Proving that $-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]=-frac12log(2-2cos2x)$












2












$begingroup$


I was looking at this answer and I was confused to see the equality
$$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
Which I am unable to prove.



I have tried
$$
S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
$$
Then with $u=e^{ix}$:
$$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
So
$$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.






algebra-precalculus proof-explanation







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    2












    $begingroup$


    I was looking at this answer and I was confused to see the equality
    $$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
    Which I am unable to prove.



    I have tried
    $$
    S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
    $$
    Then with $u=e^{ix}$:
    $$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
    So
    $$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
    Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.






    algebra-precalculus proof-explanation







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I was looking at this answer and I was confused to see the equality
      $$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
      Which I am unable to prove.



      I have tried
      $$
      S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
      $$
      Then with $u=e^{ix}$:
      $$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
      So
      $$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
      Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.






      algebra-precalculus proof-explanation







      share|cite|improve this question











      $endgroup$




      I was looking at this answer and I was confused to see the equality
      $$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
      Which I am unable to prove.



      I have tried
      $$
      S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
      $$
      Then with $u=e^{ix}$:
      $$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
      So
      $$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
      Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.






      algebra-precalculus proof-explanation




      algebra-precalculus proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 13 at 1:15









      Blue

      49.7k870158




      49.7k870158










      asked Jan 13 at 1:03









      clathratusclathratus

      5,1441439




      5,1441439






















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          $begingroup$

          This step is wrong.




          $$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$




          Correct way:



          begin{align}
          -frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
          &=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
          &=-frac12log(2-2cos2x)
          end{align}






          share|cite|improve this answer









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            $begingroup$

            This step is wrong.




            $$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$




            Correct way:



            begin{align}
            -frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
            &=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
            &=-frac12log(2-2cos2x)
            end{align}






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              This step is wrong.




              $$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$




              Correct way:



              begin{align}
              -frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
              &=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
              &=-frac12log(2-2cos2x)
              end{align}






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                This step is wrong.




                $$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$




                Correct way:



                begin{align}
                -frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
                &=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
                &=-frac12log(2-2cos2x)
                end{align}






                share|cite|improve this answer









                $endgroup$



                This step is wrong.




                $$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$




                Correct way:



                begin{align}
                -frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
                &=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
                &=-frac12log(2-2cos2x)
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 1:12









                Thomas ShelbyThomas Shelby

                4,7382727




                4,7382727






























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