From help-octave-request at bevo dot che dot wisc dot edu Fri Nov 14 21:05:59 2003
Subject: RE: 0^0 = ?
From: Mike Miller <mbmiller at taxa dot epi dot umn dot edu>
To: Randy Gober <GoberR at Bellsouth dot net>
cc: help-octave at bevo dot che dot wisc dot edu
Date: Fri, 14 Nov 2003 21:05:51 -0600 (CST)

On Fri, 14 Nov 2003, Randy Gober wrote:

> 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) )

Randy-- Tell me if I'm wrong:  x approaches zero quickly enough to offset
the rate at which log(a*x) approaches -Inf.  Let x=exp(y), then

x*log(a_1*x) = exp(y)*(a_1 + y)

What happens to that expression as y approaches -Inf?  It seems to me that
it approaches zero.  Same for lim(x->0) x*log(a_1*x).

What is lim(x->0) (x^2)*(1/x)?  Is it 0*Inf, and therefore indeterminate?
Of course not.  It converges to zero.

Mike


Goldberg's proof:

> 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."



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