analytical function, that coincides with zêta.This point of view allows us to introduce a quantity of interest that seems to give a characterization ofthe non-trivial zeros of the Riemann zêta function.For: z = x + iy, x ∈ R ∗ + , y ∈ R ∗ , zêta can be defined as: ζ ( z ) = lim N →∞ N (cid:88) n =1 n z + 1( N + 1) z − z (cid:32) − x − y tan( y ln( N +2 N +1 )) (cid:33) From here, we can show that if z is a non trivial zero of zêta, the following quantity converges toa relative integer seemingly even. Though the reciprocal is false, it is natural to compute it for theknown non-trivial zeros of zêta: ζ ( z ) = 0 ⇒ lim n →∞ − y ln( n + 1) − arg sum ( ζ n ( z )) + arctan( y − x ) π = U z , U z ∈ Z arg sum ( ζ n ( z )) = (cid:80) nm =1 arg (cid:16) (cid:80) m +1 l =1 1 lz (cid:80) ml =1 1 lz (cid:17) At the end of this document you can find the table of these values computed for the first 30 knowzeros of zêta. This last identity seems to be deeply correlated with the Riemann hypothesis.In this paper, we are taking full advantage of formal calculation algorithms, especially to computesome asymptotic expansions. Here is the GitHub link to all codes used for this work:https://github.com/UlysseREGLADE/Zeta K eywords Riemann · Zêta · Summation method a r X i v : . [ m a t h . G M ] M a r PREPRINT - M
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For convenience, we define for z ∈ Ω : ζ n ( z ) = n (cid:88) j =1 j z , n ∈ N ∗ ζ ( z ) = 0 (1)In the whole paper, z is defined as a complex number of Ω an open set of C . x and y are defined with no ambiguity asits real and complex parts: x = (cid:60) zy = (cid:61) z (2)We are not going to make a real difference between the complex representation of z and its vectorial form ( x, y ) .Therefore, we will allow ourselves to use these representations quite freely.Let’s now discuss the intuition that gave rise to this paper. I’m uncomfortable with the usual way of defining the zêtafunction [1]. ζ ( z ) = ∞ (cid:88) n =1 n z , x > (3)In this notation, information is lost on how the partials sums converge to zêta. To me, a more meaningful, but still notperfect way to define zêta would be something like: ζ ( z ) = { ζ n ( z ) } n ∈ N (4)This definition is not equivalent to the usual one, it is indeed defined even if x = 1 . Though it is possible to do muchbetter. What we are going to show is that, in a way, the zêta function can be defined as the center of a logarithmicasymptotic spiral described by the partial sums of the Riemann series { ζ n ( z ) } n ∈ N . In fact, this sequence describes aconverging spiral for x > , a circle for x = 1 , and a diverging spiral for x < . In each case, the value of zêta is thecenter of these spirals. We are going to take advantage from this fact to define zeta for: x > .Figure 1: Plot of the Riemann series, for: z = − i , z = 1 − i , ans z = − i PREPRINT - M
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27, 2019 z ∈ { iy, y ∈ R ∗ } . In this subsection, Ω is defined as: Ω = { iy, y ∈ R ∗ } (5)Let us define: ∆ n ( z ) = (cid:26) ζ n ( z ) + te i π ( n + 1) z , t ∈ R (cid:27) , z ∈ Ω , n ∈ N (6) ∆ n ( z ) is a straight line of the complex plan. Considering n z as a vector, it is clearly not collinear to n +1) z . Indeed: arg( nn +1 z ) = y ln( nn +1 ) (cid:54) = 0 . Therefore, ∆ n ( z ) and ∆ n +1 ( z ) must have an unique intersection point. Let us call it c n ( z ) : c n ( z ) = ∆ n ( z ) ∩ ∆ n +1 ( z ) , n ∈ N (7)Let us adopt the following notation: δ n ( z ) = c n ( z ) − c n − ( z ) , n ∈ N ∗ (8)Figure 2: Plot of { c n ( z ) } n ∈ N for: z = 1 − i Figure 2 is a plot { c n ( z ) } n ∈ N , and we can enunciate our first theorem: Theorem 2.1.
