@MarcGato Si j'essaie dans le sens direct, j'obtient : soit $x\in E,$, il existe $\epsilon>0$ tq $B_o(x,\epsilon)\in A$. On veux $\forall\varpi>0,\exists a,\forall w\in B_o(x,a),f(w)\in B_o(f(x),\varpi)$ mais je suis bloqué la
@robjohn One more question, the policy site says that the minimum age for the user is 13, but how if there's a user who signed up this site when he was under 13 and now he is above 13?
@MarcGato Je reformule ce que j'ai écrit plus haut :On veux $\forall\varpi>0,\exists a,\forall w\in B_o(x,a),f(w)\in B_o(f(x),\varpi)$. J'ai : soit $x\in E,$, on pose $B=B_o(f(x),\epsilon),A=f^{-1}(B)$, alors $B_o(x,\epsilon)\in A$.
@MarcGato So you need at least 150 reputation to vote. Coming from another site would automatically give you 100 reputation here (assuming you have 100 reputation there), so you'd still need to do some work here to be able to vote.
@Balarka Okay, I will ask! So I have the Metric space $X$ over the interval $[0,2]$, and the distance metric $d(x,y)=|\lfloor x \rfloor +x - y -\lfloor y \rfloor |$. I have $f:X\to \Bbb R$, $f(x)=x, x\le 0.5; f(x)=x+1, x>0.5$. I want to show discontinuty, so is the following reasoning correct:
@Balarka Actually, we know $f$ is continious in $x_0$ iff for every $\epsilon$ there is some $\delta$ so that $f(B(x_0,\delta ))\subseteq B(f(x_0),\epsilon)$
@KellyBlunie Saying "every linear combination of A produces V" is the same as saying "any linear combination of A produces V". However, a linear combination of $0$ elements of $A$ does not produce $V$. Contradiction.
Does anyone know some (as basic as possible a) paper or article/blog post with uses of the Lefschetz number? I've seen some examples already but it would be nice to see it 'in the wild'.
@KellyBlunie You wrote " lets say the set A spans a vector space V only If every linear combination of A produces V, then Span(A)=V", i'm just telling you that it's false
@Studentmath $G$ be a a group with subgroups $H$ and $K$. Under the action of $H \times K$ on $G$ via $(h, k) \cdot g \mapsto hgk^{-1}$, the orbit of some element $g$ of $G$ is $HgK = \{hgk : (h, k) \in H \times K\}$. This is called an $(H, K)$-double coset of $g$ in $G$. The set of all double coset is denoted $H\setminus G /K$. Prove that $|H \setminus G /K| = |K| \cdot |H \cap gKg^{-1}|$
OK, I gotta go. The internet connection here is a bit troubling.
@Hippalectryon what is wrong? And besides that being wrong ,ignoring it, can you say if the determinant is zero why a general vector from V is not contain in A
My book states:there is at least one choice of u for which this system will not have a solution and so u can not be written as a linear combination of these vectors. And that there are in fact infinitely many choices of u that will not yield solutions.
@Hippalectryon the matrix you have it is the transpose i have you formed a linear combination of vectors and then forme a system of equations based on the components then forme the matrix and that is the transpose of your matrix lets call this transpose B however the determinant is zero in either
there is at least one choice of u for which this system will not have a solution and so u can not be written as a linear combination of these vectors. And that there are in fact infinitely many choices of u that will not yield solutions.
there is at least one choice of u for which this system will not have a solution and so u can not be written as a linear combination of these vectors. And that there are in fact infinitely many choices of u that will not yield solutions.
Hi guys, I know that $\ell_P^*$ is isometrically isomorphic to $\ell_q$. But I am having a hard time understanding this. An element in $\ell_p^*$ is a function from $\ell_p \rightarrow \mathbb{R}$ .. so how a function be the same as an element of $\ell_q$ (which are sequences).
Okay, so I actually went with how H acts on $gK$, and I defined it as I said: $h(gK)=hgK$. So I wanna know the size of the orbit of $gK$, it is naturally the size of $hgK| h\in H$. I will mark it by $A$. Now we want to compute the order $HgK$. It is the union of $A$ disjoint sets of the form $hgK$, each of order $|K|$ so $A*|K|$ @Balarka
Next I want to compute $H:H\cap gKg^{-1}$ (the order thereof), and show it is equal to $A$.
But then a stabilizer $h\in H$ upholds $hgK=gK$ which leads me to $h\in gKg^{-1}$
Erm.. no wait, that's fine.
Yeah, and then I know that the the stabalizer is $H\cap gKg^{-1}$, and since the orbit of $gK$ is the index of the stabilizer, we complete the proof. I think.
practical question: I'd like to see an English translation of a certain arXiv paper which is in Russian i.e. Cyrillic. At this point I really just want to figure out if there's anything in that version which isn't in the English language papers by the same author, so I don't need high quality. Are there any obvious tools?
