I have managed to automatically build an HTML table based on a output from my ERP system, and managed to use CSS to format it easily.  The last step is conditional formatting in a Date Field.  If the date is <  from now make the field RED if it's between 4 <> 7 days from NOW make it YELLOW  else "no formatting.  I have gotten this far on dozens of examples from this site, and I am hoping you can help me do some basic DATE MATH which for some reason I just can't get my head around.    Right now my date field is in  the td:nth-child(4) and I want to compaire it to the current date.      <script>      $(document).ready(function() {              $('table td:nth-child(4)').each(function() {                 var D1 = $(this).text();                 var D2 = new Date();                  if ((D1 - D2) < 4) {                     $(this).css('backgroundColor', 'RED');                  }                 else if((D1 - D2) < 7) && ((D1 - D2) > 4) {                     $(this).css('backgroundColor', 'YELLOW');                  }                 else {                     $(this).css('backgroundColor', '#99faa0');                  }              });             event.preventDefault();     });      </script>

How do I get the number of days between two dates in JavaScript? For example, given two dates in input boxes:  <input id="first" value="1/1/2000"/> <input id="second" value="1/1/2001"/>  <script>   alert(datediff("day", first, second)); // what goes here? </script>


For a data set $\bigcup_{i=1}^nx_i$ of $n$ values with mean $\bar{x}$ the variance is defined as $$\sigma^2=\frac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n}$$  My textbook says that “the square ensures that each term in the sum is positive, which is why the sum turns out not to be zero.” However wouldn’t $$\sigma^2=\frac{\displaystyle\sum_{i=1}^n|x_i-\bar{x}|}{n}$$ also prevent a sum of zero?

The mean absolute deviation is:   $$\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}$$   The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$   So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value. Why isn't it $|x_i-\bar x|^2$, then? Squaring just enlarges, why do we need to do this?  A similar question is , but mine is a little different. Thanks.


I randomly found the  tag. It seems meaningless both relative to the submitter's code (of course there is a bug in it) and third-party code (only a bug if an answer says so).  I am not sure who has the permission and what the process for this, but I suggest to delete and block that tag.  Actually I just found a similar question, , however there is no authoritative answer on why it still exists.

It seems most of the questions tagged  on Stack Overflow are just intended to indicate "there's something wrong with my program", which would cover probably at least 80% of all questions on Stack Overflow.  There are some questions where it seems to refer to a possible bug in a framework or utility. I'm not sure the tag is really all that useful in that case either.  What does meta think?


I have a MacBookAir5,1 connected to a Thunderbolt Display.  When it gets very hot, PID 0 "kernel_task" uses ever more CPU until reaching 340% and slowing the Mac down sufficiently to make me restart it.  I found a proposed solution  but it says to "move the ‘plist’ file for the model identifier we discovered earlier out of that directory" which does not apply to me because my installation of Mountain Lion does not have a plist for my 2012 MacBook Air. system_profiler tells me that I have a MacBookAir5,1.  I tried a few things. I figured that since overheating appears to cause this, I should try not to overheat the poor Mac. However, I neither understand why Mac OS X' reaction to overheating is to use 340% of CPU time nor did I manager to keep the lid open and still convince the Airbook to use the Thunderbolt Display as its only display.  I tried running from battery, assuming this would make the CPU run slower and not overheat, but when I remove the power cable from the Airbook, the Thunderbolt Display just turns black, only recovering after several minutes when I reconnect the power cable.  What's going on?

I'm trying to solve the mystery of a crazy high cpu usage on kernel_task.  MacBook Pro (Retina, Mid 2012), 2.7 GHz Intel Core i7, Yosemite 10.10.2  I saw some advice to download the Intel Power Gadget. That showed me that the my CPU is slowing way down. How do I fix this? When this happens, I can sometimes cycle the power on the MBP. But this is driving me crazy.


Let $R$ be the ring of all continuous functions from the closed interval $[0,1]$ to $\mathbb {R}$  and for each $c \in [0,1]$ let $M_c = \{f \in R |f(c) = 0\}.$       (a) Prove that if $M$ is any maximal ideal of $R$ then there is a real number $c \in [0,1]$ such that $M = M_c.$   (b) Prove that if $b$ and $c$ are distinct points in $[0,1] $ then $M_b\neq M_c.$      (c) Prove that $M_c$ is not equal to the principal ideal generated by $x — c.$      (d) Prove that $M_c$ is not a finitely generated ideal.    (a) I know that $M_c$ is maximal ideal of $R.$  Since $M$ is a maximal ideal of $R$ then $M$ is prime ideal. I am unsure how to proceed.  (b) Assume that  $b\neq c$ and $M_b=M_c.$ Let $f \in M_c.$  Then  $f \in M_b.$ Thus, $ M_c$ is not  maximal ideal of $R.$  (c) and (d) I have no idea how to start it.  Help me, please!

Let $C[0,1]$ denote the ring of continuous functions on $[0,1]$. Consider the maximal ideal $M_a =\{ f\in C[0,1] | f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated?   Note: It may seem that $M_a = \langle x-a\rangle$ which is certainly not true because $ \vert x-a \vert $ is in $M_a$ but not in $\langle x-a\rangle$. Any ideas for solving this question?


was about to download and install Ubuntu a week or so ago, then Spectre and Meltdown issues came to the fore and I decided to wait. Will the LTS ISO image available for download be patched? If so, I'd prefer to wait for a short while until that happens. If it's not going to be patched, and I install from the LTS iso downloaded from Ubuntu site, will the update come automatically, or will I be on a learning curve to manually update? Thanks

Any questions relating to status updates, or asking if anything is going to be patched for these vulnerabilities should be closed as duplicates of this question.    are in the news right now and sound pretty severe. I don't see any security updates from Ubuntu that cover these vulnerabilities.  What is Ubuntu doing about these vulnerabilities, and what should Ubuntu users be doing?  These are CVE-2017-5753, CVE-2017-5715 and CVE-2017-5754.


When it is said, for example, that grand unification took place , how are these figures of time derived?  Moreover, do these figures have any meaning, considering that during these epochs, spacetime was incredibly warped, and immediately after inflation, matter was at an incredibly high energy state and moving at relativistic speeds?

I just finished watching Into the Universe with Stephen Hawking (2010). Specifically the third episode titled 'The Story of Everything.'  In the episode Hawking is explaining the mainstream theories of the events just after the Big Bang. What confuses me is that he just said at the beginning that the laws of physics were different and time (among other things) did not exist as we know it during this early stage in the life of the universe, but then goes on to say certain events took certain amounts of time.  I don't know if the numbers match, but the same issue exists on .  How can we even guess the time frames of these events if time itself is largely dependent on the current state of the universe?


Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be a function $f(r,\theta)=(r\cos\theta,r\sin\theta).$Then for which of the open subsets $U$ of $\mathbb{R}^2$ given below, $f$ restricted to $U$ admits an inverse?   $U=\mathbb{R}^2$ $U=\{(x,y)\in\mathbb{R}^2:x>0,y>0\}$ $U=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$ $U=\{(x,y)\in\mathbb{R}^2:x<-1,y<-1\}$   Here the Jacobian of $f$ is $r=\sqrt{x^2+y^2}$ and it is not vanishing on option 2 and 4. Can I say option 2 and 4 are correct?

Let  $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$  be the function defined by $f(r,\theta)=(r\cos\theta,r\sin\theta).$ Then for which of the open subset $U$ of $\mathbb{R}^2$ given below, $f$ restricted to $U$ admit an inverse?   $U=\mathbb{R}^2$ $U=\{x,y \in\mathbb{R}^2:x>0,y>0\}$ $U=\{x,y \in\mathbb{R}^2:x^2+y^2<1\}$ $U=\{x,y \in\mathbb{R}^2:x<-1,y<-1\}$   It is clear 1 is not true as $\sin ,\cos$ are periodical function. Similarly 3 is not true. What about 2nd and 4th ? please help.Thanks in advance.


The sounds in Minecraft will not work in any version of MC I have tried. Playing with mods, the mods' sounds will work but not the vanilla sounds. I have tried multiple things including:   Restarting my computer Reinstalling Minecraft Using the F3 key method Looking at the sound files on my PC Making sure MC isn't blocked by an antivirus

I hadn't played Minecraft in a few days prior to this, but when I fired it up, I had no sounds whatsoever, either in the menu or the game. First thing I checked was in-game, and all audio settings were set to 100%. I then tried the "Hold-F3-and-push-S" thing, and that had no effect. I then deleted the entire .minecraft folder (after having copied my saves out) and reinstalled, and even deleted my Minecraft.exe and re-downloaded it from minecraft.net. None of those had any effect.  I tried downloading a sounds set, with no effect. I looked in the assets/indexes/1.7.10.json file to see what files were there, and verified that the sounds exist. For example, I checked the walking on grass sound, found the file, made a copy to .ogg and played it in VLC, and it played fine.  I rebooted into Fedora and played the java version (same computer) and it plays the sound just fine, so I'm wondering what the issue is with Windows 7.  Anyone have any suggestions?


Given: K and T are sub-sets of Linear Space V  and also $Sp(K) + Sp(T)$ is a direct sum, and $Sp(K \cup T) = Sp(K) \oplus Sp(T)$  I need to do show that $ K \cup T$ isn't necessary independent.  I said that: $ V = R^2, $ $K = \{(1,0),(0,1)\}$ ,$ T = \{(1,1)\}$  From here I can say that $Sp(K) + Sp(T)$ is a direct sum because their sum is V and their intersection is zero. and also, $Sp(K \cup T) = Sp(K) \oplus Sp(T)$ because their sum is $Sp(K \cup T)$ and their intersection is zero.  From here we see that $K \cup T = \{(1,0),(0,1),(1,1)\}$ but $(1,1)$ is dependent by $(1,0) $and $(0,1)$ which makes it not necessary independent.  The question is, did I prove it right, maybe my sub-sets are wrong and I need other ones? Or just adding $(0,0)$ to K and T is enough to show that there isn't necessary independence(which is a another solution)?

Given, K and T are sub-sets of linear space V.  How can I prove that if $ V = Sp(K) + Sp(T)$ then $ K \cup  T$ is NOT a base of V.  I proved that it IS a base of V but the answer says that it's not, can you give me an example of 2 sub-sets K,T that will make their union not a base of V?  Here how I tried:  $ V = R^2 $ , $K = \{(0,1)\} $,$ T = \{(1,0)\}$ then  $V = Sp(K) + Sp(T) = Sp\{(K \cup T)\} = Sp\{(1,0),(0,1)\} = R^2$ , then  $K \cup T = \{(1,0),(0,1)\} = R^2$  Which makes the union of K,T a base of V, but I need your help to find 2 sub-sets that will make it wrong.


