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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?

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    $\begingroup$ This is not correct. Replace $x^*$ by $x_k$. Then you should be able to compute the iteration. ($x^*$ is what you want to find!) $\endgroup$ – Jas Ter Oct 3 '13 at 7:45
  • $\begingroup$ @SimenK. So my comment was too long, I posted it as an answer. The main issue is that I miss one thing, the $f(x_0)$.. What is this $f$? $\endgroup$ – onimoni Oct 3 '13 at 7:59
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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 $$\begin{eqnarray*} x_{1} &=& x_0 - J_f(x_0)^{-1}f(x_0) \\ &=& \begin{bmatrix} 0 \\ 1 \end{bmatrix} - \dfrac{1}{4}\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} \end{eqnarray*}$$

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  • $\begingroup$ Anyone! Could someone help me out with the last term $f(x_0)$! Im getting votes for an incomplete answer :p (the answer is actually a too long comment). So I still do not know what $f(x_0)$ is... $\endgroup$ – onimoni Oct 3 '13 at 8:50
  • $\begingroup$ See @user1772257's answer below. Your original equation is $f(x)=0$, so the left hand side is $f(x)$! $\endgroup$ – Jas Ter Oct 4 '13 at 7:29
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Hint: Your $f(\overrightarrow{\mathbf{x}})$ is $$f(\overrightarrow{\mathbf{x}})=\left[\begin{array}{c} x_{1}^{2}-x_{2}^{2}\\ 2x_{1}x_{2}-1 \end{array}\right]$$ So $$f(\overrightarrow{\mathbf{x_0}})=f\left(\left[\begin{array}{c} 0\\ 1 \end{array}\right]\right)=\left[\begin{array}{c} -1\\ -1 \end{array}\right]$$

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