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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
>
>
>
>
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