#### 6.69 Assume facility with functions, Assume and functional forms (11.2.97)

##### 6.69.1 Mark S. Broski

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.

##### 6.69.2 Robert Israel

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



##### 6.69.3 Brian Blank

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