What is the maximum value of $n$ for which $3^n$ divides 1000!  Am I supposed to look at it as a congruence... or apply euclid's algorithm somewhere? I'm lost  I can't use simple arithmetics... Someone help.

Given numbers $k$ and $n$ how can I find the maximum $x$ where: $n! \equiv\ 0 \pmod{k^x}$?  I tried to compute $n!$ and then make binary search over some range $[0,1000]$ for example  compute $k^{500}$ if $n!$ mod $k^{500}$ is greater than $0$ then I compute $k^{250}$ and so on but I have to compute every time value $n!$ (storing it in bigint and everytime manipulate with it is a little ridiculous) And time to compute $n!$ is $O(n)$, so very bad. Is there any faster, math solution to this problem? Math friends?:) Cheers Chris


Which is correct? Is it that you are signing "into" your account or "in to" your account?

When writing an instruction about connecting to a computer using ssh, telnet, etc., I'm not sure what spacing to use in this familiar spoken phrase:   "Log in to host.com" "Log into host.com" "Login to host.com"   Maybe this is entirely subjective or the realm of industry jargon, but I couldn't think of anywhere else to ask.  Any insight?


If a Spacecraft Was To Launch From An Asteroid Would it Be Easier To Minimize The Required Escape Velocity By Launching At A 45 degree angle? Or Would It Be Better To Head Straight Up?

Comparing the work required for a mass to escape the Earth's gravity to the necessary initial kinetic energy gives us the escape velocity from the surface of the Earth of around $11\;\mathrm{km\cdot s^{-1}}$.  But assuming no air friction and hence no loss in energy on the axis perpendicular to gravity it shouldn't matter what angle the projectile is projected at.  Is this true?  If there were no air and we fired a projectile at $1^\circ$ to the horizontal at a velocity of $11\;\mathrm{km\cdot s^{-1}}$ should it still escape the Earth's gravity despite such a low ( $11000\sin(01^\circ)\;\mathrm{m\cdot s^{-1}}$) vertical velocity?  Could it be that firing a projectile horizontally at the escape velocity in a frictionless medium would result in a perfect orbit?  Also: Taking escape velocity of $11200\;\mathrm{m\cdot s^{-1}}$ and radius af Earth of $6371000\;\mathrm{m}$ and $g=9.80665\;\mathrm{m\cdot s^{-2}}$. On a flat Earth if you fired an object horizontally at $112000\;\mathrm{m\cdot s^{-1}}$ it would fall  $0.039\;\mathrm{m}$ after $1000\;\mathrm{m}$. But using trigonometry the Earth would have curved away by $0.078\;\mathrm{m}$. Twice the value necessary for an orbit. Can someone check this?


I want to perform raster smoothing on a DEM file. I want to to assign the elevation of each pixel using the average elevations of its 8 neighbors. The ESRI's "Python scripting for ArcGIS" has a chapter on using python for raster but it did not help me to write a code for this. I'm looking for any useful literature, tutorial, or sample code...! please note that r.buffer and r.neighbors is not a good answer for me since I want to learn how to write this code completely.   P.S: I already took a look at  and . they did not help me...may be since my knowledge is limited I need more comprehensive answer. Thanks in advance for any help you are able to provide.

I have a DEM that I would like to smooth or generalize to remove topographic extremes (chop off peaks and fill-in valleys). Ideally, I would also like to have control over the radius or level of "blurriness". In the end, I will need a set of rasters that range from slightly blurry to really blurry. (Theoretically, the blurriest would be a constant raster of the arithmetic mean of all values).  Are there any tools or methods I can use (based on Esri, GDAL, GRASS)?   Do I need to home bake my own  routine?   Could I use a low-pass filter (e.g. ArcGIS's ), and if so, would I need to run it a bunch of times to get the effect of a large radius?


I am using Ubuntu 18.04. I tried installing Wine. Afterwards following error showed. E:Type 'cleadeb' is not known on line 1 in source list /etc/apt/sources.list, E:The list of sources could not be read.  Now I cannot install any other app. How do I solve this error?

I have unistalled and reinstalled the Ubuntu Software Center as per info I found in a similar thread and I got the same response about line 91 or something like that.  I just tried to upload a screen shot but since I'm new it won't allow me to.  I also can not figure out how to cut and paste anything so I have to hand type what the error screen says, both when I attempt to open the software center and nothing happens, when I try to enter commands into the terminal to uninstall, reinstall, whatever I get the same following:  COULD NOT INTITIALIZE THE PACKAGE INFORMATION An unresolvable problem occured while initializing the package information Please report t:his bug against the 'update-manager' package and include the following error message: 'E: Malformed line 91 in source list/etc/apt/sources.list (dist parse)  E: The list of sources could not be read.,  E: The package list of status file could not be parsed or opened.   How do I report bugs?  What can be done about this.  I have searched and everything everyone says to do leads me back to the same line error message.  So, I don't know how to get to line 91 in the source list; to tell you what it says. Sorry, I'm really new to this. That is what I need is to find out how to get there and fix what it says. I would really like to NOT have to re partition my hard drive and start from scratch, so I'm really looking forward to getting this problem solved. I need to be able to install new software.


I sometimes see stuff that is of the following notation:   $$ \int f(x) dG(x) $$   I reckon a specific example would be:  $$ \int x \operatorname{d \sin(x)} $$  How can one intuitively interpret this. Does this mean that we're compressing and stretching the x-axis by a sin function?

What is the intuition/idea behind integration of a function with respect to another function? Say $$\int f(x)d(g(x)) \;\;\;\;\;?$$ or may be a more particular example $$\int x^2d(x^3)$$  My concern is not at the level of problem solving. To solve we could simply substitute $u=x^3$ and then $x^2=u^{2/3}$. My concern is rather about what meaning physical/geometrical does this impart?  ADDED if you ask what kind of meaning I seek, integration of a function w.r.t a variable gives the area under the curve and above x-axis.


I recently decided to declare as an applied math major and I've realized that I don't think I fully understand what derivative notation actually means. I'm kinda worried that I might have missed something and it may affect my progress in more advanced math classes like ODE which I'm taking right now. I have always assumed that $\frac{dy}{{dx}}$ is just notation for a specific operation: $$ \frac{dy}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{y\left( {x + h } \right) - y\left( x \right)}}{h} $$ I understand that this is the definition of a derivative but this doesn't seem to be the full story. For example what does $dx$ mean by itself? In practice it seems like  $\frac{dy}{{dx}}$ is a ratio between two values $dy$ and $dx$ that have a mathematical meaning on their own. For example, its common to multiply both sides of an equation by $dx$. People say that $dx$ is just a small change in $x$ but what does that actually mean? One possible answer that I've just sorta assumed without thinking about it too much was $$dx = \mathop {\lim }\limits_{a \to 0} (x + a) - x $$ But this seems wrong because it just evaluates to $dx =  \mathop {\lim }\limits_{a \to 0} a = 0$. Reading wikipedia articles about differential notation isn't helping much but it sounds like the meaning of $dx$ has changed over time? The specific thing that prompted my question was a problem in my ODE class. Take a linear differential equation that looks something like this: $$ \frac{dy}{{dx}} + y = x^2 $$ I know what to do in order to solve this but the solution feels hand wavey. If you multiply both sides by the integrating factor $e^{x}dx$ and then integrate both sides you get $$ \int e^{x}dy + e^{x}ydx = \int e^{x}x^2 dx $$ I just don't know how to interpret the RHS of the equation. Do I integrate with respect to x or y? Do I just integrate the $dx$ and $dy$ terms separately? Doing that doesn't give you $e^{x}y$ which is the actual answer. Edit: some people are suggesting that in integral notation $dx$ is just sort of syntactical, it just shows where the end of the integration operation is. I am unsatisfied with this answer. If its just syntactical, then you would evaluate the ODE like this $$ \int (e^{x}\frac{dy}{{dx}} + e^{x}y) dx \\ $$ If I am understanding the syntax argument correctly, you should be able to split this integral into two parts like this: $$ \int e^{x}\frac{dy}{{dx}}dx + \int e^{x}y dx =  e^xy + e^xy = 2e^xy $$ But this doesn't appear to be consistent with my ODE textbook, which says the actual answer is $e^xy$. They justified that by saying that the derivative of $e^yx$ is $e^{x}\frac{dy}{{dx}} + e^{x}y$ by the chain rule. This feels handwavey and leads me to believe that $ \int e^{x}\frac{dy}{{dx}}dx + \int e^{x}y dx $ is not notationally equivalent to $\int (e^{x}\frac{dy}{{dx}} + e^{x}y) dx$

I see an equation like this:  $$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$  and solve it by "separating variables" like this:  $$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2  = e^x + c$$  What am I doing when I solve an equation this way?  Because $\textrm{d}y/\textrm{d}x$ actually means  $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$  they are not really separate entities I can multiply around algebraically.    I can check the solution when I'm done this procedure, and I've never run into problems with it.  Nonetheless, what is the justification behind it?  What I thought of to do in this particular case is write  $$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$  then by the fundamental theorem of calculus  $$y^2/2 = e^x + c$$  Is this correct?  Will such a procedure work every time I can find a way to separate variables?


Suppose $G$ is a group and $N$ is its normal subgroup, such that every subgroup of $N$ is normal in $G$ and $C_G(N)\subset N $. Is $\frac{G}{N}$ always abelian?  I do not know such $G$ and $N$ where it is not, but neither I know general proof of this statement.  Any help will be appreciated.

Let $N$ be a normal subgroup of a group $G$ such that every subgroup of $N$ is normal in $G$ and $C_G(N)\subseteq N $. Prove that $G/N$ is abelian.   Here, as usual, $C_G\left(N\right)$ means the centralizer of $N$ in $G$ (i.e., those elements of $G$ that commute with everything in $N$).  I think we need to use that every subgroup of $N$ is normal in $G$ but i can't use .Please help me with Hints.


So, it has been a few days since my computer started having problems to open the google chrome browser. I checked the activity monitor and there is a process called "accountsd" which is using around 200% of the CPU power. Any suggestions? Thanks :D

Why are the accountsd and securityd processes using so much CPU on OS X 10.10.3? I am on a Early 2011 Macbook Pro.  What are these processes for? Is it safe to kill them? The machine has only been up 4 hours so has been rebooted recently.


"No candidate" when I try to install Gnome Classic on 15.04. Will that change? If not, what other "low-tech," lower-demanding UI options do I have for my older, weaker computer with 15.04?

