I got the new Titan X Pascal GPU today installed it, and I am in an infinite login loop. Do I need to install drivers somehow? I have a 4k screen and the resolution it is showing is very low, I think 1024 x 780. Has this happened to anyone else and do you have any ideas?  I also have been trying to install the drivers in Windows, and I have been unable to do so.
I just ordered the Nvidia GTX card. I have a dilemma, though. Should I keep using the driver which is available in "additional drivers" in Ubuntu, or should I install the driver from the Nvidia site?  So which driver is the best for me?
Using Regex to generate Strings rather than match them How to compare two TeX codes and mark differences with color I would like to find easy tool, which compares two TeX codes and generates a PDF file of the first code, in which the differences from the second code are marked by a given color. I use winedt 7.0 under windows 7.  I just noticed that latexdiff provides what I need. As far as I understand it is included in Miktex. Is there an interface (menu commands) in winedt 7.0, which allow to work with latexdiff ? I remember seeing an upper bound for the binomial $\binom{n}{k}$ with an exponential function, something like $\binom{n}{k}\leq \left(ne/k\right)^k$. What exactly is it, and are there other similar good upper bounds for $\binom{n}{k}$?  Edit: As the link in Macavity's comment shows, the bound is indeed $\binom{n}{k}\leq \left(ne/k\right)^k$. How can we prove this? I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks          Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.      Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.  This is where I got stuck. Hope some one can help me solve this. Thanks. Drag and drop not working inside Nautilus on Ubuntu 17.10 I'm having a strange problem on Ubuntu 17.10: when I try to drag and drop something inside Nautilus it simply doesn't work, it doesn't even select the item, it seems like it's ignoring the action... Also, I cannot select things inside nautilus by drawing rectangles with the mouse.  All of these actions work only on the Desktop. Here is a demo:    I'm using the Xorg session with nVidia proprietary drivers. I also noticed that Nautilus seems to hang when I try to copy, paste or cut a file...  EDIT 1: I solved the problem for the slow copy/paste by disabling the clipboard sync plugin of KDEConnect. I also discovered that if I open Nautilus as root, I can drag and drop or select whatever I want... My primary server with SQL Server 2005 broke down, so I've started up the log-shipped database on the secondary server. The database is live on the secondary server, and I just need to remove the log shipping settings for it, while the primary server is no longer accessible.  It seems that the sp_delete_log_shipping_secondary_database stored procedure should be executed on the secondary server to do this, which "removes a secondary database and removes the local history and remote history", as it is stated on .  Would this also remove the database itself? Or only the log shipping settings for it? And what will it do when the primary server is not accessible?  I must not loose the database on the secondary server, as it is already in use. $a$ and $b$ are natural numbers and $p$ is a prime number. Prove that if $a^{p}-b^{p}$ is divisible by $p$, then it is also divisible by $p^{2}$.  My attempt: Based on Fermat's theorem $(a^{p}-a)$ and $(b^{p}-b)$ are divisible by p, therefore their difference is divisible by $p$. I.e. $(a^{p}-a)-(b^{p}-b)=(a^{p}-b^{p})-(a-b)$ is divisible by $p$. Since the left hand side of this equation is divisible by $p$, the right hand side also should be divisible by $p$. Based on the problem's assumption $(a^{p}-b^{p})$ is divisible by $p$ therefore we conclude $(a-b)$ must be divisible by $p$. Now We can factorize $(a^{p}-b^{p})$ $$(a^{p}-b^{p})=(a-b)(a^{p-1}+a^{p-2}b+....+b^{p-1})$$ we have to prove $(a^{p}-b^{p})$ is multiple of $p^2$ which means we have to show the right hand side should be multiple of $p^{2}$. But $(a-b)$ is multiple of p therefore we have to show  $(a^{p-1}+a^{p-2}b+\cdots+b^{p-1})$ is multiple of $p$ and I am stuck here. Any help would be appreciated. Assuming I have no infinite scroll or anything else going on in my theme: Is there a way to enqueue my custom Gutenberg block styles and scripts so they are only (lazy) loaded in the front-end when the block is really used? Something like is_active_widget, just for blocks?  If I enqueue them inside a enqueue_block_assets function they're added to any site.  On github, I found this  , but it sounds like it's more on bundling than conditional load, so I still hope they didn't leave that optimization opportunity out - otherwise any other site will soon be junked with 20 additional scripts from numerous Gutenberg block plugins people install and only use once on a page (think of large image gallery block scripts for example).  Thanks &amp; Regards! Assume a table structure of MyTable(KEY, datafield1, datafield2...).  Often I want to either update an existing record, or insert a new record if it doesn't exist.  