How to do this in latex
How to look up a symbol or identify a letter from a math alphabet or other character?
Some common problems with displaying routes created from street data, such as bus routes, include:   Routes may overlap themselves, such as when a route takes the same road in opposite directions Routes may overlap other routes that take the same street. Routes are often overly detailed, or "noisy", owing to their derivation from street data. Indicating a route's direction unambiguously without overloading it with arrows can be a challenge.   What are some techniques for dealing with these issues in order to create a practical and usable route map, such as this ? Do you know of any such public transport maps that were produced by automatic techniques rather than manual illustration/digitization?  Specifically I am using ArcGIS 10, but solutions involving other software are welcome as well.  One technique I have had some success with for dealing with 1) is using ArcMap's  with a negative offset -- assuming the lines are oriented in the direction of travel, this will offset the lines such that the right and left-hand sides are distinct and oriented as you would expect (at least for  countries, for left-hand traffic countries one would use a positive offset). However the result tends to be a bit wonky around turns, interchanges and and U-turns, and is inflexible when dealing with multiple routes.  Another possibility that I am not very familiar with but which seems promising is  -- however, I have not had much luck in finding examples of using representations for transportation maps and whether it might fit the bill. I'm trying to prove that  $$ \left| \frac{1}{x}-\frac{1}{y} \right| $$  is a metric on $X = \mathbb R^+$.  The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e.  $$ d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. $$  So, ok, I have that  $$ d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| $$  ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by $ \left| \frac{1}{x}-\frac{1}{z} \right| $, but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks. What do the numbers in a man page mean? So, for example, when I type man ls I see LS(1). But if I type man apachectl I see APACHECTL(8) and if I type man cd I end up with cd(n).  I'm wondering what the significance of the numbers in the parentheses are, if they have any. Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that      $$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$   I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$  Doing some manipulation we get:           $$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$      How can I prove this limit? What is the VC dimension of a decision tree? What is the  of a decision tree with k splits in two dimensions? Let us say the model is CART and the only allowed splits are parallel to the axes.  So for one split we can  and then for any labeling of the points we could get perfect prediction (i.e.: shattered points)  But what about 2 splits, or any general k? Can I successfully bake previously-frozen bread dough? Why am I getting "size mismatch" errors from the Iceland mirror? After installing Ubuntu 12.04 (clean install) I fetch a bunch of program via synaptic (installed specially), all programs worked except Gimp 2.8. So I tried purging Gimp, install again (no luck). Checked broken packages and reinstalled them. When using the Software center I get this MSG:  Failed to fetch http://is.archive.ubuntu.com/ubuntu/pool/main/o/openexr/libopenexr6_1.6.1-4.1_amd64.deb Size mismatch   Tried also this:  sudo apt-get purge gimp* sudo add-apt-repository ppa:otto-kesselgulasch/gimp sudo apt-get update sudo apt-get install gimp   And get this MSG:  Failed to fetch http://is.archive.ubuntu.com/ubuntu/pool/main/o/openexr/libopenexr6_1.6.1-4.1_amd64.deb  Size mismatch E: Unable to fetch some archives, maybe run apt-get update or try with --fix-missing?   I am wondering if the Icelandic Archive has an outdated version or something?   would really like to install Gimp 2.8 but Gimp 2.6 is OK to.   I am also wondering if I am doing something wrong? I recently setup a new computer for my parents with windows 10 home. I made two local accounts for them - no microsoft accounts. I also would like to keep them without a password. Problem is now that windows automatically signs in with the last used account. No lock/login screen.  I tried a number of answers from similar questions here but none worked. Since it's a home version there are no group policies and the method using task scheduler does not work because accounts with no password are not allowed to create such tasks.  Are there other options to force windows to show a user list on startup without setting passwords for the accounts? How to get exact camera view? I'm currently a beginner when it comes to 3D modelling but doing okay so far. I'm struggling a bit when it comes to the final renders of my scene, specifically on the camera view I want. Sometimes its to zoomed in and sometimes to zoomed out that it's gone out of the environment of the blender. I realized the settings for the camera on blender are for actual cameras / photography and I'm no expert on that too. A tip on how to control them precisely would be nice. Thank you. This has come up    : migrating old questions causes problems.   The most popular solution to this to date has been to . This is do-able, but frankly it feels like treating a symptom, and doing so in a way that penalizes folks who answer what are by all appearances valid, on-topic questions.   