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seen Oct 8 at 8:44

May
31
comment Simple non-closed geodesic.
I don't understand why you accepted Neal's answer. As Daniel Rust commented, it didn't answer your question, because it didn't show whether the non-closed geodedics could be simple or not.
May
31
comment Simple non-closed geodesic.
Could you please be more specific to explain how Hedlund's paper implies that there exists a simple dense geodesic?
May
29
revised Find all polynomials $P(x)$ satisfying this functional equation
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May
29
answered Find all polynomials $P(x)$ satisfying this functional equation
May
26
comment How can I show whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges?
Denote $a_n=\frac{(-1)^n}{n(2+(-1)^n)}$. Note that $a_{2n-1}+a_{2n}=-\frac{1}{2n-1}+\frac{1}{6n}<-\frac{1}{3n}$.
May
25
comment Sequences with the following properties…
@Chung.J: Thank you!
May
25
answered Sequences with the following properties…
May
24
awarded  Good Answer
May
22
revised $(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$
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May
22
answered $(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$
May
21
revised What is my operator norm (cannot get good enough bounds).
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May
21
comment What is my operator norm (cannot get good enough bounds).
@Norbert: I did use it. I just want to emphasize that since $B$ is compact, when can say more than $\|A\|^2=\|B\|$.
May
20
answered What is my operator norm (cannot get good enough bounds).
May
20
comment Let$ f : [0, 1]^2 \to R$ such that $f(x, y)$ is continuous in $x$ for each fixed $y$ and conversely also. Is $f $ continuous?
You may consider a function like $f(x,y)=\frac{xy}{x^2+y^2}$ when $(x,y)\ne (0,0)$ and $f(0,0)=0$.
May
19
comment A slight variation on the Pythagorean theorem
Hint: note that $(a-1)(a+1)=(c-b)(c+b)$ and $\gcd(a-1,a+1)=1$ or $2$.
May
19
comment Continuous function differentiable on $[0,1]\setminus\mathbb{Q}$, but nondifferentiable on all of $\mathbb{Q}\cap[0,1]$?
Perhaps "by use of mean value theorem" can be understood in this way: since $f$ is strictly increasing, for any $a<b$, there exists $c\in (f(a^+),f(b^-))$, such that $F(b)-F(a)=c(b-a)$.
May
19
comment Continuous function differentiable on $[0,1]\setminus\mathbb{Q}$, but nondifferentiable on all of $\mathbb{Q}\cap[0,1]$?
Related question.
May
19
answered If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$
May
19
comment Topological manifolds (dimension)
You are right. This is called invariance of domain.
May
19
comment Which real functions have their higher derivatives tending pointwise to zero?
Some uniformly convergent example: $f(x)=\sin(a x)$, where $0<a<1$.