Why didn't Itachi kill Sasuke? Why did Itachi not kill Sasuke but instead poked his head during their last battle that led to Itachi's death?  Was it because of the clan symbol behind him? Or because he loved him?
Why did Itachi poke Sasuke's forehead? From the beginning of the series, even during his last moments, Itachi poked Sasuke's forehead as a way of showing affection and apologizing.  But is that the only reason? Did he perhaps try to transfer some knowledge or memories to Sasuke? Or is it simply unknown?
How to input integer value to an array, based preceeding row + column values? Background  Suppose we have an Ordinary Least Squares model where we have $k$ coefficients in our regression model, $$\mathbf{y}=\mathbf{X}\mathbf{\beta} + \mathbf{\epsilon}$$  where $\mathbf{\beta}$ is an $(k\times1)$ vector of coefficients, $\mathbf{X}$ is the design matrix defined by  $$\mathbf{X} = \begin{pmatrix} 1 &amp; x_{11} &amp; x_{12} &amp; \dots &amp; x_{1\;(k-1)} \\ 1 &amp; x_{21} &amp; \dots  &amp;       &amp;   \vdots \\ \vdots &amp;   &amp; \ddots &amp;       &amp;   \vdots \\ 1 &amp; x_{n1} &amp; \dots  &amp; \dots &amp;   x_{n\;(k-1)}     \end{pmatrix}$$ and the errors are IID normal,  $$\mathbf{\epsilon} \sim \mathcal{N}\left(\mathbf{0},\sigma^2 \mathbf{I}\right) \;.$$  We minimize the sum-of-squared-errors by setting our estimates for $\mathbf{\beta}$ to be $$\mathbf{\hat{\beta}}= (\mathbf{X^T X})^{-1}\mathbf{X}^T \mathbf{y}\;. $$   An unbiased estimator of $\sigma^2$ is  $$s^2 = \frac{\left\Vert \mathbf{y}-\mathbf{\hat{y}}\right\Vert ^2}{n-p}$$ where $\mathbf{\hat{y}} \equiv \mathbf{X} \mathbf{\hat{\beta}}$ ().   The covariance of $\mathbf{\hat{\beta}}$ is given by $$\operatorname{Cov}\left(\mathbf{\hat{\beta}}\right) = \sigma^2 \mathbf{C}$$ where $\mathbf{C}\equiv(\mathbf{X}^T\mathbf{X})^{-1}$ () .  Question  How can I prove that for $\hat\beta_i$, $$\frac{\hat{\beta}_i - \beta_i} {s_{\hat{\beta}_i}} \sim t_{n-k}$$ where $t_{n-k}$ is a t-distribution with $(n-k)$ degrees of freedom, and the standard error of $\hat{\beta}_i$ is estimated by $s_{\hat{\beta}_i} = s\sqrt{c_{ii}}$.    My attempts  I know that for $n$ random variables sampled from $x\sim\mathcal{N}\left(\mu, \sigma^2\right)$, you can show that  $$\frac{\bar{x}-\mu}{s/\sqrt{n}} \sim t_{n-1} $$ by rewriting the LHS as $$\frac{ \left(\frac{\bar x - \mu}{\sigma/\sqrt{n}}\right)  }   {\sqrt{s^2/\sigma^2}}$$ and realizing that the numertor is a standard normal distribution, and the denominator is square root of a Chi-square distribution with df=(n-1) and divided by (n-1) ().  And therefore it follows a t-distribution with df=(n-1) ().  I was unable to extend this proof to my question...  Any ideas?  I'm aware of , but they don't explicitly prove it, they just give a rule of thumb, saying "each predictor costs you a degree of freedom". Event button as a module or as user program? Parametric vs non-parametric machine learning methods I looked-up many references and websites and researched on how to determine if a method is between parametric or non-parametric. I came up with below definitions,     A parametric algorithm has a fixed number of parameters. In contrast, a non-parametric algorithm uses a flexible number of parameters, and the number of parameters often grows as it learns from more data.   From .  Moreover, I found,     A parametric model, we have a finite number of parameters, and in nonparametric models, the number of parameters is (potentially) infinite.    From .  And many other methods of determination, though the problem is none of them can helo ones to determine if a certain hypothetical method is parametric or not. (For instance, why k-means' number of parameters are constant but KNN is variable or basically what do we call a parameter and what we do not?) Uncertainty principle and multiple observers My understanding is that an observer can measure the precise location of a particle so long as the corresponding uncertainty in momentum measurement is not an issue and vice-versa.    Say there is such an observer, interested in the precise position of a particular particle.  Now, consider a second, independent observer, unbeknownst to the first, who is trying to measure the exact momentum of the same particle without caring about the position.  As a thought experiment, we assume that the two observers are somehow able to access the same particle at the same time in some way without being aware of each other.  Can both observers get their desired results? Check for internet connection with Swift This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted  previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows:     Let $U$ be bounded with a $C^1$ boundary. Show that a ''typical''   function $u \in L^p(U) \ (1 \leq p &lt; \infty)$ does not have a trace on   $\partial U$. More precisely, prove there does not exist a bounded   linear operator      \begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation}      such that $Tu = \left. u \right|_{\partial U}$ whenever $u \in C(\overline{U}) \cap L^p(U)$.   So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then $Tu = \left. u \right|_{\partial U}$ is undefined.  I like the idea of using the $\mathrm{dist}(x, \partial U)$ function to do this. I was thinking about defining the function  \begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation}  which must be in $L^p(U)$ since $U$ is bounded (hence integral of $\varepsilon$ over $U$ is finite, right?). But this function is not continuous at the boundary, and so $u \notin C(\overline{U})$. Can this be modified somehow?  