https://en.m.wikipedia.org/wiki/Koenigs_function
So the Koenigs function satisfies $f(a(x)) = a f(x)$.
But when i take the derivative i get the wrong result !?
Let $^{[*]}$ denote composition.
$$ \frac{d}{dz} \lim_{n \to \infty} a^z a^n f^{[n]}(x) =^? \lim_{n\to \infty} ln(a) a^z a^n f^{[n]}...
lol @Socrates, I'll never understand upvoters around here
e.g. I thought this was super clever and basically got ignored :P http://math.stackexchange.com/questions/1876350/ways-of-showing-sum-n-1-infty-ln11-n-to-be-divergent/1876355#1876355.
Hey everyone, I'm stuck on this question above, I'm not sure exactly how to approach it, any hints would be appreciated
What I've done so far essentially is to define $\mathbb{R}^{\infty}$ by associating the set of all sequences for which after a given $N$ are zero, and then taking the set of all possible $N$
$$S_N = \left\{(x_i)_{i \in J} \ \middle| \ x_i \in \mathbb{R}^{\omega} \ \text{and } x_i = 0 \ \forall i \geq N \right\}$$ and then $$R^{\infty} = \left\{S_N \ | \ N \in \mathbb{Z_+}\right\}$$
I'm not stupid. :p I know it was deleted because people actually willingly deleted it.
im asking more as to why it was deleted.
like, it was closed. It would've deleted on its own for starters, but strangely the instant it got to around 4 reopen votes (including the guy who originally voted to close it) it all of sudden got a bunch of delete votes.
ok that's very strange (cause I never had any deleted question deleted due to delete votes, mine (in PSE or MSE), are either deleted by community or deleted by a single moderator, thus that's unfamilar scenario to me)
Are the delete votes new (i.e. they are not one of many of the reopen votes?)
It seems only a moderator or above can cast a delete vote according to meta, so this is a moderator thing. But if there is suddenly a huge number of them in such a short time, then it could be some kind of serial deletion, which is breaking the rules. I don't have enough experience to comment any further. Try to check with some moderators that you trusted about what is really happening
Vote to delete has a cooldown, if I recall (please correct me?), the same user cannot single handily cast all 3 votes. If he do take advantage of such cooldowns to cast 3 delete votes, then that is a serious offense.
Alternately, it might just happened to be 3 different users decided to cast a delete vote. Of course, there is not enough info here to rule or a possible malicious deletion intent (if the 3 users are somehow related, either by sockpuppety or decided to band together for the purpose), which is once again breaking the rules and require moderator investigation
@Secret it actually says there were three different users who voted to delete.
but it also says what order the votes were cast, and it's that the first one is one of the same guys who voted to close and that it happened within the same time the post was being reopened. That's what is disturbing. I thought maybe you could see deleted posts.
I probably shouldn't worry about it though.
I know the mods don't really like me right now anyway.
my reputation is not 10K (because I am too lazy and not knowledgeable to answer questions if it is not perfect in the main sites). Moderator privillages starts at 10K. This is the reason I cannot really help other than wondering about the possible scenarios that is involved.
do you have any moderator or high rep users that you trust, perhaps you can ask them to check out more details (since only them can look into a deleted link)?
Well I got in trouble for having too many questions that were essentially useless to the site. I'd rather not go into it. It's just not best for me to ask mods about it. They'll probably just tell me to go jump in a lake and that I probably deserved it anyway for causing trouble.
@Secret anyway changing the subject... Do you know amWhy by any chance?
Not very much, she is around when idomathart was around, and sometimes after that. She sometimes ask questions and converse with tobias and co. I have neutral chemistry with her
@Perturbative let's start with the product topology, what does an open set around $x\in\Bbb R^\infty$ look like?
Or better what does a basic open set look like?
Same question for the box topology. After answering it take $y=(1,1,1,1,\cdots)$. Does an open set around it intersect $\Bbb R^\infty$ in the product topology? What about the box topology?
@AlessandroCodenotti An open set around $x \in \mathbb{R}^{\infty}$ would be a union of products of open intervals of $\mathbb{R}$ containing $0$
For the box topology that is
For the product topology An open set around $x \in \mathbb{R}^{\infty}$ would be a union of products of finitely many open intervals of $\mathbb{R}$ containing $0$, as well $\mathbb{R}$ itself infinitely many times
An open set around $y$ should definitely intersect $\mathbb{R}^{\infty}$ in the product topology
It's neighbourhoods should (if I'm not mistaken) consist of unions of products of of finitely many open intervals of $\mathbb{R}$ containing the point $x \in \mathbb{R}^{\omega}$ to which it converges, as well as $\mathbb{R}$ itself infinitely many times.
Hmm, I'm not sure what do you mean with the point to which it converges. I'd start by thinking about basic open sets, then you can just take a union to get all the neighbourhoods
Write down an explicit example, what's an open set around $(1,2,3,4,5,\cdots)$ in the product topology?
