puzzle with bfloat and radcan



Not sure why any of this is a "puzzle".  The value of an expression
evaluated to some number of digits depends on the *form* of the expression.
 Consider

product(q-i,i,1,16)


Obviously this is 0 for q in 1..16 (even 1.0 .. 16.0):

subst(q=15.0,product(q-i,i,1,16)) => 0


but

subst(q=15.0,expand(product(q-i,i,1,16))) => -215040.0

This is not a puzzle or an error.  It is the nature of finite-precision
calculation.

            -s

On Mon, May 14, 2012 at 6:42 PM, Richard Fateman
<fateman at eecs.berkeley.edu>wrote:

> xnum : ((6-4*sqrt(2))*log(3-2*sqrt(2)**)+(3-2*sqrt(2))*log(17-12***
> sqrt(2))+32-24*sqrt(2));
> xden :(48*sqrt(2)-72)*(log(sqrt(2)+**1)+sqrt(2))/3;
> x    : xnum/xden;
>