I am stuck at this problem:
Let $g:[a,b]\to[a,b]$ be a 2 times continuously differentiable function that satisfy: for all $x\in [a,b]$, $g''(x)\neq 0$
And let $s$ be an arbitrary fixed-point of $g$ (I.e. $g(s)=s$) such that $g'(s)=0$
(There exists at least one fixed-point since $g$ is continuous and $g$ maps $[a,b]$ into $[a,b]$)
Now define a sequence $x_{k+1}=g(x_k)$ for all $k\in\Bbb{N}$ and let $x_0$ be an arbitrary point in $[a,b]$.
Show that that there exists some $\zeta\in [a,b]$ such that for all $k\in \Bbb{N}$ we have $x_{k+1}-s=\frac{1}{2}g''(\zeta)(x_k-s)^2$
I've tried using Roll's Theorem and the Mean Value Theorem to show it but it doesn't worked.
Thanks for any hint/help.
We can use Taylor exapesion with remainder to expand $g$ around $x=s$ and we get:
$g(x)=g(s)+g'(s)(x-s)+\frac{1}{2}g''(\zeta)(x-s)^2$ where $\zeta$ is between $x$ and $s$.
If we substitute $x=x_k$ we get:
$g(x_k)=g(s)+g'(s)(x_k-s)+\frac{1}{2}g''(\zeta)(x_k-s)^2=s+0\times (x_k-s)+\frac{1}{2}g''(\zeta)(x_k-s)^2=$
$ = s+\frac{1}{2}g''(\zeta)(x_k-s)^2$
And so $g(x_k)-s=\frac{1}{2}g''(\zeta)(x_k-s)^2$
But since $x_{k+1}=g(x_k)$ we get $x_{k+1}-s=\frac{1}{2}g''(\zeta)(x_k-s)^2$
As was to be shown.