✔ 最佳答案
By Euler-Maclaurin summation formula (with remainders)
|Σ[a~b] f(x) -∫[a~b] f(x) dx -[f(a)+f(b)]/2-1/12 [f'(b)-f'(a)] |
< 1/4! * 7/240 * [f'''(a) - f'''(b) ] (此為最大誤差) ----(A)
Note: 7/240 = max |B4({x})| , B4(x) is the Bernoulli poly. of degree 4
取f(x)=√x, f'(x)= 1/(2√x), f'''(x)= (3/8)/(x^2√x), a=1, b=n^2, 則
|最大誤差| (A) = (1/24)*(7/240) * (3/8) ( 1- 1/n^5)
= 7/15360 (1- 1/n^5) < 7/15360
故
|Σ[1~n^2] √x - [2/3 (n^3 -1) + (n+1)/2 +(1/24) (1/n - 1)] | < 7/15360
設S(n)= 2/3 (n^3 - 1) + (n+1)/2 + (1/24) (1/n - 1) , err= 7/15360, 則
(1) S(n)= (2/3) n^3 + n/3 + (n-1)/6 + (1/24) (1/n - 1)
Note: (2/3) n^3 + n/3為整數
(2) Σ[1~n^2] √x = S(n) + R(n) -----(B)
其中
(n-1)/6 +(1/24)(1/n - 1)- err < R(n) < (n-1)/6 + (1/24)(1/n -1) + err ---(C)
=> (n-2)/6 + 1/8 + 1/(24n) - err < R(n) < (n-2)/6 + 1/8 + 1/(24n) + err
而 n>= 2 時, 1/8 + 1/(24n) + err < 1/6
=> [R(n) ]=[ (n-2)/6 ]
由(B)=>
[ Σ[1~n^2] √x ] = [S(n)+R(n)] (Note:S(n)為整數)
= S(n) + [R(n)] = S(n) + [ (n-2)/6 ]得證
Note: n, x與原題互換
Q2:
由(B), n= 20時,
√1+√2+...+√400= S(20) + R(20) = 5340 + R(20)
又由(C):
(n-1)/6 +(1/24)(1/n - 1)- err < R(n) < (n-1)/6 + (1/24)(1/n -1) + err
19/6 + (1/24)(1/20 - 1) - 7/15360 < R(20) < 19/6 + 1/24 (1/20 -1)+ 7/15360
=> [R(20)]= 3
故 [ √1 + √2 + ...+ √400 ] = 5340 + 3 = 5343
或代Q1之公式結論: S(20)+ [ (20 - 2)/ 6 ] = 5343
2009-06-04 01:23:20 補充：
更正:
B4(x)=x^4-2x^3+x^2 - 1/30 => -1/30 <= B4(x) <= 7/240 , for x in [0,1]
=>|B4({x})|<= 1/30 , {x}= x-[x] =>|最大誤差| (A) < 1/1920= err
Q2:19/6 + (1/24)(1/20 - 1) - 1/1920 < R(20) < 19/6 + 1/24 (1/20 -1)+ 1/1920
=>[R(20)]= 3
