4,812 reputation
11541
bio website
location
age
visits member for 2 years, 11 months
seen 18 hours ago

An expert is a man who has made all the mistakes, which can be made, in a very narrow field. (Niels Bohr)


Sep
11
reviewed Reviewed Printing long lines in MAGMA without line break
Sep
9
reviewed Reject suggested edit on Are matrix p-norms unitary invariant?
Sep
8
reviewed Approve suggested edit on Simplifying expression and finding indefinite integral
Sep
6
reviewed Looks OK Showing if $n \ge 2c\log(c)$ then $n\ge c\log(n)$
Sep
5
reviewed Approve suggested edit on gaussian and mean curvatures
Sep
5
reviewed Reviewed equation I don't understand
Sep
5
reviewed Looks OK What is the smallest $d$ such that $4$ has more than one distinct factorization in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$?
Aug
31
answered Finding roots with seemingly no algebraic way
Aug
31
reviewed Leave Open Find the sum, if exists $\sum\limits_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2(n+1)}$
Aug
31
reviewed Reviewed How do I find a function given the conditions as follows :
Aug
28
reviewed Close Evaluate $\lim_{x\to0}\left(x^2\left(1+2+3+\ldots+\left[\frac{1}{|x|}\right]\right)\right)$
Aug
28
reviewed Approve suggested edit on 2000 Olympiad in Informatics Question on Box
Aug
28
reviewed Close Proof of absolute convergence
Aug
28
reviewed Reject suggested edit on prove that $T$-cyclic subspace of $V$ generated by $x$ is $T$-invariant
Aug
28
revised Fréchet normal cone
edited title
Aug
28
revised If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous
edited body
Aug
28
revised If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous
edited body
Aug
27
reviewed Approve suggested edit on Is $|1-i|$ larger than $1$?
Aug
27
reviewed Approve suggested edit on Angle between two planes in four dimensions
Aug
27
reviewed Approve suggested edit on Evaluate trig functions without a calculator