Residue of the exponential function

[Rob Beezer]

My complex analysis is very rusty, so I can't tell if the problem here is me or Sage:

z = var('z')
f = e^(z/2)
f.residue(z)

Result: 0

x = var('x')
g = e^x
t = g.taylor(x, 0, 5)
t(x = 2/z)

Result: 2/z + 2/z^2 + 4/3/z^3 + 2/3/z^4 + 4/15/z^5 + 1

So the coefficient of 1/z is 2, and the residue should be 2?

Thanks,
Rob

[Vincent Delecroix]

z/2 vs 2/z

[Rob Beezer]

On Wednesday, November 4, 2015 at 7:37:58 PM UTC-8, vdelecroix wrote:

z/2 vs 2/z

Ouch!  Sorry.  Wrong input, same output.

z = var('z')

f = e^(2/z)
f.residue(z)

Result: 0

[John Cremona]

Why are you giving the argument z to the residue function ?  If you wanted the residue at 0 put that.  But exp (2/z)  is not meromorphic at 0 so the residue there is surely undefined ?

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[Rob Beezer]

Thanks, John.  Allowable syntax is  residue(z)  or  residue(z == 0).  The former defaults to using 0, so they should be equivalent and both forms give the same result.

LIke I said, my complex analysis is rusty.  And I am really asking for somebody else.  Is the chunk of code that produces 2 (above) not following the definition of the residue?

[Nils Bruin]

On Thursday, November 5, 2015 at 5:11:42 PM UTC-8, Rob Beezer wrote:

LIke I said, my complex analysis is rusty.  And I am really asking for somebody else.  Is the chunk of code that produces 2 (above) not following the definition of the residue?

No it's not. If t=1/z then you don't get a truncated taylor expansion around t=0 by substituting t=1/z in the taylor expansion around z=0.

In general, if you define the residue by a contour integral, then the residue of exp(1/z) around z=0 is 0: Any contour around z=0 is a boundary of a simply connected domain of the Riemann sphere on which the function is regular, so integrating along the contour gets you 0.

exp(1/z) has an essential singularity at z=0, so there is no laurent expansion to read off the residue from.

[Rob Beezer]

Dear Nils,

Thanks for the careful explanation.  There's a good reason I don't teach this stuff (even if I enjoy it!).

Rob