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Summary

1 Dirichlet Beta Generating Functions

sech x , sec x and csc x can be expanded to Fourier series and Taylor series. And if the termwise higher order integration of these is carried out, Dirichlet Beta at a natural number are obtained.

Where, these are automorphisms which are expressed by lower betas. However, in this chapter, we stop those so far.

The work that obtain the non-automorphism formulas by removing lower betas from these is done in the next chapter

" 2 Formulas for Dirichlet Beta " .

In this chapter, we obtain the following polynomials from the beta generating functions of each family of sech, sec and csc . Where, Dirichlet Beta and Dirichlet Lambda are as follows.

( ) x = Σ

r=0

( 2 r +1 )

x

( -1 )

r

, ( ) x = Σ

r=1

( 2 r -1 )

x

1

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 = Σ

r=0

( 2 r +1 )

n

( -1 )

r

e

-(2r+1)x

- 2 ( -1 )

n

Σ

r=0

( 2 r + n ) ! E

2r

x

2r+n

- Σ

s=1 n-1

s !

( -1 )

s

x

s

( n - s )

Σ

r=0

( 2 r +1 )

2n+1

( -1 )

r

sin {( 2 r +1 ) x }

- 2 ( -1 )

n

Σ

r=0

( 2 n +1+2 r ) !

  E

2r

x

2n+1+2r

= Σ

s=0 n-1

( 2 s +1 ! )

( -1 )

s

x

2s+1

( 2 n -2 s )

Σ

r=0

( 2 r +1 )

2n

( -1 )

r

cos {( 2 r +1 ) x }

- 2 ( -1 )

n

Σ

r=0

( 2 n +2 r ) !

  E

2r

x

2n+2r

= Σ

s=0 n-1

( 2 s ) !

( -1 )

s

x

2s

( 2 n -2 s )

Σ

r=0

( 2 r +1 )

2n+1

( -1 )

r

cos {( 2 r +1 ) x }

= Σ

s=0 n

( 2 s ) !

( -1 )

s

x

2s

( 2 n +1-2 s )

Σ

r=0

( 2 r +1 )

2n

( -1 )

r

sin {( 2 r +1 ) x }

= Σ

s=0 n-1

( 2 s +1 ! )

( -1 )

s

x

2s+1

( 2 n -1-2 s )

( 2 n ) =

2( 2 n -1 ! ) ( -1 )

n

  2

2n-1

log 4 - H

2n-1

+ 2 ( -1 )

n

Σ

r=1

2 r ( 2 r +2 n -1 ! )

 2

2r

-2    B

2r

  2

2n-1+2r

- Σ

s=1 n-1

( 2 s -1 ! ) ( -1 )

s

  2

2s-1

( 2 n +1-2 s )

Furthermore, if the termwise higher order differentiation of the Fourier series of each family of sech and sec are carried out, the following expressions are obtained.

( - n ) = 2

n+1

1 Σ

r=0 n

( -1 )

rn

K

r

n =1 , 2 , 3 ,

( -2 n ) = 2

2n+1

1 Σ

r=0 2n

( -1 )

r2n

K

r

n =1 , 2 , 3 ,

= 2 n E

2n

n =1 , 2 , 3 ,

Where, n

K

r is a kind of Eulderian Number and is defined as follows.

n

K

r

= Σ

k=0 r

( -1 )

k

  n +1 k ( 2 r +1-2 k )

n

n =1 , 2 , 3 ,

(2)

2 Formulas for Dirichlet Beta

Here, removing the lower betas from the the automorphism formulas in the previous chapter, we obtain the following non-automorphism formulas. 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, gamma function and incomplete gamma function were as follows.

( ) p = 

0

t

p-1

e

-t

dt , ( p,x ) =

x

t

p-1

e

-t

dt

2.1 Formulas for Beta at natural number For 0 < x   /2

,

( ) n = Σ

r=0

Σ

s=0 n-1

s ! ( 2 r +1 )

s

x

s

( 2 r +1 )

n

( -1 )

r

e

-(2r+1)x

+ 2 x

n

Σ

r=0

  - 2 n r ( n +2 r ) ! E

2r

x

2r

Especially,

( ) n = Σ

r=0

Σ

s=0 n-1

s !2

s

( 2 r +1 )

s

( 2 r +1 )

n

( -1 )

r

e

-(r+1/2)