For z ∈ Ω , (cid:80) δ n ( z ) converges by Riemann sommation, and we can define with no ambiguity: c ( z ) = lim n →∞ c n ( z ) (9) Proof.
Using geometrical relationships, we can write from the
Figure 3 : | δ n ( z ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n + 1) tan( − y ln( n +2 n +1 )) − (cid:115) n + 1 n tan( y ln( n +1 n )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (10)3 PREPRINT - M
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27, 2019Figure 3: Geometrical view of the situationFrom
Equation 10 , we can calculate the asymptotic expansion of δ n ( z ) . This calculation is quite tedious, but at theend of the day one can show that: δ n ( z ) = n →∞ n (cid:12)(cid:12)(cid:12)(cid:12) y + 1 y (cid:12)(cid:12)(cid:12)(cid:12) + o ( 1 n ) (11)The series converges by Riemann sommation. This proves the theorem.In addition, we can observe that: | c n ( z ) − ζ n ( z ) | = 1( n + 1) tan( | y | ln( n +2 n +1 )) = n →∞ | y | + o (1) (12)Which leads us to our second theorem: Theorem 2.2 (Asymptotic circle) . Lets note C ab the circle of radius b and center a. We have: { ζ n } n ∈ N ⊂ C c ( z ) | y | (13) Proof.
By triangular inequality, we can write: | ζ n ( z ) − c n ( z ) | − | c n ( z ) − c ( z ) | ≤ | ζ n ( z ) − c ( z ) | ≤ | ζ n ( z ) − c n ( z ) | + | c n ( z ) − c ( z ) | (14)Let us denote d ( C c ( z ) | y | , ζ n ( z )) the distance between the circle and the sequence: d ( C c ( z ) | y | , ζ n ( z )) = (cid:12)(cid:12)(cid:12)(cid:12) | ζ n ( z ) − c ( z ) | − | y | (cid:12)(cid:12)(cid:12)(cid:12) (15)We have: | ζ n ( z ) − c n ( z ) | = 1 | y | + o (1) | c n ( z ) − c ( z ) | = o (1) (16)4 PREPRINT - M
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27, 2019From
Equation 14 we can conclude that: lim n →∞ d ( C c ( z ) | y | , ζ n ( z )) = 0 (17)which finishes the proof.From the Figure 2 , one can expect { ζ n ( z ) } n ∈ N to populate densely the circle, which is a reasonable conjecture. Itwould lead to an equality and not a simple inclusion in Theorem 2.2 . Though, the demonstration of this conjecture isnot necessary for the rest of the proof.An other reasonable hypothesis is that c is actually ζ . But as Equation 10 only applies for x = 1 , we need first to extendthe definition of c to a bigger open set and show that it is analytical and coincides with the Riemann zêta function. z ∈ { x + iy, x > , y ∈ R ∗ } For the rest this section, we define Ω and ω as: Ω = { x + iy, x > , y ∈ R ∗ } ω = { x + iy, x > , y ∈ R ∗ } (18)To understand the rest of the proof, we first need to make the following observation:Figure 4: Representation of a logarithmic spiraleLet us consider the integral test for convergence of ζ n ( z ) : (cid:90) n − z dn = n − z − z = 11 − z e (1 − x ) ln( n ) e − iy ln( n ) = 11 − z e x − y θ e iθ θ = − y ln( n ) (19)We recognize the definition of a logarithmic spiral [2] in the complex plan. Such a spiral intersects its radius with aconstant angle α , like it is shown in Firgure 4 . In this case, we can compute this value, and we find: α ( z ) = (cid:26) π − arctan( − xy ) y > π − arctan( − xy ) y < (20)5 PREPRINT - M
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27, 2019We can see that x = 1 implies α = π [ π ] . Therefore, it is natural to give a new definition to ∆ n ( z ) : ∆ n ( z ) = (cid:26) ζ n ( z ) + te iα ( n + 1) z , t ∈ R (cid:27) , z ∈ Ω , n ∈ N (21)For the same reason than in the first section, we can define with no ambiguity the sequences { c n ( z ) } n ∈ N and { δ n ( z ) } n ∈ N ∗ . Figure 5: Plot of the convergence for z = 0 . − i From the
Figure 5 , it seems reasonable to expect { c n ( z ) } n ∈ N to converge over the critical strip. It can even alreadyconjecture that their limit coincides with ζ ( z ) . c : Ω → C First, let us remind the definition of a domination function:
Definition 2.1. D d : R → R dominates f n : C → C over Ω d if and only if: ∃ n ∈ N |∀ z ∈ Ω d , ∀ n > n , D d ( n ) > | f n ( z ) | (22)We are now ready to formulate the weak version of what will be called from now the radial convergence for the Riemannzêta function. Theorem 2.3.