For a while I departed from Ubuntu to use Linux Mint because I really don't like Unity, or Gnome 3's most recent interface.  I was fine with what was called "Gnome Classic", but I feared it was being neglected because there were some significant bugs that lingered without improvement for a while.  Unfortunately, while I like the MATE interface in Linux Mint, Mint itself lacks the level of community support that Ubuntu has. Mint has development and support issues. Not to disparage their efforts, I think they've done great work, but it's tough making a UI, and unfortunately it's just not working for me.  I'm thinking of returning to Ubuntu with the 14.04 release, but not if it means using Unity. I have looked into the current situation, and I'm a little confused about the terms being thrown around. I have seen referens to "fallback", "flashback", and "classic", without any clear idea of which is available and what they offer.  Is there still an option to get an interface that resembles the way Gnome used to look?  Can I get an interface that looks like what I have with Mint, but with the support and features of Ubuntu? Can I get something close to that?


This is what I am trying to solve:  $n!>n^{\frac {n}{2}}$  I tried with induction and somehow it doesn`t work this way, I tried with logarithms and also didn´t find a way. Is there an elementary(if possible) way to prove this?  Or is it the case that for some large $n_0$ this inequality reverses sign?

I proved this inequality in the following way:   Lemma: $r \in \Bbb N, r \geq 3$. We have $r^r \gt (r+1)^{r-1}$.  Proof: We apply the AM-GM inequality to the $r$ positive integers where there are $r-1$ $(r+1)$'s and one $1$. We obtain $$\frac {1+(r-1)(r+1)}{r} \gt  ((r+1)^{r-1})^{\frac {1}{r}}$$ wherefrom we get (since the $r^{{\rm th}}$ power function is increasing), $r^{r} \gt (r+1)^{r-1}$.   Now, I used mathematical induction to prove the statement. We have from the lemma, $(k!)^{2} \gt k^k \implies (k!)^{2} \gt (k+1)^{k-1}$ and multiplying this inequality by $(k+1)^{2}$, we have $((k+1)!)^{2} \gt (k+1)^{k+1}$ and obviously $(3!)^{2} \gt 3^3$.  Is there any direct proof of the statement that does not use induction and calculus?


I know the must have/must had thing has been asked before but I have yet to see it in the context of this question.  Does "have" make grammatical sense in the following? It sounds like it conflicts with the "at that point in time."  "I remember clearly my seventh birthday (and my subsequent trip to the hospital). At that point in time, I must have been told a thousand times not to stick things in the toaster."  The main problem I'm having in my head is that if you take out the "must" then it would be grammatically correct to say "At that point in time, I had been told a thousand times..." Yet once you put the "must" in, it seems incorrect to use "had." This is my main question.

If someone says that they had been to Florida on a recent trip, which one of the following would be a correct respone:   That must had been fun.  That must have been fun.    Or is there a better way of saying this?


i have one layer published in geoserver and i added 4 styles(sld1, sld2, sld3, sld4) to the same layer, now how could i to switch between 4 styles each style with the legend use leaflet ? this wms of my layer: var mylayer = L.tileLayer.wms("http://localhost:8081/geoserver/map/wms", {     layers:'map:mylayer',     format: 'image/png',     transparent: true,     attribution: "mylayer" }); mylayer.addTo(map);

I have a Leaflet WMS Layer  layer = new L.tileLayer.wms('https://{s}.url.com/wms?', {     layers: 'Workspace:Layer',     format: 'image/png',     version: '1.3.0',     transparent: true,     styles: 'style1', });   If I choose 'style2', it works as expected, but I want to change it dynamically with a select box. I tried the following but it does not work:  <select id="styles">     <option selected="selected" onselect="layer.styles='style1';">Style1</option>     <option onselect="layer.styles='style2';">Style2</option> </select>   I also tried with layer.options.styles=''


I’m not able to execute the files in a mounted directory in my Ubuntu 16.04 VM. I get the error “command not found”:  dseltz-VirtualBox:/local> ls -al /nfs/local/gnu/i86Linux2.0/bin/makeheader -rwxr-xr-x 1 systemd-network uucp 49379 Oct 20  1999 /nfs/local/gnu/i86Linux2.0/bin/makeheader dseltz-VirtualBox:/local> /nfs/local/gnu/i86Linux2.0/bin/makeheader /nfs/local/gnu/i86Linux2.0/bin/makeheader: Command not found.  My VM is a x64 running Ubuntu 16.04 and so is the mounted system. I looked at the mounts and don’t see the noexec property(on either system). Here is the mount command:  sudo mount -o exec 10.15.1.203:/local /nfs/local  I made sure my users matched on my VM machine and the build system. Any ideas what I'm doing wrong?

I have currently a strange problem on debian (wheezy/amd64).  I have created a chroot to install a server (i can't give any more detail about it, sorry). Let's call its path /chr_path/. To make things easy, I have initialized this chroot with a debootstrap (also wheezy/amd64).  All seemed to work well inside the chroot but when I started the installer script of my server I got : zsh: Not found /some_path/perl (the installer includes a perl binary for some reasons)  Naturally, I checked the /some_path/ location and I found the "perl" binary. file in chroot environment returns :  /some_path/perl ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.2.5, not stripped   The file exists, seems ok, has correct rights. I can use file, ls, vim on it but as soon as I try to execute it - ./perl for example - I get : zsh: Not found ./perl.  This situation is quite understandable for me. Moreover :   I can execute other basic binaries (/bin/ls,...) in the chroot without getting errors I have the same problems for other binaries that came with the project When I try to execute the binary from the main root (/chr_path/some_path/perl), it works. I have tried to put one of the binaries with a copy of my ls. I checked that the access rights were the same but this didn't change anything (one was working, and the other wasn't)


When creating dungeons, I like to try to give them a history beyond "old place with traps and monsters and gold", since I like everything to feel real and have a purpose. However, whenever I try to map out one of these, they either end up with planning that doesn't make sense for what it is, or wouldn't be fun to play through.  How do I create a dungeon layout that has/had purpose and would still be fun and challenging to explore?

One of the typical location that is used as an excuse to build a dungeon is The Ancient Tomb (tm).  I often struggle building dungeons that are credible tombs, in terms of rooms disposition and type of rooms contained. It seems to me that either my dungeons are realistic but boring, or they are entertaining but quite unrealistic as tombs.  Is it even possible to build a realistic tomb that is still entertaining for players?  I would like to know if someone could provide some examples/ideas for realistic-but-enjoyable tombs (possibly based on real tombs).   I'm doing a bit of research myself, but I'm not finding workable ideas. Eg, the tomb of Tutankhamen would be rather disappointing as a dungeon:     This tomb consists of three small rooms and a hallway; this would allow for only a few encounters/traps/riddles. An action-oriented group might find a dungeon like this a bit disappointing. Also, I feel that a small dungeon makes harder to build a sense of mystery and exploration.  Also, since I was asked to better define "realistic" vs "entertaining":  For the purpose of this question, I will say that a dungeon is realistic when its features (room types, room dispositions, the furniture) make sense in an imaginary world loosely based on historical civilizations, or periods and places in history. As an example, it is conceivable that an ancient tomb would contain defenses against intruders. In an imaginary/fantasy world, these defenses becomes traps, animated guardians, riddles, curses. It is coinceivable that an ancient tomb would contain a treasure chamber, and in certain cultures, it is conceivable that a tomb would contain vast amounts of goods that belonged to the deceased in life; in an imaginary/fantasy world, a tomb containing magical items, esoteric and forbidden text, weapons etc. does not break the suspension of disbelief. On the other hand, why would a tomb contain, say, a torture chamber? While it would be possible to imagine more or less elaborate explanations ("this culture believed that torture chambers were needed by the deceased in the afterlife, to punish the intruders, and thus were routinely built inside of tombs), I feel that a torture chamber in a tomb does not readily "make sense" and would break immersion.  A dungeon is entertaining when challenges the players and allow room to build a sense of danger, mystery, and exploration of ancient and forgotten places. In an entertaining dungeon I expect a number of combat encounters, a number of clever traps, several rooms and hallways to explore, and a treasure.   While I asked the question with D&D in mind (I'm currently DMing a 3.5 campaing), I would like to keep the question as system-agnostic as possible.


Yes, I know all about perspective (I'm an artist). I even have some basic knowledge of descriptive geometry. I know how it works. My question is more about why it works.   I have a sneaking suspicion it has something to do with the curvature of space. However, I'm not certain whether the two concepts are related.   Another thing that's been bothering me for a while (this may or may not be related to the question: please enlighten me) is the relation of the vanishing point to the horizon:  Strictly speaking, the geometrical horizon is where the vanishing point is. The actual horizon is a bit lower, but the difference is negligible.   Thus, if you sketch a perfectly straight street lined with buildings of equal size stretching all the way to the horizon, you shouldn't even be able to see anything beyond 2.9 miles (the distance to the horizon): all parallel lines will merge - and vanish. And yet, if the buildings are large enough, they will be visible beyond that point.     Which would suggest that smaller objects become a point source before they reach the true vanishing point, and that the true vanishing point is a lot further than 2.9 miles (which is why we can see the actual shape of the moon, as well as some of the planets: they're not point sources; they're actual disks). I would even go so far as to posit that the reason we see stars as point sources is they aren't large enough; and the Andromeda Galaxy, which appears to the naked eye as a cloud, is not a point source; which leads me to believe that the absolute vanishing point is at an infinite distance from the observer.  With all of the above in mind, my question still is - WHY do objects appear smaller in the distance?  P.S. "That's how perspective works" is not a good answer in this case unless physics has nothing more to contribute at this juncture.  P.P.S. Yes, about those angles (the light hitting the eye, etc). Why should the angle get narrower as the distance increases? Well, for one thing, that's how our brain interprets the signal it receives from the eye. I daresay if it were the other way around (i.e. if the angle got broader as the distance increased), our brain would find a way to adjust, and we'd all be saying "It's bigger because it's further away, that's how it works").

Why does the apparent size of an object change when it is farther away from me compared to when it is closer to me?  Can someone perhaps explain this through an optics perspective?


