8 ms·
But if you define 0^0=1 in general, it doesn't cause a problem here -- that definition never disagrees with x^y=exp(ylog(x)), it just defines it at the point 0^
by Sniffnoy 13y ago
But if you define 0^0=1 in general, it doesn't cause a problem here -- that definition never disagrees with x^y=exp(ylog(x)), it just defines it at the point 0^0, while the latter leaves it undefined. In other words, it's possible to make a common extension of the two; they don't actually give different values in any case.
Of course, doing this makes exponentiation discontinuous at (0,0), but seeing as it already had an essential singularity there, this isn't really a loss.
- stephencanon 13y agoRight, but when you're working with the function on C x C, it doesn't really add anything either to arbitrarily pick 0^0 = 1, except for consistency with that other function that happens to be written the same way. I'm not really opposed to saying 0^0 = 1; it's the only reasonable choice if we're going to insist on using the same notation for these two functions, and I don't expect that's going to change.