How to install Ubuntu alongside Windows 8.1?
How do I install Ubuntu alongside a pre-installed Windows with UEFI?
How to break out of multiple loops? Given the following code (that doesn't work):  while True:     #snip: print out current state     while True:         ok = get_input("Is this ok? (y/n)")         if ok.lower() == "y": break 2 #this doesn't work :(         if ok.lower() == "n": break     #do more processing with menus and stuff   Is there a way to make this work? Or do I have do one check to break out of the input loop, then another, more limited, check in the outside loop to break out all together if the user is satisfied? Excel text alignment with custom formatting Is there anything in a custom formatted formula which defines the text alignment?  I know the different parts of the formula is numbers, negative numbers and so on, but I need to set the alignment.  Example:  _*# ##0_;_*(# ##0)_;_*"-"??_;_@_ The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. I've got an iPhone app with icon file Icon.png.  This icon shows up properly when the app is on the phone itself, but it doesn't show up in the applications pane in iTunes.  What do I need to do to get it to show up properly? .NET create scheduled task on server fails with E_ACCESSDENIED I have an ASP.NET website (in C#) that takes in user data and then attempts to create a windows scheduled task.  Of course, this works great on the DEV machine, but fails to run on the server.  I'm trying to figure out what permission(s) are required on the ASPNET user (or anonymous web user) to create tasks.  The error is:  Access is denied. (Exception from HRESULT: 0x80070005 (E_ACCESSDENIED))  Stacktrace:      at MyScheduler.NewWorkItem(String TaskName, Guid&amp; rclsid, Guid&amp; riid, Object&amp; obj)      at MyScheduler.CreateTask(String name)   I've done some searching, and the suggested resolution is to use the web.config 'impersonate' flag to force the application to run as a user with sufficient permissions, as opposed to the ASPNET account which may not have those permissions.  Example:  &lt;system.web&gt;     &lt;identity impersonate="true" /&gt; &lt;/system.web&gt;    Unfortunately, this does not seem to resolve the issue.  From the documentation I read, this should run as the anonymous web user, but it seems that user does not have enough permissions.  I altered the setting to specify a specific domain user that happens to be an administrator on the machine.  Example:  &lt;system.web&gt;     &lt;identity impersonate="true" userName="WindowsDomain\YourUserName" password="YourPassword" /&gt; &lt;/system.web&gt;    Doing this allowed the application to successfully create the Windows Scheduled Task.  So, obviously, with the correct set of Windows 2003 permissions I can get the app to perform as it does in the development environment.  However, I'm not about to place the network or machine administrator account's user credentials in plain text on a Web.config file.  Does anybody happen to know what permissions exactly need to be set in order to get the ASPNET account to behave as desired?  EDIT:  The Win32 API is being used to create scheduled tasks. Copy files from Network Mapped Drive I am trying to copy a file over from a mapped drive on a network to the current computer through a batch file.   I've already attempted the following below and still no results. The plan is to run a batch file to copy a file from a network drive down to the local C:\ drive.   @echo off copy H:\Videos\TargetVideo\TargetVideo.mp4 C:\Video\ Im trying to craft a regex that only returns &lt;link&gt; tag hrefs  Why does this regex return all hrefs including &lt;a hrefs?       (?&lt;=&lt;link\s+.*?)href\s*=\s*[\'\"][^\'\"]+        &lt;link rel="stylesheet" rev="stylesheet"      href="idlecore-tidied.css?T_2_5_0_228" media="screen">     &lt;a href="anotherurl">Slash Boxes&lt;/a>   thank you We (Russian citizens) are planning to go to Munich in late June. We would like to book our accommodation on AirBnB, but it is not very clear whether it will be accepted by the German embassy when we apply for a Schengen visa.  I can't find such information on the German embassy site or on the VFS-Global site. I've also searched for it on the Internet, and, while most sites agree that AirBnB booking is generally OK for a visa, I don't see a clear answer for Germany specifically, and there seems to be some exceptions for other European countries. Currently, once an edit by a user without full edit privilege is suggested, it can be either approved or rejected.  There is a known edge case that cause suggested edits to get auto rejected: (aka edit conflict)   User with full edit privilege click "edit" and start typing. Let's name this user Joe. Since the user from previous step did not yet send the edit, users without full edit privilege can still click the "edit" as well, oblivious to the fact someone else is editing the same post in those very seconds. Such user click "edit" too and start typing. Let's name this user Kenny. Kenny is faster than Joe: he send his edit first, creating new suggested edit. Joe is still typing. Joe finally done typing. He send his edit to its merry way oblivious to the consequences. As result of the full edit made by Joe, Kenny's suggested edit is kicked away and rejected. Boom.   This is all known and reported many times.  where the user got it hard with at least two auto rejections in short period of time. Some concrete examples: , , .  For edge case, it became pretty thick edge. This is not just minor hassle: enough rejections, and the user is banned from suggesting for a whole week: and remember we're talking here about good edits made by good users. The above used to count for the ban, but . Thanks Gilles for requesting such a change!   I know it was already suggested to lock edits and similar solutions, but I suggest whole different approach: instead of auto rejecting the suggested edit, dismiss it, i.e. cancel it like it never existed same way a bounty can be revoked by moderator.  In the edit suggestion page it will show "Dismissed" instead of Approved or Rejected maybe with a tooltip explaining what caused this. This will soften the blow from those edge cases, taking the edge out of them. :) Is there any way to recover unsaved notepad++ docs? How can I reuse existing materials, objects, etc from an existing .blend? How do I ensure that a Python while-loop takes a particular amount of time to run? How do you find the transfer function of a controller? Below is a block diagram of the system. I thought the circle symbol with the plus and minus sign was an op-amp but I'm not sure whether that's the case. I want to find the closed loop transfer function. If there was no feedback (open loop), then I think I could find the output as Y(s) = Vin*G. This would mean that the transfer function is Y(s)/Vin = G.  Any ideas for how to find the closed loop transfer function and what the circle means? I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series   $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$  Does it converge?   If so, what is its sum? How do I use itertools.groupby()? I haven't been able to find an understandable explanation of how to actually use Python's itertools.groupby() function.  What I'm trying to do is this:  Take a list - in this case, the children of an objectified lxml element Divide it into groups based on some criteria Then later iterate over each of these groups separately.  I've reviewed , but I've had trouble trying to apply them beyond a simple list of numbers. So, how do I use of itertools.groupby()?  Is there another technique I should be using?  Pointers to good &quot;prerequisite&quot; reading would also be appreciated. 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? Let $p &gt; 3$ be a prime such that $p \equiv 2 \pmod 3$. Define the elliptic curve $E$ over $\mathbb{F}_p$ by $y^2 = x^3 + 1$. Prove that $E(\mathbb{F}_p)$ consists of $p+1$ points.  Using Fermat's little theorem you can prove that $x^3 + 1$ is a bijection on $\mathbb{F}_p$. Hence, $$\#E(\mathbb{F}_p) = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x^3 + 1\} + \#\{\infty\} = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x\} + 1.$$ But then I am stuck trying to prove that $y^2 = x$ has $p$ solutions $(x,y) \in \mathbb{F}_p^2$. Normally, when a user edits one of my posts, I receive a notification like on the following depending on the case   User has editing privilege (click to visit revision)   Users' edit is submitted to review queue (click to visit review queue)     The above notifications were received instantaneously (or less than 30 minutes, in case caching is supposed to be the answer) as the edit was done/suggested.  But I did not receive any such notification today (yet) on an edit made on my . I think it is a possible bug and should be looked at. The recent notifications list I have is shown below: Strange characters when the output of `ls` is redirected to a file I want to list the components of the current working directory in a text file  ls -1 &gt; textfile   Output looks fine with more.   1010661085645 1010729039145 1010747080245 1010849051345 1010859053445 1011046075845   However when I view this textfile with emacs, several strange characters appear  [0m[01;34m1010661085645[0m [01;34m1010729039145[0m [01;34m1010747080245[0m [01;34m1010849051345[0m [01;34m1010859053445[0m [01;34m1011046075845[0m   Can anyone explain what is going on here? Prove $\limsup_{n \rightarrow \infty} a_n\geq1$ if $\lim_{n \rightarrow \infty} a_na_{n+1}=1$ Let $\{ a_n \}_{n=1}^{\infty}$ be a bounded sequence s.t. $a_n&gt;0,\ \forall n \in \mathbb{N}$. Suppose $\lim_{n \rightarrow \infty} a_na_{n+1}=1$, prove $\limsup_{n \rightarrow \infty} a_n\geq1$.   I tried to prove by contradicition using subsequence, but couldn't work it out.  Any help appreciated.