Mixing
Laurent Expansion $frace^zz-1$
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Expand $frace^zz-1$ at $|z-1|>1$
$t=z-1iff t+1=z$
$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$
$frac1t=frac11-1+t$
Is it the way?
complex-analysis laurent-series
asked Jul 16 at 14:19
newhere
759310
add a comment |Â
up vote
0
down vote
favorite
Expand $frace^zz-1$ at $|z-1|>1$
$t=z-1iff t+1=z$
$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$
$frac1t=frac11-1+t$
Is it the way?
complex-analysis laurent-series
asked Jul 16 at 14:19
newhere
759310
1
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
1
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
1
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Expand $frace^zz-1$ at $|z-1|>1$
$t=z-1iff t+1=z$
$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$
$frac1t=frac11-1+t$
Is it the way?
complex-analysis laurent-series
asked Jul 16 at 14:19
newhere
759310
Expand $frace^zz-1$ at $|z-1|>1$
$t=z-1iff t+1=z$
$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$
$frac1t=frac11-1+t$
Is it the way?
complex-analysis laurent-series
asked Jul 16 at 14:19
newhere
759310
asked Jul 16 at 14:19
newhere
759310
asked Jul 16 at 14:19
newhere
759310
asked Jul 16 at 14:19
newhere
759310
759310
1
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
1
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
1
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44
add a comment |Â
1
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
1
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
1
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44
1
1
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
1
1
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
1
1
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Simpler:
Don't forget the radius of convergence of the exponential function if infinite), so you can write
$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$
answered Jul 16 at 14:33
Bernard
110k635103
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Simpler:
Don't forget the radius of convergence of the exponential function if infinite), so you can write
$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$
answered Jul 16 at 14:33
Bernard
110k635103
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
add a comment |Â
up vote
3
down vote
accepted
Simpler:
Don't forget the radius of convergence of the exponential function if infinite), so you can write
$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$
answered Jul 16 at 14:33
Bernard
110k635103
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Simpler:
Don't forget the radius of convergence of the exponential function if infinite), so you can write
$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$
answered Jul 16 at 14:33
Bernard
110k635103
Simpler:
Don't forget the radius of convergence of the exponential function if infinite), so you can write
$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$
answered Jul 16 at 14:33
Bernard
110k635103
answered Jul 16 at 14:33
Bernard
110k635103
answered Jul 16 at 14:33
Bernard
110k635103
answered Jul 16 at 14:33
Bernard
110k635103
110k635103
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
add a comment |Â
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36
1
1
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54
add a comment |Â
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1
Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38
@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41
1
Yes that's all you need to do here.
– Winther
Jul 16 at 14:42
@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44
1
Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44