Expand $operatorname{Log}frac{z^2}{z^2-1}$ into Laurent series for $|z|>1$ [closed]












1












$begingroup$



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?










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$endgroup$



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.
















  • $begingroup$
    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
















1












$begingroup$



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?










share|cite|improve this question











$endgroup$



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.
















  • $begingroup$
    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














1












1








1





$begingroup$



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?










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 18:45









Lord_Farin

15.6k636108




15.6k636108










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


















  • $begingroup$
    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










1 Answer
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$begingroup$

Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$






        share|cite|improve this answer









        $endgroup$



        Use $$log(1-x)=-sum_{n=1}^infty frac{x^n}{n}$$ to get $$log left(1-frac{1}{x^2}right)=-sum_{n=1}^infty frac{1}{nx^{2n}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 15:52









        aledenaleden

        2,499511




        2,499511















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