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!

Residue of a ratio of polynomials

  1. Mar 14, 2010 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    The problem is to find the inverse laplace of [tex]\frac{s^2-a^2}{(s^2+a^2)^2}[/tex]

    I am supposed to use the residue definition of inverse laplace (given below)

    The poles of F(s) are at ai and at -ai and they are both double poles.

    2. Relevant equations

    [tex]f(t) = \sum_{j=1}^{N}Res[F(s)*e^{st};s_{j}][/tex]

    Where [tex]s_{j}[/tex] are the poles of F(s)

    [tex]Residue(f(z)) = \frac{1}{(m-1)!}lim_{z->z0}\frac{d^{m-1}}{dz^{m-1}}((z-z0)^{m}f(z))[/tex]

    where z0 is a pole of order m of f(z)

    3. The attempt at a solution

    The poles of F(s) are at ai and at -ai and they are both double poles.

    If you can just help find the residue of ai I'm sure I can figure out -ai.

    [tex]Residue[F(s);ai] = lim_{s->ai}[/tex][tex]\frac{d}{ds}\frac{(s-ai)^2(s^2-a^2)}{(s-ai)^2*(s+ai)^2}[/tex]

    [tex]= lim_{s->ai}[/tex][tex]\frac{d}{ds}\frac{s^2-a^2}{(s+ai)^2}[/tex]

    using quotient rule I get:

    [tex]= lim_{s->ai}[/tex][tex]\frac{2s(s+ai)^2-2(s+ai)(s^2-a^2)}{(s+ai)^4}[/tex]

    substitute ai for s


    [tex]=\frac{2ai(-4a^2)+8a^3i}{16a^4} = 0[/tex]

    But it can't equal zero right? I also get zero when I try to calculate the residue at -ai.
  2. jcsd
  3. Mar 14, 2010 #2
    To compute residues the limit formula is (in general) not recommended. It is messy and you can easily make mistakes that you then can't correct easily by backtracking the calculations. Instead, you should compute the Laurent expansion around the poles using known series expansions.

    E.g. to find the residue at s = i a, you put

    s = i a + t

    and compute the coefficient of 1/t:

    [2iat + t^2 - 2a^2]/[t^2 + 2iat]^2 =

    1/(2iat)^2 [2iat + t^2 - 2a^2] 1/[1+t/(2ia)]^2 =

    1/(2iat)^2 [2iat + t^2 - 2a^2] [1-t/(ia) + O(t^2)]

    So, we see that the residue is:

  4. Mar 14, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, the residue of F(s) at s=ia is zero. Like you said, it has a double pole there; it will only have a non-zero residue if the pole is a simple pole.

    To find the inverse Laplace transform, however, you need to calculate the residue of F(s)est, and that function does have a simple pole at s=ia. Like Count Iblis, I suggest you expand the function in a Laurent series about s=ia.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Residue ratio polynomials Date
Finding ratio of volumes of cube and sphere Today at 9:11 AM
Residue at poles of complex function Feb 2, 2018
Residue of an integral Apr 24, 2017
Residue of a series. Apr 23, 2017