Binary expansion of $frac{1}{pi}tan^{-1}left(frac{5}{12}right)$












8












$begingroup$



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?






number-theory trigonometry decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    where did you find this question ?
    $endgroup$
    – mercio
    Nov 6 '16 at 17:59






  • 1




    $begingroup$
    @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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
    $endgroup$
    – mercio
    Dec 2 '16 at 14:10
















8












$begingroup$



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?






number-theory trigonometry decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    where did you find this question ?
    $endgroup$
    – mercio
    Nov 6 '16 at 17:59






  • 1




    $begingroup$
    @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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
    $endgroup$
    – mercio
    Dec 2 '16 at 14:10














8












8








8


4



$begingroup$



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?






number-theory trigonometry decimal-expansion







share|cite|improve this question











$endgroup$





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?






number-theory trigonometry decimal-expansion




number-theory trigonometry decimal-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 16:47









Martin Sleziak

45.2k11123278




45.2k11123278










asked Nov 6 '16 at 14:05









user19405892user19405892

7,64131057




7,64131057








  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    where did you find this question ?
    $endgroup$
    – mercio
    Nov 6 '16 at 17:59






  • 1




    $begingroup$
    @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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
    $endgroup$
    – mercio
    Dec 2 '16 at 14:10














  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    where did you find this question ?
    $endgroup$
    – mercio
    Nov 6 '16 at 17:59






  • 1




    $begingroup$
    @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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
    $endgroup$
    – mercio
    Dec 2 '16 at 14:10








1




1




$begingroup$
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




$begingroup$
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




3




3




$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59




$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59




1




1




$begingroup$
@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




$begingroup$
@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




1




1




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10




$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10










1 Answer
1






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oldest

votes


















0












$begingroup$

NOT A SOLUTION:



Something that may be of use



begin{equation}
arctanleft(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[arctanleft(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.






share|cite|improve this answer









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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    NOT A SOLUTION:



    Something that may be of use



    begin{equation}
    arctanleft(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[arctanleft(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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      NOT A SOLUTION:



      Something that may be of use



      begin{equation}
      arctanleft(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[arctanleft(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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        NOT A SOLUTION:



        Something that may be of use



        begin{equation}
        arctanleft(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[arctanleft(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.






        share|cite|improve this answer









        $endgroup$



        NOT A SOLUTION:



        Something that may be of use



        begin{equation}
        arctanleft(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[arctanleft(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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 4:09







        user150203





































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