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Standard graded algebra Retrieving date in sql server, CURRENT_TIMESTAMP vs GetDate() Using SQL Server - which is the fastest or best practice method to use for date retrieval? Is there a difference? Is there any difference in meaning between 'efficacy' and 'efficiency'? How can I use an array modifier to create individually manipulatable objects? I often use an array modifier (or two) to quickly create a grid of identical or symmetric objects, which can be really useful when modeling all sorts of man-made objects. Once I've made the grid, though, is there any way for me to make it so that each of the objects are individually manipulatable?  Simply applying the modifier does not seem to be the full answer, since it still only allows me to select the grid-of-objects as a whole in object mode. Is there any way to separate them so I work with them each individually? A positive polynomial is the sum of two squares in $\mathbb{R}[X]$ Let $P\in\mathbb{R}[X]$ be a positive polynomial.  I want to show that there exists $A,B\in\mathbb{R}[X]$ so that $P=A^2+B^2$    $\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)$ where $a$ is the leading coefficient, $a_k$ are the real rooks, $b_k$ are the complex roots.  What can i do now ? I know that the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ should come in handy, but I can't figure out :  1) How to show that the product real roots is a square  2) How to use the above identity for the complex roots.  If I can solve those two points, then I can solve the problem. To show that ${x_n}$ is convergent where $|x_{n+1} - x_n|< \frac{1}{n^2}$ Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|&lt; \frac{1}{n^2}$ for all $n \in N$.  Show that the sequence is convergent.  If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.  But what to do here? Please help. Thanks in advance. Where to place 'only' This is NOT an installation problem question. I've had both op sys working fine side-by-side for a couple of weeks now. I'm using Ubunutu 12.04 64-bit.  This morning when I tried to switch from Win8 to Ubuntu, it just rebooted into Win8. No grub screen whatsoever. This is the 3rd machine I've had dual-booting &amp; have never seen this, but then Win8 is a 1st for me &amp; has thrown some pretty awesome blocks in the path to get to here. Atrocious op sys...  I'm assuming the ASUS updater did this yesterday. (ASUS R503U)  So, do I just put the Ubuntu disk in &amp; reinstall grub? Down Arrow Key is not working could you help me on this integral ? $$\int \frac{1}{\cos(x)}\,\mathrm dx$$ Here's what I've started : $$\int \frac{1}{\cos(x)}\,\mathrm dx = \int \frac{\cos(x)}{\cos(x)^2}\,\mathrm dx = \int \frac{\cos(x)}{1-\sin(x)^2}\,\mathrm dx$$ Now, I did : $u = \sin(x)$, so $\mathrm du = 1$. Now I have : $$\int \frac{\text{???}}{1-u^2}\,\mathrm du$$ But at this point, I think I did the most of the job but I'm stuck. Could you help me to solve this integral please (to the integration by substitution at the end) ? Thanks EDIT : Now I follow the steps and I got : $$\int \frac{1}{1-u^2}\,\mathrm du$$ Doing the partial fraction I got $A = 1/2$ and $B = 1/2$. So basically I have : \begin{align} &amp; \int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du \\[8pt] = {} &amp; \frac 1 2 \left(\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du\right) \\[8pt] = {} &amp; \frac 1 2 \ln\left(\frac{1+u}{1-u}\right) = \ln\left(\left(\frac{1+\sin(x)}{1-\sin(x)}\right)^{1/2}\right) \\[8pt] = {} &amp; \ln \left(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}\right) = \ln\left(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}\right) \\[8pt] = {} &amp; \ln\left(\frac{\sqrt{1+\sin(x)}}{\cos(x)}\right) \end{align} Here's what my teacher got :  What's wrong with what I did ? Did I miss something ? Why did Palpatine wait so long before giving Order 66? Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$ How can I permanently enable cheating? align vs equation Let's say when I do ls -li inside a directory, I get this: 12353538 -rw-r--r-- 6 me me 1650 2013-01-10 16:33 fun.txt  As the output shows, the file fun.txt has 6 hard links; and the inode number is 12353538. How do I find all the hard links for the file i.e. files with the same inode number? I want to write $ k = \{k1,k2,k3,...,kn\} $. How could I do this? Just that. By good I mean something at the level of SolidWorks, AutoCAD, or similar. This is a complete limitation for me (and many others) to use Ubuntu. Java Server Socket not connecting I need to show that for any countable ordinal $\alpha$ there is a set A $\subseteq \mathbb{Q}$ such that (A, &lt;) is isomorphic to ($\alpha, \in$).  To do it I am supposed to show the following stronger statement by induction on ordinals $\alpha$ &lt; $\omega_{1}$:  Let P($\alpha$) be the statement "For every interval (a, b) with rational endpoints, there is an A $\subseteq$ (a,b) such that (A, &lt;) is isomorphic to ($\alpha, \in$)."  I think that the purpose of breaking $\mathbb{Q}$ into intervals (a,b) is to make $\mathbb{Q}$ a countable set, but I don't see the significance of doing this or how it would make the proof easier.  I feel like I need to define a well-ordered set in (a,b) but I don't know how to do this since if a and b are both negative then there is no least element, so no well-ordering.  I believe that I am supposed to be using transfinite induction? But I don't know how P($\alpha$) implies P(S($\alpha$)) for S($\alpha$) successor of $\alpha$.  Any help would be greatly appreciated... thanks in advance. What does a two overlapping circles symbol mean in an electrical schematic/diagram? In the datasheet for a A3503 Hall Effect Sensor, I saw a circuit diagram labeled 'Functional Block Diagram', and it shows a symbol I have not seen before. It's shown here between 3 - Output and 2 - Ground.  What is this supposed to represent?    The full datasheet is available