Using R, I would like to import shapefiles that are in a directory on a geometric table I have on PostgreSQL/PostGIS.  What is the instruction that allows me to do it?  EDIT: here is my code to generate contour lines from the PostgreSQL data, the result is a shapefiles I would save .shp on my table PostGIS  already create before:  library(ggplot2) library(gstat) library(sp) library(maptools) library(rgdal) library(zoo) library(xts) library(RPostgreSQL) library(spacetime) library(raster) library(foreign) dbname = "postgis" user = "postgres" password = "***" #password = ""  drv <- dbDriver("PostgreSQL") con <- dbConnect(drv, dbname="BDDMeteo", user=user, password="***",host='localhost', port='5432') estonia_air_temperature_2=dbGetQuery(con,"select lat,lon,temperature from observation, station, date where id_s=nom_ville AND id_d=id_date")  estonia_air_temperature_2_test <- estonia_air_temperature_2  # duplicate air temp. data file estonia_air_temperature_2_test$x <- estonia_air_temperature_2_test$lon  # define x & y as longitude and latitude estonia_air_temperature_2_test$y <- estonia_air_temperature_2_test$lat coordinates(estonia_air_temperature_2_test) = ~x + y  plot(estonia_air_temperature_2_test) x.range <- as.numeric(c(-8.5, 9.5))  # min/max longitude of the interpolation area y.range <- as.numeric(c(26.97, 37))  # min/max latitude of the interpolation area grd <- expand.grid(x = seq(from = x.range[1], to = x.range[2], by = 0.1), y = seq(from = y.range[1],                                                                                to = y.range[2], by = 0.1))  # expand points to grid coordinates(grd) <- ~x + y gridded(grd) <- TRUE plot(grd, cex = 1.5, col = "grey") points(estonia_air_temperature_2_test, pch = 1, col = "red", cex = 0.4)  idw <- idw(formula = temperature ~ 1, locations = estonia_air_temperature_2_test,             newdata = grd)  # apply idw model for the data residual_grid = raster(idw, "var1.pred") plot(residual_grid) x <- rasterToContour(residual_grid) class(x) plot(residual_grid) plot(x, add=TRUE) writeSpatialShape(x, "/interpolate/contour/contourline")   Edit:  > ogrDrivers()              name write 1      AeronavFAA FALSE 2          ARCGEN FALSE 3          AVCBin FALSE 4          AVCE00 FALSE 5             BNA  TRUE 6             CSV  TRUE 7             DGN  TRUE 8             DXF  TRUE 9          EDIGEO FALSE 10 ESRI Shapefile  TRUE 11     Geoconcept  TRUE 12        GeoJSON  TRUE 13       Geomedia FALSE 14         GeoRSS  TRUE 15            GML  TRUE 16            GMT  TRUE 17       GPSBabel  TRUE 18  GPSTrackMaker  TRUE 19            GPX  TRUE 20            HTF FALSE 21         Idrisi FALSE 22            KML  TRUE 23   MapInfo File  TRUE 24         Memory  TRUE 25   MSSQLSpatial  TRUE 26           ODBC  TRUE 27            ODS  TRUE 28        OpenAir FALSE 29    OpenFileGDB FALSE 30         PCIDSK  TRUE 31            PDF  TRUE 32            PDS FALSE 33         PGDump  TRUE 34           PGeo FALSE 35            REC FALSE 36            S57  TRUE 37           SDTS FALSE 38       SEGUKOOA FALSE 39           SEGY FALSE 40            SUA FALSE 41            SVG FALSE 42            SXF FALSE 43          TIGER  TRUE 44        UK .NTF FALSE 45            VRT FALSE 46           Walk FALSE 47           WAsP  TRUE 48           XLSX  TRUE 49         XPlane FALSE

I'm producing some Spatialpolygondataframes with R that I would like to upload to my PostGIS Database. I've found out that the way to do this is via the  function using the .   Sadly, my version of rgdal and GDAL/OGR doesn't support the PostgreSQL function (as seen when listing the available drivers using ogrDrivers()). Is there a way to make this driver available? My sessionInfo() is as follows:  R version 3.2.3 (2015-12-10) Platform: x86_64-w64-mingw32/x64 (64-bit) Running under: Windows >= 8 x64 (build 9200)  [...]  other attached packages: rgdal_1.1-10 gdalUtils_2.0.1.7 RPostgreSQL_0.4-1


enemyArray = [Ork, Goblin]  class Ork:      name = "Mountain Dweller"`     health = 10     armor = 1     damage = 5   class Goblin:      name = "Looter"     health = 5     armor = 0     damage = 5

Assume I have the following list:  foo = ['a', 'b', 'c', 'd', 'e']   What is the simplest way to retrieve an item at random from this list?


$\sqrt {-3}$ multiplied by $\sqrt{-3}$ is $-3$. But this can also be written as $\sqrt {-3} \cdot \sqrt{-3} = \sqrt {(-3).(-3)}= \sqrt{9} =3$ So my question is why is this not possible?

I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &= i \\\\\ \frac1{\sqrt{-1}} &= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\  \sqrt{\frac1{-1}} &= \frac1i \\\\  \sqrt{\frac{-1}1} &= \frac1i \\\\  \sqrt{-1} &= \frac1i \\\\  i &= \frac1i \\\\  i^2 &= 1 \\\\  -1 &= 1 \quad !!? \end{align}


In Harry Potter,  are generally known as people who cannot perform magic, and all the examples from the books appear to be descended from wizards.   In reality, we see various levels of magical control exhibited by Squibs -- they can see magically hidden things (such as ghosts, ), use magical items, develop magical relationships with magic beasts (such as -cat hybrids), etc.  Moreover, there are many Muggles who claim abilities (such as seeing ghosts) which require magical sensitivity as well.  While Fantastic Beasts and Where to Find Them claims that these people are either lying or Squibs, it's also purposefully written with an unreliable narrator.  The Muggle-born wizards we see, however, are all of standard wizarding capability.  Can Muggles give birth to a magic sensitive (but incapable of casting spells) child, equivalent to a Squib (since the term 'Squib' itself only applies to the children of wizards); a dud wizard?

‘And what on earth’s a Squib?’ said Harry.   To his surprise, Ron stifled a snigger.   ‘Well – it’s not funny really – but as it’s Filch ...’ he said. ‘A Squib is someone who was born into a wizarding family but hasn’t got any magic powers. Kind of the opposite of Muggle-born wizards, but Squibs are quite unusual. If Filch’s trying to learn magic from a Kwikspell course, I reckon he must be a Squib. It would explain a lot. Like why he hates students so much.’ Ron gave a satisfied smile. ‘He’s bitter.’   Chamber of Secrets - pages 110-111 - British Hardcover   If a Squib is 'kind of the opposite of a Muggle-born wizards' wouldn't that make a Squib more Muggle than wizard? Does a Muggle strictly have to be born to Muggle parents to be considered a Muggle, or can the lack of magic render a person, even of wizarding parentage, a Muggle?  Hermione can say in all truthfulness that she is a 'Muggle-born witch.' The nickname (albeit derogatory) for a Muggle-born is 'Mudblood.' Filch can say in all truthfulness that he is a 'wizarding-born non-magic person,' which, according to Hagrid is a Muggle.  'Squib' may be derogatory in the same way as 'Mudblood' is. Neville laments his family thought he was a Squib when he showed no magical tendencies as a child; the Longbottoms were so rattled at the thought of Neville being a Squib that they, among other things, threw him out a second-story window to see if he would bounce. So could 'Squib' actually be a derogatory term meaning the opposite of 'Mudblood'? After all, the Ministry doesn't allow Squibs to register as magical beings, and many Squibs are schooled and trained in Muggle professions. Squibs don't get a Hogwarts letter.    Wizards can be born to Muggles. Can Muggles be born to wizards?  ETA 11.22.14: In regard to this question, I was glancing through Philosopher's Stone, and I came across this, that Neville said:     ‘Well, my gran brought me up and she’s a witch,’ said Neville, ‘but the family thought I was all Muggle for ages. My great-uncle Algie kept trying to catch me off my guard and force some magic out of me – he pushed me off the end of Blackpool pier once, I nearly drowned – but nothing happened until I was eight.’   Philosopher's Stone - page 93 - Bloomsbury - chapter seven, The Sorting Hat   Could what Neville says contribute to the theory and/or conclusion that Squibs are, or are not, Muggles? How do we apply what Neville said?


I have created a fragment in which an animation for rotation is there.      Animation rotation = AnimationUtils.loadAnimation(getContext(), R.anim.rotate);   When loading the fragment for the first time the animation is finely loaded.  But when coming back to the activity from another activity using intent, it loading the activity fine, loading the fragment fine but giving      Attempt to invoke virtual method 'android.content.res.Resources android.content.Context.getResources()' on a null object reference   pointing     Animation rotation = AnimationUtils.loadAnimation(getContext(), R.anim.rotate);   Please help. Thanks in advance.

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


I want a lamp to emit as much light as a fire. I have a property in my fire material that says "flame". It seems like a good value to copy to my point light's material strength value. I've considered drivers but I'm too new to the concept and I'm not sure how I can pull it off. If anyone has an idea...  I understand that this question has already been answered in the case of constant input values, but the Attribute node I want to use is linked to a mesh, making it trickier to use for another object.  Here is a blend file with what I want to do: . Light and fire on separate layers so it is possible to see the effect on the plane behind.

Let's say I have two materials "material1" and "material2". How do I achieve that material1's roughness(or anything else) is the same as material2's and changing material2's roughness will change both?


I'm currently using ViewPager to slide through different Layouts.  I have two so far and these two Layouts have Buttons.  I have a main Layout and two layouts which are loaded in the main layout.  <Button         android:layout_width="120dp"         android:layout_height="120dp"         android:text="@string/when"         android:id="@+id/when_button"         style="@style/ShadowText"         android:background="@drawable/button_background_states"         android:gravity="center"         android:padding="10dp"         android:adjustViewBounds="true"         android:textColor="@color/text"         android:textSize="20sp"         android:textStyle="bold"         android:layout_marginBottom="20dp"         android:layout_alignParentBottom="true"         android:layout_centerHorizontal="true" />   This is one of my buttons. But I want every button on every layout to be clicked. And when the buttons clicked to be played a sound:  final Button button = (Button) findViewById(R.id.when_button); button.setOnClickListener(new View.OnClickListener() {     public void onClick(View v) {         MediaPlayer mp = MediaPlayer.create(getBaseContext(),                 R.raw.when);         mp.start();     } });   This used to be in my onCreate, but now it makes my app crash. I'm a ultra newbie, I just cannot think of a way to set different events on those different buttons.  Process: com.phakeapps.mladmerindzhey.app, PID: 18970     java.lang.RuntimeException: Unable to start activity ComponentInfo{com.phakeapps.mladmerindzhey.app/com.phakeapps.mladmerindzhey.app.MainActivity}: java.lang.NullPointerException             at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2184)             at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2233)             at android.app.ActivityThread.access$800(ActivityThread.java:135)             at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1196)             at android.os.Handler.dispatchMessage(Handler.java:102)             at android.os.Looper.loop(Looper.java:136)             at android.app.ActivityThread.main(ActivityThread.java:5001)             at java.lang.reflect.Method.invokeNative(Native Method)             at java.lang.reflect.Method.invoke(Method.java:515)             at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:785)             at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:601)             at dalvik.system.NativeStart.main(Native Method)      Caused by: java.lang.NullPointerException             at com.phakeapps.mladmerindzhey.app.MainActivity.onCreate(MainActivity.java:24)             at android.app.Activity.performCreate(Activity.java:5231)             at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1087)             at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2148)             at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2233)             at android.app.ActivityThread.access$800(ActivityThread.java:135)             at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1196)             at android.os.Handler.dispatchMessage(Handler.java:102)             at android.os.Looper.loop(Looper.java:136)             at android.app.ActivityThread.main(ActivityThread.java:5001)             at java.lang.reflect.Method.invokeNative(Native Method)             at java.lang.reflect.Method.invoke(Method.java:515)             at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:785)             at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:601)             at dalvik.system.NativeStart.main(Native Method)   I'm following   I do believe the problem is that the buttons are in different files, but I'm not sure how to find them when a layout is changed.


