simple proof $f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$












3














$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
So the thing that I need to do is to show that $f(2019) = pi$.



So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.



Any tips?






analysis functions







share|cite|improve this question





























    3














    $f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
    So the thing that I need to do is to show that $f(2019) = pi$.



    So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.



    Any tips?






    analysis functions







    share|cite|improve this question



























      3












      3








      3







      $f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
      So the thing that I need to do is to show that $f(2019) = pi$.



      So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.



      Any tips?






      analysis functions







      share|cite|improve this question















      $f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
      So the thing that I need to do is to show that $f(2019) = pi$.



      So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.



      Any tips?






      analysis functions




      analysis functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago

























      asked 2 days ago









      Paul Keseru

      494




      494






















          2 Answers
          2






          active

          oldest

          votes


















          2














          hint
          For $z>1$,



          $$f'(z)=$$
          $$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$



          thus $f$ is constant at $[1,+infty)$ and



          $$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$



          or
          $$f(2019)=f(1)=$$
          $$arcsin(1)+2arctan(1)=pi$$






          share|cite|improve this answer























          • and how does the derivative helps me in this case ?
            – Paul Keseru
            2 days ago










          • @PaulKeseru If it is zero, the function is constant.
            – hamam_Abdallah
            2 days ago



















          1














          Start by writing



          $$z=tan(theta)$$



          So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.



          Now the second term will become



          $$=2arctan(tan(theta))=2theta$$



          This is allowed as $theta$ falls within the range of arctan$(x)$



          Now taking the first term



          $$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
          $$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$



          Since $pi/4lttheta lt pi/2$,



          $pi/2lt2theta lt pi$



          This falls outside the range of arcsin$(x)$



          Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$



          Therefore



          $$arcsinleft(sin(2theta)right)=pi - 2theta$$



          Adding the first and second terms together



          $$=pi - 2theta + 2theta = pi$$



          I am attaching a link for more explanation
          $arcsin(sin x)$ explanation?






          share|cite|improve this answer





















            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            hint
            For $z>1$,



            $$f'(z)=$$
            $$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$



            thus $f$ is constant at $[1,+infty)$ and



            $$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$



            or
            $$f(2019)=f(1)=$$
            $$arcsin(1)+2arctan(1)=pi$$






            share|cite|improve this answer























            • and how does the derivative helps me in this case ?
              – Paul Keseru
              2 days ago










            • @PaulKeseru If it is zero, the function is constant.
              – hamam_Abdallah
              2 days ago
















            2














            hint
            For $z>1$,



            $$f'(z)=$$
            $$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$



            thus $f$ is constant at $[1,+infty)$ and



            $$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$



            or
            $$f(2019)=f(1)=$$
            $$arcsin(1)+2arctan(1)=pi$$






            share|cite|improve this answer























            • and how does the derivative helps me in this case ?
              – Paul Keseru
              2 days ago










            • @PaulKeseru If it is zero, the function is constant.
              – hamam_Abdallah
              2 days ago














            2












            2








            2






            hint
            For $z>1$,



            $$f'(z)=$$
            $$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$



            thus $f$ is constant at $[1,+infty)$ and



            $$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$



            or
            $$f(2019)=f(1)=$$
            $$arcsin(1)+2arctan(1)=pi$$






            share|cite|improve this answer














            hint
            For $z>1$,



            $$f'(z)=$$
            $$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$



            thus $f$ is constant at $[1,+infty)$ and



            $$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$



            or
            $$f(2019)=f(1)=$$
            $$arcsin(1)+2arctan(1)=pi$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            hamam_Abdallah

            37.9k21634




            37.9k21634












            • and how does the derivative helps me in this case ?
              – Paul Keseru
              2 days ago










            • @PaulKeseru If it is zero, the function is constant.
              – hamam_Abdallah
              2 days ago


















            • and how does the derivative helps me in this case ?
              – Paul Keseru
              2 days ago










            • @PaulKeseru If it is zero, the function is constant.
              – hamam_Abdallah
              2 days ago
















            and how does the derivative helps me in this case ?
            – Paul Keseru
            2 days ago




            and how does the derivative helps me in this case ?
            – Paul Keseru
            2 days ago












            @PaulKeseru If it is zero, the function is constant.
            – hamam_Abdallah
            2 days ago




            @PaulKeseru If it is zero, the function is constant.
            – hamam_Abdallah
            2 days ago











            1














            Start by writing



            $$z=tan(theta)$$



            So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.



            Now the second term will become



            $$=2arctan(tan(theta))=2theta$$



            This is allowed as $theta$ falls within the range of arctan$(x)$



            Now taking the first term



            $$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
            $$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$



            Since $pi/4lttheta lt pi/2$,



            $pi/2lt2theta lt pi$



            This falls outside the range of arcsin$(x)$



            Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$



            Therefore



            $$arcsinleft(sin(2theta)right)=pi - 2theta$$



            Adding the first and second terms together



            $$=pi - 2theta + 2theta = pi$$



            I am attaching a link for more explanation
            $arcsin(sin x)$ explanation?






            share|cite|improve this answer


























              1














              Start by writing



              $$z=tan(theta)$$



              So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.



              Now the second term will become



              $$=2arctan(tan(theta))=2theta$$



              This is allowed as $theta$ falls within the range of arctan$(x)$



              Now taking the first term



              $$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
              $$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$



              Since $pi/4lttheta lt pi/2$,



              $pi/2lt2theta lt pi$



              This falls outside the range of arcsin$(x)$



              Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$



              Therefore



              $$arcsinleft(sin(2theta)right)=pi - 2theta$$



              Adding the first and second terms together



              $$=pi - 2theta + 2theta = pi$$



              I am attaching a link for more explanation
              $arcsin(sin x)$ explanation?






              share|cite|improve this answer
























                1












                1








                1






                Start by writing



                $$z=tan(theta)$$



                So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.



                Now the second term will become



                $$=2arctan(tan(theta))=2theta$$



                This is allowed as $theta$ falls within the range of arctan$(x)$



                Now taking the first term



                $$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
                $$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$



                Since $pi/4lttheta lt pi/2$,



                $pi/2lt2theta lt pi$



                This falls outside the range of arcsin$(x)$



                Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$



                Therefore



                $$arcsinleft(sin(2theta)right)=pi - 2theta$$



                Adding the first and second terms together



                $$=pi - 2theta + 2theta = pi$$



                I am attaching a link for more explanation
                $arcsin(sin x)$ explanation?






                share|cite|improve this answer












                Start by writing



                $$z=tan(theta)$$



                So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.



                Now the second term will become



                $$=2arctan(tan(theta))=2theta$$



                This is allowed as $theta$ falls within the range of arctan$(x)$



                Now taking the first term



                $$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
                $$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$



                Since $pi/4lttheta lt pi/2$,



                $pi/2lt2theta lt pi$



                This falls outside the range of arcsin$(x)$



                Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$



                Therefore



                $$arcsinleft(sin(2theta)right)=pi - 2theta$$



                Adding the first and second terms together



                $$=pi - 2theta + 2theta = pi$$



                I am attaching a link for more explanation
                $arcsin(sin x)$ explanation?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Sauhard Sharma

                73216




                73216






























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