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How to use downloaded fonts with pdfLaTeX on Ubuntu I want to use the  font in my pdfLatex document.  What steps should I do for using a downloaded font   Should I download the TTF or the OpenType version? Where should I store the font? Should any font processing be done? How can pdfLatex find this font?   I am using TexLive 2011 installed in a custom location (/somepath/texlive/2011) on Ubuntu Can a java file have more than one class? How can I lock a file using java (if possible) I am a relatively recent graphic design graduate. I have been working at a small printing house for about two years, but will soon be interviewing for a different position in another city.  My question is about what to include in my design portfolio. When I was in school, my professors all insisted that work included in a portfolio should be 100% original: my original photography, my original logos, my original layout, my original written content etc. However, at my current office, we rely heavily on client-supplied logos, type content, and photography—and stock photography (shutterstock, etc).  What do other designers include in their portfolios? Has anyone else had this issue and can I reasonable include these more recent projects in my portfolio if I label them appropriately (something like "layout design")? When writing a book and referring to a female and male aged 18–25, what term would be most appropriate? Boy and girl seem too young, whereas man and woman seem too old.  Example sentence:     A boy/man emerged from the shadows.   Edit: I think the word I am looking for is similar to 'guy' but less American and colloquial. Any thoughts? Need a MP3 ID3 tagger, and cover fetcher I need to tag my MP3 library, and have tried kid3 (which was manual tagging), when I used Kubuntu 9.10 (I now use Ubuntu Meerkat) Here are the features I am hoping for:   A good and clean UI. Tagging should be automatic, like Winamp's autotag feature, which rocks, btw! It should also embed the cover-art in the mp3, not copy a jpeg file in the folder, because now-a-days all players support displaying cover art. But acceptable if not possible. Rename the files as per some regular expression like %TrackNo - %Artist - %Title.  Should be accurate, and more importantly smart. I want to start tagging at night, and hopefully my collection should be done by the morning, w/o it being stuck at a user prompt at 1%.   If one app cant do all, I am willing to use 3, wouldn't mind exposure to a few more apps ;)  I have used picard or someting, and I didnt like it quite a lot. But wouldn't mind using it, if there is no other alternative.   Thanks for your time! The sup norm on $C[0,1]$ is not equivalent to another one, induced by some inner product Let $\mathrm{C}[0,1]$ be the space of continuous functions $[0,1]\rightarrow \mathbb{R}$ endowed with the norm $||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for $x(t)=t$ and $y(t)=1$). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties... prominently sports purple hair.    Is this her natural hair color?  Natural extensions in the answer depending on the correct answer: Is she wholly human and humans in the Galaxy Far Far Away can have this hair color?   Or is she partly human or not human?   Or does she dye her hair?   Or is there another hair color changing technology? Is spacetime absolute? As I understand it Newton's Laws imply that space is relative, as the laws of physics are the same in all inertial frames and as such there is no way, even in principle, to distinguish a frame that is truly at rest (absolute space). Hence the concept is physically meaningless, and the positions of (and distances between) physical objects, events, etc. are relative to the frame in which one is observing them from. Advancing on to Special Relativity, by postulating that the speed of light (in vacuum) is constant in all frames of reference, we are forced to conclude that time is relative also (as if it weren't then different observers would observe a different speed of light (in vacuum)). This leads to the result that the concepts of space and time are no longer completely separate and independent of one another and instead intertwined (as an event that occurs at rest over some time period in one frame, will occur over some spatial interval and a different time interval in another frame). Hence they should be considered as should be considered as a single entity, called spacetime.  Sorry for the waffling so far, just want to check that my understanding is correct up to this point?!  My main question is, given that space and time are (individually) relative quantities, is spacetime itself relative, or can it be considered absolute (as after all, it is the mathematical space of all possible events and exists independently of the physical events that occur within it)? Please have a look at this sentence:     We haven't gotten a lot of Finnish groups that we liked here at Bananas but I'm pretty happy with this four tracker from Helsinki, apparently recorded in the same place where the Ramones played in the 70's.     The first verb is in present perfect (up to now) and the second is in simple past tense. I don't understand the simple past for the second verb.  Why "like" is not in its present perfect form as the first verb is? Given a joint density function, what is the conditional expectation $E[Y|x]$? The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)= \left\{\begin{matrix}e^{-y}, \mbox{ } 0\leq x \leq y \le \infty \\ 0,  \mbox{   otherwise}\end{matrix}\right.$$ Evaluate the conditional expectation $E[Y|x]$.    I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.  My solution: (My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)  $$E[Y|x] = \int_{-\infty}^{\infty}yf_{Y|X}(y|x)dy.