Let $X$ be a random variable which takes values in $\{0,1,...,10\}$. What are the minimum and maximum possible values of the variance of $X$?  I know $\mbox{Var}(X)=E[X^{2}]-E[X]^{2}$, but I'm unable to think how I can maximize or minimize it.
Let $X$ be an arbitrary random variable  takes values in $\{0,1,2,...,10\}$. What are the minimum and  maximum values of variance of random variable $X$?
I have a client that has a Links page in SharePoint 2013. On the page is a column called URL that is set as a hyperlink column, but they want the column to accept pw://example.com/... instead of just ...  Is there a way to change the hyperlink column to allow for this behavior? I considered creating it as a single line of text and then making that text box a hyperlink-enabled box. Is there another native option? If not, which of these would be easiest? I am new to SharePoint so any help would be appreciated! Thanks in advance!   URL    Type the Web address: (Click here to test)  pw://test  Invalid URL: pw://test I'm trying to use biblatex and when I compile my document I get an error like the following:  Windows  data source C:\Users\&lt;username&gt;\AppData\Local\Temp\par-5061756c\ cache-890efc00b3ca6b775c7d44a325c1349fb2a3a3bd\inc\lib/Biber/LaTeX/recode_data.xml not found in .   Mac  data source /var/folders/v2/rld0ls7d2935gkvqv5hfr1x00000gn/T/par-616c616e/ cache-cdd483146f82a9655ce063f848d5139480fbf872/inc/lib/Biber/LaTeX/recode_data.xml not found in .   Linux  data source /tmp/par-6963617269756d/cache-f37ab610b7d79b2720a8ee3732849c6821705520 /inc/lib/Biber/LaTeX/recode_data.xml not found in . nginx location match issue auth_basic I have this configuration :  location / {     root /etc/nginx/euchet; }  location /app/ {     proxy_pass http://172.16.184.2:8080;     proxy_redirect off;     proxy_set_header Host $host;     proxy_set_header X-Real-IP $remote_addr;     proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;     proxy_set_header REMOTE_ADDR $remote_addr; }   And I need to add another location with auth_basic enabled only for the request: /app/f?p=3000  I tried with rexexp:     location ~ \/appt\/f\?p=3000 {             auth_basic "Hello! Enter password";             auth_basic_user_file /etc/nginx/euchet.htpasswd;              proxy_pass http://172.16.184.2:8080;             proxy_redirect off;              proxy_set_header Host $host;             proxy_set_header X-Real-IP $remote_addr;             proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;             proxy_set_header REMOTE_ADDR $remote_addr;     }   Nginx use previous location /app/.  I tried another way:  location ~* /euchet/f {      if ($arg_p = "4550") {         return 403;     }      error_page 403 = @hidden; }     location @hidden {             auth_basic "Hello! Enter password";             auth_basic_user_file /etc/nginx/euchet.htpasswd;              proxy_pass http://172.16.184.2:8080;             proxy_redirect off;              proxy_set_header Host $host;             proxy_set_header X-Real-IP $remote_addr;             proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;             proxy_set_header REMOTE_ADDR $remote_addr; }   And I get 403 error on f?p=3000, and 404 errors on f?p=&lt;another_number&gt;  How can I fix this problem? I read on the :     Focal-length determines the angle-of-view seen through a lens for a   given sensor-size. With a full-frame sensor a lens gives the same   angle-of-view as it would on a 35mm film camera. With a smaller   sensor, the angle-of-view becomes smaller. The crop-factor, also   called FLM, is the ratio representing the difference in equivalent   focal-lengths. So a 150mm on a full-frame DSLR such as the Nikon D700   gives the same angle-of-view as a 100mm on a D7000 since its FLM is   1.5X.      Short focal-lengths show a greater angle-of-view compared to longer   ones. [...]   In layman's terms, what is angle-of-view? Is it the same as focal length? If not, how is it different? How is it used? Why do I need to know about it? I'm developing a messenger application with encrypted chats.   In the first version of the app I've used PBKDF2 (10000 iterations, SHA1, random salt) to extend a short user password and generate keys to encrypt (AES256) and sign message (HMAC).  In the current version of the app I generate a random 130-bit key and use it instead of a user password. I think that using PBKDF2 with such a key has some overhead.   I see 2 possible ways to improve the current approach:   Keep using PBKDF2 and decrease iterations to 1 or 10. Don't use PBKDF2 and use HKDF instead.   What are the advantages and disadvantages of each approach? Are there any other options? How to drive multiple LEDs from a Li-ion battery (with varying voltage)? I want to supply constant voltage to my LED's but the supply varies from 3.7V to 4.2V (Li-ion battery). I cannot use 3.3V voltage regulator as it will need at least 4.3V voltage at the input but my battery can supply only 4.2V max. so is there any other way I can supply regulated 3.3V to my LED.   RGB LED = 3.3V, 60mA Battery = 3.7V - 4.2V, 3000mA Number of LEDs = 24 The LED's are connected in parallel $$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$. I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient? Assume toward contradiction that $f$ is differentiable at $0$. Then by definition we know that $f$ is continuous at $0$. Therefore $\lim _{x\to 0}f\left(x\right)=f\left(0\right)$ We know that $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$ Therefore we can conclude $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$ But we can also rewrite this as $\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$ Therefore $f(0) = 0$. Since $f$ is differentiable at $0$ we have $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$ Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$ $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$ Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$. Therefore $f$ is not differentiable at 0. Is Tactical Down-Voting ever valid? Has anyone noticed any tactical downvoting of their, or other people, answers?  For example, in a recent answer to a question, my answer described a possible solution.  Shortly afterwards, another answer appeared following the solution I had mentioned.    