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A Laurent series at infinity

  1. Oct 31, 2016 #1
    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]?
     
  2. jcsd
  3. Oct 31, 2016 #2

    BvU

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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.
     
  4. Oct 31, 2016 #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?
     
  5. Oct 31, 2016 #4

    mfb

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    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.
     
  6. Oct 31, 2016 #5

    Svein

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    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].
     
  7. Oct 31, 2016 #6

    BvU

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    My bad - looking at real ##z## only isn't the idea with Laurent series - sorry :frown:
     
  8. Oct 31, 2016 #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?
     
  9. Oct 31, 2016 #8

    mfb

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    The full series converges everywhere apart from z=0.
    Right.
     
  10. Nov 1, 2016 #9

    BvU

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    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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