0^0 = 1?

[Claudio L.]

(04-23-2017 02:58 PM)Thomas Okken Wrote:  So, lim (r->0) (r sin(φ))^(r cos(φ)) = 1 for φ ∈ (0, π) ∪ (π, 2π) and undefined for φ = 0 and φ = π.

Make of that what you will. :-)

Ahhh... I promised myself not to get involved in this one, but since nobody mentioned this... here it goes:

There's one missing piece of the puzzle here.
when 0 < r cos(φ) < 1, you could express that number r cos(φ) as a fraction: m/n (I know, not always but bear with me).

Now a^(m/n) has infinite solutions spinning all around the complex plane. When r->0 this fraction will have a huge denominator 'n', which means a huge number of different solutions very closely spaced. So when the limit goes to zero, turns out the solution is not (real) 1, but the entire (complex) unit circle, with infinite solutions of magnitude 1 in all directions.
So we know the magnitude is 1, but the argument in the complex plane is undefined, so for newRPL I chose 0^0=1 in real mode, and NaN in complex mode.

Does it make sense?