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### Transcript of Euler-Maclaurin Summation Formula - Fractional Calculusfractional- Euler-Maclaurin Summation Formula...

• 4 Euler-Maclaurin Summation Formula

4.1 Bernoulli Number & Bernoulli Polynomial

4.1.1 Definition of Bernoulli NumberBernoulli numbers Bk ( )k =1,2,3, are defined as coefficients of the following equation.

ex-1x

=k=0

k!Bk xk

4.1.2 Expreesion of Bernoulli Numbers

ex-1x

= 0!B0 + 1!

B1 x + 2!B2 x2 + 3!

B3 x3 + 4!B4 x4 + (1,1)

xex-1

= 1!1

+ 2!x

+ 3!x2

+ 4!x3

+ 5!x4

+ (1.2)Making the Cauchy product of (1.1) and (1.1) ,

1 = B0 + 1!1!B1 + 0!2!

B0 x + 2!1!B2 + 1!2!

B1 + 0!3!B0 x2 +

In order to holds this for arbitrary x ,

B0 =1 , 1!1!B1 + 0!2!

B0 =0 , 2!1!B2 + 1!2!

B1 + 0!3!B0 =0 ,

The coefficient of x n is as follows.

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

Bn-1 + ( )n-2 !3!Bn-2 + + 1!n !

B1 + 0!( )n +1 !B0 = 0

Multiplying both sides by ( )n +1 ! ,

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

Bn-1( )n+1 ! + ( )n-2 !3!Bn-2( )n+1 ! + + 1!n!

B1( )n+1 ! + 0!B0 = 0

Using binomial coefficiets,

n +1 n Bn + n +1 n-1 Bn-1 + n +1 n-2 Bn-2 + + n+1 1 B1 + n +1 0 B0 = 0Replacing n +1 with n ,

n n-1 Bn-1 + n n-2 Bn-2 + n n-3 Bn-3 + + n 1 B1 + n 0 B0 = 0 (1.3)Substituting n =2, 3, 4, for this one by one,

2 1 B1 + 2 0 B0 = 0 B2 = - 21

3 2 B2 + 3 1 B1 + 3 0B0 = 0 B2 = 61

Thus, we obtain Bernoulli numbers. The first few Bernoulli numbers are as follows.

B0 = 1 , B2 = 61

, B4 = - 301

, B6 = 421

, B8 = - 301

, B10 = 665

,

B1 = - 21

, B3 = B5 = B7 = = 0

- 1 -

• 4.1.3 Calculation of Bernoulli number

(1) Method of substitution one by one Generally, it is performed by the method of substituting n =2, 3, 4, for (1.3) one by one. Since it is necessary to calculate from the small number one by one, it is difficult to obtain a large Bernoulli number directly. This method is suitable for the small number.

(2) Method by double sums. It is calculated by the following double sums of binomial coefficients

Bn = k=0

n

k+11

r=0

k ( )-1 rk r rnIf the first few are written down, it is as follows.

B0 = 110 0 00

B1 = 110 0 01 + 2

1 1 0 01 - 1 1 11

B2 = 110 0 02 + 2

1 1 0 02 - 1 1 12 + 3

1 2 0 02 - 2 1 12 + 2 2 22

Using this, a large Bernoulli number can be obtained directly. However, since the number of terms increaseswith accelerating speed, this is not suitable for manual calculation.

4.1.4 Bernoulli Polynomial

(1) Definition When Bn are Bernoulli numbers, polynomial Bn( )x that satisfies the following three expressions is calledBernoulli Polynomial .

B0( )x = 1

dxd

Bn( )x = nBn-1( )x ( )n1

01Bn( )x dx = 0 ( )n1

(2) Properties The following expressions follow directly from the definition.

Bn( )x = n0xBn-1( )t dt + Bn ( )n1

Bn( )x = k=0

n

nk

Bn-k xk =

k=0

n

nk

Bk xn-k

Bn( )0 = BnBn( )1 = Bn( )0 ( )n2Bn( )x+1 - Bn( )x = n xn-1 ( )n1Bn( )1-x = ( )-1 nBn( )x ( )n1

- 2 -

• Furthermore, the followings are known although not directly followed from the definition.

