Re: Solve bug !!

*To*: mathgroup at smc.vnet.net*Subject*: [mg31205] Re: [mg31196] Solve bug !!*From*: Andrzej Kozlowski <andrzej at bekkoame.ne.jp>*Date*: Fri, 19 Oct 2001 03:11:54 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

This is no bug. You should be able to check yourself that both answers are correct. In fact suppose c satisifes (E^(-c))^2 - A*E^(-c) + 1 == 0 i.e. 1 + E^(-2*c) - A/E^c == 0 multiply the equation by the non zero expression E^(2c) and you will get E^(2c)+A E^c +1==0 Thus any c that satisifeds one equation also satisifes the other. In other words, they have the same set of roots. It is true that Mathematica does not give you all the roots, but this can't be called a bug since it has to use inverse functions and it does issue a warning to this effect. Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Wednesday, October 17, 2001, at 06:35 PM, Marcel wrote: > Where is the minus sign whe must obtain in the second case?? > >> Solve[(E^c)^2 - A*E^c + 1 == 0, c] > > {{c -> Log[(1/2)*(A - Sqrt[-4 + A^2])]}, > {c -> Log[(1/2)*(A + Sqrt[-4 + A^2])]}} > >> Solve[(E^(-c))^2 - A*E^(-c) + 1 == 0, c] > > {{c -> Log[(1/2)*(A - Sqrt[-4 + A^2])]}, > {c -> Log[(1/2)*(A + Sqrt[-4 + A^2])]}} > > > Mathematica 4.1, Windows 2000 SP2, PII400. > > Marcel Aguilella > > > > >