You don't have to do that BTW, that's kinda the point. What I do @Rememberme is write [11- when I start at page 11, then [11-15] when I'm done, then I put a line through it when I've added it to the system on my notes
Let $X$ denote a topological space and suppose that $\mathcal{O}$ is an open cover of $X$. Assume $\emptyset \notin \mathcal{O}.$ (Thanks Niels!) Now make $\mathcal{O}$ into an (undirected) graph as follows:
Vertexes: elements of $\mathcal{O}$.
Edges: we draw an arc between two vertexes iff the...
@DanielFischer Could I ask one question. I know in a first countable space a point is in the closure of a set iff there exists a sequence in the set which converges to that point. I know that this can be restated for an arbitrary topological space by considering nets rather than sequences. But in the original statement with regard to sequences, for an arbitrary topological space, does one of the implications still hold ?
@LucioD If there is a sequence in $A$ converging to $x$, then $x$ belongs to the closure of $A$. That direction holds in all spaces. The other direction fails in general because a sequence just isn't "long enough to get through all neighbourhoods unless it has a tail on the point itself".
@Lucas You cannot. Either it must be stated, or you have to decide which branch of the arcus tangent to use. Typically, if nothing is stated, one assumes the principal branch, and that would give you $\arctan \bigl(-\frac{1}{\sqrt{3}}\bigr) = -\frac{\pi}{6}$.
So we know the angle must - modulo $2\pi$ - lie in $\bigl(-\frac{\pi}{2},0\bigr)$. If you normalise your angles to lie in $[0,2\pi)$, you must add $2\pi$ to the value of the principal branch of $\arctan$ at $-\frac{1}{\sqrt{3}}$.
@Lucas We are given $z = \sqrt{3} + i\cdot (-1)$. Writing it in polar form as $z = r\cos \varphi + ir\sin \varphi$, since $r \geqslant 0$, we get the signs of $\cos \varphi$ and $\sin\varphi$ from the signs of the real and imaginary part of $z$.
@Danielfischer For the proof of the first implication is the following fine: Assume that there is a sequence in $A$ converging to $x$ but $x \notin cl(A)$. Hence $x$ is neither in $A$ nor a limit point of $A$. Since $x$ is not a limit point, there exists some neighbourhood $U$ of $x$ such that $U \cap A \setminus \{ x\} = \emptyset$, but this contradicts the assumption that $x_{n} \rightarrow x$, where $\{x_{n}\} \subset A$ unless $x_{n} = x$ for all $n \in \mathbb{N}$, but we also have that $x \notin A$, hence there is a contradiction. Therefore $x \in cl(A)$.
@SohamChowdhury I let my depression eat away at my academics, and flunked my classes last semester. Parents are none too happy, but I'll be able to go back to UT in the spring if it all pans out.
if you haven't gotten into counseling/therapy services, I -strongly- recommend it. i avoided it for a long time, and that's made it all the harder to sort through my various issues.
@SohamChowdhury Besides working on the transfer, ring theory and differential geometry to keep my math sharp. I'll likely see a counselor. Maybe get involved in extracurriculars, do something politically motivated or something. Galois had already been in jail for being a radical by the time he was my age!
(Granted, he was dead by the time he was my age. Still.)
I don't intend to put it off any longer, @Semiclassical. Family and friends have been recommending it to me as well. It's been hitting me in the guy for about a year, it's time I worked it all out with someone.
@Fargle Ah. I've finally reached the stage where there's soooo much (math-related, otherwise I was born like that) math I'm curious about and want to know that it's almost frightening. (Well, that's probably not the right word . . .)
@SohamChowdhury I'm right there with you on that. I have literally no clue how I want to specialize. Every single branch is a different type of fascinating.
