Summary
1 Zeta Generating Functions
Both of hyperbolic functions and trigonometric functions can be expanded to Fourier series and Taylor series. And if the termwise higher order integration of these is carried out, Riemann Zeta Functions are obtained.
Where, these are automorphisms which are expressed by lower zetas. However, in this chapter, we stop those so far.
The work that obtain the non-automorphism formulas by removing lower zetas from these are performed subsequent to Chapter2 .
In this chapter, we obtain the following polynomials from the zeta generating functions Where, Riemann Zeta, Dirichlet Eta and Dirichlet Lambda are as follows.
( ) x = Σ
r=1
r
x1 , ( ) x = Σ
r=1
r
x( -1 )
r-1, ( ) x = Σ
r=1
( 2 r -1 )
x1
Bernoulli numbers and Euler numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 ,
Harmonic number is
H
s= Σ
t=1 s
1/t = ( 1+s ) +
( ) n =
( n -1 ! ) ( - x )
n-1 log x - H
n-1 + 2 ( -1 )
nn ! x
n+ Σ
r=1
r
ne
-rx- ( -1 )
nΣ
r=1
2 r ( 2 r + n -1 ! ) B
2rx
2r+n-1- Σ
s=1 n-2
s !
( -1 )
sx
s ( n - s )
( ) n = - 2 ( -1 )
nn ! x
n- Σ
r=1
( -1 )
rr
ne
-rx
+ ( -1 )
nΣ
r=1
2 r ( 2 r + n -1 ! )
2
2r-1 B
2rx
2r+n-1- Σ
s=1 n-1
s !
( -1 )
sx
s ( n - s )
( ) n = 2 ( -1 )
n-1( n -1 ! ) x
n-1 log 2 x - H
n-1 + Σ
r=1
( 2r -1 )
ne
-(2r-1)x
+ 2 ( -1 )
nΣ
r=1
2 r ( 2 r + n -1 ! )
2
2r-2 B
2rx
2r+n-1- Σ
s=1 n-2
s !
( -1 )
sx
s ( n - s )
Σ
r=1
r
2n-1cosrx
- ( 2 n -2 ! ) ( -1 )
nx
2n-2 log x - H
2n-2 +( -1 )
nΣ
r=1
2 r ( 2 r +2 n -2 ! )
B
2rx
2r+2n-2= Σ
s=0 n-2
( 2 s ) !
( -1 )
sx
2s ( 2 n -1-2 s )
Σ
r=1
r
2nsin rx
- ( 2 n -1 ! ) ( -1 )
nx
2n-1 log x - H
2n-1 +( -1 )
nΣ
r=1
2 r ( 2 r +2 n -1 ! )
B
2rx
2r+2n-1= - Σ
s=1 n-1
( 2 s -1 ! )
( -1 )
sx
2s-1 ( 2 n +1-2 s )
Σ
r=1
( -1 )
r-1r
2n-1cosrx
-( -1 )
nΣ
r=1
2 r ( 2 r +2 n -2 ! )
2
2r-1 B
2rx
2r+2n-2= Σ
s=0 n-1
( 2 s ) !
( -1 )
sx
2s ( 2 n -1-2 s )
Σ
r=1
( -1 )
r-1r
2nsin rx
-( -1 )
nΣ
r=1
2 r ( 2 r +2 n -1 ! )
2
2r-1 B
2rx
2r+2n-1= - Σ
s=1 n
( 2 s -1 ! )
( -1 )
sx
2s-1 ( 2 n +1-2 s )
Σ
r=1
( 2r -1 )
2n-1cos {( 2r-1 ) x }
- 2( 2 n -2 ! ) ( -1 )
nx
2n-2 log 2
x -H
2n-2- 2 ( -1 )
nΣ
r=1
2r ( 2 r +2 n -2 ! )
2
2r-2
B
2r x
2r+2n-2= Σ
s=0 n-2
( 2s ) !
( -1 )
sx
2s ( 2n -1-2s )
Σ
r=1
( 2r -1 )
2nsin {( 2r-1 ) x }
- 2( 2n -1 ! ) ( -1 )
nx
2n-1 log 2
x -H
2n-1- 2 ( -1 )
nΣ
r=1
2r( 2r +2n -1 ! )
2
2r-2
B
2r x
2r+2n-1= - Σ
s=1 n-1
( 2 s -1 ! ) ( -1 )
sx
2s-1 ( 2n +1-2s )
Furthermore, if the termwise higher order differentiation of the Fourier series of each family of tanh, cot and tan are carried out, the following expressions are obtained.
