Two Proofs of Euler-Maclaurin Formula, Its Generalizations and Applications

Two Proofs of Euler-Maclaurin Formula, Its Generalizations and Applications

Team Member Yanjie Jing

Teacher Hongliang Shi

School

No.2 High School Attached to East China Normal University

Abstract

In this paper, we first give two proofs of Euler-Maclaurin formula in section 1. The estimation of the remainder term and some relevant theorems are also stated in this section. In section 2, we apply similar method to get some further results, which generalized Euler-Maclaurin formula. In section 3, we show how Euler-Maclaurin formula is applied to deal with some elementary summations. In section 4, we deal with some infinite series to prepare for the work in next section. In section 5, we apply Euler-Maclaurin formula to some series and give the orders of some finite summations.

Key Words: Euler-Maclaurin formula, infinite series, finite summations

Introduction

In eighteenth century, Euler and Maclaurin both obtained independently a formula linking discrete summations with continuous integrals almost at the same time. Maclaurin applied it to the numerical computation of definite integrals, while Euler used it to calculate series. It has an extensive application in many subjects of mathematics, such as number theory and combinatorics.

And relevant research followed continuously since then.

Definitions

The difference of a function f(x) is defined as∆f(x)= f(x+1)− f(x). If there exists a function φ(x) such that φ(x+1)−φ(x)= f(x), then we call φ(x) the inverse difference of f(x), and denote it as

∑

f⁽x⁾∆x. We know if φ(x) is the inverse difference of f(x), so is φ(x)+C^, where C is a function of period 1 (including a constant), and vice versa^[1]. In this paper, the

‘constant’ C is always omitted, since it always vanishes in a certain summation.

Bernoulli numbers B_n are defined as the coefficients of the Taylor expansion of

−1 ex

x , i.e.

∑

^∞

=

−1=_n 0 !

n n

x x

n B e

x .

∑

^∞

=

⋅ =

− 0 !

) ( 1 _n

n n x

n x B e

e ξ

ξ

ξ ξ

gives Bernoulli polynomials B_n(x).

Main Results

1. Proofs of Euler-Maclaurin Formula In order to calculate a discrete summation

∑

⁻

= 1

) (

b

a x

x

f , the method of inverse difference is effective. If one has found a function φ(x) such that φ(x+1)−φ(x)= f(x)^{, then}

∑

⁻

= 1

) (

b

a x

x

f is simply equal to φ(b)−φ(a). This is the main idea of the following theorem. And let us see how it is

achieved.

Before we have a rigorous statement, we show the method in a rough way so that the idea will be showed more clearly.

Since our purpose is to find a function φ(x) such that φ(x+1)−φ(x)= f(x), recall the Taylor expansion of a function at a point, we have

! ...

) (

! 2

) ) (

( ) ( ) (

) (

2 +…+ ∆ +

′′ ∆ +

′ ∆ +

=

∆

+ ⁿ xⁿ

n x x

x x x x

x

x φ φ φ φ

φ

Let ∆x=1, we have ...

! ) (

! 2

) ) (

( ) ( ) 1 (

)

( +

+

…

′′ +

′ + +

=

+ n

x x x

x x

φ n

φ φ φ

φ

Equivalently, ...

! ) (

! 2

) ) (

( ) (

)

( +

+

…

′′ +

′ +

= n

x x x

x f

φ n

φ φ

We wish to express φ(x) ^{via f(x)}, therefore, derivative either side of the equality respect to x,

and we obtain ...

! ) (

! 2

) ) (

( ) (

) 1

( +

+

…

′′′ +

′′ +

′ = ⁺

n x x x

x f

φ n

φ φ

Similarly, ...

! ) (

! 2

) ) (

( ) (

) 2 ( )

4

( +…+ +

′′′ +

′′ = ⁺

n x x x

x f

φ n

φ φ

…

Then we multiply every of these identities by a coefficient, and add them together. The coefficients are chosen so properly that all derivatives of f(x) vanish except f′(x).

Therefore, we obtain ( ) ( )

0 )

( x

f a x

n

n

∑

^∞ n

=

′ =

φ ^{, and ( ) ( )}

0

) 1

( x

f a x

n

n

∑

^∞ n

=

= −

φ is followed.

