Does a Laurent series expansion exist?...

**chisigma** · January 3rd 2011, 06:50 AM

Prove [or even disprove...] that the principal branch of the function

cannot be expanded in Laurent series around the point

, i.e. a coefficient set

such that for a definite region of the complex plane which is not limited to only one point is...

(1)

... doesn't exist.

Kind regards

$\chi$ $\sigma$

**Jose27** · January 14th 2011, 10:41 AM

How about this: If the logarithm had an (holomorphic) extension in an annulus around zero then, by a continuity argument we could conclude, since logarithm is a primitive of $\frac{1}{z}$ in the slit plane, that $\int_C \frac{dz}{z} =0$ which is a contradiction.

**chisigma** · January 14th 2011, 11:57 AM

An useful preliminary is that a complex variable z can be written as...

$\displaystyle z= x + i\ y = r\ e^{i\ \theta}$ (1)

... where...

$r= \sqrt{x^{2} + y^{2}}$

$\displaystyle \theta = \tan^{-1} \frac{y}{x} + 2\ k\ \pi\\ , k \in \mathbb{Z}$

... so that is...

$\displaystyle \ln z= u(x,y) + i\ v(x,y) = r(x,y) + i\ \theta(x,y)$ (2)

Now we compute the partial derivatives of u(*,*) and v(*,*) obtaining...

$\displaystyle \frac{\partial{u}}{\partial{x}} = \frac{x}{x^{2}+ y^{2}}$

$\displaystyle \frac{\partial{u}}{\partial{y}} = \frac{y}{x^{2}+ y^{2}}$

$\displaystyle \frac{\partial{v}}{\partial{x}} = - \frac{y}{x^{2}+ y^{2}}$

$\displaystyle \frac{\partial{v}}{\partial{y}} = \frac{x}{x^{2}+ y^{2}}$ (3)

... and we observe that the Cauchy-Riemann conditions...

$\displaystyle \frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}$

$\displaystyle \frac{\partial{u}}{\partial{y}}= - \frac{\partial{v}}{\partial{x}}$ (4)

... are satisfied for any $z \in \mathbb{C}$ with the only exception of $z=0$ , where none of the (3) exists. So $z=0$ is a singularity of the function $f(z)= \ln z$ and that means that no 'holomorphic extensions' of it including $z=0$ may exist . The 'core' of the question is why $\ln z$ has no Laurent series expansion around $z=0$ like, for example, $\displaystyle \frac{1}{z}$ ...

Kind regards

$\chi$ $\sigma$

**Bruno J.** · January 18th 2011, 07:18 AM

The reason is that the region of convergence of a Laurent series is a (possibly degenerate) annulus. If this region contained more than one point, it would actually contain a whole circle around the origin, which would imply that it is possible to define $\mbox{arg }z$ continuously on a circle around the origin, which is obviously impossible.

**chisigma** · January 18th 2011, 08:03 AM

Originally Posted by Bruno J.

The reason is that the region of convergence of a Laurent series is a (possibly degenerate) annulus. If this region contained more than one point, it would actually contain a whole circle around the origin, which would imply that it is possible to define $\mbox{arg }z$ continuously on a circle around the origin, which is obviously impossible.

If $z= x + i\ y$ then [as in my previous post...] is...

$\displaystyle \theta= \mbox{arg}\ z = \tan^{-1} \frac{y}{x} + 2\ k\ \pi\\ ,\ \ k\in \mathbb{Z}$ (1)

... and...

$\displaystyle \frac{\partial{\theta}}{\partial{x}}= - \frac{y}{x^{2}+y^{2}}$

$\displaystyle \frac{\partial{\theta}}{\partial{y}}= \frac{x}{x^{2}+y^{2}}$ (2)

... so that both the partial derivatives of $\mbox{arg}\ z$ exist everywhere with the only exception of $z=0$ . That means that on any circle around $z=0$ of radius $r>0$ the function $\mbox{arg}\ z$ exists and is continous...

