Simplify $y = frac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$












4












$begingroup$


I'm trying to simplify the following equation:



$y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$



I suspect that a simpler form using complex exponents exists, but I can't find it.



For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.






trigonometry complex-numbers







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    4












    $begingroup$


    I'm trying to simplify the following equation:



    $y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$



    I suspect that a simpler form using complex exponents exists, but I can't find it.



    For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.






    trigonometry complex-numbers







    share|cite|improve this question









    New contributor




    Maarten Baert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      0



      $begingroup$


      I'm trying to simplify the following equation:



      $y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$



      I suspect that a simpler form using complex exponents exists, but I can't find it.



      For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.






      trigonometry complex-numbers







      share|cite|improve this question









      New contributor




      Maarten Baert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I'm trying to simplify the following equation:



      $y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$



      I suspect that a simpler form using complex exponents exists, but I can't find it.



      For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.






      trigonometry complex-numbers




      trigonometry complex-numbers






      share|cite|improve this question









      New contributor




      Maarten Baert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Maarten Baert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      TheSimpliFire

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      asked Mar 31 at 2:20









      Maarten BaertMaarten Baert

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      New contributor





      Maarten Baert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          2 Answers
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          13












          $begingroup$

          $$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
            $endgroup$
            – Maarten Baert
            Mar 31 at 3:21








          • 1




            $begingroup$
            If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
            $endgroup$
            – coreyman317
            Mar 31 at 3:25



















          9












          $begingroup$

          After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
          $$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$.



          Moreover, for large $x$, $coth(x)sim 1+2e^{-2x}$.






          share|cite|improve this answer











          $endgroup$














            Your Answer





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            2 Answers
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            2 Answers
            2






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            13












            $begingroup$

            $$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
              $endgroup$
              – Maarten Baert
              Mar 31 at 3:21








            • 1




              $begingroup$
              If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
              $endgroup$
              – coreyman317
              Mar 31 at 3:25
















            13












            $begingroup$

            $$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
              $endgroup$
              – Maarten Baert
              Mar 31 at 3:21








            • 1




              $begingroup$
              If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
              $endgroup$
              – coreyman317
              Mar 31 at 3:25














            13












            13








            13





            $begingroup$

            $$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$






            share|cite|improve this answer









            $endgroup$



            $$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 31 at 2:30









            coreyman317coreyman317

            1,254522




            1,254522












            • $begingroup$
              Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
              $endgroup$
              – Maarten Baert
              Mar 31 at 3:21








            • 1




              $begingroup$
              If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
              $endgroup$
              – coreyman317
              Mar 31 at 3:25


















            • $begingroup$
              Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
              $endgroup$
              – Maarten Baert
              Mar 31 at 3:21








            • 1




              $begingroup$
              If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
              $endgroup$
              – coreyman317
              Mar 31 at 3:25
















            $begingroup$
            Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
            $endgroup$
            – Maarten Baert
            Mar 31 at 3:21






            $begingroup$
            Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
            $endgroup$
            – Maarten Baert
            Mar 31 at 3:21






            1




            1




            $begingroup$
            If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
            $endgroup$
            – coreyman317
            Mar 31 at 3:25




            $begingroup$
            If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
            $endgroup$
            – coreyman317
            Mar 31 at 3:25











            9












            $begingroup$

            After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
            $$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$.



            Moreover, for large $x$, $coth(x)sim 1+2e^{-2x}$.






            share|cite|improve this answer











            $endgroup$


















              9












              $begingroup$

              After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
              $$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$.



              Moreover, for large $x$, $coth(x)sim 1+2e^{-2x}$.






              share|cite|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
                $$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$.



                Moreover, for large $x$, $coth(x)sim 1+2e^{-2x}$.






                share|cite|improve this answer











                $endgroup$



                After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
                $$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$.



                Moreover, for large $x$, $coth(x)sim 1+2e^{-2x}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered Mar 31 at 3:41









                Claude LeiboviciClaude Leibovici

                125k1158136




                125k1158136






















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