In many documents about oriented matroids, it is stated as an observation that every rank 2 oriented matroid is realizable. However, I did not find any document where the proof is given.  Do you know any reference where this proof is written?  Thank you!  P.S. This question has already been asked here:  However, the link provided in the answer is not valid anymore

Does anyone know where I can find the proof for why every rank $2$ oriented matroid is realizable?   Thank you!


I had a very large value for Exp(B) in SPSS binary logistic regression. What is wrong and what should I do?

If you have a variable which perfectly separates zeroes and ones in target variable, R will yield the following "perfect or quasi perfect separation" warning message:  Warning message: glm.fit: fitted probabilities numerically 0 or 1 occurred    We still get the model but the coefficient estimates are inflated.   How do you deal with this in practice?


I have these shortcodes which I am using as links in my front end to get the next post and previous post. When I get to the last post it just keeps reloading the last post again. How can I get it to loop back to the first post? function prev_shortcode($atts) {     $prev_post = get_previous_post();     $permalink = get_permalink($prev_post);     return ($permalink); } add_shortcode( 'prev', 'prev_shortcode' );  function next_shortcode($atts) {     $next_post = get_next_post();     $permalink = get_permalink($next_post);     return ($permalink); } add_shortcode( 'next', 'next_shortcode' );

I have a custom post type and used get_next_post() and get_previous_post() to create a simple pagination. However i need to create a pagination loop (if that is the correct term). If i'm on the first post i need to browse back to the last post and if i'm on the last post i need to browse to the first post. Is there a simple way to do this that i haven't seen but Google, this site and the codex has not been very helpfull :/  The current code i have is. I've tried to use wp_list_pages and other functions but with little luck.  $next_post = get_next_post(); $prev_post = get_previous_post(); $nav = array(     "next_post" => array(         "url"       => get_permalink($next_post->ID),         "id"        => $next_post->ID,         "titill"    => get_field("vinstri_titill", $next_post->ID)." - ".get_field("vinstri_undirtitill", $next_post->ID)     ),     "prev_post" => array(         "url"       => get_permalink($prev_post->ID),         "id"        => $prev_post->ID,         "titill"    => get_field("vinstri_titill", $prev_post->ID)." - ".get_field("vinstri_undirtitill", $prev_post->ID)     ), );


I just checked the price of the stock I bought and is is down on Xetra -0,88 % ( where I bought it, as I wanted it in Euro) and up +0,38% on the nasdaq. Can the same stock actually go further this way and how far: up in one exchange and down in the other? I hope my question is clear enough. Thanks anyone for some insight. I am a beginner. Be kind :)

I live in Germany and I am trading stocks for around 6 months. My favorite stocks so far are mainly the big tech companies that we all know (Alphabet, Amazon) as well as the according indices (e.g. Nasdaq 100).  Due to the time lag between Europe and the US, I can trade for around 6.5 hours until the NYSE opens at 9.30 am. And when the NYSE opens, I often see a strong movement in stock prices. To me it seems that the price for a certain financial instrument being generated in Europe during the day, is not adopted by the US but more or less negated.  Do you know if this has more a technical reason or is it the sheer dominance of the US market?


One of my friends wants to visit me in Austria. We have made a plan to visit other countries too (Poland, Italy, CZ, SK, etc.). In which consulate should he apply for the Schengen visa?  He is going to land in Bratislava, Slovakia. The German consulate is near to him compared to other consulates.  Does it matter in which consulate he applies for his visa?

I'm planning to go to Germany on Sept 6-8, then Netherlands on Sept 8-11 and exit Switzerland on Sept 11-14.  My question is:  Which embassy do I need to apply for the Schengen visa? The number of days are the same for Netherlands and Switzerland, but I'm planning to apply in Netherlands since it will be my second point of entry.


Hair particles do not appear straight as like in the render preview. The mesh doesn't have any subdivision modifier and hair dynamics has no effect. Render Viewer  Final Render (Note the distortion on the streams)

Hair particle renders are not same as in 3dView(Eevee). Hair is using Principle Bsdf shader and  camera is set to Orthographic. Force and turbulence fields are applied to it. How to resolve this?


I created a zpool, called zstorage, on two mirrored drives, on Ubuntu 18.04.  The OS is installed on an SSD, with the /home directory on its own partition on that same SSD.  I want to migrate /home to the ZFS pool.  As you can see, the pool has been set up and ready - as far as I know.  Where do I go from here?    Ordinarily, when moving /home to another drive, one can edit fstab, but ZFS doesn't use fstab.  I'm rather stuck at knowing how to move my existing /home to /zstorage/home and how to mount /zstorage/home. The command  sudo zfs set mountpoint=/home zstorage/home  gives me the error   cannot mount '/home': directory is not empty property may be set but unable to remount filesystem.

ubuntu 14.04 installed on 128Gb ssd zfs pool ("zfshome") across 3x3TB disks, containing audio/video/tv etc.  I originally built this as a nas4free NAS serving my windows boxes around the house.I'm now tryin to turn it into a media server,and being always on,wanted the stabilty of ubuntu,(now that it's more user friendly, in terms of normal users...) Tt seems to me that putting /home/user on the zpool made sense,(being a newby I do frequent reinstalls), but after a week  I cannot find clear instructions how to achieve it, and after so many attempts am thoroughly confused.  Any help appreciated, Ian


My samsung galaxy tablet is low on storage and i have no idea where the storage is being used.   When I go into setting and look up storage I get a reading for the various sources documents,apps,pictures, etc.  When I add up the usage I get about 5G.  The usaged defined on the tablet is 15G.  Where is the other 10G being consumed.  Cache has all been cleared so I know it is not that and that should have been in the 5G anyways.  Is it the operating system?

I have critical low storage level and can't understand what takes much of it and how that's possible that the installed applications can use 7 GB? I only use a few social media clients without any games or soft like Photoshop. Also on the second screenshot it indicates that the apps are using just 1.13 GB and so I can't understand the situation. Also what is the "Other"?


How can I know all of the invertible elements? Is it just all of the numbers that are relatively prime to 30?

Can anyone show me how to find invertible elements in $\mathbb{Z}/m\mathbb{Z}$, $m$ is say $28$?  Also I'm not very clear about what it mean by 'invertible element in $\mathbb{Z}/m\mathbb{Z}$'


As you may know the market is in panic today. However if you look for VTTHX, you will see it frozen since Aug 21. Did this fund magically escaped the panic ? what is going on ?

How can I get intraday prices of a mutual fund? (Mine are Oppenheimer funds.)


I'm running Ubuntu as my only OS right now. I added elementary OS and decided I didn't like it. I then deleted the partition and added the space back to my main Ubuntu partition. This proceeded to wipe grub. I did the ls command on grub recovery mode found the ext2 partition my computer was running on but insmod came up with a error every single time. The only way I was able to get my machine running again was to install a second copy of ubuntu on a 20gb partition on my same drive and leave that on there. I'm going to be installing windows as a second OS in a day or two here and need the extra ubuntu gone cause its wasting space. I need help finding a way to ensure grub remains function after I delete the extra partition with ubuntu and before I restart my system and it loses track of my main ubuntu. I'm not sure what to do. Thanks in advance!

I installed Windows 7, which ate Ubuntu's boot file. When starting up the computer, it now goes straight to Windows, without giving me the option of booting Ubuntu.   How can I get Ubuntu back?


We want to rent a car in Las Vegas in February, travel across California and Arizona, then return it to Las Vegas approximately 35 days later. When I check car rentals it shows that at the 30 day mark we have to return to Vegas to have it checked then rent again. This doesn't work for us since we won't be anywhere near Las Vegas at the 30 day mark.   Does anyone know a company that lets you rent for 35 days, without having to go back to rental location at the thirty day mark?

I would like to visit a few places in the states sometime next year, and I was thinking of doing a road trip while I'm at it. I could start in either Los Angeles or Montreal, and then I would like to visit Philadelphia, Phoenix and Las Vegas during my trip; plus whatever there is on the way between - national parks would be on top of the list. I had about $2000 in mind for transport (excluding fuel).  So my question is: how would I get around? I could take the trains or public transport between the States and Montreal because I don't want to have any hassle with importing/exporting a car, and only drive a car in the states.   The 2 options I see here are either buying a used car and then sell it at the end of the trip, or rent a car.   I had a look at hertz.com, and they quoted me ~$3500 (including insurance) for 1 month and this car:   But it seems like I have to return the car to the pickup location with Hertz, because if I choose a different drop off location it shows "not available" for this car. Did I just pick the wrong drop off location here of is A->B really not an option? Do I get different rates at different locations? Has anyone advice on how to get this cheaper? I searched for "relocation" on their site, but didn't find anything.  The second option would be to buy a car for $3000, and then sell it at the end of my trip, and hopefully get more than $1000 back for it. Has anyone experience with this? Are there any places better for buying or for selling a car? In which state would I get "more car" for the same amount of money? Would there be any paperwork involved if I buy a car in one state and sell it in another? Anything else I need to know when buying or selling a car?


I have recently observed an undocumented state "I" displayed in htop output.  The default help message doesn't disclose any information about this hidden state.  Given all processes that hold this state are kernel processes(threads), there might be reasons that the "I" state is not documented in the default help message. However, I'm still curious what the "I" means and whether there are more special states for kernel threads. Any suggestions?

In man page of ps under Process State Codes, the I flag (capital i) is not mentioned but ps aux shows the I flag in some processes as shown by the image below.    What does the I flag mean?


