Non linear Newton iteration method

Can anyone tell me problem in this code? i want to solve newton Iteration method where

X(i+1) = x(i)-Jcbi inv*F(i)

function NL_Newtons_Iteration(x0,y0,TC)

Tstart = cputime;

if nargin<4, x0=0; y0=0; z0=0; TC=10^-4; end

x1(1)=x0; x2(1)=y0; x3(1)=z0; i=0; error=TC+1;

X{1}=[x1(1); x2(1); x3(1)];

while(error>TC)

Jcbi=jacobian([F1,F2,F3],[x1,x2,x3]);

Jcbi_inv=inv(Jcbi);

X{i+2}=X{i+1}-Jcbi_inv*[F1(x1(i+1), x2(i+1), x3(i+1)); F2(x1(i+1), x2(i+1), x3(i+1)); F3(x1(i+1), x2(i+1), x3(i+1))];

Ex=100*abs((x1(i+1)-x1(i))/x1(i+1));

Ey=100*abs((x2(i+1)-x2(i))/x2(i+1)); %if 3 variables: E=[Ex Ey Ez];error=max(E)

Ez=100*abs((x3(i+1)-x3(i))/x3(i+1));

E=[Ex Ey Ez];

error=max(E);

i=i+1;

end

fprintf('After %d iterations an approimate solution is',i);

soln=[x1(i+1); x2(i+1); x3(i+1)]

TEnd = cputime - Tstart

end

function F1=F1(x1,x2,x3)

F1=((-3/5)*x1)+((1/4)*x2)+((1/4)*cos(x3))+1.43;

end

function F2=F2(x1,x2,x3)

F2=((1/4)*x1)-((3/5)*x2)-((1/4)*sin(x3))-1.24;

end

function F3=F3(x1,x2,x3)

F3=((1/4)*sin(x1))-((1/4)*exp(-x2))-((3/5)*x3)-1.17;

end

I am getting errors when I run the code. The x1 x2 x3 are x y z.

[Syed Abdul Rafay]

Not enough input arguments.

Error in test>F1 (line 22)

F1=((-3/5)*x1)+((1/4)*x2)+((1/4)*cos(x3))+1.43;

Error in test (line 7)

Jcbi=jacobian([F1,F2,F3],[x1,x2,x3]);

These are the errors

Answer 1

[Jan]

Similar to your other question, [F1,F2,F3] is not a function handle, but a vector. Here you call F1, F2 and F3 without inputs.

Better:

function Y = F123(x1,x2,x3)
Y = [((-3/5)*x1)+((1/4)*x2)+((1/4)*cos(x3))+1.43; ...
     ((1/4)*x1)-((3/5)*x2)-((1/4)*sin(x3))-1.24; ...
     ((1/4)*sin(x1))-((1/4)*exp(-x2))-((3/5)*x3)-1.17];
end

And in the loop:

Jcbi = jacobian(@F123, [x1,x2,x3]);  % Failing!

But this still does not match the needs of the jacobian function, which requires symbolic input.