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Couple of Laurent Series I'm having trouble finding

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm stuck on the following problems:

    I have to find the Laurent series around 1 of

    [tex]f(z)=\frac{z^{3}}{z^{2}-1}[/tex]

    and the Laurent Series around 2 and on the annulus 1 < |z| < 2 of

    [tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}[/tex]

    2. Relevant equations

    I am familiar with the geometric series:

    [tex]\frac{a}{1-(z-c)}=a\sum{(z-c)^{k}}[/tex]

    I also know how to calculate partial fractions.

    3. The attempt at a solution

    For the first problem I get this far:

    [tex]f(z)=\frac{z^{3}}{z^{2}-1}=z^{3}\frac{1}{z^{2}-1}=z^{3}(\frac{-1}{2(z+1)}+\frac{1}{2(z-1)})[/tex] I can rewrite the (z+1) term to a series like this:
    [tex]\frac{1}{2(z-1)}=\frac{-1}{4(1-\frac{(z-1)}{(-2)})}=\frac{-1}{4}\sum{(z-1)^{k}(-2)^{-k}}[/tex] I can't do the same to the (z-1) term however so I'm stuck.

    For the second problem I get this far:

    [tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}=\frac{1}{z-2}-\frac{-2}{z^{2}+1}=\frac{1}{z-2}-\frac{i}{z+i}+\frac{i}{z-i}[/tex] Then I'm stuck because of the i's in the terms.
     
  2. jcsd
  3. Aug 6, 2011 #2

    vela

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    I didn't check the rest of your work, but I wanted to mention the z-1 term is already in the correct form for a series about z=1.
     
  4. Aug 6, 2011 #3
    It has to be of the form 1/(1-(z-1)) to give a series...
     
  5. Aug 6, 2011 #4

    vela

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    Remember the Laurent series has both positive and negative powers of (z-1). That term already is a Laurent series about z=1. [tex]\frac{1}{z-1} = \cdots + \frac{0}{(z-1)^3} + \frac{0}{(z-1)^2} + \frac{1}{z-1} + 0 + 0(z-1) + 0(z-1)^2 + \cdots[/tex]
     
  6. Aug 6, 2011 #5
    Ah, I see, actually that thought had crossed my mind but I wasn't sure one term would really "count" as a series. Thank you.
     
  7. Aug 6, 2011 #6

    HallsofIvy

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    Since you want a series in z- 1, the first thing I would do is let u= z- 1 so that z= u+ 1. Then [itex]z^3= (u+1)^3= u^3+ 3u^2+ 3u+ 1[/itex] and [itex]z+ 1= u+ 2[/itex]
    Then you can write
    [tex]\frac{z^3}{z+ 1}= \frac{u^3+ 3u^3+ 3u+ 1}{u+ 2}[/tex]
    which you can use "long division" to write as a second degree polynomial plus a constant over u+ 2.
    Whatever that constant is,
    [tex]\frac{A}{u+ 2}= \frac{A}{2}\frac{1}{1+ \frac{u}{2}}= \frac{A}{2}\frac{1}{1- \left(-\frac{u}{2}\right)}[/tex]
    which can be written as geometric series in [itex]-u/2[/itex]

    You can combine the polynomial in u with the first few terms of that to get a power series in u. Finally, divide each term by u to get a series with first term [itex]u^{-1}[/itex] and then, of course, replace u by z- 1 to get a Laurent series in z- 1.
     
    Last edited: Aug 6, 2011
  8. Aug 6, 2011 #7
    Thanks a lot people.

    So, anyone has some advice about the second problem?
     
  9. Aug 6, 2011 #8

    vela

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    Is the second one two separate problems? The series centered at z=2 and a series centered at z=0 on the annulus 1<|z|<2?
     
  10. Aug 7, 2011 #9
    I have to find the Laurent Series around z=2 and on the annulus 1<|z|<|2|, so yes, two different problems.

    Btw, I think we made a mistake with the the first problem: I'm not allowed to just factor out the z^3 and put it back in later because it's supposed to be a series around z=1, not z=0. It looks like I have to go with Ivy's more complicated method after all (although I don't understand what to do with the polynomial to the second degree).
     
  11. Aug 7, 2011 #10

    HallsofIvy

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    It's just part of the sum. If you have, for example,
    [tex]x^2- 3x+ 2+ \sum_{n=0}^\infty \frac{2}{(n+1)^2}x^n[/tex]
    (I just made this up, of course.)

    that is just
    [tex]x^2- 3x+ 2+ 2+ \frac{1}{2}x+ \frac{2}{9}x^2+ \sum_{n= 3}^\infty \frac{2}{(n+1)^2}x^n[/tex]
    [tex]= 4- \frac {5}{2}x+ \frac{11}{9}x^2+ \sum_{n=3}^\infty \frac{2}{(n+1)^2}x^n[/tex]
     
  12. Aug 7, 2011 #11
    Ok, I get[tex]\frac{z^{3}}{z^{2}-1}=\frac{u^3+ 3u^3+ 3u+ 1}{u+ 2}=3u^{2}-3u+9-\frac{17}{u+2}=\frac{7}{8}(z-1)^{2}+\frac{5}{4}(z-1)+\frac{1}{2}-17\sum_{k=3}^\infty(-1)^{k}2^{-k-1}(z-1)^{k}[/tex]

    Now I just need to solve the second problem.
     