For z ∈ Ω , the sequence { c n ( z ) } n ∈ N converges uniformly to c ( z ) . c : Ω → C is continuous over Ω , andwe have: c ( z ) = ζ ( z ) , z ∈ ω (23) c is complex-differentiable (i.e. holomorphic) over ω ⊂ Ω , and it coincides with ζ over this open set.Proof. First, we observe from
Figure 6 that we can write δ n ( z ) as: δ n ( z ) = 1 n x (cid:113) − xy ) (cid:32) − y ln n +1 n ) − n ) x (cid:32) − y ln( n +2 n +1 )) + 1 − xy (cid:33)(cid:33) e i ( α ( z ) − y ln( n +1)) (24)To obtain this result, we have first computed the algebrical module of δ n ( z ) only using geometrical relations in the Figure 6 . Then, we compute the algebrical argument of δ n ( z ) by induction and show that it is in fact: α ( z ) − y ln( n +1) .We also work out the expression of c ( z ) by hand. It can be neatly written as: c ( z ) = 1 − − z y − iy sin( y ln 2) (25)6 PREPRINT - M
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27, 2019Figure 6: Geometrical view of the situation for α ∈ ]0 , π [ ∪ ] π, π [ This enable us to write: c n ( z ) = c ( z ) + n (cid:88) i =1 δ i ( z ) (26)Then, just like in the first section, we compute the asymptotic expansion of | δ n ( z ) | as n goes to infinity: | δ n ( z ) | = 1 n x (cid:113) − xy ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y ln n +1 n ) − n ) x (cid:32) y ln( n +2 n +1 )) + x − y (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (27)We find: | δ n ( z ) | = n →∞ n x +1 x + y (cid:112) (1 − x ) + y = 1 n x +1 | z | | − z | + o ( 1 n x ) (28) c ( z ) and δ n ( z ) are continuous with respect to z over Ω . It can be even say right away that they are real-differentiable.Now, a domination function needs to be chosen. Though, it is not that easy in this case. Indeed, for large values of y ,because of the presence of y ln n +1 n ) and y ln n +2 n +1 ) in the expression of | δ n | , we can expect | δ n | to explode for thefirst values of n .From Equation 27 , we can give the following expression for | δ n | : | δ n ( z ) | = 1 n x | − z | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y sin( y ln n +1 n ) − y (1 + n ) x tan( y ln( n +2 n +1 )) + 1 − x (1 + n ) x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (29)We introduce: d x,y : R ∗ + → R , the function defined by: d x,y ( n ) = n (cid:32) y sin( y ln n +1 n ) − y (1 + n ) x tan( y ln( n +2 n +1 )) + 1 − x (1 + n ) x (cid:33) (30)The limit of d x,y is: lim n →∞ d x,y ( n ) = x + y (31)7 PREPRINT - M
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27, 2019The detailed study of d x,y is necessary in order to be able to properly dominate δ n .Let us denote Ω d a bounded open set of Ω : a = inf z ∈ Ω d xA = sup z ∈ Ω d xb = inf z ∈ Ω d | y | B = sup z ∈ Ω d | y | (32) Theorem 2.4.
Let O b (Ω) the set of the bounded open sets of Ω . We can define: n : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) O b (Ω) → N Ω d (cid:55)→ n (Ω d ) (33) This function is such as: ∀ Ω d ∈ O b (Ω) , ∀ z ∈ Ω d , n > n (Ω d ) ⇒
14 ( x + y ) < d x,y ( n ) <
34 ( x + y ) (34) Proof.
