ljr_syn's Friends

[Most Recent Entries] [Calendar View] [Friends View]

Below are the most recent 25 friends' journal entries.

[ << Previous 25 ]
 Friday, March 23rd, 2018 lj_scholar_vit 1:59a Хорошая мнемоника Я всегда забывал, какие ошибки первого рода, а какие второго. Сегодня узнал хорошее мнемоническое правило. Крестьяне в басне про мальчика, который кричал "Волки!", сначала делали ошибку первого рода, а потом ошибку второго рода — в правильном порядке.This entry was originally posted at https://scholar-vit.dreamwidth.org/532473.html. Please comment there using OpenID.comments syn_mathoverflo 3:34a Positive root of a polynomial Is anything in general known about the positive root $x_{+}$ of the following polynomial of degree $(n+1)$, as a function of the coefficients $\alpha_{1}, \alpha_{2}, ..., \alpha_{n} > 0$: $$\displaystyle\sum_{i=1}^{n}\left(1 - x^{2}\alpha_{i}\right)\prod_{j=1\\j\neq i}^{n}\left(1 + x\alpha_{j}\right) = 0.$$ Is this polynomial well-known? At this point, I know that $x_{+}$ is unique, but would like to know its dependence on $\alpha_{i}>0, i=1,...,n$. Of course, I can employ root-finding algorithms, but currently my focus is understanding the qualitative nature of $x_{+}$. I could not even tackle $n=2$ case. Please tag as appropriate. syn_slashd_hard 1:30a South Korea To Shut Off Computers Past 19:00 Hours To Stop People Working Late dryriver shares a report from the BBC: The government in South Korea's capital is introducing a new initiative to force its employees to leave work on time -- by powering down all their computers at 20:00 on Fridays. It says it is trying to stop a "culture of working overtime." South Korea has some of the longest working hours in the world. Government employees there work an average of 2,739 hours a year -- about 1,000 hours more than workers in other developed countries. The shutdown initiative in the Seoul Metropolitan Government is set to roll out across three phases over the next three months. The program will begin on March 30, with all computers switched off by 20:00. The second phase starts in April, with employees having their computers turned off by 19:30 on the second and fourth Friday that month. From May on, the program will be in full-swing, with computers shut off by 19:00 every Friday. According to a SMG statement, all employees will be subjected to the shutdown, though exemptions may be provided in special circumstances. However, not every government worker seems to be on-board -- according to the SMG, 67.1% of government workers have asked to be exempt from the forced lights-out. Earlier this month, South Korea's national assembly passed a law to cut down the maximum weekly working hours to 52, down from 68.' Read more of this story at Slashdot. syn_mathoverflo 3:00a Are there Type III codes with small but nonzero "index"? Recall that a Type III code of rank $r$ is a linear subspace $C \subset \mathbb F_3^r$ which is self-dual for the standard inner product. (These occur only when $r$ is divisible by $4$.) Elements of $C$ are called code words. The Hamming weight of a code word is its number of non-zero entries. I will call a code word maximal if its Hamming weight is $r$. (Maximal codewords clearly only occur when $r$ is divisible by $12$.) The set of maximal codewords can be partitioned into two subsets as follows: two maximal words $w_1,w_2$ are in the same subset if $w_1-w_2$ has even Hamming weight, and are in opposite subsets if $w_1-w_2$ has odd Hamming weight. Call these subsets $M_+$ and $M_-$. I suggest the following definition: Definition: The index of $C$ is $|M_+| - |M_-|$. Of course, this is only defined up to sign. I could take its absolute value if I cared. The name is because this is related to the "supersymmetric index" of a certain supersymmetric field theory. The discussion in the comments of this question implies the following when $r$ is divisible by $24$: Proposition: The index of any Type III code of rank $r=24k$ is divisible by $24$. Conjecture: This is true also when $r \equiv 12 \mod 24$. (It is trivially true for $r \equiv \pm 4 \mod 12$, as then $M_\pm$ are empty.) Moreover, I suspect that both $|M_+|$ and $|M_-|$ are necessarily divisible by $24$. Example: The Ternary Golay code, with rank $12$ has index $24$. It is the unique rank-$12$ code with non-zero index. Example: The complete classification of Type III codes of rank $24$ is