I have a computer with Linux installed. And I want to be able to read and write data to its disk from Windows 7. Is there any software or does Windows 7 have any functionality for it built-in? I've seen some website describing how to, but they are quite old:   So I wonder if there is any better way than they ones they have described there.
Is there any way I can use  partitions on Windows&nbsp;7?  I have tried to install , but it doesn't seems to run on Windows&nbsp;7.
Apparently recurrences like this cannot be solved with the Master Theorem:  $T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log(n)}$  Because $n^{\log_b(a)} = n^1$ is not a polynomial multiple of $f(n) = \frac{n}{\log(n)}$ since $\frac{n}{\frac{n}{\log(n)}} = \log(n)$, which is less than $n^\epsilon$ for any positive $\epsilon$.  But when is this really even an issue?   For example why wouldn't the time complexity be $n \log n$ using case 2 of the Master Theorem? Unrolling the recursion:  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{\log_2(2^k)}$$  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{k}$$  Doesn't this suggest $\Theta(n \log n)$ even though it supposedly cannot be done with Master Theorem? How to convert Wubi install into regular install? How long does it take for a black hole to form for an external observer? The well-known fable of an astronaut sending signals out to an external observer while falling toward an event horizon states that the time lapse between such signals becomes greater even if in the astronaut sends them out periodically (as judged in his inertial frame). When viewed from earth and weighing time-dilation due to the gravitational field of the collapsing star, how is it possible that a black hole can form in finite time (for any external observer) if it takes an infinite amount of time to "see" events occurring at the event horizon? Unable to start XAMPP using sudo in Desktop Entry Say I create an object as follows: let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot;,   &quot;regex&quot;: &quot;^http://.*&quot; };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot; }; Why is "can I get" replacing "could I have"? I noticed the other day when serving the public that when asking for something, people were saying "Can I get an xyz, please". The previous time I had such a job it was "Could I have an xyz", or "May I have..." if they were being very polite.  Does anyone know when this trend started and what caused it? Removing unwanted appearance of underlying mesh How do I set a default host for nginx? Uniform integrability of a backward submartingale Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]&lt;\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).  If we know $l=\lim_{n\rightarrow\infty}E[X_n]&gt;-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!  I have an idea, it is enough to show:    $\sup_nE[|X_n|]&lt;\infty$;  $\forall\epsilon&gt;0$, $\exists\delta&gt;0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)&lt;\delta$, we have $\sup_nE[1_{A}|X_n|]&lt;\epsilon$.    For $|X_n|=2X_n^+-X_n$, we have  $$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$  which implies $\sup_nE[|X_n|]&lt;\infty$.  Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.  $$E[1_{A}|X_n|]=2E[[1_{A}X_n^+]-E[[1_{A}X_n]$$  Does someone have an idea for this term? Thanks a lot..... Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct? ICS devices support creating backups using adb backup, and restoring them using adb restore. Using adb backup, one can select that everything should be backed up, or only a certain package.  However, adb restore lacks all the options of adb backup. When I call adb restore backup.ab, where backup.ab is only the backup of a single package, the phone shows me a warning dialog saying that all settings are going to be lost after restoring the backup. Is it true that restoring a single package backup wipes all data on the phone? If yes, is it somehow possible to restore a single package using adb restore? Interaction of Invoke Duplicity and Misty Step Multiple one to many relationships in series I am making a DB for storing reports that have following fields:  Ad_system, Ad_Campaing, Ad_Group, Keyword, Date, Impressions, Clicks, Conversions   Logic is that, Ad_system can have multiple Ad_Campaings, Ad_Campaing can have multiple Ad_Groups and Ad_Groups can have multiple Keywords, and for every Keyword there is a report. Should It be like this? Because it kind of makes sense to me, but it doesn't look correct. Primary key can also be:  ad.system.name for first table  ad.system.name ad.campaing.name for second table  ad.system.name ad.campaing.name ad.group.name for third table  ad.system.name ad.campaing.name ad.group.name keyword for fourth table  They will always be unique, but then Report will always have all DB columns in it, so it's kind of redundant.  Is the approach that I have in the diagram below correct? Or is a better one what I just described? If I am not mistaken both should satisfy 3rd NF. Is time travel possible? Is it possible to go back in time? I read somewhere that according to relativity, it is possible - involving black holes and other stuff - to jump into the past. Is it possible for anything to go back in time either continuously or by jumping? If $X_n \stackrel{d}{\to} X$ and $c_n \to c$, then $c_n \cdot X_n \stackrel{d}{\to} c \cdot X$ Let $X_n$, $X$ random variables on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and $(c_n)_n \subseteq \mathbb{R}$, $c \in \mathbb{R}$ such that $c_n \to c$ and $X_n \stackrel{d}{\to} X$. Then $c_n \cdot X_n \stackrel{d}{\to} c \cdot X$.   