Is there a simple way to merge equivalent conditions in Piecewise? For example, in the following we have (trivially) pw==x.
pw = Piecewise[{{x, x != 0}}, 0];
Is there some way to force this without cheating? I would call the following cheating:
Assuming[x != 0, Refine[pw]]
(* x *)
As a further example, can I get the simplification (for x real) pw2==Abs[x]?
pw2 = Piecewise[{{x, x > 0}}, -x];
In the first case, we have
Simplify[Piecewise[{{x, x != 0}}, 0], x ∈ Reals]
(* x *)
In the second case, Abs[x] is a function of a complex variable and will not be treated as equivalent to Piecewise[{{x, x > 0}}, -x]. (This seems correct to me.) If one wants the standard real absolute value, one can add a domain to PiecewiseExpand:
PiecewiseExpand[Abs[x], Reals]
(* Piecewise[{{-x, x < 0}}, x] *)
x was supposed to be complex?
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Commented
Sep 28, 2016 at 22:30
x real) helps with the first case? In the second example, I was interested in the case with x real (I think this is implicit in any expression involving an inequality). In this case, I think the expression Abs[x] is a reasonable simplification result - just as Abs[x] simplifies to x when I assumes x>0.
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Abs[] example). Note that UnsameQ instead of Unequal works without the assumption: Simplify[Piecewise[{{x, x =!= 0}}, 0]]. -- What I've noticed about Abs[] is that M will simplify it to something else when x is real or positive, etc., but it will not go from an expression to Abs[]. I suppose there's not a strong argument why this behavior is preferably, but I almost never want Abs[x] when x is real (differentiating it is a pain).
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Commented
Sep 29, 2016 at 18:33