Hi,
On Wed, Mar 14, 2012 at 7:56 AM, Laurent moky.m...@gmail.com wrote:
I'm searching for a function
f(x)=1 if x in [a,b], =0 otherwise
how about using the Heaviside step function?
a=0
b=1
f(x) = heaviside(x-a) - heaviside(x-b)
Cheers,
--
Vegard
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Hi all
I've (at least) two ways of plotting a function.
The first one is the classic :
f=(x**2).function(x)
Q=plot(f,x,-1,1)
I've an other one :
==
class CallFunction(object):
def __init__(self,f):
self.f=f
def __call__(self,x):
return
Il 14/03/2012 08:09, Vegard Lima ha scritto:
Hi,
On Wed, Mar 14, 2012 at 7:56 AM, Laurentmoky.m...@gmail.com wrote:
I'm searching for a function
f(x)=1 if x in [a,b], =0 otherwise
how about using the Heaviside step function?
oh yes, of course it works. Thanks.
...
well
My residual
On Wed, Mar 14, 2012 at 12:08 AM, Laurent moky.m...@gmail.com wrote:
My question in the following : is it normal that the second one is *much*
slower than the first one ?
The reason why the second one is much faster is Sage has something
called fast_callable which can optimize the evaluation of
Ashwin Ganesan:
I am forwarding your email to sage-support.
For info on isomorphism_to you can look at the section of the
Sage reference manual
http://www.sagemath.org/doc/reference/groups.html
or use the on-line help as in this example:
sage: G = CubeGroup()
sage: G.isomorphism_to?
If
On Mar 13, 8:51 pm, FOAD KHOSHNAM mako...@gmail.com wrote:
Hello Dear
how can I calculate the integral point for this curve:
[0,0,0,-3568202637461440265241263457,0]
Define the elliptic curve by E =
EllipticCurve([0,0,0,-3568202637461440265241263457,0]) and then use
E.integral_points()
On Feb 20, 12:35 pm, Burcin Erocal bur...@erocal.org wrote:
Or you can do this:
sage: t = -2/3*x + 4/3
sage: t._convert(RR)
-0.667*x + 1.33
Cheers,
Burcin
This is perfect - thank you! For the record, I've used that idea to
write the following bits:
def
