Singularity and behaviour at infinity for complex functionHow to compute the residue of a complex function with essential singularityWhat type of singularity does $exp(fract2 (z - frac1z))$ have on $z = 0, infty$?Find principal part of Laurent expansion of $f(z) = frac1(z^2+1)^2$ about $z=i$.Expansion of a function analytic at infinityClassifying singularities of $f(z)=frac11-sin z$Laurent series for $exp(-x)$ centered at infinityLaurent Series - Self teachinguse Laurent expansion of a function to categorize singularityClassification of singularities of complex multivalued functionRemovable singularity at infinity implies the Laurent series has no positive powers
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Singularity and behaviour at infinity for complex function
How to compute the residue of a complex function with essential singularityWhat type of singularity does $exp(fract2 (z - frac1z))$ have on $z = 0, infty$?Find principal part of Laurent expansion of $f(z) = frac1(z^2+1)^2$ about $z=i$.Expansion of a function analytic at infinityClassifying singularities of $f(z)=frac11-sin z$Laurent series for $exp(-x)$ centered at infinityLaurent Series - Self teachinguse Laurent expansion of a function to categorize singularityClassification of singularities of complex multivalued functionRemovable singularity at infinity implies the Laurent series has no positive powers
$begingroup$
I'm suppose to check the singularities and behaviour at infinity. However, I've never seen that and couln't find something about it online.
So i have a function
$ f(z) = frac1exp(z) -1)-frac1z$
My attempt:
expand in the variable: $u = frac1z$.
So we have:
$f(u) = frac1exp(frac1u)-1-u$
Now I think I am supposed to do an Laurent expansion around $u=0$? But I am not sure.
Would love your input!
complex-analysis infinity
edited Mar 22 at 13:59
Eman
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
$endgroup$
add a comment |
$begingroup$
I'm suppose to check the singularities and behaviour at infinity. However, I've never seen that and couln't find something about it online.
So i have a function
$ f(z) = frac1exp(z) -1)-frac1z$
My attempt:
expand in the variable: $u = frac1z$.
So we have:
$f(u) = frac1exp(frac1u)-1-u$
Now I think I am supposed to do an Laurent expansion around $u=0$? But I am not sure.
Would love your input!
complex-analysis infinity
edited Mar 22 at 13:59
Eman
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
$endgroup$
$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23
add a comment |
$begingroup$
I'm suppose to check the singularities and behaviour at infinity. However, I've never seen that and couln't find something about it online.
So i have a function
$ f(z) = frac1exp(z) -1)-frac1z$
My attempt:
expand in the variable: $u = frac1z$.
So we have:
$f(u) = frac1exp(frac1u)-1-u$
Now I think I am supposed to do an Laurent expansion around $u=0$? But I am not sure.
Would love your input!
complex-analysis infinity
edited Mar 22 at 13:59
Eman
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
$endgroup$
I'm suppose to check the singularities and behaviour at infinity. However, I've never seen that and couln't find something about it online.
So i have a function
$ f(z) = frac1exp(z) -1)-frac1z$
My attempt:
expand in the variable: $u = frac1z$.
So we have:
$f(u) = frac1exp(frac1u)-1-u$
Now I think I am supposed to do an Laurent expansion around $u=0$? But I am not sure.
Would love your input!
complex-analysis infinity
complex-analysis infinity
edited Mar 22 at 13:59
Eman
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
edited Mar 22 at 13:59
Eman
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
edited Mar 22 at 13:59
Eman
4811
edited Mar 22 at 13:59
Eman
4811
edited Mar 22 at 13:59
Eman
4811
4811
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
asked Mar 22 at 8:40
Pernk DernetsPernk Dernets
386
386
$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23
add a comment |
$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23
$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The easiest way goes as follows:
First rewrite the function $f(z)=frac1exp(z)-1-frac1z$ as a Laurent series.
Notice that the $frac1z$ part is already in the form of a Laurent series so we just need to rewrite the $frac1exp(z)-1$ part. This is done by solving the equation:
$frac1exp(z)-1=g(x)$ iff $g(x)cdot(exp(z)-1)=g(x)cdot(1+z+fracz^22+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get:
$(frac1z-frac12+fracz4+O(z^2))(z+fracz^22!+fracz^33!+O(z^4))=1$.
Now you put this back into the function $f$ and you get:
$f(z)=frac1z-frac12+fracz4+O(z^2)-frac1z=-frac12+fracz4+O(z^2)$.
Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!
answered Mar 22 at 13:26
EmanEman
4811
$endgroup$
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The easiest way goes as follows:
First rewrite the function $f(z)=frac1exp(z)-1-frac1z$ as a Laurent series.
