w**n
发帖数: 88 | |
w**n
发帖数: 88 | 2 Newton's Method fails when
(i). f(x[n])=0;
(ii).| x[n+1]-x[n]| or f'(x[n])/f(x[n] gets larger instead of smaller as n
increases. No matter how we choose the estimate. |
w**n
发帖数: 88 | 3
some comments :
1.for (i) x[n] is not the root , in your example , x=0 is happen to be the root
so newton's method just stucks in the root.
2. (i), (ii) have the relation "or" instead of "and".
3.for case (i) you may change the test point to skip the x[n] , but for case
(ii) there is no way to apply the newton's method correctly.
【在 w**n 的大作中提到】
: Newton's Method fails when
: (i). f(x[n])=0;
: (ii).| x[n+1]-x[n]| or f'(x[n])/f(x[n] gets larger instead of smaller as n
: increases. No matter how we choose the estimate.
|
m*****e
发帖数: 4193 | 4 基本上和Newton一样，但是不是用x(i)的切线，而是采用x(i)和x(i+1)的连线。
所以需要两个初始值。 |
w**n
发帖数: 88 | 5 Are you sure you are using "Pure Newton Mehtod", becuause if you mixed
the bisection with newton method , there will be no problem for any solvable
case.
Here is the proof that f(x)=x^(1/3 ) is not solvable with newton's method
x[n+1]=x[n]-f(x[n])/f'(x[n])
=x[n]-( x[n]^(1/3) )/ ( (1/3)*x[n]^(-2/3) )
=x[n]-3x[n]
=-2x[n]
This implies that | x[n+1]-x[n] |=3x[n]
Hence Newton's Method fails.
【在 m*****e 的大作中提到】
: 基本上和Newton一样，但是不是用x(i)的切线，而是采用x(i)和x(i+1)的连线。
: 所以需要两个初始值。
|
w**n
发帖数: 88 | 6 Check your code , make sure you are testing simple newton method. I have
tried on my computer , newton's method definetely failed on x^(1/3)=0
【在 m*****e 的大作中提到】
: 基本上和Newton一样，但是不是用x(i)的切线，而是采用x(i)和x(i+1)的连线。
: 所以需要两个初始值。
|
