Residue at infinity of $frac{e^{z^2}}{z^{2n+1}}$












0












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$$frac{e^{z^2}}{z^{2n+1}}$$
Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to $c_{-1}$ of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?



So $$e^{z^2}z^{-2n-1}$$ should somehow get to $$sum_{k=-infty}^{+infty}{A_k(z-z_0)^k}$$






complex-analysis residue-calculus







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    0












    $begingroup$


    $$frac{e^{z^2}}{z^{2n+1}}$$
    Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to $c_{-1}$ of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?



    So $$e^{z^2}z^{-2n-1}$$ should somehow get to $$sum_{k=-infty}^{+infty}{A_k(z-z_0)^k}$$






    complex-analysis residue-calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $$frac{e^{z^2}}{z^{2n+1}}$$
      Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to $c_{-1}$ of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?



      So $$e^{z^2}z^{-2n-1}$$ should somehow get to $$sum_{k=-infty}^{+infty}{A_k(z-z_0)^k}$$






      complex-analysis residue-calculus







      share|cite|improve this question









      $endgroup$




      $$frac{e^{z^2}}{z^{2n+1}}$$
      Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to $c_{-1}$ of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?



      So $$e^{z^2}z^{-2n-1}$$ should somehow get to $$sum_{k=-infty}^{+infty}{A_k(z-z_0)^k}$$






      complex-analysis residue-calculus




      complex-analysis residue-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 17 '18 at 19:07









      George ZorikovGeorge Zorikov

      273




      273






















          2 Answers
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          $begingroup$

          The residue at infinity of an analytic function $f$ is the residue at $0$ of $frac{-1}{z^2}fleft(frac1zright)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$frac{-1}{z^2}times z^{2n+1}e^{frac1{z^2}},$$which is easy to compute, since$$e^{frac1{z^2}}=1+frac1{z^2}+frac1{2!z^4}+cdots$$






          share|cite|improve this answer









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          • $begingroup$
            Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
            $endgroup$
            – Student
            Dec 18 '18 at 9:55





















          3












          $begingroup$

          Strictly speaking, one finds residues of differentials, not of functions.
          Here the differential is
          $$alpha=frac{exp(z^2)}{z^{2n+1}},dz$$



          To find the residue at $infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1},dw$ is the residue.



          Applying this to $alpha$ gives
          $$alpha=w^{2n+1}exp(1/w^2)left(-frac{dw}{w^2}right)
          =-left(sum_{k=0}^inftyfrac{w^{2n-1}}{k!w^{2k}}right),dw.$$
          The coefficient of $w^{-1},dw$ is $-1/n!$, and that is the residue
          of the differential $alpha$ at $infty$.






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            The residue at infinity of an analytic function $f$ is the residue at $0$ of $frac{-1}{z^2}fleft(frac1zright)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$frac{-1}{z^2}times z^{2n+1}e^{frac1{z^2}},$$which is easy to compute, since$$e^{frac1{z^2}}=1+frac1{z^2}+frac1{2!z^4}+cdots$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
              $endgroup$
              – Student
              Dec 18 '18 at 9:55


















            3












            $begingroup$

            The residue at infinity of an analytic function $f$ is the residue at $0$ of $frac{-1}{z^2}fleft(frac1zright)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$frac{-1}{z^2}times z^{2n+1}e^{frac1{z^2}},$$which is easy to compute, since$$e^{frac1{z^2}}=1+frac1{z^2}+frac1{2!z^4}+cdots$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
              $endgroup$
              – Student
              Dec 18 '18 at 9:55
















            3












            3








            3





            $begingroup$

            The residue at infinity of an analytic function $f$ is the residue at $0$ of $frac{-1}{z^2}fleft(frac1zright)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$frac{-1}{z^2}times z^{2n+1}e^{frac1{z^2}},$$which is easy to compute, since$$e^{frac1{z^2}}=1+frac1{z^2}+frac1{2!z^4}+cdots$$






            share|cite|improve this answer









            $endgroup$



            The residue at infinity of an analytic function $f$ is the residue at $0$ of $frac{-1}{z^2}fleft(frac1zright)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$frac{-1}{z^2}times z^{2n+1}e^{frac1{z^2}},$$which is easy to compute, since$$e^{frac1{z^2}}=1+frac1{z^2}+frac1{2!z^4}+cdots$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 19:14









            José Carlos SantosJosé Carlos Santos

            156k22125227




            156k22125227












            • $begingroup$
              Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
              $endgroup$
              – Student
              Dec 18 '18 at 9:55




















            • $begingroup$
              Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
              $endgroup$
              – Student
              Dec 18 '18 at 9:55


















            $begingroup$
            Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
            $endgroup$
            – Student
            Dec 18 '18 at 9:55






            $begingroup$
            Dear Teacher, ıs it possible to check my solution, if You have a few minutes..?? Thank you very much.. math.stackexchange.com/q/3043357/460967
            $endgroup$
            – Student
            Dec 18 '18 at 9:55













            3












            $begingroup$

            Strictly speaking, one finds residues of differentials, not of functions.
            Here the differential is
            $$alpha=frac{exp(z^2)}{z^{2n+1}},dz$$



            To find the residue at $infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1},dw$ is the residue.



            Applying this to $alpha$ gives
            $$alpha=w^{2n+1}exp(1/w^2)left(-frac{dw}{w^2}right)
            =-left(sum_{k=0}^inftyfrac{w^{2n-1}}{k!w^{2k}}right),dw.$$
            The coefficient of $w^{-1},dw$ is $-1/n!$, and that is the residue
            of the differential $alpha$ at $infty$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Strictly speaking, one finds residues of differentials, not of functions.
              Here the differential is
              $$alpha=frac{exp(z^2)}{z^{2n+1}},dz$$



              To find the residue at $infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1},dw$ is the residue.



              Applying this to $alpha$ gives
              $$alpha=w^{2n+1}exp(1/w^2)left(-frac{dw}{w^2}right)
              =-left(sum_{k=0}^inftyfrac{w^{2n-1}}{k!w^{2k}}right),dw.$$
              The coefficient of $w^{-1},dw$ is $-1/n!$, and that is the residue
              of the differential $alpha$ at $infty$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Strictly speaking, one finds residues of differentials, not of functions.
                Here the differential is
                $$alpha=frac{exp(z^2)}{z^{2n+1}},dz$$



                To find the residue at $infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1},dw$ is the residue.



                Applying this to $alpha$ gives
                $$alpha=w^{2n+1}exp(1/w^2)left(-frac{dw}{w^2}right)
                =-left(sum_{k=0}^inftyfrac{w^{2n-1}}{k!w^{2k}}right),dw.$$
                The coefficient of $w^{-1},dw$ is $-1/n!$, and that is the residue
                of the differential $alpha$ at $infty$.






                share|cite|improve this answer









                $endgroup$



                Strictly speaking, one finds residues of differentials, not of functions.
                Here the differential is
                $$alpha=frac{exp(z^2)}{z^{2n+1}},dz$$



                To find the residue at $infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1},dw$ is the residue.



                Applying this to $alpha$ gives
                $$alpha=w^{2n+1}exp(1/w^2)left(-frac{dw}{w^2}right)
                =-left(sum_{k=0}^inftyfrac{w^{2n-1}}{k!w^{2k}}right),dw.$$
                The coefficient of $w^{-1},dw$ is $-1/n!$, and that is the residue
                of the differential $alpha$ at $infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 19:14









                Lord Shark the UnknownLord Shark the Unknown

                103k1160132




                103k1160132






























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