Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$ [closed]
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Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$
I have simplified the expression to:
$$ operatorname{Log}frac{1}{1-frac{1}{z^2}} $$
but I am not sure what to do next. Is this a good approach to this problem?
complex-analysis logarithms laurent-series
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
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closed as off-topic by Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer Jan 11 at 0:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$
I have simplified the expression to:
$$ operatorname{Log}frac{1}{1-frac{1}{z^2}} $$
but I am not sure what to do next. Is this a good approach to this problem?
complex-analysis logarithms laurent-series
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
$endgroup$
closed as off-topic by Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer Jan 11 at 0:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
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Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
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– José Carlos Santos
Jan 10 at 15:50
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sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55
add a comment |
$begingroup$
Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$
I have simplified the expression to:
$$ operatorname{Log}frac{1}{1-frac{1}{z^2}} $$
but I am not sure what to do next. Is this a good approach to this problem?
complex-analysis logarithms laurent-series
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
$endgroup$
Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$
I have simplified the expression to:
$$ operatorname{Log}frac{1}{1-frac{1}{z^2}} $$
but I am not sure what to do next. Is this a good approach to this problem?
complex-analysis logarithms laurent-series
complex-analysis logarithms laurent-series
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
edited Jan 10 at 18:45
Lord_Farin
15.6k636108
15.6k636108
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
asked Jan 10 at 15:41
s.kovalskas.kovalska
267
267
closed as off-topic by Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer Jan 11 at 0:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer Jan 11 at 0:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, José Carlos Santos, Adrian Keister, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
$endgroup$
– José Carlos Santos
Jan 10 at 15:50
$begingroup$
sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55
add a comment |
$begingroup$
Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
$endgroup$
– José Carlos Santos
Jan 10 at 15:50
$begingroup$
sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55
$begingroup$
Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
$endgroup$
– José Carlos Santos
Jan 10 at 15:50
$begingroup$
Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
$endgroup$
– José Carlos Santos
Jan 10 at 15:50
$begingroup$
sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55
$begingroup$
sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55
add a comment |
1 Answer
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Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$
answered Jan 10 at 15:52
aledenaleden
2,499511
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$
answered Jan 10 at 15:52
aledenaleden
2,499511
$endgroup$
add a comment |
$begingroup$
Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$
answered Jan 10 at 15:52
aledenaleden
2,499511
$endgroup$
add a comment |
$begingroup$
Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$
answered Jan 10 at 15:52
aledenaleden
2,499511
$endgroup$
Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$
answered Jan 10 at 15:52
aledenaleden
2,499511
answered Jan 10 at 15:52
aledenaleden
2,499511
answered Jan 10 at 15:52
aledenaleden
2,499511
answered Jan 10 at 15:52
aledenaleden
2,499511
2,499511
add a comment |
add a comment |
$begingroup$
Why do you think that $operatorname{Log}left(frac abright)=operatorname{Log}a-operatorname{Log}b$?
$endgroup$
– José Carlos Santos
Jan 10 at 15:50
$begingroup$
sure. That shouldn't work because of the argument.
$endgroup$
– s.kovalska
Jan 10 at 15:55