Kind regards

$\chi$ $\sigma$

**chisigma** · January 18th 2011, 08:42 AM

May be it is useful to consider the case of the complex function $\displaystyle f(z)=\frac{1}{z}$ . Setting $z= x + i\ y$ is...

$\displaystyle f(z)= u(x,y)+i\ v(x,y) = \frac{x}{x^{2}+y^{2}} - i\ \frac{y}{x^{2}+y^{2}}$ (1)

Is easy to verify from (1) that the Cauchy-Riemann conditions...

$\displaystyle \frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}$

$\displaystyle \frac{\partial{u}}{\partial{y}} = - \frac{\partial{v}}{\partial{x}}$ (2)

... are satisfied everywhere with the only exception of $z=0$ , exactly as in the case of $\ln z$ . So both the functions have a singularity in $z=0$ but $f(z)=\frac{1}{z}$ has the Laurent expansion around $z=0$ [and is the function itself...], $f(z)=\ln z$ don't have it...

Kind regards

$\chi$ $\sigma$

**Bruno J.** · January 18th 2011, 09:10 AM

Originally Posted by chisigma

If $z= x + i\ y$ then [as in my previous post...] is...

$\displaystyle \theta= \mbox{arg}\ z = \tan^{-1} \frac{y}{x} + 2\ k\ \pi\\ ,\ \ k\in \mathbb{Z}$ (1)

... and...

$\displaystyle \frac{\partial{\theta}}{\partial{x}}= - \frac{y}{x^{2}+y^{2}}$

$\displaystyle \frac{\partial{\theta}}{\partial{y}}= \frac{x}{x^{2}+y^{2}}$ (2)

... so that both the partial derivatives of $\mbox{arg}\ z$ exist everywhere with the only exception of $z=0$ . That means that on any circle around $z=0$ of radius $r>0$ the function $\mbox{arg}\ z$ exists and is continous...

Kind regards

$\chi$ $\sigma$

Sure, the derivatives of $\mbox{arg}$ exist because any two branches only differ by an additive constant. You can define it locally in any small enough neighborhood of a point not the origin. However, it is impossible to define $\mbox{arg}$ continuously on a circle around the origin.

**chisigma** · January 20th 2011, 04:10 AM

We all [I suppose...] agree on the fact that a 'proof' must have a sort of 'rigorous background', so that I explain now my 'proof' and You all will decide if it is 'rigorous' or not...

It is well known that the function $\ln z$ is defined unless a constant $2\ k\ i\ \pi$ and it is not a limitation to 'indagate' on the 'principal branch', i. e. supposing $k=0$ . At this scope let's consider the complex function...

$\displaystyle \varphi(z)= z\ \ln z$ (1)

... in $z=1$ , where it is analytic, so that it can be represented in Taylor series...

$\displaystyle \varphi(z)= \sum_{n=0}^{\infty} \frac{\varphi^{(n)} (1)}{n!}\ (z-1)^{n}$ (2)

The computation of the derivatives of (1) in $z=1$ can be performed considering that $\varphi(*)$ is the solution of the DE...

$\displaystyle \varphi^{'} = 1 + \frac{\varphi}{z}\ \ ,\ \ \varphi(1)=0$ (3)

From (3) we derive in sequence...

$\displaystyle \varphi(1)=0$

$\displaystyle \varphi^{(1)}= 1 + \frac{\varphi}{z} \implies \varphi^{(1)} (1)= 1$

$\displaystyle \varphi^{(2)}= - \frac{\varphi}{z^{2}} + \frac{\varphi^{(1)}}{z} \implies \varphi^{(2)} (1)= 1$

$\displaystyle \varphi^{(3)}= 2\ \frac{\varphi}{z^{3}} -2\ \frac{\varphi^{(1)}}{z^{2}} + \frac{\varphi^{(2)}}{z} \implies \varphi^{(3)} (1)= -1$

$\displaystyle \varphi^{(4)}= -6\ \frac{\varphi}{z^{4}} +4\ \frac{\varphi^{(1)}}{z^{3}} - \frac{\varphi^{(2)}}{z^{2}} + \frac{\varphi^{(3)}}{z^{3}} \implies \varphi^{(4)} (1)= 2$

... and in general for $n \ge 2$ ...

