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Re: 0^0
Posted:
Oct 20, 2006 12:05 AM


On Thu, 19 Oct 2006 21:31:52 0600, Virgil <virgil@comcast.net> wrote:
> >The whole problem in trying to define 0^0 is that >lim_{x > 0+} lim_{y > 0+} x^y =\= lim_{y > 0+} lim_{x > 0+} x^y. > >lim_{x > 0+} lim_{y > 0+} x^y = 1 >but >lim_{y > 0+} lim_{x > 0+} x^y = 0.
No, that isn't the problem in trying to define 0^0. It is only a problem if you wish to define 0^0 in such a way as to make x^y a continuous function at (0,0).
Consider the expression (1  x)^2 = 1 2x + x^2
If we expand by the binomial theorem we get:
sum[k=0..2] C(2,k)*1^k *(1)^(2k)
= (2! / (0! 2!))*1^0*(x)^2 + (2! / (1! 1!)) *1^1*(x)^1 + (2! / (2! 0!))*1^2*(x)^0
We want this expression to identically equal 1 2x + x^2.
In particular, we would like this identity to hold if x = 0. So obviously we need to get 1 when we put x = 0 in the binomial expansion. But we get:
(2! / (0! 2!))*1^0*(0)^2 + (2! / (1! 1!)) *1^1*(0)^1 + (2! / (2! 0!))*1^2*(0)^0
You probably don't have any difficulty with the fact that we defined 0! = 1 to make this work, and you will note that if we define 0^0 = 1 the last term works. It is just a convenience of notation and the fact that x^y has an essential discontinuity at (0,0) is irrelevant.
Lynn



