Suppose I have a function $f$ $$f(x) = 1-\frac{a}{x^2}$$ then the NR substitution method is $$F(x) = \frac{3}{2}x -\frac{x^3}{2a}$$
I now need the know for which values of $x^0$ there is convergence. I.e., a range of starting values. The answer (without any explanation) is apparently $$x \in [-\sqrt{5a},\sqrt{5a}] $$ But I wouldn't know how to begin. (FYI the roots of $f(x)$ is $\pm \sqrt{a}$)
On
Hint.
Assuming $f(x)$ regular as needed, the Newton-Raphson iteration scheme for $f(x) = 0$ is
$$ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = \phi(x_k) $$
so we have
$$ x_{k+1}-x_k = \phi(x_k)-\phi(x_{k-1}) = \phi'(\theta)(x_k-x_{k-1}) $$
with $x_{k-1}\lt \theta\lt x_k$ hence in those conditions (almost geometrically) the sequence $x_k$ converges as long as
$$ |x_{k+1}-x_k| > |x_k-x_{k-1}|\Leftarrow |\phi'(\theta)| < 1 $$
Here $\phi(x) = \frac{3}{2}x -\frac{x^3}{2a}$
You check the usual conditions for a fixed-point iteration, that is, determine the range where $|F'(x)|<1$. As $$ F'(x)=\frac32\left(1-\frac{x^2}a\right)\implies |x^2-a|<\frac23a $$ you get the interval $[\sqrt{\frac a3},\sqrt{\frac{5a}3}]$ and its opposite mirror image.