Question: sum_k=0^k=n binom(n,k) exp(a k) has a nice closed form, (1 + exp(a))^n. Is there a chance that sum_k=0^k=n binom(n,k) exp(a k + b k^2) would also have a somewhat nice closed form that I could do calculations with?
Does the set {-1,0,1} count as a set with exactly 3 limit points in $\mathbb{R}^1$? Or do I have to come up with some sort of sequence like $1/n$? Just trying to grasp the definition.
@Peter If $S\subset M$, a point $x\in M$ is an adherent point of $S$ if every ball $B_M(x;r)$ contains at least 1 point of $S$. ($x$ may or may not be in $S$).
Try to prove the following: Let $S\subseteq \Bbb R$. Then $x$ is a limit point of $S$ if and only if for each $\epsilon >0$, $B(x,\epsilon)\cap S$ is infinite, that is, there are infinitely many $s\in S$ for which $|s-x|<\epsilon$.
@AlanH Heheh, well. I can write the solution here, it is more about the elements in $S$ rather than $\epsilon$. Would you like that?
First, let's do the following: for $S$ a set of real numbers, define $x$ to be an accumulation point of $S$ if each open ball about $x$ has infinite intersection with $S$. Define limit points as usual. The claim is then that $x$ is a limit point $\iff$ it is an accumulation point. (this is true in more generality for what are called Hausdorff spaces)
Suppose first $x$ is an accumulation point of $S$. Then for each $\epsilon >0$ there exists infinitely many $s\in S$ such that $|x-s|<\epsilon$. Since they are infinitely many, they cannot be all equal, in particular they cannot be all equal to $x$, that is, for some $s'$, $s'\neq x$ so $0<|x-s'|<\epsilon$ and $x$ is a limit point.
Suppose now $x$ is not an accumulation point of $S$. Then there exists an $\epsilon >0$ such that $B(x,\epsilon)\cap S=\{s_1,\dots,s_n\}$ is finite.
We now choose balls $B_i$ of $a$ and $V_i$ of $a_i$ such that $$B_i\cap V_i=\varnothing$$
(This is possible, because $\Bbb R$ is a metric space)
(Thus Hausdorff)
Then $B(x,\epsilon)\cap \bigcap B_i$ is a nbhd of $x$ that contains no points of $S$ other than possibly $x$. So $x$ is not a limit point of $S$.
hmmm......... Fourier transform takes f:R->R an gives out F:R->C. where is the duality. it should give out a function defined on same domain and range strictly speaking, for the duality. right?
are you defining f & F to be 0 to the left of 0? are you saying every f of that form is mapped to an F of that form, or are you just saying there exist a fourier pair of that form?
But if the "set of all statements" existed (including paradoxes), the "set of all true statements" would be a subset of the set of all statements, correct?
once you coe up with a specific definition of what true means, so that you can decide if any given statement is true or false, then yes: the set of true statements is a set
@MarianoSuárez-Alvarez I should really read something about model theory, because I find that Interesting. Hoever, I fear I will have difficulties understanding it if it is scientific enough. I am only in my 2nd year of maths studies
@GustavoBandeira It is. But sometimes it's really a good idea to explain what's wrong with a different, usually older notation. Just throwing unmotivated critique is not of any value.
Let $T$ be a triangle in $\mathbb{C}$ with vertices at $0$, $w_1$, $w_2$. By applying the mapping $z\mapsto \bar{w_2}z$, show that the area of $T$ is $\frac{1}{2}|Im[w_1\bar{w_2}]|$.
but yeah, you have some triangle T, and you apply the map $f(z):=w_2z$ to it - this amounts to stretching and rotating it. the stretching will scale areas of any region by a factor of $|w_2|^2$. the vertices of the resulting triangle will be $0$, $\bar{w_2}w_1$ and $\bar{w}_2w_2=|w_2|^2$, which has area = (1/2)*base*height = $\frac{1}{2}|w_2|^2{\rm Im}(\bar{w}_2w_1)$.
writing $w=re^{i\theta}$ and $z=se^{i\phi}$, we see that $wz=rse^{i(\theta+\phi)}$. thus, the map $f(z):=wz$ applied to the complex plane amounts to stretching everything by a factor of $r$ and rotating anticlockwise by an angle of $\theta$.
I thought this was an interesting question, albeit posed poorly and PSQ to boot. Got downvoted to oblivion, closed, and even on the verge of being deleted.
it seems to me that many are willing to take out collateral damage on interesting mathematics and answerers in order to punish bad question-askers as a vent for frustration with them
@anon Downvoting answers is still a no-go area as far as I'm concerned; but downvoting/closing a question should be fine. There is the possible issue of destroying interesting mathematics by this procedure, of course. This has to be weighed in. Generally, I don't consider things a big loss when the problem discusses a specific case of a nice, general phenomenon.
Of course, in the ideal world, one would compose an abstract question to deal with the general case, provide a nice, conclusive answer to that one and refer to it on the PSQ.
@robjohn My downvotes are scarce because I think closing is a more apt measure against the unwanted behaviour; both will go with a comment (or an upvote to an existing comment). Indeed, I think this is important, so as to ensure that we actually provide the OP with the necessary information.
@robjohn A quick question about site mechanics, if I may: What happens when a question is closed duplicate to a question, which in turn is at a later point also closed duplicate? Will this resolve automagically?
@robjohn No I mean the following situation: A is closed as a duplicate of B. B is subsequently closed as a duplicate of C. Will A now automatically start pointing to C as its duplicate?
I have an statement including a poistive integer $k$. It has two rules depending on $k$, while $k$ is odd and while $k$ is even. Can I proof the statement by induction on k? Thanks
But applied maths is divided into many fields that are best studied at uni so I think it is best to teach pure maths at high school so students have the skill to do well in uni :(
@PeterTamaroff: 2 years ago when I was new here and didin't know much about the privacy of the site, I was used to read the answers of missed Aurtor and made them +.
Suddenly, I found out that, i was making a mistake and I remmeber he was chatting you under one of his answers about the action circling about the answers.
I didn't do that again and give up. I always admire him. and miss him here. I know that from that time.
