What is the Laurent series expansion of \(\frac{e^{z^2}}{z^3}\) at \(z=0\)?

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Homework Help Overview

The discussion revolves around the expansion of the function \(\frac{e^{z^2}}{z^3}\) into a Laurent series centered at \(z=0\). Participants are exploring the nature of the series and its classification as a Laurent series versus a Maclaurin series.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question whether the provided series is indeed a Laurent series or simply a Maclaurin expansion of the exponential function. There is discussion about the presence of negative exponents and the definition of Laurent series compared to power series.

Discussion Status

Participants are actively engaging with the definitions and characteristics of series expansions. Some have begun to recognize the implications of negative powers in the context of Laurent series, while others are still seeking clarity on the distinctions between series types.

Contextual Notes

There is an emphasis on the formal definitions of series expansions, with references to specific formulas for Laurent series. Participants are also considering the implications of homework constraints regarding the expansion of functions in complex analysis.

MissP.25_5
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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
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$

$$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$
 
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MissP.25_5 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.
 
MissP.25_5 said:
Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.

Does a MacLaurin expansion have negative exponents?
 
SammyS said:
Does a MacLaurin expansion have negative exponents?

No it doesn't but that definitely is a power series of the exponential function. Is Laurent series the same as power series? I am confused.
 
MissP.25_5 said:
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
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$

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

Just write out the first 3 or 4 terms of your series. Does that not look like a Laurent series to you?
 
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Ray Vickson said:
Just write out the first 3 or 4 terms of your series. Does that not look like a Laurent series to you?

Aha, now I see it. So, can I conclude that a Taylor/maclaurin series is a Laurent series with only positive exponents?
 
Yes, of course. But what does that have to do with this problem? This series has negative powers.
 
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