On 9/1/2012 11:04 AM, Barton Willis wrote:
> I would think that 0^(-a) and 1/0^a behave the same, but they don't:
yes.
>
> (%i2) 0^a;
> (%o2) 0
>
> (%i3) 0^(-a);
> (%o3) 0
>
> (%i4) 1/0^a;
> expt: undefined: 0 to a negative exponent.
How did it know that a>0 and hence -a was negative?
In some sense, 0^a could be simplified to "if a>0 then 0 else
undefined" but is that simpler?
And changing it to 0 seems to be an invitation to some construction
proving that 0=1 etc.
The culprit is probably different embedded code in simplifying
'mquotient' vs 'mexpt'.
RJF
>
>