@DanielFischer After having shown by induction that $h(k) \geq k$ we have that $k \leq h(k)<n \rightarrow k<n \rightarrow k \in n=\{0,1,2, \dots, n-1 \}$.. But what do we conclude from that?
@evinda That the domain of $h$ is a natural number not larger than $n$. And since by construction $h$ is a bijection between its domain and $f(B)$, that shows that $B$ is a finite set, whose cardinality is less than or equal to the cardinality of $A$ (which is $n$).
@DanielFischer So since $k$ is an arbitrary element of the domain, we know that each element must be smaller than n and that means that the greatest natural number that can be the domain is n, right?
Testing the final digit by itself works for divisible by 2, testing the final two digits for divisible by 4, you need to test final three digits for divisibility by 8
To test divisibility by 3, add together all of the digits. If the number is larger than 9, add the digits of the newly formed number. Repeat this process until you get a number between 1 and 9
user61230
For 3, add the digits together; if that number's divisible by 3, the original number's divisible by 3.
If at the end you get a 3,6, or 9 then it is divisible by 3
for divisibility by 6., first test if divisible by 3, and then test if divisible by 2. If it is divisible by 6 it must be both divisible by 3 and 2 simultaneously
Example. $12345678\mapsto (1+2+3+4+5+6+7+8)=36 \mapsto (3+6)=9$ and 9 is divisible by 3, so 12345678 is in fact divisible by 3
Furthermore, 12345678 is even, so it is also divisible by 2. Since divisible by both, it is divisible by 6
You can do similarly for divisibility by 9, but instead of getting a 3,6, or 9 at the end of repeated summation of digits, you want only to get a 9 (or 0)
Trying to reach a contradiction, but all I got to was a nonlinear system of three equations and a finite (but potentially large) number of unknowns. I'm thinking I should reverse direction away from proof by contradiction and look for some result which can go more directly
I suppose I need to think further to understand why the rupture fields for a splitting polynomial are isomorphic and how we can apply results from one to the other. It seems an incredibly obvious statement when considering $\mathbb{Q}(\sqrt[3]{2})$, but it is still not immediately apparent how to transfer results from the one to the other
If $x$ and $y$ are roots of an irreducible $f\in K[T]$ then $K(x)\cong K[T]/(f(T))\cong K(y)$
since $\sqrt[3]{2}$ and $\sqrt[3]{2}e^{2\pi i/3}$ are both roots of $T^3-2\in\Bbb Q[T]$, irreducible, then $\Bbb Q(\sqrt[3]{2}e^{2\pi i/3})\cong\Bbb Q[T]/(T^3-2)\cong\Bbb Q(\sqrt[3]{2})$. this isomorphism preserves $-1$, squares and sums of squares, so any solution in the first field would be transported by the isomorphism to a solution in the last field.
@JMoravitz I said minimal (=> irreducible) polynomial, not splitting polynomial
Well, thank you again for taking the time to explain. It is nice that there are users like you who can answer upper level math questions (though still early grad level, probably not upper level to you).
I get to contribute by helping the undergrads at least. In another few more years hopefully I'll get more algebra under my belt enough to return the favor to the next early grad students down the line.
it seems to differ from place to place, what generally counts as undergraduate vs. graduate material? where do most people draw the line in various subjects?
If you have it the other way around, then let $\epsilon\to 0$, I'm pretty sure you get that within a neighborhood $(x-\delta, x+\delta)$ there is essentially no change
implying that the function is constant
Continuity is defined for a specific point (continuous at $c$). Absolute continuity is when the same $\delta$ works for the epsilon regardless which $c$ it is in question
whoops! look up uniform continuity, I think that's the one you want
actually even then I think I was quite wrong. see: @jmoravitz or the definition of uniform continuity on wikipedia. shows how much of this stuff I remember
Well, I was forgetful as well. Absolute continuity gets into measure theory, which says that for each $\epsilon$ there exists a $\delta$ such that breaking the domain into several disjoint subintervals of total measure less than $\delta$ the total variation on those intervals is less than $\epsilon$
but yea., wikipedia is quite helpful for remembering definitions instead of having to flip through the index of a textbook
(or relying on the faulty memory of others)
But yea, to summarize., for function $f:X\to Y$: everywhere Continuous: $(\forall c\in X)(\forall \epsilon > 0)(\exists \delta >0)~ s.t.~ ((\|x-c\|_X<\delta)\Rightarrow (\|f(x)-f(c)\|_Y<\epsilon)$. Uniformly continuous $(\forall \epsilon > 0)(\exists \delta >0)(\forall c\in X)~ s.t.~ ((\|x-c\|_X<\delta)\Rightarrow (\|f(x)-f(c)\|_Y<\epsilon)$
In a comment, I asked for something which was clearly in the OPs message. Should I delete my comment(s)? http://math.stackexchange.com/questions/1102127/equivalent-dfn-of-filtered-categories
Your comment might help other users which have the same problem as you have. (I personally would prefer to leave it. If it is the comment I linked to at all.)
