Looking at http://trac.sagemath.org/sage_trac/ticket/3980, I formed
the opinion that a lot of times where .variables() or .arguments() is
used, it may be a mistake.
On that ticket, it was pointed out that find_root(sin,-1,1) throws an
error. Maybe this is a bug, maybe not, but in any case,
plot(sin,-1,1) certainly works, and I'd hope for consistency between
the two. The reason find_root throws an error in this case is because
it asks sin what its variables are, finds there are none, and
therefore assumes sin is a constant. But find_root doesn't need to
ask an expression about its variables--rather it just needs to know
whether the expression returns a number when called with one argument.
I suspect this is true more generally. Hopefully I will back up this
assertion by showing I can use it to clean up some code, but I thought
I'd share now to see if anyone had any thoughts.
One additional consideration is that it is useful in this case to know
whether an expression is constant, as a performance consideration.
The regular algorithm works (slowly) on constant input, but it's
faster if we can just short circuit in the case of a constant.
Clearly, asking an expression for its variables is not the right way
to find out whether it's constant. How about something like this
instead:
def is_constant(f):
try:
return f.isConstant()
except AttributeError:
if not hasattr(f,'__call__'):
return True
x = sage.calculus.calculus.var('x')
try:
return x not in f(x).variables()
except (AttributeError, TypeError, ValueError), e:
# Kludge to work around seemingly broke behavior of
SymbolicComposition
if e.message == 'the number of arguments must be less than
or equal to 0':
return True
# At this point, we don't know if the expression is
constant
return None # or throw an error
sage: sin.variables()
()
sage: is_constant(sin)
False
sage: is_constant(sin(3))
True
sage: is_constant(x)
False
sage: is_constant(x - x)
True
sage: is_constant(lambda x: x + 7)
False
# Could we do better with this case without messing anything else up?
sage: is_constant(lambda x: 1) is None
True
Regards,
JM
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