On Oct 22, 2009, at 3:03 PM, Francis Clarke wrote:
The following article has interesting remarks on this question, particularly pages 407--408: \bib{MR1163629}{article}{ author={Knuth, Donald E.}, title={Two notes on notation}, journal={Amer. Math. Monthly}, volume={99}, date={1992}, number={5}, pages={403--422}, } Among the arguments given in favour of 0^0 = 1 are (1) that we might like the binomial expansions to hold in general; (2) that there is precisely one function from the empty set to itself.
I know of at least one case where 0^0 needs to be undefined in order to get the right answer to a problem. In Mechanics of Solids, there are singularity functions where, <x-a>^n = 0 , x < a = (x-a)^n , x >= a if n > 0. n = 0 gives the Heaviside function, n = -1 is the Dirac deltaand n = -2 is the unit doublet which is the derivative of the delta function. There's also integration rules, but they're unimportant for this discussion.
So, 0^0 amounts to defining that the Heaviside function = 1 at x = a. I prefer to think of it as undefined and define it strictly in terms of left and right
limits. Cheers, Tim. --- Tim Lahey PhD Candidate, Systems Design Engineering University of Waterloo http://www.linkedin.com/in/timlahey
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