Newton's method for $\bigl(\begin{smallmatrix}x^2-y^2+1\\ 2xy\end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}0\\0\end{smallmatrix}\bigr)$?

Question

How do I find first four steps of Newton's method if $f(x,y)=\begin{pmatrix}x^2-y^2+1\\ 2xy\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ and $(x_0,y_0)=(1,1)$ ?

Let $x^2-y^2+1$ be g(x) $\frac{\partial g}{\partial x}=2x$ and $\frac{\partial g}{\partial y}=-2y$.

$h(x)=2xy$, so $\frac{\partial h}{\partial x}=2y$ and $\frac{\partial h}{\partial y}=2x$.

Is the rate of convergence to the solution in accordance with the theory?

Do you understand the exercise? What does the identity $$f(x,y)=\begin{pmatrix}x^2-y^2+1\ 2xy\end{pmatrix}=\begin{pmatrix}0\0\end{pmatrix}$$ even mean? — Did, Jun 18 '16 at 11:10
Since the functions are conjugate harmonics, can't we write the system in complex variables as $z^2+1=0$ and solve that? — Oscar Lanzi, Jun 18 '16 at 11:17
@Oscar, well sure, but I think you're spoiling the point of the exercise... ;) June, again: since you have the partial derivatives, can you now construct the Jacobian, and then its inverse? — J. M. ain't a mathematician, Jun 18 '16 at 11:19
OK. Still has a bearing on what the Jacobian should look like. — Oscar Lanzi, Jun 18 '16 at 11:22
"Is the rate of convergence to the solution in accordance with the theory?" – I hope so. — Christian Blatter, Jun 18 '16 at 19:19

Behrouz Maleki · Accepted Answer · 2016-06-18T19:05:52.133

$$\left\{ \begin{align} & {{f}_{1}}(x,y)={{x}^{2}}-{{y}^{2}}+1 \\ & {{f}_{2}}(x,y)=2xy \\ \end{align} \right.$$ we have $$J=\left| \begin{matrix} \frac{\partial {{f}_{1}}}{\partial x} & \frac{\partial {{f}_{1}}}{\partial y} \\ \frac{\partial {{f}_{2}}}{\partial x} & \frac{\partial {{f}_{2}}}{\partial y} \\ \end{matrix} \right|=\left| \begin{matrix} 2x & -2y \\ 2y & 2x \\ \end{matrix} \right|=4{{x}^{2}}+4{{y}^{2}}\ne 0 , \quad (x,y)\in R^2-{(0,0)}$$ and \begin{align} & {{g}_{1}}(x,y)=x-\frac{\left| \begin{matrix} {{f}_{1}} & \frac{\partial {{f}_{1}}}{\partial y} \\ {{f}_{2}} & \frac{\partial {{f}_{2}}}{\partial y} \\ \end{matrix} \right|}{\left| \begin{matrix} \frac{\partial {{f}_{1}}}{\partial x} & \frac{\partial {{f}_{1}}}{\partial y} \\ \frac{\partial {{f}_{2}}}{\partial x} & \frac{\partial {{f}_{2}}}{\partial y} \\ \end{matrix} \right|}=x-\frac{2{{x}^{3}}+2x{{y}^{2}}+2x}{4{{x}^{2}}+4{{y}^{2}}} \\ & {{g}_{2}}(x,y)=y-\frac{\left| \begin{matrix} \frac{\partial {{f}_{1}}}{\partial x} & {{f}_{1}} \\ \frac{\partial {{f}_{2}}}{\partial x} & {{f}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} \frac{\partial {{f}_{1}}}{\partial x} & \frac{\partial {{f}_{1}}}{\partial y} \\ \frac{\partial {{f}_{2}}}{\partial x} & \frac{\partial {{f}_{2}}}{\partial y} \\ \end{matrix} \right|}=y-\frac{2{{y}^{3}+2{{x}^{2}}y}-2y}{4{{x}^{2}}+4{{y}^{2}}} \\ \end{align} as result \begin{align} & {{x}_{k+1}}={{g}_{1}}({{x}_{k}},{{y}_{k}})={{x}_{k}}-\frac{2{{x}_{k}}^{3}+2{{x}_{k}}{{y}_{k}}^{2}+2{{x}_{k}}}{4{{x}_{k}}^{2}+4{{y}_{k}}^{2}} \\ & {{y}_{k+1}}={{g}_{2}}({{x}_{k}},{{y}_{k}})={{y}_{k}}-\frac{2y_{k}^{3}+2x_{k}^{2}{{y}_{k}}-2{{y}_{k}}}{4{{x}_{k}}^{2}+4{{y}_{k}}^{2}} \\ \end{align}

now let $k=0,1,2,3$ and set $(x_0,y_0)=(1,1)$

In case it wasn't immediately apparent: Behrouz calculated the Jacobian determinant and proceeded to use Cramer's rule. — J. M. ain't a mathematician, Jun 18 '16 at 12:09

Newton's method for $\bigl(\begin{smallmatrix}x^2-y^2+1\\ 2xy\end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}0\\0\end{smallmatrix}\bigr)$?

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