$\displaystyle \varphi^{(n)} (1)= (-1)^{n}\ (n-2)!$ (4)

... so that is...

$\displaystyle \varphi (z)= z\ \ln z = (z-1) + \sum_{n=2}^{\infty} (-1)^{n} \frac{(z-1)^{n}}{n\ (n-1)}$ (5)

The (5) is a very interesting formula and more than one 'surprise' is hidden behind it. One of these is visible when we devide both terms of (5) by $z$ obtaining...

$\displaystyle \ln z = \frac{z-1}{z} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{z}\ \frac{(z-1)^{n}}{n\ (n-1)}$ (6)

It is remarkable the fact that neither (5) nor (6) are reported in any of the 'sacred books' I consulted... I wonder why!... anyway (6) offers the possibility to verify if it exists on not a couple of sequences $a_{n}$ and $b_{n}$ such that is...

$\displaystyle \ln z = \sum_{n=0}^{\infty} a_{n}\ z^{n} + \sum_{n=1}^{\infty} b_{n}\ z^{-n}$ (7)

From (6) it is immediate to verify that for $n >1$ is $b_{n}=0$ . For $n=1$ we derive from (6) that is...

$\displaystyle b_{1}= -1 + \sum_{n=2}^{\infty} \frac{1}{n\ (n-1)} =0$

... so that all the $b_{n}$ vanish... a little unusual for a function that has a singularity in $z=0$ !... Now a look to the $a_{n}$ ... for $n=0$ we have...

$\displaystyle a_{0}= 1 - \sum_{n=2}^{\infty} \frac{1}{n-1} = -\infty$

... and that is sufficient to demonstrate that a series like (6) doesn't exist...

Kind regards

$\chi$ $\sigma$

**Bruno J.** · January 20th 2011, 01:31 PM

I don't know... to go from (6) to (7), are you just rearranging the series? Because you can only rearrange a power series expansion at a point inside the region of convergence, and the point $z=0$ surely isn't inside the region of convergence of (6).

There is nothing wrong with the proof I gave. Such a Laurent series, if it existed, would converge and would be continous on an annulus centered around 0 (do you agree with this?). But this is impossible, because the logarithm has a branch cut from the north pole to the south pole, so the annulus of convergence would intersect the branch cut somewhere (agree with this?). And this would contradict the continuity of the power series expansion inside its region of convergence (agree with this?). It's the same reason why $\sqrt z$ doesn't have a Laurent series expansion about the origin, or, in general, why any function having a branch cut from $0$ to $\infty$ does not have a Laurent series expansion around the origin.

**chisigma** · January 23rd 2011, 05:06 AM

In most complex analysis textbooks it is written that the function $\ln s$ is 'non regular' along the negative real axis... according to my [very modest...] opinion that is false!...

To demostrate that it is sufficient consider the well known Taylor series expansion...

$\displaystyle \ln (1+z)= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ z^{n}\ \ ,\ \ |z|<1$ (1)

Operating tha change of variables $z=s-1$ and taking into account that $\ln s= \ln s + 2\ k\ \pi\ i$ the (1) becomes...

$\displaystyle \ln s = 2\ k\ \pi\ i - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ (s-1)^{n}\ , \ |s-1|<1$ (2)

Now if we set in (2) $-s$ instead of $s$ and consider that is $\ln (-s)= \ln s + \pi\ i$ we obtain...

