We were set the following question before we had learned about systems of non-linear equations and I'm confused as to how we're meant to answer it using the fixed point method theorems that we have been provided with.
Consider the fixed point iteration $\underline{x}^{n+1} = \underline{\Psi}(\underline{x}^{n})$ where
$\Psi(\underline{x}) = \Psi((x_1,x_2,x_3)^T))$ =
Analyse the convergence of the method to the fixed point $\alpha$ = $(0,\frac{1}{3},0)^T$.
Are we meant to show that $|\Psi(\underline{x})|<1$ and continously differentiable? And if so, do we do this using a Jacobian matrix?
When $x$ is near $\alpha$, the map $\Psi$ can be approximated by the first-order Taylor series $$ \Psi(x) \approx \Psi(\alpha) + J(\alpha) (x - \alpha) = \alpha + J(\alpha) (x - \alpha) $$ where $J(\alpha)$ is the Jacobian matrix evaluated at $x=\alpha$. Note that we have used the fact that $\alpha$ is a fixed point, i.e., $\Psi(\alpha)=\alpha$. From this you can show that that the recursion $$ x^{n+1} = \Psi(x^n) $$ is equivalent to $$ x^{n+1} - \alpha = J(\alpha) (x^n - \alpha) $$ So if you define $y^n = x^n - \alpha$ then $$ y^{n+1} = J(\alpha) y^n $$ so $$ y^n = J(\alpha)^n y^0 $$ where the superscript on $J(\alpha)$ is an exponent rather than a time index. Now the critical question is whether and how quickly $J(\alpha)^n$ shrinks with $n$. This depends on the eigenvalues of $J(\alpha)$.
Basically, the fixed point is stable if all of the eigenvalues of $J(\alpha)$ are less than 1. The convergence error is roughly $\lambda_1^n$, where $\lambda_1$ is the largest eigenvalue.