Approssimazione in serie di Laurent
Buona sera, ho problemi con il seguenete esercizio:
-determinare lo sviluppo in serie di Laurent delle seguenti funzioni relativamente ai punti indicati:
$f(z)=(z-3)sen (1/(2+z)) , z_0=-2$. Non capisco perche' nella soluzione l'approssimazione nel punto indicato e'data da: $f(u(z))=(u-5)*(1/u-1/(3!u^3)+1/(5!u^5)+...)$, avendo applicato la sostituzione $u(z)=z+2$. Quella presentata dovrebbe essere la soluzione per $z=oo$ o forse mi sto confondendo? Grazie.
-determinare lo sviluppo in serie di Laurent delle seguenti funzioni relativamente ai punti indicati:
$f(z)=(z-3)sen (1/(2+z)) , z_0=-2$. Non capisco perche' nella soluzione l'approssimazione nel punto indicato e'data da: $f(u(z))=(u-5)*(1/u-1/(3!u^3)+1/(5!u^5)+...)$, avendo applicato la sostituzione $u(z)=z+2$. Quella presentata dovrebbe essere la soluzione per $z=oo$ o forse mi sto confondendo? Grazie.
Risposte
The Laurent series of $sin z$ at $infty$ is that one you get, but your function is not $sin z$ but $sin(1/z)$ (up to that change of variables you do).
So is the expansion wrong or not?
No, it's not wrong. As I told you, the series you get would be the Laurent expansion of $sin z$ at $infty$, but your function is $sin \frac{1}{z}$, so its expansion at $0$ is the same as the expansion at $\infty$ of $sin z$ because of definition of Laurent series expansion of a function at $\infty$.
Summarising: your result is correct and your confusion comes from the definition of Laurent expansion of a function $f(z)$ at $\infty$ (which is the Laurent expansion of $f(\frac{1}{z})$ at $0$) and the fact that your function is $sin \frac{1}{z}$ instead of $sin z$.
Summarising: your result is correct and your confusion comes from the definition of Laurent expansion of a function $f(z)$ at $\infty$ (which is the Laurent expansion of $f(\frac{1}{z})$ at $0$) and the fact that your function is $sin \frac{1}{z}$ instead of $sin z$.
Ok... so this expansion of $f(z)=sen(1/z)$ is true for all point in $ CC $ with the exception of $z_0=0$; therefore ,until we exlude this point from the circular crown (in wich f is olomorphic), the approximation is valid (thanks to the laurent's expansion theorem) and suggest to us what is his comportament nearby $z_0=0$, in analogy with $sen (z) $ in an infinite point. Is it right?
Yes, it is.
Thank you for the answer,javi.
Prego.
another two questions:
1) if i have a $z_0$ point in wich $f(z_0)$ has got a n order pole (for exemple $n=1$) so the Laurent's series will be expressed ever like $f(z)=c_(-1)/(z-z_0)+sum_(n=0)^(oo)a_n(z-z_0)^n$?
2) What the differences between study a function in a point and in a intervall?
Take for exemple $f(z)=1/[(z+1)(z+3)]$ and the $A:={1<|z-0|<3}$. It's easy to see that the function has not problems in $A$ beacuse we exclude the singular points and the expansions are definited in $z_0=0$ that is not a singular point (like we can see calculating $f(z_0)$. If we decompose the $f(z)=f_1(z)+f_2(z)=A/(z+1)+B/(z+3)$ why in one case (for $|z|>1$) we do: $f_1(z)=A/[z(1+1/z)]$ while in the second (for $|z|<3$): $f_2(z)=B/[3(1+z/3)]$? And how much important is decompose in fraction? Is there any method more directly than the decomposition? Thank you.
1) if i have a $z_0$ point in wich $f(z_0)$ has got a n order pole (for exemple $n=1$) so the Laurent's series will be expressed ever like $f(z)=c_(-1)/(z-z_0)+sum_(n=0)^(oo)a_n(z-z_0)^n$?
2) What the differences between study a function in a point and in a intervall?
