Proofs related to Newton's Method We have the inductive sequence defined by $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
I need to prove the following:
(b) Suppose $f', f'', f(x_1)>0$. Show that $x_1 > x_2 > x_3 >...>c$, where c is the x-intercept of f.
(c) Let $$\delta_k=x_k-c=\frac{f(x_k)}{f'(\epsilon_k)}$$ for some $\epsilon_k\in(c,x_k)$. Show that $$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)}-\frac{f(x_k)}{f'(x_k)}$$ Conclude that:$$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)f'(x_k)}f''(\eta_k)(x_k-\epsilon_k)$$
(d) Let $m=f'(c)=inf(f')$ on $[c,x_1]$ and $M=sup(f'')$ on the same interval. Show that Newton's method works if $x_1-c<\frac{m}{M}$
What I know:
(b) $f(x)$ is certainly positive for all $x\in(c,x_1]$, and so is $f'$, so thus $\frac{f(x)}{f'(x)}$ is positive from $x_1$ to $c$. The rest of the proof should come easily once I can show that any $x_n>c$, as I can then do the rest inductively. So what I'm having trouble with here is proving that c is the lower bound for the sequence.
(c) I know how to prove the first part of the question, as I can simply substitute the expressions for $\delta_{k+1}$ with the corresponding ones for $\delta_k$. However, for the second part of the question, while I can see that it's clearly a proof making use of the mean value theorem derived from the expression for $\delta_{k+1}$, the exact mechanics of the proof are lost on me.
Edit 2: I solved that portion of the proof, but I also need to show that $\delta_{k+1}\leq \frac{f''(\eta_k)}{f'(x_k)}\delta_k^2$
(d) I think I'm supposed to run a convergence test based on the conclusion in (c), but I'm not sure.
Edit: (b) has been solved
 A: Sound like you have already shown that this is true:
$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)}-\frac{f(x_k)}{f'(x_k)}$
And you have figured out that you need to use the mean value theorem to get to:
$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)f'(x_k)}f''(\eta_k)(x_k-\epsilon_k)$.
$\delta_{k+1}=\frac{f(x_k)f'(x_k)}{f'(\epsilon_k)f'(f_k)}-\frac{f(x_k)f'(\epsilon_k)}{f'(\epsilon_k)f'(x_k)}$
$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)f'(x_k)}(f'(x_k)-f'(\epsilon_k))$
$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)f'(x_k)}\frac{f'(x_k)-f'(\epsilon)}{x_k-\epsilon_k}(x_k-\epsilon_k)$
Mean value theorem says $\exists \eta \in(x_k,\epsilon_k)$ such that $f''(\eta) = \frac{f'(x_k)-f'(\epsilon)}{x_k-\epsilon_k}$.
$\delta_{k+1}=\frac{f(x_k)}{f'(\epsilon_k)f'(x_k)}f''(\eta_k)(x_k-\epsilon_k)$.
$\frac{f(x_k)}{f'(\epsilon_k)} = \delta_k$
$c<\epsilon_k<x_k$
$(x_k-\epsilon_k)<(x_k-c)$
$(x_k-\epsilon_k)<\delta_k$
$f''(\eta_k)<M,f'(x_k)>m$
$\delta_{k+1}<\frac{M}{m} \delta_k^2$.
if$\frac{M}{m} \delta_1<n<1$ then $\delta_k<n^k$ and $\delta_k$ converges to $0$.
$\frac{M}{m} \delta_1<1$
$ \delta_1<\frac{m}{M}$
If $x_1-c<\frac{m}{M}$, $x_k$ converges to $c$.
QED
