作者yueayase (scrya)
看板Math
標題[微積] 數題問題
時間Fri Jul 20 23:17:31 2012
n
1. Xn = 1 - 1/2 + 1/3 -1/4 + ... + (-1) / n
想證明它是一個Cauchy sequence
Apostol 的書上是這樣寫的:
If m > n >= N, we find that
|Xm - Xn| = |1/(n+1)-1/(n+2)+...±1/m| < 1/n <= 1/N
可是 |Xm - Xn| = (1/(n+1)(n+2)+1/(n+3)(n+4)+...±1/m)
約有(m-n)/2項, 怎麼得到小於1/n?
2. If a, b > 0, prove that the equation
a b
------------ + ----------- = 0
3 2 3
x + 2x - 1 x + x - 2
has at least one root in the interval (-1, 1)
一開始我覺得上述方程式等價於
3 3 2
a(x + x - 2) + b(x + 2x - 1) = 0
3 3 2
Let f(x) = a(x + x - 2) + b(x + 2x - 1)
=> f(-1) = -4a < 0
f(1) = 2b > 0
則根據Intermediate Value Theorem,
there is at least one root c in (-1,1) such that
f(c) = 0
但是原先的方程式因為都在分母且
3 2 2
x + x - 2 = (x - 1)(x + x + 2) = (x - 1)[(x+1/2) + 7/4] < 0 for -1 < x < 1
3 2 2
x + 2x - 1 = (x + 1)(x + x -1)
如果 x = (-1+√5)/2 , 則原函數沒有定義, 且 0 < (-1+√5)/2 < 1
必落於 (-1, 1)中
這樣好像不能套Intermediate Value Theorem 吧 ?
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◆ From: 111.251.170.119
→ suhorng :1/(n+1) - [1/(n+2) - 1/(n+3)] - [1/(n+4) - 1/(n+5 07/20 23:29
→ suhorng :)] - ... < 1/n; 1/(n+1)-1/(n+2)+... > 0. 07/20 23:30
→ yueayase :會不會1/m無法配對,且是正號啊? 07/20 23:48
→ yueayase :應該問且是負號 07/20 23:48
→ yueayase :弄錯了正號 07/20 23:49
→ suhorng :1/(n+1) ... + 1/m 奇數項的話 配對沒問題 07/21 00:12
→ yueayase :謝謝,我試試看 07/21 00:13
→ suhorng :抱歉 我還沒弄完orz 07/21 00:19
→ suhorng :偶數項的話, 1/(n+1) ... - 1/(m-1) < 1/(n+1) 07/21 00:20
→ suhorng :所以再減 1/m 還是 < 1/(n+1) 07/21 00:21
→ suhorng : + 1/(m-1) 07/21 00:25
推 yasfun :考慮( (-1+根號5)/2 , 1 ) 07/21 02:09
→ yasfun :a/b+(x^3+2x^2-1)/(x^3+x-2)=0 07/21 02:12
→ yasfun :(x^3+2x^2-1)/(x^3+x-2)可從0到負無限(又連續XD) 07/21 02:14