Is This the Correct Laurent Series Expansion for $\frac{1}{z^3-z^4}$?

In summary, a Laurent series is a mathematical representation of a function as a power series, including both positive and negative powers of the variable. It differs from a Taylor series in that it has a larger range of powers and is used to expand functions around an annulus rather than a specific point. To find the Laurent series of a function, the region of convergence must be identified and the coefficients can be calculated using a formula and partial fraction decomposition. In complex analysis, a Laurent series is used to expand functions in the complex plane, providing information about singularities and allowing for evaluation within the annulus. While it can be used to approximate a function within its region of convergence, it cannot approximate a function at a specific point.
  • #1
aruwin
208
0
Hello.
Can you check if my answer is correct please?

For the region ${\{z\inℂ\big|0<|z|<1\}}$, expand $\frac{1}{z^3-z^4}$ that has a center z=0 into Laurent series.

My solution:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$
 
Physics news on Phys.org
  • #2
aruwin said:
Hello.
Can you check if my answer is correct please?

For the region ${\{z\inℂ\big|0<|z|<1\}}$, expand $\frac{1}{z^3-z^4}$ that has a center z=0 into Laurent series.

My solution:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$

Yes, all right!...

Kind regards

$\chi$ $\sigma$
 

FAQ: Is This the Correct Laurent Series Expansion for $\frac{1}{z^3-z^4}$?

What is a Laurent series?

A Laurent series is a mathematical representation of a function as a power series, consisting of both positive and negative powers of the variable. It is used to expand functions that cannot be represented by a Taylor series, such as those with singularities.

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. Additionally, a Taylor series is used to approximate a function around a specific point, while a Laurent series is used to expand a function around an annulus or ring-shaped region.

What is the process for finding the Laurent series of a function?

To find the Laurent series of a function, one must first identify the region of convergence, which is the annulus in which the function is analytic. Then, the coefficients of the series can be calculated using the formula for a Laurent series and the partial fraction decomposition of the function. Finally, the series can be written out using sigma notation.

How is a Laurent series used in complex analysis?

In complex analysis, a Laurent series is used to expand a complex function around an annulus in the complex plane. This allows for the evaluation of the function at points within the annulus, and can also provide information about the behavior of the function near singularities.

Can a Laurent series be used to approximate a function?

Yes, a Laurent series can be used to approximate a function within its region of convergence. However, unlike a Taylor series, it cannot be used to approximate a function at a specific point, as it includes both positive and negative powers of the variable.

Similar threads

A Laurent series for algebraic functions
Replies
9
Views
2K
MHB Understanding the Laurent Series of $\frac{1}{z(z-1)(z-2)}$
Replies
4
Views
2K
I On a remark regarding the Cauchy integral formula
Replies
2
Views
1K
MHB Find coefficiant of Laurent series, without using residue
Replies
2
Views
2K
Calculating Laurent series of complex functions
Replies
3
Views
774
MHB Complex Analysis: Does $\int_{C(0,10)} f(z)$ Equal 0?
Replies
2
Views
1K
I Functions.wolfram.com accepted and published my formula for Gamma Fn
Replies
11
Views
2K
MHB Find Laurent Series for $\frac{1}{z^2(1 - z)}$ in $0 < |z| < 1$ & $|z| > 1$
Replies
2
Views
2K
MHB Finding the Laurent series for $f(z)$ in an annulus containing $z = 2$
Replies
1
Views
2K
MHB Calculating Residues Using Laurent Series
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top