A buglet in FunctionExpand
- To: mathgroup at smc.vnet.net
- Subject: [mg27665] A buglet in FunctionExpand
- From: Jack Goldberg <jackgold at math.lsa.umich.edu>
- Date: Sat, 10 Mar 2001 00:49:55 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
hi group FunctionExpand is a very powerful tool and like all powerful tools it should be used with some care. One use I have made of FunctionExpand is simplifing UnitStep[***]. In doing so I have discovered one bug reported here a few months ago and now report a buglet, a result that is more or less correct but not in reasonable form. Try, FunctionExpand[ UnitStep[x^2] ] and you will get UnitStep[-x]+UnitStep[x] on a Mac or Unix system running ver 4.0. The output fails to agree with the input at x = 0. The correct answer is, of course FunctionExpand[ UnitStep[x^2] ] -> 1 since x^2 >= 0 for all x. Similar incorrect answers occur when the argument of UnitStep is any even power of x+a. Now some additional comments directed at those with a serious interest in Piecewise Continuous functions: My experience with identities involving UnitStep suggests that one must either give up some simplifications in order that the input and output agree for all real x. Example, if you want UnitStep[-x] to simplify to 1-UnitStep[x] then as above these two functions disagree only for x=0. Redefining UnitStep[0] = 1/2 leads to the failure of UnitStep[x]^n = UnitStep[x] at x = 0 for every positive n. Mathematica provides an interesting partial solution which I came across quite accidentally: FunctionExpand[ UnitStep[-x]UnitStep[x] ] -> DiscreteDelta[x] So, UnitStep[-x] -> 1-UnitStep[x]+DiscreteDelta[x] saves this desired identity. My suggestions to the gurus at Mathematica: Clean up these peculiarities. Either except the fact that FunctionExpand[ ... something involving UnitStep...] may lead to an expression differing from the argument of FunctionExpand at a finite number of points, or use DiscreteDelta systematically. In any case, UnitStep[x^(2r)] is identically 1! Jack