Prove that the binary expansion of $\dfrac{1}{\pi}\tan^{-1}\left(\dfrac{5}{12}\right)$ has strings of $0$s or $1$s of arbitrary length.

I didn't see how we can calculate the binary expansion of $\tan^{-1}(x)$ or $\pi$. Is there some other way of solving this question?

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    $\begingroup$ I do not know if this is really relevant, but since $\arctan(z)=\text{Im}\log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part). $\endgroup$ – Jack D'Aurizio Nov 6 '16 at 15:30
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    $\begingroup$ where did you find this question ? $\endgroup$ – mercio Nov 6 '16 at 17:59
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    $\begingroup$ @mercio I was solving a problem and needed to prove this as a step in the solution. $\endgroup$ – user19405892 Nov 6 '16 at 22:39
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    $\begingroup$ this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove. $\endgroup$ – mercio Nov 6 '16 at 22:46
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    $\begingroup$ I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/… $\endgroup$ – mercio Dec 2 '16 at 14:10


Something that may be of use

\begin{equation} \arctan\left(\frac{x + y}{1 - xy}\right) = \arctan(x) + \arctan(y) \end{equation}


\begin{equation} \frac{x + y}{1 - xy} = \frac{5}{12} \end{equation}

Which has the integer solutions $x,y = -5$

and so,

\begin{equation} \frac{1}{\pi}\left[\arctan\left(\frac{5}{12}\right)\right] = \frac{1}{\pi}\left[\arctan(-5) + \arctan(-5)\right] = -\frac{2}{\pi}\arctan(5) \end{equation}

As before, unsure if this will be of help.

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