Notice that the $frac1z$ part is already in the form of a Laurent series so we just need to rewrite the $frac1exp(z)-1$ part. This is done by solving the equation:
$frac1exp(z)-1=g(x)$ iff $g(x)cdot(exp(z)-1)=g(x)cdot(1+z+fracz^22+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get:
$(frac1z-frac12+fracz4+O(z^2))(z+fracz^22!+fracz^33!+O(z^4))=1$.
Now you put this back into the function $f$ and you get:
$f(z)=frac1z-frac12+fracz4+O(z^2)-frac1z=-frac12+fracz4+O(z^2)$.
Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!
answered Mar 22 at 13:26
EmanEman
4811
$endgroup$
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
add a comment |
$begingroup$
The easiest way goes as follows:
First rewrite the function $f(z)=frac1exp(z)-1-frac1z$ as a Laurent series.
Notice that the $frac1z$ part is already in the form of a Laurent series so we just need to rewrite the $frac1exp(z)-1$ part. This is done by solving the equation:
$frac1exp(z)-1=g(x)$ iff $g(x)cdot(exp(z)-1)=g(x)cdot(1+z+fracz^22+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get:
$(frac1z-frac12+fracz4+O(z^2))(z+fracz^22!+fracz^33!+O(z^4))=1$.
Now you put this back into the function $f$ and you get:
$f(z)=frac1z-frac12+fracz4+O(z^2)-frac1z=-frac12+fracz4+O(z^2)$.
Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!
answered Mar 22 at 13:26
EmanEman
4811
$endgroup$
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
add a comment |
$begingroup$
The easiest way goes as follows:
First rewrite the function $f(z)=frac1exp(z)-1-frac1z$ as a Laurent series.
Notice that the $frac1z$ part is already in the form of a Laurent series so we just need to rewrite the $frac1exp(z)-1$ part. This is done by solving the equation:
$frac1exp(z)-1=g(x)$ iff $g(x)cdot(exp(z)-1)=g(x)cdot(1+z+fracz^22+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get:
$(frac1z-frac12+fracz4+O(z^2))(z+fracz^22!+fracz^33!+O(z^4))=1$.
Now you put this back into the function $f$ and you get:
$f(z)=frac1z-frac12+fracz4+O(z^2)-frac1z=-frac12+fracz4+O(z^2)$.
Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!
answered Mar 22 at 13:26
EmanEman
4811
$endgroup$
The easiest way goes as follows:
First rewrite the function $f(z)=frac1exp(z)-1-frac1z$ as a Laurent series.
Notice that the $frac1z$ part is already in the form of a Laurent series so we just need to rewrite the $frac1exp(z)-1$ part. This is done by solving the equation:
$frac1exp(z)-1=g(x)$ iff $g(x)cdot(exp(z)-1)=g(x)cdot(1+z+fracz^22+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get:
$(frac1z-frac12+fracz4+O(z^2))(z+fracz^22!+fracz^33!+O(z^4))=1$.
Now you put this back into the function $f$ and you get:
$f(z)=frac1z-frac12+fracz4+O(z^2)-frac1z=-frac12+fracz4+O(z^2)$.
Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!
answered Mar 22 at 13:26
EmanEman
4811
answered Mar 22 at 13:26
EmanEman
4811
answered Mar 22 at 13:26
EmanEman
4811
answered Mar 22 at 13:26
EmanEman
4811
4811
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
add a comment |
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
Thank you for Helping me. I havent seen that method before. What is it called? I would love to read more about it
$endgroup$
– Pernk Dernets
Mar 24 at 9:49
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
$begingroup$
I don't think this "recipe" has a name or if it does I'm not aware of it. I suggest that you don't read up about this but find some similar exercises and practice solving them. Note (the rest of the recipe): I mentioned that you do not always end up with a Laurent series with no $z^k$ for $k<0$. If there are finitely many $z^k$ with $k<0$ the singularity is a pole of order $k'$ where $k'$ is the smallest $k$ for which you have a $z^k$ term. The last case is when you have infinitely many $z^k$ terms for $k<0$ . In that case your function has a essential singularity in 0.
$endgroup$
– Eman
Mar 25 at 9:58
add a comment |
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$begingroup$
It has a singularity at each of the points $2npi i$ so $infty$ is not an islolated singularity of $f$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:45
$begingroup$
Thank you. How do you do questions like this? is there a general method?
$endgroup$
– Pernk Dernets
Mar 22 at 9:20
$begingroup$
The usual classification of sigularities is for isolated singularities. So it is not clear as to what kind of answer is expected here.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 9:23