Answer this Newtons Method problem

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Answers (1)

why not use fsolve or fzero? The following is newton's method
function main
f = @(x) [
sin(x(1)*x(2)) + x(1) - x(2);
x(2)*cos(x(1)*x(2)) + 1;
];
TOL = 1e-3;
err = 1;
x0 = [1;2];
iter = 0;
fprintf('Iteration\tx1\tx2\tf1(x1,x2)\tf2(x1,x2)\n');
while err>TOL
iter = iter + 1;
x1 = x0 - diffmat(f,x0,1e-3)\f(x0) ;
err = norm(x1-x0);
x0 = x1;
fval = f(x1);
fprintf('%d\t%f\t%f\t%f\t%f\t\n', iter, x1(1), x1(2), fval(1), fval(2));
end
end
function m = diffmat(fun, x, h)
f0 = fun(x);
m = zeros(numel(f0), numel(x));
for i = 1:1:numel(x)
x0 = x;
x0(i) = x0(i) + h;
fi = (fun(x0) - f0)/h;
m(:,i) = fi;
end
end
Result printed:
Iteration x1 x2 f1(x1,x2) f2(x1,x2)
1 1.079754 1.945311 -0.002579 0.017148
2 1.086147 1.943707 -0.000034 0.000076
3 1.086187 1.943685 0.000000 -0.000000

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Asked:

on 13 Aug 2021

Answered:

on 13 Aug 2021

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