Newton's method

Question

ilaria scanu 2022-6-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739020-newton-s-method

回答： Chunru 2022-6-13

URGENT HELP!!

"Write a matlab program to calculate R^(1/3), for all R >0 with a use of iterative Newton method. Examine graphically the chosen function f, in order to check for which points this method will converge."

This is my code, but I think it's not correct:

x = linspace(0,10)
f=@(x) x^(1/3);
df=@(x) (1/3)*x^(-2/3);
n=10; %number of iterations 
for i=2:n
    x(i)=x(i-1)-f(x(i-1))/df(x(i-1));
end 

2 个评论
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Torsten 2022-6-12

在 MATLAB Online 中打开

Hint: In Newton's method, take

f(x) = x^3 - R

as the function.

Sam Chak 2022-6-12

编辑：Sam Chak 2022-6-12

在 MATLAB Online 中打开

If you need to find R^(1/3) and the value of R is given, then why do you create a function x^(1/3)? Is R a variable, something like the user input?

The rhetorical question.

If you want to find R^(1/3) = x, which is x exactly, then you can make both sides

[R^(1/3)]^3 = x^3
R = x^3

Now, this is a cubic equation. You can solve the polynomial problem.

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Answer 1

Chunru 2022-6-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739020-newton-s-method#answer_984395

在 MATLAB Online 中打开

https://en.wikipedia.org/wiki/Newton%27s_method

R = 5;              % input
f=@(x) x.^3 - R;    % f(x) = 0
df=@(x) 3*x.^2;     % df/dx
n=10;               %number of iterations 
x(1) = 1;           % initial vaule
for i=2:n
    x(i) = x(i-1) - f(x(i-1))/df(x(i-1));
end 
        
x
x = 1×10
    1.0000    2.3333    1.8617    1.7220    1.7101    1.7100    1.7100    1.7100    1.7100    1.7100
R.^(1/3)
ans = 1.7100