Re: Piecewise inside a Module or Block, I don't understand this behavior.

*To*: mathgroup at smc.vnet.net*Subject*: [mg83341] Re: Piecewise inside a Module or Block, I don't understand this behavior.*From*: Peter Pein <petsie at dordos.net>*Date*: Sat, 17 Nov 2007 05:38:40 -0500 (EST)*References*: <fhjs9v$4vg$1@smc.vnet.net>

Hi Craig, have a closer look at the syntax of Piecewise[] and remember LISP. Similar to the double parentheses in (cond ((condition1 action1) ... (t default))) in LISP, you should write Piecewise[{{action_1, condition_1},...,{action_n, condition_n}},default] in Mathematica. That's the way I memorized the double curly braces Peter W. Craig Carter schrieb: > Hello, > > I have a Piecewise function calculated in a module: > Here is a simplifed example of something which has a > behavior that puzzles me. > > a[c_, d_] := > Module[{e, f}, e = d^2; f = c^2; > Return[Piecewise[{e, 0 < x < 1/2}, {f, 1/2 < x <= 1}]]]; > a[1,2] (*doen't return what I had anticipated*) > > (*However this does*) > a[c_, d_] := > Module[{e, f}, e = d^2; f = c^2; > Return[Piecewise[{d^2, 0 < x < 1/2}, {c^2, 1/2 < x <= 1}]]]; > a[1,2] > > (*The same goes for Block, and putting an explicit Evaluate > inside the local function*) > > (*This behavior seems to be unique to Piecewise*) > a[c_, d_] := > Module[{e, f}, e = d^2; f = c^2; > Return[MyFunction[{e, 0 < x < 1/2}, {f, 1/2 < x <= 1}]]]; > a[1,2] > > I can't find anything about Piecewise, or in Module and > Block documentation that gives me a hint. > > Anyone know what is going on? > > Thanks, Craig > > PS: I have a work-around: > > aAlt[c_, d_] := > Module[{e, f}, e = d^2; f = c^2; > {{e, 0 < x < 1/2}, {f, 1/2 < x <= 1}}]; > Piecewise@aAlt[1,2] > > but, I am still curious.... >