Suppose I want to change from numbering my \sections, \subsections, and so on to not numbering them.  As far as I know, the only way is to manually change (e.g.) \section{Foo} to \section*{Foo} and so on, all through the document... one by one.  That seems at odds with the LaTeX philosophy.  It is also a very manual and tedious process, especially when one has a master document and sub-files.  I know there are tricks within editors and the command line to globally replace text with replacement text... but a cleaner solution would be a LaTeX option.  In other words:  Is there a global LaTeX option one can set to swap from numbered to unnumbered sectioning (and vice verse)?  P.
Every time I call \section{**Section Name**}, a number appears to the left of the name of the section.  How do I hide this number?
How to compare Unicode strings in Javascript? Usage of :- (colon dash) in bash Redirecting stdout to a file you don't have write permission on When you attempt to modify a file without having write permissions on it, you get an error:  &gt; touch /tmp/foo &amp;&amp; sudo chown root /tmp/foo &gt; echo test &gt; /tmp/foo zsh: permission denied: /tmp/foo   Sudoing doesn't help, because it runs the command as root, but the shell handles redirecting stdout and opens the file as you anyway:  &gt; sudo echo test &gt; /tmp/foo zsh: permission denied: /tmp/foo   Is there an easy way to redirect stdout to a file you don't have permission to write to, besides opening a shell as root and manipulating the file that way?  &gt; sudo su # echo test &gt; /tmp/foo What is the reason the log transformation is used with right-skewed distributions? I once heard that      log transformation is the most popular one for right-skewed distributions in linear regression or quantile regression   I would like to know is there any reason underlying this statement? Why is the log transformation suitable for a right-skewed distribution?   How about a left-skewed distribution? I am trying to make folder that everyone can view and can't copy,modify.    For all answers.I thinking about specially ms office files. I tried following methods  Software  - copy protect    - anti copy    This method can't protect save as and upload  MS office macro    Users can enable,disable macros.  MS office restriction    This is best choice but can't protect from file uploading.   Almost I found the solution. Way is combination of MS office restriction and restrict access from browsers using firewall or antivirus. But I don't exact how to do this. How to understand intuitively the Stolz-Cesaro Theorem for sequences? In Windows I used command-line clipboard copy-and-paste utilities...      pclip.exe and gclip.exe     These were UnixUtils ports for Windows (but they only handled plain text). There were a couple of other native Windows utilities which could write/extract any format.    I've looked for something similar in Synaptic Package Manager, but I can't find anything.    Is there something there, that I've missed? ... or maybe this is available in Bash scripting?    The type of utility I'd like will be able to read/write via std-in/std-out or file-in/file-out, and handle Unicode, Rich Text Format, picture, etc. clipboard formats...    NB: I'm not after a clipboard manager. Why Process Builder doesn't work on Delete Scenario? As most of the time we have to think of requirement to accomplished using out-of-Box feature like e.g process builder which we think of an alternative to trigger, but suppose I have to handle scenario of Delete using Process Builder then there is limitation!(e.g calculation of roll-up on Delete) so I am curious about what is that limitation because of which Delete scenario is not handled in Process Builder. Limit of $x\left(\left(1 + \frac{1}{x}\right)^x - e\right)$ when $x\to\infty$ I am stuck on how to calculate the following limit: $$\lim_{x\to\infty}x\left(\left(1 + \frac{1}{x}\right)^x - e\right).$$  Definitely, it has to be through l'Hôpital's rule. We know that $\lim_{x\to\infty} (1+(1/x))^x = e$. So, I wrote the above expression as $$\lim_{x\to\infty}\frac{\left(1 + \frac{1}{x}\right)^x - e}{1/x}.$$Both numerator and denominator tend to zero as x tends to infinity. I applied l'Hôpital's rule twice, but I got the limit equal to infinity which is clearly wrong. In the book, it says that the limit should be $-e/2$.  Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your l'Hôpital's rule to in both cases? Thanks a lot  Also, guys can you tell me how to write equations in this forum? Thanks. How to show the inequality for $\int_{0}^{\pi/4}\tan^n(x) dx$ I can't seem to figure out the inequality option, I substituted n=2 for the answer bit is there any elegant way of getting it? Curve Bevel function, scale? Is there a way to scale up the bevel object on one end of a curve, a little bit over each point?  I notice the vertex "weight" slider will tweak how much influence each vertex has on the others, is there something like this that can scale the bevel object up a little bit?  Preferably with the option to constrain X,Y,Z independently newcommand vs. DeclareMathOperator As far as I understand, whatever can be done using DeclareMathOperator can also be achieved using newcommand, but not the other way around. If this is the case, what are the circumstances where DeclareMathOperator is the better choice? When should I take solely newcommand?  