( - n ) =
2
n+1 1-2
n+1
1 Σ
r=1 n
( -1 )
r-1nD
rn =1 , 2 , 3 ,
( 1-2 n ) =
2
2n 1-2
2n ( -1 )
n-1T
2n-1n =1 , 2 , 3 ,
= ( -1 )
2n-12 n B
2nn =1 , 2 , 3 ,
Where, n
D
r are the Eulderian Numbers andT
n-1 are the tangent numbers. These are defined as follows respectively.n
D
r= Σ
k=0 r-1
( -1 )
k n +1 k ( r - k )
n, T
n-1= 2
n 2
n-1 n
B
nBy-products
- log 0 = ( ) 1 = ( ) 1 - 2
i
2 Formulas for Riemann Zeta at natural number
In this chapter, removing the lower zetas from automorphism formulas in the previous chapter, we obtain non-automorphism formulas for Riemann Zeta at natural number.
Where, Bernoulli numbers and Euler numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 ,
And Harmonic number is
H
s= Σ
t=1 s
1/t = ( 1+s ) +
For
0 < x 2
, ( ) n =
( n -1 ! ) x
n-1 n -1
1 - 2 n
x + Σ
r=1
Σ
s=0 n-1
s ! ( ) xr
sr
ne
xr1 - Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) ! B
2rx
n-1+2r ( ) n =
( n -1 ! ) x
n-1 n -1
1 + 2 n ( n -1 ) x
- log x
+ Σ
r=1
Σ
s=0 n-2
s ! ( ) xr
sr
ne
xr1 - Σ
r=1
1- 2 r n 2 r ( n -1+2 r ) ! B
2rx
n-1+2rFor
0 < x
, ( ) n =
2
n-1- 1 2
n-1 2 x n
n! - Σ
r=1Σ
n-1s=0( ) xr s !
sr
ne
xr
( -1 )
r+ Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) !
2
2r-1 B
2rx
n-1+2r ( ) n =
2
n-1 2
n 2( n -1 !( ) n -1 ) x
n-1+ Σ
r=1
Σ
s=0 n-1
s !
x ( 2 r -1 )
s( 2 r -1 )
ne
-x(2r-1)+ 2 1 Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) !
2
2r-2 B
2rx
n-1+2r ( ) n = 2
n-1
2
n 2( n -1 ! ) x
n-1 n -1
1 - log 2
x + Σ
r=1
Σ
s=0 n-2
s !
x ( 2 r -1 )
s( 2 r -1 )
ne
-x(2r-1)
+ 2 1 Σ
r=1
1- 2 r n 2 r ( n -1+2 r ) !
2
2r-2 B
2rx
n-1+2rEspecially,
( ) n =
2 n !( n -1 ) n +1
+ Σ
r=1
Σ
s=0 n-1
s ! r
sr
ne
r1 - Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) ! B
2r ( ) n =
n !( n -1 ) 2
n-1+ Σ
r=1
Σ
s=0 n-1
s ! ( 2 r )
sr
ne
2r1 - Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) ! B
2r2
n-1+2r ( ) n =
2 n !( n -1 ) n
2+1
+ Σ
r=1
Σ
s=0 n-2
s ! r
sr
ne
r1 - Σ
r=1
1- 2 r n 2 r ( n -1+2 r ) ! B
2r ( ) n =
2
n-1- 1 2
n-1 2 n !
1 - Σ
r=1
Σ
s=0 n-1
s!
r
sr
ne
r( -1 )
r+ Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) !
2
2r-1 B
2r ( ) n =
2
n-1 2
n 2( n -1 !( ) n -1 )
1 + Σ
r=1
Σ
s=0 n-1
s ! ( 2 r -1 )
s( 2 r -1 )
ne
-(2r-1)+ 2 1 Σ
r=1
2 - r -1 n 2 r ( n -1+2 r ) !
