       Re: Simplifying an expression in light of relationships between variables?

• To: mathgroup at smc.vnet.net
• Subject: [mg54283] Re: [mg54263] Simplifying an expression in light of relationships between variables?
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 16 Feb 2005 14:36:02 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```cond={d==a+b,e==c+d};

Simplify[#,cond]& /@
{a+b,d-b,a+b+c,e-c,e-c-b}

{d,d-b,e,d,d-b}

FullSimplify[#,cond]& /@
{a+b,d-b,a+b+c,e-c,e-c-b}

{d,d-b,e,d,d-b}

For reasons unknown to me, d-b won't simplify.  Hit it with a hammer.

Simplify[#,cond] /. d-b:>a& /@
{a+b,d-b,a+b+c,e-c,e-c-b}

{d,a,e,d,a}

Bob Hanlon

>
> From: "Steve W. Brewer" <ste-ve at ka-tech.com>
To: mathgroup at smc.vnet.net
> Date: 2005/02/14 Mon PM 09:51:04 EST
> To: mathgroup at smc.vnet.net
> Subject: [mg54283] [mg54263] Simplifying an expression in light of relationships
between variables?
>
> Suppose I have a few variables that are related in the following way:
>
> d == a + b
> e == c + d
>
>
>
> I want Simplify and FullSimplify to consider these relationships when
> performing simplification.  For example:
>
> FullSimplify[a + b]
>
> d
>
>
>
> FullSimplify[d - b]
>
> a
>
>
>
> FullSimplify[a + b + c]
>
> e
>
>
>
> FullSimplify[e - c]
>
> d
>
>
>
> FullSimplify[e - c - b]
>
> a
>
>
>
> Is there a straightforward way to get this behavior?  I've tried
> experimenting with TransformationFunctions, but it looks like I need to
> include every possible permutation for it to work, for example:
>
> TransformationFunctions ->
>    {Automatic,
>     # /. (d -> a + b)&,
>     # /. (a + b -> d)&,
>     # /. (a -> d - b)&,
>     # /. (d - b -> a)&,
>     # /. (b -> d - a)&,
>     # /. (d - a -> b)&,
>     ... }
>
>
>
> This is unwieldy even in this simple example, and it rapidly becomes
> unmanageable as the number of variables and relationships increases.
>
>
> Is there an easier way?
>
>

```

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