MaplePrimes - answers and comments on Question, f(x) = x^2 ln x

Check your answers, not ours

Robert Israel — Sun, 09 Mar 2008 23:29:10 Z

So, what are your answers?

FunctionAdvisor

lemelinm — Wed, 12 Mar 2008 23:02:34 Z

The FuctionAdvisor is very interesting. But it only work with known functions as you can see with:

FunctionAdvisor(known_functions)

but not for a contruction like x^2*ln(x).

If it could be algorithmic, I would be very interested to know how if this is not too complicated for no too nasty nesting.

Mario Lemelin

mario.lemelin@cgocable.ca

Domains

Alex Smith — Thu, 13 Mar 2008 03:02:51 Z

Implicit in all of this is that the domain of a function is subset of the complex numbers.

But...if I define

f:=x->x^2;

and then ask for f(Wisconsin)

then Maple returns Wisconsin^2. Since an output is associated with the input, I must conclude that Wisconsin is in the domain of f.

Of course, to some extent, this is silly, but it points out why the "domain" question really is subtle.

When mathematicians define a function, they know that they are obligated to state the domain. But when we use Maple to define a function, we are glad that we are under no such obligation.

Maple functions are not functions

Alex Smith — Thu, 13 Mar 2008 06:37:44 Z

So what is the domain of this Maple function?

f:=proc(s) :: type(s,set);

combinat[powerset](s);

end;

A powerful aspect of Maple is that we do not have to define the domain of a proc in order to properly define a proc.

And so we should not expect an algorithm to exist that will determine the domain of a proc.

If you read

jakubi — Thu, 13 Mar 2008 08:21:40 Z

you will realize that the subject of this thread is not procedures with parameters of type set or procedures in general. So, nobody has written here about an algorithm that will determine the domain of a general proc.

Strange return type assertion

Joe Riel — Thu, 13 Mar 2008 20:10:09 Z

Is the type assertion of the return value intentional? That is really strange, and I haven't figured out why it doesn't generate an error (with kernelopts(assertlevel=2)). I could understand

f := proc(s) :: set; combinat:-powerset(s); end proc:

but how can a set have the type 'type(s,set)'? That makes no sense, it seems equivalent to doing

type({{},{a}}, 'type'({a},set))

postpost:

I now understand what is going on. The return assertion `type(s,set)' is, indeed, equivalent to

ASSERT(type(output, 'type'(s,set)))

assuming output is the returned value. So the assertion, which must be true to avoid an error, is that

type(output, 'type'(s,set))

Maple's 'type' type ignores arguments (s and set) and interprets this as

type(output, 'type')

That is, it is true whenever the output is a valid type. I had only tested the procedure with a set of integers. The generate output from that is a set of sets of integers, which is a valid Maple type.

type({{},{1},{1,2}}, 'type');
                                              true

If a symbol is included, then the resulting powerset is not a valid Maple type, and an error is generated:

kernelopts(assertlevel=2):
f := proc(s) :: type(s,set); combinat:-powerset(s); end proc:
f({1,a});
Error, (in f) assertion failed: f expects its return value to be of type
type({1, a},set), but computed {{}, {1}, {a}, {1, a}}