2
2r-2 B
2r ( ) n = 2
n-1
2
n 2( n -1 ! )
1 n -1
1 + log 2 + Σ
r=1
Σ
s=0 n-2
s ! ( 2 r -1 )
s( 2 r -1 )
ne
-(2r-1)+ 2 1 Σ
r=1
1- 2 r n 2 r ( n -1+2 r ) !
2
2r-2 B
2r ( ) n = 2
n-1
2
n ( n -1 !( ) n -1 ) 2
n-2+ Σ
r=1
Σ
s=0 n-2
s ! ( 4 r -2 )
s( 2 r -1 )
ne
-(4r-2)+ 2 1 Σ
r=1
1- 2 r n 2 r ( n -1+2 r ) !
2
2r-2 B
2r2
n-1+2rExample
( ) 5 = 2 5!4
6 + Σ
r=1
1+ 1! r
1+ 2! r
2+ 3! r
3+ 4! r
4 r
51 e
r- Σ
r=1
2r -5 -1 2r( 4+2r ) ! B
2r ( ) 5 = 5!4
2
4+ Σ
r=1
1+ 1! 2 r + ( 2 2! r )
2+ ( 2 3! r )
3+ ( 2 4! r )
4 r
51 e
2r- Σ
r=1 2r -5 -1 2r( 4+2r ) ! B
2r2
4+2r ( ) 5 = 2 5!4
5
2+1 + Σ
r=1
1+ 1!
r
1+ 2!
r
2+ 3!
r
3r
5e
r1 - Σ
r=1
-4 2r 2r( 4+2r ) ! B
2r ( ) 5 = 2
4- 1
2
4 2 1 5! - Σ
r=1
1+ 1!
r
1+ 2!
r
2+ 3!
r
3+ 4!
r
4r
5e
r( -1 )
r+ Σ
r=1
2 -5 r -1 2r( 4+2r ) !
2
2r-1 B
2r ( ) 3 = 7 8 8
1 + Σ
r=1
1+ 1!
2r -1
+ 2!
( 2r -1 )
2( 2r -1 )
3e
-(2r-1)+ 2 1 Σ
r=1
2 -3 r -1 2r( 2+2r ) !
2
2r-2 B
2r ( ) 3 = 7 8 4
1 2
1 +log 2 + Σ
r=1
1+ 1!
2r -1
( 2r -1 )
3e
-(2r-1)+ 2 1 Σ
r=1
-2 2r 2r( 2+2r ) !
2
2r-2 B
2r ( ) 3 = 7 8 2
1 + Σ
r=1
1+ 1!
4r -2
( 2r -1 )
3e
-(4r-2)+ 2 1 Σ
r=1
-2 2r 2r( 2+2r ) !
2
2r-2 B
2r2
2+2r3 Formulas for Riemann Zeta at odd number
In this chapter, we obtain non-automorphism formulas for Riemann Zeta at odd number.
Where, Bernoulli numbers, Euler numbers and tangent numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 , T
1=1 , T
3=2 , T
5=16 , T
7=272 , T
9=7936 ,
And Harmonic number is
H
s= Σ
t=1 s
1/t = ( 1+s ) +
For
0 < x < 2
, ( 2n +1 ) = - x 1 Σ
r=1
Σ
s=0 n
( -1 )
s( 2 s ) !
2
2s-2 B
2s( rx )
2sr
2n+2sin rx
+( -1 )
nx
2n Σ
s=0
n
( 2s ) !( 2n +1-2s ) !
2
2s-2 B
2sH
2n+1-2s+ Σ
r=1
Σ
s=0 n
( 2s ) !( 2n +1+2r -2s ) !
2
2s-2 B
2s2r
B
2rx
2r ( 2n +1 ) = Σ
r=1
Σ
s=0 n
( 2s ) !
E
2s ( rx )
2sr
2n+1cos rx
-( -1 )
nx
2n Σ
s=0
n
( 2s ) !( 2n -2s ) ! E
2sH
2n-2s+ Σ
r=1
Σ
s=0 n
( 2s ) !( 2n +2r -2s ) ! E
2s2r
B
2rx
2rFor
0 < x
, ( 2n +1 ) = 2
2n-1
2
2n x 1 Σ
r=1
Σ
s=0 n
( 2s ) !
