Compatability with Intel XEON E3-1505L v6 (7th Gen) Processor
Will my device work with Ubuntu?
I've read some great on-topic example questions on Area 51.  My questions:   Should a proposed site's "Great On-Topic Example" questions be automatically included in its beta? Will a proposed site's "Great On-Topic Example" questions be automatically included in its beta? Everything was working fine for a long time. But suddenly the device is getting slower than usual (e.g. apps react slowly when starting an action -- like the RSS reader takes "ages" to open an article). Also, several apps start crashing: force-closes as well as unexpected restarts. Sometimes this even involves some automatic reboot of the device.  Is there anything I can do, besides doing a factory-reset? Preferably it should work on un-rooted devices.  Remark: not all described problems may occur together, so on one device e.g. there might be no automatic reboot involved. Question remains the same, nonetheless. There are two vectors called $\vec{a}$ and $\vec{b}$. Vector $\vec{c}$ is the bisector and it can be given as  $$\vec{c} = |b|\vec{a} + |a|\vec{b}$$  How to prove that? I have used the dot product method. But there I can't find the angle between them. Then I tried to do this by using unit vectors. Then I got an answer like in the given picture. . Can you help me to get the correct answer I am trying to figure out how I create a downloadable PDF template of a wedding invitation design that my customers can edit themselves. They will be approx 2 fonts used on each template and the customer will need to be able to edit all text themselves.  e.g. change their names, wedding date, location etc. etc.  Basically I want to be able to sell an editable downloadable template.  So I will need to lock the artwork and "open" the text so they can edit the text only.  If you think of something like this image. That has artwork that I want to be uneditable and then has text that I want the customer to be able to edit and print out themselves at home etc. What is the center of $D_{2n}/Z(D_{2n})$ What is the center of $D_{2n}/Z(D_{2n})$. I see that when $n=2^k$ then I have a $p$-group so the center is not trivial. But when $n$ is not power of $2$ how can i know what is the center of this group?. Surely when $n$ is odd then the center is trivial. Factorial of $i$ With $i^2=-1$, what is $i!$ ?  I'm not sure if it is $i$ or if it is an infinite product $i(i-1)(i-2)(i-3)(i-4)(i-5)\dots$  It would make sense either way.  Is it not defined? PC Blue Screens and then instantly reboots. How do I stop this or find the error later? Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$ Although we didn't do any changes, this new error appeared in the GSC Structured Data section:    I don't really get what it means. There are many fields for Schema.org, is homeLocation suddenly a required field? Does that make any sense? Difference between spin and polarization of a photon I understand how one associates the spin of a quantum particle, e.g. of a photon, with intrinsic angular momentum. And in electromagnetism I have always understood the polarization of an EM wave as the oscillations of the E and M field, not necessarily being  aligned with the direction of propagation of the wave.     Questions:   But when one talks about the polarization of a photon in Quantum Mechanics, how does it really differ from its spin?  Are they somehow related and what's the physical idea behind photon polarization in contrast to photon-spin? Feel free to use mathematical reasonings as well if you see it fit! Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. In .NET, which loop runs faster, 'for' or 'foreach'? In C#/VB.NET/.NET, which loop runs faster, for or foreach?  Ever since I read that a for loop works faster than a foreach loop a  I assumed it stood true for all collections, generic collections, all arrays, etc.  I scoured Google and found a few articles, but most of them are inconclusive (read comments on the articles) and open ended.  What would be ideal is to have each scenario listed and the best solution for the same.  For example (just an example of how it should be):   for iterating an array of 1000+ strings - for is better than foreach for iterating over IList (non generic) strings - foreach is better than for   A few references found on the web for the same:               [Edit]  Apart from the readability aspect of it, I am really interested in facts and figures. There are applications where the last mile of performance optimization squeezed do matter. Recently I needed to do weighted random selection of elements from a list, both with and without replacement. While there are well known and good algorithms for unweighted selection, and some for weighted selection without replacement (such as modifications of the resevoir algorithm), I couldn't find any good algorithms for weighted selection with replacement. I also wanted to avoid the resevoir method, as I was selecting a significant fraction of the list, which is small enough to hold in memory.  Does anyone have any suggestions on the best approach in this situation? I have my own solutions, but I'm hoping to find something more efficient, simpler, or both. Why with two classes, LDA gives only one dimension? I am working with dimensionality reduction algorithms. Linear Discriminant Analysis (LDA) is a supervised algorithm that takes into account the class label (which is not the case of PCA for example). I am using Python to do a comparative study between some algorithms.  Why with two classes (k = 2), regardless of the data dimensionality, LDA gives one dimension, i.e. the new subspace is composed of only one dimension? For example if I try to project my dataset (of 100 attributes and 2 classes) to 10-dimensional space by using LDA(n_components=10) I only obtain 1 dimension?  For k&gt;2 I can obtain the number of discriminant vectors that I specified as n_components. How can I get n_components dimensions for k=2? how to get line spacing greater than one and half spacing and less than double space Does anyone know how to get line spacing in a paragraph to be greater than one and half spacing and less than double spacing? I tried \renewcommand{\baselinestretch}{value} but it did not work. IF statement inside a LINQ SELECT Prove Inverses not need be Unique if Associativity fails Let $M=\{x, y, z\}$ be a set and let $\circ$ be a binary operation on $M$ such that the identity element is $x$ and $y \circ y = x = y \circ z$. Prove that inverses need not be unique if associativity fails given the multiplication table below.  $x \circ x = x, x \circ y = y, x \circ z = z$  $y \circ x = y, y \circ y = x, y \circ z = x$  $z \circ x = z, z \circ y = x, z \circ z = x$ Associativity of symmetric difference of sets By $A\oplus B$ we denote the symmectric difference of two sets. The definition is $A\oplus B =(A\setminus B) \cup (B\setminus A)$. Now I hope to show that $A\oplus (B\oplus C) = (A\oplus B)\oplus C$. I remember that there's an elegant proof, but I forget its detail. (The first step is to show $A\oplus (A\oplus B) = B$, and maybe applying this formula, we can obtain associativity.) How to sample from a discrete distribution? Assume I have a distribution governing the possible outcome from a single random variable X.  This is something like [0.1, 0.4, 0.2, 0.3] for X being a value of either 1, 2, 3, 4.  Is it possible to sample from this distribution, i.e. generate pseudo random numbers upon each of the possible outcomes given the probability of that outcome. So if I wanted to know what the probability of getting a 2 is, the sampling operation may return 0.34 or something like that.  The reason I ask is that I'm trying to implement an action selection policy for a reinforcement learning method based on a research paper. From what I gather from the paper, the author is able to sample the distribution by "mapping the uniform distribution U[0,1] through cumulative probability density functions obtained by adaptive numerical integration". From this he then samples the transition probabilities for each trial...  I would be grateful for any info on this...  Thanks in advance Lazy evaluation in C++ C++ does not have native support for lazy evaluation (as Haskell does).  I'm wondering if it is possible to implement lazy evaluation in C++ in a reasonable manner. If yes, how would you do it?  EDIT: I like Konrad Rudolph's answer.  I'm wondering if it's possible to implement it in a more generic fashion, for example by using a parametrized class lazy that essentially works for T the way matrix_add works for matrix.  Any operation on T would return lazy instead. The only problem is to store the arguments and operation code inside lazy itself. Can anyone see how to improve this?