Perform two iterations of Newton's Method on this system. $4{x_1}^2={x_2}^2$ and $4x_1{x_2}^2-x_1=1$.

Question

Perform two iterations of Newton's Method on this system.

Starting with $(0,1)$, $4{x_1}^2={x_2}^2$ and $4x_1{x_2}^2-x_1=1$.

I'm not sure I understand "starting with $(0,1)$". I know how to use Newton's Method and that the following system can be solved for either variable. Any solutions/hints are greatly appreciated.

Answer 1

Write $f(x) = \begin{bmatrix} 4x_1^2-x_2^2 \\ 4 x_1 x_2^2 -x_1 -1\end{bmatrix}$. You want to solve $f(x) = 0$.

Start with $x^0 = (0,1)^T$. Then $x^{n+1} = x^n - f'(x^n)^{-1} f(x^n)$. Repeat twice.

(You can solve the equation any way you want, for example, solve $f'(x^n) \delta = -f(x^n)$, then $x^{n+1} = x^n +\delta$).

Answer 2

Not using matrices in order to stay close to the one-dimensional problem.

You want to solve two equations $$F=4x_1^2-x_2^2=0$$ $$G=4x_1{x_2}^2-x_1-1=0$$ So $$F'_{x_1}=8x_1$$ $$F'_{x_2}=-2x_2$$ $$G'_{x_1}=4x_2^2-1$$ $$G'_{x_2}=8x_1x_2$$ So, just as for first order Taylor expansion around a given point, we can write at a point $(a,b)$ $$0=(4a^2-b^2)+8a (x_1-a)-2b (x_2-b)$$ $$0=(4ab^2-a-1)+(4b^2-1)(x_1-a)+8ab(x_2-b)$$

Solve for $x_1,x_2$ and now replace $a$ by $x_1$, $b$ by $x_2$ and repeat until convergence (or just for two iterations as you are required).