Solve trigonometric equation $\tan x = \arctan x$

Question

What are the non-trivial solutions of $$\tan x = \arctan x$$

Can these solutions be expressed e.g. in terms of $\pi$ or in radicals? I mean are they some "nice" numbers?

E.g. do we know if these solutions are irrational, or rational, some rational multiple of $\pi$, or say something like e.g. $\frac{\sqrt{2}}{7}$?

What about $\cot x = \text{arccot} (x)$? I became curious about these problems after graphing the four functions.

Answer 1

In the range $[0, 2 \pi]$ the graph shows two solutions, $x =0$ and $x \approx 4$:

and a "closeup":

and numerical methods give $x = 4.06759.$

Answer 2

The solutions can not be expressed in elementary functional values. But, approximate analytic solutions are available.

There are infinite number of roots as $\tan^{-1}x$ cuts across the periodic function $\tan x$. Note that $\tan^{-1}x$ approaches $\frac{\pi}{2}$ as $x\rightarrow \infty$, which approximately matches $\tan\left( \frac{\pi}{3}+n\pi\right)$. Therefore, we may assume all positive roots $r_n=\frac{\pi}{3}+n\pi + \delta_n$ and expand the equation

$$\tan x=\tan^{-1}x$$

around $s_n=\frac{\pi}{3}+n\pi$ as follows,

$$\tan s_n + \sec^2 s_n\delta_n = \tan^{-1}s_n + \frac{ \delta_n}{1+s_n^2} $$

which leads to the approximate roots $r_n = s_n + \delta_n$,

$$r_n = s_n -(\sqrt 3 - \tan^{-1}s_n) \frac{1+s_n^2}{3+4s_n^2 }$$

For illustration, the first a few roots are listed below with the exact solutions in parentheses.

$$r_1 = 4.088 \>(4.068) $$ $$r_2 = 7.256 \>(7.245) $$ $$r_3 = 10.408 \> (10.400) $$ $$...$$

For completeness, $r_0 = 0$ is a special root and the negative roots are just the mirror images of the positive ones due to symmetry.

Answer 3

To have some rather good approximations, considering that you look for the zero's of function $$f(x)=\tan (x)-\tan ^{-1}(x)$$ Perform one single iteration of Newton method with $x_0=\frac \pi 3+n \pi=t$ (as @Quanto showed$ to get the estimate $$x_1=t+\frac{2 \left(t^2+1\right) \cos (t) \left(\cos (t) \tan ^{-1}(t)-\sin (t)\right)}{2 t^2-\cos (2 t)+1}$$ Doing the same with Halley method $$x_1=t+\frac{\left(t^2+1\right) \left(\tan ^{-1}(t)-\tan (t)\right) \left(\left(t^2+1\right) \sec ^2(t)-1\right)}{\left(t^4-1\right) \sec ^2(t)+\tan ^{-1}(t) \left(\left(t^2+1\right)^2 \tan (t) \sec ^2(t)+t\right)-t \tan (t)+1}$$

You could do the same with Householder method.

For the first roots, some results

$$\left( \begin{array}{ccccc}n & \text{Newton} & \text{Halley}& \text{Householder} & \text{exact} \\ 1 & 4.088538670 & 4.067078038 & 4.067589392 & 4.067588866 \\ 2 & 7.255833334 & 7.244717199 & 7.244917158 & 7.244916979 \\ 3 & 10.40771565 & 10.39964606 & 10.39976940 & 10.39976930 \\ 4 & 13.55484465 & 13.54818508 & 13.54827702 & 13.54827695 \\ 5 & 16.69989506 & 16.69403891 & 16.69411435 & 16.69411429 \\ 6 & 19.84385223 & 19.83851226 & 19.83857769 & 19.83857764 \\ 7 & 22.98716378 & 22.98218244 & 22.98224120 & 22.98224116 \\ 8 & 26.13006234 & 26.12534424 & 26.12539826 & 26.12539822 \\ 9 & 29.27268077 & 29.26816391 & 29.26821440 & 29.26821437 \\ 10 & 32.41510047 & 32.41074234 & 32.41079012 & 32.41079008 \end{array} \right)$$