Calculating Laurent expansions using geometric series in regions defined by inequalities
$begingroup$
So I am a bit confused about Laurent series and their relation to geometric series. I will give an example but my doubts are more general. The function:
$$f(z) = {frac {{{rm e}^{z}}}{
left( z-1 right) ^{2}}}
$$
has a singularity at $z = 1$. I know that the function converges to a Laurent series only in the region $ left| z-1 right| > 0$ which is an "infinite circle" centered at 1 (right?). So relating to the geometric series:
$ left( z-1 right) ^{-2}$ which is obtained by differentiating $ left( 1-z right) ^{-1}$. However, this geometric series is only valid for $$ left| z right| <1$$ inside the circle with radius $ left| z right|=1$ which is not the same as $ left| z-1 right| > 0$.
My question is: how can I relate a geometric series with a correct Laurent expansion and find the correct geometric series in "the correct" regions defined by a certain inequality?
sequences-and-series complex-analysis laurent-series
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So I am a bit confused about Laurent series and their relation to geometric series. I will give an example but my doubts are more general. The function:
$$f(z) = {frac {{{rm e}^{z}}}{
left( z-1 right) ^{2}}}
$$
has a singularity at $z = 1$. I know that the function converges to a Laurent series only in the region $ left| z-1 right| > 0$ which is an "infinite circle" centered at 1 (right?). So relating to the geometric series:
$ left( z-1 right) ^{-2}$ which is obtained by differentiating $ left( 1-z right) ^{-1}$. However, this geometric series is only valid for $$ left| z right| <1$$ inside the circle with radius $ left| z right|=1$ which is not the same as $ left| z-1 right| > 0$.
My question is: how can I relate a geometric series with a correct Laurent expansion and find the correct geometric series in "the correct" regions defined by a certain inequality?
sequences-and-series complex-analysis laurent-series
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
$endgroup$
$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
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@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32
add a comment |
$begingroup$
So I am a bit confused about Laurent series and their relation to geometric series. I will give an example but my doubts are more general. The function:
$$f(z) = {frac {{{rm e}^{z}}}{
left( z-1 right) ^{2}}}
$$
has a singularity at $z = 1$. I know that the function converges to a Laurent series only in the region $ left| z-1 right| > 0$ which is an "infinite circle" centered at 1 (right?). So relating to the geometric series:
$ left( z-1 right) ^{-2}$ which is obtained by differentiating $ left( 1-z right) ^{-1}$. However, this geometric series is only valid for $$ left| z right| <1$$ inside the circle with radius $ left| z right|=1$ which is not the same as $ left| z-1 right| > 0$.
My question is: how can I relate a geometric series with a correct Laurent expansion and find the correct geometric series in "the correct" regions defined by a certain inequality?
sequences-and-series complex-analysis laurent-series
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
$endgroup$
So I am a bit confused about Laurent series and their relation to geometric series. I will give an example but my doubts are more general. The function:
$$f(z) = {frac {{{rm e}^{z}}}{
left( z-1 right) ^{2}}}
$$
has a singularity at $z = 1$. I know that the function converges to a Laurent series only in the region $ left| z-1 right| > 0$ which is an "infinite circle" centered at 1 (right?). So relating to the geometric series:
$ left( z-1 right) ^{-2}$ which is obtained by differentiating $ left( 1-z right) ^{-1}$. However, this geometric series is only valid for $$ left| z right| <1$$ inside the circle with radius $ left| z right|=1$ which is not the same as $ left| z-1 right| > 0$.
My question is: how can I relate a geometric series with a correct Laurent expansion and find the correct geometric series in "the correct" regions defined by a certain inequality?
sequences-and-series complex-analysis laurent-series
sequences-and-series complex-analysis laurent-series
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
edited Dec 31 '18 at 20:23
daljit97
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
asked Dec 31 '18 at 20:02
daljit97daljit97
178111
178111
$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
$begingroup$
@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32
add a comment |
$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
$begingroup$
@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32
$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
$begingroup$
@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32
add a comment |
2 Answers
2
active
oldest
votes
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Use the fact thatbegin{align}frac{e^z}{(z-1)^2}&=efrac{e^{z-1}}{(z-1)^2}\&=efrac{displaystylesum_{n=0}^inftyfrac{(z-1)^n}{n!}}{(z-1)^2}.end{align}
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
add a comment |
$begingroup$
Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Use the fact thatbegin{align}frac{e^z}{(z-1)^2}&=efrac{e^{z-1}}{(z-1)^2}\&=efrac{displaystylesum_{n=0}^inftyfrac{(z-1)^n}{n!}}{(z-1)^2}.end{align}
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
add a comment |
$begingroup$
Use the fact thatbegin{align}frac{e^z}{(z-1)^2}&=efrac{e^{z-1}}{(z-1)^2}\&=efrac{displaystylesum_{n=0}^inftyfrac{(z-1)^n}{n!}}{(z-1)^2}.end{align}
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
add a comment |
$begingroup$
Use the fact thatbegin{align}frac{e^z}{(z-1)^2}&=efrac{e^{z-1}}{(z-1)^2}\&=efrac{displaystylesum_{n=0}^inftyfrac{(z-1)^n}{n!}}{(z-1)^2}.end{align}
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
Use the fact thatbegin{align}frac{e^z}{(z-1)^2}&=efrac{e^{z-1}}{(z-1)^2}\&=efrac{displaystylesum_{n=0}^inftyfrac{(z-1)^n}{n!}}{(z-1)^2}.end{align}
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 31 '18 at 20:14
José Carlos SantosJosé Carlos Santos
176k24136245
176k24136245
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
add a comment |
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
$begingroup$
Yes that is a more "clever" way of doing this however I was more interested in using the "geometric series" approach.
$endgroup$
– daljit97
Dec 31 '18 at 22:29
add a comment |
$begingroup$
Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
$endgroup$
add a comment |
$begingroup$
Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
$endgroup$
add a comment |
$begingroup$
Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
$endgroup$
Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
edited Dec 31 '18 at 21:13
Davide Giraudo
128k17156268
128k17156268
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
answered Dec 31 '18 at 20:19
Swapnil Swapnil
515
515
add a comment |
add a comment |
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$begingroup$
Couple things: The region $|z-1| > 0$ is the region $zne 1$, it's the punctured plane. Second thing, what is an infinite circle? Also yes, you can find Taylor series for the function by centering the Taylor series around 0 as you're suggesting, and yes it's only valid in a circle with radius the distance to the nearest singularity. Also you simultaneously said the geometric series is and is not valid inside the circle $|z|=1$. The Laurent series is nicer because it's globally valid.
$endgroup$
– jgon
Dec 31 '18 at 20:12
$begingroup$
$frac1{1-z}=1+z+z^2+cdots$ for $|z|<1$. Then replacing $z$ with $z-1$: $frac1{1-(z-1)}=1+(z-1)+(z-1)^2+cdots$, valid for $|z-1|<1$.
$endgroup$
– user170231
Dec 31 '18 at 20:23
$begingroup$
@user170231 yes but that is not valid for $left| z-1 right| > 0$
$endgroup$
– daljit97
Dec 31 '18 at 22:30
$begingroup$
@jgon by infinite circle I mean the entire plane (excluding the singularity).
$endgroup$
– daljit97
Dec 31 '18 at 22:32