Find the roots of equation involving $\arctan x$

Question

I try to find the roots of the equation:

$$y=x-2\arctan\left(x\right)$$

I know that one of them is $(0,0)$ but there are two others that should solve

$$\dfrac{x}{2}=\arctan\left(x\right).$$

Is there any easy way to find them?

Thank you

Answer 1

See OEIS A$257451$. There is no known closed form for this number.

Answer 2

I presume you want to sketch and analyse the function. The graph looks like this:

You know that as $x \to \infty$ we have $y \to \infty$ as $x$ dominates the $\arctan x$ term. Similarly, as $x \to -\infty$ we have $y \to -\infty$ for the same reason.

Secondly, you can identify the extrema by solving $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - \dfrac{2}{x^2 + 1} = 0$ yielding the maximum at $x = -1$ and a minimum at $x = 1$.

You can identify an interval for the roots by looking at where $y$ changes sign given that $y$ is a continuous function. The roots occur at $x = 0, x \approx \pm 2.33112$ and that's all you need to sketch the function.

Answer 3

By the Shafer-Fink inequality we have: $$ \frac{3}{1+2\sqrt{1+x^2}}\leq\frac{\arctan x}{x}\leq \frac{\pi}{1+2\sqrt{1+x^2}} \tag{1}$$ so the only positive solution of our equation is between $\frac{1}{2}\sqrt{21}$ and $\frac{1}{2}\sqrt{4\pi^2-4\pi-3}$.

Since the second derivative of $\frac{\arctan x}{x}$ is positive over such interval, the sequence given by the Newton's method:

$$ x_0=\frac{1}{2}\sqrt{21},\quad x_{n+1}=x_n\left(1+\frac{\frac{\arctan x_n}{x_n}-\frac{1}{2}}{\frac{\arctan x_n}{x_n}-\frac{1}{1+x_n^2}}\right)\tag{2}$$ is an increasing sequence that converges very fast to the solution: $$ \xi = 2.33112237\ldots\tag{3} $$