Re: [考古] 台大98

[rygb (再生)]

※ 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!!!

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→ kkgfdsaa:第6題b錯了 是√π / 2 才對
感謝提醒 謝謝
※ 編輯: rygb            來自: 114.34.122.244       (07/05 20:51)
→ maywen:謝謝你!
※ 編輯: rygb            來自: 114.34.122.244       (07/06 14:24)