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Re: Reduce in Ver 6

  • To: mathgroup at smc.vnet.net
  • Subject: [mg85543] Re: Reduce in Ver 6
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 13 Feb 2008 04:28:11 -0500 (EST)
  • References: <fopb8t$bkk$1@smc.vnet.net>


Hi Dana,

Reduce uses results from earlier solutions to simplify expressions. You 

may prevent this by the option:

Backsubstitution -> True

hope this helps, Daniel





Dana DeLouis wrote:

> Hi.  I was wondering if anyone knows of a more efficient way to do this.

> This issue came up in ver 6.0 with these equations:

> 

> equ = {

> a + b + c == 3,

> a^2 + b^2 + c^2 < 10,

> a^3 + b^3 + c^3 == 15,

> a^4 + b^4 + c^4 == 35

> }

> 

> I use Reduce, but 'c is returned as a function of a & b.

> What I would like is for c to replace a & b with the appropriate values.

> 

> r = {ToRules[Reduce[equ, {a, b, c}]]}

> 

> {{a -> 1, b -> 1 - Sqrt[2], c -> 3 - a - b},

>  {a -> 1, b -> 1 + Sqrt[2], c -> 3 - a - b},

>  {a -> 1 - Sqrt[2], b -> 1, c -> 3 - a - b},

>  {a -> 1 - Sqrt[2], b -> 1 + Sqrt[2], c -> 3 - a - b},

>  {a -> 1 + Sqrt[2], b -> 1, c -> 3 - a - b},

>  {a -> 1 + Sqrt[2], b -> 1 - Sqrt[2], c -> 3 - a - b}}

> 

> My workaround is rather tough, especially if this was much larger.

> 

> convert = {a -> aa_, b -> bb_, c -> cc_} ->

>           {a -> aa, b -> bb, c -> 3 - aa - bb};

> 

> new = r /. convert

> 

> {{a -> 1, b -> 1 - Sqrt[2], c -> 1 + Sqrt[2]},

>  {a -> 1, b -> 1 + Sqrt[2], c -> 1 - Sqrt[2]},

>  {a -> 1 - Sqrt[2], b -> 1, c -> 1 + Sqrt[2]},

> 

> Etc..

> 

> Any help is much appreciated.  I hope I'm not overlooking something obvious.

> 

> Dana

> ddelouis at gmail.com

> 

> 

> 

> 




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