$\displaystyle \ln s = (2\ k +1)\ \pi\ i - \sum_{n=1}^{\infty} \frac{(s+1)^{n}}{n}\ , \ |s+1|<1$ (3)

Observing (2) and (3) we derive that the first is the Taylor expansion of $\ln s$ in $s=1$ and the second the Taylor expansion of $\ln s$ in $s=-1$ so that we can conclude that the function $\ln s$ is analytic both in $s=1$ and $s=-1$ ...

Kind regards

$\chi$ $\sigma$

**Jose27** · January 23rd 2011, 09:18 AM

Originally Posted by chisigma

In most complex analysis textbooks it is written that the function $\ln s$ is 'non regular' along the negative real axis... according to my [very modest...] opinion that is false!...

To demostrate that it is sufficient consider the well known Taylor series expansion...

$\displaystyle \ln (1+z)= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ z^{n}\ \ ,\ \ |z|<1$ (1)

Operating tha change of variables $z=s-1$ and taking into account that $\ln s= \ln s + 2\ k\ \pi\ i$ the (1) becomes...

$\displaystyle \ln s = 2\ k\ \pi\ i - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ (s-1)^{n}\ , \ |s-1|<1$ (2)

Now if we set in (2) $-s$ instead of $s$ and consider that is $\ln (-s)= \ln s + \pi\ i$ we obtain...

$\displaystyle \ln s = (2\ k +1)\ \pi\ i - \sum_{n=1}^{\infty} \frac{(s+1)^{n}}{n}\ , \ |s+1|<1$ (3)

Observing (2) and (3) we derive that the first is the Taylor expansion of $\ln s$ in $s=1$ and the second the Taylor expansion of $\ln s$ in $s=-1$ so that we can conclude that the function $\ln s$ is analytic both in $s=1$ and $s=-1$ ...

Kind regards

$\chi$ $\sigma$

Sure, you can make the logarithm analytic in a region containing 1 and -1, the point is you can never construct one defined on a whole circle around the origin (The point of restricting to the principal branch is that it's the easiest domain in which the function can be defined and indeed once defined this way it cannot be extended, which is what I believe authors mean when they say it isn't "regular"). Also it would be helpful if you could adress Bruno J.'s and my responses, do you think they're flawed? why? As it is, the way you're responding makes me think this is the case but it feels as if you're talking to yourself.

**Bruno J.** · January 23rd 2011, 12:29 PM

Originally Posted by chisigma

In most complex analysis textbooks it is written that the function $\ln s$ is 'non regular' along the negative real axis... according to my [very modest...] opinion that is false!...

To demostrate that it is sufficient consider the well known Taylor series expansion...

$\displaystyle \ln (1+z)= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ z^{n}\ \ ,\ \ |z|<1$ (1)

Operating tha change of variables $z=s-1$ and taking into account that $\ln s= \ln s + 2\ k\ \pi\ i$ the (1) becomes...

$\displaystyle \ln s = 2\ k\ \pi\ i - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ (s-1)^{n}\ , \ |s-1|<1$ (2)

Now if we set in (2) $-s$ instead of $s$ and consider that is $\ln (-s)= \ln s + \pi\ i$ we obtain...

$\displaystyle \ln s = (2\ k +1)\ \pi\ i - \sum_{n=1}^{\infty} \frac{(s+1)^{n}}{n}\ , \ |s+1|<1$ (3)

Observing (2) and (3) we derive that the first is the Taylor expansion of $\ln s$ in $s=1$ and the second the Taylor expansion of $\ln s$ in $s=-1$ so that we can conclude that the function $\ln s$ is analytic both in $s=1$ and $s=-1$ ...

Kind regards

$\chi$ $\sigma$

See what Jose wrote!

What is meant by "the logarithm is non-regular along the negative real axis" is that a given branch of the logarithm, if you have chosen the branch cut to run along the negative real axis, is not analytic there. Of course, you can put the branch cut anywhere.

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