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 APM  Vol.11 No.9 , September 2021
Geometric Proof of Riemann Conjecture (Continued)
Abstract: This paper will prove Riemann conjecture(RC): All zeros of ξ(τ) lie on critical line. Denote , and on critical line. We have found two mysteries in Riemann’s paper. The first mystery is the equivalence: is uniquely determined by its initial value u (t). The second mystery is Riemamm conjecture 2 (RC2): Using all zeros tj of u (t) can uniquely express . We find that the proof of RC is hidden in it. Our basic idea as follows. Consider functional equation . It is known that on critical line and , then we have the upper bound of growth To prove RC2 (or RC), by contradiction. If ξ(τ) has conjugate complex roots t'±iβ'’, β'>0, R2=t'2+β'2, by symmetry ξ(τ)=ξ(-τ), then -(t'±iβ'') do yet. So ξ must contain four factors. Then u(t) contains a real factor and ln|u(t)| contains a term (the lower bound) which contradicts to the growth above. So ξ can not have the complex roots and u(t) does not have the factor p(t). Therefore both RC2 and RC are proved. We have seen that the two-dimensional problem is reduced to one-dimension and the one-dimensional u(t) is reduced to its product expression. Perhaps this is close to the original idea of Riemann. Other results are also discussed by geometric analysis in the last section.

ln p ( t ) 4 ln t + O ( 1 ) , t T 0 ,

which contradicts to the growth above. So ξ can not have the complex roots and u ( t ) does not have the factor p ( t ) . Therefore both RC2 and RC are proved. We have seen that the two-dimensional problem is reduced to one-dimension and the one-dimensional u ( t ) is reduced to its product expression. Perhaps this is close to the original idea of Riemann. Other results are also discussed by geometric analysis in the last section.

Keywords:

1. Introduction

Riemann hypothesis (RH) is one of the most difficult problems in mathematics, which is reviewed in [1] [2] [3] [4] [5]. We shall consider ξ . Although ξ = u + i v and ζ = U + i V have the same zeros, but their properties are quite different. ξ has the symmetry, i.e. v 0 on the critical line, and { u , v } are alternative oscillation with single peak outside critical line, which geometrically implies RC true. Whereas the property of ζ is bad, even if on critical line { U , V } are not alternative oscillation, sometimes almost tangent and multiple peak. Studying ζ is very hard.

To study Riemann conjecture (RC), we have proposed a framework of geometric analysis for ξ in previous papers. If three theorems are proved, then RC holds, also see section 3. Firstly we have proved theorems 1 and 2 by the symmetry of ξ . But to prove theorem 3: “on critical line u ( t ) is single peak”, we have met essential difficulty. The symmetry is not enough and the stronger tool is needed. Thus we have to focus our attention on u ( t ) and re-investigate Riemann’s thought. We have found two mysteries in it finally proved RC by method of analysis.

Denote β = σ 1 / 2 , τ = β + i t . Riemann ξ -function has an integral expression [4], p.17,

{ ξ ( τ ) = 1 ( x τ / 2 + x τ / 2 ) x 3 / 4 f ( x ) d x = u ( t , β ) + i v ( t , β ) , f ( x ) = n = 1 ( 3 ( n 2 π x ) 2 2 n 2 π x ) e n 2 π x . (1.1)

But Riemann did not use (1.1), he had directly taken τ = i t to get the real function [4], p. 301,

ξ ( i t ) = 2 1 cos ( t 2 ln x ) x 3 / 4 f ( x ) d x = u ( t ,0 ) = u ( t ) , v ( t ,0 ) = 0, (1.2)

Why Riemann preferred (1.2) rather than (1.1)? This is the first mystery.

Riemann also regarded t as a complex variable (very important!). Taking τ = β + i t = i ( t i β ) and using the uniqueness of analytic function, we get

ξ ( τ ) u ( z ) , τ = i z , z = t i β , ξ ( τ ) = ξ ( τ ) . (1.3)

Thus the first mystery is formulated as:

Equivalence. ξ ( τ ) u ( t i β ) is uniquely determined by its initial value u ( t ) , i.e., two dimensional problem is reduced to one dimension.

