Uninstalling Window and using Ubuntu as the only OS
How do I remove Windows but keep Ubuntu?
In episode 10 of Full Metal Alchemist: Brotherhood, during the events in the Research Lab 5, Edward Elric is severely hurt. Later in the hospital, he writes down everything he had come to know, and he discusses with Alphonse, Hughes, and Armstrong about what to do with the information to continue the investigation.  At this point, King Bradley arrives with lemon into the room. He finds out what they are doing but acts very calm and chilled out and allows them to continue investigation while being careful.  Why did he not get Hughes and Armstrong killed at that point? Or at least tell them to not investigate at all? How can I get Firefox to open torrent files with Transmission? Every time I download a torrent file (.torrent), Firefox thinks I should open it with a text editor. How can I get it to open with Transmission? In Edit -> Preferences -> Applications, there's no setting for torrent files, and no apparent way to add a new filetype. In the Stack Exchange Android app, Achievements section I can see only Today's achievements. No Yesterday, Last 7 days, Last 30 days, ... details.  Previously in the 1.0.85 version itself I saw all the achievements.   I  refreshed the page many time, but still seeing Today's achievement only.  Screenshot for reference:     App Version: 1.0.85 Device Manufacturer: Motorola Device Model: XT1033 OS Version: 5.1 (2) Prove that $ \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z} $ Show that for $|z|&lt;1$, one have $$ \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots+\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots = \frac{z}{1-z} $$ and $$ \frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots+\frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots = \frac{z}{1-z} $$ Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots+2^k$. How do I run a Perl script from within a Perl script? I've got a Perl script that needs to execute another Perl script. This second script can be executed directly on the command line, but I need to execute it from within my first program. I'll need to pass it a few parameters that would normally be passed in when it's run standalone (the first script runs periodically, and executes the second script under a certain set of system conditions).  Preliminary Google searches suggest using backticks or a system() call. Are there any other ways to run it? (I'm guessing yes, since it's Perl we're talking about :P ) Which method is preferred if I need to capture output from the invoked program (and, if possible, pipe that output as it executes to stdout as though the second program were invoked directly)?  (Edit: oh, now SO suggests some related questions.  is close, but not exactly the same as what I'm asking. The second program will likely take an hour or more to run (lots of I/O), so I'm not sure a one-off invocation is the right fit for this.) How to parse hex values into a uint? Change default code page of Windows console to UTF-8 Currently I'm running Windows 7 x64 and usually I want all console tools to work with UTF-8 rather than with default code page 850.  Running chcp 65001 in the command prompt prior to use of any tools helps but is there any way to set is as default code page?  Update:  Changing HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\Nls\CodePage\OEMCP value to 65001 appear to make the system unable to boot in my case.  Proposed change of HKEY_LOCAL_MACHINE\Software\Microsoft\Command Processor\Autorun to @chcp 65001&gt;nul served just well for my purpose. (thanks to ) Can a possessive S be attached to nouns that are not human beings (or animals)? Can a possessive S be attached to nouns that are not human beings (or animals)?  For example, instead of saying "the back of the chair", can I say "the chair's back"?   I remember learning that not everything can get a possessive s, but this issue is not clear to me. Can we clarify to the OP that their question is not yet closed and the duplicates are only suggestions? How to fix lsregister failing associations after reboot? iPhone App CrashReport What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$  which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:  $$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &amp;xy &lt; 1 \\[1.5ex]  \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &amp;x&gt;0,\; y&gt;0,\; xy&gt;1 \\[1.5ex]  -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &amp;x&lt;0,\; y&lt;0,\; xy &gt; 1\end{cases}$$  Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you. Deciding whether two metrics are topologically equivalent in the space $C^1([0,1])$ Consider the space $C^1([0,1])$ and the function $d:C([0,1])\times C([0,1]) \to \mathbb R$ defined as $d(f,g)=|f(0)-g(0)|+sup_{x \in [0,1]}|f'(x)-g'(x)|$. Decide whether the metrics $d$ and $d_{\infty}$ are topologically equivalent in $C^1([0,1])$ (where $d_{\infty}=sup_{x \in [0,1]}|f(x)-g(x)|)$  My attempt at a solution:  If two metrics are topologically equivalent, then they have the same convergent sequences. Honestly, I couldn't do anything. I am trying to define a sequence of functions $\{f_n\}_{n \in \mathbb N}$ such that $f_n \to f$ in, for instance, $(C^1([0,1]),d)$ but $f_n \not \to f$ in $(C^1([0,1]),d_{\infty})$. Could it be this two metrics are topologically equivalent? If this is the case, how could I prove it? If not, I would appreciate any hint to find an adequate sequence of functions that works for what I am trying to prove. Why do I have to enter my password to log in? Using the Stack Exchange login, why do I have to enter my password again and again to log in? Wasn’t the whole purpose of OpenID to do away with this?  