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Euler-Maclaurin Summation Formula

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Euler-Maclaurin Summation Formula 4 Euler-Maclaurin Summation Formula 4.1 Bernoulli Number & Bernoulli Polynomial 4.1.1 Definition of Bernoulli Number Bernoulli numbers Bk ()k =1,2,3, are defined as coefficients of the following equation. x Bk k x =Σ x e -1 k=0 k! 4.1.2 Expreesion of Bernoulli Numbers B B x 0 1 B2 2 B3 3 B4 4 = + x + x + x + x + (1,1) ex-1 0! 1! 2! 3! 4! ex-1 1 x x2 x3 x4 = + + + + + (1.2) x 1! 2! 3! 4! 5! Making the Cauchy product of (1.1) and (1.1) , B1 B0 B2 B1 B0 1 = B + + x + + + x2 + 0 1!1! 0!2! 2!1! 1!2! 0!3! In order to holds this for arbitrary x , B1 B0 B2 B1 B0 B =1 , + =0 , + + =0 , 0 1!1! 0!2! 2!1! 1!2! 0!3! n The coefficient of x is as follows. Bn Bn-1 Bn-2 B1 B0 + + + + + = 0 n!1! ()n -1 !2! ()n-2 !3! 1!n ! 0!()n +1 ! Multiplying both sides by ()n +1 ! , Bn()n +1 ! Bn-1()n+1 ! Bn-2()n +1 ! B1()n +1 ! B0 + + + + + = 0 n !1! ()n-1 !2! ()n -2 !3! 1!n ! 0! Using binomial coefficiets, n +1Ｃn Bn + n +1Ｃn-1 Bn-1 + n +1Ｃn-2 Bn-2 + + n +1Ｃ 1 B1 + n +1Ｃ 0 B0 = 0 Replacing n +1 with n , (1.3) nnＣ -1 Bn-1 + nnＣ -2 Bn-2 + nnＣ -3 Bn-3 + + nＣ1 B1 + nＣ0 B0 = 0 Substituting n =2, 3, 4, for this one by one, 1 21Ｃ B + 20Ｃ B = 0 B = - 1 0 2 2 1 32Ｃ B + 31Ｃ B + 30Ｃ B = 0 B = 2 1 0 2 6 Thus, we obtain Bernoulli numbers. The first few Bernoulli numbers are as follows. 1 1 1 1 5 B = 1 , B = , B = - , B = , B = - , B = , 0 2 6 4 30 6 42 8 30 10 66 1 B = - , B = B = B = = 0 1 2 3 5 7 - 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 n k 1 r n Bn = Σ Σ()-1 krＣ r k=0 k+1 r=0 If the first few are written down, it is as follows. 1 0 B = 00Ｃ 0 0 1 1 1 1 1 1 B = 00Ｃ 0 + 10Ｃ 0 - 11Ｃ 1 1 1 2 1 2 1 2 2 1 2 2 2 B = 00Ｃ 0 + 10Ｃ 0 - 11Ｃ 1 + 20Ｃ 0 - 21Ｃ 1 + 22Ｃ 2 2 1 2 3 Using this, a large Bernoulli number can be obtained directly. However, since the number of terms increases with 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 called Bernoulli Polynomial . B0()x = 1 d B ()x = nB ()x ()n 1 dx n n-1 1 Bn()xdx = 0 ()n 1 0 (2) Properties The following expressions follow directly from the definition. x Bn()x = n Bn-1()tdt + Bn ()n 1 0 n n n k n n-k Bn()x = Σ Bn-k x = Σ Bk x k=0 k k=0 k Bn()0 = Bn Bn()1 = Bn()0 ()n 2 n-1 Bn()x+1 - Bn()x = nx ()n 1 n Bn()1-x = ()-1 Bn()x ()n 1 - 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 1 1 3 1 B ()x = x- ,B()x = x2-x+ ,B()x = x3- x2+ x , 1 2 2 6 3 2 2 1 5 5 1 B ()x = x4-2x3+x2- ,B()x = x5- x4+ x3- x , 4 30 5 2 3 6 5 1 1 7 7 7 1 B ()x = x6-3x5+ x4- x2+ ,B()x = x7- x6+ x5- x3+ x , 6 2 2 42 7 2 2 2 6 14 7 2 1 B ()x = x8-4x7+ x6- x4+ x2- , 8 3 3 3 30 9 21 3 B ()x = x9- x8+6x7- x5+2x3- x , 9 2 5 10 15 3 5 B ()x = x10-5x9+ x8-7x6+5x4- x2+ , 10 2 2 66 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 Bmx -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 , 1 m Bmx-x = -2m!Σ m cos 2 sx- x 0 s=1 ()2 s 2 Proof According to Formula 5.1.2 (" 05 Generalized Bernoulli Polynomials ") , the following expression holds. 1 m Bm()x = -2 m!Σ m cos 2 sx- 0 x 1 s=1 ()2s 2 Since Bmx -x = Bm()x on 0 x <1, 1 m Bmx-x = -2 m!