Take for exemple $f(z)=1/[(z+1)(z+3)]$ and the $A:={1<|z-0|<3}$. It's easy to see that the function has not problems in $A$ beacuse we exclude the singular points and the expansions are definited in $z_0=0$ that is not a singular point (like we can see calculating $f(z_0)$. If we decompose the $f(z)=f_1(z)+f_2(z)=A/(z+1)+B/(z+3)$ why in one case (for $|z|>1$) we do: $f_1(z)=A/[z(1+1/z)]$ while in the second (for $|z|<3$): $f_2(z)=B/[3(1+z/3)]$? And how much important is decompose in fraction? Is there any method more directly than the decomposition? Thank you.
1) Yes.
2) I don't understand your question. I mean, "decompose in (partial) fractions" for what?
2) I don't understand your question. I mean, "decompose in (partial) fractions" for what?
I try to be more specific. The exercise's request is:
expande with Laurent's series the function $f(z)=1/[(z+1)(z+3)]$ in the following cases:
1)$1<|z|<3$
2)$|z|<1$
3)$|z|>3$
For the first intervall the solution is showed like i wrote in the last message: decomposing $f(z)=f_1(z)+f_2(z)=A/(z+1)+B/(z+3)$ and distinguishing: for $|z-z_0|=|z|>1 => f_1(z)=A/[z(1+1/z)]=A/z*(sum_0^(oo)(-1)^n(z)^(-n))$ and for $|z-z_0|<3 => f_2(z)=B/(3[1+z/3])=B/3(sum_0^(oo)(-1)^n(z/3)^n)$; so i didn't undestand why he follow this method and not expande directly $f(z)$ in $z_0=0$? Furthermore the function is olomorphic in the point $f(z_0)$,where $z_0=0$ is the center of expansion, and is olomorphic in all the region between $1<|z|<3$ so i would use only the Taylor's rappresentation while he use for $f_1(z)$ the principal part of L.S but not for $f_2(z)$.
I hope to have been clear.
P.s: sorry for my bad english.
expande with Laurent's series the function $f(z)=1/[(z+1)(z+3)]$ in the following cases:
1)$1<|z|<3$
2)$|z|<1$
3)$|z|>3$
For the first intervall the solution is showed like i wrote in the last message: decomposing $f(z)=f_1(z)+f_2(z)=A/(z+1)+B/(z+3)$ and distinguishing: for $|z-z_0|=|z|>1 => f_1(z)=A/[z(1+1/z)]=A/z*(sum_0^(oo)(-1)^n(z)^(-n))$ and for $|z-z_0|<3 => f_2(z)=B/(3[1+z/3])=B/3(sum_0^(oo)(-1)^n(z/3)^n)$; so i didn't undestand why he follow this method and not expande directly $f(z)$ in $z_0=0$? Furthermore the function is olomorphic in the point $f(z_0)$,where $z_0=0$ is the center of expansion, and is olomorphic in all the region between $1<|z|<3$ so i would use only the Taylor's rappresentation while he use for $f_1(z)$ the principal part of L.S but not for $f_2(z)$.
I hope to have been clear.
P.s: sorry for my bad english.
is it about radius of convergence?
Well, if I'm not mistaken, he's trying to expand $f$ around $0$ (note that the terms depending on $z$ in that sums are powers of $z$). So the point is only tom make your $z$ to be of modulus less than $1$, so that you can sum as a geometric series. If you check it, the common point in those decompositions is the fact that what you obtain is something like $C\frac{1}{1+w}$, with $|w|< 1$, so you are able to write it as a geometric sum depending on $w$. To be able to do this, you have (of course) to know how the modulus of the points of your domain are. That's why the expansion is not the same for all those domains.
So if the circle crown was defined like $0<|z|<1$, it would been licit to expand the first term too ($f_1 (z) $) like a Taylor' serie becouse, in this case, we have $|w|=|z|<1$ for all point in the region?
Yes, in that region your function is holomorphic (and the functions $f_1$ and $f_2$ as well), so the Laurent series around 0 is actually the Taylor series around 0.
thank you for your time, Javi
Prego
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