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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?

Martin Sleziak
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user19405892
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    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). – Jack D'Aurizio Nov 06 '16 at 15:30
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    where did you find this question ? – mercio Nov 06 '16 at 17:59
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    @mercio I was solving a problem and needed to prove this as a step in the solution. – user19405892 Nov 06 '16 at 22:39
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    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. – mercio Nov 06 '16 at 22:46
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    I knew this was a hidden duplicate : http://math.stackexchange.com/questions/1615708/is-x-n1-fracx-n2-frac2x-n-bounded – mercio Dec 02 '16 at 14:10
  • I don't understand the problem. Don't all numbers in base 2 contain only 0s and 1s? Also this value in binary is about $0.0010000000101011$ according to wolfram alpha. – 000 May 20 '17 at 23:35

1 Answers1

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NOT A SOLUTION:

Something that may be of use

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

Here,

\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.