Migration works best as a way to salvage good but off-topic questions and give them new life on a site where they can thrive. But shuffling old, answered, on-topic questions around for the hell of it is a waste of everyone's time.   Therefore, I think the real solution is much simpler: just disable the migration of old questions. They're rarely welcomed or heavily improved post-migration, often overtly disliked by folks on the destination site, and can end up causing a rather bad experience for folks who were participating in good faith on the original.   I suggest that questions older than 601 days should not be eligible for migration, either by normal user close-votes or by moderators. In extreme cases, we can perhaps provide an alternate means of moving old questions, but under normal circumstances these should be done quickly or not at all.   1 60 days is actually based on a  introduced this past spring: migrated and then rejected questions older than 60 days cause the authors to suddenly gain (permanent) reputation on the destination. $t^3$ is never equal to Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to: (A). $4t+3$ (B). $8t+5$ (C). $10t+3$ (D). $6t+5$  We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction? Prove that matrix can be square of matrix with real entries Prove that matrix   \begin{bmatrix}1&amp;0&amp;0\\0&amp;-1&amp;0\\0&amp;0&amp;-1\end{bmatrix}  can be square of matrix with all real entries.  I have found one such matrix to be   \begin{bmatrix}1&amp;0&amp;0\\0&amp;1&amp;-1\\0&amp;2&amp;-1\end{bmatrix} but is there an elegant way to do it without any trial and error? Updating .class file in jar I'm looking for an algorithm which when given a latitude and longitude pair and a vector translation in meters in Cartesian coordinates (x,y) would give me a new coordinate. Sort of like a reverse Haversine. I could also work with a distance and a heading transformation, but this would probably be slower and not as accurate. Ideally, the algorithm should be fast as I'm working on an embedded system. Accuracy is not critical, within 10 meters would be good. My question is about one particular usage of static keyword. It is possible to use static keyword to cover a code block within a class which does not belong to any function. For example following code compiles:  public class Test {     private static final int a;         static {         a = 5;         doSomething(a);     }     private static int doSomething(int x) {         return (x+5);     } }   If you remove the static keyword it complains because the variable a is final. However it is possible to remove both final and static keywords and make it compile.   It is confusing for me in both ways. How am I supposed to have a code section that does not belong to any method? How is it possible to invoke it? In general, what is the purpose of this usage? Or better, where can I find documentation about this? Show that if the smallest prime factor $p$ of the positive integer $n$ exceeds $\sqrt[3]{n}$, then $\frac{n}{p}$ must be prime or 1.  I'm stuck trying to prove this. I tried this using contradiction, but I haven't been able to prove it.  My approach:  Let $p$ be the smallest prime factor of the positive integer $n$.  Let $p &gt; \sqrt[3]{n}$  Then we must show that $\frac{n}{p}$ is either a prime number or 1.  We know that $n = pe$ for some $e \in \mathbb{Z}$.  Thus, $\frac{n}{p} = e$  We also know that $p$ > $\sqrt[3]{n}$, so $p^3 &gt; n$, and $p^2 &gt; \frac{n}{p}$  So, assume $\frac{n}{p} = e$ is not prime nor 1, meaning it's a composite number. Let's try to get a contradiction.  If $\frac{n}{p}$ is composite, then $\exists a,b \neq 1: \frac{n}{p} = ab $  Now, here's where I get stuck. What can I do with this info? $\frac{n}{p} = ab $ and $p^2 &gt; \frac{n}{p}$. Derivative of $f(x)=|x|$ I'm trying to figure out a good workflow for developing scripts and add-ons in Blender. Here's what I'm currently doing as I try my luck with a PyNode add-on:   Open blender Create a text editor panel Open my script (external text file) Edit my script in a nice external editor Reload the script Save the .blend file Run the script   Now that I can see what my code does, this Blender instance has been modified, so I can't (make changes to my script and) run my script again. So I do this:   Close blender Open blender with the .blend from step 6 and repeat steps 4-7, over and over and over..   Is there a better workflow for this sort of thing? What causes the permittivity and permeability of vacuum? When light travels through a material, it gets "slowed down" (at least its net speed decreases).  The atoms in the material "disturb" the light in some way which causes it to make stops on its path.  This is expressed in the material's permittivity and permeability:  Its ability to transmit electromagnetic waves.  But then what causes the limited permittivity and permeability of vacuum?  There are no atoms there to disturb the light, so what is keeping it from travelling at infinite speed?  Is there a special property of space itself which determines the speed of light?  Or is it the virtual particles in vacuum which interact with the light? How can I recover my account? I tried to log into Minecraft today to find that my password isn't working. I tried to change it, but I can't login to my email because I forgot the answer to my security question. I tried changing the email to a different one, but that didn't work for some reason (it kept saying "invalid email"). I don't have my transation ID because I forgot which credit card I used.  Does this mean that I have to buy a new Minecraft account?