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[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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- Thread starter Diophantus
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- #1

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[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

Gib Z

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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.

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

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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.

[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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- #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

[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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- #8

marcusl

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Oops! I took the derivative wrong! Very sorry

- #9

mathwonk

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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!

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]

[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!

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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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

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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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- #16

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What do you mean by primary branch?

- #17

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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

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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.

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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1-z=0. z=1. "

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

- #21

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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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Ill edit my previous post so its not so capital, that was stupid i realize now.

- #25

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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.

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.

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