Πk=1. z e - z/2. 2 /e. e zk. 1-z. 1+ 2n. 1-z - 2. () z = -z( ) () z = e. = e. e - z 2 () z ( z/2) z is expressed by the Hadamard product as follows.

Transcription

1 08 Factoriation of Completed Riemann Zeta 8. Hadamard product of Formula 8..0 ( Hadamard 893 ) When is Riemann eta function and the the eros are k k =,,3,, the following expression holds. /e () = ( -) k= - k e k n = n e - n Using this formula, we can easily obtain factoriation of a completed eta function. Formula 8.. ( Hadamard product of ) Let completed eta function be as follows. - - () = -( ) () When non-trivial eros of are k = x k iy k k =,,3, and is Euler-Mascheroni constant, is expressed by the Hadamard product as follows. () = e () = e log log -- log log -- k= n = - k From the Weierstrass expression on the gamma function, n = n e - n Substituting this for Formula 8..0, From this, - / e = ( /) /e e - / () = ( -) ( / ) k= = e k x n xn - e xn yn xn y n xn y n e log e - e - / - ( / ) k= = e ( -) ( /) - -( ) () = e - k - k log-- k= log-- k= - - e k e k - k - k e k e k (.0) (.)

2 Multiplying both sides by - i.e ( ) () = e = e () log log -- log -, = e k= log log k e k k= - k This formula means that the completed eta function is incompletely factored at the non-trivial eros. Now, let the non-trivial eros are k = x k iy k k =,,3,. Then, e k (.0) k= - k e k = n = = n = - xn iy n By using this, (.0) is expressed as follows. () = e log log -- e xn iyn - xn iy n x n xn - e xn yn xn y n xn y n n = x n xn - e xn yn xn y n xn y n e xn iyn (.) If x n = / n =,,3,, (.) becomes () = e log log -- n = - /4 yn /4 yn e /4 yn Althogh a general formula for generating eros whose real part is / is not known, Mathematica has a function y n = Im[ZetaZero [n ]] that generates this numerically. When both sides of (.') are drawn overlapping using known non-trivial,500 eros, it is as follows.. (.') - -

3 As a special value for (.), an important formula used in the next section is obtained. Formula 8.. ( Special value ) When non-trivial eros of Riemann eta function are x n iy n n =,,3,, the following expression holds. n = xn x n - - e xn yn xn y n = e log -log - = (.) Giving = to (.), From this, ( ) = e log log -- n = xn x n - - e xn yn xn y n n = xn x n - - e xn yn xn y n = e log -log - = (. ) = (.) If x n = / n =,,3,, (.) becomes r= From this, n = /- - ( /) y r /4 y n / e ( / ) y r r = = e /4 y r = e log -log - log = - log- = (.' ) As a special value for (.0), the following formulas is obtained. Formula 8..3 ( Special value ) When non-trivial eros of Riemann eta function are x k iy k k =,,3,, the following expression holds. - - = (.3) n = x n iy n x n iy n 3 Giving =-, to (.0) respectively, e - e log log -- log log -- k= k= - e k k - e k k = ( -) = 3 = ( ) = - 3 -

4 Multiplying both sides respectively, k= - k k = 3 Let k = x k iy k k =,,3,. Then, k= k= From these, k= Therefore, n = - k k - k = n = = n = k - x n iy n - xn iy n xn iy n = n = n = - xn iy n - xn iy n xn iy n - x n iy n - x n iy n xn iy n = 3 - x n iy n - xn iy n xn iy n (.3) Both sides of (.3 ) were computed using known non-trivial 0000 eros in an attempt. Both sides was equal up to 3 decimal places

5 8. Non-trivial eros whose real part is not / According to Theorem 7.4. in " 07 Completed Riemann Zeta ", if Riemann eta function has non-trivial ero whose real part is not /, the one set have to consist of the following four. / s i s, /- s i s 0< s </ In this section, we will consider how the formulas in the previous section are expressed when non-trivial eros whose real part is / and non-trivial eros whose real part is not / are mixed. Lemma 8.. Let be Euler-Mascheroni constant, non-trivial eros of Riemann eta function are x n iy n n =,,3,. Among them, eros whose real part is / are / iy r r =,,3, and eros whose real parts is not / are / s i s 0< s </ s =,,3,. Then Formual 8.. (.) is expressed as follows. () = e log log -- Formula 8.. (.) was as follows. () = e r= - /4 yr /4 yr e /4 yr s s= - e / s s / s s / s s s s s= - e / s s / s s / s s log log -- n = x n xn - e xn yn xn y n xn y n s Regarding non-trivial eros / iy r r =,,3, whose real part is /, a part of the right side is expressed as follows. r= - /4 yr /4 yr e /4 yr On the other hand, regarding non-trivial eros / s i s 0< s </ whose real part is not /, a part of the right side is expressed as follows. s= s= s - e / s s / s s / s s s s s - e / s s / s s / s s Multiplying these, we obtain the desired expression. (.) (.) - 5 -

