Precision check on [pentation.gp] SOLVED JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 06/29/2011, 02:42 AM (This post was last modified: 06/29/2011, 02:44 AM by JmsNxn.) That's funny, I was just thinking about $exp^{\pi/2}$. Naively I wanna say it's because $\lim_{n\to\infty} ^n(e^{\pi/2}) = \lim_{n\to\infty} ^n(i^{1/i})= i$ which is what usually occurs in the self root function. However this would imply real to imaginary which isn't very pretty. Funny that the code fails around it tho. I think there is something special about $exp^{\pi/2}$, I'm just not sure what. « Next Oldest | Next Newest »

 Messages In This Thread Precision check on [pentation.gp] SOLVED - by Cherrina_Pixie - 06/28/2011, 07:17 AM RE: Precision check on [pentation.gp] fails? - by sheldonison - 06/28/2011, 02:33 PM RE: Precision check on [pentation.gp] fails? - by JmsNxn - 06/29/2011, 02:42 AM RE: Precision check on [pentation.gp] fails? - by sheldonison - 06/29/2011, 05:05 AM RE: Precision check on [pentation.gp] fails? - by sheldonison - 06/29/2011, 10:36 PM RE: Precision check on [pentation.gp] fails? - by sheldonison - 06/30/2011, 06:00 AM RE: Precision check on [pentation.gp] fails? - by sheldonison - 07/01/2011, 10:56 PM RE: Precision check on [pentation.gp] fails? - by Cherrina_Pixie - 07/02/2011, 01:39 AM

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