@DanielFischer For the proof of the first implication is the following fine: Assume that there is a sequence in $A$ converging to $x$ but $x \notin cl(A)$. Hence $x$ is neither in $A$ nor a limit point of $A$. Since $x$ is not a limit point, there exists some neighbourhood $U$ of $x$ such that $U \cap A \setminus \{ x\} = \emptyset$, but this contradicts the assumption that $x_{n} \rightarrow x$, where $\{x_{n}\} \subset A$ unless $x_{n} = x$ for all $n \in \mathbb{N}$, but we also have that $x \notin A$, hence there is a contradiction. Therefore $x \in cl(A)$.
Are we aware of an elementary way of proving that?
$$\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$
Of course, with the help of Mathematica it can be done, but I wonder if there exists an elementary, simple, easy way of finishing it.
$$\int \frac{...
i'll admit that, whenever i see integrals like that, the first thing I wonder is: "Hmm, I wonder if I could find physics problem where that'd show up" :)
@LucioD You don't have $x_n = x$ for all $n$, only for $n \geqslant n_0$. But mucking around with limit points is suboptimal (I wonder where that fad comes from, using limit points to characterise closedness etc. is rather inconvenient due to the many case distinctions it requires). Use the characterisation that $\overline{A} = \bigl\{ x\in X : \bigl(\forall U \in \mathscr{V}(x)\bigr)\bigl(U \cap A \neq \varnothing\bigr)\bigr\}$.
And then don't assume that there is a sequence in $A$ converging to $x$, that characterisation immediately shows that every sequence in $A$ cannot converge to $x$.
@TedShifrin Still much easier to say that every neighbourhood intersects $A$ or there is a neighbourhood not intersecting $A$. Of course when you do something that actually is about limit points, it's natural to use them.
@DanielFischer But can't I assume that there is a sequence converging to $x$ with $x \notin cl(A)$ and then by contradiction conclude that there can be no such convergent sequence?
i imagine there's some function of $a_k$ that could be plotted so that all of those chopped sections would arrange themselves into a thick straight line. but i haven't managed to find that yet.
@DanielFischer We have by definition that $\overline{A} = \bigl\{ x\in X : \bigl(\forall U \in \mathscr{V}(x)\bigr)\bigl(U \cap A \neq \varnothing\bigr)\bigr\}$. Assume that $x_{n} \rightarrow x$ in $X$, hence for every neighbourhood $U$ of $x$ it follows that there exists a $N$ such that for all $m \geq N$ we have $x_{m} \in U$. It clearly follows that for any neighbourhood $U$ of $x$ we have $U \cap A \neq \emptyset$ since we have $x_{m} \in U \cap A$ for some $m \in \mathbb{N}$. Hence $x \in \bar{A}$.
woops, forgot to be explicit. it's the coefficients of $G(x)$ in the question i linked above, generating using the recurrence relation in Robert Israel's answer
i'm forgetting. i remember there's a way in mathjax to allow additional text to be revealed when a link is clicked, but i can't seem to track down how that works.
@Chris'ssistheartist now I have a Knuth's problem solution v3.0 too !!! :D Just plain integration by parts and series manipulation :D No generating functions :-)
@BenDover and how do you define homology groups without CW complexes? (Are you talking about singular homology like Mike was?) Recall what my original comment was:
one thing I don't like about most topological invariants is that they're first defined on representatives of things (like CW complexes on spaces, or planar diagrams of knots) and then shown the be the same on representatives of the same thing
@Chris'ssistheartist The last series $\displaystyle \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)}$ looks fun! lemme try it :)
I'd take $r = n+k$ and change variables of summation .. (but since this is possibly where it originated from .. I guess there'd be no use in doing that)
@Chris'ssistheartist okay ,, the problem reduces to calculating $\displaystyle \int_0^1 \log^2 \left(1+\frac{x}{2}\right)\,\frac{dx}{x}$ and $\displaystyle \int_0^1 \log^2 \left(1-\frac{x}{2}\right)\,\frac{dx}{x}$ .. then we can kill it by symmetry :-)
@Chris'ssistheartist wait .. I'll try that .. replacing $2$ with some parameter $a$, my odd sense tells me there's something we can do about those on pen and paper
@Chris'ssistheartist In the mean time lemme hog some illicit reputation by posting a solution (with aid of wolfram to compute the two integrals) to the last series :P <- ^ spoken like a true rep rogue =P
@Chris'ssistheartist yes! those people who posts answers that often have very little to do with the query of the OP itself .. just for the sake of earning reputations .. (happens a lot with the series sequence and crazy integral questions :P)
@r9m I sent to more magazines tough series I attended so far, like the ones in Flajolet, but only to the extent I couldn't explore more those tools in the solutions such that I get other results.
