This proposal tries to solve   If a new answer to an old question with lots of answers gets a few upvotes, it is a strong indicator that the answer is good:   it is much more recent than the fastest guns, and so had less time to get upvotes (consider how old it is) it is way further down than the fastest guns, so much fewer people have viewed it (consider answer count)   For questions less than 8 short answers or so, it does not matter much to me: my attention span is enough to scan them all.  But for larger questions, it would help. Specially on the "new technology came out that perfectly answers the question" scenario.  One may argue that such questions should be closed as too broad, but let's not touch that question here and suppose that they are valid (I think many are). In any case, as time tends to infinity, the answer count per question also does, so we either deal with it some day, or start closing them at some point.  I know it is not easy to come up with a sensible metric, and I'm not going to propose a precise equation on this question, but please do so in answers / comments if you have an idea.  The most obvious metric is to simply divide upvotes / time, but that has a few downsides:   may give too much importance to new answers that get that single upvote because they showed up on the front page. It would be cool to have a coarser method with a little more statistical significance. does not take into account the "start at bottom" (number of answers) bias   Many similar metrics also take recency into account, e.g. our , Facebook  and Reddit's . Of course, I see that their need is much greater since their updates happen much more frequently than an old questions. But maybe we could take some ideas from those metrics.  Full disclosure: I might get tons of rep if this were implemented :-)
I appreciate the mechanisms which help bring attention to new answers to old questions - front page sorting, and the moderation tool explicitly designed for it. But I wonder if we can do more. I'm particularly interested in addressing answers whose validity has changed. In the worst case, they were correct when they were posted, but no longer are, so they mislead all the current users who find their way to them looking for an answer. More benignly, perhaps the best answer was good, but slightly flawed.  At this point, a helpful user may post a new answer, which is much better than existing upvoted and accepted answers. It can be difficult to overcome that momentum. What if the default sorting method took this into account? A naive suggestion: if a new answer has a large positive vote velocity, and an old answer has a negative vote velocity, perhaps the default sorting should place the new answer above the old answer. This could help make new answers visible to users who need a correct answer, and also help garner more votes to increase the confidence of the sorting decision.  I know this would be a fairly low-impact change, but nothing is permanent, and I'm afraid of answers with 100 upvotes being pinned at the top for years after they've become obsolete.  Edit: I would welcome other proposals that accomplish the same goal - my naive suggestion wasn't intended to be a complete solution, just one thing that a better sorting method would do.
My storage was not updated Ubuntu 18.04 No WiFi adapter found [Dell Inspiron 3542.] Just installed Ubuntu 18.04 but my WiFi adapter Qualcomm Atheros QCA9565 / AR9565 Wireless Network Adapter doesn't work or even appear in the system settings.  output of lspci -knn | grep Net -A3:  &gt; rahul@rahul-Inspiron-3542:~$ lspci -knn | grep Net -A3 06:00.0 Network controller [0280]: Qualcomm Atheros QCA9565 / AR9565 Wireless Network Adapter [168c:0036] (rev 01)     Subsystem: Dell QCA9565 / AR9565 Wireless Network Adapter [1028:020c]     Kernel driver in use: ath9k     Kernel modules: ath9k 07:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8101/2/6E PCI Express Fast/Gigabit Ethernet controller [10ec:8136] (rev 07)   output of lspci -knn | grep Net -A2  rahul@rahul-Inspiron-3542:~$ lspci -knn | grep Net -A2 06:00.0 Network controller [0280]: Qualcomm Atheros QCA9565 / AR9565 Wireless Network Adapter [168c:0036] (rev 01)     Subsystem: Dell QCA9565 / AR9565 Wireless Network Adapter [1028:020c]     Kernel driver in use: ath9k     Kernel modules: ath9k   output of sudo rfkill list  rahul@rahul-Inspiron-3542:~$ sudo rfkill list [sudo] password