On Wed, Apr 16, 2008 at 7:45 AM, Reckoner <[EMAIL PROTECTED]> wrote:
>
> For posterity, the subtle point that Stein made is that instead of
> substituting the integer 5, he used the the float 5.0 to get the
> result.
Thanks. Sorry for not writing out more about what I was doing
but I was in a hurry when I wrote that. Thanks for adding that.
It would be good to put an example like this in the tutorial.
William
>
> On Apr 15, 11:26 am, "William Stein" <[EMAIL PROTECTED]> wrote:
>
>
> > On Tue, Apr 15, 2008 at 11:22 AM, Reckoner <[EMAIL PROTECTED]> wrote:
> >
> > > Suppose I have something like:
> >
> > > sage: x,y,z=var('x y z')
> > > sage: eq = cos(x)*sin(y)*tan(x)
> >
> > > and I want to substitute a value for x, say, x=5. Then,
> > > I would have
> >
> > > eq = cos(5)*sin(y)*tan(5)
> >
> > > which is what I get when I do .substitute() in sage. However, I want
> > > this partially numerically evaluated to obtain
> >
> > > eq = 0.28366*sin(y)*(-3.38051)
> >
> > > How can I do this?
> >
> > sage: x,y,z=var('x y z')
> > sage: eq = cos(x)*sin(y)*tan(x)
> > sage: eq(x=5.0)
> > -0.958924274663139*sin(y)
>
>
> >
>
--
William Stein
Associate Professor of Mathematics
University of Washington
http://wstein.org
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