1. Not finding help here? Sign up for a free 30min 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 (COMPLICATED, )

  1. Oct 24, 2009 #1
    Laurent series (COMPLICATED, URGENT)

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

    Find the Laurent series for [tex]f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}})[/tex] around z = 0, in which the series absolutely converges for z = -i.

    3. The attempt at a solution

    I have several problem with this. I tell you what I tried.

    First: I separate the polinomial fraction in partial fractions. I don't know if the procedure is OK: [tex]\frac{{2{z^2}}}{{2{z^2} - 5z + 2}} = 2{z^2}\left( {\frac{2}{{3(z - 2)}} - \frac{2}{{3(z - 1/2)}}} \right)[/tex]. I leave the 2z2 there because it's its own Laurent Series around z=0, for all z.

    Second: I compare the function with the geometrical function (considering that I have to find the Laurent series in a vecinity of infinite for the fraction with the (z - 1/2)) and I use a variable change for the sine:

    a) [tex]\frac{2}{{3(z - \frac{1}{2})}} \to \frac{2}{{3z(1 - \frac{1}{{2z}})}}
    [/tex]

    [tex]\frac{2}{{3z(1 - \frac{1}{{2z}})}} = \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} {\rm{ }}\forall z/\left| z \right| > \frac{1}{2}[/tex]


    b)
    [tex]\frac{{ -1}}{{3(z - \frac{1}{2})}} = \frac{2}{{3(\frac{1}{2} - z)}} = \frac{4}{{3(1 - 2z)}}[/tex]

    c) Since [tex]\sin (u) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{u^{2n + 1}}}}{{(2n + 1)!}}}[/tex], then, using the change of variable u = 3/z2, I get that:

    [tex]\sin (\frac{3}{{{z^2}}}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{ }}\forall z/\left| z \right| > 0[/tex]


    Then, I have that:

    [tex]f(z) = 2{z^2}\left( {\frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} - \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} } \right) + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{, }}\forall z/\frac{1}{2}< \left| z \right|< 2[/tex]

    My question is, how can I come up with a sole expresion for the coefficients? I couldn't even find one for the series inside the parenthesis...

    Thanks
     
  2. jcsd
  3. Oct 24, 2009 #2
    Re: Laurent series (COMPLICATED, URGENT)

    I don't know why, but LaTeX doesn't let me put the expressions in order...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laurent series (COMPLICATED, )
  1. Laurent series (Replies: 2)

  2. Laurent series (Replies: 12)

  3. Laurent Series (Replies: 2)

  4. Laurent series (Replies: 2)

  5. Laurent series (Replies: 6)

Loading...