known. Assuming I read it correctly, there are precisely five codes of rank $24$ and index $24$. Question: In higher rank, do there exist Type III codes with index exactly $24$? For example, what about rank $36$? At best, there would be some general algorithm that produces a code with index $24$ for each rank $r = 12k$. By the way, I know how to prove: Proposition: If the code $C$ contains words of Hamming weight $3$, then its index vanishes. So if you are looking for such a code, you know not to look at such codes. It's pretty easy to show: Lemma: The index multiplies when you take the direct sum of codes. Since $24^2$ is pretty big, you probably will want to work with indecomposable codes. syn_mathoverflo 3:00a Generating prime knots (in order) In this really cool paper https://arxiv.org/abs/1612.03368, A. Malyutin shows that the probability that a random prime knot of up to $N$ crossings (as $N$ goes to infinity) is not generically hyperbolic. Now, the question is, how would one go about generating prime knots, either in order of crossing number (preferred) or in some randomized way? syn_mathoverflo 3:00a a necessary and sufficient condition for a space curve to lie on a ellipsoid Since any (arc-length parametrized) space curve is uniquely determined (up to rigid motions) by its curvature  and its torsion . Show a necessary and sufficient condition for a space curve to lie on a ellipsoid(a equation of its curvature and torsion) syn_ycombinator 1:09a Investigation Confirms IBM Layoffs Targeted Older Workers Comments syn_ycombinator 12:41a Charles Lazarus, the founder of retail chain Toys ‘R’ Us, has died Comments syn_potsnew 1:06a Упитанные какие Угадай город по фото lj_kincajou 1:27a Очень хорошая новость 22 марта вечером также стало известно, что из больницы выписали полицейского, оказывавшего Скрипалям первую помощь.Сержант полиции Солсбери Ник Бейли пробыл в больнице около двух недель. Некоторое время он находится в тяжелом состоянии.А хорошая она потому, что некоторым образом противоречит страшилкам про "Новичьок" syn_dahr 12:19a Моя лекция на ютубе. Так, вот я и залил запись своей лекции на ютуб. Сама лекция вышла не плохой, так что я не жалею что мы её записали. Она была, так сказать, вводная и обзорная. Я бегло охватил широкий круг базисных комиксных вопросов (чему в немалой степени способствовала помощь зала). При этом я не стал править при монтаже мелкие неточности, лишь снабжая их, в самых необходимых моментах, комментариями. Все таки это живая запись живой лекции, а не специально сделанный под запись видеоролик. Однако поскольку на меня налетели братья с криками “лекция должна быть со слайдами!!!”, я добавил в видео немного “слайдов”. Залил все два с половиной часа одним куском. Не хилая получилась полнометражка, но смотреть её членённой и обсуждать кусками, нет никакого смысла. Спасибо Михаилу, это он подорвался все это записывать, и записал таки. Есть ему ещё куда развиваться как оператору.) Но без него бы этой записи не было вовсе! Спасибо добрым властям библиотеки 227. Уютнейшая площадка! Неоценимая помощь, в том числе и непосредственно во время выступления! Спасибо друзьям что пришли. Без них в зале и их дополнений с мест, так как надо не получилось бы. Спасибо слушателям, что не поленились прийти ( и отдельно спасибо за подаренные лакомства)). Ну вот вроде и всё. Ах да, ещё само видео. syn_slashdot 1:30a South Korea To Shut Off Computers Past 19:00 Hours To Stop People Working Late dryriver shares a report from the BBC: The government in South Korea's capital is introducing a new initiative to force its employees to leave work on time -- by powering down all their computers at 20:00 on Fridays. It says it is trying to stop a "culture of working overtime." South Korea has some of the longest working hours in the world. Government employees there work an average of 2,739 hours a year -- about 1,000 hours more than workers in other developed countries. The shutdown initiative in the Seoul Metropolitan Government is set to roll out across three phases over the next three months. The program will begin on March 30, with all computers switched off by 20:00. The second phase starts in April, with employees having their computers turned off by 19:30 on the second and fourth Friday that month. From May on, the program will be in full-swing, with computers shut off by 19:00 every Friday. According to a SMG statement, all employees