We also consider the initial value problem of Cauchy-Riemann system

{ u β = v t , v β = u t , Ω = { β [ 0 , 1 / 2 ] , t [ 0 , ) } , u ( t , 0 ) = g ( t ) , v ( t , 0 ) = 0 , t [ 0 , ) . (1.4)

As g ( t ) is analytic, Cauchy-Kovalevshkaya theorem confirms that it has a unique analytic solution. This solution just is ξ ( τ ) = u + i v = g ( t i β ) . Actually, by direct verification,

g t i g β = u t + i v t i ( u β + i v β ) = u t + v β + i ( v t u β ) = 0 ,

then (1.3) and the equivalence hold yet. Here ξ ( τ ) = u ( t i β ) resembles a traveling-wave solution of the wave equation, where u ( t ) as an initial value. We had used it in previous papers.

We see that Riemann had studied ξ rather than ζ , his thought can be formulated as:

Riemann conjecture(RC). All zeros of ξ ( τ ) lie on critical line β = 0 .

To study u ( t ) , we find the second mystery in Riemann’s paper [4] (pp. 301-302).

Riemann conjecture 2 (RC2). Using all zeros { t j } of u ( t ) can uniquely determine

u ( t ) = u ( 0 ) j ( 1 t 2 t j 2 ) , 0 t < . (1.5)

For k-ple zeros, should take k-ple products.

The greatest mystery is that RC2 implies RC. Actually, if β 0 , replacing t by z = t i β in (1.5), each factor 1 z 2 / t j 2 0 , then ξ ( τ ) u ( z ) 0 and RC holds.

We rigorously have proved RC2 by the method of analysis in section 2. Here we briefly formulate our basic idea as follows. Consider functional equation

ξ ( s ) = G ( s ) ζ ( s ) , G ( s ) = 1 2 s ( s 1 ) π s / 2 Γ ( s / 2 ) . (1.6)

It is known that on critical line

| G ( t ) | = C t 7 / 4 e t π / 4 ( 1 + O ( t 1 ) ) and | ζ ( t ) | C t 1 / 6 ,

we get the upper bound of growth

ln ( | u ( t ) | / | G ( t ) | ) 1 6 ln t + O ( 1 ) , t T 0. (1.7)

To prove RC2, by contradiction. If ξ ( τ ) has conjugate complex roots t ± i β , β > 0 , by symmetry ξ ( τ ) = ξ ( τ ) , then ( t ± i β ) do yet. Thus by equivalence ξ ( τ ) must contain four factors

p ( z ) = ( 1 z t + i β ) ( 1 z t i β ) ( 1 + z t + i β ) ( 1 + z t i β ) , z = t i β .

Letting β = 0 , u ( t ) contains a real factor

p ( t ) = ( 1 t 2 R 2 ) 2 + 4 t 2 β 2 R 4 > 0 , R 2 = t 2 + β 2 , (1.8)

and then ln | u ( t ) | contains a term (the lower bound)

ln p ( t ) 4 ln t + O ( 1 ) , t T 0 , (1.9)

which contradicts (1.7). So ξ can not have complex roots and u ( t ) does not have the factor p ( t ) . Therefore both RC2 and RC are proved.

Our main contribution is that we for the first time have regarded RC as an initial value problem, found these mysteries and proposed the newest method to prove Riemann conjecture.

Similar work has not been found in other papers and books.

We shall continue to complete the geometric analysis of ξ in section 3.

2. Analytical Proof of RC and RC2

2.1. Two Holes in Riemann’s Analysis

Riemann denoted u ( t ) by ξ ( t ) in his paper. He pointed out, see [4] (pp. 301-302).

“If one denotes by α the roots of the equation ξ ( α ) = 0 , then one can express log ξ ( t ) as

log ( 1 t 2 α 2 ) + log ξ ( 0 ) , (R)

because, since the density of roots of size t grows only like log ( t / 2 π ) as t grows, this expression converges and for infinite t is only infinite like t log t ; Thus it differs from log ξ ( t ) by a function of t2 which is continuous and finite for finite t and which, when divided by t2, is infinitely small for infinite t. This difference is therefore a constant, the value of which can be determined by setting t = 0 .”

Because Riemann did not use α ¯ , we must regard α to be real roots, this is extremely important! But then it is misunderstood.

