I am seeing http get requests in my servers access log for a site that is not hosted on my server and also the dns records of the site do not point to my server the domain points to 3 ip addresses in china. Anyone have any ideas as to why this would happen?
When looking at my Apache2 access logs, I occasionally see GET requests asking my server for a page on a completely different server. For example:  xx.xx.xx.xx - - [18/Jan/2012:07:12:10 -0500] "GET http://someotherserver.com/?sometimesmorestuff=likethis HTTP/1.1" 200 389 "-" "Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.8.1.12) Gecko/20080201 Firefox/2.0.0.12"   I remembered reading something about that in the past, but at the time I had never seen that on my server, so I didn't worry about it.   Now that I am seeing it on my server, I just ran a test and it appears that my site does indeed serve requests for content from other servers. This seems like a bad thing. So, two questions:  A) Is this something I should be worrying about?  B) How can I fix it?  I've tried a little Googling, but I don't even know what this sort of thing is called, and that makes it difficult.
Comparison theorem for ODE Here is something I'm trying to prove:     Conjecture: Suppose $f'(x) \leq \phi(f(x), x)$ and $f(a)=\alpha$. Suppose $g'(x)=\phi(g(x),x)$ and $g(a)\geq \alpha$. Then $f(x)\leq  g(x)\,\,\forall x$.   It definitely seems like it should be true, and I don't think we even need continuity of $\phi$.  (Edit: a user has correctly pointed out that we should require $\phi$ be locally Lipshitz in the first variable, uniformly with respect to the second variable. Let's add that assumption.)   I can prove it in the case that the inequality is strict. I'll place my proof below the fold. How can I extend it to weak inequality?  If it's wrong, I'd love to see a counterexample. A reference is fine; I have the book by Teschl, for example.  A similar question with stronger assumptions was asked .     Proposition: Suppose Suppose $f'(x) &lt; \phi(f(x), x)$ and $f(a)=\alpha$. Suppose $g'(x)=\phi(g(x),x)$ and $g(a)\geq \alpha$. Then $f(x)\leq  g(x)\,\,\forall x$.  Proof: Suppose not. Suppose WLOG there is a $b&gt;a$ such that $f(b)&gt;g(b)$. Let $c:=\inf \{x&gt;a:f(x)&gt;g(x)\}$. By definition of the derivative we have $f'(c)\geq g'(c)$, a contradiction, since $f'(c)&lt;\phi(f(c),c)=\phi(g(c),c))=g'(c)$. I've seen the full proof of the Triangle Inequality  \begin{equation*} |x+y|\le|x|+|y|.  \end{equation*} However, I haven't seen the proof of the reverse triangle inequality:  \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?  Thank you very much. Is the claim that functions of independent random variables are themselves independent, true?   I have seen that result often used implicitly in some proofs, for example in the proof of independence between the sample mean and the sample variance of a normal distribution, but I have not been able to find justification for it. It seems that some authors take it as given but I am not certain that this is always the case. A basic grammar rule is to use an instead of a before a vowel sound. Given that historic is not pronounced with a silent h, I use “a historic”. Is this correct? What about heroic? Should be “It was a heroic act” or “It was an heroic act”?  I remember reading somewhere that the h is sometimes silent, in which case it’s an, and when the h is pronounced, it’s a. But then I also remember reading that it depends on which syllable is stressed. And I also think I read somewhere that it might differ between British and American English.  Personally, I pronounce the h, and believe that a is correct. I find that it sounds incorrect to use an and pronounce heroic without the h.  So how do I know when to use a and when to use an with a word beginning with the letter h? Are both acceptable or is there one that is correct? About compactness and completeness of set $X = \mathbb{R}$ with the metric $d(x, y) = \frac{|x-y|}{1+|x-y|}$ Accepting an answer should automatically upvote it as well Is there ever a case where you would want to accept an answer, but not upvote it?  I can't think of a single situation where this would be the case.. So would it not make sense to have the "Accept" button upvote the answer too? Counting integer partitions of n into exactly k distinct parts size at most M How can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$?  The number of partitions $p_k(\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$  \binom{M+k}{k}_{x} = \prod_{j=1}^{k}\frac{1-x^{M+k-j+1}}{1-x^j}= \sum_{n=0}^{kM} p_{k}(\leq M,n) x^n $$  For the number of the partitions $p_k(\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$   p_{k}(\mathcal{D},n) = p_{k}(\mathcal{D},n-k) + p_{k-1}(\mathcal{D},n) $$  But what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size?  