A bit of background I have an Android phone, which unfortunately cannot usb-tether while it is connected to the internet (using wifi or mobile data, but it works when the device is not connected to the internet. Wifi and Bluetooth tethering have this problem too). Now I want to connect my PC to the internet using my android phone (using only wired connection). The unideal solution I came up with a rather hacky solution which has a minor problem: I compiled  - a proxy server - for my device (FYI, I used GOOS=linux GOARCH=arm environmental variables to build the binary) and ran the binary on my Android phone on port 8080 (I placed the binary in /data/local/tmp. I ran it the obvious way using adb shell without superuser privileges). Then I used adb forward tcp:8080 tcp:8080 to forward device's port 8080 to my PC's port 8080. After that I used localhost:8080 as a socks5 proxy on my PC and could access the internet. The problem is that I cannot use certain applications as they don't support connecting through proxy (some of them run on a lower network layer than a socks5 proxy, so there is no way for them to work). My question Is there any way to run a VPN server on my Android phone (which conveniently has root access too) and use it like the way I currently do? If so, I'm only interested in solutions using well-maintained opensource software.
Can we make a Vpn server run on Android. Can applications do this? If yes, which protocols are supported and is this a good idea to use such apps? If not, what are some issues that we may face?
How do I add permissions via command line for "everyone" on external HDD I have an external HDD and I kind of messed up the file permissions but when fixing it I thought it is ok bc with my username I can access the files perfectly fine. Now that I use this with two PC (actually ATM I don't have access to my other PC) I can't access these files.  The problem is this directory has hundreds of folders with no permission for "everyone". I would like to give it the default permissions including have all access for the user "everyone". How do I do that via command line for these hundreds of folders? LaTeX to plain text for e.g. generation of statistics I would like to convert a large LaTeX project (i.e. spanning multiple files) into plain text. The purpose is generation of statistics, so representing mathematics is not an issue. In fact, all mathematics is ideally ignored.  I have found  but could not get it to run. It seems unfinished, but is exactly what I am looking for otherwise. What's a big-vocabulary word for someone with a big vocabulary?  I'd like to say I'm looking for a "_____". We have the following points: $$ (0,0)(1,51.8)(1.9,101.3)(2.8,148.4)(3.7,201.5)(4.7,251.1)  \\ (5.6,302.3)(6.6,350.9)(7.5,397.1)(8.5,452.5)(9.3,496.3) $$ How can we find the best fitting line $y=ax$ through the points? My calculator has the option to find the best fitting line $y=ax+b$ through these points, which is:  $$y = 53.28x + 0.37$$  How can I find the best fitting $y=ax$? It seems to me we can't just remove the $0.37$ without compensating in the $a$? ${\rm Aut}(G)$ is cyclic $\implies G$ is abelian I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK. Proof: Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian. The proof seems to be quite straightforward, but I'd rather ask for advice. Can I enter Croatia with single entry visa Schengen? Identification of particles and anti-particles The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become the normal particle and the rare and unusual positrons become the .  My question is, when you have made the choice of the electron and positron as particle and anti-particle does this automatically identify every other particle (every other fermion?) as normal or anti?  For example the proton is a particle, or rather the quarks inside are. By considering the interactions of an electron with a quark inside a proton can we find something, e.g. a conserved quantity, that naturally identifies that quark as a particle rather than an antiparticle? Or do we also just have to extend our convention so say that a proton is a particle rather than an antiparticle? To complete the family I guess the same question would apply to the neutrinos. I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.   In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:  $$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$  Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$  This is a convergent geometric series as $r = \frac{2}{3}\lt1$  now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?  