( -1 )
sB
2s 2
2s-2 ( rx )
2sr
2n+2( -1 )
rsin rx
- ( -1 )
nx
2nΣ
r=1
Σ
s=0 n
( 2s ) !( 2n +1+2r -2s ) !
2
2s-2 B
2s2r
2
2r-1
B
2rx
2r ( 2n +1 ) = - 2
2n-1
2
2n Σ
r=1
Σ
s=0 n
( 2s ) !
E
2s ( rx )
2sr
2n+1( -1 )
rcos rx
- ( -1 )
nx
2nΣ
r=1
Σ
s=0 n
( 2s ) !( 2n +2r -2s ) ! E
2s2r
2
2r-1
B
2rx
2r ( 2n +1 ) =
2
2n+1-1 2
2n+1Σ
r=1
Σ
s=0 n
( 2s ) !
E
2s ( 2r -1 ) x
2s( 2r -1 )
2n+1cos
( 2r -1 ) x
-
2
2n+1-1 ( -1 )
n( 2 x )
2n Σ
s=0
n
( 2 s ) !( 2 n -2 s ) ! E
2sH
2n-2s- Σ
r=1
Σ
s=0 n
( 2 s ) !( 2 n +2 r -2 s ) ! E
2s 2
2r-2
B
2r2 r x
2rEspecially,
( 2n +1 ) = ( -1 )
n
2n Σ
s=0
n
( 2s ) !( 2n +1-2s ) !
2
2s-2 B
2sH
2n+1-2s+ Σ
r=1
Σ
s=0 n
( 2s ) !( 2n +1+2r -2s ) !
2
2s-2 B
2s2r
B
2r
2r ( 2n +1 ) =
2
2n-1 ( -1 )
n-1( 2 )
2nΣ
r=1
Σ
s=0
n
( 2s ) !( 2n +1+2r -2s ) !
2
2s-2 B
2sT
2r-1 2
2r ( 2n +1 ) =
2
2n+1-1
( -1 )
n-1
2n Σ
s=0
n
( 2 s ) !( 2 n -2 s ) ! E
2sH
2n-2s- Σ
r=1
Σ
s=0 n
( 2 s ) !( 2 n +2 r -2 s ) ! E
2s 2
2r-2
B
2r2 r 1 2
2rExample
( ) 5 =
4 21600 269 + Σ
r=1
- ( 2r +5 ! )
1 +
6( 2r +3 ! )
1 -
360( 2r +1 ! ) 7
2r
B
2r
2r ( ) 5 = - 15 16
4Σ
r=1
- ( 2r +5 ! )
1 +
6( 2r +3 ! )
1 -
360( 2r +1 ! ) 7
2r
2
2r-1
B
2r
2r ( ) 5 = 31
4 288 83 + Σ
r=1
( 4+2r ) !
1 -
2( 2+2r ) !
1 +
24( 2 r ) ! 5
2 r
2
2r-2
B
2r 2
2r4 Formulas for Riemann Zeta at even number
In this chapter, we obtain non-automorphism formulas for Riemann Zeta at even number.
Where, Bernoulli numbers and Euler numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 ,
For
0 < x < 2
, ( 2n ) = - x 1 Σ
r=1
Σ
s=0 n-1
( 2s ) !
( -1 )
sB
2s 2
2s-2 ( ) rx
2sr
2n+1sin rx
- 2( 2n ) !
B
2nx
2n 2
2n-2 - x
2
2n+1-2
( 2n ) = Σ
r=1
Σ
s=0 n-1
( 2s ) !
E
2s( ) rx
2sr
2ncos rx
- 2( 2n ) !
E
2nx
2n+ 2
( 2n ) !
2
2n 2
2n-1
B
2nx
2n-1For
0 < x
, ( 2 n ) =
2
2n-2 2
2nΣ
r=1
Σ
s=0 n-1
( 2 s ) !
( -1 )
sB
2s 2
2s-2 ( ) rx
2s-1r
2n( -1 )
rsin rx
+ 2( 2 n ) !
B
2n ( 2 x )
2n ( 2 n ) = -
2
2n-2 2
2n Σ
r=1Σ
ns=0-1 E
2s( 2 s ( ) ) rx !
2s( -1 ) r
r2ncosrx - 2( E
2n2 n ) x !
