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Singularity approaching towards pi

  1. Apr 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Locate & name the singularity of the function

    Sinz /z- ∏


    2. Relevant equations



    3. The attempt at a solution

    The function has singularity at z= ∏ but am not able to figure out what type of singularity it is , at z=∏ it gives 0/0 form which is indeterminate form , therefore it is not a pole , neither can the singularity be removed by approaching towards pi there fore it must be an essential singularity
     
  2. jcsd
  3. Apr 4, 2008 #2

    Hootenanny

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    The singularity is not essential. What is the order of the function?
     
  4. Apr 4, 2008 #3
    Order is 1
     
  5. Apr 4, 2008 #4

    Hootenanny

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    No it isn't :wink:

    What is the order of the numerator at [itex]z=\pi[/itex]?

    What is the order of the denominator at [itex]z=\pi[/itex]?
     
  6. Apr 4, 2008 #5
    d/dz (sinz ) = cos z & at ∏ it gives value -1 so the order should be 1 & also denominatoar is 1 .


    Order is that value of d/dz which give the value of function not euqal to zero ?
     
  7. Apr 4, 2008 #6

    Hootenanny

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    Correct!
    Correct again!

    So, to determine the order of the ratio as a whole one subtracts the oder of the denominator from the numerator. Hence the order of the ratio is zero, which means that the Laurent Series has no principle part.
     
  8. Apr 4, 2008 #7
    removable singularity , also plz tell me more in detail about order , what exactly an order of a complex function represents as in my course i have not studied this part
     
  9. Apr 4, 2008 #8

    Hootenanny

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    Correct :approve:
    The order is a very useful tool to determine the type of a [isolated] singularity. As you already know the order of a function at a given point is the order of the derivative where the function is first non-vanishing. As you also know, one can treat ratios separately by determining the orders of the numerator and denominator separately.

    The order of a function at a given point denotes the index of the first non-zero coefficient of the Laurent series expansion about that point.

    So if the order is greater or equal to zero, this means that the Laurent series expansion has no terms with negative indicies, or no principle part, which means that the singularity is removable.

    If the order (n) is less than zero, then the principle part of the Laurent series has n terms and therefore the singularity is a pole of order n.
     
    Last edited: Apr 4, 2008
  10. Apr 4, 2008 #9
    Thnx for ur help

    But how will i expand the above function if terms of laurent series & show that it has removable singularity

    Sinz = z- (z3)3!+(z5)5!....../z-∏ & as z approaches ∏ this approaches infinity ??
     
  11. Apr 4, 2008 #10
    so this will be taylor series expansion , but how will i expand it
     
  12. Apr 4, 2008 #11

    Hootenanny

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    One doesn't need to use the Laurent Series to show that the singularity is removable, one merely needs to show that the limit of [itex]f(z)[/itex] as [itex]z\to\pi[/itex] exists.
     
  13. Apr 4, 2008 #12
    What can we say about the laurent series exapnsion for an isolated singularity ?
     
  14. Apr 4, 2008 #13

    Hootenanny

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    mkbh_10,

    Please do not PM me every five minutes and every time you post a reply, whilst I appreciate that you want an answer quickly you must appreciate that I am usually working on something else, whether that be reading other threads or doing work of my own. PM'ing me won't get your question answered any faster, in fact it only serves to annoy me.

    With reference to your question, consult your notes, course text or Laurent's Theorem.
     
  15. Apr 4, 2008 #14

    HallsofIvy

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    By the way, the original problem could have been done very simply in the following way:

    Let u= z-[itex]\pi[/itex]. Then z= u+ [itex]\pi[/itex] and the function becomes
    [tex]\frac{sin(z)}{z-\pi}= \frac{sin(u+\pi)}{u}[/tex]

    But [itex]sin(u+\pi)= sin(u)cos(\pi)+ cos(u)sin(\pi)= -sin(u)[/itex]. Thus,
    [tex]\frac{sin(z)}{z- \pi}= -\frac{sin(u)}{u}[/itex]
    As z goes to [itex]\pi[/itex], u goes to 0 and the fraction clearly goes to -1. Since the limit exists that is a removable singularity.

    Oh, and please use parentheses. It was not at all clear to me, at first, whether you meant sin(z)/(z-[itex]\pi[/itex]) or sin(z/(z-[itex]\pi[/itex])).

    The Laurent series of an analytic function or a function with a removeable singularity does not have any negative powers of z, of course.

    Since, as I said before,
    [tex]\frac{sin(z)}{z- \pi}= -\frac{sin(u)}{u}[/itex]
    You can find the "Laurent" series for sin(z)/(z-[itex]\pi[/itex]) by writing out the Taylor's series for sin(u), dividing by u, and then replacing u by z-[itex]\pi[/itex].

    Since sin(u) does not have a "constant" term, sin(u)/u will not have any negative powers of u, sin(z)/(z-[itex]\pi[/itex]) will not have any negative powers of z.
     
    Last edited: Apr 4, 2008
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