Re: Simplify
- To: mathgroup at smc.vnet.net
- Subject: [mg14814] Re: [mg14780] Simplify
- From: Jurgen Tischer <jtischer at col2.telecom.com.co>
- Date: Wed, 18 Nov 1998 01:29:17 -0500
- Organization: Universidad del Valle
- References: <deda4d7b.364ef5bb@aol.com>
- Sender: owner-wri-mathgroup at wolfram.com
Bob,
you made me try it once more and I found it(?)!
In[1]:= FullSimplify[ExpToTrig[(-7253*2^(1 + 2*n)*5^n -
523*4^n*5^(1 + n) + 17121*20^n + 130321*20^n*n^4)/
130321]]
Out[1]= 20^n*n^4
Thank you for insisting!
Jurgen
BobHanlon at aol.com wrote:
>
> In a message dated 11/15/98 1:16:55 AM, jtischer at col2.telecom.com.co writes:
>
> >you just showed me that it's hopeless. I'm searching for a way that will
> >
> >cook down formulas of this type, your method is ok but needs human
> >
> >skill.
> >
> >Anyway, thanks a lot.
> >
> >
>
> Jurgen,
>
> I am not convinced that it is hopeless. You might investigate along these
> lines:
>
> Identify the integers that are raised to powers of n:
>
> baseIntegers[eqn_, n_Symbol:n] :=
> Cases[eqn, i_Integer^n->i, {1, Infinity}]//Union
>
> Then automate the substitutions given the integers
>
> simplifyMore[eqn_, i1_Integer, i2_Integer, n_Symbol:n] :=
> Module[{g = GCD[i1, i2], m, eqn2}, If[g==1, eqn,
> eqn2 = FullSimplify[FullSimplify[eqn] /.
> {i1^n -> m (i1/g)^n, i2^n -> m (i2/g)^n}];
> FullSimplify[eqn2 /. m -> g^n]]]
>
> For example
>
> a = (-7253*2^(1 + 2*n)*5^n - 523*4^n*5^(1 + n) + 17121*20^n +
> 130321*20^n*n^4)/130321;
>
> baseIntegers[a]
>
> {4,5,20}
>
> simplifyMore[a, 4, 20]
>
> 20^n*n^4
>
> or
>
> simplifyMore[a, 5, 20]
>
> 20^n*n^4
>
> The liberal use of FullSimplify will tend to make this relatively slow.
>
> Bob Hanlon