I tried to post an answer in the   I got this:    ...but I am able to  in the formatting sandbox    ...and I have more than 10 rep on MSE:    One needs 10 reputation to post an answer in a protected post, but I have more than 10, and I am allowed to post comments, which is a 50 rep privilege.  Then why am I not allowed to post an answer in the formatting sandbox?
Questions can be protected (i.e. marked "highly active").   What does it mean for a question to be protected? Why are some questions protected? Who can protect and unprotect questions? Who can answer a protected question? How are protected questions displayed? When should I protect or unprotect a question? When does the Community user protect a question?
dropping trailing '.0' from floats I'm looking for a  way to convert numbers to string format, dropping any redundant '.0'   The input data is a mix of floats and strings. Desired output:  0   --> '0'  0.0 --> '0'  0.1 --> '0.1'  1.0 --> '1'  I've come up with the following generator expression, but I wonder if there's a faster way:  (str(i).rstrip('.0') if i else '0' for i in lst)   The truth check is there to prevent 0 from becoming an empty string.  EDIT: The more or less acceptable solution I have for now is this:  ('%d'%i if i == int(i) else '%s'%i for i in lst)   It just seems strange that there is no elegant way to handle this (fairly straightforward) case in python. What is a Generic Handler in asp.net and its use? If I pass PHP variables with . in their names via $_GET PHP auto-replaces them with _ characters. For example:  &lt;?php echo "url is ".$_SERVER['REQUEST_URI']."&lt;p&gt;"; echo "x.y is ".$_GET['x.y'].".&lt;p&gt;"; echo "x_y is ".$_GET['x_y'].".&lt;p&gt;";   ... outputs the following:  url is /SpShipTool/php/testGetUrl.php?x.y=a.b x.y is . x_y is a.b.   ... my question is this: is there any way I can get this to stop? Cannot for the life of me figure out what I've done to deserve this  PHP version I'm running with is 5.2.4-2ubuntu5.3. Post for clarifications on the updated pronouns FAQ Error installing Apache2 Server Does a preprint include revisions made in response to peer review? This is related to but not answered in .  After an author has submitted a manuscript to a journal for peer-review, very often the journal asks the author to revise (and resubmit) the manuscript as a precondition for its publication. I am wondering whether the term preprint refers to the manuscript before or after its authors made these revisions.  If a preprint includes revisions made in response to peer review, then there is only a slight difference, mostly regarding layout and style, between the preprint and the article that is published in the journal. From a reader’s perspective, there would be no point in paying for access to the article, since they can simply download the preprint free of charge. A publisher would probably want to prevent their authors (by publication agreements etc.) from publishing preprints that differ from the respective journal article only in form.  As a side question, what kind of preprints do publishers tolerate, as a rule? Word for phrase Good for you, bad for us There is a phrase I have heard quite a bit recently dealing with people that are leaving the department for promotions, etc. When a person leaves, it is good for them, but bad for us.  It is good for them because they will be pursuing a new opportunity, but not so good for the department since they will have to be replaced, and others will have to pick up the slack until someone else can be hired and get up to speed.  Is there a word that means that a deal is good for one party, but not necessarily the other? "Has been" vs "was" Would you please explain the difference between these two sentences?     Spain ceded (or gave) Puerto Rico to the United States. Since then, Puerto Rico has been a U.S. territory.      Spain ceded(or gave) Puerto Rico to the United States. Since then, Puerto Rico was a U.S. territory.   I am stuck between 'has been' and 'was' in the above sentences. I recently installed Ubuntu 14.04, then I installed lamp-server and placed my webpages in /var/www directory, but when I opened localhost in the browser there was nothing. I think that happened because Zend updated Apache.  Whatever the reason may be, I want to know where to keep my php files so that I can access them from my browser. Fold a plane. then fold again Imagine a piece of paper and you fold it in half and then fold it again. that's what I am trying to do and cant. If I use bones the influence of the bones won't allow me to because I need 2 bones to control the same area. If I try shape keys it just deforms the mesh if I try to make it in multiple steps or does the wrong animation because it draws the shortest path instead of rotating from a point like a piece of paper would. I'm try to change to my Desktop directory from the terminal.  Running  cd /home/desktop   gives an error in the terminal saying no directory found.  What am I doing wrong? Intuition behind logarithm change of base I try to understand the actual intuition behind the logarithm properties and came across  that explains the multiplication and thereby also the division properties very nicely:     Suppose you have a table of powers of 2, which looks like this: (after revision)      $$\begin{array}{rrrrrrrrrr} 0&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10\\ 1&amp;2&amp;4&amp;8&amp;16&amp;32&amp;64&amp;128&amp;256&amp;512&amp;1024 \end{array}$$      Each column says how many twos you have to multiply to get the number in that column.  For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.      Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$.  Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.   