6 Theorem 8.. Let be Euler-Mascheroni constant, non-trivial eros of Riemann eta function are x n iy n n =,,3,. Among them, eros whose real part is / are / iy r r =,,3, and eros whose real parts is not / are / s i s 0< s </ s =,,3,. Then the following expressions hold. n = n = n = x n - - xn y n x n x n y n = r = = (.) /4 y r s = s / s s s / s (.3) s x n log = x - log - = (.4) n y n Substituting = for Formula 8.. (.), ( ) = e log log -- n = Substituting = for Lemma 8.. (.), ( ) = e log log -- From these, n = xn e xn yn x n - - xn y n = e Here, conveniently, So, r = /4 y r s= xn x n - - e xn yn xn y n = (. ) s s / s s - / s s e r = /4 y r s s s = / s s / s s s s = - s= / s s / s s s s = / s s s / s s s s / s s - / s s s s s s = - - / s s / s s / s s / s s = = s s - / s s / s s s s / s s / s s (. ) - 6 -

7 n = n = x n - - xn y n x n x n y n = r = And, from (. ) and (.), = (.) /4 y r s = s / s s s / s (.3) s n = x n log = x - log - = (.4) n y n Although the story goes a little aside, using (.), we obtain the following special values. Formula 8..3 ( Special values ) When non-trivial eros of Riemann eta function are x k iy k k =,,3,, the following expressions hold. n = n = - xn iy n xn iy n - xn iy n xn iy n = (.5 ) = 3 From the proof of Formula 8..3 in the previous section, k= k= - k k = 3 k= From Theorem 8.., n = = n = = n = - k x n - - xn y n Substituting this for ( ), k= - k = n = - xn iy n xn iy n k - xn iy n xn iy n = n = = n = x n - - xn y n x n xn y n (.5 - ) = (.) - xn iy n Substituting this and ( - ) for ( 0 ) sequentially, = 3 k= k = n = xn iy n - xn iy n ( ) ( - ) ( 0 ) = (.5 ) xn iy n (.5 - ) (.5 ) and (.5 - ) were calculated using known non-trivial eros. It was as follows

8 Well, let us return to the subject. By using Theorem 8.., the very important following theorem is obtained. Theorem 8..4 Let non-trivial eros of Riemann eta function are x n iy n n =,,3, and be Euler-Mascheroni constant. If the following expression holds, non-trivial eros whose real parts is not / do not exist. r = /4 y r log = - log - = (. ' ) Although non-trivial eros / iy r r =,,3, exist in fact, assume non-trivial eros / s i s, /- s i s (0 < s </)exist in addition. Then, the following expression holds from Theorem 8.. (.3), (.4). r = /4 y r s = s s log = / s s / s - log - s Here, the following inequality holds for 0< s </ and arbitrary real number s, So, Thus, i.e. s / s s / s s s = r = r = s = / s s / s s / s s s s > 0 for 0< / s s / s s </ s /4 y r /4 y r log < - log - log - log - As the contrapositive to the above, this theorem holds. > 0 Note This theorem shows that the equation (. ' ) is equivalent to the Riemann hypothesis. However, for the proof of (.'), the imaginary part y r of the non-trivial eros / iy r r =,,3, have to be obtained as a formula

9 Both sides of (. ' ) were calculated with the formula manipulation software Mathematica using known nontrivial eros. Both sides coincided with four decimal places

10 8.3 Factoriation of ( ) Formula 8.. ( Hadamard product ) is what the completed eta function is incompletely factored at the non-trivial eros. However, using Theorem 8.., the compensation terms disappear and is completely factoried at the non-trivial eros. Theorem 8.3. ( Factoriation of ( )) Let Riemann eta function be, the non-trivial eros are n = x n iy n n =,,3, and completed eta function be as follows. - - () = -( ) Then, is factoried as follows. () = n = From Formula 8.. (.), () = e () x n - xn y (3.) n xn y n log log -- n = On the other hand, from Theorem 8.. (.4), n = From this, x n log = x - log - n y n xn n= xn e yn = e log - log- Substituting this for the right side of the above, () = e = n = log log -- In addition, this formula is known. n = x n n= - e xn y n xn y n x n - xn y n xn y n xn xn yn e log - log- x n - xn y (3.) n xn y n If x n = / n =,,3,, (3.) becomes () = n = - /4 yn /4 yn (3.') When both sides of (3.') are drawn overlapping using known non-trivial,500 eros, it is as follows.. This is exactly the same as the one in Formula 8.. (.')

