That's what it ought to be, but I don't see a direct approach. I don't want finitely many open sets to cover something, I want infinitely many open sets to intersect non-trivially. I'll think about this again tomorrow.
so $x/\log(x)$ is a good asymptotic formula as well
for the prime counting function, but $\text{Li(x)}$ is "the best"
possible asymptotic formula apparently
I'm wondering if there's any use in constructing an asymptotic, analytic function, that outperforms $x/\log(x)$ but is not quite as accurate as $\text{Li(x)}$
in terms of an approximation of $\pi(x)$ (prime counting function)
The offset logarithmic integral is defined as $$ \text{Li}(x)=\int_2^x\frac{1}{\log(t)}~dt. $$
It can be shown that $\text{Li(x)}\sim\pi(x)$ where $\pi(x)$ is the prime counting function. It can also be shown that $\frac{x}{\log(x)}\sim \pi(x).$
Is there any practical use in number theory f...
Goal
Write a program in C/C++ to find the smallest prime \$k\$ such that for all \$n\in \Bbb{N}\$, \$n!+k\$ is never prime. You can use any number-theoretical methods to achieve this but must explain in-depth to why it works.
Basic Concept
It is elementary to see that for any \$n\ge k\$, we ha...
@robjohn I understand what you were saying before. You mentioned that taylor series and now I understand it just adds and gets binomial theorem thus is not superfluous 😂😂😂
I was busy learning other than real analysis so my last two problem remains.
I am proving $f(x)=\sum_{n=0}^\infty (a\choose n)x^n$ for real $a$. It wants me to prove that $(1+x)f'(x)=a f(x)$.
$f'(x)=\sum (n+1)(a\choose {n+1})x^n$ was it's derivative.
After then I gave up.
And look answer since I am stupid.
And book has this answer: $n((a(a-1)...(a-n+1))\over n!)+(n+1)((a(a-1)...(a-n+))\over (n+1)!=((a(a-1)...(a-n+1))\over (n-1)!(1+(a-n)\over n)$
and it directly says: so $(1+x)f'(x)=a f(x)$
And I don't know how it is related
So can somebody with big brain install the explanation on my brain
How can that information be used to conclude $(1+x)f'(x)=a f(x)$ ?
What ways can I relate an $R$-module $M$ to ideals of $R$? For instance, if $M$ is of finite length then supposedly $M$ is finitely generated and $\operatorname{Ann}_R(M) \subset R$ is non-zero. I get that $M$ is finitely generated (finite length $\iff$ $M$ noetherian and artinian, submodules of noetherian modules are f.g. and $M$ is a submodule of itself), but why does $M$ have non-zero annihilator?
I guess I should also mention that $R$ is a PID lol
Something like this: suppose nothing annihilates M. let a be a nonzero element in R, if aM is a strict submodule of M then you get an unbounded chain M > aM > a^2M > a^3M > ... contradicting finite length (a^n is never zero because R is a domain)
what is unclear about what i said? you have identified f'(x) as a power series, (1+x)f'(x) is also a power series. so for any natural number n, x^n has some coefficient. write it down
"Mochizuki developed inter-universal Teichmüller theory which, due to its nature and applications, has attracted a high level of attention of non-mathematicians"
Does anyone know why it has attracted a high level of attention of non-mathematicians?
sanity check: Whats the negation of there exists infinitely many ideals $I_1,I_2..$ of $R$ such that $I_{n+1}$ is a proper subset of $I_n$ for all $n$?
I have a compact topological group $G$ acting on a metric space $X$. Is it true that $\dim X\leq \dim(X/G)+\dim G$? Or equivalently what is the relationship between $\dim G$ and the dimension of the orbits? ($\dim$ always refer to the Lebesgue covering dimension)
Or is there some combination of adjectives in front of $G$, $X$, or the action making this true?
Yeah, for any $C^\infty$-function $X : \Bbb R^n \to \Bbb R^n$ there exists $\varepsilon, \delta > 0$ such that there exists a $C^\infty$ function $F : B_\delta(0) \times (-\varepsilon, \varepsilon) \to \Bbb R^n$ such that $dF(x, t)/dt = X(F(x, t))$ for all $x \in B_\delta(0), -\varepsilon < t < \varepsilon$ and $F(x, 0) = x$.
Hm, the Hausdorff dim is kinda weird in that it is not a topological invariant, so it can disagree with the covering dimension. Still interesting though
If $G$ is a compact Lie group acting on a smooth manifold $M/G$ is a stratified space. The top stratum is of dimension $\dim(M) - \dim(G)$
So that case is particularly well behaved
Well
I guess the top stratum can be of lesser dimension, I am imagining the case where non-identity elements of $G$ does not fix every element of $M$. That is to say $\bigcap_{x \in M} \text{Stab}(x) = \{e\}$
Anyway the point is stratified spaces are triangulable and for simplicial complexes Lebesgue covering dimension is the affine dimension of the complex.
ok, I guess I should look at $C^{\infty}(B_{\delta}(0)\times(-\varepsilon,\varepsilon),\mathbb{R}^n)\rightarrow C^{\infty}(B_{\delta}(0)\times(-\varepsilon,\varepsilon),\mathbb{R}^n),\,F\mapsto((x,t)\mapsto x+\int_0^tX(F(x,t))\mathrm{d}t)$ and pray it's a contraction for small $\delta,\varepsilon$
@Alessandro Right, so there the quotient is $[0, \infty)$, obtained from quotienting the open submanifold of $\Bbb C$ on which the action of $S^1$ is free, namely, $\Bbb C \setminus \{0\}$ and quotienting the left out bit, $\{0\}$, on which the action of $S^1$ is the very opposite of free.
