1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laurent series

  1. Apr 10, 2009 #1
    Hi,
    can u pls help me on this?

    1. The problem statement, all variables and given/known data
    Find Laurent series that converges for

    [tex]
    \, 0 < |z - z_0| < R }
    [/tex] and determine precise region of convergance

    [tex]
    \, \frac {1}{z^2 + 1} \,\,

    [/tex]

    2. Relevant equations



    3. The attempt at a solution
    I tried to spilt this into fractions
    i.e
    [tex]
    f(x) \, = \, \frac{A}{z-i} + \, \frac{B}{z+i}

    [/tex]

    as I would have done for

    [tex]
    \frac {1}{z^2 - 1} \,\,

    [/tex]

    But in that case I would expand it with a geometrical series.
    The problem rises with [tex] i [/tex] instead of [tex] 1 [/tex]



    2. The problem statement, all variables and given/known data
    Can u pls explain how can I choose the method of expansion (Laurent, Taylor) given a function f(x) ?






    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 10, 2009 #2
    can't you just say [itex]w=z^2[/itex] then you can make

    [itex]\frac{1}{w-1}=\frac{1}{w} \frac{1}{1-\frac{1}{w}}[/itex] and expand as a geometric series...
     
  4. Apr 10, 2009 #3
    Thanks.

    In fact the question is
    [tex]

    \, \frac {1}{z^2 + 1} \,\,

    [/tex]

    not

    [tex]

    \, \frac {1}{z^2 - 1} \,\,

    [/tex]




    I'll restate the question

    Find Laurent series that converges for

    [tex]

    \, 0 < |z - z_0| < R }
    [/tex]

    and determine precise region of convergance for
    [tex]

    \, \frac {1}{z^2 + 1} \,\,

    [/tex]

    where
    [tex]
    z_0 = i
    [/tex]



    Additionally, I would like to know how u use [tex]

    \, 0 < |z - z_0| < R }
    [/tex]
    in coming up with a solution
     
    Last edited: Apr 10, 2009
  5. Apr 10, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The Laurent series will be an expansion in powers of (z-i). Use your partial fractions form. The A/(z-i) is the (-1) power. Write the B/(z+i) part as B/(2i+(z-i)) and do the geometric series expansion with (z-i) as the variable.
     
  6. Apr 10, 2009 #5
    why can we not just change variable here?
     
  7. Apr 10, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can. Isn't that basically what I did when I wrote z+i=(z-i)+2i?
     
  8. Apr 10, 2009 #7
    Thank you,

    I have few questions
    1. Is it that you choose [itex] (z - i) [/itex] terms because the question specifies it should converge in [itex] 0 < |z-i| < R [/itex] ?

    2. So [itex]\frac{B}{z + i} [/itex] will be expanded like

    [itex]\frac{1}{2i} * \frac{1}{1+\frac{z-i}{2i}} = \sum_{n=0}^{\infty}{s^n}[/itex]

    where
    [itex] s = \frac{z-i}{2i}[/itex]


    so the answer to the solution would be
    [itex]\frac{A}{z-i} + \{expansion\, in\, 2\}[/itex] ?

    Thanks again
     
  9. Apr 11, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I chose (z-i) because you specified that you want to expand around z0=i in post 3. I couldn't tell where you wanted to put the expansion from the first post. And yes, you've got the expansion right.
     
  10. Apr 11, 2009 #9
    Thank you very much Dick

    I shall summarize here, so that anybody else following this thread can get the point.


    [itex]
    \frac{1}{1+z^2} = \frac{A}{z-i} + \frac{B}{z+i}
    [/itex]

    expand
    [itex]
    \frac{B}{z+i} = \frac{B}{2i} * \frac{1}{1-(-1 * \frac{z-i}{2i})}
    [/itex]

    then by expanding in geometric series
    [itex]
    \frac{1}{1-(-1 * \frac{z-i}{2i})} = \sum_{n=0}^\infty{[(\frac{-1}{2i})^n * (z-i)^n]}
    [/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laurent series
  1. Laurent series (Replies: 12)

  2. Laurent Series (Replies: 2)

  3. Laurent series (Replies: 2)

  4. Laurent series (Replies: 6)

Loading...