In the example above, an open set around the given sequence should be a union of the product of finitely many open intervals of $\mathbb{R}$, each containing sequence elements along with $\mathbb{R}$ itself infinitely many times
But because the open set is the product of those finitely many open intervals along with $\mathbb{R}$ itself infinitely many times, we can always find open sets that intersect both sequences
Ohhh, but that cannot happen in the box topology
So the sequence you gave as an example would not intersect $\mathbb{R}^{\infty}$ in the box topology
Hi all. Anybody around who could explain to me what kind of matrix structures are achieveable by left matrix multiplication by unitary matrices alone, which require the help of right matrix multiplication by unitary matrices and which simply can't be achieved and why? (c.f. http://math.stackexchange.com/questions/2102756/matrix-structures-obtainable-by-multiplication-by-unitary-matrices )
Background
I am a college pass out. I have done my B.Tech in computer science and masters in quantum information. I am soon going to take up an industrial job. In the free time I got, after passing out from my college till now, ( few months ) I started reading Sipser's book on "Complexity and C...
I thought that using a few like N, 2N, sqrt(N), N^3 ect... as primitives would expand into a large number of possibilities, but that despite being infinite, they would limiting.
I've read : "Let $A$ be a matrix of which the similarity class is closed. Let $D = Diag(1,2,\dots, n)$. Since $[D^k A D^{-k}]_{i,j} = \left({i\over j}\right)^k a_{i,j}$ goes to $Diag(a_{1,1}, \dots, a_{n, n})$ as $k$ goes to infinity, by hypothesis we have that $A$ can be digonalised." Is it not a blunder to say that $D^k A D^{-k}$ converges to $Diag(a_{1,1}, \dots, a_{n, n})$ since for $i\gt j$, the terms diverge ?
I'm more looking for the theorem that would prove that for a function $F(\vec{\mathcal R})=\sum\frac{\vec r_i-\vec{\mathcal R}}{||\vec r_i-\vec{\mathcal R}||^3}$, where each $\vec r_i$ is unique, there must exist some value of $\vec{\mathcal R}$ such that $F=0$
I want to prove, assuming I can, that an error function has unique global max.
For $x \geq 0$ consider the function $f(x) = \tanh(x)$, such a function is always positive, strictly increasing and convex. Let $h$ be a real number strictly positive, and let also $0 \leq y < + \infty$, in the interv...
Let $f \in C^{\infty}([0,+\infty])$ such that.
$f(0) = 0$, $f(x) \geq 0$ for all $x \in (0,+\infty)$, $f(+\infty) = 1$
$f'(x) > 0$ for all $x \in (0,+\infty)$, $f'(0) = 1$
$f''(x) < 0$ for all $x \in (0,+\infty)$, $f''(0) = f''(+\infty) = 0$
There's a unique $x_0 \in (0,+\infty)$ such that $f''...
[Integral symmetries] The integral: $$\int \sin x e^x dx$$ obeys the following integro-differential equation: $$\frac{1}{2}(-\cos x e^x)"=\int \sin x e^xdx$$ Proof is easy by starting with integration by parts two times in the general case, then substitute the symmetries for $\cos x$ and $e^x$ under integration
$$\begin{array}{c|c|c|c} a \mathcal R a & a \mathcal R b \implies b \mathcal R a & a \mathcal R b \land b \mathcal R c \implies a \mathcal R c & \mathcal R \\\hline ✘&✓&✘&\ne \end{array}$$
$$\begin{array}{c|c|c|c} a \mathcal R a & a \mathcal R b \implies b \mathcal R a & a \mathcal R b \land b \mathcal R c \implies a \mathcal R c & \mathcal R \\\hline ✘&✓&✘&\ne \\\hline ✓&✘&✓&\le,\ge \\\hline ✘&✘&✓&\lt,\gt \end{array}$$
$$\begin{array}{c|c|c|c} a \mathcal R a & a \mathcal R b \implies b \mathcal R a & a \mathcal R b \land b \mathcal R c \implies a \mathcal R c & \mathcal R \\\hline ✘&✘&✓&\lt,\gt \\\hline ✘&✓&✘&\ne,\perp \\\hline ✓&✘&✓&\le,\ge \end{array}$$
@Alessandro You should get something like $(-1)^{k\ell}$, but I believe that you can work it out
Note that this in general implies $\omega_1 \wedge \omega_2 = (-1)^{k\ell} \omega_2 \wedge \omega_1$ where $\omega_1, \omega_2$ are $k$- and $\ell$- alt. mult. forms.
Can someone help me out with this probability question? There is a group of $9$ males and $6$ females. How many ways are there to choose the president and secretary? Would the correct approach be $15*14$ which then equals to $210$?
To fire a stupid question at you: can you explain why geodesics have $\omega_{12} = 0$ along them, given a Darboux frame? Intuitively it's just because the tangent field ($e_1$) is parallel along so change of it shouldn't have an $e_2$-components.
But why really?
I taught him the abstract vector space story. Next up is differential forms on manifolds/R^k.
Maybe I'll introduce them through vector calculus first.
When you say along, you mean restricted to ... But your explanation is exactly right. If $e_1$ has no tangential twisting along the curve, that means precisely that $\omega_{12}(e_1)$ has only an $e_3$ component.
https://en.m.wikipedia.org/wiki/Koenigs_function
So the Koenigs function satisfies $f(a(x)) = a f(x)$.
But when i take the derivative i get the wrong result !?
Let $^{[*]}$ denote composition.
$$ \frac{d}{dz} \lim_{n \to \infty} a^z a^n f^{[n]}(x) =^? \lim_{n\to \infty} ln(a) a^z a^n f^{[n]}...
Background
I am a college pass out. I have done my B.Tech in computer science and masters in quantum information. I am soon going to take up an industrial job. In the free time I got, after passing out from my college till now, ( few months ) I started reading Sipser's book on "Complexity and C...
