Jacobain with 3 functions

Hello,

Can someone help me to see what is wrong with my jacobian?

I reviewed it several time byt matlab still says there is a syntax error.

Additionally, why jacobian(f,x) does not work?

f=@(x) [3*x(1) - cos(x(2)*x(3)) - 3/2; 
    4*x(1)^2 - 625*x(2)^2 + 2*x(3) - 1; 
    20*x(3)^3 + exp(-1*(x(1)*x(2))) + 9]

J=@(x) [3, -1*(x(3)*cos(x(2)*x(3))), -1*(x(2)*cos(x(2)*x(3))); 
    8*x(1) , 1250*x(2), 2; 
    -1*(x(2)exp(-x(1)*x(2))),  -1*(x(1)exp(-x(1)*x(2))), 20]

[am]

I would also like to understand what is wrong with my syntax?

J=@(x) [3, -1.*(x(3).*cos(x(2)*x(3))), -1.*(x(2).*cos(x(2)*x(3)));

8*x(1) , 1250*x(2), 2;

-1*(x(2)exp(-x(1)*x(2))), -1*(x(1)exp(-x(1)*x(2))), 20]

[madhan ravi]

Correct syntax would be:

J=@(x) [3, -1.*(x(3).*cos(x(2)*x(3))), -1.*(x(2).*cos(x(2)*x(3))); 
    8*x(1) , 1250*x(2), 2; 
    -1*(x(2)*exp(-x(1)*x(2))),  -1*(x(1)*exp(-x(1)*x(2))), 20]

Answer 1

[madhan ravi]

https://in.mathworks.com/help/symbolic/jacobian.html - the function has to have symbolic arguments as mentioned in the link above , where as what you created was a function handle (@(x)).

x = sym('x',[1 3]);
syms(x) % the reason I used this is even you can access the elements of the vector x as like x1 or x(1) unlike x = sym('x',[1,3]) where you can only access the elements as x(1) but not as x1! 
f= [3*x(1) - cos(x(2)*x(3)) - 3/2; 
    4*x(1)^2 - 625*x(2)^2 + 2*x(3) - 1; 
    20*x(3)^3 + exp(-1*(x(1)*x(2))) + 9]
jacobian(f,x)

if you type

>> which jacobian -all
/Applications/MATLAB_R2018b.app/toolbox/symbolic/symbolic/@sym/jacobian.m  % sym method
>> 

It shows that jacobian() belongs to symbolic math toolbox.

[am]

Thank you!

OK, this is weird, for it works in the command window but not in the live script

I get:

Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.

[madhan ravi]

Upload the code that you are trying in live script.

J=@(x) [3, -1.*(x(3).*cos(x(2)*x(3))), -1.*(x(2).*cos(x(2)*x(3))); 
    8*x(1) , 1250*x(2), 2; 
    -1*(x(2)*exp(-x(1)*x(2))),  -1*(x(1)*exp(-x(1)*x(2))), 20]
       %    ^------ missing             ^----- missing

[am]

thank you <3

[madhan ravi]

If you just want to do the operation => J(x)\f(x) then

x = sym('x',[1 3]);
f(x) = [3*x(1) - cos(x(2)*x(3)) - 3/2       ;... 
        4*x(1)^2 - 625*x(2)^2 + 2*x(3) - 1  ;... 
        20*x(3)^3 + exp(-1*(x(1)*x(2))) + 9];
J(x) = [3                        , -1.*(x(3).*cos(x(2)*x(3))), -1.*(x(2).*cos(x(2)*x(3))) ;...
        8*x(1)                   , 1250*x(2)                 , 2                          ;... 
        -1*(x(2)*exp(-x(1)*x(2))), -1*(x(1)*exp(-x(1)*x(2))) , 20                        ];
 Calculation = J\f;
 Calculation(2,4,5)
 %           ^^^^^---- example values of x1 , x2 and x3 , this line of code indicates now you can substitute 3 values in the place of x1, x2 and x3
%  you then double the result using double() for instance
 double(Calculation(2,4,5))