I can't believe that I've just today gotten the bronze contour-integration badge. I thought I'd done a lot of contour integration. Perhaps I used contour integration on questions not tagged contour-integration.
is there a chat room protocol? i mean as part of a comment trail (as in 'continue this discussion in chat'). is the chat meant to be contemporaneous or continued over an extended period of time? (guess my generation.)
@copper.hat There are a set of chat guidelines listed in the sidebar. It is pretty free-form, but use of the comment linking helps sort out conversations when many people are talking.
If you hover over someone else's comment, a bent arrow should show up in the lower right corner. Click on that and a link to that comment should be added to your comment
@robjohn copper is talking about those one-on-one rooms that start when you click "continue this discussion in chat" in a comment thread on the main site
@robjohn: i was wondering about the social protocol aspect. even when i im with long-distance family on say skype, i wonder if i should write 'bye' or just leave it hanging in the air. the latter seems the norm, but my need for old-school hang-up-the-phone closure is strong :-).
@copper.hat sometimes people don't know when others may have to leave the keyboard, so there may be long pauses, or even apparently abrupt interruptions. This seems inevitable in such an asynchronous means of communication such as chat.
@copper.hat people will also enter into a conversation that has ended or paused a long time ago because they read back in the transcript and respond to an old comment.
For example, the comment to which I am replying is 2 hours old :-)
Also, since it may take a while for someone to respond, it is rare that someone will not move to another window to take care of something else. Pinging is the way to get their attention back. Even then, they may have had to leave the keyboard and may not be able to respond for quite a while.
@copper.hat I left the chat, but didn't say "goodbye" when I did. Sometimes it is okay to say "bye" or BBL if you know you are going to be gone for a while, but it is not necessary.
homework has now 28792 questions. So it went down by 400 questions in two days. If all those questions were bumped, than some complaints are certainly justified.
Hey, @Chris'ssis I conjectured a limit and I would like your help to prove/disprove it :-) it is $$\lim_{\sigma\to 0} \frac{1}{\sigma} \int_{-\infty}^{\infty} e^{-x^2/2 \sigma^2} \arcsin \left(1-2\left|\lfloor x\rceil-x\right|\right)\,dx= \sqrt{\frac{\pi}{2}}$$ (it is related to this question of mine; it would be very nice to have a sweet closed form for that integral)
@Chris'ssis I think I found a probabilistic proof of this fact, but I would like to see a direct one.
@robjohn the other day, someone (maybe Chris's Sister?) was asking you to evaluate limits of sums of things involving binomial coefficients - did you get those?
and if you did, did you do it in chat? (and if so, you don't happen to have a quick link back to where you did it in chat?)
@mixedmath Here is a link to a point in the transcript where I explain what I did. The answers to a couple of Chris's sis's questions are around there.
Hi. Off topic: If $f,g:\mathbb{R} \to \mathbb{R}$ have limits $L,M$, respectively as $x\to a$, then to show that $f+g $ has limit $L+M$, we use the fact that 1) $\vert (f+g) - (M+L) \vert $ can be written as $\vert (f-L) + (g-M) \vert$ and 2) The triangle inequality. However, what if I replace the absolute value with an arbitrary metric $d$? Is there an analogue to 1) in this case?
Probably you would find some analogues in topological groups or topological vector spaces.
If the target space is just a metric space, it is not enough. The metric should be in some sense "compatible" with the binary operation. IIRC continuity of the binary operation should be enough.
This means that $+ \colon \mathbb R\times\mathbb R \to \mathbb R$ should be continuous w.r.t. the metric you are using.
But if you want to discuss this in this generality, I don't see a reason to restrict this to metrics on $\mathbb R$.
I was doing an exercise that has the domain n-dimensional space and codomain $\mathbb{R}.$ Now that I think of it the author probably assumed that we should use the standard distance of $\mathbb{R}$ here.
Say 200 questions get's retagged a day, then it will only take 12 squared days to retag the 28792 homework questions. Thats like only a third of a year and certainty less than or equal to infinity.
@N3buchadnezzar The majority of those questions are in decent shape; no manual retag is needed. The hw tag will be removed silently and automatically, without a bump.
