aa r X i v : . [ m a t h . N T ] S e p BERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION
YIPING YU
Abstract.
We introduce a Bernoulli operator,let B denote the operator sym-bol,for n=0,1,2,3,... let B n := B n (where B n are Bernoulli numbers, B =1 , B = 1 / , B = 1 / , B = 0...).We obtain some formulas for Riemann’s Zetafunction,Euler constant and a number-theoretic function relate to Bernoullioperator.For example,we show that B − s = ζ ( s )( s − ,γ = − log B , where γ is Euler constant.Moreover,we obtain an analogue of the RiemannHypothesis (All zeros of the function ξ ( B + s ) lie on the imaginary axis).Thishypothesis can be generalized to Dirichlet L-functions,Dedekind Zeta func-tion,etc.In particular,we obtain an analogue of Hardy’s theorem(The function ξ ( B + s ) has infinitely many zeros on the imaginary axis).In addition,we obtain a functional equation of log Π( B s ) and a functionalequation of log ζ ( B + s ) by using Bernoulli operator. Introduction
We introduce a Bernoulli operator,let B denote the operator symbol, for n=0,1,2,3,... let B n := B n (where B n are Bernoulli numbers, B = 1 , B = 1 / , B =1 / , B = 0...).Despite the fact that Bernoulli defined B = 1 /
2, some authorsset B = − / B = 1 / − B ) n = B n (n is non-negativeinteger),and the following equation e − B z = ze z − . First we introduce some definitions.If a function’s (equation’s) expression contains the operator symbol ” B ”,thenwe say the function(equation) is a Bernoulli operator function (equation) orsimply a function (equation).For example,the function f ( z ) = e − B z is a Bernoullioperator function.If a Bernoulli operator function (equation) equal to another Bernoulli oper-ator function (equation) without taking Bernoulli operator,we say the function(equation) is primitive equivalent .For example,the following equation is primi-tive equivalent:sin( π B /
2) = sin( π B / π ) . If a Bernoulli operator function (equation) equal to another Bernoulli operatorfunction (equation) by taking Bernoulli operator, we say the function (equation)is
Bernoulli operator equivalent .For example,the following equation is Bernoulli
Mathematics Subject Classification.
Primary 11M06, Secondary 11M26,11B68.
Key words and phrases.
Bernoulli operator,Ramanujan summation,divergent series,Bernoullioperator equivalent,Bernoulli operator zero(pole),Riemann Hypothesis. operator equivalent:sin[ π (1 − B ) /
2] = sin( π B /
2) = π/ − B ) n = B n ,n is non-negative integer).The zeros (poles , singularity) of a Bernoulli operator function without takingBernoulli operator are called primitive zeros (poles , singularity) .For exam-ple,the primitive zeros of function f ( z ) = sin[ π ( B + z ) /
2] are - B +4k (k is aninteger);The zeros (poles , singularity) of a Bernoulli operator function by taking Bernoullioperator are called Bernoulli operator zeros (poles , singularity) .For exam-ple,the Bernoulli operator zero of function f ( z ) = 1 + sin π B z = 1 + πz is − π .Whenthere is no confusion,we say a Bernoulli operator zero (pole,singularity) is a zero(pole,singularity).In the following paper we must to distinguish between primitive equivalent andBernoulli operator equivalent. Otherwise,we will obtain many wrong equations.Forexample,we taking the logarithms on both sides of a primitive equivalent equationwhich remains equivalent; But we taking the logarithms on both sides of a Bernoullioperator equivalent equation,then we obtain a wrong equation(e.g.,sin( π B /
2) = π/ π B / = log π/ B · B = B = 0 = 1 / · / . From the above discussion, we know the value of e − B is e − , we naturally wantto know what is the value of a general Bernoulli operator function? Using theEuler-Maclaurin formula (see [2] p.104): N P n = M f ′ ( n ) = N R M f ′ ( x ) dx + [ f ′ ( M ) + f ′ ( N )] + B f ′′ ( x ) | NM + B f ′′′′ ( x ) | NM + ... + B v (2 v )!! f (2 v ) ( x ) | NM + R v , where R v = − v )! N Z M B v ( x ) f (2 v +1) ( x ) dx. Let M = 1 , N → ∞ and v → ∞ ,if f ( N ) → f ′ ( N ) →
0, and f (2 v ) ( N ) → ∞ X n =1 f ′ ( n ) = − f (1) + 12 f ′ (1) − B f ′′ (1) − B f ′′′′ (1) + ... − B v (2 v )!! f (2 v ) (1) + ... = − f (1 − B ) . Since (1 − B ) n = B n ,we obtain(1.1) f ( B ) = − ∞ X n =1 f ′ ( n ) . A Taylor expansion of Bernoulli operator function is f ( B + a ) = ∞ P n =0 f ( n ) ( a ) ( B − a ) n n ! ,if the right of the expansion does not converge in a certain areas of the complexplane, we use (1.1) that is f ( B + a ) = − ∞ P n =1 f ′ ( n + a ) to expand the domain of thefunction f ( B + a ); Conversely, if the right of the expansion f ( B + a ) = − ∞ P n =1 f ′ ( n + a )does not converge in a certain areas of the complex plane, we use the expansion ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 3 f ( B + a ) = ∞ P n =0 f ( n ) ( a ) ( B − a ) n n ! to expand the domain of the function f ( B + a ).Thisis a principle of analytic continuation.For example,the following Bernoulli operator function is analytic at all points ofthe complex z-plane except for some poles at z = 2 πki,e − B z = ∞ X n =0 ( − n B n z n n ! = ∞ X n =1 ze − nz = ze z − . From the above ideas and (1.1),we can obtain the formula,(1.2) B − s = ζ ( s )( s − . Therefore B − = ζ (2) = π / . Similarly,we can obtain the following formulas(1.3) ( B + n ) − s = [ ζ ( s ) − − s − − s − ... − n − s ]( s − , (1.4) ( B + α ) − s = ζ ( s, α )( s − , where α > , ζ ( s, α ) is Hurwitz Zeta function.Moreover,using a formula of [3] p.249and above (1.4), we can obtain the following formula L ( s, χ ) = k − s k X r =1 χ ( r ) ζ ( s, rk ) = k − s k X r =1 χ ( r ) ( B + rk ) − s s − . We taking the derivative of an equation of Bernoulli operator equavilent carriesa new equation of Bernoulli operator equavilent(the Bernoulli operator B can bethought of as a constant).For example,( e − B z ) ′ = ( ze z − ′ , we obtain the equation − B e − B z = e z − − ze z ( e z − . Similarly, we taking the derivatives on both sides of (1.2),we obtain(1.5) − B − s log B = ζ ( s ) + ( s − ζ ′ ( s ) . And now,we have the following equation(1.6) lim s → [ ζ ( s ) + ( s − ζ ′ ( s )] = γ, where γ is Euler constant, therefore(1.7) γ = − log B . Using (1.7) ,we give a proof of formula γ = ∞ P n =2 ( − n ζ ( n ) n (see [4] p.5). Proof.
