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UnitStep and Simplify
*To*: mathgroup at smc.vnet.net
*Subject*: [mg19980] UnitStep and Simplify
*From*: Jack Goldberg <jackgold at math.lsa.umich.edu>
*Date*: Thu, 23 Sep 1999 23:26:26 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Hi Group,
I'm happy to see that UnitStep has moved from the
"AddOns" to the "BuiltIns". This was long overdue
and, at least for me, a significant step - so to speak.
Having said that let me add a mild criticism and what
I think is an improvement to the present situation.
I think the UnitStep[x] should have been defined as
UnitStep[x,0] and UnitStep[x-a] as UnitStep[x,a]. This
is a small point but after experimenting for years (literally!)
I came to the conclusion that UnitStep ought to be a function
of two variables, the second variable the point at which the
function jumps. But, in any case, it is not hard to convert
back and forth by replacement rules. A more serious issue is
the lack of a real simplifier for complicated functions of
UnitStep. I provide this also. So here goes. (Suggestions,
comments and praises are welcome!)
First let us define Step[_,_] as the function Mathematica should have
defined! :-).
Step[x_,a_?NumericQ] /; x<a :=0
Step[x_,a_?NumericQ] /; x>=a := 1
Now a new function which simplifies functions of Step.
Later on I provide an explanation.
StepSimplify[f_,x_] /; MemberQ[f,_Step,{0,-1}] :=
Module[ {brkpts,A0,line1},
brkpts = Cases[f,Step[x,a_]->a,{0,-1}]//Union;
A0 = f - (f/. Step[x,a_]->0);
line1 = Fold[(#1-(#1/.Step[x,a_]
->Step[#2,a])Step[x,#2)Step[x,#2])&,
A0,bkpts];
f-line1 ]
Try it on some simple examples. Say,
ex1 = (x Step[x,1]-Step[x,2])^2, ex2 = Cos[Step[x,0],
ex3 = Sqrt[1+Step[x,1]*Step[x,2]-Step[x,3]].
Now try FullSimplify!
The idea behind the code given in StepSimplify is this:
Supposing that f contains various Step functions say
Step[x,0], Step[x,-1] and Step[x,1]. I assert that
f may be expanded as follows
(1) f = A[x] + B[x]* Step[x,-1] + C[x]*Step[x,0] + D[x]*Step[x,1]
(and this expansion is unique if f has certain "reasonable"
properties) where A[x], B[x] etc. does not contain any Step[x,_].
(2) brkpts (breakpoints) gives the list of jumps due to the Step
functions.
(3) A0 simply forms A0 = f-A[x] by replacing all Step's in
(1) by zero.
(4) line1 uses the iterated nature of (1) to obtain B[x],
C[x] and D[x] in turn. For example, to obtain B[x] we
take f-A[x] and set the arguments of all steps to x=-1
(the first entry in brkpts). All Steps, Step[x,a], a>-1
vanish and Step[-1,-1]->1. It then follows that we have
B[x]. We form f-A[x]-B[x] Step[x,-1] to set up the
Fold function to find C[x], etc.
(5) I think I need {0,-1} in both MemberQ and Cases to capture
the trivial case f=Step[x,a]. Also, Union was used both to
order the brkpts (breakpoints) and to remove repeats. Ordering
seems essential to the scheme and this is one minor reason
I prefer Step[x,a] to UnitStep[x-a].
By the way, it also simplifies correctly, Step[Step[x,a],b]
and replaces g[x] Step[x,a] + h[x] Step[x,a] by
(g[x]+h[x])Step[x,a], a task which I found required some
vary careful programming, particularly if Expand is used
to simplify the final coefficients of each Step.
Again, feedback is appreciated and if I am correct in
my statements about StepSimplify, perhaps someone could
suggest improvements in the code or alternatives. By the way,
FullSimplify does a litte simplification on simples expressions
containing UnitStep but is woefully
inadequate most cases.
Jack
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