I am serializing following class to JSON and returning through API:  public class HighStockChart {     public string name { get; set; }     public IEnumerable<HighStockChartData> data { get; set; } }  public class HighStockChartData {     public DateTime x { get; set; }     public double y { get; set; } }   Now when I am doing a console log  for data , I am getting following format:  {x: "/Date(1519154100000)/", y: 392.61} {x: "/Date(1519179300000)/", y: 392.732} {x: "/Date(1519212600000)/", y: 392.732} {x: "/Date(1519173000000)/", y: 392.733} {x: "/Date(1519187400000)/", y: 392.757} {x: "/Date(1519169400000)/", y: 392.761}   But I need the data without /Date. Example format below:  {x: 1519154100000), y: 392.61}  {x: 1519179300000), y: 392.732}  Edit:  To build the date String I am doing ,  IEnumerable<HighStockChartData> stockChartXY = tag.Select(t => new HighStockChartData                     {                         x = DateTime.ParseExact(t.Date, "dd MMM yyyy HH:mm:ss:fff", null),                         y = double.Parse(t.Value)                     });  HistorianTagDetail historianTagDetailsresult = new HistorianTagDetail                     {                          Average = average,                         HighStockChartData = stockChartXY                     };  chartData = ChartHelper.CreateHighStockChart(tagDetails);  return Json(chartData);   By using jQuery to loop through and strip all data will work only if I am using q webClient. I have to find another solution , If I will try to send that as an email , using a service or use the chart in desktop application , then have to strip off the characters using C#.  Can I serialize the dates as Long (we have currently) but without "/Date()"?  Is there any custom attribute to set? Or Can I loop through object after serialization , remove the characters and then return?  ex:   chartData = ChartHelper.CreateHighStockChart(tagDetails); var jsonData =  Json(chartData);   var jDataList = new List<object>();   //Not actual code, just a theory I can think of  foreach(jData in jsonData) { //Do something to strip of the brackets //Add to a list  jDataList.Add(jData); } return jDataList;

I'm taking my first crack at  with jQuery. I'm getting my data onto my page, but I'm having some trouble with the JSON data that is returned for Date data types. Basically, I'm getting a string back that looks like this:  /Date(1224043200000)/   From someone totally new to JSON - How do I format this to a short date format? Should this be handled somewhere in the jQuery code? I've tried the jQuery.UI.datepicker plugin using $.datepicker.formatDate() without any success.  FYI: Here's the solution I came up with using a combination of the answers here:  function getMismatch(id) {   $.getJSON("Main.aspx?Callback=GetMismatch",     { MismatchId: id },      function (result) {       $("#AuthMerchId").text(result.AuthorizationMerchantId);       $("#SttlMerchId").text(result.SettlementMerchantId);       $("#CreateDate").text(formatJSONDate(Date(result.AppendDts)));       $("#ExpireDate").text(formatJSONDate(Date(result.ExpiresDts)));       $("#LastUpdate").text(formatJSONDate(Date(result.LastUpdateDts)));       $("#LastUpdatedBy").text(result.LastUpdateNt);       $("#ProcessIn").text(result.ProcessIn);     }   );    return false; }  function formatJSONDate(jsonDate) {   var newDate = dateFormat(jsonDate, "mm/dd/yyyy");   return newDate; }   This solution got my object from the callback method and displayed the dates on the page properly using the date format library.


If in Maschke's Theorem, for group ring KG where K is any field s.t. char $K \nmid |G|$, I take G to be finite , then I know Maschke implies that KG will be semisimple so it is isomorphic to a direct sum of Matrices over division rings, but how does adding that G is also abelian implies that KG will be direct sum of fields?

Let $G$ be a finite abelian group and $k$ a field in which the group order is invertible. Then the group algebra $k[G]$ is a semisimple ring by Maschke's theorem. As $k[G]$ is also abelian, the Artin-Wedderburn theorem implies, that $$ k[G]=K_1\oplus\ldots\oplus K_n $$ for fields $K_i$.  I have two questions and I have to admit that they are both a little awkward:   Is $K_i$ an algebraic field extension of $k$? I am sure that this is true but I can't find an argument. If (1) is true and $k$ is algebraically closed, I can deduce, that each irreducible representation of $G$ is one-dimensional. However, this is quite easy to prove directly and the structure of $k[G]$ (obtained by non-trivial theorems) seems to contain more information. Can I deduce something about the number of irreducible representations? What can I say for $k$ not algebraically closed?


Sorry guys, I am new to this site and new to ffmpeg. I am a using ffmpeg to concat videos by taking input from a file as shown below.  mistakes.txt  file VID_20191031_030050.mp4 inpoint 2222 outpoint 2233 file VID_20191031_030050.mp4 inpoint 2350 outpoint 2364   I am using the following command to concat multiple clips whose inpoint and outpoints are read from a single file(mistakes.txt). I am writing the output to VID_20191031_030050_mistakes.mp4.  ffmpeg -f concat -i mistakes.txt VID_20191031_030050_mistakes.mp4   Basically I have sports video and I am trying to see my mistakes in a single clip. I also want to display a text on the screen which includes the details of that specific mistake. For now lets say that I want the name of the file i.e. mistake to be displayed on the bottom right corner of the screen. Is there an efficient way to do this.  I have gone through this link(), but I couldn't map it to my problem, as I am new to ffmpeg.  P.S. I am using python3 to automate the editing of my sports clips.

What I want to do: I have a folder with a great bunch of images. Using these files I want to create a slideshow video, while I want the filename of each of the original images burned into the bottom right corner.  What my current state is: I made a script to fetch all .jpg files in the folder and make a ffmpeg-readable list and I know how to concat the images to a slideshow and everything works as intended. I also figured out how to use FFmpeg's drawtext to get a handsome filename stamp.   Now the problem: Currently FFmpeg grabs the last filename from the file list (for example IMG_9935.jpg) and burns it into the entire video. But what I want to happen is, that it changes with each 'new' image - just like a timecode would change during the course of the video.  Is any of this possible?  As a side note: "Why don't you burn the filename beforehand?" I'd like to keep it as flexible as possible. So if there's another way round, I'd like to go that way.  for i in *.jpg; do echo "file '$i'" >> files.txt; done  ffmpeg -r 1/3 -f concat -safe 0 -i files.txt -vf "eq=text=$(echo $i | rev | cut -d / -f 1 | rev): fontcolor=0x808080: fontsize=14: x=w-tw- 10:y=h-th-10" -c libx264 filename.mp4


How do i prove that if $$3|a^2+b^2\Rightarrow  3|a ,\, 3|b \,?$$ The book (Problem-Solving Strategies by Arthur Engels, page 43, E9) does not prove this theorem and leaves it as an excercise. I'm relatively new to this kind of problems and I just can't get any ideas. If anyone could prove that or, even better, just give me some hints it would be great. Thanks for your time

This is a part from a homework. I solved all examples apart from this one. So the task is: We know that $3$ divides $a^2 + b^2$. Prove that $3$ divides $a$ and $3$ divides $b$. I cannot think of anything useful. I know that $a^2 + b^2 = (a + b)^2 - 2ab$, but I don't see how it can help me :(  Best regards, Petar


In Linux based operating system we can find Kernel release and Kernel version. So what is the exact different between Release and Version ?

I am using CentOS 6.2. In the man page of uname (i.e., man uname), it says:   -r, --kernel-release               print the kernel release   -v, --kernel-version               print the kernel version   While trying the command, it shows   [max@localhost ~]$ uname -r  2.6.32-220.17.1.el6.i686  [max@localhost ~]$ uname -v  #1 SMP Tue May 15 22:09:39 BST 2012   -v should show the version, right? But where is it showing the version? -r is showing the version detail.  Why is this so?


If $\{u_{k}\}_{k=1}^{\infty}$ is a sequence in an Banach algebra (and more specifically, in the set of all the bounded linear operators of a Banach space $X$). If $$\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0,\quad \text{for every $\lambda\in \mathbb{D}(0, 1/3)$},$$ where the $\mathbb{D}(0, 1/3)~$ denotes an open disc centered at $0$ and radius $1/3$.   Then can we conclude that $\boldsymbol{ u_{n}=0}$, for every $\boldsymbol{n}$?

Let $\{u_{k}\}_{k=1}^{\infty}$ be a complex number sequence. If $\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0$, for each $\lambda\in \mathbb{D}(0, 1/3)$(where the $\mathbb{D}(0, 1/3)~$denotes an open disc centered at $0$ and radius $1/3$). Then can we conclude $u_{n}=0$, for $n=1, 2, 3...$ ?


I was wondering if it was ok to downvote or upvote all the threads made by one person for personal reasons regardless of the threads content ?

I just noticed that I lost a bunch of points from my reputation score, and I used the "reputation" tab on my user profile page to try and track down the cause.  During my investigation, I noticed there was an unusual event of type "reversal". In the normal place of a question title, it says "voting corrected".     What does this mean, and what caused it?  Did I do something wrong? Why did I lose all of that reputation? Is the system punishing me for leaving too many good answers?  Is there any way to earn it back?


I have a system running Ubuntu 13.10 in a single partition. I need to create another ext4 partition, is it possible without reinstalling Ubuntu?  Please show me the procedure, if it is possible.

Previously, I have installed Windows 7 on my 320 GB laptop with three partitions 173, 84 and 63 GB each. The 63 GB partition was where the Windows was installed. The rest were for file containers.  Now I changed my OS to Ubuntu 12.04 LTS. I installed Ubuntu by replacing the entire Windows 7 on the 63 GB partition. The rest of the partitions remain as an NTFS Windows partition and I can still access them both (the 173 and 84 GB partitions).  Now I want to change the two partitions of Windows into an Ubuntu format partitions plus most importantly, I want to extend the 63 GB partition to more than a 100 GB because at the moment I am running out of disk space. Whenever I try to install any application, especially using wine, it always complains for a shortage of disk space.  How do I do the extending activity before I entirely format my laptop again and lose all the important files on my partitions?


For the community of inequalities.  If $z\leq x+y$  ,And  $x, y, z \neq -1$, prove :  $$ \frac{z}{1+z} \leq \frac{x}{1+x} + \frac{y}{1+y}.  $$  I have this proof, but I feel there should be other simpler (direct) proofs  My attempt:  $$ z(1-z+z^2-\dots)\leq (x+y)(1-z+z^2-\dots), $$  now the idea is that either $(1-z+z^2-\dots)\leq(1-x+x^2-\dots)$ and $(1-z+z^2-\dots)\leq(1-y+y^2-\dots)$ are true, and thus  $$ z(1-z+z^2-\dots)\leq x(1-x+x^2-\dots)+y(1-y+y^2-\dots), $$  which is exactly  $$ \frac{z}{1+z}\leq\frac{x}{1+x} + \frac{y}{1+y} $$.  Note: I have added the above condition for $x,y,z$ to  define The LHS and RHS of inequality

Show that $\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}$ , for $c \le a+b$ and $a,b,c \ge 0$  So need to show $\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}$  We have $\frac c {1+c} \le \frac {a+b} {1+c}\le \frac {a+b+2ab} {1+c}$  So the whole thing reduces to showing $c \ge a+b+ab$ ?  Can anyone help me with this, I'm rubbish at inequalities...


