Radius of Convergence

**CSM** · July 1st 2011, 02:21 PM

So I had to calculate the radius of convergence of
$\frac{(z-i)}{z^3-z}$ around the point $z=2i$

http://www.sb.hik.se/personal/sanno/.../Chapter_5.pdf
V.4.1.f) The Answer is 2.

This is how I tried it. I'd love to hear what is wrong with my attempt:
First I rewrote it
$\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}$
Then with partial fractioning:
$\frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2$
So we can write are original function as :
$\frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}$
And use the geometric series formula:
$=2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n$
And finally apply the ratio test to the coefficients:
$a_n=(2i)^{n-1}, a_m=i^{n-1}$
$|\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2$
$|\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1$

So I guess R would be one.
Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)

**Also sprach Zarathustra** · July 1st 2011, 02:38 PM

Originally Posted by CSM

So I had to calculate the radius of convergence of
$\frac{(z-i)}{z^3-z}$ around the point $z=2i$

This is how I tried it. I'd love to hear what is wrong with my attempt:
First I rewrote it
$\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}$
Then with partial fractioning:
$\frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2$
So we can write are original function as :
$\frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}$
And use the geometric series formula:
$=2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n$
And finally apply the ratio test to the coefficients:
$a_n=(2i)^{n-1}, a_m=i^{n-1}$
$|\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2$
$|\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1$

So I guess R would be one.
Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)

Mistake:
$\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}$

z^3-z=z(z^2-1)=z(z-1)(z+1)

**CSM** · July 1st 2011, 03:51 PM

Ouch! Of course. I'll give it a re-run.

**CSM** · July 1st 2011, 03:56 PM

Hmmmm but how do I proceed with that?
$\frac{(z-i)}{z^3-z}$ around the point $z=2i$
$=\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-1)(z+1)}$

**Prove It** · July 1st 2011, 07:57 PM

Partial Fractions.

**chisigma** · July 1st 2011, 10:04 PM

Originally Posted by CSM

So I had to calculate the radius of convergence of
$\frac{(z-i)}{z^3-z}$ around the point $z=2i$

This is how I tried it. I'd love to hear what is wrong with my attempt:
First I rewrote it
$\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}$
Then with partial fractioning:
$\frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2$
So we can write are original function as :
$\frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}$
And use the geometric series formula:
$=2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n$
And finally apply the ratio test to the coefficients:
$a_n=(2i)^{n-1}, a_m=i^{n-1}$
$|\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2$
$|\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1$

So I guess R would be one.
Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)

The function $f(z)= \frac{1}{z\ (z+i)}$ has two singularities in $z=0$ and $z=-i$ , so that the Taylor series expansion of $f(*)$ around point $z=z_{0}$ has radious of convergence equal to the distance from $z_{0}$ of the 'more near' singularity. If $z_{0}= 2 i$ then clearly is $r=2$ ...

Kind regards

$\chi$ $\sigma$

**CSM** · July 2nd 2011, 02:43 AM

Thanks. Although I made I mistake so the function your looking at isn't correct.
But I could say $\frac{z-i}{z^3-z}=\frac{z-i}{z(z+1)(z-1)}$ has poles in $z=0, z=-1, z=+1$ and then reason like you did...

I did a partional fractioning and got:
$\frac{i}{z}+\frac{\frac{-1}{2}-\frac{i}{2}}{z+1}+\frac{\frac{1}{2}-\frac{i}{2}}{z-1}$
And then rewrote it to a form with an easy geometric series approach.
So for example $\frac{i}{z}=i\frac{1}{z}=-i\frac{1}{-z}=-i\frac{1}{2i-(z-2i)}=\frac{-i}{2i}\frac{1}{1-(\frac{z-2i}{2i})}$
Gives $\frac{-1}{2}\sum(\frac{z-2i}{2i})^n=\frac{-1}{2}\sum(\frac{1}{2i})^n(z-2i)^n$
And that's just for the first part.... that's a lot of work....

**FernandoRevilla** · July 2nd 2011, 03:05 AM

Originally Posted by CSM

Thanks. Although I made I mistake so the function your looking at isn't correct. But I could say $\frac{z-i}{z^3-z}=\frac{z-i}{z(z+1)(z-1)}$ has poles in $z=0, z=-1, z=+1$ and then reason like you did...

The nearest singularity of $f$ to $2i$ is $0$ , so by a well known result the radius of convergence for the series expansion is $r=d(2i,0)=2$ . You needn't use partial fractions .

**CSM** · July 2nd 2011, 07:26 AM

Another one, -note this is not my homework I am trying to grasp it and got the solutions- which I found hard.

What is the radius of convergence of $f(z)=\frac{z^2-1}{z^3-1}$ around $z=2$

Well I tried and I tried, but couldn't rewrite it nicely.

Solution:

Rewrite $f(z)=\frac{z+1}{(z-e^{2\pi i/3})(z+e^{2\pi i/3})}$
The singularities are $z=e^{2\pi i/3}. z=-e^{2\pi i/3}$ and the distance from 2 to the nearest is $\sqrt{7}$

Well this seems very farfetched to me. How could I have come up with this rewriting myself?

**FernandoRevilla** · July 2nd 2011, 07:49 AM

Originally Posted by CSM

Well this seems very farfetched to me. How could I have come up with this rewriting myself?

$f(z)=\frac{z^2-1}{z^3-1}=\frac{(z-1)(z+1)}{(z-1)(z^2+z+1)}=\dfrac{z+1}{(z-z_0)(z-\bar{z_0})}$ with $z_0=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\;i$ . Then, $r=d(2,z_0)=d(2,\bar{z_0})=\sqrt{7}$ .

**CSM** · July 2nd 2011, 08:06 AM

Originally Posted by FernandoRevilla

$f(z)=\frac{z^2-1}{z^3-1}=\frac{(z-1)(z+1)}{(z-1)(z^2+z+1)}=\dfrac{z+1}{(z-z_0)(z-\bar{z_0})}$ with $z_0=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\;i$ . Then, $r=d(2,z_0)=d(2,\bar{z_0})=\sqrt{7}$ .

Gracias! Makes perfect sense now.

**FernandoRevilla** · July 2nd 2011, 08:23 AM

Originally Posted by CSM

Gracias!

Avec plaisir.

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