+ 2

n+1

1 Σ

r=0

  - 2 n r ( n +2 r ) !2

2r

E

2r

( ) n = Σ

r=0

Σ

s=0 n-1

s ! ( 2 r +1 )

s

( 2 r +1 )

n

( -1 )

r

e

-(2r+1)

+ 2 1 Σ

r=0

  - 2 n r ( n +2 r ) ! E

2r

Example

 ( ) 4 = Σ

r=0

1+ 1!2

1

2r+1 +

2!2

2

( 2r+1 )

2

+

3!2

3

( 2r+1 )

3

( 2r +1 )

4

( -1 )

r

e

-r-2 1

+ 2

5

1 Σ

r=0

  -4 2r ( 4+2r ) !2

2r

E

2r

 ( ) 4 = Σ

r=0

1+ 1! 

2r+1 + 2!

(2r+1)

2

+ 3!

(2r+1)

3

( 2r +1 )

4

( -1 )

r

e

-2r-1

+ 2 1 Σ

r=0

  -4 2r ( 4+2r ) ! E

2r

2.2 Formulas for Beta at even number For 0 < x /2

,

( 2 n ) = Σ

r=0

Σ

s=0 n-1

( 2 s ) !

  E

2s

{( 2 r +1 ) x }

2s

( 2 r +1 )

2n

( -1 )

r

cos {( 2 r +1 ) x }

- 2 ( -1 )

n

Σ

r=0

Σ

s=0

n-1

2 n 2 +2 s rE

2s

( 2 n +2 r ) !

  E

2r

x

2n+2r

( 2 n ) = - x 1 Σ

r=1

Σ

s=0 n-1

( 2 s ) !

( -1 )

s

B

2s

 2

2s

-2 {(  2 r +1 ) x }

2s

( 2 r +1 )

2n+1

( -1 )

r

sin {( 2 r +1 ) x }

+ ( -1 )

n

2 x

2n

Σ

r=1

Σ

s=0

n-1

( 2 s ) !( 2 n +1+2 r -2 s ) !

 2

2s

-2  B

2s

2 r

  E

2r

x

2r

Especially,

( 2 n ) = 2 ( -1 )

n-1

Σ

r=0

Σ

s=0

n-1

2 n 2 +2 s rE

2s

( 2 n +2 r ) !

  E

2r

  2

2n+2r

Example

 ( ) 4 = - 2 1 Σ

r=0

   4+2 0 r E

0

+   4+2 2 r E

2

 ( 4+2 r ) !

  E

2r

  2

4+2r

(3)

 ( ) 6 = 2 1 Σ

r=0

   6+2r 0 E

0

+   6+2r 2 E

2

+   6+2r 4 E

4

( 6+2 r ) !

  E

2r

  2

6+2r

2.3 Formulas for Beta at odd number For 0 < x /2

,

( 2 n -1 ) = Σ

r=0

Σ

s=0 n-1

( 2 s ) !

  E

2s

{( 2 r +1 ) x }

2s

( 2 r +1 )

2n-1

( -1 )

r

cos {( 2 r +1 ) x }

( 2 n -1 ) = - x 1 Σ

r=0

Σ

s=0 n-1

( 2 s ) !

( -1 )

s

B

2s

 2

2s

-2 {(  2 r +1 ) x }

2s

( 2 r +1 )

2n

( -1 )

r

sin {( 2 r +1 ) x }

Especially,

( 2 n -1 ) = 4

( 2 n -2 ! )

E

2n-2

  2

2n-2

2.4 Formulas for Beta at complex number

When

p

is a complex number such that

p 1 , 0 , -1 , -2 ,

,

For x = u+ v i s . t . 0 < | | x 2 , u0

,

( ) p = Σ

r=0

( ) p

p , ( 2 r +1 ) x

( 2 r +1 )

p

( -1 )

r

+ 2 x

p

Σ

r=0

  - 2 p r( p +1+2 r ) E

2r

x

2r

Especially,

( ) p = Σ

r=0

( ) p

( p , 2 r +1 )

( 2 r +1 )

p

( -1 )

r

+ 2 1 Σ

r=0

  - 2 p r( p +1+2 r ) E

2r

3 Global definition of Dirichlet Beta and Generalized Euler Number

Diriclet beta function is defined on the whole complex plane with patches as follows.