This proof is very technical, and is not detailed in this paper. But here are the mains ideas: • First, we observe that the sign of d x,y ( n ) does not depend on the sign of y . We can limit our study to y > . • Then, we observe that for n > n B = − e πB e πB − , the signs of sin( y ln( n +1 n )) and tan( y ln( n +2 n +1 )) do not changeanymore. • We take advantage from the fact that the series expansion of sin( x ) , cos ( x ) , ln( x ) , and x ) α> , whentruncated, gives majorations or minorations of these function for x > . • From here, we define two multivariate rational fractions that give upper and lower bounds for d x,y ( n ) startingfrom n B , Figure 7 . • The dominant coefficient of the denominator of these two fractions is an integer.Using this last argument, it is possible to show that from a certain value of n , d x,y ( n ) must lie in a certain range aroundits limit. The code used to produce these upper and lower bounds of d x,y is accessible via the GitHub link at thebeginning of the document.From here, it will be assumed such a domination function exists: D Ω d : R ∗ + → R D d ( n ) = 1 n a A + B )4 | a − | (35)This function is integrable by Riemann sommation over any Ω d in Ω . We can even specify from when this dominationis effective for Ω d . It is from n (Ω d ) , which does not depend on the choice of z in Ω d .To summarize, we have: • δ n : Ω d → C is continuous. • (cid:80) | δ n ( z ) | converges over Ω d by Riemann sommation.8 PREPRINT - M
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27, 2019Figure 7: Graph of the framing for d x,y ( n ) • D d ( n ) : R ∗ + → R dominates δ n : Ω d → C over Ω d , and is integrable.Therefore, c + (cid:80) δ n converges normally and in consequence uniformly to c over Ω d . c is continuous over any Ω d , so c is continuous over Ω .Finally, we have to prove that this expression coincides with ζ on ω ⊂ Ω . It comes from the fact that: ζ n ( z ) → n →∞ ζ ( z ) , z ∈ ω (36)From Figure 6 , we can show that: | ζ n ( z ) − c n ( z ) | = n →∞ | − z | n x − + o ( 1 n x − )1 | − z | n x − → n →∞ , z ∈ ω (37)Therefore: z ∈ ω ⇒ c ( z ) = ζ ( z ) (38)This implies that c is complex-differentiable over ω . We now need to show that c is complex-differentiable over Ω , inparticular on the critical strip. c : Ω → C Showing that c is complex-differentiable over the critical strip is quite challenging because c and δ n are not holomorphicfunctions. Therefore, the first step is to show that the Jacobian of δ n is well defined and continuous (real-differentiability).Only then we can verify it satisfies the Cauchy-Riemann equation (complex-differentiability).9 PREPRINT - M
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27, 2019A way to show that c is indeed real-differentiable is to show that (cid:80) J δ n converges normally for the infinity norm. Thisworks, but it leads to very tedious calculations that are not described here. Again, the domination is by far the mosttechnical aspect of this proof, though the formal calculation of J δ n gives immediately the following result: ∂δ nx ∂x = O ( ln( n ) n x ) ∂δ nx ∂y = O ( ln( n ) n x ) ∂δ ny ∂x = O ( ln( n ) n x ) ∂δ ny ∂y = O ( ln( n ) n x ) (39)Therefore: (cid:107) J δ n (cid:107) ∞ = O ( ln( n ) n x ) (40)We can conclude that (cid:80) J δ n converges for any ( x, y ) in Ω . We now need to find a domination function for J δ n , for theinfinity norm over Ω d .It’s possible to reuse the technique described in the last section to find a domination function for J δ n , but I think theremust be a more elegant solution to this problem. For the rest of the proof, we are just going to assume that we have thefollowing sum to converge normally for the infinity norm: J c (( x, y )) = J c (( x, y )) + ∞ (cid:88) i =1 J δ n (( x, y )) , x > , y ∈ R ∗ (41) c : Ω → C , equality with ζ : Ω → C First, we can rewrite
