A Quirky Function

Re:A Quirky Function by fisherjeff on Saturday June 25, 2016 @09:27AM (#11976232)

> And it equals its reciprocal, because 0^0 = 0^-0 = 1/0^0

Doesn't this already assume that 0^0 evaluates to 1 and not 0?

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Re:A Quirky Function by qf303rjr3 on Saturday June 25, 2016 @09:34AM (#11976262)

No - you have the following identity because 0 = -0

  0^0 = 0^(-0)

and by the definition of negative exponents,

  0^(-0) = 1/(0^0)

So whatever number 0^0 is, it is equal to its own reciprocal.

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Re:A Quirky Function by fisherjeff on Saturday June 25, 2016 @09:40AM (#11976278)

Ah, ok. I didn't have an issue with the first half, I just misunderstood: I thought he was stating that 0^0 == 1/0^0, not deriving it.

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