The error $|x_{k+1}-x^*|=|g(x_k)-g(x^*)|=|g'(\xi)(x_k-x^*)| $ with fixed point iteration

Question

Today I had a lecture about fixed point iteration. We have a $g(x)$ with one fixed point $x^*$ and a $x_{k+1}=g(x_k)$ on interval $[a,b]$.

During the lecture my professor talked about the error $|x_{k+1}-x^*|=|g(x_k)-g(x^*)|=|g'(\xi)(x_k-x^*)|$. I don't really see how he made the last step.

I tried to find it myself but I didn't get the same outcome. What I tried:

$|x_{k+1}-x^*|=|g(x_k)-g(x^*)|$ because of Taylor we have that, $g(x_k)=g(x^*)+\frac{g'(x*)}{1}(x_k-x*)+...+\frac{g'^r(x^*)}{r!}(x_k-x^*)^r+\frac{g'^{r+1}(\xi)}{\xi!}(x_k-x^*)^{r+1}$

$g(x^*)=x^*$ so $g'(x^*)=0$, so a lot of terms are $0$. This leaves us $g(x_k)=g(x^*)+\frac{g'^{r+1}(\xi)}{\xi!}(x_k-x^*)^{r+1}$

I thought $r=1$ because we have $1$ fixed point.But that gives me $g(x_k)=g(x^*)+\frac{g"(\xi)}{\xi!}(x_k-x^*)^{2}$. And than we would get $ |g(x_k)-g(x^*)|= \frac{g"(\xi)}{\xi!}(x_k-x^*)^{2}$

So something went wrong here. Any help is appreciated!

Answer 1

He assumes differentiability of $g$ and uses the Mean Value theorem, so $\xi$ is a point between $x_k$ and $x^*$.