After some calculations we can figure out that

! n

a_n = Bⁿ , where B_n denote Bernoulli number.

Hence,

^{∑ ∑}

^∞

( )

=

−

− −

=

−

=

0

) 1 ( )

1 ( 1

) ( )

! ( )

(

n

n n

n b

a x

a f b n f

x B f

Then we prove it rigorously. First, a lemma is needed.

Lemma 1: dt

n t t x

f n dt

t t x

dx f

d ^x

x

n x m

x

n

m

∫

∫

^{+1 ( )( )( +1}_!^{− ) = +1 ( +1)( )( +1}_!^{− ) .}

Proof:







 +∆ + − − + −

= ∆

−

+

∫ ∫

∫

⁺ _{∆→ +}⁺_∆^{∆ + + dt}

n t t x

f n dt

t x t x

x f n dt

t t x

dx f

d ^x

x

n x m

x x x

n m

x x

x

n

m 1

) 1 (

) ( 0

1 ( )

! ) 1 )(

! ( ) 1 )(

1 (

! lim ) 1 )( (







 +∆ + − − + −

=_∆^lim_x→0∆¹_x

∫

_x^x₊⁺_∆^∆_x^{x+1f(m)(t)(x x}_n!^{1 t)ndt}

∫

_x^x₊⁺_∆^∆_x^{x+1 f(m)(t)(x 1}_n! ^t)ndt







 + − − + −

+_∆^lim_x→0∆¹_x

∫

_x^x₊⁺_∆^∆_x^{x+1 f(m)(t)(x 1}_n! ^t)ndt

∫

_x^{x+1 f(m)(t)(x 1}_n! ^t)ndt

= dt

n t x t x f

x n x

m

! ) 1 ) (

1 ( )( + −

∂

∫

⁺ ∂ 







 + − − + −

+_∆^lim_x→0∆¹_x

∫

_x^x₊⁺₁^{∆x+1 f(m)(t)(x 1}_n! ^t)ndt

∫

_x^{x+∆x f(m)(t)(x 1}_n! ^t)ndt

= dt

n t t x

f

x n x

m 1 1

) (

)!

1 (

) 1 )( (

+ −

−

−

∫

+

! ) 1 )( (

! lim ) 1 )( (

lim ^{( ) ( )}

1 n

t t x

n f t t x

f

n m

x t n m

x t

−

− +

− + +

→ +

→

! ) (

! ) 1 )(

! ( ) 1 )(

! ( ) (

! ) 1 ) (

(

) 1 ( 1) (

1 )

( )

1 ( ) (

n x dt f

n t t x

n f t t x

n f x f n

t d x

t f

n m x

x m x

x n m

n m x

x

m + − − =− + − + + − −

−

=

∫ ∫

^{+ +}

+ +

= dt

n t t x

n f x

f ^{x n}

x m m

! ) 1 )(

! ( )

( ¹ ( 1)

)

( +

∫

^{+ +} + − ^{− f(m}_n⁾_!^{(x) =}

∫

_x^{x+1f(m+1)(t)(x+}_n¹_!^−t)ndt.

Then we turn to the main theorem of this paper.

Theorem 1 (Euler-Maclaurin): If f(x) and its derivatives to n−1 are continuous in the interval

[ ]

^{a,b , then} n

k n

k k k b

a b

a x

R a f b k f

dx B x f x

f = + − ⁻ +

=

− −

=

∫ ∑

∑

^{( ( ) ( ))}

) ! ( )

( ^{( 1)}

1

) 1 ( 1

, where R_n is the remainder term.

Proof: Refer to Taylor formula, we have

n dt t x t x

F n x

x x F

x x F

x F x F x x

F ^{x x}

x

n n

n

n⁻₋ ^{∆ − +}

∫

^{+∆ +∆}₋^{− −}

+

… +

′′ ∆ +

′ ∆ +

=

∆

+ ( 1)!

) )(

)! ( 1 (

) (

! 2

) ) (

( ) ( ) (

1 )

( 1

) 1 (

2 .

