Ok, I kinda can guess what that set looks like. Still there seemed to be no way to generate all the elements
All liuoville numbers are very close to the rationals of arbitrary high denominator, which means they lie very close to the rationals where the band of most irrationals are clustered at
The numbers are located in a set such that the intervals as shown, being functions of $q^n$ simply cannot shrink fast enough to exclude them
So $\frac{p}{q}$ determines which rational we are computing the irrational measure on and the power $n$ controls the shrinking of the intervals. The reason factorial works is because it grows faster than any polynomial function, thus the shrinkage of the interval $(0,\frac{1}{q^n})$ will never caught up as $n \to \infty$
@Semiclassic @Abcd: Just look at the definition of the adjugate. The $ij$ cofactor of $A$ is the $ji$ cofactor of $A^\top$. Remember that transposing doesn't change determinant.
@Manolis: If you're planning on graduate work, probably an F on your record is not good. (Certainly that is the case in the US.) If retaking a course replaces (or at least averages) the grades, an A on your transcript and a C averaged into your grade average is far better than an F. Again, the folks from Europe might have different thoughts.
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this way:
To...
A Liouville number is a number which can be approximated very closely be a sequence of rational numbers (here is the rigorous definition I am working off of: http://en.wikipedia.org/wiki/Liouville_number).
I'm looking for an example of a Liouville number which cannot be approximated by a sequenc...
What's the motivation behind using linear algebra between these three problems ?
A pair $(m,n)$ is called nice if there is a directed graph with (self edge are allowed, but multiple edge are not allowed) $n$ vertices such that for every pair of vertices is connected by exactly $m$ paths of len...
so that means, even if we restrict to truncations in order to plot them on a computer, there will be so many of them that they cannot be plotted to any representative amount since there is no way to enumerate all such functions $f_j$
@LeakyNun Freiburg is famous for being a green (organic, left wing, hippie-esk) young city. Car drivers might feel bullied at times. @ÍgjøgnumMeg precis
All liouville numbers of the form $0.a_1a_2000a_300...$ (can only plot to $a_3$ as the inherent limitation of the map() function means it failed to map different $a_4$ to different pixels
and one can easily imagine this tree continues down countably many levels for each line with the exact same pattern repeated throughout, at zooms of $10^{n!}$
This is how just this subset of it is uncountable
Plot of the above liouville numbers along with the rationals at 100 denominators. Note how they fit nicely in between the gaps, thus ilustrating its irrational nature
Reasking a question I posed earlier, for which kind of measure spaces $(X,\mathcal A,\mu)$ are the $L^p(X)$ spaces finite dimensional? In particular for $p=\infty$
cosets show up sometimes, depending on what you do (more in group theory than in ring theory from my experience), but you don't need them for the most part
my intuition about quotients is not based on cosets
@LeakyNun wenn $L$ der Zefällungskörper von $f$ über $K$ ist, dann gilt für jeden algebraischen Abschluss $\overline{L}\supset L$ und jeden $K$-Algebrenhomomorphismus $\sigma$ von $\overline{L}$: $\sigma(L)=L$. Beweis dafür: sei $L=K(\alpha_1, \dots, \alpha_n)$, wobei $\alpha_i$ die Nullstellen von $f$ sind, dann permutiert $\sigma$ die Nullstellen von $f$ sind, also gilt $\sigma(L)=\sigma((K(\alpha_1, \dots, \alpha_n))=K(\sigma(\alpha_1), \dots, \sigma(\alpha_n))=K(\alpha_1, \dots, \alpha_n)=L$
@Leaky yeah, I'm doing the class this fall and I should learn stuff like fundamental group and covering spaces properly. Now I can kinda mumble about fibrations and the LES but haven't done too many problems on the stuff
sei jetzt $g \in K[x]$ irreduzibel, sodass $g$ eine Nullstelle $\beta$ in $L$ hat. Sei $\beta'$ eine weiter Nullstelle. Dann gibt es einen $K$-Algebrenisomorphismus $\sigma$ $K(\beta)\cong K(\beta')$ mit $\sigma(\beta)=\beta'$ und der setzt sich zu einem $K$-Algebrenautomorphismus von $\overline{L}$ fort, daher gilt $\beta' \in K(\beta')=\sigma(K(\beta)) \subset \sigma(L) =L$
lol, the translation for "emertieren" is apparently "to give emeritus status to someone", this exemplifies how English is shorter and simpler than German
$a$ is algebraic iff $K(a)/K$ is algebraic, so it suffices to define algebraicity for extensions. Let's go for an obscure one: $L/K$ is algebraic iff every intermediate ring $R$ with $L \supset R \supset K$ is a field
@MatheinBoulomenos To make it shorter, just say give someone emeritus status. Also that day I saw you discussing a proof that a disjoint union of open intervals in R cannot have uncountably many intervals. I think a quick way to see it would be to note that there is a rational in each open interval.
@LeakyNun or how about this as you like simple and down-to-earth definitions: $a$ is algebraic iff the orbit of $a$ under the action of the $K$-algebra-endomorphism monoid of $K[a]$ is finite (fun fact: this doesn't work if you use automorphisms instead of endomorphisms)
I picked the latter because I wanted a proof of Runge's Theorem, Bloch's Theorem, Schottky's Theorem, and the Cauchy-Green Formula which are not found in the former.
However, Mergelyan's Theorem is found in only very few books, such as Rudin's Real and Complex Analysis or Krantz and Greene's Function Theory.
@LeakyNun I know I told you this before, but I'm still amazed that you can define "algebraic and inseparable extension" as "epimorphism in the category of fields"
the reason everything works: if $f \in F[X]$ is separable and splits in $K$ then $\operatorname{Hom}_F(F(\alpha),K) = \operatorname{Hom}_F(F[X]/(f),K) = \{ \beta \in K \mid f(\beta) = 0 \} = \partial f$
does anyone know if all real polynomials can be written as the sum of periodic functions? https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/
the author of the blog links to a proof, but I couldn't understand it, and I haven't been able to find this result elsewhere
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this way:
To...
What's the motivation behind using linear algebra between these three problems ?
A pair $(m,n)$ is called nice if there is a directed graph with (self edge are allowed, but multiple edge are not allowed) $n$ vertices such that for every pair of vertices is connected by exactly $m$ paths of len...