For arbitrary natural number m and interval [ ]0 ,1 , B2m( )x B2m B2m+1( )x ( )2m+1 B2m

Examples

B1( )x = x- 21

, B2( )x = x2-x+ 61

, B3( )x = x3- 23

x2+ 21

x ,

B4( )x = x4-2x3+x2- 301

, B5( )x = x5- 25

x4+ 35

x3- 61

x ,

B6( )x = x6-3x5+ 25

x4- 21

x2+ 421

, B7( )x = x7- 27

x6+ 27

x5- 27

x3+ 61

x ,

B8( )x = x8-4x7+ 314

x6- 37

x4+ 32

x2- 301

,

B9( )x = x9- 29

x8+6x7- 521

x5+2x3- 103

x ,

B10( )x = x10-5x9+ 215

x8-7x6+5x4- 23

x2+ 665

,

4.1.5 Fourier Expansion of Bernoulli Polynomial Bernoulli Polynomial Bm( )x can be expanded to Fourier series on 0 x 1 . This means that Bernoulli Polynomial Bm x - x can be expanded to Fourier series on x 0 .

Formula 4.1.5 When m is a natural number, x is a floor function and Bm are Bernoulli numbers ,

Bm x- x = -2m!s=1

( )2 s m1

cos 2 sx- 2m

x 0

Proof According to Formula 5.1.2 (" 05 Generalized Bernoulli Polynomials ") , the following expression holds.

Bm( )x = -2 m!s=1

( )2s m1

cos 2 sx- 2m 0 x 1

Since Bm x - x = Bm( )x on 0 x < 1 ,

Bm x- x = -2 m!s=1

( )2s m1

cos 2 sx- 2m 0 x < 1

On 1 x < 2 ,Left: Bm x+1- x+1 = Bm x+1- x - 1 = Bm x- xRight: -2m!

s=1

( )2s m1

cos 2 s( )x +1 - 2m = -2m!

s=1

( )2s m1

cos 2 sx - 2m

Bm x- x = -2 m!s=1

( )2s m1

cos 2 sx- 2m 1 x < 2

- 3 -

http://fractional-calculus.com/generalized_bernoulli_number.pdf

• Hereafter by induction, the following expression holds for arbitrary natural number n .

Bm x- x = -2 m!s=1

( )2s m1

cos 2 sx- 2m

n x < n+1Q.E.D.

Examples If the left side and the right side are illustrated for m=1,2 , it is as follows. Blue is m=1 and Red is m=2 .Left side

Right side

- 4 -

• 4.2 Euler-Maclaurin Summation Formula

Formula 4.2.1

When f( )x is a function of class Cm on a closed interval [ ]a ,b , x is the floor function, Br areBernoulli numbers and Bn( )x are Bernoulli polynomials, the following expression holds.

k=a

b -1f( )k = a

bf( )x dx +

r=1

m

r!Br f( )r-1 ( )b - f( )r-1 ( )a + Rm (1.1)

Rm = m!( )-1 m+1a

bBm x- x f( )m ( )x dx (1.1r)

= ( )-1 m 2ab

s=1

( )2s m1

cos 2 sx- 2m

f( )m ( )x dx (1.1r')

