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618
bio website mrcactu5.herokuapp.com/…
location New York, NY
age 29
visits member for 3 years, 8 months
seen 15 hours ago

Data Scientist @ Explorer Media


May
9
awarded  Revival
May
9
awarded  Mortarboard
May
9
answered Trying to prove that there are no p and q such that $|\sqrt5 - p/q| < 1/(7q^2)$.
May
9
revised Motivation for the study of amoebas.
citation
May
9
revised Motivation for the study of amoebas.
added 293 characters in body
May
9
answered Motivation for the study of amoebas.
May
9
reviewed Approve suggested edit on polynomial with integer coefficients divided by $x^3 -x$
May
9
asked block matrix multiplication
May
9
comment How prove $\sum_{1\le i<j\le n}(a_{j}-a_{i}+1)^2+4\sum_{i=1}^{n}a^2_{i}\le \frac{5n^2+6n+4}{4}$
If I set $a_1 = \dots a_n = 0$ the left side is $\tfrac{1}{2}n^2$.
May
9
comment How prove $\sum_{1\le i<j\le n}(a_{j}-a_{i}+1)^2+4\sum_{i=1}^{n}a^2_{i}\le \frac{5n^2+6n+4}{4}$
trivially, this is bounded by $2n^2$. You are trying to improve to slightly more than $\tfrac{5}{4}n^2$ using correlations in the $a_1, \dots, a_n$.
May
9
reviewed Reject suggested edit on partitioning numbers from 1 to n in 4 non-empty subsets so no subset has 2 consecutive numbers.
May
8
asked Oscillatory integral and Van der Corput
May
8
comment $|z_{1}- z_{2}| \leq |w_{1}- w_{2}| \implies |c_{1}z_{1}- c_{2}z_{2}| \leq |c_{1}w_{1}- c_{2} w_{2}|$?
How about $c_2 = 0.125$ ?
May
8
comment $|z_{1}- z_{2}| \leq |w_{1}- w_{2}| \implies |c_{1}z_{1}- c_{2}z_{2}| \leq |c_{1}w_{1}- c_{2} w_{2}|$?
$c_2 = 0.0000000001$
May
8
answered $|z_{1}- z_{2}| \leq |w_{1}- w_{2}| \implies |c_{1}z_{1}- c_{2}z_{2}| \leq |c_{1}w_{1}- c_{2} w_{2}|$?
May
7
revised Proof of “Japanese Theorem” — Triangulation of Cyclic Polygon
added 7 characters in body
May
5
comment How prove $|S-10^k\cdot AB|\le 9k$
@CalvinLin It looked interesting. I have attempted to salvage the problem. Please see if my edit is correct.
May
5
revised How prove $|S-10^k\cdot AB|\le 9k$
added 29 characters in body; edited tags
May
5
comment Proof of “Japanese Theorem” — Triangulation of Cyclic Polygon
@MaMing Ok, so insides cancel? And we get $(\# \text{sides} ) \times (\text{circumradius}) + \sum_\Delta (\text{inradus} ) = \sum \overline{OA}_i$. We still have to prove Carnot's result. Maybe using Barycentric coordinates?
May
5
revised Proof of “Japanese Theorem” — Triangulation of Cyclic Polygon
added 19 characters in body