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Exam problem essiential singulairity or not complex function

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    1/(1-cos(z))


    2. Relevant equations
    taylor expansion at 0 for cos(z)=1-x^2/2+x^4/24 and so on.
    1/(x^2/2+x^4/24.......)


    3. The attempt at a solution

    Because all of the powers are negative wouldn't that make it a essential singularity and 0 + 2∏n. Also it just explodes at those points.
     
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  3. May 15, 2012 #2
    Oh btw the exam was already graded this isn't a take home. For the exam I got this question wrong and I don;t understand.
     
  4. May 15, 2012 #3

    micromass

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    You haven't computed the Laurent series at 0 yet. Right now you have something of the form

    [tex]\frac{1}{a_0+a_1x+a_2x^2+...}[/tex]

    But the Laurent series is supposed to be something of the form

    [tex]...+b_{-2}\frac{1}{x^2}+b_{-1}\frac{1}{x}+b_0+b_1x+b_2x^2+...[/tex]

    Notice the difference between the two??

    In fact, it is easy to see that

    [tex]\frac{z^2}{1-cos(z)}[/tex]

    has a removable singularity at 0 (why?). What does that say about the original function?
     
  5. May 15, 2012 #4
    But in this case the Laurant series of cos(z) is equivalent to it's Taylor Series and it has all negative powers. That one has a removable singularity at 0 because it's limit as z approaches 0 is zero. Wouldn't that say that the original function doesn't have a removable singularity at 0? Also its not a pole so wouldn't it be a essential singularity like I think it is?
     
  6. May 15, 2012 #5

    micromass

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    Sure, the Laurent series of cos(z) is equal to its Taylor series, but I don't see how the Taylor series of cos(z) has all negative powers??

    Let me give another example. Consider the function

    [tex]\frac{1}{1-z}[/tex]

    What is its Laurent series at 0?? You seem to argue that 1-z is the Taylor series of 1-z and thus that

    [tex]\frac{1}{1-z}[/tex]

    is its own Laurent series and that it apparently has a pole of order 1 there.
    But actually, the Laurent series of that function is

    [tex]1+z+z^2+z^3+...[/tex]

    and it has a removable singularity at 0.


    Certain that the limit is 0?

    It indeed means that the original function does not have a removable singularity at 0. It means that it actually has a pole of order 2.
     
  7. May 15, 2012 #6
    cos(z) doesn't have negative powers. But for 1/(1-cos(z)) does because it equals as stated above 1/(z^2/2+z^4/24 and so on) You're right my bad it's 2 lol. I tried doing L'hospital rule in my head. I was thinking of 1/(1-z) and if you sub in Cosine it would be 1+cos(z)+coz(z)^2 and so on. Did I make my error by only taking the Laurent series of Cos(z)?
     
  8. May 15, 2012 #7

    micromass

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    Yes, but

    [tex]\frac{1}{z^2/2 + z^4/24+...}[/tex]

    is not the Laurent series of 1/(1-cos(z)). Do you see that? Compare to my example of 1/(1-z).
     
  9. May 15, 2012 #8
    Ah I see where I went wrong. so 1+cos(z)+cos(z)^2+cos(z)^3+cos(z)^4 .......is the Laurent Series and it at z=0 it goes to infinity and this is not a essential singularity.


    I'll be posting another question up also in a minute or two.
     
    Last edited: May 15, 2012
  10. May 15, 2012 #9

    micromass

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    No, that is not the Laurent series either. The Laurent series is of the form

    [tex]...+b_{-2}\frac{1}{z^2}+b_{-1}\frac{1}{z}+b_0+b_1 z + b_2 z^2 +...[/tex]

    Your series is not of that form.
     
  11. May 15, 2012 #10
    Whoops no wonder I got it wrong if I messed up the 2nd time. I thought you were typing a general form for all of them. And now I see why it has a removable singularity because z^2/z^2 is of course 1 so the limit would be whatever the b_2. Well I see how it is a pole or order 2 when you give me the general form of this series for 1/(1-cos(z)). But I don't see how you got that form.
     
  12. May 15, 2012 #11

    micromass

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    To find the actual Laurent series you need to make the division

    [tex]\frac{1}{z^2/2 + z^4/24+...}[/tex]

    We can write

    [tex]\frac{1}{1-cos(z)}=\frac{1}{z^2} \frac{1}{1/2 + z^2/24 + ...}[/tex]

    The left factor is now defined at 0 so it has a Taylor series. So we can write

    [tex]\frac{1}{1/2 + z^2/24 +...}=a_0+a_1z+a_2z^2+...[/tex]

    or equivalently

    [tex](1/2 + z^2/24 +...)(a_0+a_1z+a_2z^2+...)=1[/tex]

    When we multiply that out, you can find the coefficients you're looking for.
     
  13. May 15, 2012 #12
    Yah that was my biggest fear. I really didn't want to divide while taking the exam so I tried to work my way around that and it didn't work. But I see how you can use that 1/1-z to get it into that form and that the power are all not negative and the highest negative power is 2 and the it eventually becomes positive thus turning it into a pole of order 2.

    Alright thanks.
     
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