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Re: FullSimplify Question

  • To: mathgroup at
  • Subject: [mg48606] Re: FullSimplify Question
  • From: Paul Abbott <paul at>
  • Date: Mon, 7 Jun 2004 05:33:41 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <c9tmvi$sek$>
  • Sender: owner-wri-mathgroup at

In article <c9tmvi$sek$1 at>,
 "Dana DeLouis" <delouis at> wrote:

> Hello.  Hate to ask, but does anyone know what the issue is in the
> following?
> This equation basically is 5 when t is less than 0, and 1 when t>=0
> equ = 5 - 4*UnitStep[t]
> If I ask to "Simplify" this, by using the Sign function, I get the
> following, which is not even close
> FullSimplify[equ, TransformationFunctions -> {Sign}]
> 1 - UnitStep[t]

From the documentation:

   TransformationFunctions is an option for Simplify and FullSimplify   
   which gives the list of functions to apply to try to t

So Sign is being _applied_ to equ. You can see this more clearly if you 

      TransformationFunctions -> {Sign[Print[{#,Sign[#]}]; #] & }]

printing out the sub-expressions that FullSimplify is attempting to 
simplify by application of Sign.

> The above is 1 when t<0, and 0 when t>=0.
> I have to use both functions "FourierTransform" and
> "InverseFourierTransform" to get the correct form.
> 3 - 2*Sign[t]

The transformation rule, UnitStep[x_] :> (Sign[x] + 1)/2 (valid 
everywhere except at x==0) will do what you want:

   Simplify[equ /. UnitStep[x_] :> (Sign[x] + 1)/2]

> My question is that I can't figure out why FullSimplify was so far off.
> Would Mathematica have been better to leave the expression unevaluated so as 
> to use other methods?  

You "told" Mathematica to apply Sign recursively in an attempt to 
simplify the expression at hand ...


Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
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