Let ∆x=1, we obtain dt

n t t x

n F x F x

x F F x F x

F ^x

x

n n

n⁻₋ ⁺

∫

^{+ +} ₋^{− −}

+

…

′′ +

′ + +

=

+ ^{( 1) 1 ( ) 1}

)!

1 (

) 1 )( )! (

1 (

) (

! 2

) ) (

( ) ( ) 1

( .

Namely, dt

n t t x

n x x x

x x

x

f ^x

x

n n

n⁻₋ ⁺

∫

^{+ +} ₋^{− −}

+

…

′′ +

′ +

=

− +

= ^{( 1) 1 ( ) 1}

)!

1 (

) 1 )( )! (

1 (

) (

! 2

) ) (

( ) ( ) 1 ( )

( φ φ φ φ φ φ ^.

Integrate and derivative either side of the equality respect tox, apply lemma 1, and make the last term before the integral remainder term in Taylor series be a multiple of φ⁽ⁿ⁻¹⁾(x), thus we obtain

n dt t t x

n x x x

C dt t

f ^x

x

n n

x n

x

∫

∫

₀ ^{( ) + =φ( )+φ′}₂⁽_!^{)+…+φ( −1)}_!^{( )+ +1φ( )( )( +1}_!^{− ) .}

n dt t t x

n x x x

x

f ^x

x

n n

n⁻₋ ⁺

∫

^{+ +} ₋^{− −}

+

…

′′′ +

′′ +

′ = ^{( 1) 1 ( ) 2}

)!

2 (

) 1 )( )! (

2 (

) (

! 2

) ) (

( )

( φ φ φ φ ^.

n dt t t x

n x x x

x

f ^x

x

n n

n⁻₋ ⁺

∫

^{+ +} ₋^{− −}

+

… +

′′′ +

′′ = ^{(4) ( 1) 1 ( ) 3}

)!

3 (

) 1 )( )! (

3 (

) (

! 2

) ) (

( )

( φ φ φ φ ^.

…

dt t x t x

x

f ^x

x n n

n^{−2) = ( −1) +}

∫

^{+1 ( ) + −}

( ( ) φ ( ) φ ( )( 1 ) ^.

dt t x

f ^x

x n n^{−1) =}

∫

^{+1 ( )}

( ( ) φ ( ) ^.

Multiply every of these identities by a coefficienta_k, and add them together. Assume a₀ =1and )! 0

(

1

0

− =

∑

⁻

= n

k k

k n

a , and we obtain dt

k n

t x t a

x x f a C dt t

f ^x

x

n

k

k n n k

n

k

k k x

x

∑ ∫ ∑

∫

⁺

=

−

=

−

−

− + +

= +

+ ¹

0 ) ( 1

) 1 (

)!

(

) 1 ) (

( )

( ) ( )

(

0

φ

φ ^.

Then we figure out the coefficients. The recurrence equation is 0 )!

(

1

0

− =

∑

⁻

= n

k k

k n

a with the initial

condition a₀ =1. Let

! n

a_n =bⁿ , then we have 0 )!

(

!

1

0

− =

∑

⁻ ⋅

= n

k

k

k n k

b . 0

1

0

 =







∑

⁻ 

= n

k

bk

k

n . Therefore, b_n is Bernoulli number.

! n

a_n = Bⁿ , dt

k n k

t x t B

x x

k f C B

dt t

f ^x

x

n

k

k n n k

n

k k k x

x

∑ ∫ ∑

∫

⁺

=

−

=

−

−

⋅

− + +

= +

+ ¹

0 ) ( 1

) 1 (

)!

(

!

) 1 ) (

( )

( )

! ( )

(

0

φ

φ ^.

Hence, ^k _n

n

k k k b

a b

a x

R a f b k f

dx B x f a b x

f = − = + − ⁻ +

=

− −

=

∫ ∑

∑

^{( ( ) ( ))}

) ! ( )

( ) ( )

( ^{( 1)}

1

) 1 (

1 φ φ ^.

Then we consider the remainder term.

Theorem 2: f x B x x dx

dx n x x

B x n f

R ^b

a n

n b n

a n

n

n ⁼⁻¹_!

