Is it possible for the assume facility to recognize functional forms?

For instance,

`assume (f(t)>0)`

`is(f(t)>0)`

gives the result TRUE

but if I enter

`is(f(t)+f(t)>0)`

an error code is returned.

Not an error code, but FAIL (indicating Maple doesn’t know) in Release 3. This particular problem is ﬁxed in Release 4:

> is(f(t)+f(t)>0); true

However, assuming `f(t) > 0`

doesn’t say anything about f(x), or any other value of the
function except literally f(t).

> is(f(x)>0); FAIL

Ideally you’d like to be able to say

> assume(f>0);

But this doesn’t work:

> is(f(x)>0); FAIL

I note that Robert Israel has already given a response to the question as asked. Perhaps the following constructive workaround might be of some use to you:

> f := t -> exp(Re(f1(t))): # Use abs(f1(t)) if you want nonnegative. > is( f(x) > 0 ); true > is(f(x)+f(t) > 0); true > is( 1/f(Pi) > 0 ); true > is(sqrt(f(t)) , real); true > is( ln(f(t)) , real ); true > assume( b < c ); is( int( f(t) , t = b .. c ) > 0); true