Re: The domain parameter of Reduce[]
- To: mathgroup at smc.vnet.net
- Subject: [mg124180] Re: The domain parameter of Reduce[]
- From: Adam Strzebonski <adams at wolfram.com>
- Date: Wed, 11 Jan 2012 17:23:23 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201201101059.FAA27797@smc.vnet.net> <E9C5C418-1769-4B02-9F9D-51DE21A131B8@mimuw.edu.pl>
- Reply-to: adams at wolfram.com
Andrzej Kozlowski wrote: > On 10 Jan 2012, at 11:59, Szabolcs wrote: > >> My question if motivated by >> http://stackoverflow.com/questions/8780068/mathematica-finding-the-conditions-for-the-real-part-of-a-complex-number-to-be >> >> >> It seems that >> >> Reduce[{ComplexExpand@Re[-1 - Sqrt[a - b] ] < 0, a > 0, b > 0}, {a, >> b}, Complexes] >> >> will return a different result from >> >> Reduce[{ComplexExpand@Re[-1 - Sqrt[a - b] ] < 0, a > 0, b > 0}, {a, >> b}] >> >> Also the result of this latter calculation seems incorrect (I >> didn't expect 'b' to be restricted to be less than 'a') >> >> How does the domain parameter of Reduce work? Isn't Complexes the >> default domain? What changes if we specify Complexes explicitly? >> Also, if the result of the second example incorrect (a bug)? >> > > > There is something pretty fishy here, I would say. > > > Reduce uses the following principle: everything appearing on an > algebraic level in an inequality will be assumed to be real,unless > the domain Complexes is used. > > > Consider first both expressions without ComplexExpand: > > In[17]:= Reduce[{Re[-1 - Sqrt[a - b]] < 0, a > 0, b > 0}, {a, b}] > > a > 0 && (b >= a || 0 < b < a) > > Reduce[{Re[-1 - Sqrt[a - b]] < 0, a > 0, b > 0}, {a, b}, Complexes] > > a > 0 && (b >= a || 0 < b < a) > > The answers are identical and both are actually equivalent to the > given assumption a>0&&b>0. Reduce did not assume that Sqrt[a-b] is > real because of the presence of Re (this is what is mean by > "algebraic level": Re[x] is not "algebraic"). Now consider: > > ComplexExpand@Re[-1 - Sqrt[a - b]] > > -1 - ((a - b)^2)^(1/4) Cos[1/2 Arg[a - b]] > > Now there is no Re here and ((a - b)^2)^(1/4) is algebraic. So it > will be assumed to be real. But this assumption, of course, does not > entail that a>=b. Yet Reduce seems to think that it does: > > Reduce[{-1 - ((a - b)^2)^(1/4) Cos[1/2 Arg[a - b]] < 0, a > 0, b > > 0}, {a, b}] > > a > 0 && 0 < b <= a > > As one would expect, with the domain Complexes it no longer assumes > that: > > Reduce[{-1 - ((a - b)^2)^(1/4) Cos[1/2 Arg[a - b]] < 0, a > 0, b > > 0}, {a, b}, Complexes] > > a > 0 && b > 0 > > The first behaviour would not, perhaps, be surprising if the > expression was > > Reduce[{-1 - ((a - b)^(1/4))^2 Cos[1/2 Arg[a - b]] < 0, a > 0, b > > 0}, {a, b}] > > a > 0 && 0 < b <= a > > for this time indeed the expression (a - b)^(1/4) is real only if > a>=b. But the really curious thing is that if you remove the > requirement that a and b be positive you get: > > > Reduce[{-1 - ((a - b)^2)^(1/4) Cos[1/2 Arg[a - b]] < 0}, {a, b}] > > (a | b) \[Element] Reals > > Reduce[{-1 - ((a - b)^(1/4))^2 Cos[1/2 Arg[a - b]] < 0}, {a, b}] > > (a | b) \[Element] Reals > > Now that's really odd, since this case clearly includes the previous > one as a sub-case. So if the inequality holds for all real a and b, > it also holds for all positive a and b? > > Andrzej Kozlowski > > > > This is a bug. A transformation produces a disjunction term involving Sec[Arg[a - b]/2], and the requirement that Sec[Arg[a - b]/2] be real valued gets applied to a too wide context. It will be fixed in the next release of Mathematica. Best regards, Adam Strzebonski Wolfram Research
- References:
- The domain parameter of Reduce[]
- From: Szabolcs <szhorvat@gmail.com>
- The domain parameter of Reduce[]