Laurent series that does not define holomorphic function: $sum_{n=-infty }^{infty} (z+3i)^n/2^n$
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I have a doubt about this, I think I am right about it, but I want to check since I couldn't find any similar question online. The problem is the following, very early introduction to Laurent Series. It goes like this.
$$sum_{n=-infty }^{infty} frac{(z+3i)^n}{2^n}$$
I want to make sure that, this series indeed does not represent any holomorphic function. Each subseries i.e.
$$sum_{n=0 }^{infty} frac{(z+3i)^n}{2^n}$$
$$sum_{n=1 }^{infty} frac{2^n}{(z+3i)^n}$$
has a radius of convergence: $|z+3i|<2$ for the first one and $|z+3i|>2$. So the set is empty in which this series converges, therefore, there is no holomorphic function defined by it. Thanks in advance.
complex-analysis proof-verification laurent-series holomorphic-functions
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
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add a comment |
$begingroup$
I have a doubt about this, I think I am right about it, but I want to check since I couldn't find any similar question online. The problem is the following, very early introduction to Laurent Series. It goes like this.
$$sum_{n=-infty }^{infty} frac{(z+3i)^n}{2^n}$$
I want to make sure that, this series indeed does not represent any holomorphic function. Each subseries i.e.
$$sum_{n=0 }^{infty} frac{(z+3i)^n}{2^n}$$
$$sum_{n=1 }^{infty} frac{2^n}{(z+3i)^n}$$
has a radius of convergence: $|z+3i|<2$ for the first one and $|z+3i|>2$. So the set is empty in which this series converges, therefore, there is no holomorphic function defined by it. Thanks in advance.
complex-analysis proof-verification laurent-series holomorphic-functions
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
$endgroup$
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Oct 19 '17 at 6:09
add a comment |
$begingroup$
I have a doubt about this, I think I am right about it, but I want to check since I couldn't find any similar question online. The problem is the following, very early introduction to Laurent Series. It goes like this.
$$sum_{n=-infty }^{infty} frac{(z+3i)^n}{2^n}$$
I want to make sure that, this series indeed does not represent any holomorphic function. Each subseries i.e.
$$sum_{n=0 }^{infty} frac{(z+3i)^n}{2^n}$$
$$sum_{n=1 }^{infty} frac{2^n}{(z+3i)^n}$$
has a radius of convergence: $|z+3i|<2$ for the first one and $|z+3i|>2$. So the set is empty in which this series converges, therefore, there is no holomorphic function defined by it. Thanks in advance.
complex-analysis proof-verification laurent-series holomorphic-functions
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
$endgroup$
I have a doubt about this, I think I am right about it, but I want to check since I couldn't find any similar question online. The problem is the following, very early introduction to Laurent Series. It goes like this.
$$sum_{n=-infty }^{infty} frac{(z+3i)^n}{2^n}$$
I want to make sure that, this series indeed does not represent any holomorphic function. Each subseries i.e.
$$sum_{n=0 }^{infty} frac{(z+3i)^n}{2^n}$$
$$sum_{n=1 }^{infty} frac{2^n}{(z+3i)^n}$$
has a radius of convergence: $|z+3i|<2$ for the first one and $|z+3i|>2$. So the set is empty in which this series converges, therefore, there is no holomorphic function defined by it. Thanks in advance.
complex-analysis proof-verification laurent-series holomorphic-functions
complex-analysis proof-verification laurent-series holomorphic-functions
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
edited Dec 14 '18 at 11:37
Brahadeesh
6,46942363
6,46942363
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
asked Oct 19 '17 at 4:17
ivanculetivanculet
907
907
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Oct 19 '17 at 6:09
add a comment |
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Oct 19 '17 at 6:09
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Oct 19 '17 at 6:09
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Oct 19 '17 at 6:09
add a comment |
1 Answer
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From the comments above.
Yes, your argument is correct!
Just one comment: it might be more appropriate to refer to the "disc of convergence" of the Laurent series rather than its "radius of convergence".
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
From the comments above.
Yes, your argument is correct!
Just one comment: it might be more appropriate to refer to the "disc of convergence" of the Laurent series rather than its "radius of convergence".
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add a comment |
$begingroup$
From the comments above.
Yes, your argument is correct!
Just one comment: it might be more appropriate to refer to the "disc of convergence" of the Laurent series rather than its "radius of convergence".
$endgroup$
add a comment |
$begingroup$
From the comments above.
Yes, your argument is correct!
Just one comment: it might be more appropriate to refer to the "disc of convergence" of the Laurent series rather than its "radius of convergence".
$endgroup$
From the comments above.
Yes, your argument is correct!
Just one comment: it might be more appropriate to refer to the "disc of convergence" of the Laurent series rather than its "radius of convergence".
answered Dec 14 '18 at 11:35
community wiki
Brahadeesh
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Your argument is correct.
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– Kavi Rama Murthy
Oct 19 '17 at 6:09