∫

^{( )( ) ([ ]+1− ) = (−1)}_!⁻¹

∫

^{( )( ) ( −[ ]) .}

Proof : t B t t dt

dt n t x B n t

k dt n k

t x

t B ^x

x n

x n

x n

x n x

n

k

k n n k

∫

∫

∫

⁺

∑

^{+ +}

=

− = + − = + −

−

⋅

−

+ ¹ _{( )} ¹ _{( )}

1

0 )

( ( ) ([ ] 1 )

! ) 1 1 ( )

! ( 1 )!

(

!

) 1 ) (

( φ φ

φ ^.



 



 + − − + −

−

= _n

∫

^{+ x B x xdx}

∫

^{+ x B x x dx}

R ^a

a n

b n

b n

n n

1 ( ) 1 ( )

) 1 ] ([

) ( )

1 ] ([

)

! (

1 φ φ



 



 + − − + −

−

=

∑ ∫ ∑∫

⁻

= + +

=

+ 1 1

) ( 1

1 ( )

) 1 ] ([

) ( )

1 ] ([

)

! (

1 ^b

a k

k

k n

n b

a k

k

k n

n x B x xdx x B x xdx

n φ φ ^.

Since B_n([x]+1−x) is a periodic function of period 1, we can rewrite the summation as



 



 + + − − + −

−

∑∫ ∑∫

⁻

=

− +

=

+ 1 1 ( )

1 1 ( )

) 1 ] ([

) ( )

1 ] ([

) 1

! (

1 ^b

a k

k

k n

n b

a k

k

k n

n x B x xdx x B x xdx

n φ φ

∑∫

⁻

=

+ + −

−

= ^{1 1 ( )}( ) ([ ] 1 )

! 1 ^b

a k

k

k n

n x B x x dx

n f f x B x xdx

n

b

a n

∫

n ^{+ −}

−

= ( ) ([ ] 1 )

!

1 _{( )}

.

If we use B_n(1−x)=(−1)ⁿB_n(x)(n≥1), we can obtain f x B x x dx

R n ^b

a n

n n

n ^{= (−1)}_!⁻¹

∫

^{( )( ) ( −[ ])}

Theorem 3: ^Rn ^O

(

^{f(n 1)(b) f (n 1)(a)}

)

−

− −

= .

Proof : Since B_n([x]+1−x) is a periodic function and has no singular point, we know )

1 ]

([x x

B_n + − is bounded. Namely, B_n(x−[x])=O(1).

Therefore, 



 



= 

− −

= _n ⁻

∫

^{f x B x x dx O}

∫

^{f x dx}

R ^b

a b n

a n

n n

n ( ) ( [ ]) ( )

! ) 1

( ¹ _{( ) ( )} ^=O

(

^{f(n−1)(b)− f(n−1)(a)}

)

^.

[5] showed that











−

−

− =

∑

∑

∞

=

−−

∞

=

−

odd k n

x k

even k n

x k n

x x B

k

n n

k n

n

n

) , 2 (

) 2 sin(

) 2 1 (

) , 2 (

) 2 cos(

) 2 1 (

! ]) [ (

1 2 1

1 1 2 1

π π π

π

, therefore,

n n

n x n

x

B (2 )

! ) ( ]) 2 [

( π

≤ ζ

− .

Lehmer^[10] showed that the maximum value of B_n(x−[x]) obeys











−

≡

=

others n

n n n M

n n n

) , 2 (

! 2

) 4 (mod 2 ) ,

2 (

! ) ( 2

π ε π ζ

,

while the minimum obeys











+

−

≡

−

=

others n

n n n m

n n n

) , 2 (

! 2

) 4 (mod 0 ) ,

2 (

! ) ( 2

π ε π ζ

.

We distinguish two cases before a further consideration. If there exists an n that f ⁽ⁿ⁻¹⁾(x) is bounded, then ^Rn =^O

( )

¹ , and the first n−1 terms gives the main part of this summation. If any of

)

)(

1

( x

f ⁿ⁻ goes to infinity, then ^Rn ^=O

(

^{f (n−1)(b)}

)

^(b→∞).

Theorem 4: If

∑

^∞

=

− 1

) 1

( ( )

n !

n

n f x

n

B converges uniformly in the interval [a,b], then

( )

∑

∑

^∞

=

−

− −

=

−

=

0

) 1 ( )

1 ( 1

) ( )

! ( )

(

n

n n

n b

a x

a f b n f

x B

f .

