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!

Find and classify the isolated singularity

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Find and classify the singularity.

    f(z) = sin(z) / z

    2. Relevant equations
    sin(z) = (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ...

    3. The attempt at a solution
    so the singularity is at z = 0.

    f(z) = (1/z) * (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ...)

    f(z) = 1 - z2 / 3! + z4 / 5! - z6 / 7! + z8 / 9! ...

    and these are terms starting at index 0, because the series for sin(z) is Σ(-1)k * (z2k + 1 / (2k+1)!), for k >= 0.

    The definition of removable singularity is if cn = 0 for n = -1, -2, -3 ... then f(z) has a removable singularity at α.

    ANSWER: So the singularity z = 0 is removable because the series for sin(z) starts at k = 0, and so any term k<0 must have coefficient 0.

    i'm really confused how to check whether the singuarlity is removable, a pole, or essential. The definition doesn't make sense to me. How can I check if ck = 0 for all k < 0, if there are no terms for k < 0 in the power series for sin(z)?
     
  2. jcsd
  3. May 4, 2017 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Why do you think you need to check? You just said that there are no terms for k < 0 in the power series.
     
  4. May 4, 2017 #3
    yeah... so "no terms for k < 0" <=> "ck = 0 for all k < 0"?
     
  5. May 4, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper


    What do YOU think?
     
  6. May 5, 2017 #5
    Ok, so in order to find out what kind of singularity it is, I need to write this as a Laurent series.

    So a Laurent series, as I understand it, is f(z) rewritten as
    Σ Ck(z-z0)k, for k >= 0 + Σ Ck(z - z0)k, for k < 0.

    I've got the first series.. (1 - z2/3! + z4/5! - z7/8! + ...

    but I need to see the values for the negative indexes.. so I rewrite the series like this

    ##
    \frac{1}{z} * \sum_{n=1}^{-\infty} (-1)^{-k-1} \frac{z^{-2k-1}}{|z|!} =\frac{1}{z} * ( \frac{1}{z^{3}3!} -
    \frac{1}{z^{5}5!} + \frac{1}{z^{7}7!} - ...)

    = \frac{1}{z^{4}3!} - \frac{1}{z^{6}5!} + \frac{1}{z^{8}7!} - ...

    ##


    Looking at this.. I think that these negative index value have coefficients of ##\frac{1}{|k|!}## which isn't 0 which is what I expected
     
  7. May 5, 2017 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    An ordinary power series ##\sum_{n \geq 0} c_n z^n## IS a Laurent series with all negative powers absent. Usually, though, we do not refer to it as a Laurent series (with no negative-power terms); it is easier to just call it a power series.
     
  8. May 6, 2017 #7
    But if there are no terms raised to z raised to a negative exponent, then how can there exist nonexistent terms that have Ck = 0?

    If a power series is a series that has all negative powers absent, then how can there be terms:
    C-1 (z - z0) + C-2 (z - z0) + .... = 0, if the series starts at k=0?

    I know 0 + Σck(z-z0) = Σck(z-z0) how do we know Ck = 0.. what if (z-z0) = 0 and Ck is some number?
     
  9. May 6, 2017 #8
    sorry for double post
     
  10. May 6, 2017 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    It's easy: ##0 (z-z_0)^{-1} + 0 (z-z_0)^{-2} + 0 (z-z_0)^{-3} + \cdots ## is just zero. I genuinely cannot understand why you have a problem with that, but at this point I am quitting this thread.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find and classify the isolated singularity
  1. Classify singularities (Replies: 1)

Loading...