436 karma · joined March 2, 2016
What company did Tao fund with VC money?
Functions define distributions, but not all distributions are defined that way, like the Dirac delta or integration over a subset.
However: Suppose that for every n > 0 there exists a holomorphic function f_n such that |f_n(z) - z| < 1/n for all z. Then |f_n(z)| <= |f_n(z) - z| + |z*| = |z| + 1/n by the triangle inequality. A consequence of Liouville's theorem is that any entire holomorphic function with polynomial growth is a polynomial; here in particular we would need to have f_n(z) = a_n z + b_n for some complex numbers a_n and b_n. For real x we would have |(a_n - 1)x + b_n| < 1/n for all x, so a_n = 1. For imaginary iy we would have |(a_n + 1)iy + b_n| < 1/n for all y, so a_n = -1, which is a contradiction.
In fact, if a sequence of holomorphic functions converges uniformely on compact sets, the limit is itself holomorphic because of Cauchy's theorem.
I dunno, man. I have a PhD in complex differential geometry and think math is pretty hard.
So a real number is a function from pairs of rationals to a two-element set (plus some sanity conditions)? Why is that better than the other constructions?
(d bar) x = 1 / 2 pi.
Maybe it's time we lay some of the blame for us idiots just not getting Alan Kay's ideas on Alan Kay. At this point he only has himself to blame if he's spent 50 years trying and failing to communicate his wonderful ideas.
For any rational number p/q (with q > 0) there exists a smallest positive natural number k such that k * p/q is natural: Clearly q works, so we check the finitely many numbers 1, ..., q and pick the smallest.
Suppose the square root of 2 is rational. Let k be this smallest number for \sqrt 2, and proceed with the rest of the proof to find 0 < k' < k that also works.
Not one of them being "edit your writing".
At least the conquistadors had to live there in the interim years.