The exercise is as below  Let $V = \{x \in \mathbb{R}^2: 2x_1^2+3x_2^2\leq4\}$ Prove: V is convex  I started the proof like this:  Let $x,y \in V$ be given. Let $\lambda \in [0,1]$. We want to show that $\lambda x + (1-\lambda) y \in V$.  Now, I know that I have to rewrite $2[\lambda x_1 + (1-\lambda)y_1]^2+3[\lambda x_2 + (1-\lambda)y_2]^2$ into something like   $\lambda(2x_1^2+3x_2^2) + (1-\lambda)(2y_1^2+3y_2^2) \leq \lambda*4+(1-\lambda)*4 = 4$   But I just can't get my head around it, can someone please make a step-by-step derivation? I am really hopeless at this point.
I'm trying to prove that the following set ($B$) defined by an ellipse is convex, but I am getting stuck.  $$B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} $$  My sad attempt:  $$ \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1]$$  $$ \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2  )$$ $$3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)$$  and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form $(\lambda x_1 -3)^2$, $((1-\lambda) y_1 -3)^2$, $(\lambda x_2 -3)^2$, and $((1-\lambda) x_2 -3)^2$, but I don't think that helps the situation any.
Can someone help me out and explain or point me in the right direction on how to check video drivers and see if my video card running okay?  I had 10.04 installed on my Thinkpad r61 with Compiz Manager and life was great, until the machine took water damage. I bought an ASUS (X54H) since and am trying out 12.04, but the desktop just doesn't look right.  I always struggled with video driver installation. There are no proprietary drivers available in the hardware manager.  When I run lspci | grep VGA:  00:02.0 VGA compatible controller: Intel Corporation 2nd Generation Core Processor Family Integrated Graphics Controller (rev 09)   I will greatly appreciate your help. I want to use Linux more, but like I said video drivers appear to be my biggest concern. I have also tried 12.04 on my desktop PC, but again failed to configure video card, so switched back to Windows 7. Which version of Ubuntu will work on my PowerPC? I'm using iMac PowerPC G5 1.8 GHz, 1GB RAM, 160GB HDD, and 10.5.4 Leopard OS.  I want to run Ubuntu on my iMac. Please can anyone tell me which version will run in my system? Why can't I tame my ocelot? How to graphically represent a language Can we use "there is" for plural nouns? Getting the location from an IP address I want to retrieve information like the city, state, and country of a visitor from their IP address, so that I can customize my web page according to their location. Is there a good and reliable way to do this in PHP? I am using JavaScript for client-side scripting, PHP for server-side scripting, and MySQL for the database. Is it "legal" to CREATE and DROP #SomeTable more than once? What antivirus programs are available for Ubuntu?  We previously used Symantec Endpoint Protection but it does not work in Ubuntu. Generating list of manually installed packages and querying individual packages While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion? Proof of $(A - B) - C = A - (B \cup C)$ I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these.  I tried asking another problem, but it turned out to be a special case problem, so hopefully this one works out normally.   The question is:  Prove $(A - B) - C = A - (B \cup C)$  I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side.   if $x \in C$, then $x \notin A$ and $x \in B$. So $x \in (B \cup C)$ So $A - (B \cup C)$   Is this the right idea? Should I then reverse the proof to prove it the other way around, or is that unnecessary? Should it be more formal?  Thanks! Let $\Gamma$ be a set of formulas and $\phi$ be a formula. Show that $\Gamma \cup \{\neg \phi\}$ is satisfiable if and only if $\Gamma\not \models \phi$.   This seemed pretty obvious but I wanted to see if my proof made sense:  Proof: $(\Rightarrow)$  To derive for a contradiction, suppose that: $\Gamma \models \phi$. That means for all truth assignments $v$, for $\gamma \in \Gamma$, if $v(\gamma) = T$, then $v(\phi) = T$.  But this contradicts our assumption that $\Gamma \cup \{\neg \phi \}$ is satisfiable by the fact that $v(\phi) = T$ cannot happen so: $\Gamma \not \models \phi$ .  $(\Leftarrow)$   So by the definition of $\Gamma \not \models \phi$, we have that there is some truth assignment $v$ which satisfies $\Gamma$ but does not satisfy $\phi$. So  that means $v(\phi) = F$ since $v(\phi) \not = T$ which implies that $v(\neg \phi) = T$ which means $v$ satisfies $\Gamma \cup \{\neg \phi\}$.    