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MathGroup Archive 1995

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Re: UnitStep Argument

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg663] Re: [mg628] UnitStep Argument
  • From: bob Hanlon <hanlon at pafosu2.hq.af.mil>
  • Date: Fri, 7 Apr 1995 23:15:34 (EDT)

On Sat, 1 Apr 1995, Roger M. Jones wrote:

> How can I define
> UnitStep[-x], so that the answer is zero
> (.i.e. assign x as being a positive number).
> Defining: 
> 
> x/:Positive[x]=True doesn't work.
> 
> 
> Many thanks.
> 
> ---
> Regards,
> 
> Roger
> (rmj at llewelyn.slac.stanford.edu)
> 

Here are two simple approaches:

In[1]:=
z /: Positive[z] = True;

In[2]:=
UnitStep1[-x_] := 1 - UnitStep1[x];
UnitStep1[x_] := If[Positive[x], 1, 0];

In[4]:=
{UnitStep1[-4], UnitStep1[-0], UnitStep1[0], UnitStep1[4], 
	UnitStep1[-z], UnitStep1[z], UnitStep1[-y], UnitStep1[y]}

Out[4]=
{0, 0, 0, 1, 0, 1, 1 - If[Positive[y], 1, 0], If[Positive[y], 1, 0]}

In[5]:=
UnitStep2[-x_] := 1 - UnitStep2[x];
UnitStep2[x_?Positive] := 1;
UnitStep2[x_?Negative] := 0;
UnitStep2[x_ /; x == 0] := 0;

In[9]:=
{UnitStep2[-4], UnitStep2[-0], UnitStep2[0], UnitStep2[4], 
	UnitStep2[-z], UnitStep2[z], UnitStep2[-y], UnitStep2[y]}

Out[9]=
{0, 0, 0, 1, 0, 1, 1 - UnitStep2[y], UnitStep2[y]}





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