I know that this is completely in the spirit of the site, however my answer had been down voted even though it was still as valid as it ever was.  Another poster also had his equally valid solution down voted.  There is no way of knowing the reason behind the down voting, as no comments were left, but I am suspicious as to the ethics behind down voting someone else's valid answer in the hope of making their own answer more prominent.  Are people really that desperate to gain reputation?  I have seen comments in other posts that suggest that tactical downvoting has occurred.  So, does anyone think that it is acceptable to downvote another post, even though it is still correct but may not provide a complete answer, to make their own appear more favourably? $\sum_{n=1}^\infty \log \frac{n+1}{n}$  $\sum_{n=1}^\infty \log \frac{n+1}{n}$ =  $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$. So, the series diverges.  Is my procedure correct? Right so seeing as the new Deadpool movie shows the origins of Deadpool: specifically, he got cancer... blah blah had to get him the healing factor from Wolverine.  But that makes no sense — they used the same actor, so I'm assuming there must be some connection, right... right? I want to know where I can find updated information about all the stable and in-development versions of Ubuntu, where I can find this information?     This question is meant to be a canonical question. It expects to answer the following questions:         Is X version stable yet?   Is X version in beta?   Is X version still supported?   What versions are still supported?   What are the versions in Beta/Alpha?   Until where X version will be supported?   Have X version ceased support?   When will be X version released?   Where I can download any release?   When will the next LTS version of Ubuntu be released?         The list is not exhaustive. How do I prove this form of mean value theorem for integral? I only knew the standard mean value theorem for integrals. (i.e. $\int_a^b f(x)dx= f(c)(b-a)$ for some $c$ between $[a,b]$ where $f$ is continuous. This is directly derived by applying mean value theorem and Fundamental theorem of calculus) I'm taking numerical analysis this year and there is one theorem stated without a proof in my text. That is:  Let $f,g$ be continuous real-valued functions on $[a,b]$ such that $g$ is nonnegative. Then there exists some $c$ between $a,b$ such that $\int_a^b fg(x) dx = f(c) \int_a^b g(x) dx$  How do I prove this? Normal text is invisible when using beamer with notes and XeLaTeX What do I need for extreme macro photography? I am wanting to venture into some extreme macro photography, specifically insects, such as . After doing some research I have believe that my 1::25 macro lens will be sufficient if I am to use some extension tubes (which I have bought).  I would like to shoot the in the wild, is this possible/likely? Or is this always a catch, then release style of shooting?  Regarding the link I've posted, would this be possible with a 1:25 macro and extension tubes? I have a set of three (21mm, 31mm and 13mm). How would you go about shoot a blue bottle for example? Surely they are going to fly away? Should I use a tripod or will that cause more of a hindrance than a help? Are there any techniques/tricks I could use to get a sharp focus on my subject? In general what is the approach to switch the desktop environment or window manager on Ubuntu? The main difference between stock Ubuntu, Kubuntu and Lubuntu is supposedly the use of Unity vs. KDE vs. LXDE.  However, in order to try various desktop environments and window managers I don't want to have to reinstall the respective flavor from scratch just to get a particular desktop environments or window manager. There probably is an easier method than that.  The question: How can I easily test different desktop environments and respective compatible window managers (say KDE, LXDE, Unity, GNOME, XFCE and StumpWM)? Definition of complex number In many situations (problems as well as solutions) I encounter the complex number $i$ which many times is defined as $i^2=-1$ instead of $i=\sqrt{-1}$, since it is "more preferred". Does anyone know why? Is it because there are two solutions to the equation $x^2+a=0, \ \ a&gt;0$? I am able to backup my app's data from the root/data/data/you.package.name directory by just moving it to SD card manually. however, when i reverse the process,                                                                                                               AKA paste the data back to simulate a restore, the app just force closes, i have tried it with a bunch of apps and noticed the same reaction. Where am i wrong? Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:  $$|z-2i| = 2|z+3| \Leftrightarrow \\ |x+yi-2i|=2|x+yi+2|\Leftrightarrow \\ \sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\ \sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\ x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\ y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\ -3y^2-4y=5x^2+24x+26 \Leftrightarrow \\ ???$$  What do I do next? Examples in which the light maximizes the optical path length I posted a similar  about geodesics on Math.SE. Many sources ( for instance) claim that the light could maximize the optical path length in some cases. But I don't think it's actually true since between two points I can always imagine a path with arbitrary length. Therefore, or the light minimizes the optical length or it's stationary (neither maximum nor minimum). My question is: is the trajectory in red in the figure below a local minimum or a stationary curve?   If I'm wrong, please provide some examples which the light propagates in trajectory of maximum length. Windows Server 2008 R2: User account set to not show on login This afternoon I set up a Win Server 2008 R2 system for use with Team Foundation source code control. In order to give my team-mates the right to connect, I had to create a user account for each person, 10 in total. (I can't, and I don't want to create a domain).  Now, after doing this, all user accounts are displayed in the login screen.  What can I do so that all the accounts are not displayed there?