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