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MathGroup Archive 2004

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Re: solve and simplify. force left hand side to be 0==

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51179] Re: [mg51152] solve and simplify. force left hand side to be 0==
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 7 Oct 2004 05:26:07 -0400 (EDT)
  • References: <200410060834.EAA24236@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 6 Oct 2004, at 17:34, sean kim wrote:

> *This message was transferred with a trial version of CommuniGate(tm) 
> Pro*
>
> consider a system such as,
>
> sde = {
> 0==v1-d1 e[0]+k2 es[0]-k1 e[0] s[0],
> 0==-k2 es[0]-k3 es[0]+k1 e[0] s[0],
> 0==k3 es[0]-d3 p[0],
> 0==v2+k2 es[0]-d2 s[0]-k1 e[0] s[0]};
>
> sv = {e[0],es[0],p[0],s[0]};
>
> Solve[sde[[1]], e[0]]
>
> sde1= sde/.%
>
> %//Simplify
>
> shows once I simplify the solution, the structure of
> 0== stuff... is lost.
>
> is there anyway to force Simplify to keep the
> structure so that left hand side is always 0== stuff?
>
> thanks in advance  for any insights.
>
>
> sean
>

You can just Map Simplify onto the third level of your expression:


Map[Simplify, sde1, {3}]


{{True, 0 == (-k2)*es[0] - k3*es[0] + (k1*(v1 + k2*es[0])*s[0])/(d1 + 
k1*s[0]), 0 == k3*es[0] - d3*p[0],
    0 == v2 + k2*es[0] - d2*s[0] - (k1*(v1 + k2*es[0])*s[0])/(d1 + 
k1*s[0])}}

The first equation turned into 0===0 and hence you got True. But the 
rest is as you wanted.



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


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