Non linear Newton iteration method

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Syed Abdul Rafay
Syed Abdul Rafay 2022-10-27
编辑： Jan 2022-10-29
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.
  1 个评论
Syed Abdul Rafay
Syed Abdul Rafay 2022-10-29
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

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回答（1 个）

Jan
Jan 2022-10-29
编辑：Jan 2022-10-29
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.

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