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MathGroup Archive 1995

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Re: Solving inequation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg2099] Re: Solving inequation
  • From: Bernd.Cebulski at e-technik.tu-chemnitz.de (Bernd Cebulski)
  • Date: Fri, 29 Sep 1995 01:15:45 -0400
  • Organization: TU Chemnitz

In article <DFCG6A.Lz6 at wri.com> uthaisom at veena.cps.msu.edu (Patchrawat Ut=
haisombut) writes:
>From: uthaisom at veena.cps.msu.edu (Patchrawat Uthaisombut)
>Subject: Solving inequation
>Date: Sat, 23 Sep 1995 05:49:21 GMT

>Hi,

>I am a new Mathematica user.
>I wonder if Mathematica can solve inequation.
>Something like

>        Solve[ k+s < s+1, k]

>If there is no such build-in function,
>suggestion to implement this is also appreciated.

--------------------------------------------------------------	=


There is no solution for such problems built in. There were several discu=
ssions
before about this topic. Here are a few solutions for your task. If you g=
et =

other interesting answers per e-mail please forward them to me. =




=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D forwarded messages follow =3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D

<from ronald at sccs.chukyo-u.ac.jp>

  In any case, I went ahead and asked WRI tech support. =

It seems that there was nothing built in after al. But,
the tech support rep, Ed Greaves, volunteered to write
a function to do the job. As I suspecte, his version
is much better than what I would have written.
  Here it is:

convertIneq[ x_[lhs_,rhs_] ] :=3D
  Module[{temp, const, result},
    temp =3D Expand[lhs - rhs]; (* move all terms to LHS *)
    const =3D If[Head[temp] =3D=3D=3D Plus, (* >1 term?*)
              Select[temp,NumberQ],
              0];
    result =3D x[temp - const, -const ];
    (* The next is needed only if you wish =

       the rhs to always be non-negative *)
    If[const>0,  (* is rhs negative? *)
        result =3D Map[Minus, result];
        result =3D result /. { LessEqual->GreaterEqual
                             GreaterEqual->LessEqual }
    ]; (* Note: The 'If' has only a 'Then' clause *)
    result
  ]

  Everything that I send out, I have to type in by hand. Once
the screen scrolls, I cannot return to check or edit. So,
there may be a type or two above. But, I do not think anything
serious.  The function itself is a forehead slapper: As soon
as I saw it, I slapped my forehead and thought "Why didn't I
write this?"  Of course, I never would have written anything
remotely as neat and clean as this.
  To repeat, the above is a very slight modification of a =

function written by Edmumd Greaves of WRI.
  I have tried it. Ed has thought of and covered cases I =

would not have. It seems to work well.

 Hope this is of help
  Ron Notestine.
-------------------------------------------------------------------------=
------

<from rubin at msu.edu>

Yes, but the only way I know how to do it is like pulling teeth.  (Questi=
ons
about inequalities come up on this list periodically, but I haven't seen
a killer answer yet.)  I can get an answer to this using two massive klud=
ges:
inserting slacks/surpluses to convert and from equations, and defining an=

"inverse" for the absolute value function.  (Yeah, I know, it's not
invertible, but humor me.  :-) )
 =

First, I'll define a bunch of substitution rules (I'm being a little more=

general than this specific example):
 =

In[1]:=3D ineq2eq :=3D
        {(a_) <=3D (b_) -> a + slack =3D=3D b,
         (a_) < (b_) -> a + strongslack =3D=3D b,
         (a_) >=3D (b_) -> a - surplus =3D=3D b,
         (a_) > (b_) -> a - strongsurplus =3D=3D b};
In[2]:=3D eq2ineq :=3D
        {(a_) =3D=3D slack + (b_) -> a >=3D b,
         (a_) =3D=3D -slack + (b_) -> a <=3D b,
         (a_) =3D=3D strongslack + (b_) -> a > b,
         (a_) =3D=3D -strongslack + (b_) -> a < b,
         (a_) =3D=3D surplus + (b_) -> a >=3D b,
         (a_) =3D=3D -surplus + (b_) -> a <=3D b,
         (a_) =3D=3D strongsurplus + (b_) -> a > b,
         (a_) =3D=3D -strongsurplus + (b_) -> a < b};
In[3]:=3D sreduce :=3D
        {slack _?Positive -> slack,
         slack _?Negative -> -slack,
         surplus _?Positive -> surplus,
         surplus _?Negative -> -surplus,
         strongslack _?Positive -> strongslack,
         strongslack _?Negative -> -strongslack,
         strongsurplus _?Positive -> strongsurplus,
         strongsurplus _?Negative -> -strongsurplus};
In[4]:=3D invabs :=3D
        {(a_) =3D=3D (b_.) + (c_.) InverseFunction[ Abs, 1, 1][d_] ->
         a =3D=3D b + c d || a =3D=3D b - c d};
 =

