Partial solution for bug 1452341: solve(x^(5/2)+1,x)
Subject: Partial solution for bug 1452341: solve(x^(5/2)+1,x)
From: Richard Fateman
Date: Thu, 25 May 2006 12:02:23 -0700
well, you also have to notice that the base is the same (here, 5), or maybe
has a common factor. e.g. if you had 15^(1/2)- 3^(1/2)*5^(1/2) -->? maybe
0.
And that is what radcan does, mostly. If you want to make sure that the
exponents and the bases are all (small?) numbers, I guess you can keep the
cost down.
RJF
----- Original Message -----
From: "Stavros Macrakis" <macrakis at gmail.com>
To: "Richard Fateman" <fateman at cs.berkeley.edu>
Cc: "Raymond Toy" <raymond.toy at ericsson.com>; <maxima at math.utexas.edu>
Sent: Thursday, May 25, 2006 11:55 AM
Subject: Re: [Maxima] Partial solution for bug 1452341: solve(x^(5/2)+1,x)
The cheap idea is to notice that gcd(1/2,2/3,1/3)=1/6, so the expression
looks like a^3*(a^4+a^2) where a=5^(1/6).
Whether this is valid depends on your interpretation of fractional
exponents, I suppose. (domain:real vs. complex)
On 5/25/06, Richard Fateman <fateman at cs.berkeley.edu> wrote:
>
> ratsimp basically renames 5^(1/2)*(5^(2/3)+5^(1/3)) to be a*(b^2+b),
> where a=sqrt(5), b= 5^(1/3), rationally simplifies.
> ...
> If you have some ideas that are cheap and effective, go to it!