Maddog Battie

Generalisations

(temp for Newtons Method which is now believed to be incorrect)

I've been looking into the generalisation issue and have come across the following problem: For a system with two variables then the Jacobian matrix looks like this:

$J_{F}(x)={\begin{pmatrix}{\delta f_{1}(x_{1},x_{2}) \over \delta x_{1}}&{\delta f_{1}(x_{1},x_{2}) \over \delta x_{2}}\\\\{\delta f_{2}(x_{1},x_{2}) \over \delta x_{1}}&{\delta f_{2}(x_{1},x_{2}) \over \delta x_{2}}\end{pmatrix}}$

Now to use this in Newton's method, you need to calculate the inverse which ends up as:

${J_{F}(x)}^{-1}={1 \over {\delta f_{1}(x_{1},x_{2}) \over \delta x_{1}}{\delta f_{2}(x_{1},x_{2}) \over \delta x_{2}}-{\delta f_{1}(x_{1},x_{2}) \over \delta x_{2}}{\delta f_{2}(x_{1},x_{2}) \over \delta x_{1}}}{\begin{pmatrix}{\delta f_{2}(x_{1},x_{2}) \over \delta x_{2}}&-{\delta f_{1}(x_{1},x_{2}) \over \delta x_{2}}\\\\-{\delta f_{2}(x_{1},x_{2}) \over \delta x_{1}}&{\delta f_{1}(x_{1},x_{2}) \over \delta x_{1}}\end{pmatrix}}$

Unfortunately, the first section ends up as a divide by zero. I assume I must be doing something wrong here. I'd love to know what as I'm keen to use Newton's method for a 12th order problem.