2n
( 2 n ) = -
2
2n-1 2
2nΣ
r=1
Σ
s=0 n-1
( 2 s ) !
( -1 )
sB
2s 2
2s-2 {( 2 r -1 ) x }
2s-1( 2 r -1 )
2nsin
( 2 r -1 ) x
+ 2
( 2 n ) ! 2
2n B
2n x
2n-1 ( 2 n ) =
2
2n-1 2
2nΣ
r=1
Σ
s=0 n-1
( 2 s ) !
E
2s{( 2 r -1 ) x }
2s( 2 r -1 )
2ncos
( 2 r -1 ) x
+ 4
( 2 n ) ! 2
4n B
2n x
2n-1Especially,
( 2 n ) =
2( 2 n ) !
B
2n ( 2 )
2n ( 2 n ) = -
2
2n-2
1 Σ
r=1Σ
ns=0-1 E
2s( 2 s ( ) ! r )
2sr
2n
( -1 )
r- 2( 2 n ) !
E
2n
2nBy-products
Σ
s=0 n-1( 2 s ) !( 2 n -2 s ) !
2
2s-2 B
2s= -
( 2 n ) !0!
2
2n+1-2 B
2nΣ
s=0 n-1( 2 s ) !( 2 n -1-2 s ) ! E
2s=
( 2 n ) ! 2
2n 2
2n-1 B
2n5 Formulas for Riemann Zeta at complex number
In this chapter, we obtain the formulas for Riemann Zeta at a complex number by processing " 2 Formulas for Riemann Zeta at natural number "
Where, Bernoulli numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 ,
And gamma function and incomplete gamma function are
( ) p =
0t
p-1e
-tdt , ( p,x ) =
xt
p-1e
-tdt
And
p
is a complex number such thatp 1 , 0 , -1 , -2 ,
. Forx = u + v i s . t . 0 < | | x 2 , u 0
, ( ) p =
( ) p x
p-1 p -1
1 - 2 p
x + Σ
r=1
( ) p r
p ( p , xr ) - Σ
r=1
2 - r -1 p 2 r ( p +2 r ) B
2rx
p-1+2r ( ) p = ( ) p x
p-1 p -1
1 + 2p ( p -1 ) x
- log x + Σ
r=1
( p -1 ) r
p ( p -1, xr ) - Σ
r=1
1-p 2r 2r ( p +2r ) B
2rx
p-1+2rFor
x = u + v i s . t . 0 < | | x , u 0
, ( ) p =
2
p-1- 1 2
p-1 2 ( p +1 )
x
p- Σ
r=1
( ) p
( p , xr ) r
p( -1 )
r+ Σ
r=1
2 - r -1 p 2 r ( p +2 r )
2
2r-1 B
2rx
p-1+2r ( ) p =
2
p-1 2
p 2( p -1 ) ( ) p x
p-1+ Σ
r=1
( ) p ( 2 r -1 )
p
p , x ( 2 r -1 )
+ 2 1 Σ
r=1
2 - r -1 p 2 r ( p +2 r )
2
2r-2 B
2rx
p-1+2r ( ) p = 2
p-1
2
p 2 ( ) p x
p-1 p -1
1 - log 2
x + Σ
r=1
( p -1 ( ) 2 r -1 )
p
p -1 , x ( 2 r -1 )
+ 2 1 Σ
r=1
1- 2 r p 2 r ( p +2 r )
2
2r-2 B
2rx
p-1+2rEspecially,
( ) p =
2 ( p -1 ) ( p +1 ) p +1
+ Σ
r=1
( ) p r
p ( p , r ) - Σ
r=1
2 - r -1 p 2 r ( p +2 r ) B
2r ( ) p =
2( p -1 ) ( p +1 ) p
2+1
+ Σ
r=1
( p -1 ) r
p ( p -1 , r ) - Σ
r=1
1- 2 r p 2 r ( p +2 r ) B
2r ( ) p =
2
p-1- 1 2
p-1 2 ( p +1 )
1 - Σ
r=1
( ) p
( p , r ) r
p( -1 )
r+ Σ
r=1
2 - r -1 p 2 r ( p +2 r )
2
2r-1 B
2r ( ) p = 2
p-1
2
p 2( p -1 1 ) ( ) p + Σ
r=1
( ) p ( 2r -1 )
p ( p ,2r -1 ) + 2
1 Σ
r=1
2 -p r -1 2r ( p +2r )
2
2r-2 B
2r ( ) p = 2
p-1
2
p 2 1 ( ) p p 1 -1 +log 2 + Σ
r=1
( p -1 ( ) 2r -1 )
p ( p -1, 2r -1 ) + 2
1 Σ
r=1
1- 2r p 2r ( p +2r )
2
2r-2 B
2r ( ) p = 2
p-1
2
p ( p -1 2 )
p-2 ( ) p + Σ
r=1
( p -1 ( ) 2r -1 )
p ( p -1 , 4 r -2 ) + 2
1 Σ
r=1
1-p 2r 2r ( p +2r )
2
2r-2 B
2r2
p-1+2r6 Global definition of Riemann Zeta, and generalization of related coefficients
From Euler to Riemann, the zeta function was defined with patches as the domain was expanded.
