Consider the function $f(x)=\arctan(x)$ defined on $\mathbb{R}$. Let $x_0\neq0$ be an initial guess for the root of $f$ and apply Newton's Method with $x_0$. Further assume that the iterations produce an oscillating sequence $(x_n)_n=(x_0,x_1,x_0,x_1,...)$ with period $2$ and $x_0,x_1$ as repeating terms.
Problem: How to prove that $x_0$ and $x_1$ are roots of the function $g(x):=\frac{2x}{x^2+1}-f(x)$ ?
As $g(x)$ is an odd function, if $x_0$ is a root, then $x_1=-x_0$ is a root as well.
The periodic sequence is
$(x_n)=x_0,-x_0,x_0,\ldots$
Apply Newton method to find the root of $f(x)=\arctan x$
$x_1=x_{0}-\frac{f(x_{0})}{f'(x_{0}}= -x_0$
$x_1=-x_0$ so we get
$-x_0=x_0-\frac{\arctan x_0}{\frac{1}{x_0^2+1}}$
expand
$2 x_0=\left(x_0^2+1\right) \arctan x_0$
Which proves that $\pm x_0$ are roots of $g(x)=\frac{2 x}{x^2+1}-\arctan x$
$x_0\approx 1.39175$
Hope this helps
Edit
They say
$x_0,x_1$ form a periodic sequence when used in Newton method to find roots of $\arctan x$. Show that exactly those two values are roots of $g(x)$.
I say: if they are roots of $g(x)$ then $x_1=-x_0$ because $g(x)$ is odd (as shown in the image below). Then I prove that those values are exactly the values found in the first part.
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