Since B n +1 = 0(n is positive integer),applying the Taylor expansion,wereadily find that B log(1 + B ) = − B log(1 − B ) , and we have B log(1 + 1 B ) = B log(1 + B ) − B log B , YIPING YU therefore(1.8) B log(1 + 1 B ) = − B log(1 − B ) − B log B . Since (1 − B ) n = B n ,we have(1.9) − B log(1 − B ) = − (1 − B ) log B . Combining (1.8)and(1.9),we obtain(1.10) B log(1 + 1 B ) = − log B . Combining (1.7) and(1.10),we obtain(1.11) γ = B log(1 + 1 B ) . Now we have(1.12) log(1 + B − ) = log(1 + B ) − log B , and applying the Taylor expansion,we readily find thatlog(1 + B ) = 1 + log(1 − B ) = 1 + log B , therefore(1.13) log(1 + B − ) = 1 + log B − log B = 1 . Combining(1.11)and(1.13),we obtain(1.14) γ = B log(1 + 1 B ) + log(1 + B − ) − ∞ X n =2 ( − n B − n n ( n − . By (1.2),we show that(1.15) B − n = ζ ( n )( n − , Combining (1.14) and(1.15),we deduce that(1.16) γ = ∞ X n =2 ( − n ζ ( n ) n . (cid:3) Similar to complex plane,we define a formal
Bernoulli operator plane orsimply Bernoulli plane,which is B := { a + b B i | a, b ∈ R } .If we define an ”ana-lytic function” on the Bernoulli plane,and define a contour integral for the analyticfunction,then we can deduce the classic Cauchy integral theorem,Cauchy integralformula and Residue theorem.We can obtain another proof of (1.2) by Euler’s integral for Π( s − s −
1) := ∞ R e − x x s − dx .Substitution of nx for x gives ∞ Z e − nx x s − dx = Π( s − n s . ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 5
We sum this over n to obtain ∞ Z x s − e x − dx = Π( s − ζ ( s ) . Therefore Π( s − ζ ( s ) = ∞ Z x s − · xe x − dx = ∞ Z x s − e − B x dx. Let B x → t ,note that B ∞ can be thought of as ∞ ,we deduce thatΠ( s − ζ ( s ) = B − s ∞ Z t s − e − t dt = B − s Π( s −
2) = B − s Π( s − s − . Therefore we obtain (1.2) again.
Remark 1.1.
I find that (1.1) is similar to Ramanujan summation,and he hasresearched divergent series by his summation(see [6] ).On the other hand,C.Vignattold me that the Bernoulli operator somewhat similar to Bernoulli umbra (see [7] ).And J.Glinas told me that (1.7) dates from 1861 in the original articles of Blissard(see [10] ). Remark 1.2.
If a point of the complex z-plane is Bernoulli operator pole( or singu-larity) of a function,then using the transformation B → − B will cause errors.Forexample, e − πi B = cos 2 π B − i sin 2 π B , if let B → − B ,then e − πi (1 − B ) = e − πi +2 πi B = e πi B = cos 2 π B + i sin 2 π B . Therefore we obtain an error equation − i sin 2 π B = i sin 2 π B . Distribution of Bernoulli operator zeros of the function ξ ( B + s ) and sin π B · ξ ( B + s ) Theorem 2.1.
The function ξ ( B + s ) and sin π B · ξ ( B + s ) satisfy the followingfunctional equations respectively (2.1) ξ ( B + s ) = ξ ( B − s ) , (2.2) sin π B · ξ ( B + s ) = sin π B · ξ ( B − s ) . The values of these two functions are positive in the real axis;and these two functionshave infinitely many Bernoulli operator zeros on the imaginary axis.