$$  $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$    $$f_{X}(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy$$   $$= \int_{0}^{\infty}e^{-y}dy $$  $$=\left [ -e^{-y} \right]\Big|_0^\infty $$  $$ = -e^{-\infty}-(-e^{0})=1. $$ $$f_{X}(x) = 1, x \geq 0.$$ Thus,   $$E[Y|x] = \int_{0}^{\infty}-ye^{-y}dy$$  $$= -\int_{0}^{\infty}ye^{-y}dy$$ After integration by parts,  $$E[Y|x] = \left [ ye^{-y}+e^{-y} \right]\Big|_0^\infty$$ $$=0-[e^{0} + 0]=-1$$ Therefore, $E[Y|x] = -1.$  Thanks in advance for reading this long problem and responding! I'm new to Ubuntu, it looks like it is better than Windows. Last day I tried to install Kget (the download manager) but during the process my computer suddenly stopped working. So I had to switch it off. After I logged into Ubuntu again I can't install any program. The Ubuntu Software Center says:  Items cannot be installed or removed until package catalog is repaired. Do you want to repair it now?   After I clicked repair it says (after few seconds):  Package operation fails - The installation or removal of a software package failed   I've tried several times, but the same problem occurred. Can anyone tell me a solution for this problem? How do I recover the Facebook account password? I have forgotten my Facebook account password and also I have no access to the number which I use to log-in to Facebook account.   On this page (see the image below) they are asking for a government photo-ID to verify my account. The problem is that I don't have a driving license, voter ID card, passport etc. All I have is a school ID card but that hasn't worked for me. Is there any other way to recover my password? How important are headings in SEO Proof: $\cos^p (\theta) \le \cos(p\theta)$ I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions! If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$ If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$. Here is my attempt at a proof. It feels very elaborate and I am not sure if it is correct. Can someone please point out any mistakes or places where I may improve. Thanks!  By the definition of continuity at $a$ of $f,g$ we have that $\lim\limits_{x\to a}f(x)=f(a)$ and $\lim\limits_{x\to a}g(x)=g(a)$. Suppose that $f(a)&gt;g(a)$. Then there exists a $\delta&gt;0$ such that $f&gt;g$ for all $x$ satisfying $|x-a|&lt;\delta$. Then for $x$ satisfying $|x-a|&lt;\delta$ we have $h(x)=f(x)$ and $k(x)=g(x)$ and thus $h$ and $k$ are continuous at $a$ because $f$ and $g$ are continuous at $a$. Now if $f(a)&lt;g(a)$ we simply relabel $f=\tilde{g}$ and $g=\tilde{f}$, so that $\tilde{f}(a)&gt;\tilde{g}(a)$ and we apply the previous result to show that again $h$ and $k$ are continuous at $a$.\  Now if $f(a)=g(a)$ then we can distinguish three cases. $f\geq g$ in a small neighborhood about $a$, $f\leq g$  in a small neighborhood about $a$ or $f\geq g$ on one side of $a$ and $f\leq g$ on the other side. In the first case we assume that $f\geq g$ in some small neighborhood about $a$. Then, since $f(a)=g(a)$ we have $h(x) = f(x)$ in this neighborhood and $k(x) = g(x)$ and again we see that $h$ and $k$ are continuous at $a$ due to the continuity of $f$ and $g$ at $a$. Similarly if $f\leq g$ in a neighborhood around $a$ then $h(x)=g(x)$ and $k(x)=f(x)$ in this neighborhood and $h$ and $k$ are continuous at $a$.   Suppose that just to the left of $a$ we have $f\geq g$ and just to the right of $a$ we have $f\leq g$. Then $\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}f(x)=f(a)=h(a)=g(a)=\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^+}h(x)$. Similarly $\lim\limits_{x\to a^-}k(x)=\lim\limits_{x\to a^-}g(x)=g(a)=k(a)=f(a)=\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^+}k(x)$. So $\lim\limits_{x\to a}h(x)=h(a)$ and $\lim\limits_{x\to a}k(x)=k(a)$ which means that $h$ and $k$ are continuous at $a$. Lastly if $f\leq g$ just to the left of $a$ and $f\geq g$ just to the right of $a$ we can relabel $f=\tilde{g}$ and $g=\tilde{f}$ and apply the last result to show $h$ and $k$ are continuous at $a$.   Again, please point out any mistakes I may have made. Thanks!! What does "65" mean? K-2SO beats up the troopers easily. Up in the sky, the bombers are preparing to attack. Just as they start the attack run, TIE fighters come in from the shield and swarm the rebel fleet.  JON VANDER (GOLD LEADER): Red Leader, this is Gold Leader, we’re starting our attack run on the  shield gate. Keep it tight and watch out for those towers. GOLD PILOT: More fighters. Form up, at 65.  What does &quot;65&quot; mean? What is the relationship between ISO, aperture, and shutter speed? I know digital cameras have ISO options, and that ISO is the camera's sensitivity to light, but if you set  higher ISO then you can get a noisy image. I also know there are two other camera options, shutter speed and aperture.   What is the relationship of between them? Is there any equation, or something like that?  For example, if I set the ISO to 640, then how should I set the shutter speed and aperture? I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:      Show that, for all $n &gt; 1$ $$\frac{1}{n + 1} &lt; \log(1 + \frac1n) &lt; \frac1n.$$    I tried to follow the below steps in order to prove the RHS inequality:   Proving that $f &lt; g$ on $I$ from $a$ to $b$:  Step $1$. Prove that $f' &lt; g'$ on $\operatorname{Int}(I)$.  Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$  Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x &gt; 1$. One has  $$f'(x) =  -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$  We note that, for every $x &gt; 1$, $f'(x) &gt; g'(x)$. Moreover, $f(1^+) = \log(2) &lt; 1 = g(1^+).$   My problem is that I got the wrong inequality sign in Step $1$.    Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case. How can I get admin access to a Mac without knowing the current password?