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  • To: mathgroup at
  • Subject: [mg128408] Solve
  • From: JikaiRF at
  • Date: Sun, 14 Oct 2012 00:13:02 -0400 (EDT)
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Dear members:
I basecally intend to solve the following equation for \[Alpha]:

g[\[Alpha]_]:=\[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho]+\[Alpha] \[Rho]-2 Sqrt[\[Alpha] \[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho])])

First of all, in order to investigate the function g, I differentiate g(\[Alpha]) with \[Alpha]. I input the following formula into Mathematica.


As a result, I obtain 

 g'(\[Alpha])=\[Rho] (-\[Delta]+\[Rho]-(-\[Alpha] \[Delta] \[Rho]+\[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho]))/Sqrt[\[Alpha] \[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho])]).

At this stage, I input the following to find \[Alpha] satisfying g'(\[Alpha])=0.

Solve[\[Rho] (-\[Delta]+\[Rho]-(\[Rho] (\[Delta]-2 \[Alpha] \[Delta]+\[Rho]))/Sqrt[\[Alpha] \[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho])])==0, \[Alpha]],

then I obtain the following  from Mathematica:


However, g'(\[Rho]/\[Delta]) = \[Rho] (-\[Delta]+\[Rho]-((\[Delta]-\[Rho]) \[Rho])/Sqrt[\[Rho]^2]) = 2(\[Rho] - \[Delta])  holds, if \[Rho] > 0. On the other hand, if \[Rho] < 0, g'(\[Rho]/\[Delta]) = 0 holds.

This diferrence is derived from the fact that the Solve interpreates Sqrt[\[Rho]^2] = - \[Rho].
But the value I sassume is \[Rho] > 0. 
In order to make sure, I define a new function, which is equals to g'(\[Alpha]), as 

y[\[Alpha]_]:=\[Rho] (-\[Delta]+\[Rho]-(\[Rho] (\[Delta]-2 \[Alpha] \[Delta]+\[Rho]))/Sqrt[\[Alpha] \[Rho] (\[Delta]-\[Alpha] \[Delta]+\[Rho])])

And when I put \[Alpha] = \[Rho]/\[Delta] into y, I obtain \[Rho] (-\[Delta]+\[Rho]-((\[Delta]-\[Rho]) \[Rho])/Sqrt[\[Rho]^2]). 
In my opinion, the Solve function is programmed to override alternative solutions. In other words, it is overdetermined in an unfortunate way. In this case, Sqrt[\[Rho]^2]) is interpreted as \[Rho] < 0.

Is this correct?

	Fujio Takata
	Kobe University, Japan.

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