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Data Scientist @ Explorer Media


May
9
comment Gradient and Swiftest Ascent
This result is not very intuitive, is it?
May
9
comment Product of Elements in SU(2)
$V$ is an arbitrary $2 \times 2$ matrix spanned in the basis of Pauli matrices? Where did you see this formula?
May
9
reviewed Approve suggested edit on Analytical approach to a quadratics problem
May
9
comment Number of Fixed Points in a Map from the Torus to itself using Lefschetz Trace
Here $f$ is my linear map $\left( \begin{array}{cc} 3 & -1 \\ 1 & 3 \end{array}\right)$, $V = \mathbb{R}^2$, $\Gamma = \mathbb{Z}^2$. In this case, $f$ extends to an action on the wedge products $\Lambda^0(\mathbb{R}^2),\Lambda^1(\mathbb{R}^2),\Lambda^2(\mathbb{R}^2)$, globally.
May
9
revised Number of Fixed Points in a Map from the Torus to itself using Lefschetz Trace
added 56 characters in body
May
9
asked Number of Fixed Points in a Map from the Torus to itself using Lefschetz Trace
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 $|z_{1}- z_{2}| \leq |w_{1}- w_{2}| \implies |c_{1}z_{1}- c_{2}z_{2}| \leq |c_{1}w_{1}- c_{2} w_{2}|$?
$c_2 = 0.0000000001$