Mean Value theorem and Newton's Method I am trying to prove that: given $x_0, x_1, x_2 \ldots$ the sequence of approximations to $\pi$, use the mean value theorem to show that $|\pi-x_{j+1}| = |\tan c_j||\pi - x_j|$, where $c_j$ is some number between $x_j$ and $\pi$. 
So I do the following, but I am not sure if they are right. Let $g(x) = \sin x$. Then let $f(x) = x - \frac{g'(x)}{g(x)} = x - \tan x$. As $f(x)$ is continuous on $(\pi/2,3\pi/2)$ and differentiable on this interval excluding the endpoints, then the mean value theorem says that
$\exists c \in [\frac{\pi}{2}+\epsilon, \frac{3\pi}{2}-\epsilon]$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$ where $b = \frac{3\pi}{2} - \epsilon$, $a= \frac{\pi}{2} +\epsilon$ for some $\epsilon>0$. Now (is this right?) that if $x_0$ is in this interval, then $\exists c_0$ such that $f'(c_0) = \frac{f(\pi)-f(x_0)}{\pi - x_0}$, as $(x_0,\pi)\subset(\frac{\pi}{2} +\epsilon,\frac{2\pi}{2}+\epsilon)$.
(b) Now as $x_0,x_1,\ldots$ are the sequence of approximations of $\pi$ then either $x_0 < x_1 \ldots$ or $x_0 > x_1 \ldots$, depending on whether the $x_i's$ approach $\pi$ from the left or right, as $x - \tan(x)$ is strictly increasing on $(\pi/2,3\pi/2)$. In other words, this means that $(x_0,\pi) \supset (x_1,\pi) \supset \ldots$. (Can I deduce this from the above?)
So here are several bits that are bothering me. If I look at (b) above, it is sort of assuming that my sequence is strictly increasing and bounded above, and hence is convergent (which is what i am supposed to prove later on on). Also, if $f(x)$ is continuous and differentiable on $(\pi/2,3\pi/2)$, then does it mean that $f(x)$ is continuous and differentiable on a subinterval of $(\pi/2,3\pi/2)$?
Thanks,
Ben
 A: Below is my interpretation of what the question might be about, and calculations that should lead you to a complete solution of all the parts of the homework question you were asked, and, I hope, to a better understanding of the situation.  In dealing with a question, it is often crucially important to find out "what's really happening." Your posts show perhaps too great a concentration on "what standard tool shall I use."  The editorial is over, on to the mathematics!
We are using the Newton Method to approximate the solution $x=\pi$ of the equation $\sin x=0$.  Where we start matters a great deal.  Suppose that we start with the estimate $x_0$, where for now we only assume that $x_0$ is in the interval $(\pi/2,3\pi/2)$. (Later we will see that the initial estimate $x_0$ needs to be closer to $\pi$.)
The usual Newton Method iteration produces a sequence $x_0$, $x_1$, $x_2$, and so on, with
$$x_{k+1}=f(x_k)$$ 
where 
$$f(x)=x-\frac{\sin x}{\cos x}=x-\tan x$$
Note that $f'(x)=1-\sec^2 x=-\tan^2 x$.  This is defined in our interval and negative, so $f(x)$ is decreasing (you had it increasing) on our interval.
By the Mean Value Theorem, 
$$\frac{f(\pi)-f(x_k)}{\pi -x_k}=f'(c_k)$$
for some $c_k$ between $x_k$ and $\pi$.
But $f(\pi)-f(x_k)=\pi-x_{k+1}$. It follows that
$$\pi-x_{k+1}=-\tan^2(c_k)(\pi-x_k)$$
There is obvious bad behaviour if $|\tan(c_k)|$ is large. For example, if $x_0$ is only a tiny bit bigger than $\pi/2$, $x_1$ will be far away from our interval.
So now assume that $3\pi/4<x_0<5\pi/4$. The square of the tangent function is less than $1$ on this interval. Since 
$$\pi-x_{k+1}=-\tan^2(c_k)(\pi-x_k)$$
it follows that the distance from $x_{k+1}$ to $\pi$ is less than the distance from $x_k$ to $\pi$.
Now we can see that if $3\pi/4<x_0<5\pi/4$, and if $x_0$ is in the interval $(3\pi/4,5\pi/4)$, then so is $x_1$, but then so is $x_2$, and so on.  From this we can see that the sequence $(x_k)$ has limit $\pi$, by a general "bounded monotone" argument that you seem to have good control of. (Note that the $x_k$ bounce from too high to too low, so be careful!) 
But we can do better than that! 
Since
$$|\pi-x_{k+1}|\le \tan^2(c_0)|\pi-x_k|$$
for all $k$, we can see that 
$$|\pi-x_n|\le \tan^{2n}(c_0)|\pi-x_0|$$
so the convergence to $\pi$ is at least as fast as the convergence of a geometric series.
(The convergence is much faster than that, for the attraction term $\tan^2(c_k)$ approaches $0$ as $x_k$ approaches $\pi$.)  When the Newton Method is good, it is very very good.
