Proof Verification: Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has...
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This question already has an answer here:
Essential singularity
2 answers
Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=infty$.
This question has been answered in the link Thanks!
My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then
$$exp(f(frac{1}{w}))=sum_{n=0}^{infty}{frac{1}{n!}(f(frac{1}{w}))^{n}}=sum_{n=0}^{infty}{frac{1}{n!}(sum_{m=0}^{infty}{frac{a_{m}}{w^m}})^{n}}$$
We can see that $exp(f(frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!
complex-analysis proof-verification
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marked as duplicate by N. S., Davide Giraudo, mrtaurho, Lord Shark the Unknown, Eevee Trainer Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 2 more comments
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This question already has an answer here:
Essential singularity
2 answers
Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=infty$.
This question has been answered in the link Thanks!
My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then
$$exp(f(frac{1}{w}))=sum_{n=0}^{infty}{frac{1}{n!}(f(frac{1}{w}))^{n}}=sum_{n=0}^{infty}{frac{1}{n!}(sum_{m=0}^{infty}{frac{a_{m}}{w^m}})^{n}}$$
We can see that $exp(f(frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!
complex-analysis proof-verification
$endgroup$
marked as duplicate by N. S., Davide Giraudo, mrtaurho, Lord Shark the Unknown, Eevee Trainer Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
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– Kavi Rama Murthy
Dec 27 '18 at 6:00
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But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
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– user570343
Dec 27 '18 at 6:03
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You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
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– Kavi Rama Murthy
Dec 27 '18 at 6:06
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@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
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– user570343
Dec 27 '18 at 13:44
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But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
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– N. S.
Dec 27 '18 at 21:22
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show 2 more comments
$begingroup$
This question already has an answer here:
Essential singularity
2 answers
Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=infty$.
This question has been answered in the link Thanks!
My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then
$$exp(f(frac{1}{w}))=sum_{n=0}^{infty}{frac{1}{n!}(f(frac{1}{w}))^{n}}=sum_{n=0}^{infty}{frac{1}{n!}(sum_{m=0}^{infty}{frac{a_{m}}{w^m}})^{n}}$$
We can see that $exp(f(frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!
complex-analysis proof-verification
$endgroup$
This question already has an answer here:
Essential singularity
2 answers
Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=infty$.
This question has been answered in the link Thanks!
My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then
$$exp(f(frac{1}{w}))=sum_{n=0}^{infty}{frac{1}{n!}(f(frac{1}{w}))^{n}}=sum_{n=0}^{infty}{frac{1}{n!}(sum_{m=0}^{infty}{frac{a_{m}}{w^m}})^{n}}$$
We can see that $exp(f(frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!
This question already has an answer here:
Essential singularity
2 answers
complex-analysis proof-verification
complex-analysis proof-verification
edited Dec 28 '18 at 1:13
asked Dec 27 '18 at 4:43
user570343
marked as duplicate by N. S., Davide Giraudo, mrtaurho, Lord Shark the Unknown, Eevee Trainer Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by N. S., Davide Giraudo, mrtaurho, Lord Shark the Unknown, Eevee Trainer Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
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– Kavi Rama Murthy
Dec 27 '18 at 6:00
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But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
$endgroup$
– user570343
Dec 27 '18 at 6:03
$begingroup$
You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:06
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@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
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– user570343
Dec 27 '18 at 13:44
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But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
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– N. S.
Dec 27 '18 at 21:22
|
show 2 more comments
$begingroup$
Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:00
$begingroup$
But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
$endgroup$
– user570343
Dec 27 '18 at 6:03
$begingroup$
You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:06
$begingroup$
@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
$endgroup$
– user570343
Dec 27 '18 at 13:44
$begingroup$
But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
$endgroup$
– N. S.
Dec 27 '18 at 21:22
$begingroup$
Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:00
$begingroup$
Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:00
$begingroup$
But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
$endgroup$
– user570343
Dec 27 '18 at 6:03
$begingroup$
But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
$endgroup$
– user570343
Dec 27 '18 at 6:03
$begingroup$
You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:06
$begingroup$
You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:06
$begingroup$
@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
$endgroup$
– user570343
Dec 27 '18 at 13:44
$begingroup$
@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
$endgroup$
– user570343
Dec 27 '18 at 13:44
$begingroup$
But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
$endgroup$
– N. S.
Dec 27 '18 at 21:22
$begingroup$
But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
$endgroup$
– N. S.
Dec 27 '18 at 21:22
|
show 2 more comments
1 Answer
1
active
oldest
votes
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If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
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I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
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I think it's another way of saying it.
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– Chris Custer
Dec 27 '18 at 5:07
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What if the singularity was a pole?
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– N. S.
Dec 27 '18 at 21:23
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I guess I overlooked that possibility.
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– Chris Custer
Dec 27 '18 at 22:36
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Apparently the link you included above has an answer.
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– Chris Custer
Dec 27 '18 at 22:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
$endgroup$
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
add a comment |
$begingroup$
If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
$endgroup$
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
add a comment |
$begingroup$
If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
$endgroup$
If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
edited Dec 27 '18 at 22:46
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
answered Dec 27 '18 at 4:58
Chris CusterChris Custer
13.8k3827
13.8k3827
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
add a comment |
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I saw those answer, but my approach is correct? Thanks
$endgroup$
– user570343
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
I think it's another way of saying it.
$endgroup$
– Chris Custer
Dec 27 '18 at 5:07
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
What if the singularity was a pole?
$endgroup$
– N. S.
Dec 27 '18 at 21:23
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
I guess I overlooked that possibility.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:36
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
$begingroup$
Apparently the link you included above has an answer.
$endgroup$
– Chris Custer
Dec 27 '18 at 22:43
add a comment |
$begingroup$
Your argument is not valid. This question has been asked many times. Apply the following to $f(frac 1 z)$ (with $z_0=0$ math.stackexchange.com/questions/127168/essential-singularity
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:00
$begingroup$
But why, my argument is not valid?? I prove this is a essential singularity. Thanks!
$endgroup$
– user570343
Dec 27 '18 at 6:03
$begingroup$
You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 6:06
$begingroup$
@KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w.
$endgroup$
– user570343
Dec 27 '18 at 13:44
$begingroup$
But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel?
$endgroup$
– N. S.
Dec 27 '18 at 21:22