急~請數學高手 幫解這題[高微] 我快想瘋了

Question

這裡所有的 X* 的 * 是在X右下方,因為我打不出來 請包容
Here is another way to explain quadratic convergence for Newton's method.
Let g(X)屬於C^2,let X* be a fixed point of g , and suppose that g'(X*)=0 .
By Theorem , we will know that if X0 is close enough to X* ,
the functional iterates Xn= g(Xn-1) will converge to X* .
(a)Using the fact that g’(X*)=0 , show that there is a constant K such that
| Xn- X*| ≤ K| Xn-1 - X* |^2
(b)Verify that Newton’s method can be viewed as functional iteration of the function
g(X) = X - f(X)/ f'(X) and that g'(X*)=0 if f(X*) = 0.

Answer 1

已知g(X*)=X* (X*為fixed point of g)
(a)
|X(n)-X*|=|g(X(n-1)-g(X*)| 利用平均值定理(MVT)
=|g'(α)[X(n-1)-X*]| 其中α介於X(n-1)與X*之間
=|[g'(α)-g'(X*)][X(n-1)-X*]| 利用MVT,且g'(X*)=0
=|g"(β)(α-X*)[X(n-1)-X*]| 其中β介於α與X*之間
<=K*|α-X*|*|X(n-1)-X*| 其中K為|g"(x)|在α與X*間的max
<=K*|X(n-1)-X*|²
註:α介於X(n-1)與X*之間,故|α-X*|<|X(n-1)-X*|

(b)
iteration of the function g(X)=X-f(X)/f'(X)
即X(n+1)=g(X(n))=Xn-f(X(n))/f'(X(n))
而X(n+1)=Xn-f(X(n))/f'(X(n))
就是Newton's method for solving f(X)=0
又
g'(X)=1-[f'²(X)-f(X)*f"(X)]/f'²(X)
so, g'(X*)=1-1=0 (因f(X*)=0)
得證

2007-12-07 11:18:51 補充：
g'(x*)=0是已知!

2007-12-07 16:08:59 補充：
由g'(α)變成下式的g'(α)-g'(X*)
是因為g'(x*)=0 (已知)
有問題嗎?

Answer 2

我是說使用MVT
MVT的定義 好像g'(x)沒有寫可以等於0
因為我看很久 上面只有寫兩種情形
一個是大於0 一個是小於0