Re: Reduce in Ver 6

*To*: mathgroup at smc.vnet.net*Subject*: [mg85543] Re: Reduce in Ver 6*From*: dh <dh at metrohm.ch>*Date*: Wed, 13 Feb 2008 04:28:11 -0500 (EST)*References*: <fopb8t$bkk$1@smc.vnet.net>

Hi Dana, Reduce uses results from earlier solutions to simplify expressions. You may prevent this by the option: Backsubstitution -> True hope this helps, Daniel Dana DeLouis wrote: > 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 > > > >