Newton's method - finding suitable starting point I have some trouble solving a problem in my textbook:

Given the following function: $$f(x) = x^{-1} - R$$
  Assume $R > 0$.  Write a short algorithm to find $1/R$ by Newton's method applied to $f$.  Do not use division or exponentiation in the algorithm.  For positive $R$, what starting points are suitable?

OK, so I've managed to solve the first part of the problem.  Using Newton's method and rearranging terms I have gotten:
$$x_{n+1} = x_{n}(2 - Rx_{n})$$
This is correct according to the book, and I can just use my standard algorithm for Newton's method with this expression.  So far so good.
The problem is the second part, where I am supposed to find suitable starting points.  I figured that if $x_{1} = -x_{0}$, then the iterations cycle.  So then I get:
$$\begin{align*}
-x_{0} &= x_{0}(2 - Rx_{0})\\
-3x_{0} &= - Rx_{0}^2\\
-3 &= -Rx_{0}\\
x_{0} &= 3/R
\end{align*}$$
Thus my answer would be that we must have $x_{0} < 3/R$.  My book, however, says:
If $R > 1$ then the starting point $x_{0}$ should be close to $0$ but smaller than $1/R$.
So what is wrong with my reasoning here?  If anyone can help me out, I would really appreciate it!
 A: I presume the restriction on division applies to choosing a starting point as well?
The function $f$ is strictly decreasing on the domain $(0,\infty)$, $\lim_{x \downarrow 0} f(x) = \infty$, $\lim_{x\to \infty} f(x) = -R$, hence there exists a unique $x^*$ such that $f(x^*) = 0$. The picture below shows the behavior for $R=10$.

Furthermore, $f$ is strictly convex on its domain, which means, in this case, if an iteration $x_n$ satisfies $x_n < x^*$, then it is easy to show that $x_n < x_{n+1} \leq x^*$. Furthermore, if $x_n>x^*$, and if $x_{n+1}$ lies in the domain of $f$, then $x_{n+1} \leq x^*$.
So, the only real restriction on a starting point is to ensure that $x_1 \in (0, \infty)$ so that subsequent iterations are well defined. In this case, you have $x_1 = x_0(2-R x_0)$, giving $x_1>0$ iff $x_0 < \frac{2}{R}$.
So the answer is that Newton's method will converge iff you start with $0 < x_0 < \frac{2}{R}$, or, since you are not allowed division, choose $x_0>0$ such that $R x_0 < 2$. 
It is instructive to look at the Newton iteration itself. In this case, $\phi(x) = x(2-Rx)$ defines the iteration scheme (ie, $\phi_{n+1} = \phi(x_n)$). We know the solution is a fixed point of $\phi$, that is $\phi(\frac{1}{R}) = \frac{1}{R}$, which gives $\phi(x) - \phi(\frac{1}{R}) = - R (x-\frac{1}{R})^2$. So we have $|\phi(x) - \phi(\frac{1}{R})| = (R |x-\frac{1}{R}|) |x-\frac{1}{R}|$.
This is a contraction whenever $x \in (0, \frac{2}{R})$. However, as $x$ gets closer to $\frac{1}{R}$, the 'error' term (ie, distance between $x_n$ and the solution $\frac{1}{R}$) drops with the square of the previous error (ignoring the $R$ for simplicity), which gives Newton's method its so called quadratic convergence rate.
A: The points you can converge to are where
$$x=x(2-Rx)$$
$$x=0\  \text{  or  }\ x=\frac 1 R$$
What you want are points where applying the function will get you closer to $\frac 1 R$ than you were previously. So what you want is where:
$$|f(x+\epsilon)-f(x)|<|\epsilon|$$
So if you move $\epsilon$ away from $x$ (here $x=\frac 1 R$), applying the function takes you closer. For this to work in a stable way, it should work for arbitrarily small $\epsilon$. Applying this to your recurrence relation gives:
$$f(x)=x(2-Rx)$$
$$(x+\epsilon)(2-Rx-R\epsilon)-x(2-Rx)<\epsilon$$
$$2\epsilon-2Rx\epsilon-R\epsilon^2<\epsilon$$
$$2-2Rx-R\epsilon<1,\ \ \epsilon>0$$
$$2-2Rx-R\epsilon>1, \epsilon<0$$
plugging in $x=\frac 1 R$, we get that
$$|\epsilon|<\frac 1 R$$
So points within $(0,\frac 2 R)$ will converge to $\frac 1 R$ as intended. From what I can tell (I checked a few values of $R$ on a computer) this seems to work.
A: It works also for $x_0 > 1/R$, but maybe convergence isn't the fastest. Look what happens for various $x-0$ in Newtons iteration, as Ed suggested.


*

*For $x_0 < 0$ solution diverges. Of course: $x_0(2-Rx_0) < x_0$.

*For $x_n \geq 2/R$ you get $x_n < 0$ and solution diverges.


(tangent will always cross x-axis, unless if $f'(x) = 0$ which is only true for $x \rightarrow \infty$)
Now, insert $x_0 = 1/R + \Delta x, \Delta x > 0$:
$$
(1/R + \Delta x)(2 - R(1/R+\Delta x))
$$
$$
1/R (1+R\Delta x)(1-R\Delta x) = 1/R - \Delta x^2 R
$$
Notice, that you get the same result if $x_0 = 1/R - \Delta x$. This proofs, that speed of convergence is the same for $1/R + \Delta x$ and $1/R - \Delta x$, because $x_1$ are the same. Also inserting $\Delta x > \pm 1/R$ gives you divergence.
Maybe there are reasons why should be $x_0 < 1/R$ (for instance speed of convergence). You can experiment with following code in Mathematica or Wolfram Alpha (insert numbers instead of R, x_0, max_n):
NestList[N [# (2 - R*#)] &, x_0, max_n]

A: The function $f(x) = x^{-1}$ is monotonically decreasing for $x > 0$. Newton's method works by following a tangent line of the function at a certain point to the x-axis, computing the zero-crossing of that tangent line, and repeating the process at that new value.
What happens if you pick a large initial value for $x_0$, say $x_0 > 1/R$? Draw that tangent line, and follow it back up to the $x$-axis. It may not cross the $x$-axis at any point $x > 0$. That puts you in a whole different region of the function $1/x$, and you may not ever converge to your root.
