Jacobain with 3 functions
显示 更早的评论
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]
2 个评论
am
2019-3-27
madhan ravi
2019-3-27
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]
采纳的回答
madhan ravi
2019-3-27
编辑：madhan ravi
2019-3-27
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.
4 个评论
madhan ravi
2019-3-27
编辑：madhan ravi
2019-3-27
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
2019-3-27
madhan ravi
2019-3-27
编辑：madhan ravi
2019-3-27
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))
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
