On Tue, Dec 22, 2009 at 09:03:44AM -0800, slemi wrote:
> this allows me to use the (^^) powering operator, which works fine with
> non-zero exponents.
> however to my surprise when i try (^^ 0) the answer is (Scalar 1), and not
> the identity matrix as expected.
> does this mean that (a ^^ 0) is not defined as (a ^^ 1 * a ^^ (-1)) (or
> better yet (a / a)) in the prelude?
> if so, can i redefine it so that it gives the right answer?
> i am also very interested in how ghci got the answer (Scalar 1), it seems
> quite magical:)
(^^ 0) == const 1, see this link:
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