Yes, and I should have thought of that!
Fernando
On 3/5/2020 12:13 PM, Dima Pasechnik wrote:
In fact, substituting x and y directly into the equation of the curve
to plot, and clearing denominators,
produces something pretty good,IMHO:
implicit_plot(v^2*3*sqrt(1-u^2-v^2)-u^3*9+u*(1-u^2-v^2),(u
More conceptually, one can use, with care, Sage's substitution facilities:
sage: var('u v x y t');
sage: f=y^2-x^3+x
sage: fs=(f.subs(x=u*3*t^(-1/2),y=v*3*t^(-1/2))*t^(3/2)).expand() #
only works with extra variable t
sage: implicit_plot(fs.subs(t=1-u^2-v^2),(u,-1,1),(v,-1,1))
On Thu, Mar 5, 2020
In fact, substituting x and y directly into the equation of the curve
to plot, and clearing denominators,
produces something pretty good,IMHO:
implicit_plot(v^2*3*sqrt(1-u^2-v^2)-u^3*9+u*(1-u^2-v^2),(u,-1,1),(v,-1,1))
On Thu, Mar 5, 2020 at 4:51 PM Dima Pasechnik wrote:
>
> On Thu, Mar 5, 2020
On Thu, Mar 5, 2020 at 2:32 PM Fernando Gouvea wrote:
>
> This works, in the sense that there's no error. One does get a bunch of
> extraneous points near the boundary of the disk. It's as if plot_points were
> trying to connect the point at (0,1) and the point at (0,-1) along the
> circle, eve
This works, in the sense that there's no error. One does get a bunch of
extraneous points near the boundary of the disk. It's as if plot_points
were trying to connect the point at (0,1) and the point at (0,-1) along
the circle, even though f_uv is 1 on the circle.
Strangely, they occur only on
The easiest way is to use Python functions rather than symbolic ones;
define a function that is 1 outside the unit disk, and implicitly plot it.
sage: def f_uv(u,v):
: if u^2+v^2>=1:
: return 1
: else:
: x=u*sqrt(9/(1-u^2-v^2))
: y=v*sqrt(9/(1-u^
On Wed, Mar 4, 2020 at 12:20 AM Fernando Gouvea wrote:
>
> But no, it doesn't work, since it gives a rectangular plot instead of one in
> polar coordinates. But maybe we are closer.
I looked at the labels on the axes, and they do match the ranges of r
and phi, so I don't udnerstand
how it's poss
But no, it doesn't work, since it gives a rectangular plot instead of
one in polar coordinates. But maybe we are closer.
I still think implicit_plot should be smarter about values that do not
make sense.
Fernando
On 3/3/2020 6:26 PM, Dima Pasechnik wrote:
even better:
sage: var('x y u v r
Nice idea. Thanks.
Fernando
On Tue, Mar 3, 2020 at 6:27 PM Dima Pasechnik wrote:
> even better:
>
> sage: var('x y u v r phi')
> : u=r*cos(phi)
> : v=r*sin(phi)
> : x=u*sqrt(9/(1-r^2))
> : y=v*sqrt(9/(1-r^2))
> : implicit_plot(y^2-x^3+x==0,(r,0,999/1000),(phi,-pi,pi))
>
> On
even better:
sage: var('x y u v r phi')
: u=r*cos(phi)
: v=r*sin(phi)
: x=u*sqrt(9/(1-r^2))
: y=v*sqrt(9/(1-r^2))
: implicit_plot(y^2-x^3+x==0,(r,0,999/1000),(phi,-pi,pi))
On Tue, Mar 3, 2020 at 10:28 PM Dima Pasechnik wrote:
>
> On Tue, Mar 3, 2020 at 10:10 PM Fernando Gouve
On Tue, Mar 3, 2020 at 10:10 PM Fernando Gouvea wrote:
>
> The whole point of this is to show the behavior of the curve near infinity,
> so changing the limits is not an option.
just paste together a number of rectangles where (u,v) stay inside the
unit circle.
(yes, this would need writing a lo
A caveat is that at the boundary, the mapping you describe becomes
non differentiable (the determinant of the differential blows up to
infinity),
so it's going to be painful for implicit_plot to work.
That being said, the following tweak runs ok but it's not ex
The whole point of this is to show the behavior of the curve near
infinity, so changing the limits is not an option.
Fernando
On 3/3/2020 4:15 PM, Dima Pasechnik wrote:
On Tue, Mar 3, 2020 at 8:20 PM Fernando Gouvea wrote:
Here's what I ended up trying, with r=3:
var('x y u v')
x=u*sqrt(9/(
On Tue, Mar 3, 2020 at 9:15 PM Dima Pasechnik wrote:
>
> On Tue, Mar 3, 2020 at 8:20 PM Fernando Gouvea wrote:
> >
> > Here's what I ended up trying, with r=3:
> >
> > var('x y u v')
> > x=u*sqrt(9/(1-u^2-v^2))
> > y=v*sqrt(9/(1-u^2-v^2))
> > implicit_plot(y^2-x^3+x==0,(u,-1,1),(v,-1,1))
> >
> >
On Tue, Mar 3, 2020 at 8:20 PM Fernando Gouvea wrote:
>
> Here's what I ended up trying, with r=3:
>
> var('x y u v')
> x=u*sqrt(9/(1-u^2-v^2))
> y=v*sqrt(9/(1-u^2-v^2))
> implicit_plot(y^2-x^3+x==0,(u,-1,1),(v,-1,1))
>
> That gives an error:
>
> /opt/sagemath-8.9/local/lib/python2.7/site-packages
Here's what I ended up trying, with r=3:
var('x y u v')
x=u*sqrt(9/(1-u^2-v^2))
y=v*sqrt(9/(1-u^2-v^2))
implicit_plot(y^2-x^3+x==0,(u,-1,1),(v,-1,1))
That gives an error:
/opt/sagemath-8.9/local/lib/python2.7/site-packages/sage/ext/interpreters/wrapper_rdf.pyx insage.ext.interpreters.wrapper_r
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