But it's you're comment and you are the one to decide what to do with it. But if you decide to delete it, I think that it is useful to wait some time whether the OP deletes comment reacting to yours. If not, you should flag it as obsolete. One of the moderators will see your flag, they will see that it is a comment reacting to a deleted comment and remove the flagged comment, too. But occasionally it might happen that such flags are declined.
The title of this question contained $\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$. I have change it to $\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$.
It the way it looks now acceptable? Or would it be better just to change dfrac to frac?
Or some completely different suggestion?
In fact, it seems that it was one of the editors who put that thing into the title.
@Committingtoachallenge I am not sure how things will unfold for me. My mental condition declined since Dec, but today I don't feel too bad. I might try to do some part time job this year and the next before studying, we will see.
Moin @DanielFischer !!! I am looking at the following: **Propostition**:
The Cartesian product of finite sets is a finite set.
**Proof**:
Let $X,Y$ finite sets and therefore there are $n,m \in \omega$ such that $X \sim n$ and $Y \sim n$, i.e. there are $f: X \overset{\text{1-1 & surjective}}{\longrightarrow} n$, $g: Y \overset{\text{1-1 & surjective}}{\longrightarrow} m$.
We define the function $h: X \times Y \to n \times n \\ \langle x,y \rangle \to \langle f(x), g(y) \rangle$
It is obvious that $h$ is well-defined and it is $1-1$ and surjective.
@Committingtoachallenge I have not. You must understand that I cannot change certain obsessions of mine, to start only in certain ways.
@Committingtoachallenge Severe OCD is a terrible condition. Most people who get it are mostly screwed for life.
@Committingtoachallenge I will figure out something for myself. I am 34 this year. I will do my best to get well and enter grad school by 40. That is the deadline I set for myself...
If anyone asks my location, just tell them I am in Antarctica, thanks.
@Oracle I spend a lot of time sorting out my confused thoughts every day. I also go for long walks, read things for fun (non-math) and chat with people online.
@evinda It helps to translate symbols into words to understand. In words, a function is defined as the set of ordered pairs making up the points on its graph.
Minitab gives very strong evidence against the null hypothesis in terms of coffee consumption. E.g. There is very strong evidence that cups of coffee consumed is correlated with amount of study done.
Today I got 11 hours of sleep and had 2 coffees, so by that equation I should have done 1.04 hours, but I have done 1.5333 whoops, I better have a coffee ;P
The Cartesian product of finite sets is a finite set.
**Proof**:
Let $X,Y$ finite sets and therefore there are $n,m \in \omega$ such that $X \sim n$ and $Y \sim n$, i.e. there are $f: X \overset{\text{1-1 & surjective}}{\longrightarrow} n$, $g: Y \overset{\text{1-1 & surjective}}{\longrightarrow} m$.
We define the function $h: X \times Y \to n \times n \\ \langle x,y \rangle \to \langle f(x), g(y) \rangle$
It is obvious that $h$ is well-defined and it is $1-1$ and surjective.
@N3buchadnezzar: The idea is to split up the sum into two, the first is finite and thus can be bounded easily, the second you can bound using the fact that $z_n$ converges.
@N3buchadnezzar: I think it's a bit easier if you pick a very certain element of the sequence at which you split the sum into two. Remember that since $z_n$ converges to $z$, for all $\varepsilon > 0$ there exists some $N \in \mathbb{N}$ such that for all $n \geq N$, we have $|z_n - z| < \varepsilon$. What happens if you split up the sum at this $N$, for an arbitrary $\varepsilon > 0$ given?
@evinda Typo: It ought to be $h \colon X\times Y \to n \times m$. One takes $h(\langle x,y\rangle) = \langle f(x), g(y)\rangle$ because given two bijections $f \colon X \to n$ and $g\colon Y \to m$, that is the most straighforward and obvious bijection $X\times Y \to n\times m$.
@N3buchadnezzar Look at $$\frac{1}{n}\sum_{k=1}^n (z_k -z).$$ You have a couple of potentially large [in absolute value] terms at the start of the sum (where "a couple" can be a huge number), and then a lot of small terms. If you let $n$ grow, the number of large terms remains constant, so the average is dominated by the small terms, and hence itself small. Making that precise is the $\varepsilon$-$N$ argument.
@N3buchadnezzar: I didn't mean you should look at the sum itself, rather at the sum minus its (supposed) limit. I thought that was clear, but I see now it wasn't.
@N3buchadnezzar: No, I was going to bound $\frac{1}{n}\sum_{k=1}^n z_n-z$.
@N3buchadnezzar: I should continue with my work. If it's still unclear, there are a number of questions about that statement on MSE, e.g. math.stackexchange.com/questions/533626/…
@N3buchadnezzar: Sure, try yourself. I think it's a good exercise if you're not used to proving convergence with epsilons, or just revising the method.
