Why would you expect sqrt(y) etc.? After all, 'y' is bound by the function
call.
On Wed, Dec 18, 2013 at 12:55 PM, Ricardo JF <rcrdjf at hotmail.com> wrote:
>
> ok...i was not clear.
>
> f(x):= x^2;
>
> i(y):= rhs(last( solve(f(x)=y,x)));
>
> i(2);i(3);i(4); --> it shoud be sqrt(y);sqrt(y);sqrt(y);
>
>
> thanks.
> ------------------------------
> Date: Wed, 18 Dec 2013 12:19:18 -0500
>
> Subject: Re: [Maxima] define and := (basic)
> From: macrakis at alum.mit.edu
> To: rcrdjf at hotmail.com
> CC: maxima at math.utexas.edu
>
>
> On Wed, Dec 18, 2013 at 11:53 AM, Ricardo JF <rcrdjf at hotmail.com> wrote:
>
> f(x):= x^2;
> i(y):= rhs(last( solve(f(x)=y,x)));
> i(2);i(3);i(4);
>
>
> I'm not entirely sure I understand what you're asking. In particular,
> saying "doesn't work as expected" isn't very explicit: what were you
> expecting?
>
> But if I'm understanding your question correctly...
>
> In your first example, the variable 'y' is bound by the function. In the
> second it isn't. Here is a simpler example:
>
> q(w):=w$
> q(2) => 2
> val: w$
> r(w):= val$
> r(2) => val (not 2)
>
>
>