I tried googling letter art, but I don't know what it is exactly called, so my research was not fruitful. I'm talking about the part "Backwoods Distilling".  How to curve the tops or bottoms of texts like that?  PS. I know many of you will give the solution in Illustrator, which is very much welcome, but if someone knows how to do it Affinity Designer, please tell as well.

In the logo I'm making, there is a circle which has text wrapped around it on the upper half. I used the warp tool but it distorts the text and doesn't give the desired effect.  I can't use the type on path tool because that is not the effect I want.   I want the individual letters to be straight and the only way I can think of doing that is to mannually edit each letter to sit around the circle. I was wondering if there was another, more efficient way to fo that.


I'm running a Windows, Mac, Linux tripleboot off of a single SSD, and am now, unsurprisingly, in need of additional storage space in the form of an external hard drive. I want to format my hard drive so that files that can be read and written from all three operating systems.  According to this  from 2009, NTFS can be used on Windows, Mac, and Linux, after installing some drivers.  However, it looks like the , and , so NTFS is no longer a viable solution for Mac OSX.  So the question is, what is the best file system to use today for an external hard drive connected to Windows, Mac, and Linux?  I'm thinking exFAT, since Windows and Mac support it natively, and (Ubuntu) Linux supports it after running sudo apt-get install fuse-exfat.  Are there any limitations I need to consider?

Aiming to have a large internal HDD to store files, and several OSes on separate disks. They all need to read and write to one storage drive.  Ubuntu (13.04+) Mac OSX (10.8+) Windows (7+)   What format should the drive be? Would like to avoid buying third party software, here's what I have discerned so far:   NTFS - Can't be written to by Mac without buying third party software?      ext3 - Windows can't read, . Mac has  HFS+ -  exFAT - , but


Suppose $f\in C^{\infty}[a,b]$, and for any $n\geq 1$, $f^{(n)}(x)$ is non-negative. Prove that $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n\,,\, \forall x\in [a,b]\,. $$ Hint: using $R_n(x)=\frac{1}{n!}\int_a^x f^{(n+1)}(t) (t-a)^n dt $ and $f^{(n)}(x)$ is increasing.  I have tried in the following way: $$\begin{aligned} R_n(x)&=\frac{1}{n!}\int_a^x f^{(n+1)}(t) (t-a)^n dt \\&= \frac{1}{n!} f^{(n+1)}(\xi) \int_a^x  (t-a)^n dt\\&\leq \frac{f^{(n+1)}(b)}{(n+1)!} (b-a)^{n+1}. \end{aligned}$$ But I don’t know how to prove last step, which is $\frac{f^{(n+1)}(b)}{(n+1)!} (b-a)^{n+1} \to 0$ as $n\to\infty$. Or there is another method to solve this problem, looking forward to your answers! Thank you!

Let $f(x)$ be a nonnegative and infinitely differentiable function on $[-a,a]$ to $\mathbb{R}$ such that $\forall x\in[-a,a]:f^{(n)}(x)\ge0$.  Prove that the series:  $$\sum_{i=1}^\infty \frac{f^{(i)}(0)}{i!}x^i$$  converges $\forall x\in(-a,a)$.


The dimension of $\mathbb{R}^3$ over $\mathbb{R}$ with the usual operations is $3$, but can I change the dimension to one by changing the operations of $\mathbb{R}^3$?

Can we define a vector space structure on $\mathbb {R}^n$ other than usual scalar multiplication and usual addition such that the dimension of $\mathbb {R}^n$ over $\mathbb {R}$ is not $n$ but some $m$ not equal to $n$?


Is it possible to install Ubuntu on a Chromebook? If so, what are the minimum system requirements? Will Ubuntu preform as well on the Chromebook as on a Windows laptop?

I have installed Ubuntu 13.04 on my desktop (runs like a charm). Then I used Crouton to install Ubuntu 12. 04 on my chromebook samsung (ARM). It also works quite well.  Here comes my question. Can we have a true Ubuntu (preferably 13.04) on my ARM chromebook?  It seems that the Crouton/Ubuntu is not 100% the real thing.  The best option would be to have the full Ubuntu 13.04 and remove the Chrome OS.


After measuring a quantum state for a particular measurable, the state collapses to one of the eigenstate of that measurable. If we measure for momentum in an infinite square well, the system will collapse to a momentum eigenstate. But uncertainty in position is not arbitrary large (it's $2L$). So we are left with finite uncertainty in momentum and position at the same time. Does this mean that we aren't supposed to make such measurement? Or something else.

Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?


I an using ASP.NET MVC 4 to dynamically load a PartialView which contains  a  option. I want to bind a jQuery dblclick event to this and get the selected option, but the event is not firing. Here is my code:   <select size="5" id="popup-list">      <option value="test">test</option>  </select>   jQuery code:  $(document).ready(function(){      $("#popup-list").live("dblclick", function () {         var name;         $("select option:selected").each(function () {             name = this;             name = this.text();         });     });  });   I am not sure as the why this is causing a problem. The code is in a document.ready(), which should take care of the dynamic loading.

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


I'm a bit confused about the usefulness of reducing dimensions before doing a k-means clustering.  Suppose you want to apply k-means to a set points $(x_i)$ with high dimension. You want to minimize the cost function $\sum_i \|x_i-c_i\|^2$ where $c_i$ is the center of the cluster $x_i$ belongs to.  You have basically two methods:   A: do a k-means (Lloyd) directly on $(x_i)$ B: reduce the number of dimensions with some dimensional reduction method (such as SVD/PCA), and then apply k-means to the points with reduced dimensions   On the one hand, A is unlikely to find the global minimum, replications will help getting closer to it. It might have a high computational cost due to handling high dimension vectors and many replications.  On the other hand, B is more likely to get close to the global minimum (or even reach it) with fewer replications. But the minimum on the reduced version is not the original minimum (it is however known to be close to it). There is of course an additional computational cost for reducing dimensions.  All in all, is it known that B is better than A in either objective: minimizing the original cost function or computational cost?

Background: I want to classify the residential areas of a city into groups based on their social-economic characteristics, including housing unit density, population density, green space area, housing price, number of schools / health centers / day care centers, etc. I want to understand how many different groups the residential areas can be divided into, and what are their unique characteristics. This information could facilitate city planning.   Based on some examples (cf., this blog post: ), I figured out the way to do the analysis is:   First do PCA analysis. Determine the number of unique groups (clusters) based on PCA results (e.g., using the "elbow" method, or alternatively, the number of components that explains 80 to 90% of total variance).  After determining the number of clusters, apply k-means clustering to do the classification.       My questions: it seemed the number of PCA components is related to clusters analysis. So is that's true, if, say, we found 5 PCA components explained more than 90% of variation of all features, then we would apply k-means clustering and get 5 clusters. So would the 5 groups exactly corresponding to the 5 components in PCA analysis?   In another words, I guess my question is: What is the connection between the PCA analysis and k-means clustering?    Updates: Thanks to Emre, xeon, and Kirill's inputs. So the current answers:   Doing PCA before clustering analysis is also useful for dimensionality reduction as a feature extractor and visualize / reveal clusters. Doing PCA after clustering can validate the clustering algorithm (reference: ). PCA is sometimes applied to reduce the dimensionality of the dataset prior to clustering. However, Yeung & Ruzzo (2000) showed that clustering with the PC's instead of the original variables does not necessarily improve cluster quality. In particular, the first few PC's (which contain most of the variation in the data) do not necessarily capture most of the cluster structure.   Yeung, Ka Yee, and Walter L. Ruzzo. An empirical study on principal component analysis for clustering gene expression data. Technical report, Department of Computer Science and Engineering, University of Washington, 2000. ()  It seemed PCA is necessary before a . Based on Ibes (2015), in which cluster analysis was run using the factors identified in the PCA.    Ibes, Dorothy C.  Landscape and Urban Planning, Volume 137, May 2015, Pages 122–137.


Show that there exists, for $p,l$ primes, an element of order $l$ in $(\mathbb Z/p\mathbb Z)^\times$ iff $p-1$ is a multiple of $l$.  Firstly, if an element $a$ of order $l$ exists, since $a^{p-1}=1$, we have $l\mid p-1$. However I struggle with the other implication.

Lemma $1.92$ in Rotman's textbook (Advanced Modern Algebra, second edition) states, Let $G = \langle a \rangle$ be a cyclic group. (i) Every subgroup $S$ of $G$ is cyclic. (ii) If $|G|=n$, then $G$ has a unique subgroup of order $d$ for each divisor $d$ of $n$.  I understand how every subgroup must be cyclic and that there must be a subgroup for each divisor of $d$. But how is that subgroup unique? I'm having trouble understanding this intuitively. For example, if we look at the cyclic subgroup $\Bbb{7}$, we know that there are $6$ elements of order $7$. So we have six different cyclic subgroups of order $7$, right? Thanks in advance.


This seems like a simple issue and I found dozens of answers around this site and the web, but none of those has fixed my issue.  I'm trying to add a shape with rounded corners. I want the colors to be solid but they are blending with the background. I'm going to be making 100s of images, so I didn't want to have to copy & paste to a new layer and try and crop the off-color parts (like some of the solutions call for).   Any simple solution will be very much appreciated. If I have to use Illustrator I could, but I would prefer not to.

Whenever I create a square shape at 100% zoom, I get perfectly sharp edges. However, once I zoom in at any amount, drawing a perfect square (without anti-aliasing) is very difficult. Is there a way I can turn this off?


On the , there is a section for Top Network Users, which links to the profiles of the askers and answerers of the top posts featured on that page:    It reads     Top Network Users      These users asked or provided top answers for the posts on this page   It may just be me, but that sounds a bit off: they "asked or provided top answers?" I understand the intent: "they asked (or provided top answers for) the posts on this page, but it sounds like it says that they asked top answers, which doesn't make any sense.  You could insert commas to emphasize the or:     These users asked, or provided top answers for, the posts on this page   However, this still sounds quite off and I think it could use some rewording.  I think that it should be something along the lines of     Top Network Users      These users provided the top questions and answers on this page   (Thanks, random)  However, anything moderately clearer will suffice.

So, I noticed last week that . Well, it turns out that the wording is actually completely incorrect.  noticed that all of the users on the front page are the OPs of the questions listed there. This area is the issue:  It reads  Top Network Users These users asked or provided top answers for the posts on this page  I found that confusing, so I proposed this wording (provided by ):  Top Network Users These users provided the top questions and answers on this page  But, again, the users listed there are the ones who asked the questions; none answered them:  (click for larger image) Since all of the "Top Network Users" are the askers, it should be something like this:  Top Network Users These users asked the questions listed on this page  Either that, or answerers of those questions should be included along with the askers.


