It should be pointed out that (x^p)^(1/p) is NOT in general x, and that
when radcan makes that simplification, it is also making some assumptions
about x.
Let p = 2
Let x = -3.
((-3)^2)^(1/2) according to your formula should be -3. Most people seem
to think this is not the right answer. Sometimes people claim it is 3.
Sometimes +-3.
RJF
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu] On Behalf Of Viktor T. Toth
> Sent: Friday, May 18, 2007 7:56 AM
> To: 'Gary O'; maxima at math.utexas.edu
> Subject: Re: [Maxima] Simplify x^p^(1/p) to r?
>
> To Maxima, x^y^z is x^(y^z) and not (x^y)^z. However, if you use
> parentheses, you get
>
> (%i1) (x^p)^(1/p);
> p 1/p
> (%o1) (x )
> (%i2) radcan(%);
> (%o2) x
>
>
> Viktor
>
>
>
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu]
> On Behalf Of Gary O
> Sent: Friday, May 18, 2007 10:30 AM
> To: maxima at math.utexas.edu
> Subject: Simplify x^p^(1/p) to r?
>
> Q: why doesn't this return x?
> ratsimp(x^p^(1/p))
>
> If maxima doesn't know that x^y^z => x^(y*z), can I teach it somehow?
>
> thanks,
>
> -- Gary Oberbrunner
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>