Noone answers my question, I don't know if they are badly asked, or if they are too specialised, does anyone have any ideas how I can increase my exposure?
i keep coming to the end of your question and not really knowing what you were asking. It might be apparent if I went back and read carefully but most people will not take the time unless they are interested.
That in $G(n,p)$ when $p=(1+\epsilon)/n$ there's a unique component of linear size to n, and when $p=(1-\epsilon)/n$ all components are of logarithmic size?
I've been wondering why everyone focus on that, whereas Erdos in his original paper didn't find it interesting enough to focus on, and, working in G(n,m), he searched for the specific probablity of the largest component having $n-k$ vertices in it. Just felt like sharing this, no real question here..
Then everyone taking his paper focused on that instead of what he thought was way more interesting (and includes in a way that result, with a bit of work)
that way people know right away what the point of the question is, and they also have a specific example to work with. Makes the answer easy to write (for those who know how to do it).
Precisely - I find his process hard to understand.
Yeah. I read a very nice proof of the double-jump result, it wasn't much simpler, it involved some rather long-to-prove lemmas and so on. But it fit my line of thinking, so I understood where they were heading the moment they wrote the theorem itself.
Okay, look at a map of India. draw an imaginary line from the long bottom of it all the way up. If you live right to that place, you are in eastern India. Otherwise, by the logic that every statement is either true or false, you are in western.
@Sush if the discriminant is nonnegative and the middle term in the quadratic polynomial (in y^2) is nonnegative, then for all x there exists a y for which (x,y) is on the curve. that can't be bounded.
@BalarkaSen i don't have a problem but i do have some sets.
Let $$\mathcal{C}_{n}=\{z\in \mathbb{C}:p(z)=0\text{ for some polynomial }p\text{ whose coefficients are }n^\text{th}\text{ roots of unity}\}$$
(so, for clarity: we'd have $\mathcal{C_n}=\bigcup_{m=1}^\infty C_{n,m}$, where $C_{n,m}=\{z\in \mathbb{C}:\exists n^\text{th}\text{ roots of unity }a_1,\ldots, a_m\text{ such that } a_0+a_1z+\cdots +a_nz^m=0\}$)
usually I'll be talking about $C_p$ for $p$ prime, and also it seems to help to view these as points in $\mathbb{R}^2$
so, here's my question to you. How can we Galois theory these?
certainly there is something to talk about.
for example, we could look at the splitting field.
i posted some pictures of $\mathcal{C}_{n,M}=\bigcup_{m=1}^MC_{p,m}$ on here before. They seem like interesting sets.
here's $\mathcal{C}_{3,8}$, for example. What can you tell me about $\mathcal{C}_3$?
(that's also a question you might be interested, @anon)
I have a little question, suppose G acts on X, a have checked an old exam and it define $X_G=\{x: gx=x for all x\} someone knows what is the name of that. X does not have structure of group necessary is that set well defined? what characteristica can have?
Well more specific. If we let $G$ a p- group and X a G set, show that $o(X) = o(X_G) mod p$ where X_G is defined as above.
o is order.
I'm a little confused about the set X_G, maybe is a typo
G acts on X and X_G is the set of all the elements in X s.t. are unchanged under the action of G. It´s like a stabilizer but the set lies in X which not necessarily is a group. That's confuse me
Now, when is it true that $\mathcal O(x)=\{x\}$; i.e. the only element in the orbit of $x$ is $x$ itself?
This has to do with $X^G$.
@JoseAntonio I know.
$\mathcal O(x)=\{x\}$ means $gx=x$ for every $g\in G$. That is, $\mathcal O(x)=\{x\}\iff x\in X^G$. So $|X|=|X^G|+\sum \mathcal O(x)$ where the sum runs through nonsingleton orbits.
@AlexanderGruber LOL. I don't disagree for a first pass. But there are some nice advantages to seeing things from a more general point of view as well, I think. I've enjoyed Aluffi's "Algebra: Chapter 0" immensely for that very reason.
Serge Lang's Algebra is well regarded among some. A very polarizing text, it seems. Many love it and swear by it, and others want nothing to do with it.
@JoseAntonio isaacs is the best possible book you could use on finite group theory, but it is too advanced for you now. It will assume knowledge of semidirect products, for example.
@JoseAntonio yeah, i don't know... it does "start from the beginning", but the text (and, in particular, the exercises) get harder than what I'd be comfortable recommending for a first pass in algebra really quickly
@TedShifrin seems like it could have some cool properties. definitely setwise invariance under various flavors of homomorphisms. i wonder if there is anything geometric, it looks sort of fractal-y.
@Ted Could you please just tell me what my bounds for my integrals are suppose to be, and if the standard coefficient forms for $a_0,a_n,b_n$ are right math.stackexchange.com/questions/939555/… . I haven't been able to work it out for three hours
I think D&F is very nice for getting your hands dirty with groups. They have a lot of lovely problems, both computational and theoretical, which make you work out some important details.
The exposition itself might be a touch lacking, I agree. You really need to do the problems alongside reading the text if you're going to get much out of D&F for sure. In fact, the best way to go is to do almost all of them.
For sure, @TedShifrin. I've found this to be particularly true of D&F though. If you go through it problem by problem, there is a very natural flow of ideas. If you don't do enough, you might not get all the key ideas from a section, and if you pick them too sparsely, some of the problems later on may be very hard to even get started on.
Perhaps not. We aren't quite done with the chapter though. Haven't gotten to Heine/Borel yet, for example. Maybe for next week?
I think it's more that I had a lot more on my plate last semester. The QUODEs homework has been really easy, so I feel like I have to make up for that somehow.
@Ted I tried with bounds $\int_{-1}^1$ and $a_n$ ended up being $-2\cos n + 2\cos{-n} + 2\sin n - 2\sin{-n}$ which seemed wrong. I assumed when you say the 'right function'. It isn't $a_0 + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right)$, $a_n=\frac{1}{L} \int_{-1}^1 f(x) \cos\left(n\pi x\right)$, $b_n = \frac{1}{L} \int_{-1}^1 f(x) \sin\left(n\pi x\right)$, $a_0 = \frac{1}{2} \int_{-1}^1 f(x) dx$
@Chris'ssis: I'm a bit rusty with Landau symbols. Does $Q(z) = 1 - \rho r^k + \mathcal{O}(r^{k+1})$ for $r \to 0$ imply that there is some $r > 0$ such that $|Q(z)| < 1$? Of course $r = |z|$.
In mathematics, big O notation describes the limiting behavior of a function when the argument tends towards a particular value or infinity, usually in terms of simpler functions. It is a member of a larger family of notations that is called Landau notation, Bachmann–Landau notation (after Edmund Landau and Paul Bachmann), or asymptotic notation. In computer science, big O notation is used to classify algorithms by how they respond (e.g., in their processing time or working space requirements) to changes in input size. In analytic number theory, it is used to estimate the "error committed" while...
@Chris'ssis: Yes, thanks for the link. It clarified it.
@Chris'ssis: I thought about it in the way of $\lim_{r \to 0} \left| \frac{f(r)}{r^{k+1}} \right| < \infty$ but dividing the whole term by $r^{k+1}$ and letting $r \to 0$ didn't quite work.