Example of meromorphic function with given proerty
up vote
2
down vote
favorite
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
asked Nov 19 at 9:50
Shubham
1,5941519
add a comment |
up vote
2
down vote
favorite
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
asked Nov 19 at 9:50
Shubham
1,5941519
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
Nov 19 at 10:06
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
asked Nov 19 at 9:50
Shubham
1,5941519
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
asked Nov 19 at 9:50
Shubham
1,5941519
asked Nov 19 at 9:50
Shubham
1,5941519
asked Nov 19 at 9:50
Shubham
1,5941519
asked Nov 19 at 9:50
Shubham
1,5941519
asked Nov 19 at 9:50
Shubham
1,5941519
1,5941519
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
Nov 19 at 10:06
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10
add a comment |
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
Nov 19 at 10:06
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
Nov 19 at 10:06
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
Nov 19 at 10:06
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10
add a comment |
2 Answers
2
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
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$
answered Nov 19 at 10:48
Martin R
26.3k33148
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
add a comment |
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.
answered Nov 19 at 10:05
Richard Martin
1,6648
add a comment |
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
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$
answered Nov 19 at 10:48
Martin R
26.3k33148
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
add a comment |
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
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$
answered Nov 19 at 10:48
Martin R
26.3k33148
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
add a comment |
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
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$
answered Nov 19 at 10:48
Martin R
26.3k33148
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
$$
sum_{n=1}^infty left (frac{n}{z-n} + 1 + frac zn right)
= sum_{n=1}^infty frac{z^2}{n(z-n)}
$$
answered Nov 19 at 10:48
Martin R
26.3k33148
answered Nov 19 at 10:48
Martin R
26.3k33148
answered Nov 19 at 10:48
Martin R
26.3k33148
answered Nov 19 at 10:48
Martin R
26.3k33148
26.3k33148
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
add a comment |
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
Ah yes this is a better answer
– Richard Martin
Nov 19 at 11:52
add a comment |
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.
answered Nov 19 at 10:05
Richard Martin
1,6648
add a comment |
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.
answered Nov 19 at 10:05
Richard Martin
1,6648
add a comment |
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.
answered Nov 19 at 10:05
Richard Martin
1,6648
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.
answered Nov 19 at 10:05
Richard Martin
1,6648
edited Nov 19 at 11:53
answered Nov 19 at 10:05
Richard Martin
1,6648
answered Nov 19 at 10:05
Richard Martin
1,6648
answered Nov 19 at 10:05
Richard Martin
1,6648
1,6648
add a comment |
add a comment |
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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
Nov 19 at 10:06
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
Nov 19 at 10:07
So it is. Then I would try the approach suggested in the answer(s).
– MPW
Nov 19 at 10:10