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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 9,163 of 10,432   
   Axel Vogt to Nasser M. Abbasi   
   Re: On simplifying (a^n)^(1/n)   
   12 Sep 16 21:35:04   
   
   From: //noreply@axelvogt.de   
      
   On 12.09.2016 06:25, Nasser M. Abbasi wrote:   
   > I was looking at some algebra cheat sheet, where it said that   
   > something I never seen before:   
   >   
   > (a^n)^(1/n) was "a" if n is odd, and |a| if n is even.   
   ...   
      
   The *general* formula (which is more usefull, I think) is:   
      
      (z^a)^b = z^(a*b) * exp(-2*Pi*I*b*K(a*ln(z)))   
      
   where K = "unwinding number" (which cares for brunch cuts   
   in a consistent (!) way),   
      
   K(z) = ceil(1/2*(Im(z)-Pi)/Pi)         # Maple   
         = 1/2*I*(-z+ln(exp(z)))/Pi        # "Blätterzahl"   
         = 1/2*(Im(z)-argument(exp(z)))/Pi # Aslaksen (*)   
      
   Note that brunch cuts for log and sqrt need to fit (there   
   is neat theorem connecting log and sqrt in complex analysis).   
      
   (*) Aslaksen: Can your computer do complex analysis (2005)   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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