Re: 0^0
- From: Dave Seaman <dseaman@xxxxxxxxxxxx>
- Date: Thu, 1 Sep 2005 23:36:57 +0000 (UTC)
On 1 Sep 2005 14:39:29 -0700, john wrote:
> a^0 = a^(1-1)
> 0^0 = 0^(1-1) = 0^1/0^1 = 0/0
0^0 = (1-1)^0
= sum_{k=0}^0 binomial(0,k) * 1^k * (-1)^k
= 1
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
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