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visits member for 2 years, 7 months
seen Dec 12 at 6:00

Oct
28
answered Trace Minimization of Covariance Matrix
Apr
27
awarded  Yearling
Apr
15
awarded  Revival
Feb
6
accepted Number-Theoretic Coin Puzzle
Feb
5
revised Number-Theoretic Coin Puzzle
deleted 14 characters in body
Feb
5
comment Number-Theoretic Coin Puzzle
@MarcvanLeeuwen Done, thanks.
Feb
5
revised Number-Theoretic Coin Puzzle
deleted 14 characters in body
Feb
5
revised Number-Theoretic Coin Puzzle
edited tags
Feb
5
comment Number-Theoretic Coin Puzzle
@MarcvanLeeuwen Yes, I put that there to avoid consideration of the trivial case.
Feb
5
comment Number-Theoretic Coin Puzzle
@KonstantinosGaitanas Yes, you may move coins from any pile $x$ to any other pile $y$, but only such that the number of coins in $y$ is doubled.
Feb
5
asked Number-Theoretic Coin Puzzle
May
6
awarded  Caucus
Apr
27
awarded  Yearling
Mar
9
accepted Can the Basel problem be solved by Leibniz today?
Mar
9
comment Can the Basel problem be solved by Leibniz today?
This is a great answer! Quite $(\heartsuit)$ly.
Mar
9
awarded  Nice Question
Mar
9
comment Can the Basel problem be solved by Leibniz today?
Thanks, I am aware of the above relations. I found a reference that establishes that given one more ingredient, (1) is sufficient to infer (2) - as far as heuristic arguments go (see here). Also, this talk seems to indicate that Newton's theorem (which must have existed pre-Leibniz?) is sufficient to infer (2), again the rigor is forsaken.
Mar
9
comment Probability that n points on a circle are in one semicircle
Why is $ \lim_{k\to\infty}k2^{-n}\left(1-\left(1-\frac2k\right)^n\right)=\lim_{k\to\infty‌​}k2^{-n}\left(\frac{2n}k\right)$ true?
Mar
9
revised Probability that n points on a circle are in one semicircle
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Mar
9
comment Probability that n points on a circle are in one semicircle
There's a slight variation of this answer that uses the inherent symmetry: instead of picking $n$ points at random, pick $n$ random diameters of the circle and pick the $n$ points by randomly picking one of the 2 poles of each diameter. By essentially the same argument, you have the probability given by $(2n)/2^n=n/2^{n-1}$.