- #1

Diophantus

- 70

- 0

[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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Diophantus
- Start date

- #1

Diophantus

- 70

- 0

[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

Homework Helper

- 3,352

- 6

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

Homework Helper

- 3,352

- 6

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.

Last edited:

- #4

Diophantus

- 70

- 0

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.

Last edited:

- #5

Zurtex

Science Advisor

Homework Helper

- 1,120

- 1

- #6

marcusl

Science Advisor

Gold Member

- 2,797

- 454

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

Last edited:

- #7

Diophantus

- 70

- 0

- #8

marcusl

Science Advisor

Gold Member

- 2,797

- 454

Oops! I took the derivative wrong! Very sorry

- #9

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

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

Science Advisor

Gold Member

- 2,797

- 454

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]

Last edited:

- #11

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

didnt i make this trivial? or did i screw up?

- #12

marcusl

Science Advisor

Gold Member

- 2,797

- 454

Can't answer 'cause I don't know what you mean by "2^(1-z) equals 1 simply when z=1"?

Last edited:

- #13

Gib Z

Homework Helper

- 3,352

- 6

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

- 70

- 0

What about [tex]z=1 + \frac{2\pi in}{log(2)}[/tex] for any integer n?

Last edited:

- #15

Gib Z

Homework Helper

- 3,352

- 6

- #16

Diophantus

- 70

- 0

What do you mean by primary branch?

- #17

Diophantus

- 70

- 0

Still, do we have poles here or not?

- #18

Gib Z

Homework Helper

- 3,352

- 6

Yes! Think about the definiton of a simple pole, and what mathwonk said.

- #19

Diophantus

- 70

- 0

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

Homework Helper

- 3,352

- 6

1-z=0. z=1. "

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

- #21

Diophantus

- 70

- 0

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

Homework Helper

- 3,352

- 6

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:

- #23

Zurtex

Science Advisor

Homework Helper

- 1,120

- 1

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

Homework Helper

- 3,352

- 6

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

- #25

Diophantus

- 70

- 0

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.

- #26

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

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.

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.

Last edited:

- #27

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

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

Science Advisor

Gold Member

- 2,797

- 454

Thanks for steering this back to the puzzling part!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.

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).

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.

- #29

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

and you can of course recognize a simple zero from the derivative.

- #30

marcusl

Science Advisor

Gold Member

- 2,797

- 454

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?

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?

Last edited:

- #31

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

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.

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.

Last edited:

- #32

Zurtex

Science Advisor

Homework Helper

- 1,120

- 1

- #33

mathwonk

Science Advisor

Homework Helper

- 11,383

- 1,608

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.

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.

Last edited:

Share:

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 8K

- Last Post

- Replies
- 10

- Views
- 908

- Last Post

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 8

- Views
- 8K

- Last Post

- Replies
- 4

- Views
- 4K

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 7K

- Last Post

- Replies
- 3

- Views
- 2K