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Prove: If a sequence converges, then every subsequence converges to the same limit. Why CTRL + C multiple times I have recently noticed that I press CTRL + C more than once, even if I know that the file has been copied to the clipboard. But I press CTRL + X or CTRL + V only once.   Why do I do this? Is this only a problem for me, or is anybody else is facing it too? Do I need to activate the firewall? I only use Ubuntu for a home desktop use? What is happening here, when I use squared loss in logistic regression setting? I am trying to use squared loss to do binary classification on a toy data set. I am using mtcars data set, use mile per gallon and weight to predict transmission type. The plot below shows the two types of transmission type data in different colors, and decision boundary generated by different loss function. The squared loss is $\sum_i (y_i-p_i)^2$ where $y_i$ is the ground truth label (0 or 1) and $p_i$ is the predicted probability $p_i=\text{Logit}^{-1}(\beta^Tx_i)$. In other words, I am replace logistic loss with squared loss in classification setting, other parts are the same. For a toy example with mtcars data, in many cases, I got a model &quot;similar&quot; to logistic regression (see following figure, with random seed 0).  But in somethings (if we do set.seed(1)), squared loss seems not working well.  What is happening here? The optimization does not converge? Logistic loss is easier to optimize comparing to squared loss? Any help would be appreciated.  Code d=mtcars[,c(&quot;am&quot;,&quot;mpg&quot;,&quot;wt&quot;)] plot(d$mpg,d$wt,col=factor(d$am)) lg_fit=glm(am~.,d, family = binomial()) abline(-lg_fit$coefficients[1]/lg_fit$coefficients[3],        -lg_fit$coefficients[2]/lg_fit$coefficients[3]) grid()  # sq loss lossSqOnBinary&lt;-function(x,y,w){   p=plogis(x %*% w)   return(sum((y-p)^2)) }  # this random seed is important for reproducing the problem set.seed(0)  x0=runif(3) x=as.matrix(cbind(1,d[,2:3])) y=d$am opt=optim(x0, lossSqOnBinary, method=&quot;BFGS&quot;, x=x,y=y)  abline(-opt$par[1]/opt$par[3],        -opt$par[2]/opt$par[3], lty=2) legend(25,5,c(&quot;logisitc loss&quot;,&quot;squared loss&quot;), lty=c(1,2)) New top bar quick reference guide / Where has the link to  gone? Since the new top bar has been rolled out - a number of different people, established and new users alike, have asked questions on where a specific link has disappeared to with the new top bar design. Where has the link to &lt;X&gt; gone? Are all oscillatory motions periodic motions? For example, if a pendulum system in the real world oscillates its amplitude and period will eventually decrease. Does that still count as a periodic motion? Since the motion doesn't have the same amplitude at equal time periods. Recently I found a claim saying that  $$  \int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2  $$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.   Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$ \int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x $$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the , but there are some subtle differences. What are bitwise operators? Render turns out plain white For some reason, when I render out the image, it is plain white. I've checked the sky settings, and I tried applying textures, but nothing seems to change.  How can I fix this? "...his parents' dream of *him* achieving a Cambridge degree." What is the function of "him" here? I have a problem analysing this sentence from the point of finite/nonfinite clauses, clause elements and their functions:     He does not want to destroy his parents' dream of him achieving a Cambridge degree.   I am especially interested in the: dream of him achieving a Cambridge degree.  I know that 'achieving a Cambridge degree' is a non-finite -ing participle clause. However what is its function? And what is the function of 'of him'? Is it a postmodification? I read here at  that holding Shift during boot can bring up the GRUB menu. However I try holding one or the other Shift keys, tapping, tapping then holding, nothing works. Am I missing something? How does one access the GRUB menu during boot time?   No dual boot here, just standard Lubuntu 14.04 install. Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}), \forall a,b\in\mathbb{Q}$ Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then   $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}),  \forall a,b\in\mathbb{Q}$   It has something to do with $\sqrt{a} = -\sqrt{b}\implies a^2 = b^2$  The inclusion $\mathbb{Q}(\sqrt{a}+\sqrt{b})\subseteq \mathbb{Q}(\sqrt{a},\sqrt{b})$ is obvious  Now take $x\in \mathbb{Q}(\sqrt{a},\sqrt{b})$, that is: $x = c + d\sqrt{a} + e\sqrt{b} + f\sqrt{ab}$. I need to show that this is in $\mathbb{Q}(\sqrt{a}+\sqrt{b})$, that is, $x = g + h\sqrt{a} + i\sqrt{b}$, but I can't get that $\sqrt{ab}$ removed Shadows along edges of mesh in 3D View I have recently started using blender and one of the models I'm working on has dark shadows along edges. I started this model from scratch. I can't remember them being there when I first started the model, but I had trouble joining areas together and think I've pressed something I shouldn't have. However, there are so many options I'm not sure what I did.  Here is an image of what I mean   The one on the left has the shadow. As a test I started a new model, which is on the right in the image, and as you can see it doesn't have any shadows.  Also, occluded vertices partially show through the model. Occlusion doesn't work properly.   In orthogonal view small darkened triangles appear along the the edges.    I don't really want to start the model again from scratch so any ideas on how to remove this shadow would be great. Thanks. Which is correct, "neither is" or "neither are"? A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.     Let be $a,b,c &gt;0$ with $abc=1$. Prove that:    $$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$   Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.   Our inequality becomes: $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$  Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$  Now applying Cauchy-Schwarz we obtain the desired result.  What I wrote can be found on this link: . But now, I don't know how to apply Cauchy-Schwarz.   Thanks:) Categorical Quotient and group actions I am trying to practice calculating categorical quotients and I ran into this example. I am unable to get the answer and was wondering if someone can help?  Let $G = Z/3Z =$ $\{1, \omega, \omega^2\}$, where $\omega$ is a cube root of unity. Let $G$ act on $𝔸^{2}$ via $\omega(x, y) = (\omega x, \omega y)$. Find the categorical quotient of $𝔸^{2}$ by G. Proving $v(s,p)=2^{p-1}(2s-1)$ is a bijection of natural numbers and $f(s)=2s-1$ is a bijection between natural numbers and odd numbers. How do I prove this function is bijective? $$ v(s,p)=2^{p-1}(2s-1). $$ The domain is natural numbers and the codomain is also the natural numbers.  And this one: $$ f(s)=2s-1. $$ The domain is the natural numbers, and the codomain is the odd numbers in the natural numbers.  With this one I would do this to show it's injective: $$ \begin{align} v(s)&amp;=v(s_1)\\ \implies 2s-1&amp;=2s_1-1\\ \implies (2s_1)/2&amp;=(2s_2)/2\\ \implies s=s_1 \end{align} $$ So it's injective since if $v(s)=v(s_1)$ then $s=s_1$.  And to show it's surjective $f(s)=y$: $$ \begin{align} y&amp;=2s-1\\ \implies s&amp;=(y+1)/2 \end{align} $$ Then the function must be surjective since every $y$ is the same as the codomain for $f$.  Am I correct? The value of a, b and c for this integral Find out the value of $a, b$ and $c$, for which $$ \lim _{x\to 0}\:\:\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\frac{1}{5} $$  My first thought was to separate the integral into multiple ones because i thought that it is pretty clear that each integral is a common one and a,b,c are constants and can be taken out . After I solved the integral, that's where I had the problem. I couldn't do the limit of the result that I've got in order to find out the value of $a,b$ and $c$ . What are the disadvantages of using a Zener diode over a linear voltage regulator? I wonder why a Zener diode is worse than linear voltage regulator.  Some of the posts online mentioned that because of the power dissipating problem. Zener diodes keep consuming a large amount of current while the load may just need a very little amount of current.   If I understand correctly, this phenomenon will happen only when the Zener diode is connected with the load in parallel. What if I connected my Zener diode with my load in series? Also, someone said that linear voltage regulators (e.g. LM7805) are not good voltage regulators as the power dissipated in them is much higher than that in switching voltage regulators. In my understanding, the power dissipated in a linear voltage regulator can be calculated by dropout voltage x current = power converted to heat.   Regarding these 2 points, other than a linear voltage regulator can handler a wider current range, what is the benefit of choosing a linear voltage regulator over a Zener diode? deleting windows system folder