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Re: 0^0
Posted:
Oct 20, 2006 12:05 AM
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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)^(2-k)
= (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
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