Harald, thank you for reminding us of this post.
Raphael,
Actually I started an answer a while ago, but thought it wouldn't be
helpful and therefore didn't post it.
On 27 Jul., 19:01, lesshaste wrote:
> I am new to sage and am attempting to solve systems of multivariate
> polys over GF(2). My
I am using Sage Version 4.1, Release Date: 2009-07-09 on an iMac Intel
core 2 duo running Mac OS X 10.5.7 to compute the rank of a large
number of elliptic curves. Computing
EllipticCurve([0,0,0,-289061442785767716,0]).rank()
produces no output after more than one hour. However, using mwrank
dir
So, here is a test for the misleading warning message:
sage: print sage0.eval("alarm(1); singular._expect_expr('1')")
Control-C pressed. Interrupting R. Please wait a few seconds...
---
KeyboardInterrupt
Hi William,
On Jul 31, 6:21 pm, William Stein wrote:
...
> Here's an example of a way to doctest something like this...
>
> D-69-91-130-10:tmp wstein$ more a.py
> def hi():
> """
> sage: sage0.eval("alarm(1); gp.eval('factor(2^997-1)')")
> 'Interrupting GP/PARI interpreter...Keyboard
On Fri, Jul 31, 2009 at 6:53 AM, Simon King wrote:
>
> On Jul 28, 3:15 am, William Stein wrote:
> ...
> > > Do you think I should post a ticket, changing it into "Control-C
> > > pressed. Interrupting interface. Please wait a few seconds"?
> >
> > Definitely, yes. Thanks!! But instead of not
bump, since there was no answer so far and despite i don't know the
answer it should be rather simple ... ?
On Jul 27, 7:01 pm, lesshaste wrote:
> I am new to sage and am attempting to solve systems of multivariate
> polys over GF(2). My first attempt with a small example is
>
> R.=GF(2)[]
> I=
On Jul 28, 3:15 am, William Stein wrote:
...
> > Do you think I should post a ticket, changing it into "Control-C
> > pressed. Interrupting interface. Please wait a few seconds"?
>
> Definitely, yes. Thanks!! But instead of not putting the interface
> name, just replace R by self.name().
It i
> Btw it seems the maximum of f(x,y) is 1/4, for x=y=1/2.
yes it is.
Thank you,
Paul
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F
what about the following ?
y=0.25
plot(lambda x : f(x, y), (0,1) )
to try different values of y on the same graph, go:
sum( [plot(lambda x : f(x,y), (0,1)) for y in srange(0, 1, 0.1)])
and of course there is
plot3d( lambda x, y : f(x, y), (0,1), (0,1) )
Btw it seems the maximum of f(x,y
nice, thank you David Joyner. For y=1/4 the function takes its maximum 3/16
and it's clear on the graph.
Interesting for me is how does the plot of f for fixed x look like if f
takes 1 as maximum?
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