作者rygb (再生)
看板trans_math
標題Re: [考古] 96中興考古
時間Sun Jun 26 20:20:53 2011
※ 引述《arebecca455 (sg)》之銘言：
: http://ppt.cc/acxU
: 想請問第7題 第11題
: 第11題讓我想到高中微積分
: 但他沒有給幾次方 我要怎麼設呢？
: 希望大家可以幫我解惑 謝謝
7.
Use integral by Parts
1 1
∫xf''(2x)dx = --- ∫xdf'(2x) = ----[f'(2x)x - ∫f'(2x)dx]
2 2
1 1
= ---- f'(2x)x - ---- f(2x) + c
2 4
and put the bound into above. f(0) = 1 ,f(2) = 4 , f'(2) = 5
get the answer 7/4.
11.
-1
x = f (y) -> y = f(x)
-1
and we want to know f (y) differntial to y =?
-1
so from x = f (y) , we differntial both side respect to x.
-1
df (y) dy dy
-------- * ----- = 1 and we know ---- = f'(x)
dy dx dx
(chain rule)
so if we differential with respect to x again and again , we get the answer.
2 -1
d f (y) -f''(x)
-------- f'(x) = ----------- ,
2 ' 2
dy [f (x)]
2
-1 ''' 3[f''(x)] -f'''(x)
[f (y)] f'(x) = ----------- + --------- and put the condition.
4 3
[f'(x) ] [f'(x)]
f(1)=4 imply x = 1 , y=4
-1 '''
[f (4)] =
-3/32
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◆ From: 114.34.122.244
1F：→ dali510313:那個...我的第11題跟你不一樣說= = 111.184.178.34 06/26 20:28
2F：→ rygb:真的假的 我檢查一下好了 114.34.122.244 06/26 20:31
※ 編輯: rygb 來自: 114.34.122.244 (06/26 20:34)
3F：推 dali510313:對了!!!:) 111.184.178.34 06/26 20:37
4F：→ dali510313:推r大的過程阿!!!!我都不會打Q_Q 111.184.178.34 06/26 20:37
5F：→ rygb:感謝你 :D 114.34.122.244 06/26 20:38
6F：→ arebecca455:謝謝r大！！我要再去把觀念弄清楚了 59.115.20.186 06/26 20:50
7F：→ arebecca455:謝謝你們 59.115.20.186 06/26 20:50