Re: imposing side conditions on Solve
- To: mathgroup at smc.vnet.net
- Subject: [mg24056] Re: [mg24029] imposing side conditions on Solve
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Thu, 22 Jun 2000 01:01:48 -0400 (EDT)
- References: <8ipn8l$anb@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej You give the following examples and explanations > In[13]:= > Solve[{x + y == 2, x == a, y == a}, {x, y}] > Out[13]= > {} > > We get no answer because the only solution forces a specific value of the > parameter, namely a=1. This is in accordance with the Mathematica Book. SNIP > But now consider a slightly different case: > > In[15]:= > Solve[{x + y == 2, x == a, y == a, a == 1}, {x, y}] > > Out[15]= > {{x -> 1, y -> 1}} > > We get an answer, even though again it only holds for a specific value of > the parameter. The difference is that the value a=1 is forced explicitely > (without using x and y) rahter than "implicitely", as in the first example. Here are three more variants: Solve[{x + y == 2, x == a, y == a}, {x, y, a}] {{x -> 1, y -> 1, a -> 1}} Solve[{x + y == 2, x == a, y == a}, {x, y}, {a}] {{x -> 1, y -> 1}} Solve[{x + y == 2, x == a, y == a}] {{a -> 1, x -> 1, y -> 1}} Any suggestions? -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Andrzej Kozlowski" <andrzej at bekkoame.ne.jp> wrote in message news:8ipn8l$anb at smc.vnet.net... >However, as is fairly often the case with the Mathematica Book, > the situation is more subtle than you realize on first reading. In fact, > whether you get a solution or not depends on the way the values of > parameters gets "forced" on them. To see this consider the following > examples. > > In[13]:= > Solve[{x + y == 2, x == a, y == a}, {x, y}] > Out[13]= > {} > > We get no answer because the only solution forces a specific value of the > parameter, namely a=1. This is in accordance with the Mathematica Book. As > explained in the Mathematica Book Reduce will give you all the answers: > > In[14]:= > Reduce[{x + y == 2, x == a, y == a}, {x, y}] > > Out[14]= > a == 1 && x == 1 && y == 1 > > But now consider a slightly different case: > > In[15]:= > Solve[{x + y == 2, x == a, y == a, a == 1}, {x, y}] > > Out[15]= > {{x -> 1, y -> 1}} > > We get an answer, even though again it only holds for a specific value of > the parameter. The difference is that the value a=1 is forced explicitely > (without using x and y) rahter than "implicitely", as in the first example. > If you now look carefully at your equations you can also see that the values > of the e's are forced explicitely, so you do get an answer. > > > Andrzej > > > -- > Andrzej Kozlowski > Toyama International University, JAPAN > > For Mathematica related links and resources try: > <http://www.sstreams.com/Mathematica/> > > >