Update: Now I know the generating function for the number of distinct restricted partitions $p_k(\leq M, \mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$    \prod_{j=1}^{M} (1+xq^{j}) = \sum_{k,n=0}^{\infty}p_k(\leq M, \mathcal{D},n)x^{k}q^{n} $$ and there is also a recurrence relation $$    p_k(\leq M, \mathcal{D},n) =         p_{k-1}(\leq M-1, \mathcal{D},n-k) +         p_k(\leq M-1, \mathcal{D},n-k) $$ How can I prove this? Could you recommend a book, where I could read about this? Lost space after removing bootcamp So I removed bootcamp and now there is a partition called "free space" that I can't remove.  Here is my output from diskutil list:    Here is the screenshot from disk utility showing the partitions:    And here is the error I get when I try to remove "free space":    Heres the output when I run   sudo diskutil apfs resizeContainer disk0s2 0 Enumeration within self defined environment My self defined environment having a symbol at the beginning and at the end works fine for text without enumeration. When including enumerate the symbols are not there where desired. Is it possible to have the first symbol directly in front of &quot;1.&quot; i. e. in the same line and the ending symbol in the same line as the last line of the enumeration? With \ignorespacesafterend and \unskip I did not succeed. \documentclass{article} \usepackage{amssymb} \newenvironment{myenv}{$\blacktriangleright$}{\hfill$\blacktriangleleft$}  \begin{document} There is a text.  \begin{myenv} And an example \end{myenv}  Here is another text.  \begin{myenv} \begin{enumerate} \item A \item B \end{enumerate} \end{myenv}  \end{document} What are the different ways that a message can be displayed to a bash shell after a user logs in? I have a CentOS 5.7 VPS using bash as its shell that displays a branded greeting immediately after logging in via SSH. I've been trying to modify it, but can't seem to find where it is in the usual places. So far I've looked in the motd file and checked sshd_config for banner file settings. A banner file is not set.  Where else can I look for where the login message might be? How can I keep Ubuntu updated automatically? I want help to  make Ubuntu 'update &amp;&amp; upgrade' itself automatically, select the settings for this automatic system.  How can I do it? Please advice about  where to find the menu (GUI) or file (command line) to do it, what settings that can be recommended (if different from the default).  As a bonus, please describe how different types of program packages are treated,  program packages from Ubuntu repositories and PPAs snap packages flatpak packages other kinds of programs/packages. Say i have this PHP code:  $FooBar = "a string";   i then need a function like this:  print_var_name($FooBar);   which prints:  FooBar   Any Ideas how to achieve this? Is this even possible in PHP? Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative? PHP: Return all dates between two dates in an array Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance! Python: access class property from string How to get a tool that can only break specific blocks and has a name? So, right now I'm using this:  /give @p minecraft:diamond_pickaxe 1 0 {CanDestroy:[minecraft:stone]}   But I need one that also has the Name value. Anyone know this one? What is the difference between a symbolic link and a hard link? Recently I was asked this during a job interview. I was honest and said I knew how a symbolic link behaves and how to create one, but do not understand the use of a hard link and how it differs from a symbolic one. Is it possible, on Debian with Klipper running, to have a separate "clipboard stack" for a particular application?  I'd like the text that I copy in application Y not to be shared with all the other applications running.  All the solutions mentioned at  seems like overkill. Theorem style with name as argument Is there some way to get a theorem type which will make a theorem with the name passed as a parameter? In other words, is there a way to define "named" such that   \begin{named}[NAME] text \end{named}   will print   "NAME: text"?  I tried using the definition \newtheorem*{named}{}, but that results in   "(NAME): text", which is quite different from what I am trying to do.  I know that I could also create a new instance of \newtheorem for each named theorem I want to use, but that is clunky and inconvenient. Is there a better way?