for rahul:  0: dell-rbtn: Wireless LAN     Soft blocked: yes     Hard blocked: yes 1: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no 2: hci0: Bluetooth     Soft blocked: yes     Hard blocked: no   output of sudo lshw -C network  rahul@rahul-Inspiron-3542:~$ sudo lshw -C network   *-network DISABLED                description: Wireless interface        product: QCA9565 / AR9565 Wireless Network Adapter        vendor: Qualcomm Atheros        physical id: 0        bus info: pci@0000:06:00.0        logical name: wlp6s0        version: 01        serial: 4c:bb:58:30:aa:07        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress bus_master cap_list rom ethernet physical wireless        configuration: broadcast=yes driver=ath9k driverversion=4.17.11-041711-generic firmware=N/A latency=0 link=no multicast=yes wireless=IEEE 802.11        resources: irq:18 memory:f7900000-f797ffff memory:f7980000-f798ffff   *-network        description: Ethernet interface        product: RTL8101/2/6E PCI Express Fast/Gigabit Ethernet controller        vendor: Realtek Semiconductor Co., Ltd.        physical id: 0        bus info: pci@0000:07:00.0        logical name: enp7s0        version: 07        serial: 74:e6:e2:14:2a:c8        size: 10Mbit/s        capacity: 100Mbit/s        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress msix vpd bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd autonegotiation        configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=half firmware=rtl8106e-1_0.0.1 06/29/12 latency=0 link=no multicast=yes port=MII speed=10Mbit/s        resources: irq:19 ioport:e000(size=256) memory:f7800000-f7800fff memory:f2100000-f2103fff   *-network        description: Ethernet interface        physical id: 2        logical name: enp0s20u1        serial: 7a:14:5b:44:b4:96        capabilities: ethernet physical        configuration: broadcast=yes driver=rndis_host driverversion=22-Aug-2005 firmware=RNDIS device ip=192.168.42.109 link=yes multicast=yes   output of dmesg | grep ath9k  rahul@rahul-Inspiron-3542:~$ dmesg | grep ath9k [   23.712505] ath9k 0000:06:00.0 wlp6s0: renamed from wlan0   output of sudo iw dev  rahul@rahul-Inspiron-3542:~$ sudo iw dev [sudo] password for rahul:  phy#0     Interface wlp6s0         ifindex 3         wdev 0x1         addr 4c:bb:58:30:aa:07         type managed         txpower 0.00 dBm   output of dmesg   I already tried so many solutions but nothing works for me , I'm looking for answer from so many days. Please Someone solve my issue . If anyone want any other information please let me know.  Any help would be greatly appreciated. Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) How to type en-dash and em-dash in Windows 7/8? I write blog posts, and I often need to type en-dash (–) and sometimes em-dash (—) in my blog posts. Currently I keep these characters saved in a text file, and copy-paste them whenever I need them. Unnecessary to say, it is a headache to move between keyboard and mouse when you type large texts.  It would be better for me if there ware a keyboard shortcuts available for these two characters. Is there a way to type en-dash and em-dash in Windows? I use Windows 7 and Windows 8 in different computers.    Here's an aligned comparison of dashes, if you need to see the difference for some reason.   Here's-dash Here's–en-dash Here's—em-dash Before Voldemort, who was the last Horcrux creator? Horcruxes are rare (and then seven come along at once...) and only a few people know about them.  But when was the last Horcrux before Voldemort's? 50 years before him? Centuries? Millennia? ubi-partman failed with exit code 10 - ubuntu 18.04 I'm trying to install Ubuntu 18.04 on my laptop. After choosing the partitionn i have the following error message: grub-efi-amd64-signed package failed to install into / target/.    Then I tried to install via "try ubuntu" and i get:  ubi-partman failed with exit code 10     What should I do? Print lines between (and including) two patterns How should I act when I see tiny "Have you tried this..?" type of answers? Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible. Is it correct to hyphenate with compound premodifiers? If so, where is the hyphen placed? Solve $x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$ How does the Ranger's Extra Attack feature work with Two-Weapon Fighting? I think I know the answer to this one but I'm double checking.  