Hi Will, On Fri, Sep 21, 2012 at 12:08 AM, <w...@ccs.neu.edu> wrote: > Alex Shinn wrote: > >> One more thing - how was (expt 0 z) extended to >> return 0 for complex z? The approaches I've tried >> would result in NaN, and indeed this seems to be >> what most implementations return. > > You may be misinterpreting the R6RS.
I may indeed :) > (expt 0 z) returns zero > if the real part of z is positive. If z is zero, then it's > supposed to return zero. If z is non-zero and its real part > is zero or negative, then implementations are free to return > whatever number object they like, or to raise an exception > with condition type &implementation-restriction. The question has to do with non-real z with positive real part, i.e. (expt 0.0 c+di), c > 0, d != 0. As a simplification I can see how it would be useful to simply define this to be 0, but it can't be derived as far as I can see from the definition of complex exponentiation, w^z = e^(z log(w)), because log(0) is undefined. Existing implementations also differ in their results here. -- Alex > Implementation is straightforward. Here is Larceny's code: > > (define (expt x y) > > ; x is nonzero, and y is an exact natural number. > > (define (e x y) > (cond ((= y 0) > 1) > ((odd? y) > (* x (e x (- y 1)))) > (else > (let ((v (e x (quotient y 2)))) > (* v v))))) > > (cond ((zero? x) > (let ((result (cond ((= y 0) 1) > ((> (real-part y) 0) 0) > (else +nan.0)))) > (if (and (exact? x) (exact? y)) > result > (exact->inexact result)))) > ((and (exact? y) (integer? y)) > (if (negative? y) > (/ (expt x (abs y))) > (e x y))) > (else > (exp (* y (log x)))))) > > Will _______________________________________________ r6rs-discuss mailing list r6rs-discuss@lists.r6rs.org http://lists.r6rs.org/cgi-bin/mailman/listinfo/r6rs-discuss
