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
asked Nov 15 at 7:46
user398843
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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 Nov 15 at 7:46
user398843
536215
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51
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down vote
favorite
up vote
0
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 Nov 15 at 7:46
user398843
536215
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 Nov 15 at 7:46
user398843
536215
asked Nov 15 at 7:46
user398843
536215
asked Nov 15 at 7:46
user398843
536215
asked Nov 15 at 7:46
user398843
536215
asked Nov 15 at 7:46
user398843
536215
536215
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51
add a comment |
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51
1
1
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51
Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51
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 Nov 15 at 9:26
Rebellos
12.2k21041
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
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 Nov 15 at 9:26
Rebellos
12.2k21041
add a comment |
up vote
1
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 Nov 15 at 9:26
Rebellos
12.2k21041
add a comment |
up vote
1
down vote
up vote
1
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 Nov 15 at 9:26
Rebellos
12.2k21041
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 Nov 15 at 9:26
Rebellos
12.2k21041
answered Nov 15 at 9:26
Rebellos
12.2k21041
answered Nov 15 at 9:26
Rebellos
12.2k21041
answered Nov 15 at 9:26
Rebellos
12.2k21041
12.2k21041
add a comment |
add a comment |
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Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
Nov 15 at 7:51