The symbols "slack," "strongslack," "surplus" and "strongsurplus" should
probably be protected, to keep you from accidentally assigning them value=
s =

(they're really placeholders).  The "invabs" pattern replaces any call to=

the (nonexistent) inverse of the absolute value function with a disjuncti=
on
of plus and minus the argument to the "inverse."  (In other words, when M=
ma
sees y =3D=3D Abs[ x ], it solves for x by setting x to the inverse of th=
e Abs
function at y, which is really x =3D=3D y || x =3D=3D -y.)  Now comes the=
 fun:
 =

In[5]:=3D prob :=3D 1 > Abs[ u - 2 ];  (* original problem *)
In[6]:=3D prob =3D prob /. ineq2eq  (* convert to equations *)
Out[6]=3D {1 - strongsurplus =3D=3D Abs[-2 + u]}
In[7]:=3D prob =3D Reduce[ prob, u ]  (* solve for u *)
Out[7]=3D
                 (-1)
     u =3D=3D 2 + Abs    [1 - strongsurplus]  (* note "inverse" abs funct=
ion *)
In[8]:=3D prob =3D prob /. invabs  (* get rid of "inverse function" *)
Out[8]=3D u =3D=3D 3 - strongsurplus || u =3D=3D 1 + strongsurplus
In[9]:=3D prob =3D prob /. sreduce  (* eliminate multipliers of surplus;
                                   not needed in this particular example =
*)
Out[9]=3D u =3D=3D 3 - strongsurplus || u =3D=3D 1 + strongsurplus
In[10]:=3D prob =3D prob /. eq2ineq  (* back to inequalities *)
Out[10]=3D u < 3 || u > 1
 =

> Help with this simple example will lead me to using it for more complex=
 =

> inequalities..
 =

Well, maybe.  It gets a lot messier with multiple variables, and it relie=
s
heavily on Reduce being able to do something with the equation in the fir=
st
place.  I tried something like Abs[ x - 3 ] < Abs[ 2 - 2 x ] (or somethin=
g
along those lines) and Reduce quit in disgust.

-------------------------------------------------------------------------=
--
<from g.gsaller at amda.or.at>

There was a module called ConvertIneq in the mathgroup conference. I dont=
 know
the author.
This module converted inequalities to standard form for linear programmin=
g.

I changed this module to SolveIneq to solve simple inequalities.

solveIneq[ x_[lhs_, rhs_] ] :=3D
Module[{temp, const, fact, result},
temp =3D Expand[lhs - rhs];
const =3D If[Head[temp] =3D=3D=3D Plus,
Select[temp, NumberQ],
0];
fact =3D If[Head[temp] =3D=3D=3D Times,
Select[temp, NumberQ],
If[Head[temp[[2]]] =3D=3D=3D Times,
Select[temp[[2]], NumberQ]],1];
result =3D x[(temp - const)/fact, -const/fact];
If[fact < 0,
result =3D result /. {Less -> Greater,
LessEqual -> GreaterEqual,
Greater -> Less,
GreaterEqual -> LessEqual}];
result
]

Sample Input:  solveIneq[2(x-1)<(10x+5)/7+38]

Guenther R. Gsaller

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D end of forwarde=
d messages =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D =


PS: There is some hope that WR will implement facilities for solving =

inequalities in the next release of MMa (perhaps 3.0?). BTW, other
packages (for example DERIVE) can solve inequalities ...

Hope that helps a bit,

	cheerio, =

			Bernd.








=2E.-----------------------------------------------.
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|--- bernd.cebulski at e-technik.tu-chemnitz.de ---|
|            Phone: +49 (371) 5313318           |
|            Fax:   +49 (371) 5313361           |
|                   DL 1 DTP                    |
`-----------------------------------------------=B4  =





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