# How to Find the Matrix in the Simple Iteration Method for Nonlinear Systems

I'm trying to write a c++ programme to find the solutions to a nonlinear system using the simple iteration method described on page 120 here. It says: Given a system of nonlinear equations

$$\left\{\begin{array}{l} f_{1}\left(x_{1}, \ldots, x_{m}\right)=0 \\ f_{2}\left(x_{1}, \ldots, x_{m}\right)=0 \\ \vdots \\ f_{m}\left(x_{1}, \ldots, x_{m}\right)=0 \end{array}\right.$$

If we let $$\mathbf{F}=\left(\begin{array}{c} f_{1}(\mathbf{x}) \\ f_{2}(\mathbf{x}) \\ \vdots \\ f_{m}(\mathbf{x}) \end{array}\right): \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$$

Then we can rewrite the first expression as $$\mathbf{F}(\mathbf{x}) = 0, \qquad \mathbf{x} = \mathbf{G}(\mathbf{x}) \qquad \mathbf{G}: \mathbb{R}^m \to \mathbb R^m$$.

Solution $$\boldsymbol{\alpha}: \boldsymbol{\alpha}=\mathbf{G}(\boldsymbol{\alpha})$$ is called a fixed point of G. Example: $$\mathbf{F}(\mathbf{x})=0 \space ,$$ $$\mathbf{x}=\mathbf{x}-A \mathbf{F}(\mathbf{x})=\mathbf{G}(\mathbf{x}) \quad$$ for some non singular matrix $$A \in \mathbb{R}^{m \times m}$$.

Iteration: initial guess $$x_{0}$$ $$\mathbf{x}_{n+1}=\mathbf{G}\left(\mathbf{x}_{n}\right), \quad n=0,1,2, \ldots$$

An example is given on page 135.

Solve $$\left\{\begin{array}{l}f_{1} \equiv 3 x_{1}^{2}+4 x_{2}^{2}-1=0 \\ f_{2} \equiv x_{2}^{3}-8 x_{1}^{3}-1=0\end{array}, \text { for } \boldsymbol{\alpha} \text { near }\left(x_{1}, x_{2}\right)=(-.5, .25)\right.$$

The iterative solution given is $$\left[\begin{array}{c} x_{1, n+1} \\ x_{2, n+1} \end{array}\right]=\left[\begin{array}{c} x_{1, n} \\ x_{2, n} \end{array}\right]-\left[\begin{array}{cc} .016 & -.17 \\ .52 & -.26 \end{array}\right]\left[\begin{array}{c} 3 x_{1, n}^{2}+4 x_{2, n}^{2}-1 \\ x_{2, n}^{3}-8 x_{1, n}^{3}-1 \end{array}\right]$$

The notes don't explain how to find the matrix A. How can I find the matrix A?

• en.wikipedia.org/wiki/… Mar 28, 2020 at 3:05
• @metamorphy the linked page explains Newton's method, not the method I have described in my post. Mar 28, 2020 at 5:15
• It gives the best possible matrix (namely, the inverse of $\mathbf{F}'(\mathbf{x})$ taken at the solution). You can take any reasonable approximation to it. Mar 28, 2020 at 5:41

$$F(x_1, x_2) = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \dfrac{\partial f_1}{\partial x_2} \\ \dfrac{\partial f_2}{\partial x_1} & \dfrac{\partial f_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 6 x_1 & 8 x_2 \\ -24 x_1^2& 3 x_2^2 \end{bmatrix}$$
Using the given starting point $$(x_1(0), x_2(0)) = (-0.5, 0.25)$$, we have
$$F(-0.5, 0.25) = \begin{bmatrix} -3. & 2. \\ -6. & 0.1875 \end{bmatrix} \implies F^{-1}(-0.5, 0.25) =\begin{bmatrix} 0.0163934 & -0.174863 \\ 0.52459 & -0.262295 \end{bmatrix}$$