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Re: Loss of precision when using Simplify


Let's step through the expansion here.


can be expanded to

9*0.546/16 + 9*x[2]/16

which becomes

.307125 + 9*x[2]/16

My question was why the 9/16 gets converted to .5625, as I see no reason to
do so. Even more troubling (to me, at least) is the following:


0.4 + x[0] + 0.75 x[1] + 0.5625 x[2]

I don't want Simplify to change my nice rational numbers to machine number

Carl Woll
Physics Dept
U of Washington

"Bill Rowe" <listuser at> wrote in message
news:angi42$q06$1 at
> On 10/2/02 at 3:32 AM, carlw at (Carl K. Woll) wrote:
> >I'm writing to report what I consider to be a bug. First, I want to
> >show a simplified example of the problem. Consider the following
> >expression:
> >
> >expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16;
> >
> >When simplified I expected to get some real number plus
> >x[0]+3x[1]/4+9x[2]/16, but instead I get the following:
> >
> >Simplify[expr] 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2]
> >
> >As you can see, for some reason Mathematica converted the fractions
> >3/4 and 9/16 to real machine numbers. I consider this to be a bug.
> You really are not seeing a loss of precision here. When simplify carries
out the indicated multiplication such as  9*.546/16 a machine precision
number is returned because on of the arguments only has machine precision.
It would be incorrect for Mathematica to return a result with greater
precision than the arguements. It would also be incorrect for Mathematica to
refuse to preform the required multiplications when simplifying this
> Or said differently, if you want an exact result from Mathematica *all* of
the information you supply Mathematica must also be exact. It is not
sensible for Mathematica to do otherwise.

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