# Newton raphson tranfsorm from 1d to 2d

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bbah on 1 Dec 2019
Edited: David Wilson on 2 Dec 2019
i have the solution for a 1d problem. How to solve it in 2d ?
1D:
function [x] = my_NR(fun, dfun, x_0,E_tol)
q = 1;
E = inf;
x(q) = x_0;
while E > E_tol
x(q+1) = x(q) - (fun(x(q)))/(dfun(x(q)));
E = abs(x(q+1)-x(q));
q = q+1;
end
end
How to do it in 2D ?
e.g
Input : f = @(x)[0.5*cos(x(1))-0.5*sin(x(2))-x(1);0.5*sin(x(1))+0.5*cos(x(2)-x(2))];
J = @(x)[-0.5*sin(x(1))-1,-0.5*cos(x(2));0.5*cos(x(1)),-0.5*sin(x(2))-1];
x_0 =[0 0]';
E_tol = 10E-04;
Output: x = [0.0 0.0 ;0.2 0.6;0.2287 0.5423;0.2291 0.5391]
David Wilson on 2 Dec 2019
Edited: David Wilson on 2 Dec 2019
The Newton-Rhapson is generally specific to 1D. If you want the more general nD case, you need to use the (multivariable) Newton scheme. See e.g. http://fourier.eng.hmc.edu/e176/lectures/NM/node21.html
or of course look up Wikipedia
If you want to just solve your particular problem then try fsolve.
By the way, do you really mean (in the 2nd equation of f(x))
cos(x(2)-x(2))
And is J above the Jacobian? (It's not what I compute!)