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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...
     
  22. Apr 14, 2007 #21
    Try [tex]n = \frac{2\pi i}{\log 2} \not = 0[/tex].
     
  23. Apr 14, 2007 #22

    Gib Z

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    Ok fine let me rephrase that then, the only value n can be, in the primary branch on the complex log function which is what we are dealing with me, is zero.
     
    Last edited: Apr 14, 2007
  24. Apr 14, 2007 #23

    Zurtex

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    Erm, o.k, I don't really understand what you are on about. You're talking about the solutions of a complex equation, there's no reason to think we should limit our search to only real numbers.

    But furthermore I don't understand why you are talking about the primary branch of the complex log function, it doesn't seem to make sense to the context of the question. I don't think stating it in capitals makes it any more relevant, perhaps you could explain your motive for talking about this better.
     
  25. Apr 14, 2007 #24

    Gib Z

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    Im not limiting the search to real numbers, just the primary branch of the complex log. In this case, the solution happens to be a real number. And We needed to bring the primary branch of the complex log into this because otherwise we would have an infinite number of solutions to choose from to solve 2^(1-z)=1, as the OP noticed.

    Ill edit my previous post so its not so capital, that was stupid i realize now.
     
  26. Apr 15, 2007 #25
    I think we may have gone off on a bit of a tangent. Thank you for trying to expain Gib Z but I think a few of us do not understand mathwonk's motives for his method and hence don't understand your motives for limiting the number of solutions. Not a clue what's going on there. I myself have only seen two ways of showing that a function has simple poles, namely the limit method that marcusl demonstrated; and finding a Laurent series in order to show that the only nonzero coefficient of a negative power is that corresponding to the power -1. I can't see how mathwonk's method falls into either of these categories.

    It would be nice to know though. And I'm still curious as to whether there is a nice way of deducing the exact form of the Laurent series since Zurtex claims that mathematica gives it fairly succinctly.


    Just don't know what he's trying to do here.
     
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