Re: Solve
- To: mathgroup at smc.vnet.net
- Subject: [mg128423] Re: Solve
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Sun, 14 Oct 2012 23:42:01 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
- Delivered-to: mathgroup-newsend@smc.vnet.net
- References: <20121014041302.448AD6868@smc.vnet.net>
g[\[Alpha]_] := \[Rho] (\[Delta] - \[Alpha] \[Delta] + \[Rho] + \[Alpha] \ \[Rho] - 2 Sqrt[\[Alpha] \[Rho] (\[Delta] - \[Alpha] \[Delta] + \[Rho])]) sol1 = Solve[g'[\[Alpha]] == 0, \[Alpha], Reals] {{\[Alpha] -> ConditionalExpression[1, \[Rho] > 0]}, {\[Alpha] -> ConditionalExpression[\[Rho]/\[Delta], \[Rho] < 0]}} Solve[{g'[\[Alpha]] == 0, \[Rho] > 0}, \[Alpha], Reals] // Simplify[#, \[Rho] > 0] & {{\[Alpha] -> 1}} g'[1] // Simplify[#, \[Rho] > 0] & 0 Solve[{g'[\[Alpha]] == 0, \[Rho] < 0}, \[Alpha], Reals] // Simplify[#, \[Rho] < 0] & {{\[Alpha] -> \[Rho]/\[Delta]}} g'[\[Rho]/\[Delta]] // Simplify[#, \[Rho] < 0] & 0 "Solve deals primarily with linear and polynomial equations. " Use Reduce sol2 = Reduce[g'[\[Alpha]] == 0, \[Alpha], Reals] // Simplify (\[Rho] < 0 && ((\[Alpha] == \[Rho]/\[Delta] && (\[Delta] < 0 || \[Delta] + \[Rho] > 0 || (\[Delta] > 0 && \[Delta] + \[Rho] < 0))) || (\[Alpha] < 0 && \[Delta] + \[Rho] == 0))) || (\[Rho] > 0 && (\[Alpha] == 1 || (\[Alpha] > 0 && \[Delta] + \[Rho] == 0))) || (\[Delta] != 0 && \[Rho] == 0 && (\[Alpha] > 1 || \[Alpha] < 0 || 0 < \[Alpha] < 1)) Simplify[sol2, \[Rho] > 0] \[Alpha] == 1 || (\[Alpha] > 0 && \[Delta] + \[Rho] == 0) Simplify[g'[1], \[Rho] > 0] 0 Simplify[g'[\[Alpha]], {\[Rho] > 0, \[Alpha] > 0, \[Delta] + \[Rho] == 0}] 0 Simplify[sol2, \[Rho] < 0] (\[Alpha] == \[Rho]/\[Delta] && (\[Delta] < 0 || \[Delta] + \[Rho] > 0 || (\[Delta] > 0 && \[Delta] + \[Rho] < 0))) || (\[Alpha] < 0 && \[Delta] + \[Rho] == 0) Simplify[g'[\[Alpha]], {\[Rho] < 0, \[Alpha] == \[Rho]/\[Delta], \[Delta] < 0}] 0 Simplify[g'[\[Alpha]], {\[Rho] < 0, \[Alpha] == \[Rho]/\[Delta], \[Delta] + \[Rho] > 0}] 0 Simplify[g'[\[Alpha]], {\[Rho] < 0, \[Alpha] == \[Rho]/\[Delta], \[Delta] > 0, \[Delta] + \[Rho] < 0}] 0 Simplify[g'[\[Alpha]], {\[Rho] < 0, \[Alpha] < 0, \[Delta] + \[Rho] == 0}] 0 Bob Hanlon On Sun, Oct 14, 2012 at 12:13 AM, <JikaiRF at aol.com> wrote: > 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. > > D[g[\[Alpha]],\[Alpha]]. > > 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: > > {{\[Alpha]->1},{\[Alpha]->\[Rho]/\[Delta]}}. > > 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. >
- References:
- Solve
- From: JikaiRF@aol.com
- Solve