In other cases I simply cannot send problems because I use a powerful tool in more problems, so once I reveal it I might lose a lot of stuff to publish.
@r9m let me know if there exists any paper with all series in Flajolet nicely done by real methods only. Since I couldn't find any so far and I looked a lot after them I might think that I'm the first that ever did that. Maybe for some it means nothing, but to me is one of the greatest achievements so far.
The factor theorem states that a polynomial with root x_1 can be represented as $g(x)(x-x_1)$, where g(x) is a polynomial. What is a rigorous way to deduce from this that if $x_1, ... x_k$ are all roots then we can write the polynomial as g(x)(x-x_1)...(x-x_k) for some polynomial g(x)? Does this require proof by infinite descent?
@r9m So, you would approach it differently than how I proceeded? $$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz $$
Start with $f(x)=(x-x_1)g(x)$. Since $g(x)$ has degree $n-1$ and $x_2,\dots,x_k$ are roots of $g(x)$ (why?), it follows by induction hypothesis that $g(x)=(x-x_2)\dots (x-x_k)h(x)$ for some polynomial $h(x)$.
@Simeon: Au contraire. There's no "and so on." The statement I'm proving is that for any polynomial $f(x)$ of degree $n$, whenever $x_1,\dots,x_\ell$ are roots, then $f(x)=(x-x_1)\dots (x-x_\ell) j(x)$ for some polynomial $j(x)$. Induction on $n$ !!
Some dude I don't know asked me to help him solve a problem on GP but left halfway -_- anyway, is there a way easier than what I sketched there ? chat.stackexchange.com/rooms/25366/…
@Chris'ssistheartist not sure I'd call that ugly .. :| seems like it's been known to people how to solve these integrals for a long time (since Lewin's book discusses it .. )
@Chris'ssistheartist he deals with the case without the a,b,c,d etc parameters .. putting them $1$ or $0$ .. in the Trilogarithm chapter page 159 onwards
@r9m yeah, but think one might get that at an exam. How to calculate that? Even the integral where I used the primitive form given in Lewin's book can be approached by Euler sums.
@r9m Hypothetically speaking, of course. If I were a professor I might think to give such questions on a test. Why to let the dust on beauty of mathematics?
@Hippalectryon seriously man!! what's it with you guys?? why are French selection tests are so difficult .. ? what is it they try to prove by making those absurdly difficult test papers?
@r9m It's not absurdly difficult. Which one are you referring to exactly ?
@r9m Keep in mind that it's for competitive exams. The goal isn't to do 100% of the test (though it's made so that the best will be able to), it's to outperform all the others
As a result, I believe that our tests are far more interesting that those of other countries (in physics for instance we see more genuinely interesting topics rather than do some theoretical exercises)
@Hippalectryon well in my opinion most of the test paper I've seen are not really genuinely created problems (rather taken from some old paper/article that discusses the topic) .. that puts people who are aware of the results in a ridiculously advantageous position in comparison to those who have not seen those results .. which in my opinion is unfair. A proper test of merit should challenge the on spot thinking of all students alike irrespective of their knowledge and experience,
@Hippalectryon well I know they are doable .. but my objection is the difference in advantage of knowing a specific results to not being aware of it, created because the problems were taken from some paper available to everyone on internet.
@r9m Papers aren't anything like tests. You won't get a single point (!) from giving the result if the reasoning is wrong. Papers may give the results but they won't develop how they got it in depth at all since it's supposed to be for advanced raeders
@r9m 1) we don't read papers at all 2) once again, you'd need to read tens of thousands of papers and master them.
