Principal part of function at a pole












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I have a function $dfrac{e^zz}{z^2-1}$. It has isolated singularities $z=pm 1$.
To find the principal part at $z=1$, I am trying to find a Laurent series expansion around $z=1$. I have the following mess:
$dfrac{ze^z}{(z-1)(z+1)}=sum_{k=0}^infty dfrac{z^kz}{k!(z+1)(z-1)}=sum_{k=0}^infty dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)}
\
=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^nrbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+1)}{k!((z-1)+2)}
\
=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+2)-lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack}{k!((z-1)+2)}=
\
sum_{k=0}^inftydfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!}- dfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$
.



How can I get a Laurent series just with powers of $(z-1)$ so that I can find the principal part at $z=1$ (that is, the part of the series with negative powers)?










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    1












    $begingroup$


    I have a function $dfrac{e^zz}{z^2-1}$. It has isolated singularities $z=pm 1$.
    To find the principal part at $z=1$, I am trying to find a Laurent series expansion around $z=1$. I have the following mess:
    $dfrac{ze^z}{(z-1)(z+1)}=sum_{k=0}^infty dfrac{z^kz}{k!(z+1)(z-1)}=sum_{k=0}^infty dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)}
    \
    =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^nrbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+1)}{k!((z-1)+2)}
    \
    =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+2)-lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack}{k!((z-1)+2)}=
    \
    sum_{k=0}^inftydfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!}- dfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$
    .



    How can I get a Laurent series just with powers of $(z-1)$ so that I can find the principal part at $z=1$ (that is, the part of the series with negative powers)?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a function $dfrac{e^zz}{z^2-1}$. It has isolated singularities $z=pm 1$.
      To find the principal part at $z=1$, I am trying to find a Laurent series expansion around $z=1$. I have the following mess:
      $dfrac{ze^z}{(z-1)(z+1)}=sum_{k=0}^infty dfrac{z^kz}{k!(z+1)(z-1)}=sum_{k=0}^infty dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)}
      \
      =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^nrbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+1)}{k!((z-1)+2)}
      \
      =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+2)-lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack}{k!((z-1)+2)}=
      \
      sum_{k=0}^inftydfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!}- dfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$
      .



      How can I get a Laurent series just with powers of $(z-1)$ so that I can find the principal part at $z=1$ (that is, the part of the series with negative powers)?










      share|cite|improve this question











      $endgroup$




      I have a function $dfrac{e^zz}{z^2-1}$. It has isolated singularities $z=pm 1$.
      To find the principal part at $z=1$, I am trying to find a Laurent series expansion around $z=1$. I have the following mess:
      $dfrac{ze^z}{(z-1)(z+1)}=sum_{k=0}^infty dfrac{z^kz}{k!(z+1)(z-1)}=sum_{k=0}^infty dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)}
      \
      =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^nrbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+1)}{k!((z-1)+2)}
      \
      =sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+2)-lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack}{k!((z-1)+2)}=
      \
      sum_{k=0}^inftydfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!}- dfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$
      .



      How can I get a Laurent series just with powers of $(z-1)$ so that I can find the principal part at $z=1$ (that is, the part of the series with negative powers)?







      complex-analysis laurent-series singularity






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      edited Dec 10 '18 at 16:22









      José Carlos Santos

      170k23132238




      170k23132238










      asked Dec 10 '18 at 15:50









      HBHSUHBHSU

      302111




      302111






















          3 Answers
          3






          active

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          $begingroup$

          You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you get the first expansion for $dfrac{ze^z}{z+1}$?
            $endgroup$
            – HBHSU
            Dec 10 '18 at 16:40










          • $begingroup$
            You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
            $endgroup$
            – John Hughes
            Dec 10 '18 at 19:05





















          2












          $begingroup$

          You can rewrite your function as



          $$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$



          where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is



          $$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
          &=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
          &=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$



          where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
            $$
            f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
            $$

            Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
            $$
            lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
            $$






            share|cite|improve this answer









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              3 Answers
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              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                How do you get the first expansion for $dfrac{ze^z}{z+1}$?
                $endgroup$
                – HBHSU
                Dec 10 '18 at 16:40










              • $begingroup$
                You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
                $endgroup$
                – John Hughes
                Dec 10 '18 at 19:05


















              3












              $begingroup$

              You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                How do you get the first expansion for $dfrac{ze^z}{z+1}$?
                $endgroup$
                – HBHSU
                Dec 10 '18 at 16:40










              • $begingroup$
                You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
                $endgroup$
                – John Hughes
                Dec 10 '18 at 19:05
















              3












              3








              3





              $begingroup$

              You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.






              share|cite|improve this answer











              $endgroup$



              You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 10 '18 at 19:04









              John Hughes

              64.9k24292




              64.9k24292










              answered Dec 10 '18 at 16:03









              José Carlos SantosJosé Carlos Santos

              170k23132238




              170k23132238












              • $begingroup$
                How do you get the first expansion for $dfrac{ze^z}{z+1}$?
                $endgroup$
                – HBHSU
                Dec 10 '18 at 16:40










              • $begingroup$
                You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
                $endgroup$
                – John Hughes
                Dec 10 '18 at 19:05




















              • $begingroup$
                How do you get the first expansion for $dfrac{ze^z}{z+1}$?
                $endgroup$
                – HBHSU
                Dec 10 '18 at 16:40










              • $begingroup$
                You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
                $endgroup$
                – John Hughes
                Dec 10 '18 at 19:05


















              $begingroup$
              How do you get the first expansion for $dfrac{ze^z}{z+1}$?
              $endgroup$
              – HBHSU
              Dec 10 '18 at 16:40




              $begingroup$
              How do you get the first expansion for $dfrac{ze^z}{z+1}$?
              $endgroup$
              – HBHSU
              Dec 10 '18 at 16:40












              $begingroup$
              You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
              $endgroup$
              – John Hughes
              Dec 10 '18 at 19:05






              $begingroup$
              You compute $f(1), frac{f'(1)}{1!}, frac{f''(1)}{2!}$, etc., perhaps by hand. It is, after all, a Taylor series.
              $endgroup$
              – John Hughes
              Dec 10 '18 at 19:05













              2












              $begingroup$

              You can rewrite your function as



              $$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$



              where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is



              $$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
              &=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
              &=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$



              where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                You can rewrite your function as



                $$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$



                where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is



                $$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
                &=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
                &=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$



                where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You can rewrite your function as



                  $$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$



                  where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is



                  $$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
                  &=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
                  &=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$



                  where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.






                  share|cite|improve this answer











                  $endgroup$



                  You can rewrite your function as



                  $$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$



                  where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is



                  $$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
                  &=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
                  &=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$



                  where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '18 at 9:52

























                  answered Dec 10 '18 at 16:16









                  MasacrosoMasacroso

                  13.2k41747




                  13.2k41747























                      1












                      $begingroup$

                      The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
                      $$
                      f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
                      $$

                      Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
                      $$
                      lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
                        $$
                        f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
                        $$

                        Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
                        $$
                        lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
                          $$
                          f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
                          $$

                          Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
                          $$
                          lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
                          $$
                          f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
                          $$

                          Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
                          $$
                          lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
                          $$







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                          answered Dec 10 '18 at 18:56









                          FormulaWriterFormulaWriter

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