Newton's method arctan Why does it oscillate? I looked up the graph of it and I think it is convergent? And when the function is $0$, the solution is also 0. What is the difference of choosing diverse starting values? Thanks!
 A: Let $\phi(x) = x - {f(x) \over f'(x)} = x - (x^2+1) \arctan x$. This is the Newton update. Define the sequence $x_n$ starting at $x_0$ by $x_{n+1} = \phi(x_n)$,
Note that $\phi$ is odd and $\phi(0) = 0$, $\lim_{x \to -\infty} (\phi(x)-x) = \infty$ and
$\lim_{x \to\infty} (\phi(x)+x) = - \infty$.
We have $\phi'(x) = -2 x \arctan x$, hence $\phi$ is strictly decreasing.
Let $x^*$ be the unique positive $x$ satisfying $\phi(x^*) = -x^*$.
For $0<|x| < |x^*|$, we have $|\phi(x)| < |x|$, and for $|x| < |x^*|$, we have $|\phi(x)| \le |x|$
Hence if $|x_0| < |x^*|$, we have $|x_{n+1}| \le |x_n|$ for all $n$, and so $|x_n|$ converges to some value $y$ and continuity shows that $|y|=|\phi(y)|$,
and so $y=0$ (since $|y| < |x^*|$). Hence $x_n \to 0$ if $|x_0| < |x^*|$.
If $|x_0| = |x^*|$, then the sequence is $x_0, -x_0, x_0,-x_0,...$, which neither
converges nor diverges.
For $|x| > |x^*|$, we have $|\phi(x)| > |x|$, hence $|x_n|$ is non decreasing.
If $|x_n|$ is bounded, then $|x_n| \to y$ for some $y$ and, as above, we have
$|y|=|\phi(y)|$ which is a contradiction since $|y| > |x^*|$. Hence $|x_n|$
diverges to infinity.
It is not hard to estimate $x^* \approx 1.3917$.
A: One of the basic properties of Newton's method is local convergence: if a function is continuously differentiable on a neighborhood of its root, then for any $x_0$ in a (generally smaller) neighborhood of the root, Newton's method converges. Examples like this one show us that it can have very erratic behavior otherwise. This is basically because linear approximation isn't actually good past a short range, and Newton's method is based on linear approximation. 
Your problem shows an example of this: if you start too far away from $x=0$, $\arctan$ is around 1 or so, while its derivative is very small (in particular, much smaller than $1/x$). Consequently the first iteration sends you to a large value of the opposite sign, and then the phenomenon repeats.