YIPING YU
Proof. By ξ ( s ) = ξ (1 − s ),and let s → B + s ,we have(2.3) ξ ( B + s ) = ξ (1 − B − s ) . Since (1 − B ) n = B n ,we have(2.4) ξ (1 − B − s ) = ξ ( B − s ) . Using(2.3)and(2.4),we obtain(2.1). Similarly,we havesin π B · ξ ( B + s ) = sin π (1 − B ) · ξ (1 − B + s ) = sin π (1 − B ) · ξ ( B − s ) . Since sin π (1 − B ) is primitive equal to sin π B ,we havesin π (1 − B ) · ξ ( B − s ) = sin π B · ξ ( B − s ) . Therefore sin π B · ξ ( B + s ) = sin π B · ξ ( B − s ) . We now prove that the function ξ ( B + s ) is positive in the real axis. We havethe following equation (see [2] p.17), ξ ( s ) = ∞ Z d [ x / ψ ′ ( x )] dx (2 x s − + 2 x − s ) dx, where ψ ′ ( x ) = d [ ∞ P n =1 e − n πx ] dx , therefore ξ ( B + s ) = ∞ Z d [ x / ψ ′ ( x )] dx (2 x s − B + 2 x − s − B ) dx. Since e − B x = xe x − , we have x − B = log x x / − . Let φ ( x ) = d [ x / ψ ′ ( x )] dx log xx / − ,then ξ ( B + s ) = ∞ Z φ ( x )( x s + x − s ) dx. Because φ ( x ) > x ∈ (1 , ∞ )) , we obtain ξ ( B + s ) > , s ∈ ( −∞ , ∞ ) . Using (1.1),weconclude that sin π B · x − B = πx +1 .Similarly,we can prove that sin π B · ξ ( B + s ) > , s ∈ ( −∞ , ∞ ) . We now prove that the function ξ ( B + s ) has infinitely many Bernoulli operatorzeros on the imaginary axis.Let G ( x ) = ∞ P n = −∞ e − πn x , H ( x ) = x d [ xG ( x ) − x − dx , then we have the following equa-tion (see [2] p. 228), H ( x ) = 1 π ∞ Z −∞ ξ ( a + it ) x a − x it dt, multiply both sides of above equation by x − a x − a H ( x ) = 1 π ∞ Z −∞ ξ ( a + it ) x it dt. ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 7
Let a → B ( B is Bernoulli operator),then x − B H ( x ) = 1 π ∞ Z −∞ ξ ( B + it ) x it dt, = 1 π ∞ Z −∞ ξ ( B + it ) ∞ X n =0 ( it log x ) n n ! dt. Denote C n = πn ! ∞ R −∞ ξ ( B + it ) t n dt ,then x − B H ( x ) = x log xx − H ( x ) = ∞ X n =0 C n ( i log x ) n . Since ξ ( B + it ) = ξ ( B − it ),we have C n = 0 (n is odd). The differential operator ix ( d/dx ) doing x log xx − H ( x ) any number of times carries x log xx − H ( x ) to a functionwhich approaches zero as x → i .Therefore, according to [2] p.228-229 ,we canprove that function ξ ( B + s ) has infinitely many Bernoulli operator zeros on theimaginary axis.Similarly,we can prove that function sin π B · ξ ( B + s ) has infinitely many Bernoullioperator zeros on the imaginary axis. (cid:3) Conjecture 2.2.
All zeros of the function ξ ( B + s ) ( sin π B · ξ ( B + s ) ) lie on theimaginary axis. The Riemann Hypothesis generalized to Dirichlet L-functions,Dedekind Zetafunction (see [3] p.176)and Zeta Functions of varieties over Finite Fields,etc.Similarly,theconjecture 2.2 can be generalized to other Zeta functions(L-functions).3.
An application of Bernoulli operator to number-theoretic functionTheorem 3.1. we define a number-theoretic function: ψ ( x ) = 12 X p n
1) + log ξ ( B ) + P ρ log(1 − sρ − B ) s ] x s ds. According to [2] p.26-31,we show that12 πi x a + i ∞ Z a − i ∞ dds [ log( s + B − s ] x s ds = 12 πi x a + i ∞ Z a − i ∞ dds [ log( s/ ( B −
1) + 1) + log( B − s ] x s ds ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 9 = 12 πi x a + i ∞ Z a − i ∞ dds [ log(1 − s/ B ) + log( B − s ] x s ds = 12 πi x a + i ∞ Z a − i ∞ dds [ log( s/ B −
1) + log(1 − B ) s ] x s ds = lim ε → ( − ε Z t B − log t dt + x Z ε t B − log t dt ) − log(1 − B )= lim ε → ( − ε Z t − B log t dt + x Z ε t − B log t dt ) − log(1 − B )(3.4) = lim ε → ( − ε Z t − dt + x Z ε t − dt ) − log(1 − B ) . If x > ε → ( − ε Z t − dt + x Z ε t − dt ) = log( x − . Therefore(3.5) 12 πi x a + i ∞ Z a − i ∞ dds [ log( s + B − s ] x s ds = log( x − − log(1 − B ) . According to [2] p.26-31,we show that12 πi x a + i ∞ Z a − i ∞ dds [ P ρ log(1 − sρ − B ) s ] x s ds = X Im ρ> [ Li ( x ρ − B ) + Li ( x − ρ − B )]= X Im ρ> [ Li ( x ρ − B ) + Li ( x B − ρ )](3.6) = lim ε → X Im ρ> ( − ε Z t ρ − + t − ρ t − dt + x Z ε t ρ − + t − ρ t − dt ) . Using a formula form [2] p.8,we show thatlog Π( s + B ∞ X n =1 (cid:20) − log(1 + s + B n ) + s + B n ) (cid:21) . Since log(1 + s + B n ) = log(1 + B n + s n ) = log(1 + B n ) + log(1 + s B + 2 n ) , we obtain 12 πi x a + i ∞ Z a − i ∞ dds [ log Π( s + B ) s ] x s ds = 12 πi x ∞ X n =1 a + i ∞ Z a − i ∞ dds (cid:20) − log(1 + B n ) − log(1 + s B + 2 n ) + s + B n ) (cid:21) x s ds = 12 πi x ∞ X n =1 a + i ∞ Z a − i ∞ dds (cid:20) − log(1 + B n ) − log(1 + s B + 2 n ) + B n ) (cid:21) x s ds = ∞ X n =1 [log(1 + B n ) − B n )] − πi x ∞ X n =1 a + i ∞ Z a − i ∞ dds (cid:20) log(1 + s B + 2 n ) (cid:21) x s ds = − log Π( B − πi x ∞ X n =1 a + i ∞ Z a − i ∞ dds (cid:20) log(1 + s B + 2 n ) (cid:21) x s ds (3.7) = − log Π( B − ∞ X n =1 H ( − n − B ) . Where the function H ( x ) defined in [2] p.28.According to [2] p.32,we show that ∞ X n =1 H ( − n − B ) = ∞ Z x dtt ( t − t − . Therefore(3.8) 12 πi x a + i ∞ Z a − i ∞ dds [ log Π( s + B ) s ] x s ds = − log Π( B − ∞ Z x dtt ( t − t − . Using(3.3),(3.5),(3.6)and(3.8),we obtain(3.9) ψ ( x ) = log( x − − log(1 − B ) + B log π + log ξ ( B ) − lim ε → P Im ρ> ( − ε R t ρ − + t − ρ t − dt + x R ε t ρ − + t − ρ t − dt ) − log Π( B ) + ∞ R x dtt ( t − t − . On the other hand,we havelog ξ ( B ) = log Π( B − B ) − B π + log[ − ζ ( B )] . By the above equation and (3.9),we complete the proof. (cid:3)
Remark 3.2. sin π B · log ζ ( B + s ) = sin π B · X p [ ∞ X n =1 n p − n ( s + B ) ] , Using (1.1) ,we conclude that sin π B · p − n B = πp n +1 ,therefore sin π B · log ζ ( B + s ) = π X p [ ∞ X n =1 n p − ns p n + 1 ] . The above series on the right is absolutely convergent for Re s > . ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 11 The values of B log B , log sin π B , log Π( B ) , ζ ′ ( B ) ζ ( B ) and sin π B · ζ ′ ( B ) ζ ( B ) Proposition 4.1.