will be subjected to the shutdown, though exemptions may be provided in special circumstances. However, not every government worker seems to be on-board -- according to the SMG, 67.1% of government workers have asked to be exempt from the forced lights-out. Earlier this month, South Korea's national assembly passed a law to cut down the maximum weekly working hours to 52, down from 68.' Read more of this story at Slashdot. syn_sciencenews 1:00a Does your kids’ DNA matter more than which school they go to? How well your kids do at school depends in part on the DNA you bequeathed them. What’s not clear is what we should do about this syn_slashd_hard 12:50a Experts Say Video of Uber's Self-Driving Car Killing a Pedestrian Suggests Its Technology May Have Failed Ever since the Tempe police released a video of Uber's self-driving car hitting and killing a pedestrian, experts have been racing to analyze the footage and determine what exactly went wrong. (If you haven't watched the video, you can do so here. Warning: it's disturbing, though the actual impact is removed.) In a blog post, software architect and entrepreneur Brad Templeton highlights some of the big issues with the video: 1. On this empty road, the LIDAR is very capable of detecting her. If it was operating, there is no way that it did not detect her 3 to 4 seconds before the impact, if not earlier. She would have come into range just over 5 seconds before impact. 2.On the dash-cam style video, we only see her 1.5 seconds before impact. However, the human eye and quality cameras have a much better dynamic range than this video, and should have also been able to see her even before 5 seconds. From just the dash-cam video, no human could brake in time with just 1.5 seconds warning. The best humans react in just under a second, many take 1.5 to 2.5 seconds. 3. The human safety driver did not see her because she was not looking at the road. She seems to spend most of the time before the accident looking down to her right, in a style that suggests looking at a phone. 4.While a basic radar which filters out objects which are not moving towards the car would not necessarily see her, a more advanced radar also should have detected her and her bicycle (though triggered no braking) as soon as she entered the lane to the left, probably 4 seconds before impact at least. Braking could trigger 2 seconds before, in theory enough time.) To be clear, while the car had the right-of-way and the victim was clearly unwise to cross there, especially without checking regularly in the direction of traffic, this is a situation where any properly operating robocar following "good practices," let alone "best practices," should have avoided the accident regardless of pedestrian error. That would not be true if the pedestrian were crossing the other way, moving immediately into the right lane from the right sidewalk. In that case no technique could have avoided the event. The overall consensus among experts is that one or several pieces of the driverless system may have failed, from the LIDAR system to the logic system that's supposed to identify road objects, to the communications channels that are supposed to apply the brakes, or the car's automatic braking system itself. According to Los Angeles Times, "Driverless car experts from law and academia called on Uber to release technical details of the accident so objective researchers can help figure out what went wrong and relay their findings to other driverless system makers and to the public." Read more of this story at Slashdot. syn_ycombinator 12:06a Luck and startups: Our journey Comments syn_ycombinator 12:15a What is Mastodon? [video] Comments syn_slashdot 12:50a Experts Say Video of Uber's Self-Driving Car Killing a Pedestrian Suggests Its Technology May Have Failed Ever since the Tempe police released a video of Uber's self-driving car hitting and killing a pedestrian, experts have been racing to analyze the footage and determine what exactly went wrong. (If you haven't watched the video, you can do so here. Warning: it's disturbing, though the actual impact is removed.) In a blog post, software architect and entrepreneur Brad Templeton highlights some of the big issues with the video: 1. On this empty road, the LIDAR is very capable of detecting her. If it was operating, there is no way that it did not detect her 3 to 4 seconds before the impact, if not earlier. She would have come into range just over 5 seconds before impact. 