  13. Aug 7, 2011 #12

    HallsofIvy

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    No, you have forgotten the "u" in the denominator.

     
  14. Aug 7, 2011 #13
    So it should be:

    [tex]\frac{z^{3}}{z^{2}-1}=\frac{u^{3}+ 3u^{2}+ 3u+ 1}{u(u+ 2)}[/tex]

    and I should work from there to get:

    [tex]\frac{7}{8}(z-1)^{2}+\frac{5}{4}(z-1)+\frac{1}{2}-17\sum_{k=3}^\infty(-1)^{k}2^{-k-1}(z-1)^{k-1}[/tex]

    Basically just dividing everything by u?
     
  15. Aug 7, 2011 #14

    vela

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    If you wanted to proceed with your original approach, you just need to write the z3 term in terms of z-1:[tex]z^3 = [(z-1)+1]^3 = (z-1)^3+3(z-1)^2+3(z-1)+1[/tex] and multiply this expression into the series you got for the denominator.

    Note that your original approach to the problem and HallsofIvy's essentially involve just rewriting everything in terms of z-1. Using the substitution u=z-1 at the start makes the algebra a bit easier overall.
     
  16. Aug 8, 2011 #15
    Thanks, do you have any thoughts on thr second problem?
     
  17. Aug 8, 2011 #16
    Please help me with the second problem: having an i in every term really makes it harder because you can't say for example z/i < 1.
     
  18. Aug 8, 2011 #17

    vela

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    Do the same thing as before. Don't let the i confuse you. It's just a constant.

    But you can say |z/i|<1.
     
  19. Aug 8, 2011 #18
    Ok, for the annulus 1 < |z| < 2 I get:

    [tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}=\frac{-1}{2(1-\frac{z}{2})}-\frac{1}{1-(\frac{-z}{i})}-\frac{1}{1-\frac{z}{i}}=-\sum_{k=0}^{\infty}{z^{k}2^{(-k-1)}}-\sum_{k=0}^{\infty}{z^{k}(-1)^{(k+1)}i^{-k}}-\sum_{k=0}^{\infty}{z^{k}i^{-k}}[/tex]I can't find the series around z=2, sure the first term is trivial but I can't find a way to rewrite the other two in terms of (z-2).
     
    Last edited: Aug 8, 2011
  20. Aug 8, 2011 #19

    vela

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    That series won't converge in the annulus because |z/i|>1. Say you have a term like[tex]\frac{1}{z-a}[/tex]What you've been doing is saying[tex]\frac{1}{z-a}=-\frac{1}{a(1-z/a)} = -\frac{1}{a}\left[1+\frac{z}{a}+\left(\frac{z}{a}\right)^2+ \cdots\right][/tex]But this series only converges when |z/a|<1. When |z/a|>1, you want to expand in powers of (a/z) instead because |a/z| will be less than 1.[tex]\frac{1}{z-a}=\frac{1}{z(1-a/z)} = \frac{1}{z}\left[1+\frac{a}{z}+\left( \frac{a}{z}\right)^2+\cdots\right][/tex]So the first series would be the Laurent series of 1/(z-a) in the region |z|<|a|, and the second series is its Laurent series for |z|>|a|.

    For the series about z=2, use HallsofIvy's technique by using the substitution u=z-2 and expanding the resulting function in a series about u=0.
     
    Last edited: Aug 8, 2011
  21. Aug 9, 2011 #20
    Yes, of course: |i/z| < 1. For the annulus 1 < |z| < 2 I now get:

    [tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}=\frac{-1}{2(1-\frac{z}{2})}-\frac{i}{z(1-(\frac{-i}{z}))}+\frac{i}{z(1-\frac{i}{z})}=-\sum_{k=0}^{\infty}{z^{k}2^{(-k-1)}}+\sum_{k=0}^{\infty}{z^{-k-1}(-i)^{(k+1)}}+\sum_{k=0}^{\infty}{z^{(-k-1)}i^{(k+1)}}[/tex]

    Around z=2 (so around u=0) I get:

    [tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}=\frac{1}{2(1-\frac{z}{2})}-\frac{i}{(2+i)(1-(\frac{-u}{2+i}))}+\frac{i}{(2+i)(1-\frac{-u}{2+i})}=\frac{1}{2(1-\frac{z}{2})}+\sum_{k=0}^{\infty}{(z-2)^{k}(-1)^{(k+1)}(2+i)^{-k-1}}+\sum_{k=0}^{\infty}{(z-2)^{k}(-1)^{k}(2+i)^{(-k-1)}}[/tex]Where I have assumed |u/(2+i)| < 1 because I'm looking at u near 0.
     
    Last edited: Aug 9, 2011
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