Trouble using jacobian inside a function

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David Hagan
David Hagan on 11 Feb 2012
Commented: Jan on 10 Nov 2017
I wrote a function that uses the Newton-Raphson method to solve a system of n equations and n uknowns (n<= 4 at the moment) but for some reason, I started getting an error concerning using the jacobian matlab function inside my function. The error reads:
??? Undefined function or method 'jacobian' for input arguments of type 'double'.
Error in ==> NewtRap at 21
J = jacobian(f0,v)
My function code is:
function [ answer ] = NewtRap( )
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
Input %%%
global syms x y z w
fprintf('\n')
disp(' 1 ) One Equation ')
disp(' 2 ) Two Equations ')
disp(' 3 ) Three Equations ')
disp(' 4 ) Four Equations ')
method = input('\n Choose the number of Equations that you have: ');
if method == 1
f1 = input('Enter your equation. Use x as the variable: ');
x0 = input('Enter x0: ');
tol = input('Enter the maximum tolerance: ');
v = [ x ]; %This is used to define the jacobian
f0 = [f1]; %This is used to define the jacobian
J = jacobian(f0,v);
J_inv = inv(J);
delta_x = -J_inv*f0;
%While loop
dx = [];
xx = [];
xx(:,1) = x0;
tol1 = 100;
i = 1;
while tol1 > tol
i = i + 1;
dx(:,i-1) = subs(delta_x,v,xx(:,i-1));
xx(:,i) = xx(:,i-1) + dx(:,i-1);
tol1 = abs(xx(1,i) - xx(1,i-1));
end
answer = xx(:,end);
elseif method == 2
f1 = input('Enter your equation. Use x and y as the variables: ');
f2 = input('Enter your equation. Use x and y as the variables: ');
x0 = input('Enter initial values as [x0 ;y0]: ');
tol = input('Enter the maximum tolerance: ');
v = [x y]; %This is used to define the jacobian
f0 = [f1;f2]; %This is used to define the jacobian
J = jacobian(f0,v);
J_inv = inv(J);
delta_x = -J_inv*f0;
%While loop
dx = [];
xx = [];
xx(:,1) = x0;
tol1 = 100;
tol2 = 100;
i = 1;
while tol1 > tol || tol2 > tol
i = i + 1;
dx(:,i-1) = subs(delta_x,v,xx(:,i-1));
xx(:,i) = xx(:,i-1) + dx(:,i-1);
tol1 = abs(xx(1,i) - xx(1,i-1));
tol2 = abs(xx(2,i) - xx(2,i-1));
end
answer = xx(:,end);
elseif method == 3
f1 = input('Enter your equation. Use x, y, and z as the variables: ');
f2 = input('Enter your equation. Use x, y, and z as the variables: ');
f3 = input('Enter your equation. Use x, y, and z as the variables: ');
x0 = input('Enter initial values as [x0;y0;z0]: ');
tol = input('Enter the maximum tolerance: ');
v = [x y z]; %This is used to define the jacobian
f0 = [f1;f2;f3]; %This is used to define the jacobian
J = jacobian(f0,v);
J_inv = inv(J);
delta_x = -J_inv*f0;
%While loop
dx = [];
xx = [];
xx(:,1) = x0;
tol1 = 100;
tol2 = 100;
tol3 = 100;
i = 1;
while tol1 > tol || tol2 > tol || tol3 > tol
i = i + 1;
dx(:,i-1) = subs(delta_x,v,xx(:,i-1));
xx(:,i) = xx(:,i-1) + dx(:,i-1);
tol1 = abs(xx(1,i) - xx(1,i-1));
tol2 = abs(xx(2,i) - xx(2,i-1));
tol3 = abs(xx(3,i) - xx(3,i-1));
end
answer = xx(:,end);
elseif method == 4
f1 = input('Enter your equation. Use x, y, z, and w as the variables: ');
f2 = input('Enter your equation. Use x, y, z, and w as the variables: ');
f3 = input('Enter your equation. Use x, y, z, and w as the variables: ');
f4 = input('Enter your equation. Use x, y, z, and w as the variables: ');
x0 = input('Enter initial values as [x0;y0;z0;w0]: ');
tol = input('Enter the maximum tolerance: ');
v = [x y z w]; %This is used to define the jacobian
f0 = [f1;f2;f3;f4]; %This is used to define the jacobian
J = jacobian(f0,v);
J_inv = inv(J);
delta_x = -J_inv*f0;
%While loop
dx = [];
xx = [];
xx(:,1) = x0;
tol1 = 100;
tol2 = 100;
tol3 = 100;
tol4 = 100;
i = 1;
while tol1 > tol || tol2 > tol || tol3 > tol || tol4 > tol
i = i + 1;
dx(:,i-1) = subs(delta_x,v,xx(:,i-1));
xx(:,i) = xx(:,i-1) + dx(:,i-1);
tol1 = abs(xx(1,i) - xx(1,i-1));
tol2 = abs(xx(2,i) - xx(2,i-1));
tol3 = abs(xx(3,i) - xx(3,i-1));
tol4 = abs(xx(4,i) - xx(4,i-1));
end
answer = xx(:,end);
elseif method ~= 1 || method ~= 2 || method ~= 3 || method ~= 4
fprintf('This option is not currently available\n');
end
end

Answers (2)

Walter Roberson
Walter Roberson on 11 Feb 2012
The function jacobian is part of the symbolic toolbox and applies only to symbolic expressions.
Perhaps you want the gradient() function?

David Hagan
David Hagan on 11 Feb 2012
Thanks for the help. I ended up just defining the symbols inside each method instead and it works fine!
  2 Comments
omkar jagan
omkar jagan on 10 Nov 2017
how did u do that. Can u explain with some small syntax.
Jan
Jan on 10 Nov 2017
@omkar jagan: David is not active in the forum for 5 years now. Most likely he will not see your question or even remember, what he did some years ago. I suggest to open a new thread and ask a specific question about your own problem.

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