Another option is to use that the boundary is $C^1$. So for each $x^0 \in \partial U$ there exists an $r &gt; 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n &gt; \gamma(x_1,\ldots, x_n) \}$. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the $C^1$ function $\gamma$. How can we use this? Can't launch graphical apps from terminal after updating to 15.10 How find the maximum value of $|bc|$ Question:     Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \leq 1$ holds true for any complex number $z, |z| \leq 1$. Find the maximum value of $|bc|$   It is said this is answer is $$|bc|\le \dfrac{3\sqrt{3}}{16}$$  My idea: let $z=1$,then $$|a+b+c|\le 1$$ let $z=-1$,then $$|a-b+c|\le 1$$ let $z=0$, then $$|c|\le 1$$  let $|\theta|=1$,then we have $$|a(z/\theta)^2+b(z/\theta)+c|\le 1\Longrightarrow |az^2+b\theta z+c\theta^2|\le 1$$ and only this can't solve this problem,Thank you I am using Adobe CC(2015) latest version and the behaviour of the smart objects isn't normal. When I paste a vector from Illustrator to Photoshop, the vector looks low quality or rough.   ORIGINAL  IN PHOTOSHOP   However, if I click over the smart object layer(in Photoshop)and this is opened in Illustrator, the vectors looks as the original one.   VECTOR OPENED IN ILLUSTRATOR FROM PHOTOSHOP SMART OBJECT   How can I solve this? How to insert an image that also acts as a link What is CTAN.org now? I just want to know if it's possible to get Cain and abel installed in ubuntu, it would really help my work instead of dual booting back to my windows 7 just to use it. THanks! I'm using Android 4.4.2 (on ). I've a Windows PC at home which has highspeed broadband internet connection. I don't have wireless router or access point. I just want to use PC's internet connection on my phone. I've tried multiple methods but none seems to work.  Method 1: Using native Android 4.4.2 option  In my phone when I connect my phone, in Settings &gt; Wireless &amp; networks &gt; USB Internet (sharing Windows PC internet via USB cable) checkbox, which I checked then it asked me to choose my Windows version (which is Windows 8.1 but since it didn't show that option) I chose Windows 8.  It gives me further instructions as showing in the following picture:    Here is the problem, there is no sharing tab (see the following screen shot)   Method 2: As described in   I learned on this site that it is called Reverse tethering. I read and followed the procedure mentioned in    I followed all the steps described in that answer, until executing the command.   Install USB drivers from Android SDK [DONE] Connect USB cable and activate USB Tethering (You should see on a new network interface.) [DONE] Bridge the 2 network interfaces [DONE] On your computer execute adb shell netcfg usb0 dhcp [PROBLEM]   When I execute adb shell netcfg usb0 dhcp it says: error: device not found. I tried the same thing with USB Debugging in Developer settings, ON &amp; OFF. The error remains the same.  Method 3: Rooting the device and using an app  I read here and there that there is another method, in which we can directly use an app but it requires rooting the device. My phone is a new phone and warranty becomes void if I root it.   EDIT: I've tried the method suggested in answer, at the 4th step of method 2, I'm facing this problem: How do I configure Apache 2 to run Perl CGI scripts? I would like to configure Apache 2 running on Kubuntu to execute Perl CGI scripts. I've tried some steps that I came across by googling, but nothing seems to work.  What is the right way of achieving this? Evaluating and proving $\lim\limits_{x\to\infty}\frac{\sin x}x$ I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist?   It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically.   $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} =  \frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x}   = ? $$  L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here? Check if email exist php How to prove that minimum of two exponential random variables is another exponential random variable? How can I prove that the minimum of two exponential random variables is another exponential random variable, i.e. Z = min(X,Y) What would you call a word that doesn't exist in or translate well into another language? I've run into this situation several times, being a native Spanish speaker.   There are some words you just can't translate into another language. Is there a particular word to describe this?  I'm not refering to idioms.  Also, if its a noun, people tend to use them by surrounding them in quotes. Is this appropiate?  Example:     I arrived at around 5 pm and had some quesadillas.   Edit:  I also meant to cover words that exist in English, but cannot be translated into another language.   Example:      There isn't a word for "drive-thru" in your language. Determining bare earth DEM from unclassified LAS file? I have data in LAS format with RGB values created from aerial photogrammetry using a UAV. I am  trying to find a solution to extract the bare earth DEM from the point cloud.   I have tried SAGA, Fusion, MCC-LIDAR, but it is seems they need the LAS file to be already classified (which it naturally isn't). Can anyone point me in the right direction with a brief explanation of the process?   Generally, I would need to process about 100 mill points at a time (can tile them if needed).