We now explain his analysis. On critical line u ( t ) is an even entire function and has infinite number of zeros ± t j . We consider formally the remainder of ln u ( t ) ,

j = n + 1 ln ( 1 t 2 / t j 2 ) j = n + 1 t 2 t j 2 = t 2 γ n , γ n = j = n + 1 1 t j 2

One knows t n = O ( n / ln n ) (We have better t n = B ( n ) 2 π n / ln n , B ( n ) = 1.46 1 , n 10 3 ). Thus

γ n = C j = n + 1 ln 2 n ( B ( n ) n ) 2 C n ln 2 x ( B ( x ) x ) 2 d x C ln 2 n B ( n ) n , C = ( 2 π ) 2 . (2.1)

This series (R) converges for finite t. (Riemann said) u ( t ) has the growth ln u ( t ) = O ( t ln t ) (This expression is not suitable. We shall use ln | u ( t ) | , which admits u ( t ) = 0, see (2.6)). The series (R) differs from ln u ( t ) by a function of t2, which, when divided by t2, is infinitely small for infinite t. (Riemann said) “This difference is therefore a constant.” This is not rigorous, in general, it should be O ( ln t ) rather than a constant. There is a hole of the uniqueness.

Hadamard in 1893 proved product formula for general entire function, see [4] (p. 20), [5] (p. 16). Denoting τ = β + i t = i z , z = t i β , the zeros of ξ are conjugate, one gets

ξ ( i z ) = e A + B z j ( 1 z 2 z j 2 ) , z = t i β , z j = t j i β j .

which is even if β = 0 , then B = 0 . Taking z = 0 , then e A = ξ ( 0 ) .

But this Hadamard’s way from β 0 to β = 0 implies a serious contradiction. If all zeros are real, then which itself assumes RC. If conjugate complex zeros are admitted, which just denies RC. Hadamard’s theorem was referred by Von Mangoldt “The first real progression in the field in 34 years”, see [4], p. 39. We think this is a misunderstanding. Actually, Hadamard’s way is independent of proving RC and far from Riemann’s thought. Our idea is to consider u ( t ) at β = 0 as an initial value. We shall directly prove RC2 by method of analysis, so this contradiction is cast off.

2.2. Analytical Proof of RC and RC2

We consider the functional equation

ξ ( s ) = G ( s ) ζ ( s ) , G ( s ) = 1 2 s ( s 1 ) π s / 2 Γ ( s 2 ) , ξ ( s ) = ξ ( 1 s ) . (2.2)

Firstly, Γ ( s / 2 ) has asymptotic expansion

Γ ( s / 2 ) = 2 π ( s / 2 ) s / 2 1 / 2 e s / 2 ( 1 + O ( 1 / s ) ) .

Take logarithm and decompose the real part, c = ln ( 2 π ) ,

ln Γ ( s / 2 ) = c + ( s / 2 1 / 2 ) ln ( s / 2 ) s / 2 + O ( 1 / s ) , s = σ + i t , = c + ( ( σ 1 ) / 2 + i t / 2 ) { ln ( t / 2 ) + i ( π / 2 arctan σ t ) } ( σ + i t ) / 2 + O ( 1 / s ) ,

R e ln Γ = c + ( ( σ 1 ) / 2 ) ln ( t / 2 ) t π / 4 + ( t / 2 ) ( σ / t ) σ / 2 + O ( 1 / t ) ,

Thus for σ = 1 / 2

| Γ ( s / 2 ) | = 2 π ( t 2 ) 1 / 4 e t π / 4 ( 1 + O ( 1 / t ) ) . (2.3)

Besides, s ( 1 s ) / 2 = ( t 2 + 1 / 4 ) / 2 and | π s / 2 | = π 1 / 4 , we get

| G ( t ) | = 2 3 / 2 π 1 / 2 1 / 4 ( t 2 ) 7 / 4 e t π / 4 ( 1 + O ( 1 / t ) ) . (2.4)

Secondly, there are growths of ζ ( s ) for large t, [4], p. 185, p. 200,

{ | ζ ( σ ) | C t ( 1 σ ) / 2 ln t , 0 σ 1, | ζ ( 1 / 2 + i t ) | C t λ , λ = 1 / 5 or λ = 19 / 116 . (2.5)

We only need the estimate of σ = 1 / 2 , also see Remark 1.

