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Simple Pole

  1. Apr 8, 2007 #1
    How exactly would one go about proving that

    [tex] \frac{1}{1-2^{1-z}}[/tex] has a simple pole at [tex]z=1[/tex]?

    I've tried writing [tex]2^{1-z}[/tex] in terms of e to get a Taylor series for the denominator but can't quite figure out where to go from there.
     
  2. jcsd
  3. Apr 8, 2007 #2

    Gib Z

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    In complex analysis, a pole of a holomorphic function is a certain type of simple singularity that behaves like the singularity 1/z^n at z = 0. A pole of order 1 is called a simple pole as you know. Essentially you need to prove
    z behaves like 1-2^(1-z) near zero. Perhaps if you showed that [tex]\lim_{z\to 0} \frac{z}{1-2^{1-z}} = 1[/tex]...
     
  4. Apr 8, 2007 #3

    Gib Z

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    Scratch that, im not sure that helps..All you should need to show is that it approaches the same limit, so the first one, z, as z goes to 0, its just zero. the second one is also 0. That should do it.

    If your not happy, show that the function Laurent series near z=1 below degree −n vanishes and the term in degree −n is not zero.
     
    Last edited: Apr 8, 2007
  5. Apr 8, 2007 #4
    OK so I've opted for the Laurant Series route. Where am I going wrong here:

    [tex]\frac{1}{1-2^{1-z}} = \displaystyle\sum_{n=0}^{\infty}(2^{1-z})^n = \displaystyle\sum_{n=0}^{\infty}(e^{(1-z)log2})^n = \displaystyle\sum_{n=0}^{\infty}(\displaystyle\sum_{m=0}^{\infty}\frac{(1-z)^m(log2)^m}{m!})^n[/tex]

    This doesn't look like it has any singilarities.
     
    Last edited: Apr 9, 2007
  6. Apr 10, 2007 #5

    Zurtex

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    I'm not sure using the geometric series is really valid for about z = 1, I can get an appropriate series on mathematica which shows it is a simple pole, but I have no idea how to generate it myself.
     
  7. Apr 10, 2007 #6

    marcusl

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    If it has a simple pole, then the residue should be bounded and given by

    [tex]R=\lim_{z\rightarrow 1} (z-1)f(z)=\lim_{z\rightarrow 1} \frac{z-1}{1-2^{1-z}}.[/tex]

    Substitute [tex]u=z-1[/tex] to get

    [tex]R=\lim_{u\rightarrow 0} \frac{u}{1-2^{-u}}[/tex]

    Evaluating using l'Hospital's rule gives R=2.
    EDIT: Accordingly, 2/(z-1) is a term in the Laurent series.
    EDIT2: define u=z-1 instead of 1-z for clarity
     
    Last edited: Apr 10, 2007
  8. Apr 10, 2007 #7
    Hmmm, it's actually 1/log2 but thanks anyway. I'm really intrigued to know whether there is a reasonable way of calculating the Laurant series though. Surely there must be a bit of trickery that will work.
     
  9. Apr 10, 2007 #8

    marcusl

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    Oops! I took the derivative wrong! Very sorry :blushing:
     
  10. Apr 10, 2007 #9

    mathwonk

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    you want to show [z-1] over that function is bounded.

    but you just want to show 2^(1-z) equals 1 simply when z=1, or that 2^z equals 1 simply when z=0, but thats sort of clear, since the derivative is not zero anywhere, 2^z has no multiple values at all.
     
  11. Apr 11, 2007 #10

    marcusl

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    I'll repeat my earlier derivation, without the silly error.

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

    has a pole if the residue

    [tex]R=\lim_{z\rightarrow 1} (z-1)f(z)=\lim_{z\rightarrow 1} \frac{2^z (z-1)}{2^z-2}[/tex].

    is bounded. To evaluate via l'Hospital's rule, write

    [tex]2^z=(e^{\ln2})^z=e^{z\ln2}[/tex]

    so the derivative is

    [tex]\frac{d(2^z)}{dz}=2^z \ln(2).[/tex]

    Then

    [tex]R=\lim_{z\rightarrow 1} \frac{2^z[1+(z-1)\ln(2)]}{2^z \ln2} = \frac{1}{\ln(2)} .[/tex]

    Hope I redeemed myself! o:)


    EDIT: It's even simpler to define u=z-1, then

    [tex]R=\lim_{u\rightarrow 0} \frac{u}{1-2^{-u}}[/tex]

    and applying l'Hospital's rule gives

    [tex]R=\lim_{u\rightarrow 0} \frac{1}{2^u \ln(2)}=\frac{1}{\ln(2)}.[/tex]
     
    Last edited: Apr 11, 2007
  12. Apr 11, 2007 #11

    mathwonk

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    didnt i make this trivial? or did i screw up?
     
  13. Apr 12, 2007 #12

    marcusl

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    Can't answer 'cause I don't know what you mean by "2^(1-z) equals 1 simply when z=1"?
     
    Last edited: Apr 12, 2007
  14. Apr 13, 2007 #13

    Gib Z

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    >.<" Perhaps the "simply" is troubling you? Thats not a mathematical term he's using, hes just saying that it is simple.

    For 2^n to equal 1, the only value n can be is 0. In this case, n is 1-z.
    1-z=0. z=1.
     
  15. Apr 13, 2007 #14
    What about [tex]z=1 + \frac{2\pi in}{log(2)}[/tex] for any integer n?
     
    Last edited: Apr 13, 2007
  16. Apr 13, 2007 #15

    Gib Z

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    No need for that double post :P And we are dealing with the primary branch only, otherwise nothing we talk about here are functions anymore.
     
  17. Apr 13, 2007 #16
    I'm sorry you've lost me now. Is it or is it not true that the function 2^(1-z) has poles at [tex]z=1 + \frac{2\pi in}{log(2)}[/tex]?

    What do you mean by primary branch?
     
  18. Apr 13, 2007 #17
    Oh I see now, the primary branch of the complex log function. I forgot about that.

    Still, do we have poles here or not?
     
  19. Apr 13, 2007 #18

    Gib Z

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    Yes! Think about the definiton of a simple pole, and what mathwonk said.
     
  20. Apr 13, 2007 #19
    I don't understand why you said

    when it is clearly wrong.

    And what's all this talk about it not being a function? I have only ever alluded to 2^n being 1 whose imaginary part, if I am not mistaken, lies between + and - pi.
     
  21. Apr 14, 2007 #20

    Gib Z

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    "For 2^n to equal 1, the only value n can be is 0. In this case, n is 1-z.
    1-z=0. z=1. "

    I don't understand why that is clearly wrong...
     
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