91Ó°ÊÓ

Start by expanding the function \(g(x)\) around the fixed point \(x^*\) using a Taylor's series expansion up to the \((r+1)\)-th derivative term: \(g\left(x\right) = g\left(x^{*}\right)+\sum_{n=1}^{r}\frac{g^{(n)}\left(x^{*}\right)}{n!}(x-x^{*})^n+\frac{g^{(r+1)}\left(\xi\right)}{(r+1)!}(x-x^{*})^{r+1}\), where \(x^{*}\) is the fixed point and \(\xi\) is some point between \(x\) and \(x^{*}\). Since \(x^{*}\) is the fixed point, it satisfies the fixed-point equation \(x^{*}=g\left(x^{*}\right)\). Also, for \(1 \leq n \leq r\), it is given that \(g^{(n)}\left(x^{*}\right)=0\). Using these two facts, we simplify the Taylor's series expansion: \(g\left(x\right) = x^{*}+\frac{g^{(r+1)}\left(\xi\right)}{(r+1)!}(x-x^{*})^{r+1}\).

The fixed point iteration is given by: \(x_{k+1}=g\left(x_{k}\right)\quad\text{for }k=0,1,\ldots\). Substitute our simplified Taylor's series expansion for \(g\left(x_{k}\right)\): \(x_{k+1}=x^{*}+\frac{g^{(r+1)}\left(\xi_{k}\right)}{(r+1)!}\left(x_{k}-x^{*}\right)^{r+1}\), where \(\xi_{k}\) is some point between \(x_{k}\) and \(x^{*}\).

To find the convergence properties, rearrange the equation to isolate the error term \(e_{k}=x_{k}-x^{*}\): \(e_{k+1}=\frac{g^{(r+1)}\left(\xi_{k}\right)}{(r+1)!}e_{k}^{r+1}\). Taking the absolute value of both sides, we get: \(|e_{k+1}|=\frac{|g^{(r+1)}(\xi_{k})|}{(r+1)!}|e_{k}|^{r+1}\). As the iteration converges, the error term \(|e_{k}|\) goes to zero, so the ratio \(\frac{|e_{k+1}|}{|e_{k}|^{r+1}}\) approaches a constant value: \(\lim_{k\to\infty}\frac{|e_{k+1}|}{|e_k|^{r+1}}=\frac{1}{(r+1)!}|g^{(r+1)}(x^*)|\).

We've found that the order of convergence of the fixed point iteration is at least \((r+1)\) if the first \(r\) derivatives of \(g\) vanish at \(x = x^*\), and the \((r+1)\)-th derivative does not vanish. The fixed point iteration is expected to converge superlinearly (faster than linear rate), and the order of convergence is at least \((r+1)\).