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 19 hours ago
Shubham
1,2111518
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 19 hours ago
Shubham
1,2111518
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago
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 19 hours ago
Shubham
1,2111518
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 19 hours ago
Shubham
1,2111518
asked 19 hours ago
Shubham
1,2111518
asked 19 hours ago
Shubham
1,2111518
asked 19 hours ago
Shubham
1,2111518
asked 19 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
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago
add a comment |
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
18 hours ago
Maybe use terms $a_n/(z-n)^2$ with an appropriate choice of $a_n$ to give the desired residues?
– MPW
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago
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 18 hours ago
Martin R
25.7k32945
Ah yes this is a better answer
– Richard Martin
17 hours ago
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 18 hours ago
Richard Martin
1,2588
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 18 hours ago
Martin R
25.7k32945
Ah yes this is a better answer
– Richard Martin
17 hours ago
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 18 hours ago
Martin R
25.7k32945
Ah yes this is a better answer
– Richard Martin
17 hours ago
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 18 hours ago
Martin R
25.7k32945
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 18 hours ago
Martin R
25.7k32945
answered 18 hours ago
Martin R
25.7k32945
answered 18 hours ago
Martin R
25.7k32945
answered 18 hours ago
Martin R
25.7k32945
25.7k32945
Ah yes this is a better answer
– Richard Martin
17 hours ago
add a comment |
Ah yes this is a better answer
– Richard Martin
17 hours ago
Ah yes this is a better answer
– Richard Martin
17 hours ago
Ah yes this is a better answer
– Richard Martin
17 hours ago
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 18 hours ago
Richard Martin
1,2588
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 18 hours ago
Richard Martin
1,2588
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 18 hours ago
Richard Martin
1,2588
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 18 hours ago
Richard Martin
1,2588
edited 17 hours ago
answered 18 hours ago
Richard Martin
1,2588
answered 18 hours ago
Richard Martin
1,2588
answered 18 hours ago
Richard Martin
1,2588
1,2588
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
18 hours ago
@MPW The residue of this function at $n$ is $0$ whatever $a_n$ is.
– Kavi Rama Murthy
18 hours ago
So it is. Then I would try the approach suggested in the answer(s).
– MPW
18 hours ago