The values of B log B , log sin π B , log Π( B ) , ζ ′ ( B ) ζ ( B ) and sin π B · ζ ′ ( B ) ζ ( B ) are − log 2 π , − log 2 , log 2 π − − γ, γ + log 2 π π and π (1 + γ + log 4 π ) respectively. Lemma 4.2.
If a function f ( z ) has the following Taylor expansion in the neigh-borhood of z = a ∞ X n =0 f ( n ) ( a ) n ! ( z − a ) n , then (4.1) f ( a + B z ) = f ( a − B z ) + f ′ ( a ) z, (4.2) B f ′ ( a + B z ) = − B f ′ ( a − B z ) + f ′ ( a ) , (4.3) B f ′′ ( a + B z ) = B f ′′ ( a − B z ) . Proof.
It is clear from the properties of Bernoulli numbers.
Remark 4.3.
We can obtain (1.1) by (4.1).Using (4.1),we have f (1 − B ) = f (1 + B ) − f ′ (1) , Since f ( B ) = f (1 − B ) ,we have f ( B ) = f (1 + B ) − f ′ (1) . Therefore f ( B ) = f (1 + 1 − B ) − f ′ (1) = f (2 − B ) − f ′ (1) . using (4.1) again, f ( B ) = f (2 + B ) − f ′ (1) − f ′ (2) . Therefore f ( B ) = f ( n + B ) − n X k =1 f ′ ( k ) . Let n → ∞ ,then we obtain (1.1) again. Now we prove that Proposition 4.1.Set s = 0 in (1.5),we obtain(4.4) − B log B = ζ (0) − ζ ′ (0) = − − ζ ′ (0) . And we have the following equation (see [2] p.135),(4.5) ζ ′ (0) = −
12 log 2 π. Therefore(4.6) B log B = 1 − log 2 π . Here we will use another way to prove (4.6). We take the logarithms of bothsides of the Riemann’ Zeta functional equation,let s → B ,we obtainlog[ − ζ ( B )]+log(1 − B ) = log Π(1 − B )+( B −
1) log 2 π +log 2+log sin π B − ζ (1 − B )]Therefore(4.7) − γ = log Π( B ) − log 2 π π B . Using the Stirling formula (see [2] p.109),(4.8) log Π( B ) = ( B + 12 ) log B − B + log 2 π ∞ X n =1 ( − n +1 B n (2 π ) n n · · (2 n )! . Note that we have used the following formula B − k = ζ (2 k )(2 k −
1) = ( − k +1 (2 π ) n B k · (2 k )! , where k is positive integer. Sincecos z sin z = 1 z − ∞ X n =1 ( − n +1 ((2 n )!) − n B n z n − , taking the integral on both sides to get(4.9) log sin z = log z − ∞ X n =1 ( − n +1 n B n n · (2 n )! z n . (see [9] p.39). Let z → B π ,we have(4.10) log sin B π = log B + log π − ∞ X n =1 ( − n +1 B n (2 π ) n n · (2 n )! . Using(4.8)and(4.10),we obtain(4.11) log Π( B ) = ( B + 12 ) log B − B + log 2 π B + log π − log sin B π . Using(4.7)and(4.11),we obtain − γ = ( B + 12 ) log B − B + log 2 π B + log π − log sin B π − log 2 π π B . Therefore B log B = 12 − log 2 − log π B π − B π
2= 12 − log 2 − log π B π cos B π ) − B π . (4.12) = 12 − log 2 − log π B π − log sin B π . And we have(4.13) log cos B π − B ) π B π . Combining(4.12) and(4.13),we obtain(4.6).
ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 13
We began to calculate the value of log sin π B ,we have B log sin π B − B ) log sin[ π − B )] = (1 − B ) log cos π B
2= log cos π B − B log cos π B π B − B log cos π B . Using the Taylor expansion,we have(4.15) B log cos π B . Using(4.9),we have B log sin π B B log π B B log π B − B log 2 = B log π + B log B − log 22(4.16) = log π − log 2 π − log 22 = 12 − log 2 . Using(4.14),(4.15)and(4.16),we obtain(4.17) log sin π B − log 2 . Using(4.17)and(4.7),we obtain − γ = log Π( B ) − log 2 π − log 2 . Therefore(4.18) log Π( B ) = log 2 π − − γ. We calculate the values of ζ ′ ( B ) ζ ( B ) and sin π B · ζ ′ ( B ) ζ ( B ) by the following equation,(4.19) ζ ′ ( s ) ζ ( s ) = − Π ′ ( − s )Π( − s ) + log 2 π + π πs/ πs/ − ζ ′ (1 − s ) ζ (1 − s ) . Let s → B ,both sides multiplied by B ,we obtain B ζ ′ ( B ) ζ ( B ) = − B Π ′ ( − B )Π( − B ) + B log 2 π + π B cos( π B / π B / − B ζ ′ (1 − B ) ζ (1 − B )= − B Π ′ ( − B )Π( − B ) + B log 2 π + π B cos( π B / π B / − (1 − B ) ζ ′ ( B ) ζ ( B ) . Hence(4.20) ζ ′ ( B ) ζ ( B ) = − B Π ′ ( − B )Π( − B ) + B log 2 π + π B cos( π B / π B / . Firstly,we calculate the value of π B cos( π B / π B / .We have(4.21) B sin( π B / π B /
2) = (1 − B ) sin( π (1 − B ) / π (1 − B ) /
2) = (1 − B + B ) cos( π B / π B / , Using the Taylor expansion,we can prove the following equations,(4.22) cos( π B / π B /
2) = 2 π B − · · π B π · π − π
12 = π , (4.23) B sin( π B / π B /
2) = 0 , (4.24) B cos( π B / π B /
2) = 2 B π B = 2 B π = 1 π . Using(4.21),(4.22),(4.23)and(4.24),yields(4.25) π B cos( π B / π B /
2) = π
16 + 14 . Let us now calculate the value of − B Π ′ ( − B )Π( − B ) ,taking the logarithmic derivative onboth sides of the equation πs Π( s )Π( − s ) = sin πs . Let s → B ,and both sides multipliedby B ,yields(4.26) B Π ′ ( B )Π( B ) − B Π ′ ( − B )Π( − B ) = BB − π B cos π B sin π B = 1 − π B cos π B sin π B . And now calculate the value of π B cos π B sin π B . Since cos π B sin π B = cos π (1 − B )sin π (1 − B ) = − cos π B sin π B ,wehave cos π B sin π B = 0and π B cos π B sin π B = π (1 − B ) cos π (1 − B )sin π (1 − B ) = − π (1 − B + B ) cos π B sin π B . Therefore π B cos π B sin π B = π B cos π B sin π B . Since π B π B sin π B = π B π B = B = ,we obtain(4.27) π B cos π B sin π B = 12 . Using(4.26)and(4.27),we obtain(4.28) B Π ′ ( B )Π( B ) − B Π ′ ( − B )Π( − B ) = 12 . Using (4.2) of Lemma 4.2 , we obtain(4.29) B Π ′ ( B )Π( B ) = − B Π ′ ( − B )Π( − B ) + Π ′ (0)Π(0) = − B Π ′ ( − B )Π( − B ) − γ, where γ is Euler constant.Using(4.28)and(4.29),we obtain(4.30) B Π ′ ( B )Π( B ) = 14 − γ , − B Π ′ ( − B )Π( − B ) = 14 + γ . Using(4.20),(4.25)and(4.30),we obtain ζ ′ ( B ) ζ ( B ) = 14 + γ B log 2 π + π
16 + 14 = 1 + γ + log 2 π π . ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 15
And now calculate the value of sin π B · ζ ′ ( B ) ζ ( B ) .Both sides of (4.19) multiplied bysin π B , and let s → B ,we obtainsin π B · ζ ′ ( B ) ζ ( B ) = − sin π B · Π ′ ( − B )Π( − B ) +log 2 π · sin π B + π π B / π B /
2) sin π B − sin π B · ζ ′ (1 − B ) ζ (1 − B )= − sin π B · Π ′ ( − B )Π( − B ) + log 2 π · π B + π cos ( π B / − sin π (1 − B ) · ζ ′ ( B ) ζ ( B )= − sin π B · Π ′ ( − B )Π( − B ) + π log 2 π π π B − sin π B · ζ ′ ( B ) ζ ( B ) . Therefore(4.31) sin π B · ζ ′ ( B ) ζ ( B ) = − sin π B ′ ( − B )Π( − B ) + π log 2 π π . Firstly, we calculate the value of sin π B · Π ′ ( − B )Π( − B ) . Using (4.1) of Lemma 4.2, wehave(4.32) sin π B · Π ′ ( B )Π( B ) = sin( − π B ) · Π ′ ( − B )Π( − B ) + π Π ′ (0)Π(0) = − sin π B · Π ′ ( − B )Π( − B ) − πγ. Taking the logarithmic derivative on both sides of the equation πs Π( s )Π( − s ) = sin πs . Let s → B ,and both sides multiplied by sin π B ,we obtainsin π B · ( Π ′ ( B )Π( B ) − Π ′ ( − B )Π( − B ) ) = sin π BB − π sin π B cos π B sin π B = sin π BB − π cos π B = sin π BB . We can prove the equation sin π BB = π log 2 by(1.1), therefore(4.33) sin π B · Π ′ ( B )Π( B ) − sin π B · Π ′ ( − B )Π( − B ) = π log 2 . Using(4.32)and(4.33),we obtain(4.34) sin π B · Π ′ ( − B )Π( − B ) = π log 2 + πγ − . Using(4.31)and(4.34),we obtainsin π B · ζ ′ ( B ) ζ ( B ) = π log 2 + πγ π log 2 π π π γ + log 4 π ) . (cid:3) Proposition 4.4.
Let λ = P ∞ n =1 [ P ∞ k =1 1(2 n + k ) − n ] , λ = P ∞ n =1 [ P ∞ k =1 ( − k − n + k − n ] , we have λ = 1 + log 22 − π ,λ = 1 − . Proof.