We want to show that the following expression is less than epsilon, for all epsilon % \begin{align*} \left| z - \frac{z_1+z_2+\cdots z_n}{n} \right| = \left| \frac{1}{n} \sum_{k=1}^n z-z_n \right| \end{align*} % Since the limit $\lim_{n\to\infty} z_n$ exists, we know that there exists some number $N\in\mathbb{N}$ such that for all $n\geq N$ we have $\left| z_n-z\right|\leq \varepsilon$. Hence % \begin{align*} \left| \frac{1}{n} \sum_{k=1}^n z-z_n \right| \leq \frac{1}{n} \sum_{k=1}^n \left| z-z_n \right|
i'm trying to understand why the map $q:S^2 \to \mathbb{R}P^2$ that sends each point on the sphere to the line through the origin and the point, is a covering map
if we have a point in the projective plane, can't we just take a small intervall around that point and then the preimage of this intervall will be exactly the union of this interval on the upper hemisphere and on the lower hemisphere
which is a disjoint union of connected open subsets
@DanielFischer Nice... Can we show as follows that $h$ is bijective?
Let $x_1, x_2 \in X $ with $x_1 \neq x_2$ and $y_1, y_2 \in Y$ with $y_1 \neq y_2$. Then $f(x_1) \neq f(x_1)$ since $f$ is 1-1 and $g(y_1) \neq g(y_2)$ since $g$ is 1-1.
@evinda You have one typo, you typed $f(x_1) \neq f(x_1)$ where you meant $f(x_1) \neq f(x_2)$. Just for the record. You have a conceptual problem, because $\langle x_1,y_1\rangle \neq \langle x_2,y_2\rangle$ means $(x_1 \neq x_2) \lor (y_1 \neq y_2)$, not $(x_1\neq x_2) \land (y_1\neq y_2)$. So your argument for injectivity is incomplete. It's fixable.
The surjectivity part is okay, but not written up well. You need equivalences instead of $\Rightarrow$, and you should say something like "since $f,g$ are surjective, such $e$ and $k$ exist".
@evinda You start by saying "Let $x_1,x_2 \in X$ with $x_1 \neq x_2$ and $y_1,y_2 \in Y$ with $y_1 \neq y_2$". So you never consider the case where one of the components is equal and the other different.
@JohnDoe You're welcome.
@JohnDoe Do you know that Lipschitz functions are almost everywhere differentiable (Rademacher)? And the distributional derivative is given by the a.e. defined gradient? So $\lVert \nabla \overline{v}_m\rVert_{L^\infty(I)} \leqslant 2\sigma$ is just Mac Shane.
@evinda For example. But typically, it is more convenient to show injectivity the other way round, suppose $h(\langle x_1,y_1\rangle) = h(\langle x_2,y_2\rangle)$ and deduce $\langle x_1,y_1\rangle = \langle x_2,y_2\rangle$.
@DanielFischer And the distributional derivative is given by the a.e. defined gradient? By distributional derivative are you referring to 'weak derivative'?
@DanielFischer In my lecture notes, it says that $h$ is obviously well-defined and it is 1-1 and surjective. And I was sondering how it could be shown.
@DanielFischer If we want to show that a set is finite and we find a bijective function, do we have to show that it is well-defined?
@JohnDoe Almost certainly "Yes" or "Yes, but". Depends on what you call "weak derivative". The derivative in the sense of distributions is know to exist, and is sufficiently nice, hence it coincides with the weak derivative - of which we may not know a priori that it exists - in all reasonable definitions of "weak derivative".
@evinda I don't even know how well-definedness of $h$ could be an issue. Maybe it would make sense in context.
@DanielFischer So a kind of generalized definition of weak derivative. Is the following in line with what you mentioned: $\frac{|v_{m}(x)-v_{m}(y)|}{|x-y|} \leq 2\sigma$ for all $x,y$, it follows then that $\nabla v_{m}(x) = \lim\limits_{x \rightarrow y}\frac{|v_{m}(x)-v_{m}(y)|}{|x-y|} \leq 2\sigma$. Therefore $\| v_{m} \|_{L^{\infty}(I)} \leq 2\sigma$.
Because $v_{m}$ is Lipschitz it is almost everywhere classically differentiable as you stated.
@DanielFischer It is just said that it $h$ is well-defined. Anyway... :) How could we show that the set $n \times m$ has the same cardinality as the natural number $n \cdot m$ and thus $n \times m \sim n \cdot m$, i.e. $Card(X \times Y)=Card(n \times m)=n \cdot m$? Do we consider the function $f: n \times m \rightarrow n \cdot m$?
hey, if I have a function f with ||f||_Lp(X) = c < infty, L(X) < infty (Lp stands for Lp norm, L for lebesgue measurement). is it true, that when I integrate f over C_eps with L(C_eps) = eps: ||f||_Lp(C_eps) --> 0 for eps --> 0
The set of permutations having $n-k$ elements in their original place $\subseteq$ set of permutations having $n$ elements in their original place, right?