( ) p =    Σ

r=1

2 ( ( 2 -1

1-p

r -1 ) cos

r-1

)

p

p 2 ( 1- p ) Σ

r=1

( 2 r -1 )

1-p

Re ( ) p 0 ( -1 )

r-1

Re ( ) p < 0

This is inconvenient. so, we focus on the following sequence.

n

B

r

= Σ

s=0 r

( -1 )

r-sr

s

  s - 2 1

n

r =0 , 1 , 2 , ,n

Using this sequence, we can define Diriclet beta function on the whole complex plane as follows.

Definition 3.2.1

( ) p = Σ

r=1

2

r+1

1 Σ

s=1 r

( -1 )

s-1

  r s ( 2 s -1 )

-p

Furthermore, by using this sequence, Euler Number can be defined on the whole complex plane.

Definition 3.3.1

E

p

= Σ

r=1

2

r

1 Σ

s=1 r

( -1 )

s-1

  r s ( 2 s -1 )

p

(4)

4 Completed Dirichlet Beta

In 4.1, symmetric functional equations are derived from functional equations.

Formula 4.1.1

  2

1+z

2 1 + 2 z ( ) z =   2

2-z

2 1 + 1- 2 z ( 1- z )

  2

2

3+z

  4 3 + 2 z2 1 + z =   2

2

3-z

  4 3 - 2 z   2 1 - z

In 4.2, we define the completed Dirichlet beta functions

 ( ) z ,  ( ) z

as follows, respectively.

( ) z =   2

1+z

  1+ 2 z ( ) z

( ) z =   2

2

3+z

  2 

1  2 

3 +z   2 

1 + z

Then, Formula 4.1.1 is expressed as follows.

( ) z =  ( 1- z )

( ) z =  ( - z )

From the latter, we can see that

( ) z

is an even function. Therefore,

 ( ) z

has the same properties as completed Riemann zeta function

 ( ) z

. ( See " 07 Completed Riemann Zeta ". ) And, as in

 ( ) z

, the following theorem holds.

Theorem 4.2.1

If Dirichlet beta function

( ) z

has a non-trivial zero whose real part is not

1/2

, the one set consists of the following four.

1/2+ 

1

i

1

, 1/2- 

1

i

1

( 0 < 

1

< 1/2 )

05 Factorization of Completed Dirichlet Beta

In 5.1, the following Hadamard product is derived.

Formula 5.1.1

Let completed beta function be as follows.

( ) z =   2

1+z

  1+ 2 z ( ) z

When non-trivial zeros of

( ) z

are

z

k

= x

k

i y

k

k =1, 2,3,

and

is Euler-Mascheroni constant,

( ) z

is expressed by the Hadamard product as follows.

( ) z = e

2

3log -2

 - log2 - 4log

 

4 3 z

Π

k=1

1- z

k

z e

zk

z

( ) z = e

2

3log -2

 - log2 - 4log

 

4 3 z

Π

n=1

1- x

n

2

+ y

n2

2x

n

z

+

x

n2

+ y

n2

z

2

e

xn

2+yn2 2xn z

And, the following special values are obtained.

Π

n=1

1- x

n2

+ y

n2

2 x

n

- 1 e

xn

2+yn2 2xn

= e

4log

 

4 3 +

2

 + log2 - 2 3log

= 1.08088915

(5)

Π

n=1

1-x

n

iy

n

2

1  1-x

n

i y

n

2

1 =  ( -1 ) = 1.16624361

In 5.2, 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 5.2.2

Let

be Euler-Mascheroni constant, non-trivial zeros of Dirichlet beta function are

x

n

+i y

n

n =1, 2, 3,

. Among them, zeros whose real part is 1/2 are

1/2  i y

r

r =1, 2,3,

and zeros whose real parts is not 1/2 are

1/2 

s

i

s

0 <

s

< 1/2

s =1, 2,3,

. Then the following expressions hold.

Π

n=1

1- x

n2

+ y

n2

2x

n

-1

= 1

Σ

n=1

x

n2

+ y

n2

2x

n

= Σ

r=1

1/4 + y

r2

1 + Σ

s=1

1/2 

s2

 

s2

1 2 

s

+

1/2 

s2

 

s2

1 2 

s

Σ

n=1

x

n2

+ y

n2

2x

n

= 4log    4 3 +

2

 + log 2 - 2 3 log

= 0.07778398

Formula 5.2.3 ( Special values )

When non-trivial zeros of Dirichlet beta function are

x

k

i y

k

k =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

1+ x

n

i y

n

1  1+ x

n

iy

n

1 =  ( -1 ) = 1.1662436

Theorem 5.2.4

Let non-trivial zeros of Dirichlet beta function are

x

n

+ i y

n

n =1, 2, 3,

and

be Euler-Mascheroni constant.