Thus, we get an upper bound of growth (note: not ln | u | = O ( t ln t ) )

ln ( | u ( t ) | / | G ( t ) | ) 1 6 ln t + C , if t T 0. (2.6)

So u ( t ) = 0 is admitted.

Finally, to prove RC2 (or RC), by contradiction. If ξ ( τ ) has conjugate complex roots t ± i β , β > 0 , R 2 = β 2 + t 2 , by symmetry ξ ( τ ) = ξ ( τ ) , then ( t ± i β ) do yet. Denoting z = t i β and using the equivalence, ξ ( τ ) u ( z ) must contain four factors

p ( z ) = ( 1 z t + i β ) ( 1 z t i β ) ( 1 z t + i β ) ( 1 z t i β ) .

Letting β = 0 , then u ( t ) must contain a real polynomial of fourth degree

p ( t ) = ( 1 t t + i β ) ( 1 t t i β ) ( 1 + t t i β ) ( 1 + t t + i β ) = ( 1 + t 2 2 t t R 2 ) ( 1 + t 2 + 2 t t R 2 ) = ( 1 t 2 R 2 ) 2 + 4 t 2 β 2 R 4 > 0 , (2.7)

and ln | u ( t ) | contains a term (as a lower bound)

ln p ( t ) 4 ln t + O ( 1 ) , t T 0. (2.8)

Its growth contradicts (2.6). Thus ξ ( τ ) can not have complex roots and u ( t ) does not have the factor p ( t ) . Therefore both RC and RC2 are rigorously proved.

Originally we want to prove only RC2, fortunately, both RC and RC2 are proved.

Remark 1. In period of Riemann, no estimates (2.5), but it is possible to prove RC. As

ζ ( s ) = n = 1 n s = n = 1 0 1 ( n s ( n + x ) s ) d x + 1 y s d y = 1 s 1 + n = 1 s 0 1 x s 1 ( 1 x ) d x , by integration by part

is already continued analytically to R e ( s ) > 0 (actually this is Euler’s method). Thus

| ζ ( s ) | | 1 s 1 | + | s | 0 x σ 1 d x C t ,

and gets an coarse estimate (e.g. | u ( t ) | C t 3 e t π / 4 in [5], p. 27)

ln ( | u ( t ) | / | G ( t ) | ) ln t + O ( 1 ) . (2.9)

By (2.8) one can still prove that ξ ( τ ) does not have complex roots. But nobody noted it. Therefore I feel, Riemann had already approached to prove RC. Our proof is completed to follow Riemann’s thought.

Numerical experiments 1. Using the data of the first 105 zeros in Odlyzko [6], we have computed w n ( t ) = u ( 0 ) j = 1 n ( 1 t 2 / t j 2 ) and u ( t ) in (1.2) for t [ 0,50 ] . We draw these curves by variable scale u / M , here M ( t ) = ( t / 2 + 1 ) 23 / 12 e t π / 4 , M ( 0 ) = 1 . We see in Figure 1 that w n ( t ) approximates u ( t ) very well, of course, larger is t, then larger is its deviation.

We have for the first time computed w n and Figure 1, which make us believe the correctness of RC2, and then consider its analytic proof as before.

3. Continuation of Geometric Analysis

In previous papers [7] [8] [9], we have proposed geometric analysis and proved three results:

Theorem 1. If u ( t ) is single peak and single zero, then the peak-valley structure for β > 0 and RC hold.

Theorem 2 (old). If two roots of u ( t ) are very close to each other (including double zeros), then the peak-valley structure for β > 0 and RC still hold.

Theorem 3. u ( t ) is single peak.

RC can be derived by these three theorems. We at present re-examine these theorems. Theorem 1 is correct, see section 3.1. Theorem 2 is also correct but uncomplete, which is generalized in section 3.4. The original proof of theorem 3 [9] has defect (see section 3.2), which is derived by RC2 in section 3.3.

3.1. A Concise Proof of Theorem 1

Consider a root-interval I j = [ t j , t j + 1 ] of u ( t , β ) , obviously t j and t j + 1 depend on β . Assume u ( t , β ) > 0 inside I j for β ( 0, 1 / 2 ] . At the left end t j , u ( t j , β ) = 0 and u t ( t j , β ) > 0 , then the slop u t ( t j , r ) > 0 , r ( 0 , β ] (which was proved in [9] ) we have

v ( t j , β ) = v ( t j , 0 ) + 0 β u β ( t j , r ) d r = 0 β u t ( t j , r ) d r < 0.

