How to find the Laurent series of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$?
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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?
complex-analysis complex-numbers
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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?
complex-analysis complex-numbers
asked yesterday
user398843
529215
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Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday
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favorite
up vote
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down vote
favorite
How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?
complex-analysis complex-numbers
asked yesterday
user398843
529215
How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked yesterday
user398843
529215
asked yesterday
user398843
529215
asked yesterday
user398843
529215
asked yesterday
user398843
529215
asked yesterday
user398843
529215
529215
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday
add a comment |
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday
1
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday
add a comment |
1 Answer
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For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.
Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.
Let the coefficient of $z^k$ be $a_k$. Then, it should be :
$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$
Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.
answered yesterday
Rebellos
11.3k21039
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.
Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.
Let the coefficient of $z^k$ be $a_k$. Then, it should be :
$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$
Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.
answered yesterday
Rebellos
11.3k21039
add a comment |
up vote
0
down vote
For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.
Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.
Let the coefficient of $z^k$ be $a_k$. Then, it should be :
$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$
Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.
answered yesterday
Rebellos
11.3k21039
add a comment |
up vote
0
down vote
up vote
0
down vote
For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.
Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.
Let the coefficient of $z^k$ be $a_k$. Then, it should be :
$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$
Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.
answered yesterday
Rebellos
11.3k21039
For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.
Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.
Let the coefficient of $z^k$ be $a_k$. Then, it should be :
$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$
Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.
answered yesterday
Rebellos
11.3k21039
answered yesterday
Rebellos
11.3k21039
answered yesterday
Rebellos
11.3k21039
answered yesterday
Rebellos
11.3k21039
11.3k21039
add a comment |
add a comment |
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Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday