Assume that $g := f' + f'''$
has only finitely many zeros in ${\bf R} / 2\pi{\bf Z}$.
Then it has an even number of sign changes. We show that there must be
more than $2$, and thus that there are at least $4$ sign changes
in each period interval, so a fortiori at least $4$ zeros.
Vladimir alrea...
We show that there must be more than 2, and thus that there are at least 4 sign changes in each period interval, so a fortiori at least 4 zeros. why is that even true ?
@Anthony do you have the book by Arthur Engel .. Problem Solving Strategies ? problems of this kind are called tiling problems/coloring problems .. you should find detailed explanations of how the coloring proof works in that book (chapter - coloring proofs)
Hi all! Just a quick question. Under the Reproducing Kernel Hilbert Spaces content. We say that the (positive definite) kernel $k\colon\mathcal{x}\times\mathcal{x}\to\mathbb{R}$ produces the entire RKHS, $\mathcal{H}$, and thus we write: $$\mathcal{H} = \overline{\operatorname{span}\{k(\mathbf{x},\cdot) | \mathbf{x}\in\mathbb{R} \}}$$, where the overbar denotes the closure of the perspective space. I Would like to give a brief explanation of the "closure". Could you help me? Thanks!
Should I explain briefly the closure notation using Cauchy convergence? What would you write after my original post, in order to explain in brief what you mean by closure, granted that the reader does not know exactly what you mean? Thanks a lot!
@nullgeppetto You want to explain it to people who presumably have no familiarity at all with topological concepts like closure? I guess then saying that everything in $\mathcal{H}$ is the limit of linear combinations of $k(\mathbf{x},\cdot)$ is the best option, presumably they have at least a vague knowledge of limits.
I've completely revisted all levels, with all the feedback I got, thanks for all your guys help. You can find it here: http://euclidthegame.org/Level1.html
@DanielFischer, first, thanks a lot! Then, no, there is no need to be so vague. I could be more rigorous, but I am not sure whether I can do so! So, besides your nice brief explanation, how could I be a bit more rigorous? Or, on the other hand, how would you explain it including all your knowledge about? :) Thanks again!
@nullgeppetto Not sure. If people know some topology, the notation $X = \overline{Y}$ tells them everything at a glance, and there's no need to explain anything (well, it's good to say something like "the $k(\mathbf{x},\cdot)$ have dense span" to draw attention to it). Otherwise, you could go into explaining what a closed subset of a metric space is (after explaining what a metric space is), and tell them that the closure of a set is the smallest closed subset containing it. But
unless you can devote several lectures to a short introduction to basic topology, I think that would be counter-productive.
I mean, telling them that every point in $\mathcal{H}$ can be approximated by linear combinations of kernel slices isn't vague, what may be vague is their understanding of limits.
This follows directly by applying Sylvester-Gallai Theorem and inversion.
Consider any collection of $n+1$ points $\{ P_1, P_2, \ldots P_{n+1}\}$. Fix $P_{n+1}$, and apply inversion (with respect to a unit circle) to the remaining $n$ points to obtain $\{Q_1, Q_2, \ldots Q_{n}\}$. Note that $Q_{...
I don't get how applying the inversion a second time gives us the result on concylic points
It has a closed form. It is part of the evaluation of $$\int_0^1 \frac{\operatorname{(Li_2}(x))^2}{x} \ dx$$ that is related to an open problem I'll solve soon.
4
@skullpatrol Hello! How are you doing? :-)
It should be clearly emphasized that problem posted by Ryan is very important.
@DanielFischer I'm sorry -- does it annoy you if I bother you in comments to other people's questions?
I also like the smell of books. Except some books smell like puke. I wonder what the publisher was thinking when they decided to use ingredients that smell like puke. Bleh.
@MattN. If you exaggerate it, it will. But not so far. In your answer, you have the small problem that the $S_k = f^{-1}(B_k)$ are not disjoint for balls $B_j$ covering the essential range. You need disjointness for $\sum c_k\chi_{S_k}$ to be a good approximation. Replace the balls with $C_k = B_k \setminus \bigcup\limits_{j < k} B_j$ to get disjointness.
@DanielFischer Thank you for your comment. Right, I didn't realise. Of course if they are not disjoint the equality in the displaystyle equation won't hold. (I'm planning to not exaggerate it.)
I'm doing some work on web site readability, and for that purpose I have developed a web crawler that crawls n number of pages and produce an overall score for the enitre web site. Since the number of pages for each web site varies and is unknown, how can I best estimate a sample size for how many pages it is reasonable to crawl?
@G.T.R he doesn't mention much about that, if i recall correctly. he just states the connection between the convergence of that and the irrationality measure of pi
you can search the paper up, it's online i believe.
@DanielFischer Btw, did you see this question? Looks like something you could answer. It sounds like undergraduate real analysis so I suppose it's low hanging fruit for you?
@TedShifrin Ah, I noticed I've overcomplicated things with 1. The function $x/4+4/x$ is 0.75-Lipschitz over $[2,\infty($. Picard fixed point theorem finishes the proof
$\mathbb D \setminus \{\frac 12(\cos t+i\sin t) : 0 \leq t \leq \pi\}$ is conformally equivalent to some annulus. But I dunno how to pass from one to the other.
(I'm interested in more general classes of slit discs, but if I can figure out what the map is for this one I should be able to generalize with some effort.)
If I were some kind of oracle, I would have known that "what does iff stand for" was actually meant to communicate "what is iff short for"
Instead I've just paid close attention to every social interaction for some years and try to incorporate all data into one unified mental model of people.
For a given prime $p$ and integer $a$, let $e_p(a)$ be the exponent of $p$ in $a$. We'll need the following property of $e_p$: for any pair $a,b$ of integers, $e_p(a+b)=\min{e_p(a),e_p(b)}$ except possibly when $e_p(a)=e_p(b)$. **How is that proved ?**
@DanielFischer Turns out the algorithm I was going to use required me to solve the Dirichlet problem for my starting domain, but I was only doing this because I wanted to move somewhere where I could solve it pretty easily. Ah well. Time for a beer instead.
So if i haven't made any error, $c_n=\sum\limits_{k=0}^{E(n/2)}\binom{n}{2k}b^ka^{n-2k}$ and $d_n=\sum\limits_{k=0}^{E((n-1)/2)}\binom{n}{2k+1}b^ka^{n-2k+1}\sqrt{b}$
Does this series converge? Root test and ratio test are inconclusive.