At the right end t j + 1 , u ( t j + 1 , β ) = 0 and u t ( t j + 1 , β ) < 0 , similarly,

Figure 1. w n ( t ) approximates u ( t ) by n = 10 5 zeros t j .

v ( t j + 1 , β ) = v ( t j + 1 , 0 ) + 0 β u β ( t j + 1 , r ) d r = 0 β u t ( t j + 1 , r ) d r > 0.

Because v ( t , β ) has opposite signs at two ends of I j , there surely exists some inner point t = t ( β ) such that v ( t , β ) = 0 . Then | v ( t , β ) | is valley and { | u | , | v | } form a peak-valley structure. There is a positive lower bound independent of t I j .

min t I j ( | u ( t , β ) | + | v ( t , β ) | ) = μ j ( β ) > 0 , β ( 0 , 1 / 2 ] .

So RC holds in I j . Because the zeros { t j } of analytical function u ( t , β ) do not have finite condensation point, then any finite t surely falls in some I j . RC holds for any t.

3.2. Defect in Original Proof of Theorem 3

To prove u ( t ) to be single peak, we discussed monotone growth of the argument ϕ + ψ of ξ in [9] and used Riemann’s estimate ψ ( t ) = π S ( t ) = O ( ln t ) . As ϕ ( t ) = t 2 ( ln t ln ( 2 e π ) ) + 7 8 π + O ( t 1 ) is of super-linear growth, when t increases to t + 1 , the increment δ ϕ = 1 2 ( ln t ln ( 2 π ) ) + O ( t 1 ) slowly increases,

whereas δ ψ = O ( 1 ) , then claimed that ϕ + ψ monotone increases. But here δ ψ = O ( 1 ) is not correct, which may be O ( ln t ) . We consider Riemann’s symmetrization

Z ( t ) = e i ϕ ( t ) ζ ( 1 / 2 + i t ) , ζ ( 1 / 2 + i t ) = U ( t ) + i V ( t ) ,

(here the decay factor | G ( s ) | 0 is reduced) and have

Z ( t ) = U cos ϕ V sin ϕ , g ( t ) = U sin ϕ + V cos ϕ 0 , V U = sin ϕ cos ϕ .

Thus the local argument π S ( t ) = arctan ( V / U ) = ϕ in root-interval I j = [ t j , t j + 1 ] .

We see in Figure 2 (also see [6] [10] ) that S ( t ) jumps by 1 at zero t j of cos ϕ , then linearly decreases in ( t j , t j + 1 ) with the slope ψ 1 2 ln t . Whereas ϕ has slope ϕ ( t ) 1 2 ln t . It seems ϕ + ψ 0 , and difficult to prove ϕ + ψ > 0 . Besides, as g ( t ) = 0 , this research is not suitable.

This defect makes us turn to u ( t ) on critical line and re-investigate RC2.

3.3. Revision Proof of Theorem 3

By RC2, take logarithm ln u ( t ) and derivation to get

ln u ( t ) = ln u ( 0 ) + j = 1 ln ( 1 t 2 t j 2 ) , u ( t ) u ( t ) = j = 1 2 t t 2 t j 2 . (3.1)

Consider root-interval I n = [ t n , t n + 1 ] and decompose the summation into three parts.

Figure 2. Curves U , V , S , cos ϕ . S jumps by 1 twice in I 126 = [ t 126 , t 127 ] .

u t ( t ) = 2 t u ( t ) { j n 1 1 t 2 t j 2 + ( 1 t 2 t n 2 1 t n + 1 2 t 2 ) j n + 2 1 t j 2 t 2 } , = 2 t u ( t ) g ( t ) , g ( t ) = g 1 ( t ) + g 2 ( t ) + g 3 ( t ) , t n < t < t n + 1 . (3.2)

Note that u ( t ) inside I n has same sign. We investigate each sum

{ g 1 ( t ) = j n 1 1 t 2 t j 2 > 0 , t > t n > t j , g 2 ( t ) = j n + 2 1 t j 2 t 2 < 0 , t < t n + 1 < t j . g 3 ( t ) = 1 t 2 t n 2 1 t n + 1 2 t 2 > 0 , t n < t < t n + 1 . (3.3)

