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

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The discussion centers on the Laurent series expansion of the function \(\frac{e^{z^2}}{z^3}\) at \(z=0\). The solution provided is \(\sum_{n=0}^{\infty}\frac{z^{2n-3}}{n!}\), which includes negative exponents, indicating it is indeed a Laurent series. Participants clarify that while a Maclaurin series consists only of non-negative powers, a Laurent series can include both positive and negative powers, thus confirming the classification of the given series.

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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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