\documentclass{article}  \usepackage[utf8]{inputenc}  \begin{document}  \maketitle  \text{I can't use the #-symbol}  \end{document}

I need to output the below text but since \ is a special character, I cannot:  [RegularExpression(@"\d+")]   Also sometimes I need use dollar sign $ as well but it seems to be a special char, too.  What is the way to escape those in LaTeX?  UPDATE:  I used \verb as suggested but I am unable to run commands inside it. Also, in the output the font and the opacity of the text is different than the default one:  \verb|[RegularExpression(\newline @"\d+")]|


<aura:component>  <aura:if isTrue="{!$Browser.isTablet}">     You are using a tablet device      ( <lightning:icon iconName="utility:tablet_portrait" size="x-small" alternativeText="Indicates approval"/> ). </aura:if> <aura:if isTrue="{!$Browser.isPhone}">     You are using a phone      ( <lightning:icon iconName="utility:phone_portrait" size="x-small" alternativeText="Indicates approval"/> ).     This message is rendered using isPhone
 </aura:if> <aura:if isTrue="{!$Browser.isIPhone}">     You are using an IPhone, If it is IPhone X, _/\_
 </aura:if>  <aura:if isTrue="{!$Browser.formFactor == 'DESKTOP'}">     You are using a DESKTOP      ( <lightning:icon iconName="utility:desktop" size="x-small" alternativeText="Indicates approval"/> )     Browser device </aura:if> <aura:if isTrue="{!$Browser.formFactor == 'PHONE'}">     You are using a Phone, This message is rendered using formFactor </aura:if> <aura:if isTrue="{!$Browser.formFactor == 'TABLET'}">     You are using a Table device </aura:if>     Can we Do this in Lightning Web Component also?

We want to perform different logic in our LWC component js controller if the user is viewing our app in mobile.  With aura components, we could do:  if ($A.get("$Browser.formFactor") != 'DESKTOP') {     //do something for mobile only here }   How can we do this with LWC?


I'm applying a filter to multiple images, and it works well but really slow. (NOTE: images, filter, filter_size, div are already loaded and assigned) for (int i = 0; i < 5; i++)     out[i] = MyFiltering((Bitmap)images[i], filter, filter_size, div);  Therefore, I tried to use a Thread for each image filtering to speed up as below: for (int i = 0; i < 5; i++) {     FilterThread[i] = new Thread(() => {out[i] = MyFiltering((Bitmap)images[i], filter, filter_size, div)}     FilterThread[i].IsBackground = true;     FilterThread[i].Start(); }  I've tried different ways (e.g., ThreadPool, Task.Factory.StartNew), but always get the NullReferenceException in MyFiltering for input img (i.e., images[i]), which should be definitely NOT NULL! Anyone has an idea?

I met an interesting issue about C#. I have code like below.  List<Func<int>> actions = new List<Func<int>>();  int variable = 0; while (variable < 5) {     actions.Add(() => variable * 2);     ++ variable; }  foreach (var act in actions) {     Console.WriteLine(act.Invoke()); }   I expect it to output 0, 2, 4, 6, 8. However, it actually outputs five 10s.  It seems that it is due to all actions referring to one captured variable. As a result, when they get invoked, they all have same output.  Is there a way to work round this limit to have each action instance have its own captured variable?


My site works fine in local system, Upload code and database in server show Error.  Fatal error: Class 'SelectQuery' not found in /*/includes/database/database.inc on line 813

Fatal error: Class 'SelectQuery' not found in /home/content/71/11241671/html/drupal/includes/database/database.inc on line 813   We built a site in a subfolder called /drupal while building the site, and then moved the files to the root directory when we were finished but it gave the above error on a WSOD.  I have tried:    flushing the cache before moving  tried the registry rebuild truncating and replacing the registry and registry_file tables with a fresh install copy, and rebuilding the registry again rewrite base in htaccess setting a new base url in settings.php   But nothing has worked.  The second we put the files back into the /drupal folder, it works again. It also looks like the error mentions the file in the /drupal directory, which isnt there anymore.   This is on 7.37   Anyone have any ideas?


I downloaded some data from my bank for some transactions.  Using MS Excel 2008 V 12.2 on a Imac 2015  I usually convert the dates to YYYY-MM-DD (System International Standard) so there is absolutely no confusion ever (as what 11-02-04 would mean), as with other formats, at any time in the future.  However, when I convert the data in the cell to YYYY-MM-DD using format, half the dates never convert. They remain 26/07/2015. The strange thing is there is no real consistency, some entries convert, some don't. It seems strange if an engineer would program their dates to different formats. I've tried everything, including manually deleting the entire contents of the cell, and replacing it with 2015-07-26, then it converts to July 7, 2015. I've researched this with no luck on the web, even looking at some answers in this forum. Would appreciate any help. I can try to post the sample excel here or a screen shot if possible.   26/04/2015  27/04/2015 29/04/2015  30/04/2015  2015-04-06  2015-05-06 14/06/2015  15/06/2015  26/07/2015  27/07/2015 26/07/2015  27/07/2015 26/07/2015  27/07/2015 28/07/2015  29/07/2015 28/07/2015  29/07/2015 2015-08-08  2015-10-08 2015-08-08  2015-10-08 2015-08-08  2015-10-08 20/08/2015  21/08/2015 20/08/2015  21/08/2015 23/08/2015  24/08/2015 23/08/2015  24/08/2015   Note that in August, only one date reformatted properly. Also in April.   Am not a techie, so would a appreciate simple step by step instructions on how to fix if you can help. Also a layman's terms to explain why this is happening would be great. it seems really mindboggling to me how/why/if an software engineer? would change date formats in mid month or whatever, and also why the date won't change properly even with complete deletion and manual input.  My other question, out of interest, is that I assume hundreds of brilliant engineers design Excel. And in decades of existence, how come the International SI standard of YYYY-MM-DD, the first format I would include on day one, is NOT even an option in the drop down field?   Thanks in advance for your kind assistance,  Stu

I constantly have problems working with dates in Excel, I must be doing something wrong but I don't understand what.  I have a spreadsheet, exported from our exchange server, that contains a column with dates on. They have come out in US format even though I'm in the UK.   The column in question looks like this  04/08/2012 04/09/2009 04/01/2010 04/21/2011 04/05/2012 08/30/2009 08/29/2010 08/28/2011   In Excel, I have highlighted the column and selected Format Cells.... In this dialog box, I have selected the Date, selected English (United States) as the locale and chosen the matching date format from the list. I hit OK and try to sort the data by this column.   In the sort dialogue I choose this column, select sort on Values but the order only gives me options for A to Z, not oldest to newest as I would expect.  This in turn sorts the date data by the first two digits.  I am aware I could re-format this data to ISO and then the A to Z sort would work but I shouldn't have too, I'm obviously missing something. What is it?  EDIT: I messed up the bounty but this should have gone to @r0berts answer, his first suggestion of Text to Columns with no delimiter and choosing 'MDY' as the type of data works. Additionally, if you have a time (i.e. 04/21/2015 18:34:22), you need to first get rid of the time data. However after that the method suggested by @r0berts works fine.


i have a table like this  ID  NAME    Amount 1   cal     100 2   cal     200 3   cal     300 4   cal     400 1   ser     500 2   ser     600 5   ser     700   i want to write a select query so that i wil get the resul like this  ID  NAME            Amount 1   cal and ser     600 2   cal and ser     800 3   cal             300 4   cal             400 5   ser             700   here i need to group by id and sum of amount and concat the string name with same id and differnet name

Consider a database table holding names, with three rows:  Peter Paul Mary   Is there an easy way to turn this into a single string of Peter, Paul, Mary?


I've been looking around a lot, asking friends etc. about how the playthrough system specifically works in Borderlands 2 I am completely new to the franchise but have a decent idea of how things work.   I simply ask what specifically happens/changes in each playthrough , such as if gear will level up with you, at what points gear levels to you and so on. I may end up asking a few questions too. (Don't take it to heart about your skill at explaining things, I'm a slow learner :P).    P.S Bear in mind this is the Handsome Collection, playing on PS4. I will have all DLCs.

In Borderlands 1 the second playthrough had a "playthrough 2.5" once you beat The Destroyer, which changed the game so enemies (and most mission rewards) scaled to your level. Because of this it was best to play only story missions until beating the Destroyer so mission loot would be the best loot possible.  How do the playthroughs work in Borderlands 2? I've unlocked True Vault Hunter mode which starts out just like playthrough 2 in BL1, but I don't know if there's a "2.5" equivalent of if it's finally a real third playthrough or what.


$$f(x) =f'(x) f''(x)$$ Find the polynomial function $f:\Bbb R\to \Bbb R$, with $\deg(f) =n$. I have found that the degree is $3$ but just replacing gives me that $f(x) =\frac{x^3}{18}$

If a polynomial of degree $n$ satisfies $f(x) = f'(x)\cdot f''(x)$ (such that $n$ belongs to $\mathbb R$) then $f(x)$ is?  A) an onto function   B) an into function   C) no such function possible   D) even function   I tried this question by letting a polynomial $f(x) = ax^n + bx^{n-1} \cdots$ and then derivated it but it did not help much , then i assumed a quadratic expression $ax^2 + bx + c$ and got $ax^2 + x(b-4a^2)$.


In the anime, it says that Ikaros Apollon can destroy a country several times throughout the manga. Her shield is strong enough to contain the blast. Hephaestus, her cannon was created to be an anti-weapon and one slight mistake could lead to the destruction of the planet.  Is there any canon source that shows if this is true because the anime sometimes contradicts itself?

I was told that this anime is non-canon, but there are things from the anime that are just as the same as the manga.


When I use regular brackets () in Overleaf, they come out a regular size, but the size doesn't change if I have larger than regular terms inside the brackets. Do you know how I can control the size of the brackets? Thank you.

I have an equation contained inside \[...\], which automatically makes a \sum with sub- and superscripts turn big--so that the summation sign looks awkward inside parenthesis.  Any idea how to make the parenthesis completely enclose the whole summation?  \documentclass{article} \begin{document} We have: \[ \sum_{i=1}^n i = (\sum_{i=1}^{n-1} i) + n = \frac{(n-1)(n)}{2} + n = \frac{n(n+1)}{2} \] \end{document}


Pardon the interruption.  Medium is tracking me and I don't care about Medium. How can I block their cookies in Safari?

I would like to configure Safari to block cookies from specific domains, but looking at the settings I couldn't find any option. Then I searched for extensions that could implement this feature, having no success with that either.  Is there any way to block specific cookies?


$\lim_{n \rightarrow  \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ where the 2 inside the roots appear n times. For example if n = 2 :  $2^{2} \sqrt{2-\sqrt{2}}$ I discovered this. Has this been already developed/made before?

Find $\displaystyle \lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$.   By geometry method, I know that this is $\pi$. But is there algebraic method to find this ?   Thank you.