RE: 0^0 = ?

[Mike Miller]

On Thu, 13 Nov 2003, John W. Eaton wrote:

> 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.

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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