How do I find out if my processor is 32-bit or 64-bit?
32-bit, 64-bit CPU op-mode on Linux
Do superheroes/villains like Quicksilver or Flash with super speed get tired due to running? I mean a average human can run at his full speed for 15 to 20 mins before getting tired. Trained athletes can go on running for much more time without tiring out.  Do Superheroes like Quicksilver or Flash ever get tired due to excessive running? How to encrypt folders (not in Home)? Is there a way to encrypt a folder, which is not in Home (/home/user/) but even in a different partition, so only my user can access/read contained files?  Alternatively, I would like to understand if it is possible to turn a complete ext4 partition into an encrypted volume, which would be mounted at user login.  If possible, I would like to make the change without reinstalling Ubuntu.  My PC has (mount output):  /dev/sda1 on / type ext4 (rw,errors=remount-ro) /dev/sda3 on /home type ext4 (rw) /dev/sda4 on /home/igor/Personale type ext4 (rw)   sda4 is the partition containing folders I would like to protect. The code below compiles, but has different behavior for the char type than for the int types.  In particular     cout &lt;&lt; getIsTrue&lt; isX&lt;int8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;char&gt;::ikIsX  &gt;() &lt;&lt; endl;   result in 3 instantiations of templates for three types: int8, uint8 and char. What gives?  The same is not true for ints: int and uint32 which result in the same template instantiation, and signed int another.  The reason seems to be that C++ sees char, signed char and unsigned char as three different types. Whereas int is the same as a signed int. Is this right or am I missing something?  #include &lt;iostream&gt;  using namespace std;  typedef   signed char       int8; typedef unsigned char      uint8; typedef   signed short      int16; typedef unsigned short     uint16; typedef   signed int        int32; typedef unsigned int       uint32; typedef   signed long long  int64; typedef unsigned long long uint64;  struct TrueType {}; struct FalseType {};  template &lt;typename T&gt; struct isX {    typedef typename T::ikIsX ikIsX; };   // This  int==int32 is ambiguous //template &lt;&gt;            struct isX&lt;int  &gt;    { typedef FalseType ikIsX; };  // Fails template &lt;&gt;            struct isX&lt;int32  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;uint32 &gt;  { typedef FalseType ikIsX; };   // Whay isn't this ambiguous? char==int8 template &lt;&gt;            struct isX&lt;char  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;int8  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;uint8 &gt;  { typedef FalseType ikIsX; };   template &lt;typename T&gt; bool getIsTrue(); template &lt;&gt;           bool getIsTrue&lt;TrueType&gt;() { return true; } template &lt;&gt;           bool getIsTrue&lt;FalseType&gt;() { return false; }  int main(int, char **t ) {    cout &lt;&lt; sizeof(int8) &lt;&lt; endl;  // 1    cout &lt;&lt; sizeof(uint8) &lt;&lt; endl; // 1    cout &lt;&lt; sizeof(char) &lt;&lt; endl;  // 1     cout &lt;&lt; getIsTrue&lt; isX&lt;int8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;char&gt;::ikIsX  &gt;() &lt;&lt; endl;     cout &lt;&lt; getIsTrue&lt; isX&lt;int32&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint32&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;int&gt;::ikIsX  &gt;() &lt;&lt; endl;  }   I'm using g++ 4.something Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}$\ $(-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$  My attempt :   Assume $f$ is strictly positive  (if not n=0 works). Let $Z(f^{(n)})$ the number of zeros of $f^{(n)}$ on $(-1,1)$. Then we have: $$ \forall n\in\mathbb{N}:\;f^{(n)}(-1)=f^{(n)}(1)=0 $$ so that iterating the Rolle theorem to the successive derivatives of f we have  $$  \;Z(f^{(k+1)}) \geq Z(f^{(k)}) +1 $$ thus, by induction : $$ \forall n\in\mathbb{N}: \;Z(f^{(n+1)}) \geq Z(f^{(n)}) +1 $$ Therefore, $$ Z(f^{(n)}) \geq n $$ How can I find an another zero ?   I'm still looking for an another proof which does not use the lemma ..,  Thank you in advance. How do I get rid of "U3 System" on my USB drive? I have a USB drive (SanDisk SDCZ40-016G) that mounts in a way I've never seen before. When inserted, two items appear in Places -> Computer:    "SanDisk Ultra Backup: 16GB" contains my data, and "CD Drive: U3 System" contains a Windows executable. The part with the Windows executable doesn't look useful to me so I'd like to remove it. I used GParted to delete the only partition listed on the device and then I created and formatted a new partition:    Strangely, the "CD Drive" containing the Windows executable was perfectly intact after this operation.  What's going on? Viruses: Adaptation to a new host through repeated host jumps Why can't I run this C program? I'm trying to run my first "process" program, but I get the following error :  ./fork.c: line 4: syntax error near unexpected token `(' ./fork.c: line 4: `int main()'   I'm pretty sure that the code is correct:   #include &lt;sys/types.h&gt;  #include &lt;stdio.h&gt;  int main() {      pid_t pid;      printf("Hello fork()\n");      switch(pid=fork())  {          case -1: printf("Error by fork().....