109,463 reputation
698199
bio website von-eitzen.de/math/tntrep.xml
location Bonn, Germany
age 48
visits member for 1 year, 11 months
seen yesterday

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


Aug
21
answered The number of divisors of any positive number $n$ is $\le 2\sqrt{n}$
Aug
21
revised What's time complexity of algorithm for “Word Break”?
added 204 characters in body
Aug
21
answered What's time complexity of algorithm for “Word Break”?
Aug
21
answered Is it true that a field is a vector space over a field?
Aug
21
comment Is infinite product of groups also a group?
A product of groups $G_i$, $i\in I$, is a group $G$ together with homomorphisms $\pi_i\colon G\to G_i$ such that for any group $H$ and homomorphisms $f_i\colon H\to G_i$ there exists a unique homomorphism $h\colon H\to G$ such that $f_i=\pi_i\circ h$ for all $i\in I$. Then: Yes, products exist in the category of groups. See mesel's answer for a reference, Vera's answer for an explicit construction.
Aug
21
comment Euler's Formula, Square root
What formula do you not know how to write down? The one you just wrote down?
Aug
21
comment Is this polynomial equation solvable? $ \alpha x^{n+2} + \beta x^{n+1} + \gamma x^3 + \delta x^2 + \epsilon x + \zeta = 0 $
For large $n$, $x_0=-\beta/\alpha$ is a good starting point for numerical methods.
Aug
21
comment The process of solving the inequality $\frac{8}{19} x\ge -1$
If you follow the next steps, you should see that by choosing $\frac{19}8$, the coefficient $\frac 8{19}$ gets happily cancelled. This would not happen with $\frac8{19}$ or any other multiplier.
Aug
21
answered For a cubic equation, prove that two critical points of the same sign imply one root
Aug
21
comment Set of points of $[0,1)$ that have a unique binary expansion
Assume a number $x$ has two distinct expansions, first differing at the $n$th place. Then my multiplying with $2^n$ and subtracting an integer, you get two representations $0.a_1a_2\ldots$ and $1.b_1b_2\ldots$ of the same number. The latter is $\ge 1$, the former is $\le 1$, hence both must equal $1$, so they must be $0.1111\ldots$ and $1.000\ldots$. Thus one of the original representations for $x$ ends in all zeroes, i.e. is finite. But a finite 0-1-string (ending in $1$) can be viewed as a reversed representation of an integer, hence there are only countably many
Aug
21
answered Estimate bias of a coin
Aug
21
comment Do journals that published a proof of an important theorem $T$ publish another proof of $T$?
It's less about $T$ than about $P'$, you might say: A substantially new proof may use different (new?) methods, from which we can possibly learn more than from $T$ itself. Also, $P'$ may be so differnt that it may suggest different generalizations of $T$. You might even publish a new(!) proof of Pythagoras, adding to an already long collection.
Aug
21
comment Is there a general way to parameterize all implicit functions?
Depends on what you precisely mean by parametrizing. What about disconnected sets like $xy=1$?
Aug
21
answered Path Connectedness argument for $SO(n, \mathbb{R})$
Aug
21
answered What does linearly equivalent mean in this context
Aug
20
comment If $\operatorname{rank}(A)=n$ then $\operatorname{rank}(AB)=\operatorname{rank}(B)$
Well, if $n$ occurs in a matrix problem without further mention, I thought it safe to assume that $n=\dim V$ ...
Aug
20
revised A continuous bijection from a Hausdorff space to a non-compact space which is not a homeomorphism
added 353 characters in body
Aug
20
answered A continuous bijection from a Hausdorff space to a non-compact space which is not a homeomorphism
Aug
20
comment A continuous bijection from a Hausdorff space to a non-compact space which is not a homeomorphism
$\mathbb R$ with discrete topology?
Aug
20
answered If $\operatorname{rank}(A)=n$ then $\operatorname{rank}(AB)=\operatorname{rank}(B)$