Problem 14 of Chapter 22, "Infinite Sequences", of Spivak's Calculus is a bit long. It has five items, and my question is about implicit assumptions in item $(d)$.
Let me show item $(d)$ straight away to cut to the chase, and then show $(a)-(c)$.
14.(d) Let $m=f'(c)=\inf(f')$ on $[c,x_1]$ and let $M=\sup(f'')$ on $[c,x_1]$. Show that Newton's method works if $$x_1-c<\frac{m}{M}\tag{1}$$.
Here is the beginning of the solution from the solution manual. My questions are about hidden/implicit assumptions.
$$\delta_1=x_1-c<\frac{m}{M}\tag{2}$$
Then
$$\frac{M}{m}\delta_1=\alpha<1\tag{3}$$
for some $\alpha$.
"For some $\alpha$" sounds weird. $\alpha$ is simply being defined as the expression $\frac{M}{m}\delta_1$, which is $<1$.
How do we know that $\frac{m}{M}>0$?
By assumption, we are on the interval $[c,x_1]$ and $f'(c)$ is the smallest slope on this interval. Hence $f''(c)$ must be $\geq 0$.
If $f''(c)=0$ then the expression $(1)$ doesn't make sense, so for the purposes of the current problem we needn't consider it (note however, that it isn't a problem in general for $f''(c)=0$).
But what if $f'(c)<0$ and $f''(c)=\sup{(f'')}>0$? Then $\frac{m}{M}<0$.
But then it seems we can't get $(3)$.
Is this line of reasoning correct?
The rest of the proof relies critically on $(3)$. Thus, it seems that the result we are trying to prove isn't a general case. It is the case in which $f'(c)>0$.
Now let me take a step back and show what happens in the previous items (usually items build results that are useful to solve later items).
Items $(a)$ and $(b)$ serve to introduce Newton's Method
In (a), we are asked to show that the tangent line to the graph of $f$ at $(x_1,f(x_1))$ intersects the horizontal axis at $(x_2,0)$, where
$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}$$
If we keep repeating this process of obtaining the tangent and figuring out where it intersects the $x$-axis, we get a sequence $\{x_n\}$ defined by
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
We hope that this sequence converges to a root $c$ of $f$.
In (b), we are asked to show that if $f',f''>0$, and we choose $x_1$ with $f(x_1)>0$ then $x_1>x_2>x_3>...>c$.
This is a special case of a convex increasing function. If we start the sequence at a value where $f$ is positive, then the sequence is decreasing. Now, it seems to me that $c$ need not exist (e.g. $e^x$). In any case, it is a simple matter to show that
$$x_1>x_2>x_3>...>x_n>...$$
Here is item (c) in full
Let $\delta_k=x_k-c$. Then
$$\delta_k=\frac{f(x_k)}{f'(\xi_k)}$$
for some $\xi_k\in (c,x_k)$. Show that
$$\delta_{k+1}=\frac{f(x_k)}{f'(\xi_k)}-\frac{f(x_k)}{f'(x_k)}$$
Conclude that
$$\delta_{k+1}=\frac{f(x_k)}{f'(\xi_k)f'(x_k)}\cdot f''(\eta_k)(x_k-\xi_k)$$
for some $\eta_k\in (c,x_k)$
and then that $$\delta_{k+1}\leq \frac{f''(\eta_k)}{f'(x_k)}\delta_k^2$$
I went through the calculations to obtain these results, though the latter aren't at this point particularly enlightening.
I've been going through the text and I ran into similar questions. I think a cleaner set of assumptions for part b) (that follow through to the later parts) is
With these assumptions part b and c work out. In part d, if $x_1-c < m/M$, then it must be that $m=f'(c) >0$, since otherwise $x_1 \leq c$. This can't occur because $f$ is increasing on $[x_1,\infty)$.