11 - -

12 8.4 Factoriation of () By replacing with / in Theorem 8.3., completed eta function that is an even function is obtained. Theorem 8.4. ( Factoriation of () ) Let Riemann eta function be, the non-trivial eros are n = x n iy n n =,,3, and completed eta function be as follows. () = - Then, is factoried as follows. () = ( 0) n = Where, ( 0 ) = n = - - x n -/ - x n -/ y n x n -/ (4.) y n x n -/ y n x n y n = - 4 /4 4 = (4. 0 ) From Theorem 8.3., - - () = -( ) () = n = () x n - xn y (3.) n xn y n Replacing with / in the first expression, = - Substituting =0 for this, ( 0 ) = - /4 4 4 Replacing with / in (3.), = n = = n = = n = - - x n - xn y n x n - xn y n = xn y n /4 x n - xn y n xn y n xn y n x n -x n /4 y n x n - - x n y n xn y n xn y n xn y n = : () - -

13 i.e. = n = Since / = (), () = n = Substituting =0 for this, ( 0 ) = n = x n -/ y n x n y n x n y n x n -/ y n x n -/ y n x n y n Substituting this for the right side of (), we obtain (4.), x n -/ - x n -/ y n x n -/ y n x n -/ - x n -/ y n x n -/ y n Lemma 8.4. Among Theorem 8.4. (4.), the product h () of the factor whose real part x n is / is expressed as follows. h () = h () 0 r= yr (4.) Where, h () 0 = r= y r /4 y r (4. 0 ) Replacing a part of x n with / in Theorem 8.4., we obtain the desired expression immediately. Note It is the Riemann hypothesis that () = h () must be. Lemma Assume that the factor whose real part is not / exists among Theorem B8.4. (4.). Then, when two real s > /8, the product () of these factors is numbers are s, s expressed as follows. () = ( 0) s= s.t. 0< s </ & s s s - s s s Where, ( 0 ) = s= / s s 4 s (4.3) s s s /- s s < (4.3 0 ) If such a product of factors exists, according to Theorem 7.4. in " 07 Completed Riemann Zeta ", the one set have to consist of the following four

14 / s i s, /- s i s 0 < s < / So, replacing a part of x n, y n with / s, s respectively in Theorem 8.4., Last, 0 s= () =( ) = s () 0 s= s () 0 = s= - = s () 0 s= = s= s s s s - s s s s / s s s s / s s s - s s s s s s s s s s s s s s - s s s / s s /- s s s s / s s s s / s s 4 s (4.3) s = / s /- s / s s /- s 4 s s s - s = 6 s s s 4 s 4 (4.3 0 ) i.e. / s s /- s s s - s = 6 s s Here, if 0< s </ & s > /8, the following inequality holds. From this, Therefore, s > s - 8 s - s 6 > 0 / s s /- s s Since, both sides are positive, From this, / s s s s / s s /- s s s s /- s s < < > s s s s - 4 -

15 If there are a plurality of such sets, s s s= / s s s s /- s s < Note The conditional expression s > /8 is valid. It is because the ero of () does not exist in the domain 0< s </ & s /8. Theorem When Riemann eta function is and the non-trivial eros sre n = x n iy n n=,,3,, If the following expression holds, non-trivial eros whose real parts is not / do not exist. r= y r /4 y r = - 4 /4 4 = (4.4 0 ) Although non-trivial eros / iy r r =,,3, exist in fact, assume non-trivial eros / s i s, /- s i s (0 < s </ & s > /8) exist in addition. Then, the following expression holds from Theorem 8.4., Lemma 8.4. and Lemma () = h () () = ( ) 0 r= yr s= ( 0= ) h () 0 ( 0) = r= y r /4 y r s s s s s s s= / s s s - s s s /- s s And, according to Lemma 8.4.3, when 0< s </ & s > /8, s s s= / s s Then, form (4. 0 '), i.e. r= r= y r /4 y r y r /4 y r > - 4 /4 - s s /- s s /4 As the contrapositive to the above, this theorem holds. 4 = - 4 /4 (4.') 4 = (4. 0 ') < (4.3 0 ) =

16 Both sides of (4.4 0 ) were calculated with the formula manipulation software Mathematica using known nontrivial eros. Both sides coincided with five decimal places. cf. If the square root of (4.4 0 ) is taken, it is as follows. This is also equivalent to the Riemann hypothesis. r= y r = /4 y r /8-4 = (4.5) Each factor on the left side is an imaginary part when non-trivial ero r = / i y r is converted to polar coordinates. That is, sinr = r= /8-4 = (4.5) Added Sec Added Sec.4 (Renewed) Alien's Mathematics Kano. Kono - 6 -