Hey can someone please help me with the Newton – Raphson method.
Let f : [a, b] → IR be twice differentiable, that is f’’(x) exists for all x [a, b] and there are positive numbers m and M

|f’(x)| ≥ m and 0 < |f’’(x)| ≤ M for all x [a, b].

We know that f’ and f’’ can’t change sign on [a,b] and we suppose that f(a)f(b) < 0 (that is, one of f(a) or f(b) is negative). Thus, there is a unique r [a, b] with f(r) = 0 and f is one-to-one on [a, b]
We define a sequence {an} by a1 = a if f’f’’< 0 otherwise a1 = b and

$\displaystyle a_n+1 = a_n + (fa_n)/f'(a_n)$ for n $\displaystyle \ge$2

I have already proved that

f(x) = f($\displaystyle x_0$) + $\displaystyle f'$($\displaystyle x_0$)(x − $\displaystyle x_0$) + (1/2)$\displaystyle f''(c)$(x − $\displaystyle x_0$$\displaystyle )^2 $

HOW DO I SHOW that
r < $\displaystyle a_n+1$ < $\displaystyle a_n$ ≤ b for all n. Thus, $\displaystyle {a_n} $ is a decreasing sequence. Let it’s limit be s