@MikeMiller @DanielFischer Let $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$ be the subfields of $\overline{\mathbb{Z}}_p$ with $p^n$ and $p^m$ elements respectively.
How can I find the field $\mathbb{F}_{p^m} \cap \mathbb{F}_{p^n}$ ??
@MaryStar The nonzero elements of $\mathbb{F}_{p^k}$ are the elements of $\overline{\mathbb{Z}}_p$ with $x^{p^k-1} = 1$. So for $x\neq 0$, we have $$x\in \mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \iff x^{p^n-1} = x^{p^m-1} = 1.$$ Suppose $n > m$. It then follows that $x^{p^n - p^m} = \bigl(x^{p^{n-m}-1}\bigr)^{p^m} = 1$. Since the Frobenius homomorphism is injective, it follows that $x^{p^{n-m}-1} = 1$, i.e. $x\in \mathbb{F}_{p^{n-m}}$.
Iterate the reasoning to find the $k$ such that $\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^k}$.
@KellyBlunie You can perform reversible column operations that don't change the determinant that diagonalize any matrix $M$. The concatenation of those column operations can be represented by an invertible matrix on the right (that doesn't change the determinant). So suppose that $MA$ is diagonal and $A$ does not change the determinant of $M$. Since $\det(MA)=\det(M)=0$, one of the diagonal elements of $MA$ must be $0$. Choose a vector $v$ with a $1$ at that position and $0$s elsewhere.
@KellyBlunie consider $Av$. Since $A$ is invertible $Av$ cannot be $0$. However, $MAv=0$
@DanielFischer I understand... But, what do you mean by "Iterate the reasoning to find the $k$ such that $\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^k}$." Isn't $k$ equal to $n-m$ ??
@MaryStar Generally, no. We have $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}},$$ but generally, not equality. Note that $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}} \cap \mathbb{F}_{p^m}.$$
Well, in the last, we do have equality, $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^{n-m}}\cap \mathbb{F}_{p^m}.$$
@Venus The idea is to compute it without pen and paper, without using the asymptotic expansion of the harmonic number $$\lim_{n\to\infty} n\left(H_n-\log(n)-\gamma\right)$$
@MaryStar We have a function, call it $f$, with $f(n,m) = f(n-m,m)$ if $n > m$, $f(m,m) = m$, and $f(n,m) = f(m,n)$, and $f(m,1) = 1$. What could $f$ be?
I don't see the appeal of computing things without pen and paper. If you have a piece of paper and a pen, why not use them? Each to their own, I guess.
My problem asks me to show that if $f$ is non-constant, then $\mathbf{V}(f)$ is finite. Assume that $f \in \mathbb{C}[x]$.
If $f$ was an ideal, this would be straightforward; however, $f$ is merely a polynomial described above. I suppose by "non-constant" the prompt means "a polynomial with max...
IIRC this thread was one of the weirdest ever on math.SE in that all answers, and they were plenty, were downvoted. One of them even got a bounty, which is verified in the comments.
Evaluate the following integral$$\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$$
My Attempt:
Let
$$\tag1 I = \int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx.\\[15pt]$$
Putting $x=\frac{\pi}{2}-x$ and using property $\displaystyle\int_{0}^{a}f(x)...
@DanielFischer Thanks! There must have been a deletion spree going on wiping out all om them at once, or at least in a quite short span of time. Strange...
@Venus I'm against banning (or banishing) anyone ! :P (especially not her .. her posts are more than half the reason I stuck to this chat for so long .. torture is just a light price you pay .. nothing severe :P)
Guys real quick, if f is a group homomorphism and bijective, how do I show f^{-1} is a homomorphism (I hate proofs that fly close to definitions), I just applied f to both sides of f(a)f(b)=f(ab) where means inverse, that's okay right? [quick yes/no, I'm not stuck]
@r9m You're a bit short on why $d_i^mw_{ij} = w_{ij}d_j^m$ for all $i,j$ implies $d_i w_{ij} = w_{ij}d_j$, but it's correct. To see that you can write $B = p(B^m)$ for some polynomial $p$, first diagonalize, then you need a polynomial with $p(d_i^m) = d_i$ for $1\leqslant i \leqslant n$. Lagrange knew that such a polynomial exists.
My problem asks me to show that if $f$ is non-constant, then $\mathbf{V}(f)$ is finite. Assume that $f \in \mathbb{C}[x]$.
If $f$ was an ideal, this would be straightforward; however, $f$ is merely a polynomial described above. I suppose by "non-constant" the prompt means "a polynomial with max...
Evaluate the following integral$$\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$$
My Attempt:
Let
$$\tag1 I = \int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx.\\[15pt]$$
Putting $x=\frac{\pi}{2}-x$ and using property $\displaystyle\int_{0}^{a}f(x)...
