($C_n$ acts on $C_n^n$ by cyclically permuting the coordinates, so we can form the semidirect product $C_n^n\rtimes C_n$, which is the wreath product $C_n\wr C_n$)
well, by sequence of abgrps I suppose you want the middle term to be abelian, which the wreath product ain't
(i) has subnormal series with cyclic factors (ii) every composition factor has prime order (iii) has subnormal series with abelian factors and each element of the series normal in $G$.
the definition : $f : X \to Y $ is a map from $X$ to $Y$ , Banach spaces . $f$ is called differentiable at say $x \ in X$ if there is a linear map $Df(x) : X \to Y$ and also a continuous map $\phi : V_0(X) \to Y$ with $\phi(0)=0$ , so that $$f(x+h)= f(x)+ Df(x)[h] +||h|| \phi(h)$$ for all $h \in V$ @robjohn i don't see the necessity of introducing the function $\phi$ .
@Complexanalysis One could also say ... such that $$\lim_{h\to 0} \frac{\lVert f(x+h)-f(x)-Df(x)[h]\rVert}{\lVert h\rVert} = 0,$$ but occasionally it's useful to have a name for the difference.
@DanielFischer But they aren't equivalent , are they ? the definition i posted will give your definition , but i don't see why would your definition give back the one i am referring to.
@DanielFischer ahh , got it i guess . the inequality says that only for $$x \in X:||x||=0$$ is possibly $T(x)=0$ , since $T$ is linear and also using the norm property only if $x=0$, $T(x)=0$ , Hence $Kern(T)=0$
@Complexanalysis That gives you $\lVert Tx\rVert = 2\lVert x\rVert$. However, indeed in general for an injective $T$ you don't have an estimate $\lVert Tx\rVert \geqslant c\lVert x\rVert$.
@WLOG It must be $D(0,1) - [0,1)$, since $D(0,1) - (0,1)$ is not an open set.
So i have to show that there exists no $c$ such that $$\Right paren(\sum {x_1}^2+{x_2/2}^2+....\left) \ge c . \right(\sum_{i=1}^{\infty}x_i^p \left) $$ @DanielFischer
@Alyosha To see that that is a contradiction, you need to show that the minimum cardinality of spanning sets is the maximum cardinality of linearly independent sets.
So with that definition of dimension, I suggest something else.
@robjohn This is unrelated to the previous question. Is it possible to prove that $$\sum_{n=0}^{\infty} S_k^n \frac{z^n}{(n-1)!} = 6\log^3(1-z)/(1-z)$$?
Where $S_m^n$ is Stirling number of the first kind
I meant $$\sum_{n=0}^{\infty} S_4^n \frac{z^n}{(n-1)!} = 6\log^3(1-z)/(1-z)$$
@Mike Note that $\text{Ext}_{\mathbb Z}^1(A,A^n)=\prod_{i=1}^n\text{Ext}_{\mathbb Z}^1(A,A).$ So there exists a nontrivial choice if and only if $A$ has a nontrivial extension by itself: $0\to A\to X\to A\to 0$. This may offer insight behind anon's example $A=\mathbb Z/2\mathbb Z$ as well as provide a source of other examples.
Is the given proof of the following correct?
Let $\{A_n\}$ be a sequence of sets, then: $$\liminf_{n\to\infty}=\bigcup_{i=1}^\infty\bigcap_{j=i}^\infty A_j=\{\text{elements that belong to all but finitely many } A_i's\}$$
Pf.
($\leftarrow$) Let $$B_i=\bigcap_{j=i}^\infty A_j$$
Then, $$x\i...
so c(1,1,0) doesn't have the z component, but all linear combinations will span a plane in x,y and then d(0,1,1) will be a plane in y,z together they can span all 3D space no? @Mike
a line has 'one free variable' - how far up or down the line you are
but a plane has two - you can move right/left or up/down (for whatever these should mean for your plane)
when you've only got c, that's only one free variable. you can only move up or down the line in the (1,1,0) direction - and c determines how far up or down it you are
If
$$y_{0}=\frac{1}{5}x^{5}+\frac{1}{2}x^{4}+\frac{2}{3}x^{3}+x^{2}+2x+1$$
and
$$y_{1}=-\frac{1}{325}x^{13}-\frac{1}{60}x^{12}-\frac{109}{1100}x^{11}-\frac{19}{150}x^{10}-\frac{593}{1620}x^{9}-\frac{3}{5}x^{8}-\frac{311}{315}x^{7}-\frac{68}{45}x^{6}-\frac{32}{15}x^{5}
-\frac{7}{3}x^{4}-\fra...
Recall the fact that
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n$ and summing both sides from $n =0$ to $\infty$, we get...
Review my answer there, please.
Also, latex on (6) has a very clunky format, maybe someone could fix that?
@BalarkaSen There are several identities that are not that common that I need to verify, but if you know that $$\sum_{n=0}^\infty H_n^3x^n=\frac{\mathrm{Li}_3(x)}{1-x}$$ couldn't you try to integrate that and evaluate at $-1$?
@BalarkaSen suppose $(a+b\theta)(c+d\theta)=1$. I only get $ac-5bd+\theta(ad+bc+bd)=1$. Evaluating $(a+b\theta)(a+b\theta')$ only gives $a^2+ab\theta' +ab \theta + 5b^2$. Now sure what to do next
let $D_1$ and $D_2$ be different distributions. Let $n,m$ be integers. Is there a way to formally decompose a data set such that $n$ of the points are drawn from $D_1$ and $m$ from $D_2$?
my question is a bit general
for example say $D_1$ and $D_2$ are different multivariate normal distributions
is the joint distribution of this data going to be some kind of elliptical distribution?
I have a small question related to the Lie Group SU(2). Supposing we define a one-parameter subgroup determined by a homomorphism from the additive reals to the Lie group, we get a solution of the form F(t) = exp tA (assuming we set t=0 as the identity). Now using a representation of quaternions we can write A as ia+jb+kc. Expressing exp (tA) in terms of sin and cos functions I am unable to show that F(t1)F(t2)=F(t2)F(t1)=F(t1+t2). What's wrong?
Is the given proof of the following correct?
Let $\{A_n\}$ be a sequence of sets, then: $$\liminf_{n\to\infty}=\bigcup_{i=1}^\infty\bigcap_{j=i}^\infty A_j=\{\text{elements that belong to all but finitely many } A_i's\}$$
Pf.
($\leftarrow$) Let $$B_i=\bigcap_{j=i}^\infty A_j$$
Then, $$x\i...
If
$$y_{0}=\frac{1}{5}x^{5}+\frac{1}{2}x^{4}+\frac{2}{3}x^{3}+x^{2}+2x+1$$
and
$$y_{1}=-\frac{1}{325}x^{13}-\frac{1}{60}x^{12}-\frac{109}{1100}x^{11}-\frac{19}{150}x^{10}-\frac{593}{1620}x^{9}-\frac{3}{5}x^{8}-\frac{311}{315}x^{7}-\frac{68}{45}x^{6}-\frac{32}{15}x^{5}
-\frac{7}{3}x^{4}-\fra...
Recall the fact that
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n$ and summing both sides from $n =0$ to $\infty$, we get...
