※ 引述《maywen (王小咩)》之銘言： : http://exam.lib.ntu.edu.tw/sites/default/files/exam/undergra/98/98026.pdf : 我想問 : 1.第二題的(b) : 為什麼我算出來的答案是f(x)=(x+2)/(x+3) : 我看之前有人算出來是f(x)=3(x+2)/(x+3) 你可能忘了積分常數 三是從那邊出現的 f'(x) = f(x) / (x+2) / (x+3) f(0) = 2 (boundary condition) df(x) dx ----- = ---------- ln f(x) = -1 ln (x+3) + 1 ln (x+2) + ln c f(x) (x+2)(x+3) x+2 f(x) = ----- c x+3 and use boundary condition to solve c , c = 3 x + 2 so f(x) = 3 * ------- x + 3 : 2.第三題的(a)和(b) (a) Set if lim an = k , so does lim an+1 = k n->∞ n->∞ 1/2 from the condition we know an+1 = ( 6 + an ) 1/2 so we get k = ( 6 + k ) 2 and k = 6 + k k = 3 or -2 but.. k = -2 put it back to the equation -2 ≠ 2 so the solution is only 3 . (b) when t approach 0 , the integral part approach 1 ∞ but 1/t approach ∞ 1 is indetermin form when t approach 0 1 1 t 0 --- ln (∫ ( 1 + x) dx) = ( ---- ) t 0 0 but you must know the above equation is differential respect to t 然後我之前好像寫錯了 羅畢達完應該是長這樣 t |x=1 ∫ln(1+x)(1+x) dx ∫ln(1+x) dx x*ln(1+x) - x +ln(1+x)| |x=0 ------------------- = ------------------ = ---------------------------- t |x=1 ∫(1+x) dx x | 1 |x=0 先把t->0 放入 方便計算 get 2ln2 -1 , too. and put it back on exp. Or你可以先對x積分 再去微分t 或著是直接 將裡面降次皆可 或著 是你一開始就把積分部分 表示成t函數 t+1 t+1 1 t (1+x) | 2 - 1 ∫ (1+x) dx = ------- | = -------- 0 t+1 | t + 1 and t -> 0 , becomes to 1 we get 0/0 , too Use L'hopital's rule t+1 1 ln2*2 1 --------- * ( --------- - ------- ) t+1 1 2 -1 t + 1 set t = 0 , we get 2ln 2- 1 (2ln2-1) -1 so answer is e = 4 e : 3.第六題的(b) 2 -x Set I = ∫e dx 2 2 2 2 -(x + y ) -r I = ∫∫e dxdy = ∫∫e r drdt x = 0 ~ ∞ --> only positive. r = 0 ~ ∞ t = 0 ~ π/2 y = 0 ~ ∞ a so solve it , we get I. Note e always positive so that I is positive. I = √π / 2 . : 麻煩大家了!!謝謝!!! --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.34.122.244 感謝提醒 謝謝 ※ 編輯: rygb 來自: 114.34.122.244 (07/05 20:51) ※ 編輯: rygb 來自: 114.34.122.244 (07/06 14:24)