Use Newton’s method (and indicate each successive approximation) to solve the nonlinear system below for $x_1$ and $x_2$ both near $0.6$:
$$x=\sinh(y)\\ 2y=\cosh(x)$$
I found the Jacobi matrix, found it's inverse and used the formula: Xo-inverse of $jacobi^{*}f()$ and tried to iterate, but I am no where close to the answer. The answer given is $x=0.6000,0.6367,0.6281,0.6424,0.6389, y=0.6000, 0.5927 0.6048,0.6019,0.6068$. Could anyone please help me out?
I suppose that you considered two equations $$F(x,y)=x-\sinh (y)=0$$ $$G(x,y)=\cosh(x)-2y=0$$ So, expanding around a point ($x_0,y_0$), you have two linear equations in ($x-x_0$) and ($y-y_0)$ which write $$0=F(x_0,y_0)+(x-x_0)-(y-y_0) \cosh(y_0)$$ $$0=G(x_0,y_0)-(x-x_0) \sinh(x_0)-2(y-y_0)$$ or $$0=F(x_0,y_0)+\Delta x- \cosh(y_0) \Delta y$$ $$0=G(x_0,y_0)-\sinh(x_0)\Delta x-2\Delta y$$ which you have to solve.
So starting at $(0.6,0.6)$, the iterates are $(0.620357,0.586253)$, $(0.631485,0.595658)$, $(0.637755,0.600937)$, $(0.641342,0.603954)$, $(0.643412,0.605693)$, $(0.644612,0.606702)$, $(0.645310,0.607288)$, $(0.645716,0.607629)$ and so on.
The almost exact solution is $(0.646285,0.608106)$.
As I mentioned in my comment, the problem would be much simpler to solve reducing it to a single equation : eliminating $x$ from the first equation leads to finding $y$ such that $$2 y-\cosh (\sinh (y))=0$$ which corresponds to a very well conditioned function; using Newton method would solve much faster that solving the two equations for the two unknowns.
