bio | website | research.microsoft.com/~cohn |
---|---|---|
location | Cambridge, MA | |
age | 41 | |
visits | member for | 5 years, 5 months |
seen | 3 hours ago | |
stats | profile views | 4,403 |
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. |
Mar
9 |
answered | Factorization when a factor is partially known |
Jan
21 |
comment |
algebraic topology and 3d/4d printing
Thanks for the link! My first thought was that it was the analogue of 3d printing for people who live in $\mathbb{R}^4$. |
Jan
21 |
comment |
algebraic topology and 3d/4d printing
The title refers to "printing" (what's 4d printing?), but the body of the question does not. What are you looking for? I don't think the question can be given a useful answer without more details. |