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MathGroup Archive 2004

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Re: fullsimplify problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52938] Re: fullsimplify problem
  • From: Robert <robert_dot_paynter_ at _eng_.ox.ac.uk>
  • Date: Fri, 17 Dec 2004 05:19:09 -0500 (EST)
  • Organization: Oxford University, England
  • References: <cprjjd$r57$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

symbio wrote:
> Given ( - a / b^2) = =  ( + a / b^2), it can be written as (-2 a / b^2), but 
> when I use FullSimplify on that equation, I get the incorrect result ( a / 
> b ) = = 0.  How is this possible?
> 
> In[2]:=
> -a/b^2 == a/b^2
> % // FullSimplify
> Out[2]=
> \!\(\(-\(a\/b\^2\)\) == a\/b\^2\)
> Out[3]=
> \!\(a\/b == 0\)
> 
> In[5]:=
> -2a/b^2 == 0 // FullSimplify
> Out[5]=
> \!\(a\/b == 0\)
> 


I think what you have found is the "trivial" solution to an
eigenvalue problem.

R


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