Equation 24 as: δ n ( z ) = y − z | − z | (cid:32) in z n +1 n iy − − i ( n + 1) z n +2 n +1 2 iy + 1 n +2 n +1 2 iy − n + 1) z − xy (cid:33) (42)The series (cid:80) δ n converges normally, therefore we can rearrange the terms in the sum. More precisely we observe thissum is actually telescopic. We can write that: N (cid:88) n =1 δ n ( z ) = N (cid:88) n =2 n z + 1 − z | − z | (cid:32) y − iy sin( y ln 2) − y ( N + 1) z y ln( N +2 N +1 )) + (1 − x ) 1( N + 1) z (cid:33) (43)Therefore, from Equation 25 , we can write: c N ( z ) = c + N (cid:88) n =1 δ n = ζ N ( z ) + 1( N + 1) z − z | − z | (cid:32) − x − y tan( y ln( N +2 N +1 )) (cid:33) (44)10 PREPRINT - M
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27, 2019Now, let us introduce: g n ( z ) = 1( n + 1) z f n ( z ) = 1 − z | − z | (cid:32) − x − y tan( y ln( N +2 N +1 )) (cid:33) r n ( z ) = g n ( z ) f n ( z ) (45)Because (cid:80) J δ n converges normally, this rearrangement is also true for the jacobian of c n ( z ) : J c n (( x, y )) = J ζ n (( x, y )) + J r n (( x, y )) (46) ζ n ( z ) is a sum of holomorphic functions. Therefore, what we have to show is that J r n (( x, y )) tends to satisfy theCauchy-Riemann equation as n goes to infinity for any ( x, y ) ∈ Ω .First, we compute the following quantities: ∂f nx ∂x − ∂f ny ∂y = n →∞ − x | − z | + o ( 1 n ) ∂f nx ∂y + ∂f ny ∂x = n →∞ − x | − z | + o ( 1 n ) (47)Now, we can write: ∂r nx ∂x = ∂∂x (cid:0) f nx g nx − f ny g ny (cid:1) = ∂f nx ∂x g nx + ∂g nx ∂x f nx − ∂f ny ∂y g ny − ∂g ny ∂y f ny ∂r ny ∂y = ∂∂y (cid:0) f nx g ny + f ny g nx (cid:1) = ∂f nx ∂y g ny + ∂g ny ∂y f nx + ∂f ny ∂y g nx + ∂g nx ∂y f ny (48)Then, we take adventage from the fact that g n is complex-differentiable: ∂g nx ∂x − ∂g ny ∂y = 0 ∂g nx ∂y + ∂g ny ∂x = 0 (49)Therefore we find: (cid:12)(cid:12)(cid:12)(cid:12) ∂r nx ∂x − ∂r ny ∂y (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) ∂f nx ∂x − ∂f ny ∂y (cid:12)(cid:12)(cid:12)(cid:12) | g nx | + (cid:12)(cid:12)(cid:12)(cid:12) ∂f nx ∂y + ∂f ny ∂x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g ny (cid:12)(cid:12) (50)Exactly the same phenomenon happens for (cid:12)(cid:12)(cid:12) ∂r nx ∂y + ∂r nx ∂y (cid:12)(cid:12)(cid:12) , and we can write: (cid:12)(cid:12)(cid:12)(cid:12) ∂r nx ∂x − ∂r ny ∂y (cid:12)(cid:12)(cid:12)(cid:12) = n →∞ n + 1) x (cid:12)(cid:12)(cid:12)(cid:12) − x − z (cid:12)(cid:12)(cid:12)(cid:12) + o ( 1( n + 1) x ) (cid:12)(cid:12)(cid:12)(cid:12) ∂r nx ∂y + ∂r nx ∂y (cid:12)(cid:12)(cid:12)(cid:12) = n →∞ n + 1) x (cid:12)(cid:12)(cid:12)(cid:12) − x − z (cid:12)(cid:12)(cid:12)(cid:12) + o ( 1( n + 1) x ) (51)11 PREPRINT - M
