Re: Reduce in Ver 6

*To*: mathgroup at smc.vnet.net*Subject*: [mg85491] Re: Reduce in Ver 6*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Mon, 11 Feb 2008 07:14:03 -0500 (EST)*References*: <fopb8t$bkk$1@smc.vnet.net>

If you use the rules repeatedly then the correct value of c is replaced. So checking the solutions by back substitution using equ/.r leaves gives an expression that still depends on a and b (from the rules expressing c in terms of a and b), but equ//.r (note //. NOT /.) gives the required result: {{True,True,True,True},{True,True,True,True},{True,True,True,True},{True,True,True,True},{True,True,True,True},{True,True,True,True}} Stephen Luttrell West Malvern, UK "Dana DeLouis" <dana.del at gmail.com> wrote in message news:fopb8t$bkk$1 at smc.vnet.net... > Hi. I was wondering if anyone knows of a more efficient way to do this. > This issue came up in ver 6.0 with these equations: > > equ = { > a + b + c == 3, > a^2 + b^2 + c^2 < 10, > a^3 + b^3 + c^3 == 15, > a^4 + b^4 + c^4 == 35 > } > > I use Reduce, but 'c is returned as a function of a & b. > What I would like is for c to replace a & b with the appropriate values. > > r = {ToRules[Reduce[equ, {a, b, c}]]} > > {{a -> 1, b -> 1 - Sqrt[2], c -> 3 - a - b}, > {a -> 1, b -> 1 + Sqrt[2], c -> 3 - a - b}, > {a -> 1 - Sqrt[2], b -> 1, c -> 3 - a - b}, > {a -> 1 - Sqrt[2], b -> 1 + Sqrt[2], c -> 3 - a - b}, > {a -> 1 + Sqrt[2], b -> 1, c -> 3 - a - b}, > {a -> 1 + Sqrt[2], b -> 1 - Sqrt[2], c -> 3 - a - b}} > > My workaround is rather tough, especially if this was much larger. > > convert = {a -> aa_, b -> bb_, c -> cc_} -> > {a -> aa, b -> bb, c -> 3 - aa - bb}; > > new = r /. convert > > {{a -> 1, b -> 1 - Sqrt[2], c -> 1 + Sqrt[2]}, > {a -> 1, b -> 1 + Sqrt[2], c -> 1 - Sqrt[2]}, > {a -> 1 - Sqrt[2], b -> 1, c -> 1 + Sqrt[2]}, > > Etc.. > > Any help is much appreciated. I hope I'm not overlooking something > obvious. > > Dana > ddelouis at gmail.com > > > >