I believe that the commercial macsyma includes a program by Bill Gosper
to determine the values of inequalities by a more principled method
involving computing how many bits of precision are necessary to
determine the sign of an algebraic expression (or the direction of an
inequality)
and then computing the result. At least when possible.
RJF
Barton Willis wrote:
>maxima-bounces at math.utexas.edu wrote on 11/20/2006 11:55:32 AM:
>
>
>
>>Hi,
>>
>>does anybody know how the floor function behaves when its argument is an
>>
>>
>
>
>
>>integer which is not obviously integer ?
>>for instance
>>floor(log(5)/log(2)) ==> 2 (correct)
>>but
>>floor(log(4)/log(2)) ==> floor(log(4)/log(2)) (I would expect 2 or 1)
>>
>>The documentation says the argument is bfoat-evaluated at first. This of
>>
>>
>
>
>
>>course could give a wrong answer (if log(4)/log(2)=1.999 instead of 2)
>>but I didn't expect such unevaluated answer. Did I miss something ?
>>
>>
>
>Given an argument that is a constant, the floor function evaluates the
>argument using three different values for fpprec. If all of these values
>basically agree *and* if all of these three values are reasonably far away
>from an integer, floor returns an integer; otherwise floor returns a noun
>form. For floor(log(4)/log(2)), floor decides that the big float values
>of log(4)/log(2) are all too close to an integer to safely determine the
>output, so floor (wisely, I think ;)) returns a noun form. Try something
>like
>
> for i : 10 thru 20 do (fpprec : i, print(i,is(rationalize(2 <
>log(4.0b0)/log(2.0)))));
>
>I wrote the current floor and ceiling functions. I wanted them to return
>noun forms more often then wrong values. I still think this is the best
>approach.
>You could do something like
>
>(%i70) radcan(floor(log(4)/log(2)));
>(%o70) 2
>
>Maybe radcan isn't quite the right function, I can't think right now.
>
>Barton
>
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