Mixing
Classification of isolated singularity /Laurent series
$begingroup$
I want to find out, what kind of singularties does
$ f(z)= frac{1}{z^3-z^5}$ have.
I would do the following steps:
$ f(z)= frac{1}{z^3-z^5} = frac{1}{z^3(1-z)(1+z)}$ so I have $ z_1=1, z_2=-1 , z_3=0$
I can argue by the limit test, that $ z_i$ are poles.
For example $ lim_{zrightarrow 1} (1-z)f(z)= lim_{zrightarrow 1} frac{1}{z^3(1+z)}= frac{1}{2} ne 0 $
Am I right?
How can I find the Laurent series of f and classify the singularties without using the Limit test.
laurent-series singularity
asked Jan 21 at 19:01
SvenMathSvenMath
377
$endgroup$
add a comment |
$begingroup$
I want to find out, what kind of singularties does
$ f(z)= frac{1}{z^3-z^5}$ have.
I would do the following steps:
$ f(z)= frac{1}{z^3-z^5} = frac{1}{z^3(1-z)(1+z)}$ so I have $ z_1=1, z_2=-1 , z_3=0$
I can argue by the limit test, that $ z_i$ are poles.
For example $ lim_{zrightarrow 1} (1-z)f(z)= lim_{zrightarrow 1} frac{1}{z^3(1+z)}= frac{1}{2} ne 0 $
Am I right?
How can I find the Laurent series of f and classify the singularties without using the Limit test.
laurent-series singularity
asked Jan 21 at 19:01
SvenMathSvenMath
377
$endgroup$
$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17
add a comment |
$begingroup$
I want to find out, what kind of singularties does
$ f(z)= frac{1}{z^3-z^5}$ have.
I would do the following steps:
$ f(z)= frac{1}{z^3-z^5} = frac{1}{z^3(1-z)(1+z)}$ so I have $ z_1=1, z_2=-1 , z_3=0$
I can argue by the limit test, that $ z_i$ are poles.
For example $ lim_{zrightarrow 1} (1-z)f(z)= lim_{zrightarrow 1} frac{1}{z^3(1+z)}= frac{1}{2} ne 0 $
Am I right?
How can I find the Laurent series of f and classify the singularties without using the Limit test.
laurent-series singularity
asked Jan 21 at 19:01
SvenMathSvenMath
377
$endgroup$
I want to find out, what kind of singularties does
$ f(z)= frac{1}{z^3-z^5}$ have.
I would do the following steps:
$ f(z)= frac{1}{z^3-z^5} = frac{1}{z^3(1-z)(1+z)}$ so I have $ z_1=1, z_2=-1 , z_3=0$
I can argue by the limit test, that $ z_i$ are poles.
For example $ lim_{zrightarrow 1} (1-z)f(z)= lim_{zrightarrow 1} frac{1}{z^3(1+z)}= frac{1}{2} ne 0 $
Am I right?
How can I find the Laurent series of f and classify the singularties without using the Limit test.
laurent-series singularity
laurent-series singularity
asked Jan 21 at 19:01
SvenMathSvenMath
377
asked Jan 21 at 19:01
SvenMathSvenMath
377
asked Jan 21 at 19:01
SvenMathSvenMath
377
asked Jan 21 at 19:01
SvenMathSvenMath
377
asked Jan 21 at 19:01
SvenMathSvenMath
377
377
$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17
add a comment |
$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17
$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=frac1{z^3}timesfrac1{1-z^2}$$and since, near $0$, you have$$frac1{1-z^2}=1+z^2+z^4+z^6+cdots$$then (again, near $0$) you have$$f(z)=frac1{z^3}+frac1z+z+z^3+cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $pm1$.
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=frac1{z^3}timesfrac1{1-z^2}$$and since, near $0$, you have$$frac1{1-z^2}=1+z^2+z^4+z^6+cdots$$then (again, near $0$) you have$$f(z)=frac1{z^3}+frac1z+z+z^3+cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $pm1$.
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
|
show 4 more comments
$begingroup$
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=frac1{z^3}timesfrac1{1-z^2}$$and since, near $0$, you have$$frac1{1-z^2}=1+z^2+z^4+z^6+cdots$$then (again, near $0$) you have$$f(z)=frac1{z^3}+frac1z+z+z^3+cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $pm1$.
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
|
show 4 more comments
$begingroup$
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=frac1{z^3}timesfrac1{1-z^2}$$and since, near $0$, you have$$frac1{1-z^2}=1+z^2+z^4+z^6+cdots$$then (again, near $0$) you have$$f(z)=frac1{z^3}+frac1z+z+z^3+cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $pm1$.
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=frac1{z^3}timesfrac1{1-z^2}$$and since, near $0$, you have$$frac1{1-z^2}=1+z^2+z^4+z^6+cdots$$then (again, near $0$) you have$$f(z)=frac1{z^3}+frac1z+z+z^3+cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $pm1$.
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
edited Jan 22 at 0:17
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 21 at 19:20
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
|
show 4 more comments
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
Thank you:) How do I ge the Laurent series for 1. I don't see how I can do that?
$endgroup$
– SvenMath
Jan 21 at 19:23
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
This one is more complex, but you have$$frac1{-z^3-z^4}=-frac{1}{2}+frac{7 (z-1)}{4}-frac{31}{8}(z-1)^2+cdots$$and therefore$$f(z)=frac1{z-1}timesfrac1{-z^3-z^4}=-frac{1}{2(z-1)}+frac74-frac{31}8(z-1)+cdots$$
$endgroup$
– José Carlos Santos
Jan 21 at 19:29
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
How do you get the series for $frac{1}{-z^3-z^4}$?
$endgroup$
– SvenMath
Jan 21 at 19:34
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I just computed the first $3$ terms of its Taylor series.
$endgroup$
– José Carlos Santos
Jan 21 at 19:38
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
$begingroup$
I see:) For the singularity at $ z=-1$ you do the same thing, but you factor $ z+1$ out.
$endgroup$
– SvenMath
Jan 21 at 19:45
|
show 4 more comments
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$begingroup$
"The Laurent sieres"...there are many: about what point?
$endgroup$
– DonAntonio
Jan 21 at 19:09
$begingroup$
That's the point. There is no explicit point or can i take the difference between singularities as radius of convergence? How would you classify the singularities?
$endgroup$
– SvenMath
Jan 21 at 19:17