Martin Rubey <[EMAIL PROTECTED]> writes:
> At least, it seems to me that from a user's perspective
>
> f: EXPR INT -> EXPR INT := x +-> res
>
> and
>
> f: EXPR INT -> EXPR INT := x +-> exp(-x)
>
> are equal. I'll look a little more into it.
Oh no, they are not!!! Sorry about my confusion:
(2) -> deq := differentiate(y x, x) + y x
,
(2) y (x) + y(x)
Type: Expression(Integer)
(3) -> res := first(solve(deq, y, x).basis)
- x
(3) %e
Type: Expression(Integer)
(4) -> f: EXPR INT -> EXPR INT := x +-> res
(4) theMap(*1;anonymousFunction;0;frame0;internal)
Type: (Expression(Integer) -> Expression(Integer))
(5) -> f a
- x
(5) %e
Type: Expression(Integer)
So, my problem actually has nothing at all to do with eval! Happy again.
Martin
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