If the following expression holds, non-trivial zeros whose real parts is not

1/2

do not exist.

Σ

n=1

1/4 + y

n2

1 = 4 log    4 3 + 2 + log 2 - 3 log 2 = 0.07778398

Incidentally, when this was calculated using 10000

y

r, both sides coincided with the decimal point 3 digits.

In 5.3, we show that

( ) z

is factored completely.

Theorem 5.3.1 ( Factorization of

( )z )

Let Dirichlet beta function be

( ) z

, the non-trivial zeros are

z

n

= x

n

i y

n

n =1, 2,3,

and completed beta function be as follows.

( ) z =   2

1+z

  1+ 2 z ( ) z

Then,

 ( ) z

is factorized as follows.

( ) z = Π

n=1

1- x

n2

+ y

n2

2 x

n

z +

x

n2

+ y

n2

z

2

In 5.4, we first derive the factorization of

( ) z

.

(6)

Theorem 5.4.1 ( Factorization of

( )z )

Let Dirichlet beta function be

( ) z

, the non-trivial zeros are

z

n

= x

n

i y

n

n =1, 2,3,

and completed beta function be as follows.

( ) z =   2

2

3+z

  2 

1  2 

3 +z   2 

1 + z

Then,

 ( ) z

is factorized as follows.

( ) z =  ( ) 0 Π

n=1

1-

x

n

-1/2

2

+ y

n2

2

x

n

-1/2

z

+

x

n

-1/2

2

+ y

n2

z

2

Where,

 ( ) 0 = Π

n=1

x

n2

+ y

n2

x

n

-1/2

2

+ y

n2

=   

2

3/2

   4

3    2

1 = 0.98071361

And, using this theorem and Theorem 4.2.1 in the previous section, we obtaine the following theorem.

Theorem 5.4.4

When Dirichlet beta function is

 ( ) z

and the non-trivial zeros are

z

n

= x

n

iy

n

n =1 , 2 , 3 ,

, If the following expression holds, non-trivial zeros whose real parts is not

1/2

do not exist.

Π

r=1

1/4 + y

r2

y

r2

=   2

3/2

  4 3   2 1 = 0.98071361

(4.40) Incidentally, when this was calculated using 10000

y

r, both sides coincided with the decimal point 4 digits.

06 Zeros on the Critical Line of Dirichlet Beta

In 6.1, substituting

z = 0+iy

for the completed Dirichlet beta

( ) z

,

h

( ) z =   2

2

3+iy

  2 12 3 + iy   2 1 + iy

We use this to calculate the zeros on the critical line. However, this function is too small in absolute value and can only find the zeros up to

y =917

.

So we normalize

h

( ) y

and define the following sign function.

sgn ( ) y = -

 

h

( ) y

h

( ) y

= -   2

i y

2

1  2 

3 + iy  2 

1 + iy

  2 

1  2 

3 + iy   2 

1 + iy

Using this sign function

sgn ( ) y

, we can find the zeros at large

y

.

(7)

In 6.2, multiplying this sign function

sgn ( ) y

by the absolute value of the Dirichlet beta

 ( 1/2+i y )

, we obtain a smooth function

B ( ) y

.

B ( ) y = sgn ( ) y2 1 + iy   = -   2

i y

2

1  2 

3 + i y

  2 

1  2 

3 + i y

  2 

1 + i y

Using this

B ( ) y

function, we can find the zeros on the critical line of

 ( ) z

by the intersection of the curve and the

y

-axis In 6.3, first, a lemma is prepared.

Lemma

When

f ( ) z

is a complex function defined on the domain

D

, the following expression holds.

e

i Im log f ( )z

=

f ( ) z

f ( ) z

Applying this lemma to the gamma function in the 6.2 ,

B ( ) y = -   2

i y

2

1  2 

3 + i y

  2 

1  2 

3 + i y

  2 

1 + i y

= -    2

i y

e

i Im log

2

1

2

3+

i y

  2 

1 + i y

= - e

i

Im log

2

 

1

2

3+

i y

- 2 ylog

4

  2 1 + i y

From this, we obtain

B ( ) y = - e

i ( )y

  2 1 + i y

where,

( ) y = Im log 2 12 3 + i y   - 2 y log 4

This is Riemann-Siegel style

B

function .

2012.04.13

2019.03.27 Updated

Kano Kono Alien's Mathematics

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