計算物理學(科學出版社)習題答案

1、 i . h 2 Î 0, ¥, f (h2 = 2 p 2 1 (h - m 2 exp- ii . p(h = 2s 2 p s h-m 1 × = f( º f (h(h × h¢(h s s h -m Þ h(h = g -1 (h = s h-m Þ h = sh2 + m . set h 2 = h(h = s 12. 證：歸納法證明之。 1 direct sampling （1） n = 1 時， f1 ( x = ae - ax ¾¾¾¾¾ 

2、4;h1 = - ln x1 ; a （2）假設 n = k 時成立，即 f k ( x = 1 抽樣 h k = - ln(x1x 2 Lx k -1x k ; a （3） n = k + 1 時， f k +1 ( x = a k +1 k - ax a k +1 - ax k -1 xe = e ò t dt k! (k - 1! 0 x e 1 2 - h2 2 ; ak x k -1e - ax 有 (k - 1! =ò x ak t k -1e- at × ae- a ( x -t dt (k - 1! 0 ¥ -¥ x = 

3、42; f k (t × f1 ( x - t dt = 0 ò f k (t × f1 ( x - t dt. 抽 ： 樣 法 1 hk +1 - h k = h1 (x k +1 ® hk +1 = hk + h1 (x k +1 = - ln(x1 Lx k +1 . a 13. 證：歸納法證明之。 數學準備： 由 復 合 b ( x, y º ò t x -1 (1 - t y -1 dt , G(z × G( z + = 2 2 0 1 1 2 1 -2 z 2p G(2 z b ( x, y = G ( x G

4、( y , G( x + y ¥ ò 0 e - x /2 p dx = 2p . （1） n = 1, f1 ( x = 1 - 1 1 x 2 e - x /2 = e - x /2 . 2× p 2p x i . y Î (-¥, +¥, f (y = ii . set x = y 2 , y = ± x ; iii . f (y ( x where 1 - y 2 /2 ; e 2p 1 - x2 /2 1 1 dy e = = f1 ( x. dx 2p 2 x 2 1 arising from the diffe

5、rent ranges of x and y. 2 1 2 n -1 2 （2）假設 n - 1 時成立，即 f n -1 ( x = n -1 G( 2 e - x /2 x n -1 -1 2 有 2 2 抽樣 xn -1 = y12 + y2 + L + yn -1 （3）當取 n 時， f n ( x = n n -1 -1 1 1 n-2 1 - x /2 2 - x /2 2 e x = e x , b( n /2 n -1 1 2 G(n / 2 2 2 n /2 2 G( G ( 2 2 1 n 1 n -3 -1 - 1 = e - x /2 x 2 ò t 2 (

6、1 - t 2 dt n -1 1 0 2n /2 G( G ( 2 2 x n -3 1 - 1 t ® xt - x/2 2 ¾¾¾ ® e ò t ( x - t 2 dt n -1 1 n /2 0 2 G( G ( 2 2 x ( x -t n -3 - 1 1 - x /2 2 2 =ò e t × e dt n -1 1 n - 2p ( x - t 0 2 2 G( 2 = ò f n -1 ( x × f1 ( x - t dt = 0 x +¥ -¥ n

7、ò f n -1 ( x × f1 ( x - t dt 2 由復合抽樣法： xn - xn -1 = x1 ( yn ® xn = xn -1 + yn = å yi2 . i =1 14. 解：MC 計算步驟： ¥ 3/2 - x ò x e dx º ò f ( xdx = ò 0 0 0 ¥ ¥ f ( x - x e dx º ò f * ( xe - x dx, f * ( x = x 3/2 . -x e 0 ¥ i . xi Î

8、0,1, f (xi = 1, i = 1,L , n; ii . 首先對偏倚密度函數 g ( x = e - x 抽樣： F1 ( x = ò e - x dx = 1 -e - x , 0 x set xi º F1 (hi Þ hi = - ln xi ; iii . 對f * ( x º x f ( x 進行抽樣： g ( x 3/2 3 5 F2 ( x = ò x dx = x 2 , 5 0 5 2 3 5 set hi º F1 (h = h 2 Þ h = ( hi 5 ; 3 5 2 5 2 5 2 -

9、- 5 1 n -1 5 3 5 iiii . h = ( ln xi = (ln xi , I= å (ln xi 3 5 . 3 n i =1 15. 解：依題，Metropolis 計算步驟： i . 選擇初始位置：h0 = x0 = 0, f max ( x = A; ii . x1 Î 0,1, f (x1 = 1, def. step d i = - L L L + Lx1 Î - , ; 2 2 2 iii . x 2 Î 0,1, f (x 2 = 1, 引入過渡概率： w(xi ® xi +1 = min1, e - ( xi

10、 +di e - xi 2 2 judge: x 2 £ w(xi ® xi +1 , if it's true, hi +1 = xi +1 = xi + d i , then goto i and walk for the next step x1 ® x0 ,xi +1 ® xi + 2 ; if not, goto ii and walk for the step xi ® xi +1 again. r 1 iiii . h =h0 ,h1 ,L ,hi ,L ,h N , calculate x 2 = N if ( x 2

11、åh i =0 N 2 i , s 2 = 1, .eq. s then 2 time stop! 16. 解：依題： f ( x = x 2 e - x º f1 ( x × g ( x, with f1 ( x = x 2 , g ( x = e- x 引入過渡概率： w(x ® x¢ = min1, xÎ0,4 g ( x¢ . g ( x Metropolis 計算步驟： i . 選擇初始位置：h0 = x0 = 0, g max ( x = 0 = 1; ii . x1 Î 0,1, f (x1 = 1,

12、 L=4, def. step d i = Lx1 Î 0, 4; iii . x 2 Î 0,1, f (x 2 = 1, judge: x 2 £ w(xi ® xi +1 , if it's true, hi +1 = xi +1 = xi + d i , then goto i and walk for the next step x1 ® x0 ,L , xi +1 ® xi + 2 ; if not, goto ii and walk for the step xi ® xi +1 again. iiii . h =h0 ,h1 ,L ,hi ,L ,h N , r 2 -x ò x e dx = I= 0 4 1 N åh i =1 N 2 i . 17. 解：依題，首先離散化區(qū)域 D : 0 £ x £ 1, 0 £ y £ 1. 1 則區(qū)域 D 內任意正則點 0 的勢值： f0 = (f1 + f2 + f3 + f4 . 4 隨機游走步驟： i . x Î 0,1, f (x = 1; 1 1 3 ii . if( < x &#