Newton's method
5 views (last 30 days)
Show older comments
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 Comments
Sam Chak
on 12 Jun 2022
Edited: Sam Chak
on 12 Jun 2022
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.
Accepted Answer
Chunru
on 13 Jun 2022
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
R.^(1/3)
0 Comments
More Answers (0)
See Also
Categories
Find more on Interpolation in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!