106,359 reputation
696194
bio website von-eitzen.de/math/tntrep.xml
location Bonn, Germany
age 48
visits member for 1 year, 11 months
seen 6 hours ago

I did study math and had a knack for it, but I am sooo out of that business now ...


Sep
13
answered Nonexistence of Limit of Sum of Prime Factors
Sep
13
comment On a Putnam's 2009 problem
@user2566092 Indeed, $n=4$ is a counterexample instead: Looking only at $\mathbb Z^2$ and squares with side lengths $1$ or $\sqrt 2$, there are 8 solutions $f(x,y)=ax+by+c+2\mathbb Z$ and 8 solutions $f(x,y)=\lfloor\frac{x+a}2\rfloor+\lfloor\frac{y+b}2\rfloor+c+2\mathbb Z$; observing that $f$ is uniquely determined by $f(0,0)$, $f(0,1)$, $f(1,0)$, $f(2,0)$, these are all solutions. Then $f$ restricted to $(4\mathbb Z)^2$ is constant. Back to the case $\mathbb R^2\to\mathbb Z_2$, any two points can be viewed as belonging to a translated/rotated/scaled $(4\mathbb Z)^2$, so $f$ is constant.
Sep
13
answered Open set of $\Bbb R$ which is not bounded below can be written as atmost countable collection of disjoint segments
Sep
13
awarded  Enlightened
Sep
13
awarded  Nice Answer
Sep
13
revised Show $g(x)=\sqrt{x}$ is continuous at x=4
added 35 characters in body
Sep
13
comment Show $g(x)=\sqrt{x}$ is continuous at x=4
@Crypto More generally, to show continuity of $x\mapsto\sqrt x$ at $x_0>0$, you can pick $\delta=\min\{\sqrt{x_0}\epsilon, x_0\}$. We shold ensure $\delta<x_0$ so that $\sqrt x$ is defined in the first place. Then again, the definition of continuity asks only for $\delta>0$ such that for all $x$ with $|x-x_0|<\delta$ for which $f(x)$ is defined - so this restriction is not really needed, you can take $\delta=\sqrt{x_0}\epsilon$.
Sep
13
answered Improper integral properies
Sep
13
comment $E=\{x:f(x)=0\}$ is measurable
Assume there exists a nonmeasurable subset $E\subset X$. Let $f(x)=\begin{cases}0&\text{if }x\in E,\\1&\text{if }x\notin E\end{cases}$. Then $E_1=E$ is not measurable.
Sep
13
comment Let $a,b$ be in a group $G$. Show $(ab)^n=a^nb^n$ $\forall n\in\mathbb{Z}$ if and only if $ab=ba$.
If the equality holds for all $n\in\mathbb Z$, it holds specifically for $n=2$. What does that imply? - And if $ab=ba$ then the subgroup generated by $a,b$ is abelian.
Sep
13
answered Show $g(x)=\sqrt{x}$ is continuous at x=4
Sep
13
answered Proving basic limit laws without finding $\delta$s.
Sep
13
answered Solution to $x^2+x-1\equiv 0$ mod $p$
Sep
12
answered Let $A$ be an $m\times n$ matrix. If $\forall \vec{x}\in\mathbb{R}^{n},\ A\vec{x}=\vec{0}$, show that $A$ is the zero matrix.
Sep
12
comment Infinite composition of automorphisms.
The answerd given so far give examples where the limit can hardly be defined or where the limit (according to pointwise convergence) exists but fails to be an automorphism. Are there also examples where the limit exists, but is not even a homomorphism?
Sep
12
answered How can I compute a similarity score between two documents with the provided algorithm?
Sep
12
comment An inconstructible quadrilateral
Why does this triangle have four vertices?
Sep
11
comment Is the following equivalent to the axiom of choice?
@AsafKaragila Well, not really. If we $|\emptyset|\le^*|B|$ is false by the simplified definition, then the case $A=\emptyset$ is trivially true, and that does not hurt. All this $\le, \le^*$ is just fancy writing for "If there exists a surjection $B\to A$, there exists an injection $A\to B$".
Sep
11
answered Roll five dice. What's the chance of rolling exactly one pair?
Sep
11
comment Computing sums of divisors in $O(\sqrt n)$ time?
@sudeepdino008 At least we can verify that (if $n$ is not a square) $p(n)-p(n-1)=2\sum (\lfloor\frac nk\rfloor - \lfloor \frac {n-1}k\rfloor) = 2\sum_{d|n, d^2<n}1$ as required. And if $n=m^2$ is a square, the additional summand $2m$ is correctly compensated by $m^2-(m-1)^2=2m-1$.