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Re: The domain parameter of Reduce[]

• To: mathgroup at smc.vnet.net
• Subject: [mg124149] Re: The domain parameter of Reduce[]
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 11 Jan 2012 04:21:31 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201201101059.FAA27797@smc.vnet.net>

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

• References:
  • The domain parameter of Reduce[]
    • From: Szabolcs <szhorvat@gmail.com>