Re: Formula Stirlinga
- To: mathgroup at smc.vnet.net
- Subject: [mg130699] Re: Formula Stirlinga
- From: DC <b.gatessucks at gmail.com>
- Date: Fri, 3 May 2013 03:52:49 -0400 (EDT)
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It looks better if you take the logs : n = 1000; N[Log[Factorial[n]] - (1/2 Log[2 Pi n] + n Log[n] - n)]