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Re: Bug in Reduce?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg60483] Re: [mg60406] Bug in Reduce?
*From*: Pratik Desai <pdesai1 at umbc.edu>
*Date*: Sat, 17 Sep 2005 02:32:22 -0400 (EDT)
*References*: <dfrhi4$g4l$1@smc.vnet.net> <dg8lfv$r8g$1@smc.vnet.net> <200509140926.FAA01590@smc.vnet.net> <200509150916.FAA15875@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Kennedy wrote:
>The source of this apparent bug could be my misunderstanding of the middle,
>"vars" parameter of Reduce, but it sure seems like the following output
>indicates that c must be 0 for my two equations to be satisfied, when in
>fact if a and b are both 0, c does not need to be 0.
>
>Regards,
>Jack
>
>In[1]:=
>Reduce[{a c - b d == 0, a d + b c == 0}, {a, b, c, d}, Reals] //
>FullSimplify
>
>Out[1]=
>c == 0 && (d == 0 || (a == 0 && b == 0))
>
>(version 5.1 for Windows)
>
>
>
I get a more richer solution when I used assumptions (c notequal 0) and
simplify
Clear[a, b, c, d]
expr1 = {a*c - b*d == 0, a*d + b*c == 0}
$Assumptions = {c â? 0, a Ïµ
Reals, b Ïµ Reals, cÏµ Reals, d Ïµ Reals, c Ïµ Reals}
s1 = Reduce[expr1, {a, b, c, d}] // Simplify
>>(a == 0 && (b == 0 || (c == 0 && d == 0))) || ((b*c)/a + d == 0 && a
!= 0 && (b == (-I)*a || b == I*a)) ||
(c == 0 && d == 0 && a^2 + b^2 != 0)
Hope this helps
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--
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134
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