Number of solution of the equation $2\tan^{-1}|x|\cdot\ln|x|=1$ is

Question

This questions can be solved by sketching graphs of both $1/\ln|x|$ and $\tan^{-1} |x|$. Can this be solved by algebra?

Answer 1

Probably not: You can check WA (link) to see the "exact" form doesn't offer anything more by way of algebraic insight. You can observe that if $x$ works then $-x$ does, too; so, the two roots are $\pm 1.6318\ldots$ (and there are no further roots as could be shown by wielding differential calculus).

Answer 2

Just for the fun of it !

As Benjamin Dickman answered, there is no hope for an analytical solution and numerical methods would be required. For example, using $x_0=1$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.00000000000 \\ 1 & 1.63661977237 \\ 2 & 1.63183858797 \\ 3 & 1.63184112271 \end{array} \right)$$ which is the solution for twelve significant figures (I suppose that is how Wolfram Alpha computes the givan solution).

Now, could we get a "nice" looking approximation of it ?

Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants.

For example, built at $x=1$ (this is a very convenient value for the $\tan ^{-1}(x)$ and the $\log(x)$ functions), the $[2,2]$ Padé approximant would be $$2 \tan ^{-1}(x) \log (x)-1\approx\frac{-1+\frac{\left(96-52 \pi -2 \pi ^2+\pi ^3\right) }{2 \left(\pi ^2-48\right)}(x-1)+\frac{\left(-432-156 \pi +22 \pi ^2+3 \pi ^3\right) }{12 \left(\pi ^2-48\right)}(x-1)^2 }{1+\frac{(\pi -6) (8+\pi ) }{\pi ^2-48}(x-1)+\frac{(\pi -6) (12+\pi ) }{6 \left(\pi ^2-48\right)} (x-1)^2}$$ You would be amazed to see how close are the two functions for $1 \leq x \leq 2$.

Now, solving the quadratic in $(x-1)$ which appears in numerator, the solution is $$x_{est}=\frac{1440-56 \pi ^2+\sqrt{1327104+12096 \pi ^2-2544 \pi ^4+36 \pi ^6}}{864+312 \pi -44 \pi ^2-6 \pi ^3}\approx 1.63238$$ that is to say within an error of $0.03$%