# Thread: Showing a removable singularity

1. ## Showing a removable singularity

http://i55.tinypic.com/wlf901.jpg

Hey guys,

I need to solve the above problem. I think I need to use Laurent's theorem but I am unsure how to proceed with it.

Thanks

2. ## Re: Showing a removable singularity

Hi,
I think your idea is good.
You can use the fact that the function is analytic in the punctured disk to obtain a Laurent series (in (z-a)) that converges in the punctured disk. You can then argue that as $\displaystyle z \rightarrow a$ and $\displaystyle |f(z) \leq {C \over \sqrt{z-a}}|$ all the negative order contribution must vanish, since they would otherwise asymptotically beat $\displaystyle {C \over \sqrt{z-a}}$.

3. ## Re: Showing a removable singularity

Another way: Since $\displaystyle |(z-a)f(z)|\leq C$ and $\displaystyle |(z-a)f^2(z)|\leq C^2$ we can say that $\displaystyle f,f^2$ have a pole of order at most one at $\displaystyle a$, this can't happen unless $\displaystyle a$ is a removable singularity for $\displaystyle f$.

4. ## Re: Showing a removable singularity

Originally Posted by mathshelpee
http://i55.tinypic.com/wlf901.jpg

Hey guys,

I need to solve the above problem. I think I need to use Laurent's theorem but I am unsure how to proceed with it.

Thanks
A simple counterexample...

$\displaystyle f(z)= \ln (z-a)$ (1)

Applying l'Hopital rule You find that...

$\displaystyle \lim_{z \rightarrow a} \ln (z-a)\ \sqrt{z-a} =0$ (2)

... but it is obvious that $\displaystyle \ln (z-a)$ don't have in a a removable singularity. The 'statement' of Your 'theorem' has to be changed... how?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: Showing a removable singularity

Originally Posted by chisigma
A simple counterexample...

$\displaystyle f(z)= \ln (z-a)$ (1)

Applying l'Hopital rule You find that...

$\displaystyle \lim_{z \rightarrow a} \ln (z-a)\ \sqrt{z-a} =0$ (2)

... but it is obvious that $\displaystyle \ln (z-a)$ don't have in a a removable singularity. The 'statement' of Your 'theorem' has to be changed... how?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
That function is not analytic in a neighbourhood of $\displaystyle a$...

6. ## Re: Showing a removable singularity

Originally Posted by Jose27
That function is not analytic in a neighbourhood of $\displaystyle a$...
Let's suppose [without loss of generality..] that $\displaystyle a=1$ so that we can use the same approach used for the 'Riemann zeta function' in...

http://www.mathhelpforum.com/math-he...on-186544.html

The function $\displaystyle f(s)= \ln (s-1)$ is analytic in $\displaystyle s=2$ and here its Taylor expansion is...

$\displaystyle \ln (s-1)= (s-2) - \frac{(s-2)^{2}}{2} + \frac{(s-2)^{3}}{3} -...$ (1)

The Taylor expansion (1) converges of course inside a circle centered in $\displaystyle s_{0}=1$ with radious r=1. Now , as illustrated in the figure...

... we can proceed as in the post about the Riemann zeta function and extend the region in which $\displaystyle \ln (s-1)$ is analytic 'everywhere around' the point $\displaystyle s=1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. ## Re: Showing a removable singularity

Originally Posted by chisigma
... we can proceed as in the post about the Riemann zeta function and extend the region in which $\displaystyle \ln (s-1)$ is analytic 'everywhere around' the point $\displaystyle s=1$...
Impossible.

8. ## Re: Showing a removable singularity

Originally Posted by FernandoRevilla
Impossible.
I don't think that, for the specific question, the word 'impossible' is written in the Coran... then a minimum of [possibly convincing...] demonstration is necessary...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

9. ## Re: Showing a removable singularity

Originally Posted by chisigma
I don't think that, for the specific question, the word 'impossible' is written in the Coran... then a minimum of [possibly convincing...] demonstration is necessary...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
By the way, with the precise purpose to avoid waste of time, I suggest to read the following recent post...

http://www.mathhelpforum.com/math-he...es-186754.html

Here for the computation of the integral....

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{(1+2x)\ (1+x+x^{2})}\ dx$ (1)

... has been proposed the integration of the complex function $\displaystyle f(z)= \frac{\ln z}{(1+2z)\ (1+z+z^{2})}$ along the path in the figure...

Now it is evident that if $\displaystyle \ln z$ is not analytic inside the 'circular proximity' of $\displaystyle z=0$ of radious r the method fails and the computation of the integral (1) becomes [really...] 'impossible'... fortunately that is not the case ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

10. ## Re: Showing a removable singularity

Originally Posted by chisigma
By the way, with the precise purpose to avoid waste of time, I suggest to read the following recent post...

http://www.mathhelpforum.com/math-he...es-186754.html

Here for the computation of the integral....

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{(1+2x)\ (1+x+x^{2})}\ dx$ (1)

... has been proposed the integration of the complex function $\displaystyle f(z)= \frac{\ln z}{(1+2z)\ (1+z+z^{2})}$ along the path in the figure...

Now it is evident that if $\displaystyle \ln z$ is not analytic inside the 'circular proximity' of $\displaystyle z=0$ of radious r the method fails and the computation of the integral (1) becomes [really...] 'impossible'... fortunately that is not the case ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
I don't know what you're trying to get at with this but your counterexample, as I said, doesn't work because the function is not analytic there: A function is analytic in a neighbourhood of $\displaystyle a$ if and only if it has a Laurent expansion around $\displaystyle a$, and as you pointed out here such an expansion of the logarithm doesn't exist.

11. ## Re: Showing a removable singularity

Originally Posted by chisigma
Now it is evident that if $\displaystyle \ln z$ is not analytic inside the 'circular proximity' of $\displaystyle z=0$ of radious r the method fails and the computation of the integral (1) becomes [really...] 'impossible'... fortunately that is not the case ...
Are you assuring that $\displaystyle f(z)=\ln z$ is analytic in $\displaystyle 0<|z|<r$ ?

12. ## Re: Showing a removable singularity

Originally Posted by FernandoRevilla
Are you assuring that $\displaystyle f(z)=\ln z$ is analytic in $\displaystyle 0<|z|<r$ ?
... the procedure requires f(z) to be analytic on the integration path if $\displaystyle R \rightarrow \infty$ and $\displaystyle r \rightarrow 0$... then, non considering the poles with negative real part of f(z), $\displaystyle \ln z$ must be analytic for $\displaystyle |z|=r>0$...

The proof that $\displaystyle \ln z$ is analytic in $\displaystyle \mathbb{C}$ with the only exception of the point $\displaystyle z=0$ is given in...

http://www.mathhelpforum.com/math-he...st-167358.html

$\displaystyle \chi$ $\displaystyle \sigma$
What for?. We are talking about a simple question: is $\displaystyle f(z)=\ln z$ analytic in $\displaystyle 0<|z|<r$? (of course when the notation $\displaystyle \ln$ appears without any particular logarithm having been specified, it is generally assumed that the principal value is intended). The answer is no and this is related with the objection of Jose27 to your counterexample, so I think that it is useless to continue with this discussion.