作者tasukuchiyan (Tasuku)
看板Math
标题[微积] 台大数研90高微考古题
时间Sun Dec 11 14:28:13 2011
4.(b)
Let a(n)>0, a(n+1)/a(n)<=(1-2/n) for n>=3.
Show that the series of a(n) is convergent.
6.
Prove that, for each integer n, there exists a C^2 function w=g(x,y)
defined in some neighborhood of (0,0) such that
x+2yw+cosw=0,
g(0,0)=nπ+π/2.
请问有任何解题的方向吗?
谢谢
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◆ From: 134.208.26.13
1F：推 bineapple :6 implicit function theorem 12/11 15:44
2F：→ tasukuchiyan:隐函数定理我有想过，不过不知道要怎麽使用 12/11 16:00
3F：→ tasukuchiyan:严格的写下证明 12/11 16:01
4F：→ Sfly :4的不等号反了吧 12/11 16:05
5F：→ bineapple :F(x,y,w)=x+2yw+cosw=0 at (0,0,nπ+π/2) 12/11 16:18
6F：→ bineapple :所以可以把w写成g(x,y)在一个neighborhood of (0,0) 12/11 16:19
7F：→ bineapple :然後对F(x,y,g(x,y))偏微 证明g是C^2 12/11 16:19
※ 编辑: tasukuchiyan 来自: 134.208.26.13 (12/11 17:40)
8F：→ tasukuchiyan:要怎麽对F(x,y,g(x,y))偏微，证明g是C^2呢? 12/11 19:17
9F：→ bineapple :F(x,y,g(x,y))在(0,0)的附近都是0 所以得到在0的附近 12/12 01:33
10F：→ bineapple :x+2yg(x,y)+cos(g(x,y))=0 对这个式子偏微後能够把 12/12 01:34
11F：→ bineapple :g的一次微分提出来写成x,y,g(x,y)的式子 然後再利用 12/12 01:34
12F：→ bineapple :g是C^1的性质 12/12 01:35
13F：→ tasukuchiyan:那g'不是从(0,0)的邻域映射到L(R^2,R)的函数吗? 12/12 12:09
14F：→ tasukuchiyan:怎麽用g是C^1的性质去证明g'可微分? 12/12 12:10
15F：→ Sfly :把w_x表成 w跟y,x的函数 12/12 13:30
16F：→ tasukuchiyan:If all mixed second order partial derivatives are 12/12 15:25
17F：→ tasukuchiyan:continuous at a point, f is termed a C^2 function 12/12 15:26
18F：→ tasukuchiyan:at that point. 12/12 15:26
19F：→ tasukuchiyan:是这一回事吗? 12/12 15:28
20F：→ bineapple :yes 12/12 16:08
21F：→ Sfly :这句话要证明 虽然只是隐函数定理的一个推论 12/12 16:27
22F：→ tasukuchiyan:感谢所有回答的人 12/13 11:05