I need a check on the following exercise, since I don't have the solution
Let $g \colon \mathbb{R} \rightarrow \mathbb{R}$ defined as $g(x)=\sqrt{1+x^2}$. Show that Newton iterates for the equation $g'(x)=0$ satisfy: $$|x_0|<1 \Longrightarrow g(x_{k+1})<g(x_k)$$ and > $$\lim_{k \rightarrow \infty} x_k = 0$$
Here's my attempt:
First I write the Newton iterates for $g'(x)=0$, which is $$x_{k+1}=\frac{x_k^3}{1+2x_k^2}$$ Now, I note that $\frac{x_k^2}{1+2x^2} \in (0,\frac{1}{2})$ for every $x_k$. Therefore, if I take an initial data $|x_0|<1$, then $|x_1|<1$ as well. Generalizing, every time I multiply for a number less than $\frac{1}{2}$ a number less than $1$, therefore this sequence is monotone.
If $x_0 \in (0,1)$, I have that such a sequence bounded from below by $0$ and it's decreasing, hence $\lim_k x_k=0$
On the other hand, if $x_0 \in (-1,0)$, then the sequence is bounded from above by $0$ and it's increasing. Therefore $\lim_k x_k = 0$.
Since $x \mapsto g(x)$ is increasing for $x \in (0,+\infty)$ and decreasing in $(-\infty,0)$, I have that:
- If $x_0 \in (-1,0)$ ( i.e $\{ x_k \}_k$ increasing and lives in $(-1,0)$): $x_{k+1} > x_k$, but if I apply $g$ (which is decreasing): $g(x_{k+1})<g(x_k)$
- If $x_0 \in (0,1)$, (i.e. $x\{ x_k \}$ decreasing): $g(x_{k+1})<g(x_k)$ since $g$ here preserves the monotonicity.
Hope it's everything fine.
The key is noting that $|x_{k+1}|\le|x_k|/2$ so that $|x_k|\le|x_0|/2^k$. From this, what can you conclude?