At the 5th level the Ranger gets Extra Attack, which lets them attack a second time during the Attack action. If they are fighting with 2 weapons, the second weapon normally uses their bonus action.  Does this mean when using the Extra Attack and using two weapons, a ranger can attack twice during the Attack action, and still make 1 extra attack as a bonus action, for a total of 3 attacks? For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$ I don't understand the proof to the a/m claim. How we know that $\eta &lt; \kappa$ and $(\alpha,\beta)&lt;_{cw} (0, \eta) $ and hence $h: \mu \to \eta \times \eta$ is injective? Appreciate if anyone could advise.  Here is an excerpt of the proof:  For an infinite ordinal $\kappa,$ the canonical well-ordering of $\kappa \times \kappa,$ denoted by $&lt;_{cw}$ is defined as follows: $(\alpha_1, \beta_1)&lt;_{cw} &lt; (\alpha_2, \beta_2)$ iff either one of the following holds.  $(1): \max\{\alpha_1,\beta_1\} &lt; \max\{\alpha_2,\beta_2\}$  $(2): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1&lt;\alpha_2$   $(3): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1 = \alpha_2$ and $\beta_1 &lt; \beta_2.$  Claim: For all infinite cardinals $\kappa, (\kappa \times \kappa, &lt;_{cw}) \cong (\kappa, \in).$  Proof: (It has been show that it is true for $\kappa = \omega$) Assume there is a counterexample  $\kappa.$ Let $A= \{\lambda \in \kappa + 1| \lambda\geq \omega \ is \ a \ cardinal \ and \ (\lambda \times \lambda, &lt;_{cw}) \not\cong (\lambda, \in)\}.$ Then, $\kappa \in A.$ Let $\mu$ be least element of $A.$ Thus, $\mu &gt; \omega$ and for all $\lambda,$ if $\omega \leq \lambda &lt; \mu$ and $\lambda$ is a cardinal, then $(\lambda \times \lambda, &lt;_{cw}) \cong (\lambda, \in).$  By Comparability theorem, $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, &lt;_{cw})$ since $|\mu \times \mu| \geq \mu$ and $\mu$ is cardinal. Let $(\alpha, \beta) \in \mu \times \mu$ such that $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, &lt;_{cw})$ given by $(\alpha, \beta).$ Let $h$ be the isomorphism. Let $\eta= \max\{\alpha, \beta\}+ \omega.$ Then $\eta &lt; \kappa$ and $(\alpha,\beta)&lt;_{cw} (0, \eta).$ HENCE, $h: \mu \to \eta \times \eta$ is injective. Biblatex ieee incorrect display of volume and issue number Is there an infinite amount of wavelengths of light? Is the EM spectrum continuous? The electromagnetic spectrum is a continuum of wavelengths of light, and we have labels for some ranges of these and numerical measurements for many.  Question: Is the EM spectrum continuous such that between two given wavelengths (e.g. 200nm and 201nm) there is an infinite number of distincts wavelengths of light?  Or is there some cut-off of precision with which light might exist (e.g. can light only have wavelengths of whole number when measured in nanometers, etc.)? Many sysv init scripts used a corresponding file in /etc/default to allow the administrator to configure it. Upstart jobs can be modified using .override files. How do I override or configure systemd units, now that systemd is the default in Ubuntu? Are there any coding style guidelines for LaTeX? X Windows special programs: do some of these still exist? What is an ultra-wide lens? What is the point at which a lens is considered "ultra-wide angle" vs just wide angle? How can I check whether a variable is defined in JavaScript? How to check whether a JavaScript variable defined in cross-browser way?  I ran into this problem when writing some JavaScript utilizing FireBug logging. I wrote some code like below:  function profileRun(f) {     // f: functions to be profiled     console.profile(f.constructor);     f();     console.profileEnd(f.constructor); }   It works fine in FireFox/FireBug, but it reports error in IE8 RC1. So, I'd like to do some checking whether console variable exists in the execution environment.  Below code works fine in FireFox, but not in IE8 RC1.   function profileRun(f) {     if (console != undefined) {         console.profile(f.constructor);     }      f();      if (console != undefined) {         console.profileEnd(f.constructor);     } }   However, if I do it this way. It works in IE8 RC1. Why?   function profileRun(f) {     if (window.console != undefined) {         console.profile(f.constructor);     }      f();      if (window.console != undefined) {         console.profileEnd(f.constructor);     } }   Is there any cross-browser way to check it?