Simple Pole

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  • #1
Diophantus
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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.
 

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  • #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]...
 
  • #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.
 
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  • #4
Diophantus
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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.
 
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  • #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.
 
  • #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
 
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  • #7
Diophantus
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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.
 
  • #8
marcusl
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Oops! I took the derivative wrong! Very sorry :blushing:
 
  • #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.
 
  • #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]
 
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  • #11
mathwonk
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didnt i make this trivial? or did i screw up?
 
  • #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"?
 
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  • #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.
 
  • #14
Diophantus
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What about [tex]z=1 + \frac{2\pi in}{log(2)}[/tex] for any integer n?
 
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  • #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.
 
  • #16
Diophantus
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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?
 
  • #17
Diophantus
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Oh I see now, the primary branch of the complex log function. I forgot about that.

Still, do we have poles here or not?
 
  • #18
Gib Z
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Yes! Think about the definiton of a simple pole, and what mathwonk said.
 
  • #19
Diophantus
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I don't understand why you said

>.<" 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.

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.
 
  • #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...
 
  • #21
Diophantus
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For 2^n to equal 1, the only value n can be is 0.

Try [tex]n = \frac{2\pi i}{\log 2} \not = 0[/tex].
 
  • #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.
 
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  • #23
Zurtex
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Ok fine let me rephrase that then, the only value n can be, IN THE PRIMARY BRANCH OF THE COMPLEX LOG FUNCTION WITH WHICH WE ARE DEALING WITH HERE, is zero.

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.
 
  • #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.
 
  • #25
Diophantus
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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.


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.

Just don't know what he's trying to do here.
 
  • #26
mathwonk
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im using "simply" in the same sense in which he used "simple pole". i.e. a pole is simple, if the reciprocal has a simple zero.

a simple zero at z=a, means a zero of order one.

or it means the taylor series starts with the term c(z-a).

and order of zeroes is a local matter, so it matters not what branch one takes.
 
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  • #27
mathwonk
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so his function has a simple pole if 1 - 2^(1-z) has a simple zero at z=1.

but that occurs iff the derivative of 2^(1-z) is not zero at z=1.

but that derivative is a constant multiple of 2^(1-z) which is never zero.

done.
 
  • #28
marcusl
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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.
Thanks for steering this back to the puzzling part!

so his function has a simple pole if 1 - 2^(1-z) has a simple zero at z=1.

but that occurs iff the derivative of 2^(1-z) is not zero at z=1.

but that derivative is a constant multiple of 2^(1-z) which is never zero.

done.
It seems to me that d/dz of 2^(1-z) cannot establish the order of the singularity. A simple pole must depend on the form of the denominator being, e.g., [1-2^(1-z)]^(-1) as opposed to [1-2^(1-z)]^(-2) or some ill-behaved function of 2^(1-z).
 
  • #29
mathwonk
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you are not listening. f has a simple pole if 1/f has a simple zero.

and you can of course recognize a simple zero from the derivative.
 
  • #30
marcusl
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I am listening but you are not explaining in a way I can understand, and in fact simply keep repeating the same answer. I am apparently not the only one in this thread who has not come across a simple zero before. Perhaps you can explain:

Why does a non-zero derivative define a "simple zero"? Is it because that's the coefficient of the first term in the taylor's series?
 
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  • #31
mathwonk
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sorry i waS CRANKY TODAY.

but it is kind of odd to know what a simple pole is and not know what a simple zero is.

a meromorphic function has a simple pole at a if it looks locally near a, like

1/(z-a) times a holomorphic function which is not zero at a.

it has a simple zero at a if it looks locally at a like (z-a) times a holomorphic function which is not zero at a.


hence obviously f has a simple pole iff 1/f has a simple zero.

as you realized, since a holomorphic fucntion has a taylor series, whose coefficient of (z-a) is its first derivative at a, a holomorphic f has a simple zero at a iff f(a) = 0 and f'(a) is not zero.


so to show the function above has a simple pole at z=1, it is much easier to turn it upside down and show the reciprocal has a simple zero, which can be checked by taking a derivative.


but somebody is teaching you amiss, if they have you doing meromorphic functions and poles and have not even taught you about using derivatives to compute the order of a zero, which is easier and more fundamental.

a holomorphic function, e.g. a polynomial, has a zero of order at least k at a, iff if is divisible by (z-a)^k, (with holomorphic quotient),

iff its first k derivatives (0'th through k-1'st) are zero at a.

a simple zero is a zero of order one. i would think this would be familiar even from high school algebra.
 
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  • #32
Zurtex
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Thanks a lot mathwonk, I know the author of the thread in real life, I don't think either of us have come across simple zero's before but I think we both understood what you were on about straight away, very useful thanks :smile:
 
  • #33
mathwonk
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you are quite welcome.

there is also a geometric version of the order of a zero or pole.

a point a is a zero of f of order k iff the inverse image of a small punctured disc D centered at 0, intersects a small nbhd of a, in a set that maps exactly k to one onto D.

since reciprocation is an isomorpism from a nbhd of 0 to a nbhd of infinity, a pole of order k at a, means some punctured nbhd of a maps exactly k to one, onto the exterior of a large disc, i.e. ointo a small opunctured disc about infinbity.
 
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