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

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Discussion Overview

The discussion revolves around the correct formulation of the Laurent series for the cosine function with inverse arguments, specifically \(\cos{\frac{1}{z}}\) at the singularity \(z = 0\). Participants explore the validity of using the Maclaurin series for this purpose and the implications of evaluating the series at different singularities.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant questions the validity of the Laurent series expansion of \(\cos{\frac{1}{z}}\) around \(z = 0\), suggesting that it is based on a Maclaurin series centered at infinity, which may not be appropriate.
  • Another participant asserts that the series diverges and questions the source of the initial claim.
  • A different participant cites multiple sources that support the original series expansion, asking how to construct a valid Laurent series if the provided one is incorrect.
  • Some participants argue that the Laurent series is correct for \(z = \infty\) and converges, but caution that many terms are needed for small \(z\) due to oscillatory behavior.
  • One participant notes that \(\cos{\frac{1}{z}}\) approaches every point in the interval \([-1, 1]\) as \(z\) approaches \(0\), indicating that it does not have a limit at that point.
  • There is a discussion about the necessity of including higher powers of \(1/z\) to accurately evaluate the function around the singularity.
  • Another participant compares the behavior of \(\cos{\frac{1}{z}}\) with that of \(e^{-1/z^2}\), suggesting that the latter behaves more decently despite also having a singularity at \(z = 0\.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the series expansion and its applicability at different singularities. There is no consensus on the correctness of the original series or the method of constructing a valid Laurent series.

Contextual Notes

Some participants highlight the limitations of evaluating the series at \(z = 0\) due to the oscillatory nature of \(\cos{\frac{1}{z}}\) and the lack of a limit at that point. The discussion also reflects dependence on the definitions and interpretations of series expansions.

Mantella
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Question 1:
Find the Laurent series of \cos{\frac{1}{z}} at the singularity z = 0.
The answer is often given as,
\cos\frac{1}{z} = 1 - \frac{1}{2z^2} + \frac{1}{24z^4} - ...
Which is the MacLaurin series for \cos{u} with u = \frac{1}{z}. The MacLaurin series is the Taylor series when u_0 = 0, however, we are interested in the "point" where u_0 = \infty! 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 \cos{\frac{1}{z-1}} at the singularity of z=1 then would the answer simply be the Taylor expansion of \cos{u} around the point u_0 =1 with u = \frac{1}{z}?
 
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The answer is clearly wrong; where did you get it ?
The cosine is limited to the range [-1,1] and the series diverges.
 
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 \cos{\frac{1}{z}} such that I could find it's residue?
 
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.
 
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 z\rightarrow 0, \cos(\frac{1}{z}) gets arbitrarily close to each point in [-1, 1] (each point (0, y) with y∈[-1, 1] is a cluster point). Thus \cos(\frac{1}{z}) has no limit as z\rightarrow 0.
 
My bad - looking at real ##z## only isn't the idea with Laurent series - sorry :frown:
 
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?
 
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.
 
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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