On Monday, February 16, 2015 at 8:33:49 AM UTC-8, Nils Bruin wrote:
f3=Piecewise([([0,1],SR(0).function(x)),([1,2],(1-x).function(x))])
Incidentally, the Piecewise documentation, which you can get with
Piecewise? , has a nice shortcut form:
sage: f3 = Piecewise([([0,1],SR(0)), ([1,2],1-x)],
I haven't gotten around to updating OSX yet, but I have no objection to it.
I guess this will be good motivation for me to get it done. Thanks!
On Saturday, February 14, 2015 at 4:28:09 PM UTC-7, Volker Braun wrote:
AFAIK you can't compile Sage on OSX 10.7 because of this bug. See also the
On Monday, February 16, 2015 at 5:46:06 AM UTC-8, pdenapo wrote:
Hi,
I'm having trouble with some piecewise constant functions.
Suppose that I define
f=Piecewise ([([0,1],0),([1,2],x-1)])
Then f.integral() works as expected, but f.derivative() will fail with
TypeError:
See also the
possibly-somewhat-related
http://cnx.org/search?q=subject:%22Mathematics%20and%20Statistics%22
--
You received this message because you are subscribed to the Google Groups
sage-devel group.
To unsubscribe from this group and stop receiving emails from it, send an email
to
Hi,
Another strange behavoir:
sage: f=Piecewise([[(1/3,1/2),x]])
sage: f.extend_by_zero_to(0,1)
Piecewise defined function with 3 parts, [[(0, 1/3), 0], [(1/3, 1/2),
x], [(1/2, 1), 0]]
sage: f.domain()
(1/3, 1/2)
extend_by_zero shouldn't have changed the domain to (0,1) ?
sage: f.integral()
On Monday, 16 February 2015 12:01:31 UTC+1, Jeroen Demeyer wrote:
On 2015-02-14 16:40, Bill Hart wrote:
Wikipedia claims that it is possible to compute gcd in R[x] for any
unique factorisation domain R. But in thinking about it last night, I
couldn't see why the algorithm doesn't work
Hi,
I'm having trouble with some piecewise constant functions.
Suppose that I define
f=Piecewise ([([0,1],0),([1,2],x-1)])
Then f.integral() works as expected, but f.derivative() will fail with
TypeError: 'sage.rings.integer.Integer' object is not callable
It seems that Sage does not
On 2015-02-14 16:40, Bill Hart wrote:
Wikipedia claims that it is possible to compute gcd in R[x] for any
unique factorisation domain R. But in thinking about it last night, I
couldn't see why the algorithm doesn't work in any GCD domain R.
Perhaps it works for GCD domains, but the proofs are
