m*****e
发帖数: 4193 | 1 我不大有把握。
Suppose f: R -> R and a is the single zero f(a) = 0. x(0) is close to a
and for i = 0, 1, 2..
z = x(i) - f(x(i))/f'(x(i)) ( Newton )
x(i+1) = z - f(z)/f'(x(i)) ( 类似于 Newton )
What is the order of convergence of this iteration?
我做了一下，因为Newton法已经有 (z-a) = o(x(i)-a)^2
最后做出来一个 (x(i+1)-a) = o(z-x(i))(z-a) + o(z-a)^2 = o(x(i)-a)(z-a)
所以order是3。不过没把握。大家帮我验证一下。:) | E*****y
发帖数: 33 | 2 x(i+1)=x(i)-f(x(i))/f'(x(i))-f(z)/f'(x(i))
=x(i)-f(x(i))/f'(x(i))-[f(x(i))-f'(x(i)) f(x(i))/f'(x(i))+f''(x(i))
(f(x(i))/f'(x(i)))^2 + O(f(x(i))/f'(x(i))^3)]/f'(x(i))
=x(i)-f(x(i))/f'(x(i))+[f''(x(i))(f(x(i))/f'(x(i)))^2
+ O(f(x(i))/f'(x(i))^3)]/f'(x(i))
x(i)-f(x(i))/f'(x(i))-a=-f''(x(i))(x(i)-a)^2/f'(x(i)) + O((x(i)-a)^3
(f(x(i))/f'(x(i)))^2=(x(i)-a)^2 + O((x(i)-a)^3)
=>
x(i+1)-a=O((x(i)-a)^3)
【在 m*****e 的大作中提到】
: 我不大有把握。
: Suppose f: R -> R and a is the single zero f(a) = 0. x(0) is close to a
: and for i = 0, 1, 2..
: z = x(i) - f(x(i))/f'(x(i)) ( Newton )
: x(i+1) = z - f(z)/f'(x(i)) ( 类似于 Newton )
: What is the order of convergence of this iteration?
: 我做了一下，因为Newton法已经有 (z-a) = o(x(i)-a)^2
: 最后做出来一个 (x(i+1)-a) = o(z-x(i))(z-a) + o(z-a)^2 = o(x(i)-a)(z-a)
: 所以order是3。不过没把握。大家帮我验证一下。:)
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