f := proc(x,y) if x>y then erf(x-y) else erf(y-x) fi end;
But when I try to plot this I get
> plot3d( f(x,y), x=-2..2, y=-2..2 ); Error, (in f) cannot evaluate boolean
Answer: The plot functions have normal evaluation rules.
What happens then is that before plot3d is called,
is evaluated at symbolic
and
. I.e. we get
> f(x,y); Error, (in f) cannot evaluate boolean
Maple gives an error because it can't decide if when
and
have no values. There are two ways of using the plotting commands
plot and plot3d as illustrated in the following example
for plotting the binomial function.
plot3d( binomial(n,k), n=-3..3, k=-3..3 ); plot3d( binomial, -3..3, -3..3 );
Most users are familiar with the first way, which is to plot a formula
in two variables, in this case and
. The second way allows you to
plot a function given by a Maple procedure or an operator expression.
Therefore the solution to the problem is
plot3d( f, -2..2, -2..2 );
Alternatively, one can use quotes to prevent from being evaluated
before the plot routine is called thus
plot3d( 'f(x,y)', x=-2..2, y=-2..2 );
Note the frustrated user tried to get around the problem
by trying to express this way
f := proc(x,y) if signum(x-y)=1 then erf(x-y) else erf(y-x) fi end;
This avoids the error but produced a different surprise.
This ends up always plotting because
> signum(x-y);
signum(x - y)
cannot be evaluted to for symbolic
and
hence the
if test always returns false !!
> if signum(x-y) = 1 then true else false fi;
false
> f(x,y);
- erf(x - y)
Note Maple simplified to
.
The solution again is to simply plot the function
directly
plot3d( f, -2..2, -2..2 );
Of course, since
we could have expressed
as a formula
plot3d( signum(x-y)*erf(x-y), x=-2..2, y=-2..2 );