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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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