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RE: 0^0 = ?
From: |
Randy Gober |
Subject: |
RE: 0^0 = ? |
Date: |
Thu, 13 Nov 2003 19:40:49 -0600 |
Thanks for the explanation John.
As a small fyi, L'Hopitals really doesn't apply.
Since we need lim x->0, to apply L'Hopitals, you need both the numerator and
the denominator to be continuous on an interval [a,b] about zero and
differentiable on an interval (a,b) about zero.
ln(x) and 1/x are neither.
Thanks,
--Randy
-----Original Message-----
From: John W. Eaton [mailto:address@hidden
Sent: Thursday, November 13, 2003 10:51 AM
To: address@hidden
Cc: address@hidden; 'Cong'
Subject: RE: 0^0 = ?
On 13-Nov-2003, address@hidden <address@hidden> wrote:
| I know that by L'Htpital's Rule you should get:
|
| ln(y)=x*ln(x) = ln(x)/(1/x) so
|
| Lim x->0+ ln(x)/(1/x) = ( 1/x )/( -1/( x^2)) =
|
| Lim x->0+ (-x) = 0
| so ln(y) = 0 and then y=1.
|
| Maybe this is the reason for the behavior?
The 0^0 == 1 behavior is part of the IEEE 754 standard for floating point
arithmetic. The paper "What every computer scientist should know about
floating point arithmetic" by David Goldberg provides a rationale for the
behavior that is a bit different than above (it's at the end of a section
titled "ambiguity"). To start with, I think you need to look at this as
y^x, not x^x. A quick google search should turn up a copy of the paper if
you want the details.
jwe
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- 0^0 = ?, Cong, 2003/11/12
- RE: 0^0 = ?, j . e . drews, 2003/11/13
- RE: 0^0 = ?, John W. Eaton, 2003/11/13
- RE: 0^0 = ?, Mike Miller, 2003/11/13
- RE: 0^0 = ?, Randy Gober, 2003/11/14
- RE: 0^0 = ?, Mike Miller, 2003/11/14
- RE: 0^0 = ?, Randy Gober, 2003/11/14
- RE: 0^0 = ?, Mike Miller, 2003/11/14
- RE: 0^0 = ?,
Randy Gober <=
- RE: 0^0 = ?, Boud Roukema, 2003/11/14
- RE: 0^0 = ?, John W. Eaton, 2003/11/14
- Re: 0^0 = ?, Geraint Paul Bevan, 2003/11/14
- RE: 0^0 = ?, Ted Harding, 2003/11/14
- RE: 0^0 = ?, John W. Eaton, 2003/11/14
- RE: 0^0 = ?, Mike Miller, 2003/11/14