Calculating Laurent expansions using geometric series in regions defined by inequalities












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 20:23







daljit97

















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


















  • $begingroup$
    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










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}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.






share|cite|improve this answer











$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    3












    $begingroup$

    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}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    3












    3








    3





    $begingroup$

    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}






    share|cite|improve this answer









    $endgroup$



    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}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    0












    $begingroup$

    Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.






        share|cite|improve this answer











        $endgroup$



        Try using: $e^z = ecdot e^{z-1}$ where $|z - 1| > 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 21:13









        Davide Giraudo

        128k17156268




        128k17156268










        answered Dec 31 '18 at 20:19









        Swapnil Swapnil

        515




        515






























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