Thanks Let $X \subset L^1(\mathbb{R})$ a closed linear subspace satisfying \begin{align} X\subset \bigcup_{p&gt;1} L^p(\mathbb{R})\end{align}    Show that $X\subset L^{p_0}(\mathbb{R})$ for some $p_0&gt;1.$   I guess the problem is that in infinite measure spaces the inclusion $L^p\subset L^q$ only holds for $p=q$. Is it maybe possbile to apply Baire's Theorem in some way? TikZ arrow emanating from node at an angle for a distance How do I draw a line of length l and at an angle of a from a circular node N in TikZ? How can I run scripts automatically when Ubuntu starts up so I don't have to run them manually after startup? $\sum_{n=1}^\infty \log \frac{n+1}{n}$  $\sum_{n=1}^\infty \log \frac{n+1}{n}$ =  $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$. So, the series diverges.  Is my procedure correct? AngularJS Directives: Change $scope not reflected in UI Inset in Blender not coming out even $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: . The context for my question might be a bit strange.   I have stuttering and therefore I have some difficulties pronouncing some words starting with particular phonetics. And I've found that if a sentence starts with a /th/ sound, it makes me feel uncomfortable.  Especially in the word is "thanks" or "thank you".  So I want to know if there's a popular alternative for "thank you" that may not seem strange and that unnatural. I've searched the website and I know there are alternatives like "I appreciate your help." but that sounds a bit overly formal for me. (Or is it? correct me if I'm mistaken.) Why did Littlefinger give Sansa to the Boltons? This question is about the TV show only.  In the books, Sansa is still with Littlefinger and everything that happens between Sansa and Ramsay in the show actually takes place between Ramsay and Jeyne Poole in the books.  However I don't recall why Littlefinger sends Sansa north. Apparently, in the books, it's safer to keep her traveling. So why did he do it in the show? How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? Specific system of differential equations I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 &lt; r \leq 1$ satisfies $4r - r^3 &gt; 0$, and thus $\frac{dr}{dt} &gt; 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive.  Can someone tell me if this is correct? Set root domain record to be a CNAME I need to create an NS record for a domain that is a CNAME, for the purpose of having two domains pointed at one IP, and not having to maintain the current IP address in two different places.  The DNS provider for this domain is DynDNS, but they block this operation:     CNAME cannot be created with label   that is equal to zone name   I can do this with another domain whose DNS is served by 1and1:  root@srv-ubuntu:~# dig myseconddomain.co.uk  ; &lt;&lt;&gt;&gt; DiG 9.4.2-P1 &lt;&lt;&gt;&gt; myseconddomain.co.uk ;; global options:  printcmd ;; Got answer: ;; -&gt;&gt;HEADER&lt;&lt;- opcode: QUERY, status: NOERROR, id: 61795 ;; flags: qr rd ra; QUERY: 1, ANSWER: 2, AUTHORITY: 0, ADDITIONAL: 0  ;; QUESTION SECTION: ;myseconddomain.co.uk.             IN      A  ;; ANSWER SECTION: myseconddomain.co.uk.      71605   IN      CNAME   myfirstdomain.co.uk. myfirstdomain.co.uk.     59      IN      A       www.xxx.yyy.zzz  ;; Query time: 298 msec ;; SERVER: 10.0.0.10#53(10.0.0.10) ;; WHEN: Tue Aug 18 14:17:26 2009 ;; MSG SIZE  rcvd: 78   Is this a breach of the RFCs or does DynDNS have a legitimate reason for blocking this action?  Followup Thanks to the two answers already posted I now know that 1and1 IS breaching RFCs to do this. However it does work and they seem to support it. For a company that hosts so many domains it seems very odd that they get away with doing this on such a massive scale without objection.  More followup  The output of "dig myseconddomain.co.uk ns" as requested.  root@srv-ubuntu:~# dig myseconddomain.co.uk ns  ; &lt;&lt;&gt;&gt; DiG 9.4.2-P1 &lt;&lt;&gt;&gt; myseconddomain.co.uk ns ;; global options:  printcmd ;; Got answer: ;; -&gt;&gt;HEADER&lt;&lt;- opcode: QUERY, status: NOERROR, id: 18085 ;; flags: qr rd ra; QUERY: 1, ANSWER: 2, AUTHORITY: 0, ADDITIONAL: 2  ;; QUESTION SECTION: ; myseconddomain.co.uk.             IN      NS  ;; ANSWER SECTION: myseconddomain.co.uk.      4798    IN      NS      ns67.1and1.co.uk. myseconddomain.co.uk.      4798    IN      NS      ns68.1and1.co.uk.  ;; ADDITIONAL SECTION: ns67.1and1.co.uk.       78798   IN      A       195.20.224.201 ns68.1and1.co.uk.       86400   IN      A       212.227.123.89  ;; Query time: 59 msec ;; SERVER: 10.0.0.10#53(10.0.0.10) ;; WHEN: Wed Aug 19 12:54:58 2009 ;; MSG SIZE  rcvd: 111