I found that very helpful to understand the actual proofs for this property.  I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:  $$\log_a x = y \implies a^y = x$$ $$\log_b a^y = \log_b x$$ $$y \cdot \log_b a = \log_b x$$ $$y = \frac{\log_b x}{\log_b a}$$  But can somehow provide a explanation in the style of the quoted answer why this actually works? Set of left cosets is a group Let $N$ be a normal subgroup of $G$. Let $G/N = \{gN : g \in G \}$. Show that the $(G/N, \circ, 1_{G/N})$ is a group, where:  $\circ : G/N \times G/N \rightarrow G/N$ is defined by $(gN \circ hN) = ghN$ and $1_{G/N} = 1*N$.  Thanks for your help. How to set mount point of boot partition to /boot/efi? Suppose I am given a morphism of schemes $f: X \longrightarrow Y$ and suppose that for any point $y \in Y$, $\kappa(y)$ denotes the residue field. I would like to show that the fibered product $$X \times_{Y} \text{Spec}(\kappa(y))$$ is homeomorphic to the preimage $f^{-1}(y)$ in the induced topology.  The following is my attempt so far:  The question is obviously affine on the target, so without loss of generality we reduce to the case of a morphism $f: X \longrightarrow \text{Spec }A$. Then we have  $$ f^{-1}(\text{Spec }A) = \bigcup_{i \in I} \text{Spec }B_{i} $$ via morphisms $$ \phi_{i}: A \longrightarrow B_{i}. $$ We can then write, \begin{align*} f^{-1}(\text{Spec } A) \times_{\text{Spec } A}\text{Spec } \kappa(y) &amp;= \bigcup_{i \in I} \left( \text{Spec } B_{i} \times_{\text{Spec }A} \text{Spec }\kappa(y) \right) \\  &amp;= \bigcup_{i \in I} \text{Spec }  \left( B_{i} \otimes_{A} \kappa(y)  \right). \end{align*} Now let $\mathfrak{p}$ be the prime ideal of $A$ corresponding to the point $y \in \text{Spec } A$. Then $\kappa(y) = (A/ \mathfrak{p})_{\mathfrak{p}}$. Then for any of the $B$ (dropping the index for brevity), we have \begin{align*} B\otimes_{A} (A/ \mathfrak{p})_{\mathfrak{p}} &amp;\simeq B \otimes_{A} A / \mathfrak{p} \otimes_{A} A_{\mathfrak{p}} \\ &amp;\simeq B_{\mathfrak{p}} \otimes_{A} A / \mathfrak{p} \\ &amp; \simeq B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}} \end{align*}  Now the prime ideals in $B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$. Further, the prime ideals of $B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B$ contained in $\phi(\mathfrak{p})B$. That is, we have \begin{align} \text{Spec }B_{\mathfrak{p}} &amp;= \{  \mathfrak{q} \subseteq B :  \phi^{-1}(\mathfrak{q})  \subseteq \mathfrak{p}  \} \\ &amp;=  \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p}      \}. \end{align} Note that in the very last line we have abused notation slightly: There is not really any notion of being "inside" $\mathfrak{p}$ when it is being treated as a point in the scheme. So really we mean that $f(\mathfrak{q})$ is a point in the scheme whose corresponding prime ideal is a subset of $\mathfrak{p}$. Now as we said above, the prime ideals of $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ are the following $$ \text{Spec }(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}) = \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p}      \text{ and } ???  \} $$ This is the point at which I get stuck. Ideally I would like to make some argument from the statement "Now the prime ideals in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$" to conclude that the question marks are the condition $\mathfrak{p} \subseteq f(\mathfrak{q})$. Could someone point me in the right direction as to how to get this condition?   Is the proof up until this point correct? Different sources online seem to do this many different ways, so it's difficult for me to check. echo text after curl command on same line #!/bin/bash echo "$(curl -s -I https://google.com|grep Server)" abc   returns:  abcer: gws   and bash -x shows:    ++ grep Server ++ curl -s -I https://google.com ' abco 'Server: gws  abcer: gws   but    echo abc "$(curl -s -I https://google.com|grep Server)"`    returns  abc Server: gws   In other words, if I add text AFTER the curl command, it gets mangled, but it's ok if I add it before curl. What is happening? I cannot figure it out. Consider the integral equation $\phi(x)-\frac{e}{2} \int_{-1}^{1} x e^{t} \phi(t) d t=f(x) .$ Then  there exists a continuous function $f:[-1,1] \rightarrow(0, \infty)$ for which solution exists there exists a continuous function $ f:[-1,1] \rightarrow(-\infty, 0)$ for which solution exists for $f(x)=e^{-x}\left(1-3 x^{2}\right)$, a solution exists for $f(x)=e^{-x}\left(x+x^{3}+x^{5}\right),$ a solution exists  what I tried $f(x)=f(x)+\frac{e}{2} \int_{-1}^{1} x e^{t} \phi(t) d t$ $\phi(x)=f(x)+\frac{e x}{2} \int_{-1}^{1} e^{t} \phi(t) d t$ $ \begin{array}{l}L e t \quad c=\int_{-1}^{1} e^{t} \phi(t) d t \\ \Rightarrow \phi(x)=f(x)+\frac{x e}{2} c\end{array} $ $c=\int_{-1}^{1} e^{t}\left(f(t)+\frac{c e}{2} t\right) d t$ $c=\int_{-1}^{1} e^{t} f(t)+\frac{\operatorname{ce}}{2} \int_{-1}^{1} t e^{t} d t$ $c=\int_{-1}^{1} e^{t}f( t)+\frac{\operatorname{ce}}{2}\left[t e^{t}-e^{t}\right]_{-1}^1$ $c=\int_{-1}^{1} e^{t} f(t)+\frac{c e}{2}\left(2 e^{-1}\right)$ $c=\int_{-1}^{1} e^{t} f(t) d t+c$ $\int_{-1}^{1} e^{t} f(t) d t=0$ Option (3) and (4) satisfies this Relation so, (3) and (4) seems true  to me.is i am right? I am stuck  for option (1) and (2)  is there any result or theorem ? Why does Stack Exchange use Imgur over Amazon S3 or Azure? What benefit does  offer over , , etc. for image hosting?   Is it the simplicity of integration? The cost model? What's better about using a "gateway"? Configuring, compiling and installing a custom Linux kernel How to create a DEB file manually? I want to create a DEB file manually. I would like to just provide a folder which contains data to install, and a script to be executed after installation.  Is this possible?