Essentially:  IF (key exists)   run update command ELSE   run insert command   What's the best performing way to write this? uname -r and available linux-headers- packages do not match Do I have to pay taxes in the US if my online store sells to US customers even though I don't live in the US? I  own a website and I sell digital content. I'm a nonresident alien with the US. Most of my customers are US based so I was wondering will I have to pay any taxes to the US?  I have read that      For nonresidents, taxable income is separated into two categories:         Income that is effectively connected with a U.S. trade or business, after the deductions that are allowed, is subject to U.S. income tax   at graduated rates - the same rates that apply to U.S. citizens and   residents - or lower rates if the provisions of a tax treaty apply.   Income that is subject to U.S. income tax, for being U.S. source income, but that is not effectively connected with a U.S. trade or   business, is subject to a flat 30% tax      Is this true and do I belong to any of these categories?  If I form an LLC or C corp am I liable for this withholding tax?  My customers are based in the US and I believe all income is considered FDAP income according to the IRS:     Fixed, Determinable, Annual, or Periodical (FDAP) income is all   income, except:         Gains derived from the sale of real or personal property (including market discount and option premiums, but not including original issue   discount)   Items of income excluded from gross income, without regard to the U.S. or foreign status of the owner of the income, such as tax-exempt   municipal bond interest and qualified scholarship income      The thing is the IRS also says:     Effectively Connected Income, after allowable deductions, is taxed at   graduated rates. These are the same rates that apply to U.S. citizens   and residents. FDAP income generally consists of passive investment   income; however, in theory, it could consist of almost any sort of   income. FDAP income is taxed at a flat 30 percent (or lower treaty   rate) and no deductions are allowed against such income. Effectively   Connected Income should be reported on page one of Form 1040NR. FDAP   income should be reported on page four of Form 1040NR.   In the last quote, it’s like they say it only applies to passive income but if they feel like it, they can apply this to all income?? How to make a Copper/Rose Gold smudge material? Zero probability and impossibility I read a comment under :     There are plenty of events that can occur that have zero probability.   This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering    if zero probability and impossibility mean the same? if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?   Thanks and regards! Are there any rules in baseball which prohibit teams from "borrowing" players via trade? Every season near the trade deadline, we see teams that are out of it trading expensive players and/or free agents-to-be, typically for younger players or prospects.  For example, Matt Garza, set to become a free agent after the season, was traded to the Rangers for a few prospects.  It made sense for the Cubs to trade Garza, because he wasn't under contract any further than this season, and thus doesn't have much value to them.  Compare Garza to someone like Mike Trout, who is likely the most valuable property in baseball, given his skill and inexpensive cost over the next several years.  There's no way the Angels would trade him, because he'll be the cost-controlled centerpiece of the team for years to come.  However, with the Angels out of it this year, what if they wanted to trade Trout to a contender (say the Pirates), but only for the remainder of this season?  That is, they'd get several top prospects now, and would have an agreement with the Pirates that they'd trade him back in the offseason (for something pretty worthless, say for a journeyman minor leaguer).  The Pirates get Trout for the stretch run and are happy to give up a few prospects.  Trout doesn't have much value to the Angels this year, but he has a ton of value for next year.  So they get a few prospects for him, in exchange for the part of his career which has very little value to them.  So in effect, the Pirates are renting or borrowing him for the remainder of 2013.  Why don't trades like this happen?  Are there any rules prohibiting trades like this? Preservation of ideals in localized rings Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then  $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$.  We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$. My daughter asked me to cosign on a car loan.  Somehow I ended up as the borrower and she a the cosigner.  Is that legal? How to replace plain URLs with links? PHP Session Security What are some guidelines for maintaining responsible session security with PHP?  There's information all over the web and it's about time it all landed in one place! If  $$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$ then prove that  $$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$  I have no idea how to solve it. Plz help. Let $z \in \Gamma (R)$ where this is the upper semi circle centred at the origin with radius $R&gt;1$. Let $z=x+iy$ with $x \in \mathbb{R}$ and $y \geq 0$. So   $$|e^{i \lambda z}|=|e^{i \lambda z}| |e^{- \lambda y}|= |e^{- \lambda y}|$$  where $\lambda \in \mathbb{R}^{+}$.  How can the last equality possibly be true?