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Sep
22
awarded  Enlightened
Sep
22
awarded  Nice Answer
Sep
17
comment electron configuration on manifolds
Nope, it's generally far from unique, even on spheres. The first case seems to be 16 points on $S^2$, for which there are (at least) two local optima. As the number of points grows, the number of local minima seems to increase exponentially, but no proof is known. In higher dimensions there are cases with arbitrarily large numbers of non-isometric global minima (the configurations in the last line of Table 1 in arxiv.org/abs/math/0607446).
Aug
27
awarded  Nice Answer
Aug
24
awarded  Good Answer
Jul
3
awarded  Good Answer
Jun
28
awarded  Nice Answer
Jun
27
comment How to explain the concentration-of-measure phenomenon intuitively?
That's a good way of putting it. Making one component small forces the others to be large in aggregate, but when there are many of them this is still compatible with each one being small individually.
Jun
27
answered How to explain the concentration-of-measure phenomenon intuitively?
May
15
awarded  Enlightened
May
15
awarded  Nice Answer
Apr
15
awarded  Nice Answer
Apr
14
comment Question on the irrationality of $e$
Incidentally, $\int_0^1 (1-x)^k e^x dx = k!(e - \sum_{0 \le i \le k} 1/i!)$, so if you use $(1-x)^k$ instead of $x^k$ you get the usual approximation to $e$ through Taylor series.
Apr
4
comment Open problems in continued fractions theory
What's the question here? The title and answers look like you are after a list of open problems or conjectures on continued fractions, but the body of the question focuses 100% on one conjecture. If you want a list, it would be clearer to ask for that in the body of the question and move the current content to an answer.
Apr
1
comment Existence of a “quasi-uniform” probablility distribution on $\mathbb{Z}$
Oops, I was being silly (and had somehow convinced myself this wasn't enough to get full independence for more than two).
Apr
1
comment Existence of a “quasi-uniform” probablility distribution on $\mathbb{Z}$
How do you get independence for different primes? They are certainly pairwise independent, but I don't see how to deduce mutual independence for more than two primes.
Mar
18
awarded  Yearling
Mar
9
comment Factorization when a factor is partially known
(I edited the answer to try to clarify this, since the original version did make it sound like the assumption of equal-sized factors might be essential.)
Mar
9
revised Factorization when a factor is partially known
added 148 characters in body
Mar
9
comment Factorization when a factor is partially known
Knowing the first 75 digits of $a$ is essentially the same as knowing them for $b$ (since you know all the digits of the product $ab$ and can do approximate division), so you can still apply Coppersmith's algorithm to $b$. It's a little less efficient if you don't know how big the numbers are: the setup I have in mind requires knowing the number of digits, but you can brute force this since there are only 125 possibilities.