\n"); exit(0);          case 0: printf("I'm the child process \n"); break;          default: printf("I'm the dad \n"); break;      }   exit(0); }   What is wrong? What is the best way to remove and prevent leaf stains on concrete pavers? Why is this command deleting all entities? I'm trying to remove a specific, named, item if it is dropped on the floor. I have the following on my command block:  kill @e[type=Item] {Item:{tag:{display:{Name:"RedFlag"}}}   But it's removing all items, not just the one I want. What am I doing wrong?  I also tried  kill @e[type=Item] {Item:{id:"red_flower",Damage:2}}   But that didn't work either, it's still removing all items, even things that aren't red flowers! Is it possible to enable System Protection from the command prompt? Mathematical operators, such as function names, should be set in roman type, not italics. Latex already has commands for some operators, including \max, \min, and \log.  How can I define additional such commands? Is there a general method (which can be implemented by hand) to finding the $n$-th roots of $2 \times 2$ matrices? Is there a similar method for a general $m \times m$ matrix? (for $n &gt; 1$ and $n\in\mathbb{Z}$) I have an enum construct like this:  public enum EnumDisplayStatus {     None    = 1,     Visible = 2,     Hidden  = 3,     MarkedForDeletion = 4 }   In my database, the enumerations are referenced by value. My question is, how can I turn the number representation of the enum back to the string name.   For example, given 2 the result should be Visible. Prove that if $n$ is a composite, then $2^n-1$ is composite. Not sure if I'm doing this correctly but this is what I've done:  Assume that $n$ is composite and suppose $2^n-1$ is a prime for $n \gt 2$. Then, $2^n-1 = 2k$ for some $k \in \Bbb Z $, $\forall n$.  But this would be a contradiction since $n \gt 2$.  I'm not sure if this a correct proof.  EDIT: I realized that it's to prove $2^n - 1$ NOT $2^{n-1}$ so my attempt above is completely wrong. In order to solve  $$y''-2y'+y=te^t$$ I did:  $$y = (at+b)e^t\implies\\ y' = ate^t+ae^t+be^t\implies \\y''= ate^t+2ae^t+be^t$$  but when I put these in the differential equation I'm getting zeros eveywhere. Does somebody know what I'm doing wrong? Teams of users rails app While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so? Where do I ask questions about the Windows operating systems? Be it Win 8, 7 or XP.  The questions that I want to ask may not be strictly related to a problem I am facing, but related to files and features of the OS itself or asking just because of my curiosity. For example, I want to know the purpose of the file "Bootvid.dll" in a Win 8 OS.  Or, for example, I want to what "hal.dll" is.  Are these questions appropriate for Stack Overflow or Super User? If not, please recommend some SE sites in which I can post such questions, because I am interested in getting into the guts of the Windows OS. Open to all suggestions. When should I use an em-dash, an en-dash, and a hyphen? I generally know how to use a hyphen, but when should I use an en-dash (–) instead of an em-dash, or when should I use a hyphen (-) instead of an em-dash (—)? Homotopy equivalence of universal cover As part of am exam question (Q21F ), I'm trying to prove that if $X$ and $Y$ are path-connected, locally path-connected spaces with universal covers $\widetilde{X}$ and $\widetilde{Y}$, respectively, then if $X \simeq Y$ then $\widetilde{X} \simeq \widetilde{Y}$.  My attempt might be correct, but it seems too complicated and amongst the jumble I may have made some incorrect assumptions. So what I'm looking for is (a) verification or correction; and (b) ideas as to how I could simplify my argument.    My argument:  Let $p:\widetilde{X} \to X$ and $q:\widetilde{Y} \to Y$ be the covering maps, and let $X \overset{f}{\underset{g}{\leftrightarrows}} Y$ be a homotopy equivalence.  Fix a point $\widetilde{x} \in \widetilde{X}$, let $x=p(x) \in X$ and $y=f(x) \in Y$, and fix $\widetilde{y} \in q^{-1}(\{y\}) \subseteq \widetilde{Y}$. Also let $x' = g(y) \in X$ and fix $\widetilde{x'} \in p^{-1}(\{x'\}) \subseteq \widetilde{X}$.  Define a map $\widetilde{f} : \widetilde{X} \to \widetilde{Y}$ as follows. For $\widetilde{z} \in \widetilde{X}$ let $\widetilde{u}:[0,1] \to \widetilde{X}$ be a path from $\widetilde{x}$ to $\widetilde{z}$. Let $z=p(z)$ so that $u=p\widetilde{u}$ is a path in $X$ from $x$ to $z$. Let $v=fu$, so that $v$ is a path in $Y$ from $y$ to $f(z)$. Lift $v$ to a path $\widetilde{v}$ in $\widetilde{Y}$ with $\widetilde{v}(0) = \widetilde{y}$. Define $\widetilde{f}(\widetilde{z}) = \widetilde{v}(1)$.  Notice that $q\widetilde{v}=v$ so that $q\widetilde{f} = fp$.  Define $\widetilde{g}:Y \to X$ analogously: For $\widetilde{z} \in \widetilde{Y}$ let $\widetilde{a}:[0,1] \to \widetilde{Y}$ be a path from $\widetilde{y}$ to $\widetilde{z}$. Let $z=q(z)$ so that $a=q\widetilde{a}$ is a path in $Y$ from $y$ to $z$. Let $b=fa$, so that $b$ is a path in $X$ from $x'$ to $g(z)$. Lift $b$ to a path $\widetilde{b}$ in $\widetilde{X}$ with $\widetilde{b}(0) = \widetilde{x'}$. Define $\widetilde{g}(\widetilde{z}) = \widetilde{b}(1)$.  Likewise, notice that $p\widetilde{g}=gq$.  Claim: $\widetilde{X} \overset{\widetilde{f}}{\underset{\widetilde{g}}{\leftrightarrows}} \widetilde{Y}$ is a homotopy equivalence.  We have $p\widetilde{g}\widetilde{f} = gq\widetilde{f}=gfp \simeq p$ and $q\widetilde{f}\widetilde{g} = fp\widetilde{g} = fgq \simeq q$.  This shows that $\widetilde{g}\widetilde{f}$ and $\widetilde{f}\widetilde{g}$ are covering translations (deck transformations) and are therefore homotopic to the respective identity maps. So we have a homotopy equivalence.    Any comments would be appreciated.