Behavior of Newton's method at a specific interval.

Question

I have the equation $f(x)=2-\frac{1}{x}=0$. Does Newton's method converge for $f(x)$ at $x_0=[1/4,3/4]$ , and what happens for $x_0 \leq 0$.

Answer 1

The immediate basin of attraction of an attracting fixed point $p$ for iteration of a continuous function $g$ of one variable is an interval whose endpoints (if finite) can only be

1. non-attracting fixed points
2. points mapped to those non-attracting fixed points
3. points on $2$-cycles, i.e. such that $g(g(x)) = x$.

The closest of these to $p$ on either side of $p$ (if any) are the boundary points of the immediate basin of attraction. If there are none on one side, the basin extends to infinity on that side

Newton's method for this function is the iteration $a(n+1) = 2 a(n) - 2 a(n)^2$. The function $N(x) = 2 x - 2 x^2$ has unstable fixed point at $0$, and stable fixed point at $1/2$. It doesn't have any real $2$-cycles. There is one other point $x$ such that $N(x) = 0$, namely $1$. Thus the immediate basin of attraction of $1/2$ is $(0,1)$, i.e. for every initial value $a_0$ with $0 < a_0 < 1$ Newton's method converges to $1/2$.