Laurent series z0=infinity

Hi there. I have to find the Laurent series for $\displaystyle \frac{1}{z-2},z_0=\infty$

So what I did was calling $\displaystyle z=\frac{1}{\omega}$, I don't know if this is right.

Then:

$\displaystyle f(\omega)=\frac{1}{\frac{1}{\omega}-2}=\frac{-1}{2-\frac{1}{\omega}}=\frac{-1}{2}\frac{1}{1-\frac{1}{2\omega}}=\frac{-1}{2}\sum_0^{\infty}\left ( \frac{1}{2\omega} \right )^n,|\omega|>\frac{1}{2}$

And then back to z:

$\displaystyle f(z)=\frac{-1}{2}\sum_0^{\infty}\left ( \frac{1}{2} \right )^n z^n,|z|<2$

I have to make the series for |z|>2, but I see that it gives the same as when I centered the series in zero, so I think I'm doing something wrong.

Re: Laurent series z0=infinity

We have $\displaystyle f(z)=\frac{1/z}{1-2/z}=\dfrac{1}{z}\sum_{n=0}^{\infty}\left(\frac{2}{z }\right)^n=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+1}}$ , expansion valid for $\displaystyle |2/z|<1$ or equivalently for $\displaystyle 2<|z|<\infty$ .

Re: Laurent series z0=infinity

Yes, thats the other part for the Laurent series, but is it centered at $\displaystyle z_0=\infty$? because I've made the expansion centered at zero, and it gives exactly the same it gives when centered at infinity.

Re: Laurent series z0=infinity

Quote:

Originally Posted by

**Ulysses** Yes, thats the other part for the Laurent series, but is it centered at $\displaystyle z_0=\infty$?

Of course, $\displaystyle 2<|z|<\infty$ is a neighborhood of $\displaystyle \infty$ .

Quote:

because I've made the expansion centered at zero, and it gives exactly the same it gives when centered at infinity.

What? I see different expansions.

Re: Laurent series z0=infinity

What's the difference for $\displaystyle z_0=0$ I just don't see it. I've proceeded exactly as you did when finding the Laurent series for this same function centered at zero, and I get the same series that we've found here, with exactly the same expressions for the modulus.

If I take $\displaystyle z_0=0$ then I get:

$\displaystyle f(z)=\frac{1}{z-2}=\frac{-1}{2-z}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}=\frac{-1}{2}\sum_0^{\infty}\left( \frac{z}{2} \right )^n,|z|<2$

That's the Taylor part, the other gives the same that you found.

Re: Laurent series z0=infinity

Let us see, you have __completed__ the Laurent expansion valid for $\displaystyle |z|<2$ that is, __centered at__ $\displaystyle z_0=0$ which is the same than the Taylor expansion (there is no principal part). On the other hand we have __completed__ the Laurent expansion valid for $\displaystyle |z|>2$ that is , in another region __centered at__ $\displaystyle z_0=0$. Right? Then, in the second case __we also say__ that we have the expansion centered at $\displaystyle z_0=\infty$ because $\displaystyle 2<|z|<\infty$ is a neighborhood of $\displaystyle \infty$ .

Re: Laurent series z0=infinity

Thank you Fernando :D I was reading some of your works by the way. That poly-dimensional time that you've proposed, and the movement of movements it sounds like a really interesting thing. It's not something that we should discuss right here, I can't discuss it anyway, but I wanted to mention it :P

Bye bye! see you around.