We have the following equation (see [5] p.158),(4.35) ζ ′ ( s ) ζ ( s ) = − s − γ + log π X ρ s − ρ + ∞ X n =1 ( 1 s + 2 n − n ) . where ρ are zeros of ξ ( s ). Let s → B ,we have(4.36) ζ ′ ( B ) ζ ( B ) = − B − γ + log π X ρ B − ρ + ∞ X n =1 ( 1 B + 2 n − n ) . Since X ρ B − ρ = X ρ − B − ρ = X ρ − ρ − B = X ρ ρ − B , we obtain(4.37) X ρ ρ − B = 0 . Using(1.3),we have(4.38) ( B + 2 n ) − = (2 − ζ (2) − − − − − − − ... − (2 n ) − ) = ∞ X k =1 n + k ) . Using(4.36),(4.37)and(4.38),we obtain(4.39) ζ ′ ( B ) ζ ( B ) = − B − γ + log π λ = 1 B + γ + log π λ = π γ + log π λ . By Proposition 4.1 and(4.39),we obtain λ = 1 + log 22 − π . Both sides of (4.35) multiplied by sin π B , and let s → B ,we obtainsin π B · ζ ′ ( B ) ζ ( B ) = sin π B − B + γ + log π π B + X ρ sin π BB − ρ + ∞ X n =1 sin π B · ( 1 B + 2 n − n )= sin π (1 − B ) B + γ + log π π B + X ρ sin π BB − ρ + ∞ X n =1 ( sin π BB + 2 n − π B n )= sin π BB + γ + log π π + X ρ sin π BB − ρ + ∞ X n =1 ( sin π BB + 2 n − π n )(4.40) = π log 2 + γ + log π π + X ρ sin π BB − ρ + ∞ X n =1 ( sin π BB + 2 n − π n ) . Similarly,we can prove that(4.41) X ρ sin π BB − ρ = 0 . Using(1.1),we have(4.42) sin π BB + 2 n = − ∞ X k =1 π (2 n + k ) cos kπ − sin kπ (2 n + k ) = π ∞ X k =1 ( − k − n + k . Using(4.40),(4.41)and(4.42),we obtainsin π B · ζ ′ ( B ) ζ ( B ) = π log 2 + γ + log π π + π ∞ X n =1 [ ∞ X k =1 ( − k − n + k − n ] ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 17 (4.43) = π log 2 + γ + log π π + πλ . By Proposition 4.1 and(4.43),we obtain λ = 1 − . (cid:3) The functional equation of log Π( B s ) Proposition 5.1.
The function log Π( B s ) satisfy the following functional equation (5.1) log Π( B s ) = s log Π( B s ) + s + 12 log s + (1 − s ) log 2 π . Proof.
Using the Stirling formula (see [2] p.109), let s → B /s , we obtain(5.2) log Π( B s ) = ( B s + 12 ) log B s − B s + log 2 π ∞ X n =1 ( − n +1 B n (2 π ) n n · · (2 n )! s n − . Note that we have used the following formula B − k = ζ (2 k )(2 k −
1) = ( − k +1 (2 π ) n B k · (2 k )! , where k is positive integer.Using(4.9),we have(5.3) ∞ X n =1 ( − n +1 B n (2 π ) n n · · (2 n )! s n − = − log sin π B s s + log π s + log B s s . Using(5.2)and(5.3),we have(5.4) log Π( B s ) = ( B s + 12 ) log B s − B s + log 2 π − log sin π B s s + log π s + log B s s . By πs Π( s )Π( − s ) = sin πs, we have(5.5) log Π( B s ) + log Π( − B s ) = log π B s − log sin π B s. Using (4.1) of Lemma 4.2 , we havelog Π( B s ) = log Π( − B s ) + Π ′ (0)Π(0) s. Therefore(5.6) log Π( B s ) = log Π( − B s ) − γs. Using(5.5)and(5.6),we have(5.7) log Π( B s ) = 12 ( − γs − γ + log πs − log sin π B s ) . Using(5.4)and(5.7),we obtain(5.1). (cid:3)
Remark 5.2.
I find that Proposition 5.1 is equivalent to Ramanujan and A.P.Guinand’s result,but the proofs are not the same(see [1] , [8] ). The functional equation of log ζ ( B + s ) Theorem 6.1.
The function log ζ ( B + s ) satisfy the following functional equation (6.1)log ζ ( B − s ) − log ζ ( B + s ) = s Π ′ ( s )Π( s ) − − s − s log 2 π + π s cos πs sin πs + π πs sin πs − cos( πs/ πs/
2) ) . Moreover the function log ζ ( B + s ) has poles at the non-positive integers(i.e. at s = 0 , − , − , − , ... ). Lemma 6.2. log( B + s ) = Π ′ ( s )Π( s ) . Proof.
By Π( s ) = s Π( s −
1) ,we havelog Π( B + s ) = log( B + s ) + log Π( B + s − B + s ) + log Π( − B + s ) . Using(4.1),we have(6.3) log Π( B + s ) = log Π( − B + s ) + Π ′ ( s )Π( s ) . Using(6.2)and(6.3),we complete the proof. (cid:3)
Remark 6.3.
In particular,set s=0 in Lemma 6.2,we obtain (1.7) again.The func-tion log( B + s ) has poles at the negative integers (i.e. at s = − , − , − , ... ). Lemma 6.4. ( B + s ) log( B + s ) − B log B − s = log Π( s ) . Proof.
By Lemma 6.2,we have Z log( B + s ) ds = Z Π ′ ( s )Π( s ) ds. Therefore ( B + s ) log( B + s ) − s = log Π( s ) + c. Let s = 0,we obtain c = B log B . (cid:3) Remark 6.5.
We can obtain a very short proof of Stirling formula by Lemma 6.4.Proof.
Using Lemma 6.4,we havelog Π( s ) = ( B + s ) log(1 + B s ) + ( B + s ) log s − s − B log B = ( B + s ) ∞ X n =1 ( − n − B n ns n + ( 12 + s ) log s − s − − log 2 π . (cid:3) Remark 6.6.
We can obtain another proof of Proposition 5.1 by Lemma 6.4.
ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 19
Proof.