2.On the dash-cam style video, we only see her 1.5 seconds before impact. However, the human eye and quality cameras have a much better dynamic range than this video, and should have also been able to see her even before 5 seconds. From just the dash-cam video, no human could brake in time with just 1.5 seconds warning. The best humans react in just under a second, many take 1.5 to 2.5 seconds. 3. The human safety driver did not see her because she was not looking at the road. She seems to spend most of the time before the accident looking down to her right, in a style that suggests looking at a phone. 4.While a basic radar which filters out objects which are not moving towards the car would not necessarily see her, a more advanced radar also should have detected her and her bicycle (though triggered no braking) as soon as she entered the lane to the left, probably 4 seconds before impact at least. Braking could trigger 2 seconds before, in theory enough time.) To be clear, while the car had the right-of-way and the victim was clearly unwise to cross there, especially without checking regularly in the direction of traffic, this is a situation where any properly operating robocar following "good practices," let alone "best practices," should have avoided the accident regardless of pedestrian error. That would not be true if the pedestrian were crossing the other way, moving immediately into the right lane from the right sidewalk. In that case no technique could have avoided the event. The overall consensus among experts is that one or several pieces of the driverless system may have failed, from the LIDAR system to the logic system that's supposed to identify road objects, to the communications channels that are supposed to apply the brakes, or the car's automatic braking system itself. According to Los Angeles Times, "Driverless car experts from law and academia called on Uber to release technical details of the accident so objective researchers can help figure out what went wrong and relay their findings to other driverless system makers and to the public." Read more of this story at Slashdot. syn_mizantrop 12:46a О том, что нужнее всего Вытрезвитель для футбольных фанатов оборудуют в Ростове comments syn_compilers 2:00a Re: Regular expression string searching & matching On 2018-03-20, Clint O wrote:> [ reposted to try to make the special characters look right ]>> q0: /Â·[*]Â·([^*] | [*]+Â·[^/])*Â·[*]+Â·/> [/] q2> ['\x00'-.0-Ã¿] q1> q1: â syn_mathoverflo 1:51a Characterization of a meromorphic function as arithmetic zeta function I'd like to know if there is a (conjectural) criterion for a meromorphic function on $\mathbb{C}$ to be the zeta function of an arithmetic scheme, i.e., a statement of the form "If a meromorphic function on $\mathbb{C}$ has the following properties, then it is the zeta function of a regular, irreducible scheme of finite type over $\mathbb{Z}$ of dimension n." One kind of criterion would say that if the function has a Dirichlet series with appropriate Euler product and condition on growth of its terms, an appropriate functional equation, zeros and poles at expected locations etc., then it comes from an arithmetic scheme. This would be an analog of the converse theorems of Hecke, Weil et al. for automorphic forms which, of course, are relevant to the question due to their conjectural association with arithmetic schemes via Galois representations. An extension of the question would ask for a characterization of the set of all arithmetic schemes whose zeta functions are equal to a given meromorphic function. I should perhaps pose the question for varieties over finite fields first where an answer is more likely to exist, or is even obvious, in view of Weil conjectures. syn_mathoverflo 1:51a Is there a dense subset on closed Jordan curve $C$ which,,, Is there a dense subset $A$ of arbitrary given closed Jordan curve of $C$ on a plane which for each point member of $A$ like $p$ we have the following property: If we rotate $C$ around $p$ by $90^{\circ}$ clockwise on the plane then the created curve $C'$ has no intersection with $C$ exept on $p$. What if we change $90^{\circ}$ by another degree? syn_mathoverflo 1:51a Minuscule weights of parabolic sub-root systems are not far from dominant Let $\Phi$ be a crystallographic root system in an $n$-dimensional Euclidean vector space $(V,\langle\cdot,\cdot\rangle)$. For a root $\alpha\in \Phi$ we use $\alpha^\vee := \frac{2}{\langle \alpha,\alpha\rangle}\alpha$ to denote the corresponding co-root. Suppose that the simple roots of $\Phi$ are $\alpha_1,\ldots,\alpha_n$. For $I\subseteq[n]$, let $\Phi_I$ denote the corresponding parabolic sub-root system, i.e., $\Phi_I := \Phi\cap \mathrm{Span}_{\mathbb{R}}\{\alpha_i\colon i\in I\}$. Use $\Phi^+_I$ for the positive roots of $\Phi_I$ (with respect to the choice of simple roots $\{\alpha_i\colon i\in I\}$). Claim: Let $I\subseteq [n]$. Let $u = \sum_{i\in I}a_i\alpha_i$ be an element of $\mathrm{Span}_{\mathbb{R}}\{\alpha_i\colon i\in I\}$ which is zero or a minuscule weight of $\Phi_I$, i.e., which satisfies $\langle u,\alpha^\vee\rangle \in\{0,1\}$ for all $\alpha\in \Phi^+_I$. (Note crucially that $u$ need not be an integral weight of $\Phi$.) Then $\langle u,\alpha_j^\vee\rangle > -2$ for all $j\in[n]$. I know that the above claim is true thanks to a careful analysis of the classical types together with exhaustive computation for the exceptional types. Question 1: Is there a conceptual, uniform proof of the above claim? Now define the quantity $$M_{\Phi} := -\mathrm{min}\{\langle u,\alpha_j^\vee\rangle\colon I\subseteq [n], u=\sum_{i\in I}a_i\alpha_i \textrm{ is zero or a minuscule weight of \Phi_I},j\in[n]\}.$$ For example, here is the value of $M_\Phi$ for all irreducible root systems, together with the minimizing $I$ and $j$, following Bourbaki's numbering of the nodes of the Dynkin diagram: $\Phi = A_{2m+1}$: $M_{\Phi}= \frac{2m}{m+1}$ with $I=\{1,\ldots,m,m+2,\ldots,2m+1\}$ and $j=m+1$; $\Phi = A_{2m}$: $M_{\Phi}= \frac{m}{m+1}+\frac{m-1}{m}$ with $I=\{1,\ldots,m,m+2,\ldots,2m\}$ and $j=m+1$ (or also the symmetric $I=\{1,\ldots,m-1,m+1,\ldots,2m\}$ and $j=m$); $\Phi = B_n$: $M_{\Phi}= \frac{2n-2}{n}$ with $I=\{1,\ldots,n-1\}$ and $j=n$; $\Phi = C_n$: $M_{\Phi}= \frac{2n-3}{n-1}$ with $I=\{1,\ldots,n-2,n\}$ and $j=n-1$; $\Phi = D_n$: $M_{\Phi}= \frac{2n-5}{n-2}$ with $I=\{1,\ldots,n-3,n-1,n\}$ and $j=n-2$; $\Phi = G_2$: $M_{\Phi}= \frac{3}{2}$ with $I=\{2\}$ and $j=1$; $\Phi = F_4$: $M_{\Phi}= \frac{11}{6}$ with $I=\{1,2,4\}$ and $j=3$; $\Phi = E_6$: $M_{\Phi}= \frac{11}{6}$ with $I=\{1,2,3,5,6\}$ and $j=4$; $\Phi = E_7$: $M_{\Phi}= \frac{23}{12}$ with $I=\{1,2,3,5,6,7\}$ and $j=4$; $\Phi = E_8$: $M_{\Phi}= \frac{59}{30}$ with $I=\{1,2,3,5,6,7,8\}$ and $j=4$. Question 2: Is there a uniform, root-theoretic formula for $M_{\Phi}$? EDIT: For context on the significance of this question, please see https://arxiv.org/abs/1803.08472, especially Remark 2.11. The inequality in the above claim is closely related to a certain integrality property of slices of $W$-permutohedra. syn_ycombinator 12:04a The FBI Says It Can't Find Hackers to Hire Because They All Smoke Pot (2014) Comments syn_slashdot 12:10a KeepVid Site No Longer Allows Users To 'Keep' Videos An anonymous reader quotes a report from TorrentFreak: For many years, KeepVid has been a prime destination for people who wanted to download videos from YouTube, Dailymotion, Facebook, Vimeo, and dozens of other sites. The web application was free and worked without any hassle. This was still the case earlier this month when the site advertised itself as follows: "KeepVid Video Downloader is a free web application that allows you to download videos from sites like YouTube, Facebook, Twitch.Tv, Vimeo, Dailymotion and many more." However, a few days ago the site radically changed its course. While the motivation is unknown at the time, KeepVid took its popular video download service offline without prior notice. Today, people can no longer use the KeepVid site to download videos. On the contrary, the site warns that using video download and conversion tools might get people in trouble. "Video downloading from the Internet will become more and more difficult, and KeepVid encourages people to download videos via the correct and legal ways," the new KeepVid reads. The site now lists several alternative options to enjoy videos and music, including Netflix, Hulu, Spotify, and Pandora. Read more of this story at Slashdot. syn_rfrl 1:52a Джон Болтон станет советником президента США по нацбезопасности Бывший посол США в ООН Джон Болтон заменит генерала Макмастера в должности помощника президента США по национальной безопасности
[ << Previous 25 ]