@900sit-upsaday yeah, the problem is the few questions only tagged with homework. But it seems this pool is far smaller than the homework pool. Due to it's nature of being a very vague tag..
@N3buchadnezzar We started with 900+ such questions. A couple hundreds were already retagged or otherwise disposed of. My post here has a SEDE query for them; unfortunately it's updated only once a week.
@5space Yeah, but have you seen self-learning and how it's used?
@N3buchadnezzar Those with hw-only? The median score is zero. The mean is positive, but I'd have to download the CSV fiile and calculate in Excel... if you want to know, you can do it...
@N3buchadnezzar There is a significant survivor bias, of course: negatively scored questions are less likely to stay on the site for long. Survival of the fittest...
@William, when I was a grad student taking a course in analytic number theory, a fellow student stopped the professor in mid-proof to ask about a funny-looking symbol on the blackboard. The answer: "It's a 5.
$F$ has dimension $2$ because it's the $0$-set of a linear form and $G$ has dimension $1$ Since $g$ does not satisfy the equation $x-y+z=0$, $g\notin F$ so that $F\cap G=\{0\}$ and the sum $F+G$ is direct.
Finally $\Bbb R^3=F\oplus G$ because $E=\Bbb R^3$ and $F\oplus G$ have both dimension $3$.
Let $x>1$ then $$ \int_0^x \left\{ \frac{1}{t} \right\}\,\mathrm{d}t = \int_0^1 \left\{ \frac{1}{t} \right\}\,\mathrm{d}t + \int_1^x \frac{1}{t} - 0 \,\mathrm{d}t = 1 - \gamma + \log x $$
If $x = 1$ then $\int_0^x \{1/t\} \,\mathrm{d}t = \int_0^1 \{ 1/t \} \mathrm{d}t = 1 - \gamma$
If $x=0$ then the integral is zero and if $0<x<1$ then $$ \int_0^x \left\{ \frac{1}{t} \right\}\,\mathrm{d}t = \int_0^1 \left\{ \frac{1}{t} \right\}\,\mathrm{d}t - \int_x^1 \left\{ \frac{1}{t} \right\}\,\mathrm{d}t = 1 - \gamma - \int_0^{1/x} \frac{t - \lfloor t \rfloor }{t^2}\,\mathrm{d}t $$
For the last integral my idea was $$ \int_0^{1/x} \frac{t - \lfloor t \rfloor }{t^2}\,\mathrm{d}t = \left( \sum_{n=1}^{\lfloor 1/x \rfloor - 1} \int_{n}^{n+1} \frac{t-n}{t^2}\,\mathrm{d}t\right) + \int_{1/x}^{\lfloor 1/x \rfloor} \frac{t-x}{t^2}\,\mathrm{d}t $$
Does the last one look okay? I ws a bit unsure about the how to split it up and the plusses and minuses. Eg making sure the intervals are alright
In case nobody knew, the cg site is going to shut down (due to various reasons). But we're not giving up. Join the new proposal here: area51.stackexchange.com/proposals/74985/…
Also, i was once told that some people don't like proofs by contradiction because they didn't quite agree with the axiom of choice. However, it seems to me that the latter is not really linked to it, and that what would conflict with it was whether we accept or not the law of excluded middle. Am I wrong ?
Yeah, its tertium non datur that is used for proof by contradiction, not choice. But it's not unreasonable that people who reject choice also reject other sensible axioms, hence their problem with proofs by contradiction may be related to their rejection of choice.
@RandomVariable I have, and I would also not like him to quit. But he's taking things a bit far: "I interpret it as systematic deliberate erasure of my work." Puh-leaze!
@DanielFischer I highly doubt he's being targeted. But considering the number of answers he posts, it might feel like to him that he is being singled out.
@BalarkaSen: You know that $Q_8$ is the smallest hamiltonian group. You can construct $Q_8\times\Z_2$ form it and ... and $Q_8\times\Z_n^k$. Since I am not a native one, so what can I call this process from $Q_8$ to $Q_8\times\Z_2$ and ....and $Q_8\times\Z_n^k$. Of course this way IS NOT an extension since this word has its own certain meaning in Group theory. Can I call it expansion? Or?
They say on wiki that the linear order of cardinal numbers is equivalent to the axiom of choice. Can anyone provide a link to this proof ? Or should I ask on MSE ?
