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Re: Simplify last version
*To*: mathgroup at smc.vnet.net
*Subject*: [mg14924] Re: [mg14868] Simplify last version
*From*: Jurgen Tischer <jtischer at col2.telecom.com.co>
*Date*: Wed, 25 Nov 1998 17:48:44 -0500
*Organization*: Universidad del Valle
*References*: <E1168E31432DD211AF990000F806EFD82E667A@nems04.nawcad.navy.mil>
*Sender*: owner-wri-mathgroup at wolfram.com
Thanks, Ted. I want to point out that this was my error, not Allan's.
Jurgen
"Ersek, Ted R" wrote:
>
> Jugen Tischer posted the following solution by Allan Hayes,
>
> >
> >
> >expr =(-7253*2^(1 + 2*n)*5^n - 523*4^n*5^(1 + n) + 17121*20^n +
> >130321*20^n*n^4)/130321
> >
> >simpl[expr_] :=
> > Simplify[Factor[
> > expr /. (x_Integer)^(k_) :>
> > Times@@(C[#1[[1]]]^(k #1[[2]]) &)/@FactorInteger[x]] /.
> > C[x_] -> x]
> >
> >simpl[expr]
> >
> >20^n n^4
> >
> >
>
> Nice solution. However, using patterns on the left side Rule(->) may give
> you a result you don't want. Consider the following.
>
> In[6]:=
> x=123;
>
> ...
> ...
> ...
>
> We still have a global value for (x).
> In[86]:=
> {C[10],C[20],C[30]}/.C[x_]->x
> Out[86]=
> {123,123,123}
> We didn't get what we wanted for Out[86].
>
> Instead used RuleDelayed (:>) as below.
> In[87]:=
> {C[10],C[20],C[30]}/.C[x_]:>x
> Out[87]=
> {10,20,30}
> That's better.
>
> If you absolutely must use a pattern on
> the left side of (->), do so inside a block as below.
> In[88]:=
> x=145;
> Block[{x},{C[10],C[20],C[30]}/.C[x_]->x]
> Out[88]=
> {10,20,30}
> That works too.
>
> _________________________
>
> With that said I recommend the following over the solution above.
>
> simpl[expr_] :=
> Simplify[Factor[
> expr /. (x_Integer)^(k_) :>
> Times@@(C[#1[[1]]]^(k #1[[2]]) &)/@FactorInteger[x]] /.
> C[x_] :> x]
>
> _________________________
>
> Note:
> Set, UpSet, and TagSet have the same trouble that Rule does (shown above)
> with patterns on the left side.
>
> Cheers,
> Ted Ersek
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