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MathGroup Archive 2005

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Re: solve doesn't solve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54285] Re: [mg54265] solve doesn't solve
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 16 Feb 2005 14:36:03 -0500 (EST)
  • References: <200502150251.VAA27363@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 15 Feb 2005, at 03:51, juejung wrote:

> hi,
>
> can anybody solve the following equation for variable s with 
> Mathematica?
>
> ((-(-1 + b))*R*beta)/(a - (-1 + b)*R*s)^sigma - (-s + w -
> w*tau)^(-sigma)==0
>
> Solve doesn't work. I get this annoying error message.
>
> Solve::tdep: The equations appear to involve the variables to be 
> solved for
> in an essentially non-algebraic way.
>
> I would like to get analytical solutions (numerical is not problem) to
> problems like this one.I know this one's easy done by hand but I have 
> more
> cumbersome ones to deal with.
> Is there a package out there that can solve this analytically, or is 
> this
> just impossible on computers?
>
> Thanks a lot.
>
> Juergen
> -- 
> Using Opera's revolutionary e-mail client: http://www.opera.com/m2/
>
>
>
There are no general methods for solving transcendental (non-algebraic 
equations) symbolically either by humans or computers. Matheamtica can 
solve some of them: those that can be reduced to solving an algebraic 
(polynomial) equation by applying an inverse function of one of the 
well known transcendental functions. Even in such cases Mathematica 
can't always solve the equation because is is sometimes unable to "see" 
that it can be reduced in this way. This is one example of such a 
situation.

In such cases there is no other way but to guide Matheamtica by 
transforming the equation until Mathematica is able to see the 
"underlying" algebraic problem:

In[1]:=
eq = -(-s - w*tau + w)^(-sigma) == ((b - 1)*R*beta)/(a - (b - 
1)*R*s)^sigma;

In[2]:=
PowerExpand[(#1^(1/sigma) & ) /@ eq];

In[3]:=
sol = FullSimplify[Solve[%, s]]

Out[3]=
{{s -> -(((b - 1)^(1/sigma)*beta^(1/sigma)*(tau - 1)*w*R^(1/sigma) + 
(-1)^(1/sigma)*a)/
       ((b - 1)^(1/sigma)*beta^(1/sigma)*R^(1/sigma) - (-1)^(1/sigma)*(b 
- 1)*R))}}

The use of PowerExpand means that the solution may not be valid for all 
values of the parameters (or sometimes for any values).


Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/


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