       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, c -> 3 - a - b},
> {a -> 1, b -> 1 + Sqrt, c -> 3 - a - b},
> {a -> 1 - Sqrt, b -> 1, c -> 3 - a - b},
> {a -> 1 - Sqrt, b -> 1 + Sqrt, c -> 3 - a - b},
> {a -> 1 + Sqrt, b -> 1, c -> 3 - a - b},
> {a -> 1 + Sqrt, b -> 1 - Sqrt, 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, c -> 1 + Sqrt},
> {a -> 1, b -> 1 + Sqrt, c -> 1 - Sqrt},
> {a -> 1 - Sqrt, b -> 1, c -> 1 + Sqrt},
>
>Etc..
>
>Any help is much appreciated.  I hope I'm not overlooking something obvious.
>
>Dana
>ddelouis at gmail.com
>
>
>
>
>

```

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