TITLE: Newton's method - finding suitable starting point
QUESTION [6 upvotes]: 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!

REPLY [4 votes]: 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.