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