For t I n , g 1 ( t ) > 0 , g 2 ( t ) < 0 are finite. If t is close to t n + 0 , then g 3 ( t ) tends to + . If t is close to t n + 1 0 , then g 3 ( t ) tends to . There surely exists some inner point t * I n such that g ( t * ) = 0 . We show this point t * is unique. For this we consider their derivatives

{ g 1 ( t ) = j n 1 2 t ( t 2 t j 2 ) 2 < 0 , t > t n > t j , g 2 ( t ) = j n + 2 2 t ( t j 2 t 2 ) 2 < 0 , t < t n + 1 < t j . g 3 ( t ) = { 2 t ( t 2 t n 2 ) 2 + 2 t ( t n + 1 2 t 2 ) 2 } < 0 , t n < t < t n + 1 . (3.4)

Thus all g 1 , g 2 , g 3 are monotone decreasing for t I n , their sum g ( t ) does yet. Therefore this zero t * of g ( t ) is unique and then u ( t ) is single peak. This proves theorem 3.

Using Lagarias’s positivity [11] (RC is assumed), we get monotone growth [8], p. 344,

| ξ ( β + i t ) | > | ξ ( β 0 + i t ) | 0 , for β > β 0 .

This is a clear description for RC. Therefore Riemann ξ -function has mathematical beauty: The symmetry, single peak and monotone growth (i.e. the ordering).

3.4. Theorem 2 Holds for m-ple Zeros

In large scale computations [6] [12] [13] [14] all zeros of u ( t ) are single, no multiple zeros are found. People believe there are only single zeros, but so far do not prove. We have to skirt round the difficulty to prove theorem 2. We shall extend theorem 2 (old) as:

Theorem 2. If u ( t ) has m-ple zeros on critical line, then { u , v } for small β > 0 will bifurcate into m alternative oscillations with single peak, and RC still holds.

Proof. Assume that there are three consecutive zeros { t j 1 , t j , t j + 1 } of u ( t ) on critical line { β = 0 , 0 < t < } , which form two root-intervals, and t j = t * is a m-ple zero, m 1 . Denote H = min ( t j + 1 t j , t j t j 1 ) and h = H / 2 . Denoting y = t t * and the origin O ( y = 0 , β = 0 ) , and fixing a small β > 0 , we discuss a circle K δ ( t * ) with r = y 2 + β 2 δ h ( δ to be defined). Assume that the real function g ( y ) = u ( t ) has m-ple zero at O

g ( y ) = a ( t * ) y m + b ( t * ) y m + 1 + , a ( t * ) 0, (3.5)

These coefficients a ( t * ) and b ( t * ) are of same order. Thus

g ( y i β ) = a ( t * ) { ( y i β ) m + K ( t * ) ( y i β ) m + 1 } , K b ( t * ) / a ( t * ) 0 , (3.6)

Below we discuss P ( y i β ) = ( y i β ) m , and temporarily omit high-order remainder.

Using y i β = r ( y / r i β / r ) = r ( cos ϕ i sin ϕ ) and De Moivre formula we have

( y i β ) m = r m ( cos ϕ i sin ϕ ) m = r m ( cos ( m ϕ ) i sin ( m ϕ ) ) = 0 ,

and discuss r m cos ( m ϕ ) = 0 and r m sin ( m ϕ ) = 0 respectively.

1) The zeros of cos ( m ϕ ) satisfy m ϕ j = ( 2 j 1 ) π / 2 , i.e.

ϕ j = 2 j 1 2 m π , 0 < ϕ j < π , j = 1 , 2 , , m ,

which are symmetric with respect to π / 2 . If m = 2 n even, then all ϕ j π / 2 . If m = 2 n 1 odd, the middle argument ϕ n = π / 2 satisfies cos ( ϕ n ) = 0 , i.e. original point y = 0 .