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27, 2019This is enough to conclude that c in holomorphic over Ω . In addition, it coincides with ζ over ω which is an open set,therefore we finally have: c ( z ) = ζ ( z ) , z ∈ Ω (52) This summation procedure can remind the reader the link between the Riemann zêta function, and the Dirichlet êtafunction [3]: η ( z ) = ∞ (cid:88) n =1 ( − n − n z , x > ζ ( z ) = 11 − − z η ( z ) , z (cid:54) = 1 , x > (53)Like the radial convergence, this definition is valid for x > . However, there is a much more interesting observation tomake here.Indeed, applying Cesàro [4] summation on the original Dirichlet series makes it converge for x > − . Applying itagain will have the series to converge for x > − . From here, one can show by inference that the Dirichlet series canbe analytically extended to the whole complex plan by applying Cesàro summation infinitely many times.It seems that we have exactly the same kind of property for the radial convergence applied on the Riemann series. Thisresult is purely numerical and will not be proven here. But it appears that applying the radial convergence infinitelymany times on the Riemann series makes it converge over the whole complex plan, private from the real axis.An observation that is consistent with this idea is the following fact: For x = 0 , one can observe that the sequence { c n ( z ) } z ∈ N converges to an asymptotic circle. In this section, Ω is now the critical strip: Ω = { x + iy, < x < , y ∈ R ∗ } (54)We are now going to change our point of view on c n ( z ) . Because it is the intersection of two straight lines in thecomplex plan, we can see c n ( z ) as the solution of the following system: (cid:20) (cid:60) c n ( z ) (cid:61) c n ( z ) (cid:21) A n = B n (55)With: A n = (cid:34) (cid:61) e iα ( n +1) z −(cid:60) e iα ( n +1) z (cid:61) e iα ( n +2) z −(cid:60) e iα ( n +2) z (cid:35) , B n = (cid:34) (cid:60) ζ n ( z ) (cid:61) e iα ( n +1) z − (cid:61) ζ n ( z ) (cid:60) e iα ( n +1) z (cid:60) ζ n +1 ( z ) (cid:61) e iα ( n +2) z − (cid:61) ζ n +1 ( z ) (cid:60) e iα ( n +2) z (cid:35) (56)We can now see explicitly that this system always has solutions: det ( A n ) = sin( − y ln( n +2 n +1 ))(( n + 1)( n + 2)) x (cid:54) = 0 (57)The complete expression of A − n is: A − n = (( n + 1)( n + 2)) x sin( − y ln( n +2 n +1 )) (cid:34) −(cid:60) e iα ( n +2) z (cid:60) e iα ( n +1) z −(cid:61) e iα ( n +2) z (cid:61) e iα ( n +1) z (cid:35) (58)12 PREPRINT - M
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27, 2019We then rewrite B n as: B n ( z ) = (cid:20) b n ( z ) b n +1 ( z ) (cid:21) (59)With: b n = | ζ n ( z ) | ( n + 1) x sin (cid:18) arg (cid:18) e iα ( n + 1) z (cid:19) − arg( ζ n ( z )) (cid:19) (60)For z ∈ Ω , it has already been proven that | ζ n ( z ) | is unbounded. Therefore, considering the Equation 55 , the followingassumption seems very reasonable, though it would deserve a explicit proof.If z is a zero of ζ over Ω , then we have: (cid:12)(cid:12)(cid:12)(cid:12) sin (cid:18) arg (cid:18) e iα ( n + 1) z (cid:19) − arg( ζ n ( z )) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) → n →∞ (61)Which with a bit of rearrangement can be formulated as the following theorem: Theorem 3.1. z ∈ Ω , ζ ( z ) = 0 ⇒ lim n →∞ − y ln( n + 1) − arg sum ( ζ n ( z )) + arctan( y − x ) π = U z , U z ∈ Z (62) With arg sum ( ζ n ( z )) the cumulated argument of ζ n ( z ) : arg sum ( ζ n ( z )) = n (cid:88) j =1 arg (cid:18) ζ j +1 ( z ) ζ j ( z ) (cid:19) (63)Again, the complete proof of this theorem is not given here, but the numerical results are very consistent.Let us call this integer U z when it is defined. For the known zeros of the Riemann zêta function, we can numericallycompute this value. Here are the first 30 values for the 30 first non trivial zeros:We can observe these values are all even, But there are no clear pattern in their distribution despite this fact. The pythoncode used to generate this table of values is on the GitHub. This sequence does