高等微積分 Newton's method

Question

我把整個題目抄下來好了~~
Here is another way to explain quadretic convergence for Newton's method.
Let g(x) 可兩次微分, let x* be a fixed point of g,and suppose that g'(x*) = 0.
We know that if x0 is close enough to x*, the functional iterates xn =g ( x(n-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 |x(n-1) - x*|2 (此2為平方).
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.

上面x(n-1) 表示第 n-1項的 x
xn 表示第 n 項的 x
x0 表示第 0 項的 x

Answer 1

X*為fixed point of g, so g(X*)=X*
(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 因|g"(x)|在[α, X*]或[X*,α]間連續,故有最大值
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))=X(n)-f(X(n))/f'(X(n))
==>X(n+1)=X(n)-f(X(n))/f'(X(n))
就是Newton's method for solving the equation 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

2007-12-04 14:19:09 補充：
f'²(x)即[f'(x)]²