Laurent attorno a infinito

[cesare14]

Ho un dubbio riguardo le singolarità all'infinito.
Come faccio a trovare il residuo all'infinito di una funzione e determinare che singolarità è o se è un punto regolare? Se non sbaglio il residuo all'infinito è il coefficiente di 1/z cambiato di segno nello sviluppo di f(z) per grandi z, quindi se io ad esempio ho $f(z)=e^(1/z)$ ho che l'espansione attorno a 0 è:
$sum_(k =0\)^oo (1/z)^k /(k!)$

quindi deduco che il residuo vale 1, e vedo che è una singolarità essenziale perché ho infinite potenze negative di $z$

mentre l'espansione attorno a infinito la farei così, sostituirei $z=1/w$ e farei

$sum_(k =0\)^oo (w)^k /(k!)$

a questo punto devo cercare il coefficiente di $1/w$, ma vale 0, quindi il residuo mi verrebbe zero, ma dovrebbe uscire -1.
Cosa sbaglio?

[javicemarpe]

How do you know it should be $-1$?

[cesare14]

Because $Res(f(z),0) + Res(f(z),oo) = 0$

[javicemarpe]

Well, I didn't remember this property to work with a general function. I only remember it is true for rational functions. In any case, if I'm not mistaken, the residue of $f$ at $\infty$ is defined as $-Res (\frac{1}{z^2}f(\frac{1}{z}),0)$.


Edit: ok, now I see the formula is correct even for non-rational functions.

[cesare14]

It's true for every function. Yeah i know that formula but id like to understand how to expand with laurent to infinity. I have to expand $e^(1/z)/(z+1)$ to discuss about the singularity that this function has. So i have to expand it for$ z=0, z=-1$ and $z=oo$. For $z=-1$ its easy and the residue is $1/e$
For $z=oo$ i did like this:
$e^(1/z)=1+1/z+1/(z^2*2)+1/(z^3*6)+o(z^3)$
$(1/(z+1)=1/z(1-1/z+1/z^2-1/z^3+o(z^3))$
So you have that the residue is -1
For 0 i can't find it because there's the product between to sum
$(1+1/z+1/(z^2*2)+1/(z^3*6)+o(z^3))*(1-z+z^2-z^3+o(z^3))$ and i have to find the coefficient of $(1/z)$

[javicemarpe]

$(1+1/z+1/(z^2*2)+1/(z^3*6)+o(z^3))*(1-z+z^2-z^3+o(z^3))=$

You have to find the coefficient of $\frac{1}{z}$ in that product. You can see easily that the coefficient of $z^{-n}$ in the expansion of $e^{\frac{1}{z}}$ corresponds to the coefficient of $z^{n-1}$ in the expansion of $\frac{1}{z+1}$. Then, if I'm right, the coefficient of $\frac{1}{z}$ in the expansion of your function is $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}$

[cesare14]

Yeah its correct because the series converges in $1-1/e$ so the sum of residues is 0! Thanks. But why is the expansion of the $e^(1/z)$ around 0 the same as around $oo$? Is it because the radius of the series is $oo$?

[javicemarpe]

As far as I know, it doesn't exist anything called expansion of a function around infinity.

[cesare14]

I mean at infinity

[javicemarpe]

I know, but I don't know what is a Laurent expansion of a function at $\infty$. Could you define what do you mean by that?

[cesare14]

Well idk how to explain. You expand the function for $z$ that get bigger. Like $1/(z-1)$, it's $1/(z(1-1/z))=sum_(z =0 \)^oo (1/z)^(k+1) $ so you can see that i used the geometrical series because z is big, so its bigger than 1. Idk how to explain you that

[javicemarpe]

I think I understand what you want to say, but that is not an expansion at $\infty$. You only used a "trick" to be able to sum when $z$ is sufficiently far away from zero.

[cesare14]

We call it expansion around 0. Also if you use wolfram alpha and ask for a series around infinite it gives you that one. Its used when you have to calculate the residue at infinity

[javicemarpe]

I didn't know that, I've always calculated the residue at $\infty$ as I just told you.

[cesare14]

yeah but if you have a function like $1/((z)(1-z))^(1/2)$ you have to expand it around infinity

[javicemarpe]

I think I eventually understood what you are trying to tell m. By "around $\infty$" you mean for $z$ with modulus bigger than some $R>0$. Right?

[cesare14]

Yeah