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Why its not removable singular point?

  1. Jul 12, 2010 #1
    why its not removable singular point??

    [tex]\frac{z-\sin z}{z^4}[/tex]

    my singular point is 0

    if i will do lhopital 4 times on the lmit i will get -sin z/6 =0

    which is singular point because the result is constant



    so why its a first order pole?
     
  2. jcsd
  3. Jul 12, 2010 #2

    Office_Shredder

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    Re: why its not removable singular point??

    If you do L'Hopital on it three times you don't get something you can do L'Hopital on again

    Try writing out sine as a power series and see if you can get some info on what the pole should be from that
     
  4. Jul 12, 2010 #3
    Re: why its not removable singular point??

    so in a complex function
    what are the conditions for me doing lhotipal
    is it just 0/0 inf/inf linke in real functions
    or are the other conditions?
     
  5. Jul 12, 2010 #4

    Office_Shredder

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    Re: why its not removable singular point??

    Same as for real functions. After doing L'Hopital three times the numerator is a cosine, that does NOT give you 0/0
     
  6. Jul 13, 2010 #5
    Re: why its not removable singular point??

    ok now i tried to determine the power of the pole
    i did
    g(x)=1/f(x)
    where x=0
    and i got that g'(0)=0/0 which means that its not a first order pole
    actualyy i should get zero but i got 0/0
    ??
     
  7. Jul 13, 2010 #6

    lanedance

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    Re: why its not removable singular point??

    have you tried the power series suggested in post #2?
     
  8. Jul 13, 2010 #7

    vela

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    Re: why its not removable singular point??

    It's a simple pole because the smallest positive integer n for which

    [tex]\lim_{z \to 0}\,(z-0)^n\,\frac{z-\sin z}{z^4}[/tex]

    exists is n=1. This is probably what you had in mind when you decided to take limits.

    From a practical standpoint, it's not a very efficient way of determining the order of a pole. It's better to find the Laurent series as the others have suggested.
     
  9. Jul 13, 2010 #8
    Re: why its not removable singular point??

    the derivative way is a way from my text book
    do you know it?
    if so can you tell me where did i got wrong
    ??
    i tool g=1/f and i put x=a if every derivative till i get to the derivative for ach
    it differs zero

    where is my mistake
    ?
     
  10. Jul 13, 2010 #9

    hunt_mat

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    Re: why its not removable singular point??

    A removable singular point is a point which if you evaluate the function you'll get 0/0 but when you expand the function via a taylor series then it vanishes. To give an example:
    [tex]
    \frac{\sin z}{z}
    [/tex]
    When you evaluate at z=0, you get 0/0, but when you expand in a taylor series you arrive at:
    [tex]
    \frac{\sin z}{z}=1-\frac{z^{2}}{3!}+\cdots
    [/tex]
    There isn't a problem on the RHS.
     
  11. Jul 13, 2010 #10

    lanedance

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    Re: why its not removable singular point??

    ok so show us your derivatives...

    as you won't try the power series, i'll show you the utility of it:
    [tex]sin(x) = x - \frac{x^3}{3!}+O(x^5)[/tex]

    then
    [tex]f(x) = \frac{z-sin(z)}{z^4}
    = \frac{z-(z-\frac{z^3}{3!}+O(z^5))}{z^4}
    = \frac{z^3}{3! z^4}+O(z)
    = \frac{1}{3! z}+O(z)[/tex]

    which shows its a 1st order pole
     
  12. Jul 13, 2010 #11

    vela

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    Re: why its not removable singular point??

    0/0 is an indeterminate form. You need to use the hospital rule to calculate the limit.
     
  13. Jul 13, 2010 #12
    Re: why its not removable singular point??

    but its not a limit
    its a derivative
     
  14. Jul 14, 2010 #13
    Re: why its not removable singular point??

    ok suppose if i have a singularity at point a in this function
    so i put (z-a) instead of z everywhere
     
  15. Jul 14, 2010 #14

    Office_Shredder

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    Re: why its not removable singular point??

    Whenever you have a function of the form 0/0 at a point and you want to know if there's a removable singularity there you can just calculate what the limit as you approach that point is to see if you get continuity
     
  16. Jul 14, 2010 #15

    i2c

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    Re: why its not removable singular point??

    Am I the only one thinking, "Graph (z-sinz)/z^4 and you will see why it's the limit does not exist (i.e. not a single removable point) at z = 0 ?" (hint hint, it has the same shape as a 1/x graph)
     
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