This picture generalizes as follows.
Look at conjugacy classes of subgroups of $G$. For any such conjugacy class $[H]$, define $M^{[H]} = \{x \in M : G_x \in [H]\}$.
These are called the "orbit types" in $M$. In the earlier examples, we had two orbit types, $\Bbb C^{[1]} = \Bbb C \setminus \{0\}$ and $\Bbb C^{[S^1]} = \{0\}$.
In general, these orbit types $M^{[H]}$ are submanifolds of $M$, and the decomposition $M = \bigcup_{[H]} M^{[H]}$ into the orbit types form a stratification of $M$, and $G$ acts on $M^{[H]}$ with stabilizers always conjugate to $H$ (by definition), so by slice theorem $M^{[H]}/G$ turns out to be smooth manifolds as well.
Then the decomposition $M/G = \bigcup_{[H]} M^{[H]}/G$ turns out to be a stratification of $M/G$
The point is, essentially, equivariant tubular neighborhood theorem - which says these $M^{[H]}$ always admit nice $G$-equivariant neighborhoods in the whole of $M$.
So when you quotient the whole thing by $G$, it descends to a set-theoretic stratification such that each manifold stratum has a nice "tubular neighborhoods" inside $M/G$.
This notion of a stratification with a "tubular neighborhood" happens to be equivalent to the definition you say, which I assume is Whitney stratification
@BalarkaSen This is not what you mean to say. For instance, take SO(3) acting on RP^2 --- then the stabilizers vary over the circle subgroups. You meant that the minimal dimension of stabilizer is 0
I think the reason nobody really cares about stratified spaces is because it quickly becomes very technical. The general idea is very clear but to do reasonable mathematics with it one has to work so hard
The limit I asked about above is so obvious, and just lies in the fact that exp grows much faster than any polynomial - but I somehow struggle to express this formally. L'Hospital also does not lead to anything useful.
Because essentially I want the top stratum of $M/G$ to be $M^{free}/G$ where $M^{free}$ is the open submanifold of $M$ on which the action of $G$ is free. Only then can I say the dimension of the stratified space $M/G$ is $\dim M - \dim G$, right?
@MikeMiller I am not sure I see why I want the minimal dimension of the isotropy group to be 0; that completely rules out most actions eg S^1 acting on S^3 by Hopf action
I also don't understand your example. SO(3) is a 3-dimensional group, RP^2 is 2-dimensional
I've a question, why $\mathbb{E}[X]$ and $\mathbb E_\rho$ written differently? I mean why some are written inside square brackets and some as subscripts..?
@abhas_RewCie I am used to seeing $E[Y]$. I don't think I have seen $E_Y$, but if it were stated in the article, I would understand and it would be correct.
Yeah your stabilizers of the SO(3)-action on RP^2 were different circles intersecting at the identity, that is clearly not enough. I was taking "intersection upto conjugation", which is really a convoluted way of demanding there to be a trivial stabilizer.
@abhas_RewCie Or better to say, I already tried. It leads to $\sum\limits_{k = 0}^\infty \frac{x^p}{x^{2k} k!}$ = $\sum\limits_{k = 0}^\infty \frac{x^{p-2k}}{k!}$
@StupidKid I would think that that might be covered in a Discrete Math course, or possibly an elementary number theory course. It all depends on the level of the course and the instructor.
I honestly forget how the notation for conditional expectation goes. $\Bbb E[X|Y]$ is the random variable which takes value $\Bbb E[X|Y = y] = \displaystyle \int x \;d\Bbb P(X = x|Y = y)$ with probability $\Bbb P(Y = y)$, right?
Ah, no, if $Y : \Omega \to \Bbb R$, I compose this with $\Bbb R \to \Bbb R$, $y \mapsto \Bbb E[X|Y = y]$, to get the RV $\Bbb E[X|Y]$ defined on the probability space $\Omega$.
Well, the goal is to understand the comparison between de Rham cohomology and étale cohomology. I am trying to understand whether the topological case could be easier.
The first cohomology group is somehow more obvious. Given a closed 1-form $\omega$, the indefinite integral $\int\omega$ corresponds to a covering space.
@Yai0Phah Can you point towards what is the comparison precisely? As far as I understand etale cohomology just means sheaf cohomology over the etale site of a scheme to me, so I am not sure what is the topological analogue that you look for.
I mean, de Rham cohomology can be viewed as the sheaf cohomology of the sheaf of forms on the site of open sets in a manifold, I imagine
The connection between $H^1_{dR}(M; \Bbb R)$ and covering spaces that you describe (given a closed 1-form, pass to a cover on which it is exact) seems specific to $n = 1$, and comes from topology, because $H^1(X; \Bbb Z) = \text{Hom}(\pi_1(X), \Bbb Z)$, which classifies regular finite-sheeted covering spaces over $X$, essentially.