Here is my proof: By Lévy's continuity theorem it suffices to show that $$\forall \xi: \quad \hat{\mu}_n(c_n \cdot \xi) \to \hat{\mu}(c \cdot \xi) \qquad (n \to \infty)$$  where $\hat{\mu}_n$ (resp. $\hat{\mu}$) denotes the characteristic function of $X_n$ (resp. $X$). Using the tightness of the distributions one can show that $\{\hat{\mu}_n; n \in \mathbb{N}\}$ is uniformly equicontinuous. We have $$|\hat{\mu}_n(c_n \cdot \xi) - \hat{\mu}(c \cdot \xi)| \leq |\hat{\mu}_n(c_n \cdot \xi)- \hat{\mu}_n(c \cdot \xi)| + |\hat{\mu}_n(c \cdot \xi)- \hat{\mu}(c \cdot \xi)|$$ The first addend converges to zero as $n \to \infty$ by the uniform equicontinuity and the second one converges to zero since $X_n \stackrel{d}{\to} X$.  I was wondering whether there is an easier proof - my own proof seems rather like overkill to me. Thanks! Simple newbie help page about avoiding "Thanks" I think everybody with a history of participation can't help but notice the explosive growth of repeated duplicate questions. The work of looking for duplicates and then closing questions has overtaken the work of actually answering questions.  It's become virtually impossible to ask something which hasn't been yet asked. Lately I feel boredom each time I see a new question and start answering it only to realize after the first written sentence that I have answered similar questions five times over or seen them answered some ten times or more. I know where it's going to be going and what answers with what votes it is going to receive before any answer is even written.  It's obviously not the administration's fault, it is simply so. The questions is, what to do about it? Engaging more moderators to close and close duplicates won't work because it's already starting to repel users who get mad at their questions closed and become afraid of asking more or perhaps just go away.  The original idea of Jeff stated the the site is for questions that can be answered. Meaning no open-ended discussions. When I first read this I had a casual thought - well, and what's gonna happen when they all are asked and answered? Will it become a museum, a monument of wisdom and knowledge or what?  I'm not sure what to do about it. Perhaps deprecation/versioning of older questions so that their new versions can sparkle the tiresome evening...  POSTSCRIPTUM.  I recall Star Trek Voyager episode  when another Q asked Captain Janeway for asylum. He was tired of immortality and wanted to die. They opened a hearing and went on to visit the continuum to evaluate the living conditions there which caused the suffering and the wish to escape it in death. A lifeless picture presented itself to the visitors... a house lone in an open space, two people going about their business, nobody talking. When asked why they were silent, the following was said. It's been so long, all has already been said, all stories told, all jokes are made... there is nothing left to do...  Are we there as well?  UPDATE: Just saw this answer in another question:       Also, some of the core questions got asked early on in the life of the site. Great questions still come around from time to time, but many have already been asked.   Doesn't that confirm my point? I have installed Ubuntu Server on an old desktop, and I'm new to this.  I had set it up and was trying to connect through Putty but that was not working.  When I tried ifconfig there was no eth0 IP address, and logging onto the router showed the only device connected by ethernet is the printer.  However, the ethernet cable is plugged into the server and is showing the green light by it.  I tried opening /etc/networking/interfaces but this shows the following:  source /etc/network/interfaces.d/*  # The loopback network interface auto lo iface lo inet loopback   I tried adding 'auto eth0' after this and saving but made no difference, and have tried sudo ifconfig eth0 up which didn't do anything either.  sudo lshw -C network says that the network is DISABLED.  Any ideas to help get this up and running?  EDIT: Running command lspci -knn | grep Eth -A3 results in the following:  02:05.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL-8110SC/8169SC Gigabit Ethernet [10ec:8167] (rev 10)          Subsytem: Gigabyte Technology Co., Ltd GA-MA69G-S3H Motherboard [1458:e000]          Kernel driver in use: r8169          Kernel modules: r8169 How can I delete a (null) app in usage? In my usage list(settings>general>storage &amp; iCloud usage>manage storage), there is an app that registers as (null) with no icon, a version number that doesn't match anything I've previously downloaded, and it's taking up storage space. I first noticed it about a week or two ago. And no, I can't download recent apps to find the one that it is, because I downloaded multiple apps recently, and didn't notice the (null) app until my storage was critically low and I was looking to delete something. I am running iOS 9.2.1 on an iPhone 5C. Is there anything canon in the Terminator Universe that picks up where Terminator: The Sarah Connors Chronicles leaves off? So, Terminator canon is pretty twisted, given all that time-travel stuff.  But in the canon that includes the T:SCC, does anything pick up where the show left off that is considered canon?