I learned this technique from some guys in a forum. So I thought I might as well write it up.
Personally, I am much fond of my second highly upvoted answer. I came up with it all by myself (though my technique is not quite very formal, unlike the technique of the guy whose answer got accepted)
@DanielFischer Is $f(z)= \frac{1}{\sqrt[n]{1-z^{n}}}$ single-valued on the complex plane if you omit the line segments that extend from the origin to the nth roots of unity?
@RandomVariable Look at $g(w) = \sqrt[n]{w^n-1}$ on $G = \mathbb{C}\setminus \{ r\cdot e^{2\pi ik/n} : r \geqslant 1, 0 \leqslant k < n\}$. $G$ is simply connected, and $w^n-1$ has no zero in $G$, hence there are $n$ branches of $g$ on $G$. Now you can take $f(z) = \frac{1}{z\cdot g(1/z)}$. For each of the branches of $g$ on $G$, you have a single-valued branch of $f$.
Also: $1-z^n$ attains the value $\infty$ with multiplicity $n$ at $\infty$, hence $\infty$ is not a branch point of $\sqrt[n]{1-z^n}$.
(And, adding $\infty$ to your domain, we get a simply connected domain.)
@DanielFischer So then $ \int_{0}^{1} \frac{1}{\sqrt[n]{1-x^{n}}} \ dx$ could be evaluated by using a contour that consists of small circles around the nth roots of unity attached to the origin by two line segments (one below the cut and one above the cut)?
@DanielFischer The case $n=2$ is a standard problem in many textbooks. But someone asked a few days ago about evaluating the general case using contour integration.
@RandomVariable hmm... the integral is $\frac{\Gamma\left(\frac1n\right)^2}{2n\Gamma\left(\frac1{2n}\right)}$. It would need reduce back to something like the Beta integral
I guess the proof will be similar to showing that there are n extensions of an automorphism, a, of F to an automorphism of K that agrees with a on F, where n is the degree of the field extension of K over F.
The above is not quite true, but it's obvious which theorem I'm referring to.
@RandomVariable Ah... I brought the $(1-x)^{-1/n}$ forward as $(1-x)^{1/n}$... Correcting that, I get $$\frac1n\int_0^1x^{1/n-1}(1-x)^{-1/n}\,\mathrm{d}x =\frac1n\mathrm{B}(1/n,1-1/n) =\frac{\pi}{n\sin(\pi/n)}$$
@RandomVariable However, I have a contour integration solution using a contour integral in one of my answers...
I'd have to work on it, @Balarka, and I'm just back (literally) from travelling for 2+ weeks. I have to check if those two quadric hypersurfaces intersect transversely and see if there's a group action. One can use adjunction formulas, etc., to figure out a lot of stuff about complete intersections.
That's the point, @Hippa. That's how addition is defined in the quotient group. For example, let $G=\Bbb Z$ and $H=6\Bbb Z\subset G$. To add $1+H$ and $3+H$ you get $4+H$. This means that if you add any representative of $1+H$ (i.e., any number whose remainder when divided by $6$ is $1$) and any representative of $3+H$, you get numbers whose remainders are $4$ when divided by $6$.
@robjohn I'd still like to see an evaluation using that contour I described. That's what I attempted to do a couple of days ago, and what I think the OP would like to see. But I don't seem to have the value of $f(z)$ above and below the cuts quite right.
@robjohn Speaking of harmonic numbers, is it well known that a closed-form expression for the alternating harmonic numbers (i.e., $H_{n}^{-} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k}) $ is $H_{n}^{-} = \frac{(-1)^{n-1}}{2} \left[\psi \left(\frac{n+2}{2}\right) - \psi \left( \frac{n+1}{2}\right) \right] + \log2 $ ?
$F$ has dimension $2$ because it's the $0$-set of a linear form and $G$ has dimension $1$ Since $g$ does not satisfy the equation $x-y+z=0$, $g\notin F$ so that $F\cap G=\{0\}$ and the sum $F+G$ is direct.
Finally $\Bbb R^3=F\oplus G$ because $E=\Bbb R^3$ and $F\oplus G$ have both dimension $3$.