If you've mastered a lot of papers in different subjects anyway, chances are that you are well prepared for the exams since it's tantamount to studying normally.
q_q do you have an example exam in mind though ? I passed such exams this year, which is why I'm surprised by what you're saying. Sure, the exams are (way) harder than what one ca find elsewhere, but they're far from impossible or favoring such behaviors.
For some reason, and I cannot understand why for the life of me, it behaves totally different on a device than it does when run in the programming simulator.
@r9m Btw about your starred message, also remember that in the prep school in France we don't have 'lectures'. We have one teacher for each subject, the same for the whole year, and we work in classes. All the necessary material (exercices, ...) is given by the teachers, we don't need any additional book.
I had that from earlier this week. I thought it would be evidence it (the code) is working correctly. Though I can't be sure.
Someone told me that maybe the floats I am using are more precise on the simulator than on the device. I doubt it had an effect, but I went ahead and changed it all to doubles. I had to make my own C types for vectors and stuff. It did not affect anything. So I can rule that out I think.
@Hippalectryon I was wondering how my editors will react when they will see the whole paper by Flajolet done by real methods, easily (and far more advanced series). :-)))
@Chris'ssistheartist Yeah, I'm pretty sure they've never heard of Flajolet. Most people (even mathematicians) haven't I think, because only a tiny fraction devote themselves to sums and integrals like that.
The publisher in charge of maths subjects most likely has a broad knowledge of maths in various topics, but no advanced knowledge in a specific topic.
@Hippalectryon Well, I'm sure no editor in the world will tell you how to present the stuff in your book, it's not their job to do that, that's the hardest part I'm afraid of at most.
@Hippalectryon They might make suggestions, new proposal about the whole thing you wanna publish, sure. In the end, it's their decision to publish or not.
@Hippalectryon Agree.
@Hippalectryon I'm not concerned about the publisher, I'm concerned about my way of presenting the problems and the solutions.
One of the objctives is to present in a very easy way solutions to the most craziest problems such that one can understand them entirely, and then fall in love with them. :-)
It's more than presenting problems and solutions, it's that special element that attracts, motivates you to start working on such stuff.
Anyone can put together a bunch of problems and then think to publish them. What's the difference then?
@Hippalectryon Well, the problems are connected to each other, well most of them, this is one of the imporant things that my reader will note throughout my book.
@Hippalectryon Here is another point: when one start mathematics and solve a problem, never let that problem be covered by dust. My book is also about this point.
I finished a problem. Really? I think it's better to say I have begun to use it
@Semiclassical it comes to the same thing as $\int_0^1\log\left(1-xt\right)\log\left(1-yt\right)\frac{\mathrm{d}t}t$, but I have not computed that integral.
@Hippalectryon The results I got so far are only because of my attitude to the work, I'm not a prodigy, a gifted person (like you or others), no, not at all, but I worked extremely hard, and with extremely hard work one can get on top, that's sure.
@robjohn is there a general formula known for integrals of the type $\displaystyle \int_0^1 \frac{\log^2 (1-ax)}{x}\,dx$ .. in terms of $a$ I mean .. the specific values given by $a = 1/2, -1/2, 1/4$ by W|A looks similar to each other in terms of the $\operatorname{Li}_s$ functions appearing atleast
Let $X$ denote a topological space and suppose that $\mathcal{O}$ is an open cover of $X$. Assume $\emptyset \notin \mathcal{O}.$ (Thanks Niels!) Now make $\mathcal{O}$ into an (undirected) graph as follows:
Vertexes: elements of $\mathcal{O}$.
Edges: we draw an arc between two vertexes iff the...
Are we aware of an elementary way of proving that?
$$\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$
Of course, with the help of Mathematica it can be done, but I wonder if there exists an elementary, simple, easy way of finishing it.
$$\int \frac{...
