To my
>> Introduce the program variables x,y... and bound them to the symbolic
>> indeterminates. [..] in DoCon program, it is arranged about like this:
>>
>> let {s = cToPol ["x","y"] 1; [x,y] = varPs s} in x^2*(x - x*y) ...
>> [..]
>> Hence, in many computations after `in', x,y denote what is needed.
D. Tweed <[EMAIL PROTECTED]> writes on 28 Nov 1999
> This sounds fascinating: is this approach powerful enough that if I have
> a definition of a haskell function purely as a definition for actual
> values and then apply it to an indeterminate I automatically get the
> expected result (no evaluation)? (ie,can I write
>
> harmonic n = (sum.map (\x->1/x)) [1..n]
> result = let {s = cToPol ["x"]; [x] = varPs s}
> in harmonic x
>
> to get result => harmonic(x)?)
> [..]
I feel, i am missing something in the question, and in the initial
notice on the Maple variables too.
Anyway, here what i wanted to tell.
The DoCon application written in Haskell supports the following
algebraic programming style:
let s = cToPol ["x","y"] 1 :: P
-- unity polynomial in x,y over Integer
-- - a sample element of domain P
[x,y] = varPs s -- images of "x","y" in P
in
x^2*(x - x*y) ...
As the polynomial constructor is supplied with the Ring
(or call it Num & Fractional) instance, the program after `in' can
do arithmetics on x,y. Because they are polynomials. And varPs
sets the correspondence between the true algebraic indeterminates x,y
and their names "x","y".
varPs makes, say, x from "x", x :: P - a value of type P.
Hence, everything applicable to type P can apply to x.
DoCon provides the standard functions
cToPol "coefficient to polynomial",
varPs "variables as polynomials".
In other algebra systems, they are easy to program too - as soon as
the polynomial representation method is chosen.
How this all relates to your question?
In your example, if you set harmonic x
after the above `in', it might have sense only if [1..x], 1/x
have sense. The domain P (polynomials!) to which x belongs, should
have Enum order. How do you put it, in what order the polynomials
"from 1 :: P up to x :: P"
have to follow?
If you had set such reasonable Enum instance, then, according to
your formula, 1/f will start to evaluate for several f :: P.
harmonic x might produce some result, but, hm... may fail at 1/x.
If you put, say, the rational functions F = Fraction P rather than
P, then harmonic (x ...::F) would have more chances ...
Again, i fear i mislead the idea.
------------------
Sergey Mechveliani
[EMAIL PROTECTED]
