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Re: Beginner--How to get a positive solution from Solve Command

• To: mathgroup at smc.vnet.net
• Subject: [mg67814] Re: [mg67791] Beginner--How to get a positive solution from Solve Command
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sat, 8 Jul 2006 04:56:56 -0400 (EDT)
• Reply-to: hanlonr at cox.net
• Sender: owner-wri-mathgroup at wolfram.com

terms=Normal[Series[1 - c *ArcTanh[x]/x, {x, 0, 2}]]

-((c*x^2)/3) - c + 1

Reduce[terms==0,x,Reals]

terms=Normal[Series[1 - c *ArcTanh[x]/x, {x, 0, 2}]]

-((c*x^2)/3) - c + 1

Reduce[terms==0,x,Reals]

Inequality[0, Less, c, LessEqual, 1] && 
  (x == -Sqrt[(3 - 3*c)/c] || x == Sqrt[(3 - 3*c)/c])

assump={0<c<=1,0<=x};

Simplify[Reduce[Append[assump,terms==0],x],assump]//
    Reverse//ToRules

{x -> Sqrt[3/c - 3]}

Simplify[Select[Solve[terms==0,x],
      Simplify[(x/.#)>=0,assump]&],assump]//Flatten

{x -> Sqrt[3/c - 3]}


Bob Hanlon

---- abdou.oumaima at hotmail.com wrote: 
> Hello mathgroup,
> 
> I'm trying to solve and equation in a limit case, so I've to use Taylor series.
> 
> In[1]: Normal[Series[1 - c  ArcTanh[&#958;]/&#958;, {&#958;, 0, 2}]]
> Out[1]: 1-c-c (&#958;^2/3)
> 
> and 
> 
> In[2]: Solve[%==0,&#958;]
> Out[2]: {{&#958;->- (3(1-c)/&#958;)^1/2},{&#958;->(3(1-c)/&#958;)^1/2}}
> 
> I need just the positive solution. How to force Solve to give me the positive value please.
> 
> Any Help please.
> 
> Thank you
> 
> Link to the forum page for this post:
> http://www.mathematica-users.org/webMathematica/wiki/wiki.jsp?pageName=Special:Forum_ViewTopic&pid=11752#p11752
> Posted through http://www.mathematica-users.org [[postId=11752]]
> 
>