Proof : dx

n x x

x B f dx

x x

B x n f

R ^b

a

n b n

a n

n

n ⁼⁻¹_!

∫

^{( )( ) ([ ]+1− ) =−}

∫

^{( )( ) ([ ]+}_!^{1− )}

∫

∫

₊^{+ − =} ₊^{+ − −} ₊^{+ −}

= ^{+ + b +}

a

n n b

a n

b n a

n n

n

x x

x B n df

x x

x B n f

x x

d B x

f ( 1)!

) 1 ] ) ([

)! ( 1 (

) 1 ] ) ([

)! ( 1 (

) 1 ] ) ([

( ^{1 ( ) 1 ( ) 1}

) (

n dx

x x

x B f

a f b n f

B ^b

a

n n

n n

n ^{− −}

∫

₊^{+ −}

= +^{+ + +}

)!

1 (

) 1 ] ) ([

( ))

( )

( )!( 1 (

1 )

1 ( )

( )

( 1

1 )

( )

(

1 ( ( ) ( ))

)!

1

( ⁺

+ − +

= +ⁿ f ⁿ b f ⁿ a R_n n

B .

If

∑

^∞

=

− 1

) 1

( ( )

n !

n

n f x

n

B converges uniformly, so does

∑

^∞

+

=

− 1

) 1

( ( )

n !

i

i

i f x

i

B , therefore,

∑

^∞

+

=

= − 1

) 1

( ( )

n !

i

i i

n f x

i

R B .

As n→∞, ( ) 0

1 !

) 1

( →

∑

^∞

+

=

− n

i

i

i f x

i

B , R_n →0. Hence,

^{∑ ∑}

^∞

( )

=

−

− −

=

−

=

0

) 1 ( )

1 ( 1

) ( )

! ( )

(

n

n n

n b

a x

a f b n f

x B

f .

Therefore, an effective estimation of the remainder term depends on the behavior of the derivatives of f(x). Namely, we need an n such that f⁽ⁿ⁻¹⁾(x) is bounded. Therefore, if a function increases so rapidly that all its derivatives tend to infinity, then Euler-Maclaurin formula doesn’t work now.

However, in this case, the first few terms are so inferior to the last one that they can just be dropped.

For instance, one can easily find that ! !

1

0

x n

x

n

∑

⁻ ≤

=

. The following theorem gives a criterion of this case.

Theorem 4: Assume f(x)≥0 and f′(x)≥0 if x> x₀^. f⁽x⁾=o

(

f′⁽x⁾

)

⁽x→∞⁾. Then ^{1 ( )}

(

^{( )}

)

^{( )}

0

∞

→

∑

⁻ =

=

x x f o n f

x

x n

.

Proof : _Q f(x)=o[f′(x)](x→∞), ∴Given ∀N >0, ∃x_N, if x>x_N, we have N x f

x f′ >

) (

)

( .

Namely, f′(x)> Nf(x). dt t f N dt t

f ^x

x x

x

∫

∫

^{′ >}

∴

0 0

) ( )

( , f x f x N ^x f t dt

∫

x

>

−

0

) ( )

( )

( ₀ , f x N ^x f t dt

∫

x

>

0

) ( )

( .

Since f(x) increases, we have

∫ ∑∫ ∑

⁻

=

−

=

+ ≥

= ^{1 1 1}

0 0

0

) ( )

( )

(

x

x n x

x n

n n x

x f t dt f t dt f n .

Hence, N

n f

x f

x

x n

∑

⁻ >

= 1

0

) (

)

( , ( ) [ ( )]( )

1

0

∞

→

∑

⁻ =

=

x x f o n f

x

x n

.

We now fix a, denote it as x₀, and consider

∑

⁻

= 1

0

) (

x

x n

n

f as a function of x. We rewrite the formula as

n n

k

k k n

x x x

x n

R x k f

C B dt t f n

f =

∫

+ +

∑

+

∑

=

− −

= 1

) 1 ( 1

)

! ( )

( )

(

0 0

, where C_n is a constant and R_n is the remainder term.

