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MathGroup Archive 2001

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Re: Solve bug !!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31215] Re: [mg31196] Solve bug !!
  • From: Rob Pratt <rpratt at email.unc.edu>
  • Date: Fri, 19 Oct 2001 03:12:06 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On Wed, 17 Oct 2001, 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

I imagine you will receive several replies about this one, but there are
(at least) two issues to be addressed:

1. Since the two equations are equivalent (multiply both sides of the
second one by Exp[2c]), the solution sets have to be the same.

2. Since the second equation is just a change of variable from c to -c,
you ought to be able to make the inverse (in this case, the same) change
of variable to get the new solutions.  Indeed, you get the same pair back:

Log[(1/2)*(A - Sqrt[-4 + A^2])] == -Log[(1/2)*(A + Sqrt[-4 + A^2])]

for real A, as you can check by using Log[u] + Log[v] = Log[u v] and
simplifying.

Simplify[Log[(1/2)*(A - Sqrt[-4 + A^2])] + Log[(1/2)*(A + Sqrt[-4 + A^2])]
/.{Log[u_]+Log[v_]->Log[u v]}]

should return 0.

So the first solution in the first equation is the same as the second
solution in the second equation,
and the second solution in the first equation is the same as the first
solution in the second equation.

In short, there is no bug.

Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill

rpratt at email.unc.edu

http://www.unc.edu/~rpratt/




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