Re: Re(2): assumption, supposition?
- To: mathgroup at smc.vnet.net
- Subject: [mg15378] Re: [mg15375] Re(2): [mg15331] assumption, supposition?
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Tue, 12 Jan 1999 03:14:43 -0500
- Sender: owner-wri-mathgroup at wolfram.com
My last message below contained a few misprints which make the code
unusable unless they are corrected. The correct definition of cond is:
cond[expr_List,var_]:=
SemialgebraicComponents[Join[expr,given],var]==
SemialgebraicComponents[given,var]
and the correct definition of sqrt consists of two statements:
sqrt[x_^2/;cond[{x>0},variables]]:=x; sqrt[x_]:=Sqrt[x]
On Sat, Jan 9, 1999, Andrzej Kozlowski <andrzej at tuins.ac.jp> wrote:
>Pattern matching clearly is unsuitable for your purpose, since it
>involves no deductions just mere "blind" substitution. However, one can
>easily implement simple "deduction" of the kind you seem to want in
>Mathematica, by making use of its ability to solve inequalities. There
>are two p-ackages for this purpose, Algebra`InequalitySolve` and
>Algebra`AlgebraicInequalities`. However, the former one is limited only
>to inequalities in one variable, so the latter one seems to be the
>right one to use here. I have written some code which, seems to me, to
>do the sort of thing that you might want. It is rather clumsy and very
>imperfect as I spent only minutes on writing it. It's only meant to
>suggest the kind of thing that can be done. I hope that some real
>expert on these matters might be induced to comment, and perhaps even
>to develop a real package that will properly do what I am only hinting
>at. Anyway, here is the idea.
>First we load the Inequality Solving package.
>
>In[1]:=
><<Algebra`AlgebraicInequalities`
>
>Next we define the function we shall use for testing conditions:
>
>In[2]:=
>cond[expr_List,var_]:=
> SemialgebraicComponents[Join[eepr,given],var]==
> SemialgebraicComponents[given,var]
>
>This function has two parameters: a list of conditions (inequalities) to
>be tested and a list of variables var. It depends on one global
>variable, given, which is also a list of inequalities. What cond does
>is to join the inequalities in expr to those in given and check if the
>solution of the joint set is the same as that of given alone. If this
>is the case than the conditions in expr forllow from those in given and
>are thus satisifed.This is how it works. We first set a value for
>given.
>
>
>In[3]:=
>given={x>0,y>1}
>Out[3]=
>{x>0,y>1}
>Next we test if x+y>1 follows from "given". In[4]:=
>cond[{x+y>1},{x,y}]
>Out[4]=
>True
>
>On the other hand x+y>2 does not follow: In[5]:=
>cond[{x+y>2},{x,y}]
>Out[5]=
>False
>
>Next, we can define sqrt a replacement for Sqrt which will do what you
>wanted. Really, of course, one should do this with Power, but I shan't
>bother. Also, I introduce another global variable, variables, which
>probably is bad programming style, but again since this is only meant
>as an illustration I shall not try to make it nicer:
>
>In[4]:=
>sqrt[x_^2/;cond[{x>0},variables]]:=x sqrt[x_]:=Sqrt[x]
>
>Now we assign values to our global variables:
>
>given={x>0,v>x,w>x+1};
>variables={x,v,w};
>
>We can see that sqrt behaves correctly on values which it knows as being
>>0: In[5]:=
>sqrt[w^2]
>Out[5]=
>w
>
>On the other hand, in the case of variables it does not know anything
>about it behaves as before:
>In[6]:=
>sqrt[u^2]
>Out[6]=
> 2
>Sqrt[u ]
>
>
>Andrzej
>
>
>
>On Fri, Jan 8, 1999, Erk Jensen <Erk.Jensen at cern.ch> wrote:
>
>>Probably I'm looking up the wrong keyword in Mathematica help, but I
>>can't find what I'm looking for. Maybe that's because I've learned
>>those expressions in german...
>>
>>My problem:
>>
>>I want Mathematica to assume that a certain condition is satisfied for
>>the following algebraic transformations. More specifically: Let rho
>>describe a radius coordinate. I know it is not negative real. So, for
>>transformations of expressions containing rho, e.g. Sqrt[rho^2], I
>>want assure Mathematica that 0 <= rho is in fact satisfied, and that it
>>is consequently allowed to replace Sqrt[rho^2] by rho. How do I do
>>that?
>>
>>My solution was
>>
>>Unprotect[Sqrt]; Sqrt[rho_^2] := rho/; 0 <= rho; Protect[Sqrt];
>>
>>rho /: 0 <= rho = True;
>>
>>and this seems to work, but I wonder whether this is really the proper
>>way. Since the TagSet I'm using here assigns the whole statement to
>>rho, and ?rho results in
>>
>>Global`rho
>>rho /: 0 <= rho = True
>>
>>so I still have my doubts ...
>>
>>Can some of you experts enlighten me?
>>
>>Thanks in advance
>> -erk-
>
>>Andrzej Kozlowski wrote:
>>>
>>> It seems to me there is nothing wrong with what you are doing, but it is
>>> rather limited. While indeed Sqrt[rho^2] will now return rho,
>>> (rho^2)^(1/2) , (rho^3)^(1/3) will all behave as before. So you need to
>>> modify the behaviour of Power rather than Sqrt. However, it seems to me
>>> that the simplest way to achieve what you want is by using Abs[rho] in
>>> place of rho in your formulas and then replacing Abs[rho] by rho in the
>>> final output. Mathematica knows that (Abs[rho]^2)^(1/2) is Abs[rho] etc.
>>>
>>
>>Thanks for your remark!
>>1st part: it's even worse: if I set "rho /: 0 <= rho = True;" it will
>>recognize
>>0 <= rho as True, but not "rho >= 0" (!) so it doesn't actually do the
>>checking, it just
>>replaces "0 <= rho" by true.
>>
>>2nd part: this wouldn't help if my allowed range for rho would be say
rho0 <=
>>rho <= rho2!
>>Which is actually what I have.
>>
>>I think I may go on with PowerExpand[Sqrt[x^2]] after all, that also
>works for
>>your
>>examples ... but it's not exactly what I want. What I really want is best
>>described by the
>>option "Assumptions" of "Integrate"!
>>
>>Best regards,
>> -erk-
>>
>
>
>Andrzej Kozlowski
>Toyama International University
>JAPAN
>http://sigma.tuins.ac.jp/
>http://eri2.tuins.ac.jp/
>
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp/
http://eri2.tuins.ac.jp/