I feel like I'm missing something in the forward direction, but at the same time... It looks pretty trivial as well. Am I missing anything crucial?  Thank you! How does order-preserving encryption work? Order-preserving encryption (OPE) is, apparently, a method of encrypting data so that it's possible to make efficient inequality comparisons on the encrypted items without decrypting them.  I've been coming across this term in various places (including ) lately, but I have no idea how such encryption schemes are supposed to work.  Any obvious methods I can think of to allow such comparisons on encrypted data would result in catastrophic security failure.  Certainly, I assume that OPE must involve some kind of security tradeoff compared to traditional encryption, but given that it seems to be actively  and , it seems that it must be possible to implement in a way that retains at least some level of useful security.  I just don't see how.  Given that  didn't turn up any convenient Wikipedia articles or other popular summaries of OPE, I figured I'd try asking for one here before diving into the academic literature.  (At worst, I may try to answer my own question later, if nobody else beats me to it.)  So, to summarize, my question is: How does order-preserving encryption work, and what security properties does it provide? class and private variables Cannot turn Bluetooth off How to install Broadcom wireless drivers offline? I have installed Ubuntu alongside Windows 8.1. But the wifi is not working in Ubuntu (perhaps the drivers are not present), which worked fine in windows. Also I cant connect with Ethernet since my laptop does not have an Ethernet port. So is there any way for me to connect to the internet in ubuntu? Edit: Output of the terminal command  lspci -knn | grep Net -A2:  vikram@vikram-Inspiron-11-3148:~$ lspci -knn | grep Net -A2 01:00.0 Network controller [0280]: Broadcom Corporation BCM43142 802.11b/g/n [14e4:4365] (rev 01)     Subsystem: Dell Device [1028:0018]     Kernel driver in use: bcma-pci-bridge vikram@vikram-Inspiron-11-3148:~$    I did what Pilot6 suggested. Here is the output:    vikram@vikram-Inspiron-11-3148:~$ sudo dpkg -i *.deb [sudo] password for vikram:  Selecting previously unselected package bcmwl-kernel-source. (Reading database ... 163804 files and directories currently installed.) Preparing to unpack bcmwl-kernel-source_6.30.223.141+bdcom-0ubuntu2_amd64.deb ... Unpacking bcmwl-kernel-source (6.30.223.141+bdcom-0ubuntu2) ... Selecting previously unselected package dkms. Preparing to unpack dkms_2.2.0.3-1.1ubuntu5.14.04_all.deb ... Unpacking dkms (2.2.0.3-1.1ubuntu5.14.04) ... Setting up dkms (2.2.0.3-1.1ubuntu5.14.04) ... Setting up bcmwl-kernel-source (6.30.223.141+bdcom-0ubuntu2) ... Loading new bcmwl-6.30.223.141+bdcom DKMS files... First Installation: checking all kernels... Building only for 3.16.0-30-generic Building for architecture x86_64 Building initial module for 3.16.0-30-generic ERROR (dkms apport): kernel package linux-headers-3.16.0-30-generic is not supported Error! Bad return status for module build on kernel: 3.16.0-30-generic (x86_64) Consult /var/lib/dkms/bcmwl/6.30.223.141+bdcom/build/make.log for more information. modprobe: FATAL: Module wl not found. update-initramfs: deferring update (trigger activated) Processing triggers for man-db (2.6.7.1-1ubuntu1) ... Processing triggers for initramfs-tools (0.103ubuntu4.2) ... update-initramfs: Generating /boot/initrd.img-3.16.0-30-generic vikram@vikram-Inspiron-11-3148:~$ How do I convert two lists into a dictionary? How to install .bundle packages in Ubuntu? Adding admin-ajax.php to the frontend. Good or bad idea? I love admin-ajax.php. But I hate having to localize in order to point frontend scripts to it, and I wish there was an equivalent, easy-to-find file for themes. (It also just bothers me to see frontend requests go through "/wp-admin/". No practical reason, just looks ugly IMO.)  So I've simply copied admin-ajax.php to the root dir at "/ajax.php", adjusted the include paths and removed the WP_ADMIN constant definition. Seems to work like gangbusters (I can now just direct all my frontend AJAX requests to /ajax.php! And I can still use the normal wp_ajax hooks in my plugins!).  But is this safe? What might go wrong? Since this isn't built into core, I assume there's a good reason as to why not. But looking through the code, I can't see any immediate problems.   You're smart--tell me if this approach is crazy. Or if there's a simpler method that I'm overlooking. java.lang.OutOfMemoryError: PermGen space Exception