I'm trying to use Newton's method to solve the following system of equations, where f and g are functions of x and y. (h,a,f,c,d,b and k are just constants).
$f(y,x)=\left[\begin{array}{c} y^{1}\\ y^{2}\\ y^{3}\\ y^{4} \end{array}\right]-\left[\begin{array}{c} y_{0}^{1}\\ y_{0}^{2}\\ y_{0}^{3}\\ y_{0}^{4} \end{array}\right]+h\left[\begin{array}{cccc} a_{1} & f_{1} & 0 & 0\\ c_{1} & a_{1} & f_{1} & 0\\ 0 & c_{1} & a_{1} & f_{1}\\ 0 & 0 & c_{1} & a_{1} \end{array}\right]\left[\begin{array}{c} y^{1}\\ y^{2}\\ y^{3}\\ y^{4} \end{array}\right]+\left[\begin{array}{cccc} b_{1} & 0 & 0 & 0\\ d_{1} & b_{1} & 0 & 0\\ 0 & d_{1} & b_{1} & 0\\ 0 & 0 & d_{1} & b_{1} \end{array}\right]\left[\begin{array}{c} x^{1}\\ x^{2}\\ x^{3}\\ x^{4} \end{array}\right]-\left[\begin{array}{c} k_{1}\\ 0\\ 0\\ k_{2} \end{array}\right]=0$
$g(y,x)=\left[\begin{array}{c} x^{1}\\ x^{2}\\ x^{3}\\ x^{4} \end{array}\right]-\left[\begin{array}{c} x_{0}^{1}\\ x_{0}^{2}\\ x_{0}^{3}\\ x_{0}^{4} \end{array}\right]+h\left[\begin{array}{cccc} a_{2} & f_{2} & 0 & 0\\ c_{2} & a_{2} & f_{2} & 0\\ 0 & c_{2} & a_{2} & f_{2}\\ 0 & 0 & c_{2} & a_{2} \end{array}\right]\left[\begin{array}{c} y^{1}\\ y^{2}\\ y^{3}\\ y^{4} \end{array}\right]+\left[\begin{array}{cccc} b_{2} & 0 & 0 & 0\\ d_{2} & b_{2} & 0 & 0\\ 0 & d_{2} & b_{2} & 0\\ 0 & 0 & d_{2} & b_{2} \end{array}\right]\left[\begin{array}{c} x^{1}\\ x^{2}\\ x^{3}\\ x^{4} \end{array}\right]-\left[\begin{array}{c} k_{1}\\ 0\\ 0\\ k_{2} \end{array}\right]=0$
Am I right in saying that the Newton Iteration equations be:
$y^{k+1}=y^{k}-\frac{f(y,x)}{\frac{\partial f(y,x)}{\partial y}}$ and $x^{k+1}=x^{k}-\frac{g(y,x)}{\frac{\partial g(x,y)}{\partial x}}$.
Or would the denominators be partial derivatives: $\left(x\frac{\partial f}{\partial y}+y\frac{\partial f}{\partial x}\right)$.
I know it was not necessary to write the entire functions but I have some more questions about the equations depending on the answer to this one.
Many Thanks!