RE: 0^0 = ?

[Randy Gober]

Goldberg's proof is flawed. He writes that f(x)^g(x)= e^[g(x)log{f(x)}], but
f(x) = 0, => log(f(x)) = -inf
So g(x)log(f(x)) is a 0*inf form, which itself is indeterment. 

(similar problem with the limit at the end: lim(x->0) x*log(a_1*x) )


I still think NaN is still the correct answer.

--Randy





-----Original Message-----
From: Mike Miller [mailto:address@hidden 
Sent: Thursday, November 13, 2003 11:14 AM
To: John W. Eaton
Cc: address@hidden; address@hidden; 'Cong';
address@hidden
Subject: RE: 0^0 = ?


[stuff deleted]
I found it here:

http://docs.sun.com/source/806-3568/ncg_goldberg.html

Here is the part we wanted to see:

"In the case of 0^0, f(x)^g(x) = e^[g(x)log{f(x)}]. Since f and g are
analytic and take on the value 0 at 0, f(x) = a_1*x^1 + a_2*x^2 + ... and
g(x) = b_1*x^1 + b_2*x^2 + .... Thus lim(x->0) g(x)*log[f(x)] =
lim(x->0) x*log[x(a_1 + a_2*x + ...)] = lim(x->0) x*log(a_1*x) = 0. So
f(x)^g(x) -> e^0 = 1 for all f and g, which means that 0^0 = 1."

Mike

-- 
Michael B. Miller, Ph.D.
Assistant Professor
Division of Epidemiology
and Institute of Human Genetics
University of Minnesota
http://taxa.epi.umn.edu/~mbmiller/



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