Set s → B ′ /s in Lemma 6.4,where B ′ is another Bernoulli operator,we havelog Π( B ′ /s ) = ( B + B ′ /s ) log( B + B ′ /s ) − B log B − B ′ /s = ( B + B ′ /s ) log( B s + B ′ ) − ( B + B ′ /s ) log s − B log B − B ′ /s. Therefore s log Π( B ′ /s ) = ( B s + B ′ ) log( B s + B ′ ) − ( B s + B ′ ) log s − B log B · s − B ′ . On the other hand,set s → B ′ s in Lemma 6.4,where B ′ is another Bernoulli oper-ator,we have log Π( B ′ s ) = ( B + B ′ s ) log( B + B ′ s ) − B log B − B ′ s. Since ( B s + B ′ ) log( B s + B ′ ) = ( B + B ′ s ) log( B + B ′ s ) , we have s log Π( B ′ /s ) = ( B + B ′ s ) log( B + B ′ s ) − s s − − log 2 π s −
12= log Π( B ′ s ) + 1 − log 2 π s − s s − − log 2 π s −
12= log Π( B ′ s ) + log 2 π s − log 2 π − s s. (cid:3) Lemma 6.7. log Π( s − B ) = s Π ′ ( s )Π( s ) − − log 2 π − s. Proof.
By Π( s ) = s Π( s −
1) ,we have B log Π( B + s ) = B log( B + s ) + B log Π( B + s − B log( B + s ) + (1 − B ) log Π( − B + s ) . Using (4.2),we have(6.5) B log Π( B + s ) = − B log Π( − B + s ) + log Π( s ) . Combining(6.4)and(6.5),we obtainlog Π( s ) = B log( B + s ) + log Π( s − B ) . By Lemma 6.2 and Lemma 6.4,we havelog Π( s ) = log Π( s ) + B log B + s − s log( B + s ) + log Π( s − B )= log Π( s ) + 1 − log 2 π s − s Π ′ ( s )Π( s ) + log Π( s − B ) . (cid:3) Lemma 6.8. log Π( B + s ′ ( s/ s/ − s s ′ ( s )Π( s ) − s ′ ( s )2Π( s ) −
14 + log π . Proof.
By Π( s ) = 2 s Π( s s −
12 ) π − / , we havelog Π( B + s ) = ( B + s ) log 2 + log Π( B + s s − B −
12 log π = ( 12 + s ) log 2 + log Π( B + s s + B −
12 Π ′ ( s/ s/ −
12 log π. Therefore2 log Π( B + s B + s ) − ( 12 + s ) log 2 + 12 Π ′ ( s/ s/
2) + 12 log π = log Π( − B + s ) + Π ′ ( s )Π( s ) − ( 12 + s ) log 2 + 12 Π ′ ( s/ s/
2) + 12 log π. By Lemma 6.7,we obtain2 log Π( B + s s Π ′ ( s )Π( s ) − − log 2 π − s + Π ′ ( s )Π( s ) − ( 12 + s ) log 2+ 12 Π ′ ( s/ s/
2) + 12 log π = ( s + 1) Π ′ ( s )Π( s ) −
12 + log π − s − s log 2 + 12 Π ′ ( s/ s/ . (cid:3) Lemma 6.9. (6.6) log sin π ( B − s )2 = − log 2 + πs πs sin πs + π πs sin πs − cos( πs/ πs/
2) ) . Proof.
Since πs Π( s )Π( − s ) = sin πs, we have log sin π ( B − s )2 = log π ( B − s )2 − log Π( B − s − log Π( − B + s π ′ ( − s )Π( − s ) − log Π( B − s − log Π( B + s ′ ( s/ s/ . By Lemma 6.8,we obtainlog sin π ( B − s )2 = log π ′ ( − s )Π( − s ) − Π ′ ( − s/ − s/ − s s ′ ( − s )Π( − s ) − s −
12 Π ′ ( − s )Π( − s )+ 14 −
12 log π −
14 Π ′ ( s/ s/
2) + s − s ′ ( s )Π( s ) + s −
12 Π ′ ( s )Π( s ) + 14 − log π ′ ( s/ s/ π ′ ( − s )Π( − s ) −
14 [ Π ′ ( s/ s/
2) + Π ′ ( − s/ − s/
2) ] + s ′ ( − s )Π( − s ) − Π ′ ( s )Π( s ) ](6.7) −
12 [ Π ′ ( s )Π( s ) + Π ′ ( − s )Π( − s ) ] + 12 − log π + 12 Π ′ ( s/ s/ . Moreover(6.8) Π ′ ( s/ s/ − Π ′ ( − s/ − s/
2) = 1 s − π πs/ πs/ , ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 21 (6.9) Π ′ ( s )Π( s ) − Π ′ ( − s )Π( − s ) = 1 s − π cos πs sin πs . Using(6.7),(6.8)and(6.9),we obtain(6.6). (cid:3)
We now prove the Theorem 6.1.
Proof.