Σ m cos 2 sx- 0 x <1 s=1 ()2s 2 On 1 x <2, Left: Bmx+1-x+1 = Bmx+1-x -1 = Bmx-x 1 m 1 m Right: -2m!Σ m cos 2 s()x +1 - = -2m!Σ m cos 2 sx- s=1 ()2s 2 s=1 ()2s 2 1 m Bmx-x = -2 m!Σ m cos 2 sx- 1 x <2 s=1 ()2s 2 - 3 - Hereafter by induction, the following expression holds for arbitrary natural number n . 1 m Bmx-x = -2 m!Σ m cos 2 sx- n x < n +1 s=1 ()2s 2 Q.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 m When f()x is a function of class C 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. b -1 b m Br ()r-1 ()r-1 Σf()k = f()xdx + Σ f ()b - f ()a + Rm (1.1) k=a a r=1 r! m+1 b ()-1 ()m Rm = Bmx-xf()xdx (1.1r) m! a b m 1 m ()m (1.1r') = ()-1 2 Σ m cos 2 sx- f ()xdx a s=1 ()2s 2 f()m ()x When limRm = m is a even number s.t. = minimum for x[]a,b m ()2 m Proof When Br are Bernoulli numbers and Bn()x are Bernoulli polynomials, x 1 B0()x =1 , Bn()xdx= Bn+1()x -Bn+1 0 n +1 Bn+1()1= Bn+1()0= Bn+1 Using these, 1 1 f()xdx = B0()xf()xdx 0 0 1 1 1 B1()x ' = B1()xf()x - f ()xdx 1! 0 0 1! 1 1 1 1 ' 1 B2()x " = B1()xf()x - B2()xf()x + f ()xdx 1! 0 2! 0 0 2! 1 1 1 ' 1 = B ()xf()x - B ()xf()x 1! 1 0 2! 2 0 1 1 " 1 B3()x "' + B3()xf()x - f ()xdx 3! 0 0 3! r m -1 1 ()-1 ()r-1 1 m Bm()x ()m = Σ Br()xf ()x + ()-1 f ()xdx r=1 r! 0 0 m! Here, 1 1 1 1 B ()xf()x = 1- f()1- 0- f()0 = f()1+ f()0 1 0 2 2 2 And r-1 r-1 ()1 ()0 ()-1 Br = ()-1 Br = -Br for r 2 Therefore, - 5 - 1 -1 r-1 B xf()r-1 x = -B f()r-1 1- f()r-1 0 () r() () 0 r () () Substitutin these for the above, 1 m 1 Br ()r-1 ()r-1 f()xdx = f()1+ f()0 - Σ f ()1- f ()0 0 2 r=2 r! m 1 ()-1 ()m + Bm()xf ()xdx m! 0 Replacing f()x with f()x +k , 1 m 1 Br ()r-1 ()r-1 f()x+kdx = f()k+1 + f()k -Σ f ()k+1 - f ()k 0 2 r=2 r! m 1 ()-1 ()m + Bm()xf ()x+kdx m! 0 That is, k+1 m 1 Br ()r-1 ()r-1 f()xdx = f()k+1 + f()k -Σ f ()k+1 - f ()k k 2 r=2 r! m k+1 ()-1 ()m + Bm()x-kf ()xdx m! k Accumulating this from k =a to k =b -1 , b b -1 m b -1 1 Br ()r-1 ()r-1 f()xdx = Σf()k+1 + f()k -Σ Σf ()k+1 - f ()k a 2 k=a r=2 r! k=a m b k+1 ()-1 -1 ()m + Σ Bm()x-kf ()xdx m! k=a k Here, b -1 Σf()r-1 ()k+1 - f()r-1 ()k = f()r-1 ()b - f()r-1 ()a k=a b -1 b Σf()k+1 + f()k + f()a +f()b = 2Σf()k k=a k=a Bm()x-k = Bmx-x () k x k+1 Therefore, b b m 1 Br ()r-1 ()r-1 f()xdx = Σf()k - f()a +f()b -Σ f ()b - f ()a a k=a 2 r=2 r! m b ()-1 ()m + Bmx-xf()xdx m! a Transposing the terms, b b m 1 Br ()r-1 ()r-1 Σf()k = f()xdx + f()a +f()b +Σ f ()b - f ()a k=a a 2 r=2 r! m b ()-1 ()m - Bmx-xf()xdx m! a Subtracting f()b from both sides, - 6 - b -1 b m 1 Br ()r-1 ()r-1 Σf()k = f()xdx - f()b -f()a +Σ f ()b - f ()a k=a a 2 r=2 r! m b ()-1 ()m - Bmx-xf()xdx m! a Since B1 = -1/2 , including the 2nd term of the right side into , we obtain b -1 b m Br ()r-1 ()r-1 Σf()k = f()xdx + Σ f ()b - f ()a + Rm (1.1) k=a a r=1 r! m+1 b ()-1 ()m Rm = Bmx-xf()xdx (1.1r) m! a Last, substituting Formula 4.1.5 1 m Bmx-x = -2m!Σ m cos 2 sx- x 0 s=1 ()2 s 2 for (1.1r) , we obtain b m 1 m ()m (1.1r') Rm = ()-1 2 Σ m cos 2 sx- f ()xdx a s=1 ()2s 2 f()m ()x When R is divergent , the amplitude is almost determined by .
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