Taking the roots y / r = cos ϕ j , i.e. y 2 = ( y 2 + β 2 ) cos 2 ϕ j , we have y / β = ± cos ϕ j / sin ϕ j = ± cot ϕ j and re-arrange the ordering of these m zeros as

y / β = { cot ( ϕ 1 ) , , cot ( ϕ n 1 ) , 0 , cot ( ϕ n 1 ) , , cot ( ϕ 1 ) } , if m = 2 n 1. y / β = { cot ( ϕ 1 ) , , cot ( ϕ n ) , cot ( ϕ n ) , , cot ( ϕ 1 ) } , if m = 2 n , (3.7)

2) The zeros of sin ( m ψ ) satisfy m ψ k = k π , i.e.,

ψ k = k m π , k = 1 , 2 , , m 1 ,

here no k = m , as ψ m = π corresponds a trivial zero β / r = sin ( π ) = 0 (i.e. β = 0 ). Thus there are only m 1 nontrivial zeros, whose arguments are symmetric with respect to π / 2 . When m = 2 n even, ψ n = π / 2 for k = n , i.e. sin ( ψ n ) = 1 , which corresponds the original y = 0 .

Taking the roots β / r = sin ψ k , i.e. β 2 = ( y 2 + β 2 ) sin 2 ψ k , we have y / β = ± cos ( ψ k ) / sin ( ψ k ) = ± cot ( ψ k ) and re-arrange the ordering of these m 1 roots as

y / β = { cot ( ψ 1 ) , , cot ( ψ n 1 ) , cot ( ψ n 1 ) , , cot ( ψ 1 ) } , if m = 2 n 1 , y / β = { cot ( ψ 1 ) , , cot ( ψ n 1 ) , 0 , cot ( ψ n 1 ) , , cot ( ψ 1 ) } , if m = 2 n . (3.8)

Comparing (3.7) and (3,8) we see that for fixing β > 0 , the maximum of these roots is | y | = β cot ( ϕ 1 ) = β cot ( π / 2 m ) , thus the radius of circle K δ ( t * ) satisfies r = ( y 2 + β 2 ) 1 / 2 = β / sin ( π / 2 m ) h , i.e. we should confine β h sin ( π / 2 m ) < h π / 2 m = H h π / 4 m = d .

For g ( y ) = a ( t * ) ( y i β ) m = u 1 + i v 1 , we have the following conclusions:

1) The real part u 1 ( y , β ) has m zeros y / β = cot ϕ j , and the imaginary part v 1 ( y , β ) has m 1 zeros y / β = cot ψ k . Due to

ϕ j = 2 j 1 2 m π < ψ j = j m π , j = 1 , 2 , , n 1 ,

obviously cot ϕ j > cot ψ j . Hence all zeros of (3.7) and (3.8) are alternatively arranged.

2) At zeros ψ j = j π / m of v 1 , the peak values u 1 = a ( t * ) r m cos ( j π ) = a ( t * ) r m ( 1 ) j alternatively change their signs. At zeros ϕ j = ( j 1 / 2 ) π / m of u 1 , the peak values v 1 = a ( t * ) r m sin ( ( j 1 / 2 ) π ) = a ( t * ) r m ( 1 ) j 1 also alternatively change their signs.

3) Because v 1 0 at zero of u 1 and u 1 0 at zero of v 1 , they all are single peak. Thus in the m root-intervals of u 1 , all { | u 1 | , | v 1 | } form local peak-valley structures, and RH locally holds.

Finally, in K δ ( t * ) with β ( 0, δ ] suitably small we discuss a general case

g ( y i β ) = a ( t * ) { ( y i β ) m + R m } , R m = O ( r m + 1 ) , r δ , (3.9)

The actual zeros of ξ ( τ ) are small perturbations of these zeros mentioned above, which do not change these m peak-valley structures in K δ ( t * ) . When β increases, they will continue to develop toward locally convex direction by

Figure 3. R e ( ξ 2 ) , R e ( ξ 3 ) at zero point t 2 bifurcate into 2 or 3 zeros.

theorem 1, so RC holds. Hence theorem 2 is proved.

Numerical experiments 2. With the suggestion of Dr.XM Jiao, we have computed ( ξ ) m = u + i v , m = 2 , 3 at the second zero t 2 = 21.0220 of ξ ( s ) . We see in Figure 3 for β = 0.05 , 0.1 , u ( t , β ) indeed bifurcate into m curves of single peak, and { u , v } are alternative oscillation. The peak of u ( t , β ) develops toward its convex direction.

