Re: Reduce in Ver 6
- To: mathgroup at smc.vnet.net
- Subject: [mg85537] Re: [mg85464] Reduce in Ver 6
- From: Carl Woll <carlw at wolfram.com>
- Date: Wed, 13 Feb 2008 04:25:06 -0500 (EST)
- References: <200802111113.GAA11303@smc.vnet.net>
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}}
>
>
Use the Reduce option Backsubstitution->True.
Carl Woll
Wolfram Research
>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
>
>
>
>
>
- References:
- Reduce in Ver 6
- From: "Dana DeLouis" <dana.del@gmail.com>
- Reduce in Ver 6