Newton's method function output.
显示 更早的评论
Hello,
I have a little problem with Newton's method. I have the code and it is working but I would like to get the results such that I have the teration number n and the value of x_n according to it. Also I need the error and f(x_n) but I am only getting the x_n values on their own. The code is as follows
function [x,n,err] = newton( f, dx, x0, tol, ITMAX )
format long
f= inline(f);
dx = inline(dx);
x(1) = x0 - (f(x0)/dx(x0));
err(1) = abs(x(1)-x0);
fx=feval(f,x);
for n=2:ITMAX
if (err(n-1) >= tol) & (n <= ITMAX)
x(n) = x(n-1) - (f(x(n-1))/dx(x(n-1)));
err(n) = abs(x(n)-x(n-1));
n = n+1;
else
break
end
end
end
It would be nice if I could get the result in a table like this
disp('_________________________________________________________________________________ ')
disp(' n x_n f(x_n) error ')
disp('_________________________________________________________________________________ ')
fprintf('\n')
However I am not quite sure how to use fprint command. Any help would be appreciated. Thank you.
采纳的回答
The code contains several problems. E.g. you do not have to check for n < ITMAX inside the FOR loop, because the FOR itself limits n to this value already. The FOR cares for increasing n also, so omit n=n+1.
As the editor warnings show already, inline is deprecated. Use anonymous functions or function handles instead as inputs.
For fprintf:
% Before the loop:
fprintf('%10s%10s%10s%10s\n', 'n', 'x_n', 'f(x_n)', 'error');
% Inside the loop:
fprintf('%10g%10g%10g%10g\n', n-1, x(n-1), f(x(n-1)), err(n-1));
I suggest to switch to a while loop, because then you can update the values and the counter index n more consistently.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Mathematics 的更多信息
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
