Laurent series (COMPLICATED, )

In summary, the problem is to find the Laurent series for f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}}) around z = 0, where the series absolutely converges for z = -i. The attempt at a solution involved separating the polynomial fraction into partial fractions and using a variable change for the sine function. However, a single expression for the coefficients could not be found.
  • #1
libelec
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Laurent series (COMPLICATED, URGENT)

Homework Statement



Find the Laurent series for [tex]f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}})[/tex] around z = 0, in which the series absolutely converges for z = -i.

The Attempt at a Solution



I have several problem with this. I tell you what I tried.

First: I separate the polinomial fraction in partial fractions. I don't know if the procedure is OK: [tex]\frac{{2{z^2}}}{{2{z^2} - 5z + 2}} = 2{z^2}\left( {\frac{2}{{3(z - 2)}} - \frac{2}{{3(z - 1/2)}}} \right)[/tex]. I leave the 2z2 there because it's its own Laurent Series around z=0, for all z.

Second: I compare the function with the geometrical function (considering that I have to find the Laurent series in a vecinity of infinite for the fraction with the (z - 1/2)) and I use a variable change for the sine:

a) [tex]\frac{2}{{3(z - \frac{1}{2})}} \to \frac{2}{{3z(1 - \frac{1}{{2z}})}}
[/tex]

[tex]\frac{2}{{3z(1 - \frac{1}{{2z}})}} = \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} {\rm{ }}\forall z/\left| z \right| > \frac{1}{2}[/tex]


b)
[tex]\frac{{ -1}}{{3(z - \frac{1}{2})}} = \frac{2}{{3(\frac{1}{2} - z)}} = \frac{4}{{3(1 - 2z)}}[/tex]

c) Since [tex]\sin (u) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{u^{2n + 1}}}}{{(2n + 1)!}}}[/tex], then, using the change of variable u = 3/z2, I get that:

[tex]\sin (\frac{3}{{{z^2}}}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{ }}\forall z/\left| z \right| > 0[/tex]


Then, I have that:

[tex]f(z) = 2{z^2}\left( {\frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} - \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} } \right) + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{, }}\forall z/\frac{1}{2}< \left| z \right|< 2[/tex]

My question is, how can I come up with a sole expresion for the coefficients? I couldn't even find one for the series inside the parenthesis...

Thanks
 
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  • #2


I don't know why, but LaTeX doesn't let me put the expressions in order...
 

Related to Laurent series (COMPLICATED, )

1. What is a Laurent series?

A Laurent series is a representation of a complex function as an infinite sum of terms, including both positive and negative powers of the variable. It is named after French mathematician, Pierre Alphonse Laurent.

2. How is a Laurent series different from a Taylor series?

A Taylor series only includes positive powers of the variable, while a Laurent series includes both positive and negative powers. This allows for a more general representation of functions, particularly those with singularities.

3. What is the region of convergence for a Laurent series?

The region of convergence for a Laurent series is the annulus, which is the area between two concentric circles on the complex plane. The inner circle represents the radius of convergence for the positive powers, while the outer circle represents the radius of convergence for the negative powers.

4. How is a Laurent series used in complex analysis?

A Laurent series is used to represent complex functions, particularly those with poles or essential singularities. It allows for the evaluation of these functions in a neighborhood of the singularity, and can also be used for analytic continuation.

5. Can a Laurent series be used to approximate a function?

Yes, a Laurent series can be used to approximate a function in a neighborhood of its singularity. However, the accuracy of the approximation is limited by the size of the annulus of convergence.

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