Remark 2. The author of this paper should sincerely thank Dr. Xiangmin Jiao (Stone Brook University, US). He, on 22 April in 2021, sent e-mail to me to discuss the highest super-convergence (see [15] ), we know for the first time. I told him I’m studying RC, and brings special interests and discuss together. He has carefully verified my papers and proposed valuable comments. He suggested the example ( ξ ( s ) ) m and sent papers [10] [16] to me. If no support from him, I very hard, at least in a shorter time, propose the newest proof.

Acknowledgements

The author expresses sincere gratitude to the reviewer for his careful remark, valuable and constructive comments. Besides, I should thank Prof. Zhengtin Hou and Prof. Xinwen Jiang for their precious opinion in discussion.

Cite this paper: Chen, C. , (2021) Geometric Proof of Riemann Conjecture (Continued). Advances in Pure Mathematics, 11, 771-783. doi: 10.4236/apm.2021.119051.
References

[1]   Bombieri, E. (2000) Problems of the Millennium: The Riemann Hypothesis. AMS, 107-124.
https://www.claymath.org/sites/default/files/official_problem_description.pdf

[2]   Conrey, J. (2003) The Riemann Hypothesis. Notices of the AMS, 50, 341-353.

[3]   Sarnak, P. (2004) Problems of the Millennium: The Riemann Hypothesis.
http://xcivzuj.claymath.org/sites/default/files/sarnak_rh_0.pdf

[4]   Edwards, H.M. (2001) Riemann’s Zeta Function. Dover Publication Inc., Mineola.

[5]   Borwein, P., Choi, S., Rooney, B. and Weirathmuller, A. (2008) The Riemann Hypothesis. Springer, New York.

[6]   Odlyzko, A. (n.d.) Tables of Zeros of the Riemann Zeta Function. URL.
http://www.dtc.umn.edu/~odlyzko/

[7]   Chen, C.M. (2020) The Symmetry of Riemann ξ-Function. Advances in Pure Mathematics, 10, 464-470.
https://doi.org/10.4236/apm.2020.108028

[8]   Chen, C.M. (2020) Local Geometric Proof of Riemann Conjecture. Advances in Pure Mathematics, 10, 589-610.
https://doi.org/10.4236/apm.2020.1010036

[9]   Chen, C.M. (2021) Geometric Proof of Riemann Conjecture. Advances in Pure Mathematics, 11, 334-345.
https://doi.org/10.4236/apm.2021.114021

[10]   Riemann Hypothesis. Wikipedia.
https://en.wikipedia.org/w/index.php?title=Riemann_hypothesis&oldid=1025138468

[11]   Lagarias, J. (1999) On a Positivity Property of the Riemann ξ-Function. Acta Arithmetica, 89, 213-234.
https://doi.org/10.4064/aa-89-3-217-234

[12]   Lune, J. and Riele, H. (1983) On the Zeros of the Riemann Zeta Function in the Critical Strip. Mathematics of Computation, 41, 759-767.
https://doi.org/10.1090/S0025-5718-1983-0717719-3

[13]   Lune, J., Riele, H. and Winter, D. (1986) On the Zeros of the Riemann Zeta Function in the Critical Strip. Mathematics of Computation, 46, 667-681.
https://doi.org/10.1090/S0025-5718-1986-0829637-3

[14]   Platt, D. and Trudgian, J. (2021) The Riemann Hypothesis is True up to 3·1012. Bulletin of the London Mathematical Society, 53, 792-797.
https://doi.org/10.1112/blms.12460

[15]   Chen, C.M. and Hu, S.F. (2013) The Highest Order Superconvergence for bi-k Degree Rectangular Elements at Nodes: A Proof of 2k-Conjecture. Mathematics of Computation. 82, 1337-1355.
https://doi.org/10.1090/S0025-5718-2012-02653-6

[16]   Schleich, W.P., Bezdekova, I., Kim, M.B., Abbott, P.C., Maler, H., Montgomery, H.L. and Neuberger, J.W. (2018) Equivalent Formations of the Riemann Hypothesis Based on Lines of Constant Phase. Physica Scripta, 93, Article No. 065201.
https://doi.org/10.1088/1402-4896/aabca9

 
 
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