Example of meromorphic function with given proerty











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I wanted to construct meromorphic function which has pole at each natural number with residue as same natural number.


I attempted as following $sum_{nin mathbb N}n/(z-n)$.

But problem is that this series is not convergent .Please can some one give suggestion to make above example work by doing some maniputation.

Any Help will be appreciated






complex-analysis examples-counterexamples







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  • Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
    – MPW
    11 hours ago










  • @MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
    – Kavi Rama Murthy
    11 hours ago










  • So it is. Then I would try the approach suggested in the answer(s).
    – MPW
    11 hours ago

















up vote
2
down vote

favorite
2












I wanted to construct meromorphic function which has pole at each natural number with residue as same natural number.


I attempted as following $sum_{nin mathbb N}n/(z-n)$.

But problem is that this series is not convergent .Please can some one give suggestion to make above example work by doing some maniputation.

Any Help will be appreciated






complex-analysis examples-counterexamples







share|cite|improve this question






















  • Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
    – MPW
    11 hours ago










  • @MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
    – Kavi Rama Murthy
    11 hours ago










  • So it is. Then I would try the approach suggested in the answer(s).
    – MPW
    11 hours ago















up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





I wanted to construct meromorphic function which has pole at each natural number with residue as same natural number.


I attempted as following $sum_{nin mathbb N}n/(z-n)$.

But problem is that this series is not convergent .Please can some one give suggestion to make above example work by doing some maniputation.

Any Help will be appreciated






complex-analysis examples-counterexamples







share|cite|improve this question













I wanted to construct meromorphic function which has pole at each natural number with residue as same natural number.


I attempted as following $sum_{nin mathbb N}n/(z-n)$.

But problem is that this series is not convergent .Please can some one give suggestion to make above example work by doing some maniputation.

Any Help will be appreciated






complex-analysis examples-counterexamples




complex-analysis examples-counterexamples






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asked 12 hours ago









Shubham

1,2111518




1,2111518












  • Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
    – MPW
    11 hours ago










  • @MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
    – Kavi Rama Murthy
    11 hours ago










  • So it is. Then I would try the approach suggested in the answer(s).
    – MPW
    11 hours ago




















  • Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
    – MPW
    11 hours ago










  • @MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
    – Kavi Rama Murthy
    11 hours ago










  • So it is. Then I would try the approach suggested in the answer(s).
    – MPW
    11 hours ago


















Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
11 hours ago




Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
11 hours ago












@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
11 hours ago




@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
11 hours ago












So it is. Then I would try the approach suggested in the answer(s).
– MPW
11 hours ago






So it is. Then I would try the approach suggested in the answer(s).
– MPW
11 hours ago












2 Answers
2






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4
down vote



accepted










For fixed $z in Bbb C$ and $n to infty$ we have (using the geometric
series)
$$
frac{n}{z-n} = -1 - frac{z}{n} + Oleft(frac zn right)^2
$$
which suggests to consider the series
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$






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  • Ah yes this is a better answer
    – Richard Martin
    10 hours ago


















up vote
2
down vote













If you can put poles in other places too: $$sum_{n=1}^infty frac{n}{z-n} + frac{n}{z+n} - frac{in}{z-in} - frac{in}{z+in} = sum_{n=1}^infty frac{4n^2}{z^4-n^4} $$ which is convergent as it's 'no worse than' $sum_{n=1}^infty n^{-2}$. But the other answer shows you don't need to.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    For fixed $z in Bbb C$ and $n to infty$ we have (using the geometric
    series)
    $$
    frac{n}{z-n} = -1 - frac{z}{n} + Oleft(frac zn right)^2
    $$
    which suggests to consider the series
    $$
    sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
    = sum_{n=1}^infty frac{z^2}{n(z-n)}
    $$






    share|cite|improve this answer





















    • Ah yes this is a better answer
      – Richard Martin
      10 hours ago















    up vote
    4
    down vote



    accepted










    For fixed $z in Bbb C$ and $n to infty$ we have (using the geometric
    series)
    $$
    frac{n}{z-n} = -1 - frac{z}{n} + Oleft(frac zn right)^2
    $$
    which suggests to consider the series
    $$
    sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
    = sum_{n=1}^infty frac{z^2}{n(z-n)}
    $$






    share|cite|improve this answer





















    • Ah yes this is a better answer
      – Richard Martin
      10 hours ago













    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    For fixed $z in Bbb C$ and $n to infty$ we have (using the geometric
    series)
    $$
    frac{n}{z-n} = -1 - frac{z}{n} + Oleft(frac zn right)^2
    $$
    which suggests to consider the series
    $$
    sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
    = sum_{n=1}^infty frac{z^2}{n(z-n)}
    $$






    share|cite|improve this answer












    For fixed $z in Bbb C$ and $n to infty$ we have (using the geometric
    series)
    $$
    frac{n}{z-n} = -1 - frac{z}{n} + Oleft(frac zn right)^2
    $$
    which suggests to consider the series
    $$
    sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
    = sum_{n=1}^infty frac{z^2}{n(z-n)}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 11 hours ago









    Martin R

    25.7k32945




    25.7k32945












    • Ah yes this is a better answer
      – Richard Martin
      10 hours ago


















    • Ah yes this is a better answer
      – Richard Martin
      10 hours ago
















    Ah yes this is a better answer
    – Richard Martin
    10 hours ago




    Ah yes this is a better answer
    – Richard Martin
    10 hours ago










    up vote
    2
    down vote













    If you can put poles in other places too: $$sum_{n=1}^infty frac{n}{z-n} + frac{n}{z+n} - frac{in}{z-in} - frac{in}{z+in} = sum_{n=1}^infty frac{4n^2}{z^4-n^4} $$ which is convergent as it's 'no worse than' $sum_{n=1}^infty n^{-2}$. But the other answer shows you don't need to.






    share|cite|improve this answer



























      up vote
      2
      down vote













      If you can put poles in other places too: $$sum_{n=1}^infty frac{n}{z-n} + frac{n}{z+n} - frac{in}{z-in} - frac{in}{z+in} = sum_{n=1}^infty frac{4n^2}{z^4-n^4} $$ which is convergent as it's 'no worse than' $sum_{n=1}^infty n^{-2}$. But the other answer shows you don't need to.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        If you can put poles in other places too: $$sum_{n=1}^infty frac{n}{z-n} + frac{n}{z+n} - frac{in}{z-in} - frac{in}{z+in} = sum_{n=1}^infty frac{4n^2}{z^4-n^4} $$ which is convergent as it's 'no worse than' $sum_{n=1}^infty n^{-2}$. But the other answer shows you don't need to.






        share|cite|improve this answer














        If you can put poles in other places too: $$sum_{n=1}^infty frac{n}{z-n} + frac{n}{z+n} - frac{in}{z-in} - frac{in}{z+in} = sum_{n=1}^infty frac{4n^2}{z^4-n^4} $$ which is convergent as it's 'no worse than' $sum_{n=1}^infty n^{-2}$. But the other answer shows you don't need to.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 10 hours ago

























        answered 11 hours ago









        Richard Martin

        1,2588




        1,2588






























             

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