Laurent Series: Exploring the Expansion of e^(z^2)/z^3 at z=0

In summary, the conversation is discussing the concept of Laurent series and its relation to Taylor/Maclaurin series. The question is about expanding a given function into Laurent series form, and the solution is presented. However, there is confusion about whether the given solution is a Laurent series or a Taylor/Maclaurin series. It is clarified that the given solution is indeed a Laurent series and the conversation ends with the understanding that Taylor/Maclaurin series can be considered a subset of Laurent series with only positive exponents.
  • #1
MissP.25_5
331
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
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • Like
Likes 1 person
  • #6
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?
 
  • #7
Yes, of course. But what does that have to do with this problem? This series has negative powers.
 
  • Like
Likes 1 person

What is a Laurent Series?

A Laurent Series is an expansion of a complex function into a series of powers of the variable z, with both positive and negative powers allowed. It is used to represent functions in a more compact form and to approximate functions that are difficult to evaluate directly.

Why is e^(z^2)/z^3 expanded at z=0?

The expansion at z=0 is known as the Maclaurin Series, which is a special case of the Laurent Series. It allows us to approximate the behavior of the function near the origin, which can be useful in many applications.

What is the significance of expanding e^(z^2)/z^3 at z=0?

The expansion of e^(z^2)/z^3 at z=0 is significant because it is related to the error function, which is commonly used in statistics and probability. It also has applications in physics, particularly in the study of quantum mechanics.

How is the Laurent Series of e^(z^2)/z^3 at z=0 calculated?

The Laurent Series is calculated by taking the derivatives of the function at z=0 and plugging them into the formula for the Laurent coefficients. This process can be repeated to find as many terms as needed for the desired level of accuracy.

What is the convergence of the Laurent Series for e^(z^2)/z^3 at z=0?

The convergence of the Laurent Series depends on the radius of convergence, which is determined by the location of the singularities of the function. In the case of e^(z^2)/z^3 at z=0, the series converges for all values of z except for z=0, where it has a pole of order 3.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
312
  • Calculus and Beyond Homework Help
Replies
9
Views
869
  • Calculus and Beyond Homework Help
Replies
6
Views
397
  • Calculus and Beyond Homework Help
Replies
6
Views
676
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
453
  • Calculus and Beyond Homework Help
Replies
2
Views
792
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
640
Back
Top