Laurent series of a fractional cosine and sine












0












$begingroup$


i was wondering how i can write the laurent series of




$$cos(dfrac{1}{z-2})$$ in z=2 ?




and if i want the laurent series of $$sin(dfrac{1}{z})$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )






complex-analysis trigonometry taylor-expansion laurent-series







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    0












    $begingroup$


    i was wondering how i can write the laurent series of




    $$cos(dfrac{1}{z-2})$$ in z=2 ?




    and if i want the laurent series of $$sin(dfrac{1}{z})$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )






    complex-analysis trigonometry taylor-expansion laurent-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      i was wondering how i can write the laurent series of




      $$cos(dfrac{1}{z-2})$$ in z=2 ?




      and if i want the laurent series of $$sin(dfrac{1}{z})$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )






      complex-analysis trigonometry taylor-expansion laurent-series







      share|cite|improve this question











      $endgroup$




      i was wondering how i can write the laurent series of




      $$cos(dfrac{1}{z-2})$$ in z=2 ?




      and if i want the laurent series of $$sin(dfrac{1}{z})$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )






      complex-analysis trigonometry taylor-expansion laurent-series




      complex-analysis trigonometry taylor-expansion laurent-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 19:40









      José Carlos Santos

      164k22131234




      164k22131234










      asked Jul 26 '18 at 13:01









      xmaionxxmaionx

      124




      124






















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          0












          $begingroup$

          Since$$cos(z)=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots,$$you have$$cosleft(frac1{z-2}right)=1-frac1{2!(z-2)^2}+frac1{4!(z-2)^4}-cdots$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:19










          • $begingroup$
            How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:31












          • $begingroup$
            so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:32








          • 1




            $begingroup$
            @xmaionx In fact.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:33






          • 1




            $begingroup$
            @xmaionx There's no simple answer for that question.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:54











          Your Answer





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          0












          $begingroup$

          Since$$cos(z)=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots,$$you have$$cosleft(frac1{z-2}right)=1-frac1{2!(z-2)^2}+frac1{4!(z-2)^4}-cdots$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:19










          • $begingroup$
            How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:31












          • $begingroup$
            so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:32








          • 1




            $begingroup$
            @xmaionx In fact.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:33






          • 1




            $begingroup$
            @xmaionx There's no simple answer for that question.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:54
















          0












          $begingroup$

          Since$$cos(z)=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots,$$you have$$cosleft(frac1{z-2}right)=1-frac1{2!(z-2)^2}+frac1{4!(z-2)^4}-cdots$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:19










          • $begingroup$
            How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:31












          • $begingroup$
            so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:32








          • 1




            $begingroup$
            @xmaionx In fact.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:33






          • 1




            $begingroup$
            @xmaionx There's no simple answer for that question.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:54














          0












          0








          0





          $begingroup$

          Since$$cos(z)=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots,$$you have$$cosleft(frac1{z-2}right)=1-frac1{2!(z-2)^2}+frac1{4!(z-2)^4}-cdots$$






          share|cite|improve this answer









          $endgroup$



          Since$$cos(z)=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots,$$you have$$cosleft(frac1{z-2}right)=1-frac1{2!(z-2)^2}+frac1{4!(z-2)^4}-cdots$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 26 '18 at 13:11









          José Carlos SantosJosé Carlos Santos

          164k22131234




          164k22131234












          • $begingroup$
            BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:19










          • $begingroup$
            How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:31












          • $begingroup$
            so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:32








          • 1




            $begingroup$
            @xmaionx In fact.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:33






          • 1




            $begingroup$
            @xmaionx There's no simple answer for that question.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:54


















          • $begingroup$
            BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:19










          • $begingroup$
            How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:31












          • $begingroup$
            so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
            $endgroup$
            – xmaionx
            Jul 26 '18 at 13:32








          • 1




            $begingroup$
            @xmaionx In fact.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:33






          • 1




            $begingroup$
            @xmaionx There's no simple answer for that question.
            $endgroup$
            – José Carlos Santos
            Jul 26 '18 at 13:54
















          $begingroup$
          BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
          $endgroup$
          – xmaionx
          Jul 26 '18 at 13:19




          $begingroup$
          BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
          $endgroup$
          – xmaionx
          Jul 26 '18 at 13:19












          $begingroup$
          How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:31






          $begingroup$
          How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:31














          $begingroup$
          so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
          $endgroup$
          – xmaionx
          Jul 26 '18 at 13:32






          $begingroup$
          so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1{z-3}right)=1-frac1{2!(z-3)^2}+frac1{4!(z-3)^4}-cdots$$
          $endgroup$
          – xmaionx
          Jul 26 '18 at 13:32






          1




          1




          $begingroup$
          @xmaionx In fact.
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:33




          $begingroup$
          @xmaionx In fact.
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:33




          1




          1




          $begingroup$
          @xmaionx There's no simple answer for that question.
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:54




          $begingroup$
          @xmaionx There's no simple answer for that question.
          $endgroup$
          – José Carlos Santos
          Jul 26 '18 at 13:54


















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