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Topic: 0^0
Replies: 23   Last Post: Oct 25, 2006 3:13 AM

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Lynn Kurtz

Posts: 1,278
Registered: 12/6/04
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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