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

[aruwin]

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

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

 

[aruwin]

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.

 

[topsquark]

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
[math]\frac{e^{z^2}}{z^3} = \frac{b_0}{z^3} + \frac{b_1}{z^1} + a_0z^1 + \text{...}[/math]

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

-Dan

 

[Euge]

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 $e^{z^2}/z^3$:

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

 

[aruwin]

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?

 

[topsquark]

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

 

[aruwin]

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?

 

[topsquark]

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