( ) p = Σ
r1-2
=1( 2 1 r 1
1-pp)
1-pΣ
r=1( -1 r )
pr-1Re 0 p ( ) p Re 1 > 1 ( ) p 1 2 ( 1- p )
sin p 2 1-2 1
pΣ
r=1
r
1-p( -1 )
r-1Re ( ) p 0 p 0
This is inconvenient. so, we focus on the following sequence.
n
B
r= Σ
s=1 r
( -1 )
r-s r s s
nr =0 , 1 , 2 , ,n
Using this sequence, we can define the Zeta function on the whole complex plane as follows.
Definition 6.2.1
We difine the Riemann Zeta Function on the complex plane as follows.
( ) p =
1- 2
1-p1 Σ
r=1
2
r+11 Σ
s=1 r
( -1 )
s-1 r s s
-pp 1
Furthermore, by using this sequence, the following various coefficients can be generalized.
Generalized Stirling Number of the 2nd kind
S
2( p , r ) = r !
1 Σ
s=1
( -1 )
s-1 r s s
pr =1 , 2 , 3 ,
Generalized Tangent Number
T
p= Σ
r=10 p = 0
2
p-rΣ
s=1 r
( -1 )
s-1 r s s
pp 0
Generalized Bernoulli Number
B
p= - 2 1 p = 1 2
p-1
p Σ
r=1
2
r+11 Σ
s=1 r
( -1 )
s-1 r s s
p-1p 1
7 Completed Riemann Zeta
In 7.1, we consider even function and odd function for complex function. Generally, we obtain the same results as for real-valued functions, but the results unique to complex-valued functions are as follows.
Theorem 7.1.3
Let
f ( ) z
be a complex function in the domainD
.(1)
Iff ( ) z
is an even function , both the real part and the imaginary part are even functions.(2)
Iff ( ) z
is an odd function , both the real part and the imaginary part are odd functions.Theorem 7.1.4
Let
f ( ) z
be a complex function in the domainD
. Then,if
f ( ) z
is an even function or an odd function , f ( ) z
2 is an even function.In 7.2, we study complex conjugate properties. Especially when the function
f ( ) z
is an even function or an odd function with complex conjugate properties, two important theorems are obtained.Theorem 7.2.3
When
f ( x,y ) = u ( x,y ) + iv ( x,y )
is a function with the complex conjugate property in the domainD
,(1)
iff ( x,y )
is an even function,u ( x,y ) = u ( x, - y ) = u ( - x,y ) = u ( - x, - y ) v ( x,y ) = - v ( x, - y ) = - v ( - x,y ) = v ( - x, - y )
(2)
iff ( x,y )
is an odd function,u ( x,y ) = u ( x, - y ) = - u ( - x,y ) = - u ( - x, - y ) v ( x,y ) = - v ( x, - y ) = v ( - x,y ) = - v ( - x, - y )
Corollary 7.2.3
Let
f ( x,y ) = u ( x,y ) + iv ( x,y )
be a function with the complex conjugate property in the domainD
.Then, the followings hold for any real number
x , y D
.(1)
Whenf ( x,y )
is an even function ,v ( x, 0 = 0 ) , v ( 0 ,y ) = 0
.(2)
Whenf ( x,y )
is an odd function ,u ( 0 ,y ) = 0 , v ( x, 0 = 0 )
.Theorem 7.2.4
When
f ( ) z
is a function with the complex conjugate property in the domainD
and has a zeroz
1= x
1+ iy
1 x
1 0
,(1)
iff ( ) z
is an even function,- x
1- iy
1, x
1- iy
1, - x
1+ iy
1 are also zeros off ( ) z
.(2)
iff ( ) z
is an odd function,- x
1- i y
1, x
1- iy
1, - x
1+ iy
1 are also zeros off ( ) z
.In 7.3, symmetric functional equations are derived from functional equations.
