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MathGroup Archive 2000

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RE: [Q] Equation solving?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23318] RE: [mg23282] [Q] Equation solving?
  • From: "David Park" <djmp at earthlink.net>
  • Date: Mon, 1 May 2000 18:00:18 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com


> From: James [mailto:research at proton.csl.uiuc.edu]
To: mathgroup at smc.vnet.net
> Hello!
> 
> I'm a newbie to Mathematica
> and have a question about what's wrong in the following
> expression.
> (The parameters are generated by the program, 
>  so there're some terms that can be cancelled out.) 
> 
> Solve[{ Q1==(2*z+1-1)^2+4*1*1 ,
>         k1==(1/Sqrt[Q1])*(v*z-(3/2)*(2*z+1+1-Sqrt[Q1])) ,
>         Q2==(2*z+1-1)^2+4*1*1 ,
>         k2==(1/Sqrt[Q2])*((6.10-v)* z-(2/2)*(2*z+1+1-Sqrt[Q2])) ,
>         (3/2)*(2*z+1+1)+(k1-(3/2))*Sqrt[Q1] ==
>               6.10*z-((2/2)*(2*z+1+1)+(k2-(2/2))*Sqrt[Q2])}, {z,v}]
> 
> I want to solve the above equations according to 'z',
> and I tried several ways in expressing {z,v} or z, {v}, etc
> to get an answer(s), but all in vain.
> 
> Is there anyone who can tell me how to express the equations?
> Sorry about the question, if this is trivial.
> Thank you.
> 
> 
> James.
> 

This seems to work:

eqns = {Q1 == (2*z + 1 - 1)^2 + 4*1*1, 
    k1 == 1/Sqrt[Q1]*(v*z - 3/2*(2*z + 1 + 1 - 
         Sqrt[Q1])), Q2 == (2*z + 1 - 1)^2 + 4*1*1, 
    k2 == 1/Sqrt[Q2]*((6.1 - v)*z - 
       2/2*(2*z + 1 + 1 - Sqrt[Q2])), 
    3/2*(2*z + 1 + 1) + (k1 - 3/2)*Sqrt[Q1] == 
     6.1*z - (2/2*(2*z + 1 + 1) + (k2 - 2/2)*Sqrt[Q2])}; 

Eliminate[eqns, {Q1, Q2}]
Solve[%, {k1, k2, z}]

-3.*k1 + k1^2 == 0. && -2.*k2 + k2^2 == 0. && z == 0.

{{z -> 0., k1 -> 0., k2 -> 0.}, {z -> 0., k1 -> 0., 
   k2 -> 2.}, {z -> 0., k1 -> 3., k2 -> 0.}, 
  {z -> 0., k1 -> 3., k2 -> 2.}}

Or... 

Eliminate[eqns, {k1, k2}]
Solve[%, {Q1, Q2, z}]

Q1 == 4. && Q2 == 4. && z == 0. && Q1 != 0. && Q2 != 0.

{{Q1 -> 4., Q2 -> 4., z -> 0.}}

It appears that v drops out of the equations.

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/ 


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