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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.

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  • $\begingroup$ Note $x_{k+1} - s = g(x_k) - g(s)$. Taylor. Lagrange. $\endgroup$ – Daniel Fischer Aug 2 '15 at 21:24
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    $\begingroup$ Form. Of. Remainder. $\endgroup$ – David C. Ullrich Aug 2 '15 at 21:36
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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.

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