Formula 7.3.1 (
Riemann)
-2z
2
z ( ) z =
- 21-z
2
1- z ( 1-z ) z 0, 1
-21
2
1+z
2
1 2
1 + z 2
1 + z =
-21
2
1-z
2
1 2
1 -z 2
1 - z
Where,
z 1/2
In 7.4, we define the completed Riemann zeta functions
( ) z , ( ) z
as follows, respectively. These are a little different from Landau's definition. ( ) z = - z ( 1- z )
-2z
2 z ( ) z
( ) z = - 2 1 + z 2 1 - z
-21
21+z 2 1 2 1 + z 2 1 + z
Then, the following equations hold from Formula 7.3.1 .
( ) z = ( 1- z )
( ) z = ( - z )
From the latter, we can see that
( ) z
is an even function. Therefore, Thoerem 7.2.3 (1) and Corollary 7.2.3 hold for the real partu ( x,y )
and the imaginary partv ( x,y )
of ( ) z
as they are. And from Theorem 7.2.4 , the following very important theorem is obtained.Theorem 7.4.1
If Riemann zeta function
( ) z
has a non-trivial zero whose real part is not1/2
, the one set consists of the following four.1/2+
1 i
1, 1/2-
1 i
1( 0 <
1< 1/2 )
08 Factorization of Completed Riemann Zeta
In 8.1, the following Hadamard product is shown.Formula 8.1.1 ( Hadamard product of ( ) z )
Let completed zeta function be as follows.
( ) z = - z ( 1- z )
-2z
2 z ( ) z
When non-trivial zeros of
( ) z
arez
k= x
k i y
kk =1, 2,3,
and
is Euler-Mascheroni constant, ( ) z
is expressed by the Hadamard product as follows. ( ) z = e
log2 + 2
log -1-2
z
Π
k=1
1- z
k
z e
zkz
( ) z = e
log2 + 2
log -1-2
z
Π
n=1
1- x
n
2
+ y
n 22 x
nz +
x
n 2+ y
n2
z
2e
xn2+yn2 2xn z
And, the following special values are obtained.
Π
n=1
1- x
n2+ y
n2
2 x
n-1 e
xn2+yn2 2xn
= e
1+2
-log2 - 2 log
= 1.02336448
Π
n=1
1- x
n iy
n
2
1 1- x
n i y
n
2
1 =
3
In 8.2, we consider how the formulas in the previous section are expressed when non-trivial zeros whose real part is 1/2 and non-trivial zeros whose real part is not 1/2 are mixed. Then, we obtain the following theorems.
Theorem 8.2.2
Let
be Euler-Mascheroni constant, non-trivial zeros of Riemann zeta function arex
n+ i y
nn =1, 2, 3,
. Among them, zeros whose real part is 1/2 are1/2 i y
rr =1, 2,3,
and zeros whose real parts is not 1/2 are1/2
si
s
0 <
s< 1/2
s =1, 2,3,
. Then the following expressions hold.Π
n=1
1- x
n
2
+ y
n22x
n-1
= 1
Σ
n=1
x
n2+ y
n22x
n= Σ
r=1
1/4 + y
r21 + Σ
s=1
1/2
s2
s2
1 2
s+
1/2
s2
s21 2
sΣ
n=1
x
n2+ y
n22x
n= 1 + 2
- log 2 - 2 log
= 0.0230957
Formula 8.2.3 ( Special values )
When non-trivial zeros of Riemann zeta function are
x
k i y
kk =1, 2, 3,
, the following expressions hold.Π
n=1
1- x
n i y
n
1 1- x
n iy
n
1 = 1
Π
n=1