I found a question regarding , and I wonder what other aspects should be taken into consideration when it comes to choosing which method to use when defining math operators. In System Info in 11.10, the graphics driver wasn't detected. I use ACER Aspire 4750 with Intel Core i3-2340M and Intel HD Graphics 3000 graphic card  How to install the driver for my Ubuntu? VirtualBox error kernel headers I'm with Ubuntu 15.04 and I'm trying to create an emulator (genymotion), but when run virtualbox, i got the error:     Stopping VirtualBox kernel modules ...done.   Uninstalling old VirtualBox DKMS kernel modules ...done.   Trying to register the VirtualBox kernel modules using DKMSError! Your kernel headers for kernel 3.16.0-30-generic cannot be found.   Please install the linux-headers-3.16.0-30-generic package,   or use the --kernelsourcedir option to tell DKMS where it's located    ...failed!     (Failed, trying without DKMS)   Recompiling VirtualBox kernel modules ...failed!     (Look at /var/log/vbox-install.log to find out what went wrong)   I performed   uname -a   And the result is:     Linux douglas 3.16.0-30-generic #40-Ubuntu SMP Mon Jan 12 22:06:37 UTC 2015 x86_64 x86_64 x86_64 GNU/Linux   Someone help me?Already I uninstalled and installed again. If $f(0)=0$ then prove that $|f(z)+f(-z)| \leq 2|z|^2$ Let , $\Delta = B(0,1)$.  If $f:\Delta \rightarrow \Delta$ is holomorphic, and $f(0)=0$, prove that $|f(z)+f(-z)| \leq 2|z|^2$.  My attempt so far:  $\displaystyle f(z) = \sum_{n=0}^{+\infty} a_n z^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} + \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}$  $\displaystyle f(-z) = \sum_{n=0}^{+\infty} a_n (-z)^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} - \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}$  Then, $\displaystyle f(z)+f(-z) = 2  \sum_{n=0}^{+\infty} a_{2n} (z^2)^n =: 2 g(z^2)$  If $g(\Delta) \subset \Delta$, as $g(0) = 0$, I can invoke Schwarz Lemma and I win. However, I don't know how to prove this. Could anyone help me? Thanks! How to convert a Collection to List? Fermat: Last two digits of $7^{355}$ I am doing this problem mentioned above and I know the answer because I know Euler's Theorem that $$a^{\varphi{(m)}}\equiv{1}\pmod{m}.$$  I used 100 as my modulus and got that the last two digits of $7^{355}$ are $43$.  My question is this.  What prime should I pick to solve this problem using Fermat's theorem $$a^{p-1}\equiv{1}\pmod{p}$$  As a precursor problem, the author (Dudley; Elementary Number Theory) asked to find the last digit of the number above; so using a modulus $5$, $7^4\equiv{1}\pmod{5}$ by Fermat.  Since the answer is $7^{355}\equiv{3}\pmod{5}$, the last digit is either $3$ or $8$.  Since $7$ is odd, the answer will clearly be odd, so $3$ is it.  The book is working linearly, building from the previous chapters and Euler's Theorem isn't mentioned for another two chapters (I remembered Euler from a problem solving class in college).    So again, what prime would be best?  How should I procede?    EDIT: I've seen a bunch of problems on the site and none that I looked at did it using Fermat.  If there is one, that could be posted and I would be happy.  EDIT 2:  it seems that the problem of finding of finding the last two digits using is rather difficult because of the implication that we need something mod 100.  Is there a way around this , like adding a step or complexifying, in order to finish the problem with Fermat? Do there Exist Proper Classes that aren't "Too Big" Some proper classes are "too big" to be a set in the sense that they have a subclass that can be put in bijection with $\alpha$ for every cardinal $\alpha$. It is implied in  post that every proper class is "too big" to be a set in this sense, however I have been unable to prove it. It's true if every two proper classes are in bijection, but it's consistent with ZFC for there to be a pair of non-bijective classes.  So, is the following true in ZFC:     For all proper classes, $C$, and $\alpha\in\mathbf{Card}$, $\exists S\subset C$ such that $|S|=\alpha$?   If not, is there something reasonable similar that preserves the intuition about classes that are "too big to be sets"? $a,b,c$ are distinct real numbers that $$(a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.$$ Prove $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$$ I have tried two different approaches to this problem but am stuck.  First I started by expanding the bottom of each fraction. $\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1$  I also tried to expand $(a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca$   $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca$ $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0$ Then I added one to both sides $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1$ Help will be appreciated. I gave a quick answer to part of  which had not been covered by previous answers, trying to clarify the reason you would say time of decoding but not decoding’s time. I said it was ’s usually indicates possession, but of course there were several counterexamples that would have occurred to me after a moment’s consideration, and these where helpfully supplied:   Britain’s climate two days’ time  a day’s work  the sun’s rays   I am still of a mind to say that possession of some sort is what allows the ’s. Even though the sun does not have title to its rays, they do belong to the sun. Now, at the risk of duplicating the original question and/or being pigheaded, I am curious as to why time of decoding but not decoding’s time is correct, if not for the reason I gave.