@Yai0Phah Finite etale maps are local diffeomorphisms, and since in the topological category inclusion of open sets are already local diffeomorphisms, I imagine the etale site is not really different from the plain vanilla site of open sets - though I don't know how to formalize this
It does not seem to be possible to talk about arbitrary "open sets" in the étale site. It is coarser.
Anonymous
2:51 PM
Hello, could someone help me a bit with this. I'm trying to see why this theorem is true: If $G$ is a group with a normal subgroup $K$ such that $G/K$ is solvable, and $H$ is a nonabelian simple subgroup of $G$, then $H \leq K$. On the main site, they're telling me to consider the homomorphism $\pi: G \to G/K$ but I'm not sure what to conclude from there.
The only relationship that I know if is that there is an injection $H^2(\pi_1 X; \Bbb R) \hookrightarrow H^2(X;\Bbb R)$. And this is the only relationship, in that you can have $H^2(X;\Bbb R)$ be whatever you want given this relation.
This relation follows from the fact that you can construct $B\pi_1 X$ from $X$ by attaching cells of dimension 3 and higher.
@robjohn Well it is for the same obvious reasons $0$, but I have the same trouble showing that.... Writing the exponential and its series expansion and multiplying by $x^p$ does not seem to be helpful
Fundamental group classifies covering spaces, yes, but it loses the information of higher homotopies. That is to, you ignores the information how you identifies different covering spaces.
You identifies it by isomorphic classes, essentially taking $\pi_0$ of some space-like object.
I suppose formally it must be that every etale morphism X -> M of smooth manifolds can be written as a fibered product of "basic etale morphisms" U -> M where U is an open subset of M. So these guys should be a basis of the Grothendieck topology, whatever that means, which implies the etale site of a smooth manifold is completely useless
So I don't think there's an answer to your question in the topological world
Certainly not every etale morphism of schemes in general can be written as a fibered product of the inclusion of the Zariski open sets!
Let me describe the "philosophy". On one hand, it is about étale maps, covering spaces, etc. On the other hand, it is about differential equations (essentially, de Rham cohomology is about diff eqs)
If you try to write down "the space of covering spaces with fiber n points", like you would "the space of fiber bundles with fiber F", this space up to homotopy will be the space of unpointed maps $[M, BS_n]$. This splits up over connected components which do not carry information about higher homotopy, because the fibration sequence $\text{Map}_*(M, BS_n) \to \text{Map}(M, BS_n) \to BS_n$ is split, so splits into short exact sequences in higher htpy & the first term is discrete up to homotopy.
@BalarkaSen Yes, you add 3-cells to kill off $\pi_2$, which kills off the image of Hurewicz in second homology --- and because you never add 2-cells that second map is a surjection
(Then higher cells to kill off higher homotopy)
So $\pi_k \text{Map}(M, BS_n) = 0$ for $k \geq 2$.
Maybe you are correct. The space-like object could also be spectrum-like object, that is to say, the correct information is "encoded" in negative degrees. Now I am not at a very good state to reflect deeply on this.
The important thing is presumably that you break up over local open charts / local covering maps, instead of global. In the smooth case this recovers the homotopy type of manifolds, because the Cech nerve of a good open cover recovers the homotopy type of the manifold.
In the singular setting I'm sure there are interesting things happening from the local etale maps and how they fit together.
But this is more than just talking about covering spaces of M
How does one characterize a topology by specifying convergence of nets? E.g., very often authors will introduce the weak and strong operator topologies on $B(H)$ (for some Hilbert space $H$) by merely specifying convergence of nets. But I've never understood how this completely characterizes the topology.
Let $G$ be any group, and let $\Bbb{C}[G]$ denote the complex group ring. Is it true that if $Z(G) = \{1\}$, then $Z(\Bbb{C}[G]) = \{1\}$; that is, if the center of the group is trivial, then the center of the group ring is trivial?
Like Leaky pointed out, think of $\Bbb CG$ as ring of set functions $G \to \Bbb C$ with obvious pointwise addition and multiplication. Then $Z(\Bbb CG)$ is just the set of functions constant on the conjugacy classes of $G$ ("class functions")
The class functions are generated by the irreducible characters by representation theory, so that gives a full description of $Z(\Bbb CG)$
The offset logarithmic integral is defined as $$ \text{Li}(x)=\int_2^x\frac{1}{\log(t)}~dt. $$
It can be shown that $\text{Li(x)}\sim\pi(x)$ where $\pi(x)$ is the prime counting function. It can also be shown that $\frac{x}{\log(x)}\sim \pi(x).$
Is there any practical use in number theory f...
Goal
Write a program in C/C++ to find the smallest prime \$k\$ such that for all \$n\in \Bbb{N}\$, \$n!+k\$ is never prime. You can use any number-theoretical methods to achieve this but must explain in-depth to why it works.
Basic Concept
It is elementary to see that for any \$n\ge k\$, we ha...