This doesn’t even only happen when visiting a new Stack Exchange site – it also happens for sites where I already have an account but which I don’t visit regularly. Here’s what I mean:    Of course I don’t remember my password (it’s a long random string). And so I have to go to my password manager and manually copy the password from openid.stackexchange.com’s account into the login form. – And incidentally, this is what my password manager looks like now:    (And, by the way, those are the only entries in my keychain so far, but by far not the only SE sites where I have an account.)  … I feel that this defeats the purpose of site-wide login and OpenID. In the olden times when MyOpenID was still working properly, I’d simply enter my OpenID on the login form, and the software would take care of the rest – namely, redirect to the OpenID service, (automatically) verify that I was authorised to use it, and redirect back to the original website, having logged me in. It was literally one button click for me. It’s substantially more work now, and I’d like to understand why this regression is necessary.  (Note: of course once a site’s password has landed in my password manager I (probably?) won’t have to enter it again. But still: wasn’t site-wide login and OpenID supposed to solve this already?) I'm interested in the properties of zero-diagonal symmetric (or hermitian) matrices, also known as hollow symmetric (or hermitian) matrices.   The only thing I can come up with is that it cannot be positive definite (if it's not the zero matrix): The Cholesky decomposition provides for positive definite matrices $A$ the existence of a lower triangular $L$ with $A=LL^*$. However the diagonal of $LL^*$ is the inner product of each of the rows of $L$ with itself. Since the diagonal of $A$ consists of zeros, so $L$ (and thus $A$) must be the zero matrix.  The sorts of questions that interest me are:   which symmetric matrices can be transformed orthogonally into a zero-diagonal matrix? what can we say about the eigen-values of a zero-diagonal symmetric matrix?. any other interesting known properties?? What is the difference between single-quoted and double-quoted strings in PHP? So, i want to make a timer that will turn on/off a circuit. I want to turn on the connected device at 08:00 and stop it at 20:00, so the interval is 12 hours. Correct me if i'm wrong but i can't use IC 555(or another IC), i'll need some huge resistors/capacitors. Is there another way to build that other than using a microcontroller? Why 'mkpasswd -m sha-512' produce different result every time it is called? I am using this function to create a password in the shadow file  $ mkpasswd -m sha-512 password $6$o50n4EfNOBzfs$K5l5AcOI5e0I9n2df0LmR11OLSPKSVM6AGBOnWVfDDfMubb46MrAj0zehdLNEoYzD2fj5q4XUdDwexj/dWLHy/ $ mkpasswd -m sha-512 password $6$ADmDOhDpW$gsYQuiKPpV1ewaUpRq2VGAvrcocffLQ8XGF94e6LNLZfKIjy6Ku.cZoR/5exeCi1ESjPB9TRM6HxCiY5BCNCV0 $ mkpasswd -m sha-512 password $6$Efj1agaI$HGkny3q1OTYT4KREh18gueHJae/3Bvil0iOEhfXj8bD.qy9Lg2UIQJMuBcq0XtG3xzueK.7cp0GfKr7tEo5YI/ $ mkpasswd -m sha-512 password $6$zdVPlCPai4Y$x3CBvlP99xZXZcr4PTiE..YLpZx39h5OHDxqazd9wFLImPuwsXF0M6KmqLzlCCrnQhI2lmEPdCzfmHA/fDiOz.   It makes me wonder how does the output value work as a password Can you explain this word to me?   Can I rephrase it as     Can you explain me this word? How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group? Orthochronous Lorentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$) $$ \Lambda^0{}_0 \geq +1.$$ How to prove they form a subgroup of Lorentz group? All books I read only give this result, but no derivation.  Why is this condition $ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be orthochronous?  The temporal component of a transformed vector is  $$x'^0=\Lambda^0{}_0 x^0+\Lambda^0{}_1 x^1+\Lambda^0{}_2 x^2+\Lambda^0{}_3 x^3,$$ the positivity of $\Lambda^0{}_0$ alone does not seem at first glance sufficient for the preservation of the sign of temporal component.  And how to prove that all Lorentz transform satisfying such simple conditions can be generated from $J_i,\ K_i$?    For those who think that closure and invertibility are obvious, keep in mind that $$\left(\bar{\Lambda}\Lambda \right)^0{}_0\neq \bar{\Lambda}^0{}_0\Lambda^0{}_0,$$ but instead $$\left(\bar{\Lambda}\Lambda \right)^0{}_0= \bar{\Lambda}^0{}_0\Lambda^0{}_0+\bar{\Lambda}^0{}_1\Lambda^1{}_0+\bar{\Lambda}^0{}_2\Lambda^2{}_0+\bar{\Lambda}^0{}_3\Lambda^3{}_0.$$  And I'm looking for a rigorous proof, not physical "intuition".