Sometimes

∫ ∑

^∞

=

+ −

+

1

) 1

( ( )

) ! (

0 k

k x k

x f x

k C B dt t

f diverges almost everywhere. But

∫ ∑

^∞

=

+ −

+

1

) 1

( ( )

) ! (

0 k

k x k

x f x

k C B dt t

f may be the asymptotic series of

∑

⁻

= 1

0

) (

x

x n

n

f . We have the following theorem:

Theorem : Assume f^(N−1)(x)=o(1)(x→∞), and f⁽ⁿ⁾(x)=o(f ⁽ⁿ⁻¹⁾(x))(x→∞) for n≥N, then )

( )

! ( )

(

~ ) (

1

) 1 ( 1

0 0

∞

→ +

+

∑

∑ ∫

^∞

=

− −

=

x x k f

C B dt t f n

f

k k k x

x x

x n

.

Proof : We know the remainder term dt

n t t

t B f

R ⁿ

x n

n !

) 1 ] ) ([

)(

( + −

=

∫

^{∞ for n≥N}, noticing that

1 )

1 ( ) 1

1 1 (

)

( ( )

)!

1 ( )!

1 (

) 1 ] ) ([

)! ( 1 (

) 1 ] ) ([

( ^{∞ + + +} ₊

∞

+ +

= + +

− + +

+

−

− +

=

∫

_x ^{n n n n n}

x n n

n f x R

n dt B n

t t

t B n f

t t

t B f

R .

Therefore, for n≥N,

) lim (

) (

)

! ( )

( )

(

lim ( 1) ( 1)

1

) 1 ( 1

0 0

x f

R x

f

x k f

C B dt t f n f

n n n x

n

k k k x

x x

x n

x − →∞ −

=

− −

=

∞

→ =

−

−

−

∫ ∑

∑

) (

)!

1 (

) 1 ] ) ([

( )

)! ( 1 lim( )

(

! ) 1 ] ) ([

(

lim ( 1)

) 1 1 ( )

1 (

) 1 ( ) (

x f

n dt t t t B

f x

n f B x

f

n dt t t

t B f

n

n x

n n n

n x n x

n

x −

∞ + + +

∞

− →

∞

∞

→

+

− + +

= +

− +

=

∫ ∫

) 0 (

)!

1 )( ( ) lim

( )!

1 (

) 1 ] ) ([

(

lim ( )

1 )

1 (

) 1 (

1 )

1 (

+ =

− + =

− +

=

+ +

∞

− →

∞ + +

∞

→

∫

x f

n x B f

x f

n dt t t t B

f

n n n

n x n x

n

x .

Hence, ( )( )

) ! (

~ ) (

1

) 1 ( 1

0 0

∞

→ +

+

∑

∑ ∫

^∞

=

− −

=

x x k f

C B dt t f n

f

k

k x k

x x

x n

.

Now let us consider the problem in another way. Since Weierstrass theorem states that any function continuous in a closed interval can be approximated by polynomials, now we use Bernoulli

polynomials to expand a function.

Similarly, we have a rough sight on the operating first.

Assume that f(x)can be expanded via Bernoulli polynomials, i.e.,

∑

^∞

=

=

0

) ( )

(

n n

nB x

a x

f , where a_n is

undetermined coefficients. To decide a₀, we integrate either side of the expansion from x=0 to

=1

x , and notice that





>

= =

∫

₀^{1Bn(x)dx 1}₀^,_,ⁿ_n ⁰₀. Therefore, we obtain ^{a0 =}

∫

₀^{1 f(x)dx}. To decide a₁, we derivative the expansion, and notice that





>

= =

′

− ( ), 0 0

, ) 0 (

1 x n

nB x n

B

n

n . We obtain

∑

^∞

= −

′ =

1

1( ) )

(

n

n

nB x

na x

f . Similarly, we integrate from x=0 to x=1, and obtain ^{a1 =}

∫

₀^1f′(x)dx.

Since

∑

^∞

= − − + −

=

m n

n n

m x n n n m a B x

f^{( )}( ) ( 1)...( 1) ₁( ), eventually we obtain ^{an =} _n¹_!

∫

₀^{1f (n)(x)dx.}