Re: [Axiom-mail] sum()

[Martin Rubey]

Martin Rubey writes:
 > Martin Rubey writes:
 >  > You tried: 
 >  > 
 >  >  >    f: Integer -> Integer;
 >  >  >    f(0)==1; f(1)==2;
 >  >  >    f(n)==sum(f(i)*f(n-i), i=0..(n-1))
 >  > 
 >  > Try
 >  > 
 >  > f: Integer -> Integer;            
 >  > f(0)==1; f(1)==2;                 
 >  > f(n)==reduce(+,[f(i)*f(n-i) for i in 1..(n-1)])
 >  > 

Now I got it:


f:EXPR INT -> EXPR INT
f(0)==1; f(1)==2;                       
f(n | number? n)==sum(f(i)*f(n-i), i=1..(n-1))
f(n | not number? n)==kernel(fop, [n])
fop := operator 'fop
evaluate(fop,f)

We declare the function to take expressions as arguments.  If the argument is a
number, go ahead.  Otherwise, return an unevaluated operator fop, which
evaluates to f...

I think that evaluating fop i does the following:

axiom looks for an %eval property of fop and finds f.
axiom applies f to i and returns kernel(fop, [i])

If we wrote

f(n | not number? n)==fop n

we produced an infinite loop...

Martin