How do I expand $\frac{e^{z^2}}{z^3}$ into a Laurent series?

In summary, the conversation discusses the concept of Laurent series and how it differs from Maclaurin series. The specific example of expanding $\frac{e^{z^2}}{z^3}$ into a Laurent series is given, along with the formula for Laurent series and how it relates to the Maclaurin series of the exponential function. It is also mentioned that Taylor series is a restricted form of Laurent series. Finally, there is a brief mention of the coefficients $a_n$ and $b_n$ representing positive and negative degrees respectively.
  • #1
aruwin
208
0
Hello.
I need explanation about this Laurent series.

The question is:
Let {$z\inℂ|0<|z|$}, expand $\frac{e^{z^2}}{z^3}$ where the centre z=0 into Laurent series.

And the solution is:
$$\frac{e^{z^2}}{z^3}=\sum_{n=0}^{\infty}\frac{\frac{(z^2)^n}{n!}}{z^3}=\sum_{n=0}^{\infty}\frac{z^{2n-3}}{n!}$$

I don't understand the solution because isn't the formula for Laurent series
$$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$

where
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$

$$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$
 
Physics news on Phys.org
  • #2
aruwin said:


And the solution is:
$$\frac{e^{z^2}}{z^3}=\sum_{n=0}^{\infty}\frac{\frac{(z^2)^n}{n!}}{z^3}=\sum_{n=0}^{\infty}\frac{z^{2n-3}}{n!}$$


Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.
 
  • #3
aruwin said:
Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.
Okay, maybe my original (deleted) post was something that should have been mentioned...

The series is of the form
\(\displaystyle \frac{e^{z^2}}{z^3} = \frac{b_0}{z^3} + \frac{b_1}{z^1} + a_0z^1 + \text{...}\)

The negative powers of z make it a Laurent, as opposed to Maclaurin, expansion.

-Dan
 
  • #4
aruwin said:
Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.

Keep in mind Laurent series are expressions of the form

$\displaystyle \sum_{n = - \infty}^\infty a_n (z - z_0)^n$,

where $a_n \in \Bbb C$. In particular, Maclaurin series are Laurent series.

The Maclaurin series of the exponential function $e^z$ is

$\displaystyle \sum_{n = 0}^\infty \frac{z^n}{n!}$

Replacing $z$ by $z^2$, we get the Maclaurin series for $e^{z^2}$:

$\displaystyle \sum_{n = 0}^\infty \frac{z^{2n}}{n!}$

Dividing by $z^3$ yields the Laurent series for $e^{z^2}/z^3$:

$\displaystyle \sum_{n = 0}^\infty \frac{z^{2n-3}}{n!} = \frac{1}{z^3} + \frac{1}{z} + \frac{z}{2} + \frac{z^3}{3!} + \cdots$.
 
  • #5
Euge said:
Keep in mind Laurent series are expressions of the form

$\displaystyle \sum_{n = - \infty}^\infty a_n (z - z_0)^n$,

where $a_n \in \Bbb C$. In particular, Maclaurin series are Laurent series.

The Maclaurin series of the exponential function $e^z$ is

$\displaystyle \sum_{n = 0}^\infty \frac{z^n}{n!}$

Replacing $z$ by $z^2$, we get the Maclaurin series for $e^{z^2}$:

$\displaystyle \sum_{n = 0}^\infty \frac{z^{2n}}{n!}$

Dividing by $z^3$ yields the Laurent series for $e^{z^2}/z^3$:

$\displaystyle \sum_{n = 0}^\infty \frac{z^{2n-3}}{n!} = \frac{1}{z^3} + \frac{1}{z} + \frac{z}{2} + \frac{z^3}{3!} + \cdots$.

So basically, Laurent series is Taylor series but with the difference that it has negative terms?
 
  • #6
aruwin said:
So basically, Laurent series is Taylor series but with the difference that it has negative terms?
Yes, but it's probably better to say that a Taylor series is a Laurent series with only positive exponents. (ie. A Taylor series is a restricted form of a Laurent series.)

-Dan
 
  • #7
topsquark said:
Yes, but it's probably better to say that a Taylor series is a Laurent series with only positive exponents. (ie. A Taylor series is a restricted form of a Laurent series.)

-Dan

One more question. Is the coefficient $a_n$ represent the coefficient for the positive degree and $b_n$ the negative degree?
 
  • #8
aruwin said:
One more question. Is the coefficient $a_n$ represent the coefficient for the positive degree and $b_n$ the negative degree?
That's how I remember learning it, but I doubt it would cause any confusion by not following it.

-Dan
 

FAQ: How do I expand $\frac{e^{z^2}}{z^3}$ into a Laurent series?

1. What is a Laurent series?

A Laurent series is a type of mathematical series used to represent complex functions as a sum of infinitely many terms. It is similar to a Taylor series, but it includes both positive and negative powers of the variable.

2. Why is it important to expand a function into a Laurent series?

Expanding a function into a Laurent series allows us to better understand its behavior, particularly near singularities or branch points. It also makes it easier to perform calculations and approximate values of the function.

3. How do I determine the coefficients in a Laurent series?

The coefficients in a Laurent series can be found by using the formula: , where z0 is the center of the series and f(z) is the given function.

4. How do I determine the domain of convergence for a Laurent series?

The domain of convergence for a Laurent series is the set of all points in the complex plane for which the series converges. To determine this, we can use the ratio test or the root test, similar to determining the radius of convergence for a Taylor series.

5. Can any function be expanded into a Laurent series?

No, not all functions can be expanded into a Laurent series. The function must have a singularity or branch point within the region of convergence for the series to exist. For example, the function cannot be expanded into a Laurent series because it has a singularity at z=0 and the region of convergence would be the entire complex plane.

Similar threads

Back
Top