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Re: Simplify and Abs


FullSimplify is giving the simplest result already.

>> the assumptions constrain the argument of Abs[] to positive values

That's just not true. 2/3 is less than one, greater than 1/2, but 2/3 - 1 is not positive.

This works, on the other hand:

FullSimplify[Abs[p-1],p>1]

-1+p

Bobby

On Thu, 24 Feb 2005 03:21:06 -0500 (EST), 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]
>
> How do I make Matheamtica notice, that the assumptions constrain the
> argument of Abs[] to positive values?
>
> 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?
>
> TIA
>    Simon
>
>
>
>



-- 
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