How to prove the error estimate of the Newton-iteration? I'm trying to get familiar with the Newton-iteration over here but I got stuck at the proof of the error estimate.
Let $f: [a,b] \rightarrow \mathbb{R}$ be continuously differentiable twice, concave or convex and $f' \neq 0 \;\; \forall x \in [a,b]$. Let $\xi$ be the root of $f$. We define the Newton-iteration for $k \in \mathbb{Z}_{\geq 0}$:
$$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$$
Also, we assume $x_1 \in [a,b]$ for $x_0 = a$ and $x_0 = b$.
I already showed that the sequence $(x_k)_{k \in \mathbb{N}}$ converges to $\xi$. Now, I want to show the following error estimate:
$$|\xi - x_{k+1}| \leq \frac{\max_{a \leq x \leq b} |f''(x)|}{2 \min_{a \leq x \leq b} |f'(x)|} (x_{k+1}-x_k)^2$$
I am quite sure I will have to combine the mean value theorem and Taylor's theorem (and Lagrange's remainder), but I have no idea, how to. I don't quite know at what point I should use Taylor's theorem, also, I don't know between which two points I should apply the mean value theorem.
I'd be very happy if somebody could give me a little hint so that I can proceed.
 A: Check out the proof on wikipedia. 
If you just want a hint: call $g(x) = x - \frac{f(x)}{f'(x)}$. Consider $g(x) - g(\xi)$ with the Taylor expansion. You will get something related to $g''(c)$, which is your desired result, upon expanding $g''$. 
A: We first apply the Taylor-expansion for $f(x_{k+1})$ around $x_k$:
$$f(x_{k+1})=f(x_k) + f'(x_k)(x_{k+1}-x_k)+R_2$$
where $R_2$ is the remainder. We'll take Lagrange's remainder and we get to:
$$f(x_{k+1}) = f'(x_k) \cdot (\frac{f(x_k)}{f'(x_k)} - x_k + x_{k+1}) + \frac{1}{2} f''(x_0)(x_{k+1} - x_k)^2$$
where per definition $\frac{f(x)}{f'(x)}-x_k = -x_{k+1}$ and $x_0$ is some point between $x_{k+1}$ and $x_k$. 
So the first term vanishes and we can write:
$$|f(x_{k+1})| \leq \frac{1}{2} \max_{a \leq x \leq b} |f''(x)| (x_{k+1}-x_k)^2$$
Also, from the mean value theorem we know that:
$$\min_{a \leq x \leq b} |f'(x)| \leq \frac{|f(x_{k+1} - f(\xi)|}{|x_{k+1} - \xi|} = \frac{|f(x_{k+1}|}{|x_{k+1} - \xi|}$$
It follows:
$$|\xi - x_{k+1}| \leq \frac{1}{2} \frac{\max_{a \leq x \leq b} |f''(x)|}{\min_{a \leq x \leq b} |f'(x)|} (x_{k+1}-x_k)^2$$
q.e.d.