When limm Rm = m is a even number s.t. ( )2 m

f( )m ( )x = minimum for x[ ]a,b

Proof When Br are Bernoulli numbers and Bn( )x are Bernoulli polynomials,

B0( )x =1 , 0xBn( )x dx = n +1

1 Bn+1( )x -Bn+1

Bn+1( )1 = Bn+1( )0 = Bn+1Using these,

01f( )x dx = 0

1B0( )x f( )x dx

= 1!1 B1( )x f( )x 0

1 - 01

1!B1( )x f '( )x dx

= 1!1 B1( )x f( )x 0

1 - 2!1 B2( )x f '( )x 0

1 +0

1

2!B2( )x f "( )x dx

= 1!1 B1( )x f( )x 0

1 - 2!1 B2( )x f '( )x 0

1

+ 3!1 B3( )x f "( )x 0

1 -0

1

3!B3( )x f "'( )x dx

= r=1

m

r!( )-1 r-1

Br( )x f( )r-1 ( )x 01 + ( )-1 m0

1

m!Bm( )x f( )m ( )x dx

Here,

B1( )x f( )x 01 = 1- 2

1f( )1 - 0- 2

1f( )0 = 2

1 f( )1 + f( )0

And

( )-1 r-1Br ( )1 = ( )-1r-1Br ( )0 = -Br for r2

Therefore,

- 5 -

• ( )-1 r-1 Br( )x f( )r-1 ( )x 01 = -Br f( )r-1 ( )1 - f( )r-1 ( )0

Substitutin these for the above,

01f( )x dx = 2

1 f( )1 + f( )0 -

r=2

m

r!Br f( )r-1 ( )1 - f( )r-1 ( )0

+ m!( )-1 m0

1Bm( )x f( )m ( )x dx

Replacing f( )x with f( )x +k ,

01f( )x+k dx = 2

1 f( )k+1 + f( )k -

r=2

m

r!Br f( )r-1 ( )k+1 - f( )r-1 ( )k

+ m!( )-1 m0

1Bm( )x f( )m ( )x+k dx

That is,

kk+1

f( )x dx = 21 f( )k+1 + f( )k -

r=2

m

r!Br f( )r-1 ( )k+1 - f( )r-1 ( )k

+ m!( )-1 mk

k+1Bm( )x-k f( )m ( )x dx

Accumulating this from k =a to k =b -1 ,

abf( )x dx = 2

1k=a

b -1 f( )k+1 + f( )k -

r=2

m

r!Brk=a

b -1 f( )r-1 ( )k+1 - f( )r-1 ( )k

+ m!( )-1 m

k=a

b -1k

k+1Bm( )x-k f( )m ( )x dx

Here,

k=a

b -1 f( )r-1 ( )k+1 - f( )r-1 ( )k = f( )r-1 ( )b - f( )r-1 ( )a

k=a

b -1 f( )k+1 + f( )k + f( )a + f( )b = 2

k=a

bf( )k

Bm( )x-k = Bm x- x ( ) k x k+1Therefore,

abf( )x dx =

k=a

bf( )k - 2

1 f( )a + f( )b -

r=2

m

r!Br f( )r-1 ( )b - f( )r-1 ( )a

+ m!( )-1 ma

bBm x- x f( )m ( )x dx

Transposing the terms,

k=a

bf( )k = a

bf( )x dx + 2

1 f( )a + f( )b +

r=2

m

r!Br f( )r-1 ( )b - f( )r-1 ( )a

- m!( )-1 ma

bBm x- x f( )m ( )x dx

Subtracting f( )b from both sides,

- 6 -

• k=a

b -1f( )k = a

bf( )x dx - 2

1 f( )b - f( )a +

r=2

m

r!Br f( )r-1 ( )b - f( )r-1 ( )a

- m!( )-1 ma

bBm x- x f( )m ( )x dx

Since B1 = -1/2 , including the 2nd term of the right side into , we obtain

k=a

b -1f( )k = a

bf( )x dx +

r=1

m

r!Br f( )r-1 ( )b - f( )r-1 ( )a + Rm (1.1)

Rm = m!( )-1 m+1a

bBm x- x f( )m ( )x dx (1.1r)

Last, substituting Formula 4.1.5

Bm x- x = -2m!s=1

( )2 s m1

cos 2 sx- 2m

x 0for (1.1r) , we obtain

Rm = ( )-1 m 2abs=1

( )2s m1

cos 2 sx- 2m

f( )m ( )x dx (1.1r')

When Rm is divergent , the amplitude is almost determined by ( )2 m f( )m ( )x

. Therefore, at this time,

even number m should be chosen such that ( )2 m f( )m ( )x

becomes minimum for x[ ]a,b .Q.E.D.

Removing the terms of larger odd number than 1 in the Formula 4.2.1 , we obtain the following formula.

Formula 4.2.2

When f( )x is a function of class C2m on a closed interval [ ]a ,b , x is the floor function,Br are Bernoulli numbers and Bn( )x are Bernoulli polynomials, the following expression holds.

k=a

b -1f( )k = a

bf( )x dx - 2

1 f( )b - f( )a

+ r=1

m

( )2r !B2r f( )2r-1 ( )b - f( )2r-1 ( )a + R2m (2.1)

R2m = - ( )2m !1 a

bB2m x- x f( )2m ( )x dx (2.1r)

= ( )-1 m 2ab

s=1

( )2s 2mcos( )2 sx

f( )2m ( )x dx (2.1r')