What Is the Correct Laurent Series for Cosine Functions with Inverse Arguments?

In summary, the Laurent series of \cos{\frac{1}{z}} at the singularity z=0 is 1 - \frac{1}{2z^2} + \frac{1}{24z^4} - ... , which is the MacLaurin series for \cos{u} with u = \frac{1}{z}. This series is valid because it converges at z=infinity and can be used to accurately evaluate the function around the singularity at z=0, as long as all terms are included. The series converges everywhere apart from z=0, and for accurate evaluation around the singularity, a high power in 1/z is needed.
  • #1
Mantella
10
0
Question 1:
Find the Laurent series of [itex]\cos{\frac{1}{z}}[/itex] at the singularity [itex]z = 0[/itex].
The answer is often given as,
[tex]\cos\frac{1}{z} = 1 - \frac{1}{2z^2} + \frac{1}{24z^4} - ...[/tex]
Which is the MacLaurin series for [itex]\cos{u}[/itex] with [itex]u = \frac{1}{z}[/itex]. The MacLaurin series is the Taylor series when [itex]u_0 = 0[/itex], however, we are interested in the "point" where [itex]u_0 = \infty[/itex]! Why is this answer considered valid if it expands the function around the wrong point?

Question 2:
If the above answer is considered correct then if I was interested in finding the Laurent series of [itex]\cos{\frac{1}{z-1}}[/itex] at the singularity of [itex]z=1[/itex] then would the answer simply be the Taylor expansion of [itex]\cos{u}[/itex] around the point [itex]u_0 = [/itex]1 with [itex]u = \frac{1}{z}[/itex]?
 
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  • #2
The answer is clearly wrong; where did you get it ?
The cosine is limited to the range [-1,1] and the series diverges.
 
  • #3
Question 1. a) here -> http://www.math.ubc.ca/~sjer/math300/s7.pdf. I found a variety of other sources saying the same thing just by googling "laurent series of cos(1/z)". If this is wrong then how would I construct a Laurent series for [itex]\cos{\frac{1}{z}}[/itex] such that I could find it's residue?
 
  • #4
The Laurent series is correct for z=infinity and it converges. For very small z, you need many terms to get close to the limit, but this is typical for a Laurent series with oscillating behavior.

For evaluation at z=0, you would need powers of z, but then your series evaluation gets very odd. You should be able to use the standard formulas for that. The result should be the same, as only negative powers will contribute.
 
  • #5
Mantella said:
Find the Laurent series of ##cos{\frac{1}{z}}## at the singularity z = 0.
Before you start thinking about the Laurent series, remember that as [itex]z\rightarrow 0 [/itex], [itex]\cos(\frac{1}{z}) [/itex] gets arbitrarily close to each point in [-1, 1] (each point (0, y) with y∈[-1, 1] is a cluster point). Thus [itex]\cos(\frac{1}{z}) [/itex] has no limit as [itex]z\rightarrow 0 [/itex].
 
  • #6
My bad - looking at real ##z## only isn't the idea with Laurent series - sorry :frown:
 
  • #7
So, even though the expansion occurs at z=infinity it is still technically accurate around the singularity of z=0 because I am hypothetically using an infinite number of terms? If I actually wanted to accurately evaluate the function around the singularity I would need to go to very high power in 1/z right? So in this way it doesn't really matter where I am centering the series provided that I include all the terms in the series?
 
  • #8
The full series converges everywhere apart from z=0.
Mantella said:
If I actually wanted to accurately evaluate the function around the singularity I would need to go to very high power in 1/z right? So in this way it doesn't really matter where I am centering the series provided that I include all the terms in the series?
Right.
 
  • #9
Compare the pictures for ##e^{-1/z^2}## which behaves a little more decently, but still has a singularity at ##z=0##.

(hehe, I learned from this thread!:smile:)
 
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Related to What Is the Correct Laurent Series for Cosine Functions with Inverse Arguments?

1. What is a Laurent series at infinity?

A Laurent series at infinity is a type of power series expansion that is used to represent functions in the complex plane. It is also known as a "Laurent expansion" or "Laurent polynomial" and is written in the form of a sum of terms with increasing powers of z and 1/z.

2. How is a Laurent series at infinity different from a Taylor series?

A Taylor series is a type of power series that represents a function in a neighborhood of a specific point. On the other hand, a Laurent series at infinity represents a function in the entire complex plane, including the point at infinity. This means that the Laurent series includes both positive and negative powers of z, while a Taylor series only includes positive powers.

3. What is the significance of "poles" in a Laurent series at infinity?

"Poles" refer to points in the complex plane where a function has a singularity, or a point at which it is undefined. In a Laurent series at infinity, poles are represented by the negative powers of z. The order of the pole is determined by the highest negative power of z in the series.

4. How is a Laurent series at infinity used in complex analysis?

A Laurent series at infinity is used to analyze the behavior of functions in the complex plane. It can help determine the location and nature of singularities, as well as the behavior of a function near these points. It is also used in contour integration and in solving differential equations in the complex plane.

5. Can any function be represented by a Laurent series at infinity?

No, not all functions can be expressed as a Laurent series at infinity. The function must have a pole at infinity, which means that it must be unbounded as z approaches infinity. Functions with essential singularities, such as e^z, cannot be represented by a Laurent series at infinity.

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