Taking the logarithms on both sides of functional equation of Riemann Zeta,let s → B − s ,we havelog ζ ( B − s ) = log Π( s − B )+log 2 π · ( − s + B − π ( B − s )2 +log ζ (1 − B + s ) . By lemma 6.7 and lemma 6.9,we havelog ζ ( B − s ) = s Π ′ ( s )Π( s ) − − log 2 π − s − ( 12 + s ) log 2 π + log 2 − log 2 + πs πs sin πs + π πs sin πs − cos( πs/ πs/
2) ) + log ζ ( B + s )= s Π ′ ( s )Π( s ) − − s − s log 2 π + πs πs sin πs + π πs sin πs − cos( πs/ πs/
2) ) + log ζ ( B + s ) . Because log ζ ( B + s ) is analytic at Re s > ζ ( B + s ) has poles at the negative integers (i.e. at s = − , − , − , ... ).Becauselog ζ ( B ) = log[ ζ ( B )( B − − log( B − , by (1.1),we havelog ζ ( B ) = − lim s → ζ ′ ( s )( s −
1) + ζ ( s ) ζ ( s )( s − − ∞ X n =2 [ ζ ′ ( n ) ζ ( n ) + 1 n − s → s − ∞ X n =2 n − − γ − ∞ X n =2 ζ ′ ( n ) ζ ( n ) + lim s → s − , therefore the function log ζ ( B + s ) has a pole at s = 0.By the way,we givelog[ − ζ ( B )] = log[ ζ ( B )( B − − log(1 − B )= − lim s → ζ ′ ( s )( s − ζ ( s ) ζ ( s )( s − − ∞ P n =2 [ ζ ′ ( n ) ζ ( n ) + n − ] − log B = − γ − ∞ P n =2 ζ ′ ( n ) ζ ( n ) + log B − log B = − γ − ∞ P n =2 ζ ′ ( n ) ζ ( n ) , thereforelog[ − ζ ( B )] = − γ − ∞ P n =2 ζ ′ ( n ) ζ ( n ) = − γ + log ζ (1 + B )= − γ + ∞ P n =2 Λ( n )(log n ) − n − − B = − γ + ∞ P n =2 Λ( n )(log n ) − n − nn − = − γ + ∞ P n =2 Λ( n ) n ( n − ≈ log[ − ζ (0)] + ζ ′ (0) ζ (0) B = − log 2 + log 2 π ≈ . , where Λ( n ) is von Mangoldt function.By calculating, we can find the value oflog[ − ζ ( B )] is close to the value of log[ − ζ (1 / (cid:3) Remark 6.10.
Set s = 1 in (3 . ,we have log ζ ( B + 1) = − log Π( B +12 ) + B +12 log π − log(1 + B −
1) + log ξ ( B ) + P ρ log(1 − ρ − B ) . On the other hand, log ξ ( B ) = log Π( B − B ) − B π + log[ − ζ ( B )] . Therefore log ζ ( B + 1) = − log Π( B +12 ) + B +12 log π − log B + log Π( B )+ log(1 − B ) − B log π + log[ − ζ ( B )] + P ρ log(1 − ρ − B )= − log Π( B +12 ) + log π + log Π( B ) + log[ − ζ ( B )] + P ρ log B +1 − ρ B − ρ = − log Π( B +12 ) + log π + log Π( B ) − γ + log ζ (1 + B ) + P ρ log B +1 − ρ B − ρ . By Lemma 6.8,we have log Π( B + 12 ) = Π ′ (1 / / −
12 log 2 + Π ′ (1)Π(1) − −
14 + log π , log Π( B ′ (0)4Π(0) + Π ′ (0)2Π(0) −
14 + log π . By Lemma 6.2,we have X ρ log B + 1 − ρ B − ρ = X ρ Π ′ (1 − ρ )Π(1 − ρ ) − Π ′ ( − ρ )Π( − ρ ) = X ρ − ρ = X ρ ρ . Therefore we deduce that − Π ′ (1 / / + log 2 − Π ′ (1)Π(1) + + − log π + log π + Π ′ (0)4Π(0) + Π ′ (0)2Π(0) − + log π − γ + P ρ ρ , we arrive at − (2 − − γ ) + log 22 − γ + − γ − γ + log π − γ + P ρ ρ . Hence, we deduce that X ρ ρ = 1 + γ − log π − log 2 . corollary 6.11. − s ζ ′ ( B ) ζ ( B ) = B s Π ′ ( B s )Π( B s ) − s + s log 2 π B πs π B s sin π B s − π s. Proof.
Set s → B ′ s ( B ′ is also a Bernoulli operator) in (6.1),we havelog ζ ( B − B ′ s ) − log ζ ( B + B ′ s ) = B ′ s Π ′ ( B ′ s )Π( B ′ s ) − − B ′ s − B ′ s log 2 π + π B ′ s π B ′ s sin π B ′ s + π π B ′ s sin π B ′ s − cos( π B ′ s/ π B ′ s/
2) )
ERNOULLI OPERATOR AND RIEMANN’S ZETA FUNCTION 23 (6.10)= B s Π ′ ( B s )Π( B s ) − − B s − B s log 2 π + π B s π B s sin π B s + π π B s sin π B s − cos( π B s/ π B s/
2) ) . Using(4.1),we have(6.11) log ζ ( B − B ′ s ) − log ζ ( B + B ′ s ) = ζ ′ ( B ) ζ ( B ) s. Applying the Taylor expansion,we find that(6.12) π π B s sin π B s − cos( π B s/ π B s/
2) ) = π π B s − B π B s ) − π B s + 2 B π B s − π s. Using(6.10),(6.11)and(6.12),we obtain Corollary 6.11. (cid:3)
In particular,set s = 1 in Corollary 6.11,combining (4.27) and (4.30),we obtainthe following equation again ζ ′ ( B ) ζ ( B ) = 1 + γ + log 2 π π . On Riemann Hypothesis
If Riemann Hypothesis is true,then log ζ (1 / s ) is analytic at all points ofRe s > s = 1 / ζ ( B + s ) is an analogue of log ζ (1 / s ).Because log ζ ( B + s ) is analytic atRe s > , this seems to believe that the Riemann hypothesis is true.In fact,if we canprove function log ζ ( B + s ) has singularities at ρ − (where ρ are zeros of ξ ( s )),thenRiemann Hypothesis is proved. Set s → ρ ′ − in (3 . ρ ′ is a zero of ξ ( s ),wehave log ζ ( B + ρ ′ − ) = − log Π( B + ρ ′ − / ) + B + ρ ′ − / log π − log( ρ ′ − + B −
1) + log ξ ( B ) + P ρ log(1 − ρ ′ − / ρ − B )= − log Π( B + ρ ′ − / ) + B + ρ ′ − / log π − log( ρ ′ − + B −
1) + log ξ ( B ) + P ρ log ρ − B − ρ ′ +1 / ρ − B . when ρ = ρ ′ ,the right of above equation will lead to a singular item log(1 / − B )(But need a rigorous proof). Hence Re ρ ′ = 1 / . References
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