On Fri, 2007-03-09 at 23:40 +0600, Andrey G. Grozin wrote:
> In some other mail, there was an even more convincing proof that
> 1^inf
> should be und:
>
> 1^inf = exp(inf*log(1)) = exp(inf*0) = exp(und) = und
that proof is not convincing. You didn't show the first
I even think it is wrong. You skipped the first step,
1^inf = exp(log(1^inf)), and jumped into the (wrong) conclusion
that log(1^inf) = inf*log(1).
Let's start by considering
s1: sum(a, i, 1, n)
where a is a real number and n is a positive integer. There is
no doubt that s1 equals n*a. Now consider
s2: sum(a, i, 1, inf)
there are three possible results: s2=inf, if a>0, s2=minf, if a<0 or
s2=0, if a=0. It means that you should be cautious with the
following simplification:
log(a^n) = log(prod(a, i, 1, n)) = sum(log(a), i, 1, n) = n*log(a)
If a=1, then, for any positive integer n:
log(a^n) = log(prod(a, i, 1, n)) = sum(log(a), i, 1, n) = 0
Thus, I would interpret log(a^inf) as 0, if a=1. That means:
1^inf = exp(log(1^inf)) = exp(0) = 1
Regards,
Jaime