Dear Nils,
Thanks for the careful explanation. There's a good reason I don't teach
this stuff (even if I enjoy it!).
Rob
On Thursday, November 5, 2015 at 7:35:47 PM UTC-8, Nils Bruin wrote:
>
> On Thursday, November 5, 2015 at 5:11:42 PM UTC-8, Rob Beezer wrote:
>>
>> LIke I said, my complex a
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
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 (ab
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 ?
On Nov 5, 2015 3:53 AM, "Rob Beezer" wrote:
>
>
> On Wednesday, November 4, 2015 at 7:37:58 PM UTC-8, vdelecroix
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
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z/2 vs 2/z
On 05/11/15 00:32, Rob Beezer wrote:
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
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 sh