not appear yet on: https://oeis.org/ [5].Let us call: θ n ( z ) = arg ζ n +1 ( z ) ζ n ( z ) (64)In the figure Figure 9 , we also introduce γ n ( z ) accordingly. One can show by induction that this γ n ( z ) value is actually: γ n ( z ) = − y ln( n + 1) − arg sum ( ζ n ( z )) (65)Now, using geometrical considerations, it is possible to give the following induction formula for sin( θ n ( z )) : sin( θ n +1 ( z )) = sin( − y ln( n +2 n +1 ) + γ n ( z ))( n + 2) x (cid:114) sin( − y ln( n +1 n )+ γ n − ( z )) ( n +1) x sin( θ n ( z )) + n +2) x + − y ln( n +1 n )+ γ n − ( z ))( n +1) x ( n +2) x sin( θ n ( z )) cos ( − y ln( n +2 n +1 ) + γ n ( z )) (66)This formula does not generalize nicely by induction. That is why the quantity arg sum ( ζ n ( z )) is not easy to compute.However, from a geometrical point of view, because the Riemann series is notv bounded if z is in the critical strip, onecan remark that we must have γ n ( z ) to converge to α ( z ) modulo π as n goes to infinity. This interpretation gives an13 PREPRINT - M
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27, 2019n y(n) U(1/2-i*y(n))1 14.1347251417346937904572519835624766 82 21.0220396387715549926284795938969162 143 25.0108575801456887632137909925627734 184 30.4248761258595132103118975305839571 245 32.9350615877391896906623689640747418 286 37.5861781588256712572177634807052984 327 40.9187190121474951873981269146334247 388 43.3270732809149995194961221654068456 409 48.0051508811671597279424727494276636 4610 49.7738324776723021819167846785638367 4811 52.9703214777144606441472966088808216 5212 56.4462476970633948043677594767060321 5613 59.3470440026023530796536486749921759 6014 60.8317785246098098442599018245240815 6415 65.1125440480816066608750542531836072 6816 67.0798105294941737144788288965220700 7217 69.5464017111739792529268575265546586 7618 72.0671576744819075825221079698261175 7819 75.7046906990839331683269167620305404 8420 77.1448400688748053726826648563046925 8821 79.3373750202493679227635928771160578 9022 82.9103808540860301831648374947705599 9423 84.7354929805170501057353112068275569 9624 87.4252746131252294065316678509191351 10025 88.8091112076344654236823480795095125 10426 92.4918992705584842962597252418104965 10827 94.6513440405198869665979258152079645 11028 95.8706342282453097587410292192466718 11429 98.8311942181936922333244201386223539 11830 101.317851005731391228785447940292361 122Figure 8: Values of U z for the first 30 non trivial zeros of zetaFigure 9: Geometrical view of the situation for θ n ( z ) explanation to why U z is always even. One can remark it is in reality well defined all over the critical strip, though itsconvergence rate appears to be slower. From here, we can plot U z for x = ( Figure 10 ).From this plot, we can clearly see that the zeros of the Riemann zêta function play a special role is the distribution ofthe values of U z . 14 PREPRINT - M
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27, 2019Figure 10: Plot of U z for z = − y References [1] H. M. Edwards,
Riemann’s Zeta Function . Dover, 2001.[2] E. W. Weisstein, “Logarithmic spiral.” http://mathworld.wolfram.com/LogarithmicSpiral.html . A Wol-fram Web Resource, Accessed on 2019-02-20.[3] T. Tanriverdi, “Notes on the riemann zeta function,”
ArXiv , 2019.[4] M. Hazewinkel,
Encyclopedia of Mathematics . Springer Science+Business Media B.V. / Kluwer AcademicPublishers, 1994.[5] “Oeis foundation inc. (2011), the on-line encyclopedia of integer sequences.” http://oeis.orghttp://oeis.org