So we have the following set of nonlinear equations
$$ x_1^2 – x_2^2 = 0, $$
$$ 2x_1x_2 = 1, $$
with starting value $x_0 = [ \: 0 ,\: 1 \: ]^T$.
So, how do I do one iteration of newton's method?
I know that for a nonlinear system of equations, Newton's Method amount to
$$ x_{k+1} = x_k – J_f(x^*)^{-1}f(x_k)$$.
But how do I find $x^*$? And thus how do I find the Jacobian?
Best Answer
I think Simen K. pointed my in the right direction.
So for a nonlinear system of equations, Newton's Method amount to
$$ x_{k+1} = x_k - J_f(x_k)^{-1}f(x_k).$$
Let us first find the Jacobian matrix,
$$ J_f(x) = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \dfrac{\partial f_1}{\partial x_2} \\\\ \dfrac{\partial f_2}{\partial x_1} & \dfrac{\partial f_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 2x_1 & -2x_2 \\\\ 2x_2 & 2x_1 \end{bmatrix}.$$
For which its inverse is given by,
$$ J_f(x)^{-1} = \dfrac{1}{4x_1^2+4x_2^2}\begin{bmatrix} 2x_1 & -2x_2 \\\\ 2x_2 & 2x_1 \end{bmatrix}.$$
So we end up with $$\begin{eqnarray*} x_{1} &=& x_0 - J_f(x_0)^{-1}f(x_0) \\ &=& \begin{bmatrix} 0 \\ 1 \end{bmatrix} - \dfrac{1}{4}\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} \end{eqnarray*}$$