Residuals in a linear model are independent but sum to zero; isn't it a contradiction?
Errors and Residuals
Let $f(x)\in C^2:[0,1]\rightarrow\mathbb{R}$ satisfy $f(0)=0,f(1)=1,f'(0)=f'(1)=0$, prove that: $$\max_{x\in[0,1]}|f''(x)|\ge4$$  By using Taylor series I can prove that $\max_{x\in[0,1]}|f''(x)|\ge2$, but I cannot go further. Edit Mail rules manually I'm trying to draw a rounded rectangle with round "holes" using  but I need the rectangle borders to be thick and "holes" borders to be thin. Is there a better way to do so than:  \begin{tikzpicture}[x=1mm, y=1mm]     % Background         \fill[fill=black!10] (1.5, 1.5) rectangle (15, -15);      % Fill     \fill[fill=black!2, even odd rule, rounded corners=0.5mm]          (0, 0) rectangle (10, -10)          (2, -2) circle[radius=1.25mm]         (2, -5) circle[radius=1.25mm];     % Rectangle border     \draw[thick, rounded corners=0.5mm] (0, 0) rectangle (10, -10);     % Holes borders     \draw[thin] (2, -2) circle (1.25mm);     \draw[thin] (2, -5) circle (1.25mm); \end{tikzpicture}   The result image should look like this: Solve the recurrent relation Real analysis derivate problem. How can I prove this? Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea. How do rays of light getting focused at a point create an image?  How are the information of object's shape and dimensions expressed as an image when light rays converge?  Is it understandable by solving wave equation for light? How do I install wine? Can someone please explain how I can install  in the following ways?   By using the package manager? By downloading and installing a pre-built package? Reconciling different definitions of left-exact functor What's the difference between the server version and the desktop version? What's the difference between the server version of Ubuntu and the desktop version? Page 105 of D. Burton's A First Course in Rings and Ideals reads     It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ if and only if $n=m$.   Well, it is getting difficult for me, any help or would be appreciated.    My progress so far: Suppose $\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ is an isomorphism. Then $\phi(u)=u$ for every $u \in \mathbb{Q}$, so that $\phi(\sqrt{n})^2=\phi(n)=n$. If $\phi(\sqrt{n})=a+b\sqrt{m}$ for some $a,b\in\mathbb{Q}$, then $(a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n$ implies $$a^2+b^2m = n$$ and $$2ab=0.$$  Then $b=0$ cannot happen since that would imply that $\sqrt{n}\in\mathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$ If $\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ were an isomorphism then there would exist some $s\in\mathbb{Q}$ such that $$s^2m=n,$$ then $s=b$ or $s=-b$. Therefore the only isomorphisms from $\mathbb{Q}(\sqrt{n})$ to $\mathbb{Q}(\sqrt{n})$ are $\phi(u+v\sqrt{n})=u+vs\sqrt{m}$ and $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.    I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :) A secant inequality for convex functions Can the most important information of reputation audit appear first? I regularly find myself having to scroll to the bottom of the reputation audit to find all the good stuff that appears last. On my computer this is a single key press, but on my mobile it takes 30 seconds of scrolling. Lord alone knows how long it would be for Jon Skeet!  Since the principles of good user experience suggest putting the most important information first, why not do the same here? I never look at my first 5 days of rep gathering. But I often look at the current day, the true rep, days with 200 rep (hunting legendary badge!), trigger recalc etc. Pretty please, can we have it up top? How do you sort du output by size? How do you sort du -sh /dir/* by size? I read one site that said use | sort -n but that's obviously not right. Here's an example that is wrong.  [~]# du -sh /var/* | sort -n 0       /var/mail 1.2M    /var/www 1.8M    /var/tmp 1.9G    /var/named 2.9M    /var/run 4.1G    /var/log 8.0K    /var/account 8.0K    /var/crash 8.0K    /var/cvs 8.0K    /var/games 8.0K    /var/local 8.0K    /var/nis 8.0K    /var/opt 8.0K    /var/preserve 8.0K    /var/racoon 12K     /var/aquota.user 12K     /var/portsentry 16K     /var/ftp 16K     /var/quota.user 20K     /var/yp 24K     /var/db 28K     /var/empty 32K     /var/lock 84K     /var/profiles 224M    /var/netenberg 235M    /var/cpanel 245M    /var/cache 620M    /var/lib 748K    /var/spool Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise? If it then what would be the sum? Youtube video lag on 18.04 I just upgrade from Ubuntu 17.10 to Ubuntu 18.04, there is an issue when playing youtube video.The video stops play but audio playing, then I move the cursor &amp; video is playing...... What should I do?? What is the best way to put a backdoor in an encryption system? How can you put a backdoor into an encryption algorithm? Are there any techniques that can be used to reduce the time it takes to break a key?  I am looking for practical examples encryption schemes that have backdoors, because they were designed to or new techniques exposed them. It is my understanding that you can use a lack of entropy and diffusion in block ciphers which will reduce the necessary time that it takes to break a key. This is especially true with s-boxes, which can be designed to not increase the strength of the key. As for one-way algorithms, using none-primes can make it easier to reverse keys. What are other decisions that can be made to make a broken encryption algorithm?   The goal is find examples that can be used for encryption pedagogy, by collecting a sample of weak ciphers to practice on. Reinstalling Windows 10 - how many digital licenses I have linked to my account I need to reinstall Win 10 (after virus). After reading online, it looks like I would be able to do it easily since my system is activated with digital license.   However, I have a question. I have at least 3 PCs in my house with the same Microsoft account on them. Presumably, all of them have digital license linked to this same Microsoft account. I wonder if there is a centralized place to check all my digital licensed connected to my Microsoft account? Is this a valid proof for the reverse triangle inequality: Given $x,\,y\in\mathbb{R}^n$, $\|x-y\|\ge |\|x\| - \|y\||$. $\def\a{{\bf a}} \def\b{{\bf b}}$ $\def\x{{\bf x}}$ $\def\y{{\bf y}}$  I've seen various methods of proving the reverse triangle inequality but I was wondering if this one is fine as well.   Prove the reverse triangle inequality:  Given $\x,\,\y\in\mathbb{R}^n$, $\|\x-\y\|\ge |\|\x\| - \|\y\||$.   $\textbf{Solution}$ Let $\x,\y \in \mathbb{R}^n$. Now, $\x=\x-\y+\y$ and so $||\x|| = ||\x-\y+\y|| \le ||\x-\y|| + ||\y||$ by triangle inequality. Thus, $||\x|| - ||\y|| \le ||\x-\y||$ [*].   Now for $\y$. Again, $\y = \y-\x+\x$ so $||\y|| = ||\y-\x+\x||$ implies $||\y|| \le ||\y-\x|| + ||\x||$ by triangle inequality. So, $||\y|| - ||\x||\le ||\x-\y||$. Hence, $-||\x-\y|| \le ||\x|| - ||\y||$ [**].   By [*] and [**], $-||\x-\y|| \le ||\x|| - ||\y|| &lt; ||\x-\y||$ implying $\bigl ||\x|| - ||\y|| \bigr\rvert \le ||\x-\y||$ for all $\x,\y \in \mathbb{R}^n$. External program from our Java program How can I write a program in Java that will execute another program? Also, the input of that program should be given from our program and the output of that program should be written into a file.  This is my small set of code to get its output:  Process p = Runtime.getRuntime().exec("C:\\j2sdk1.4.0\bin\\helloworld.java"); BufferedReader input =         new BufferedReader(new InputStreamReader(p.getInputStream())); while ((line = input.readLine()) != null)    System.out.println(line);  input.close();   This was my set of code but this throws an IOException. How run a .jar file with a double-click?