On Wed, 2009-09-23 at 10:31 -0400, Marc Feeley wrote:
> On 22-Sep-09, at 5:48 PM, Aubrey Jaffer wrote:
> > And exact-rational support is halfbaked: (expt 27 1/3) should return
> > exact 3, but doesn't in the implementations I have tried.
> At least Gambit and MIT-Scheme do this computation exactly.
> 1 ]=> (expt 27 1/3)
> ;Value: 3
As an aside, I'm interested in how this function is implemented.
My methods for taking roots (or raising to fractional exponents)
are generally iterative and work by successive interval division.
This does not lead one to an exact answer, unless one is
unusually lucky.
So, if expt ought to take rational exponents, what does an
example implementation of it look like?
Bear
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