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Re: Simplify and Abs


Simon Anders <simon.anders at uibk.ac.at> wrote:
> Hi,
>
> can it really be that this is already beyond Mathematica?
>
>     In :=  FullSimplify[Abs[p - 1], p < 1 && p > 1/2]
>
>     Out := Abs[-1 + p]

Strange.

> How do I make Matheamtica notice, that the assumptions constrain the
> argument of Abs[] to positive values?

Huh?
The assumptions constrain the argument of Abs[] to _negative_ values, and
thus the expression should simplify to  1 - p.

> Any suggestions how to treat these kinds of problems? Specifically, I
> have a list of products of absolute values of simple polynomials in p
> and I know that p is in the interval [0,1].
>
> I would like to know whether the polynomials have constant sign over the
> interval so that the Abs[] can be removed. Can this be done
> automatically?

How about using Refine instead? For example,

In[14]:= Refine[Abs[p - 1], 1/2 < p < 1]

Out[14]= 1 - p

But I don't understand why Simplify (or FullSimplify) doesn't give that
also. After all, the documentation states that
"Refine is one of the transformations tried by Simplify."

David Cantrell


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