I just noticed that trying to visit one of the per-site metas using HTTPS will (after confirming the security exception due to the ) give a 403 error:    Yes, I know that .  But the per-site metas used to work over HTTPS, so this is a regression.  I did some extra testing, and it appears that this error shows up only for the per-site metas with a four-part hostname of the form meta.sitename.stackexchange.com.  That is, the following meta sites with custom domains work fine over HTTPS:           but any other per-site metas, (like, say, ) are broken.  From the error message, I suspect this regression has something to do with the use of the CloudFlare CDN.  Perhaps something needs to be (re)configured in SE's CloudFlare settings to make this work again?
Clicking on  doesn't connect me with the meta S.Ex.SE site.  EDIT: For some entry points, . Observation courtesy Gilles.  I consider it a pity that the devs have not paid attention to this issue nor have indicated that they noted the problem. While HTTPS is not yet official on this network, I refrain from using unencrypted connections. I can't participate in metas, oh well.  EDIT #2 (2015-11-07) - with the switch to CloudFlare SSL the problem looks finally resolved.  EDIT #3 (2015-11-17) - still getting 403 Forbidden from CloudFlare. D'rats!..  EDIT #4 (2016-01-11) - We're still there. It's 2016 already... Starting a bounty.
Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ Proving that $G$ is a group if $a*x=b$ and $y*a=b$ have solutions. 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! How do you reindex an array in PHP but with indexes starting from 1? I have the following array, which I would like to reindex so the keys are reversed (ideally starting at 1):  Current array (edit: the array actually looks like this):  Array (  [2] =&gt; Object     (         [title] =&gt; Section         [linked] =&gt; 1     )  [1] =&gt; Object     (         [title] =&gt; Sub-Section         [linked] =&gt; 1     )  [0] =&gt; Object     (         [title] =&gt; Sub-Sub-Section         [linked] =&gt;      )  )   How it should be:  Array (  [1] =&gt; Object     (         [title] =&gt; Section         [linked] =&gt; 1     )  [2] =&gt; Object     (         [title] =&gt; Sub-Section         [linked] =&gt; 1     )  [3] =&gt; Object     (         [title] =&gt; Sub-Sub-Section         [linked] =&gt;      )  ) Methods for autocompletion in TexStudio don't work I've made a mymacros.sty file which I would like my LaTeX files to be able to access with commands that I've defined. I've managed to get it to work by putting \usepackage{mymacros} in my preamble.   However when I try to autocomplete I can't. I looked at this question:     ), which recommends 3 ways of doing this, none of which work for me.    The first it recommends is to have mymacros.sty open in the editor. But this doesn't work.  The second is to activate the option Editor -> Automatically load included files in the configure TeXstudio options panel. However while I have an Editor tab , there is no such option there.  The third is to using .cwl files, however I'm not sure how to go about this. In the completion tab I've noticed that there is a mymacros.sty option that has come up, which I ticked but that has had no effect. Do I need to make my own mymacros.cwl file? I've tried to look this up but haven't had any luck.  Anyone have any ideas? Thanks in advance! Do free falling observers see gravitational blueshift? Observers at rest in gravitational fields will see infalling light signals as blueshifted.  Do inertial observers in free fall in a gravitational field see the same shift?  If someone is standing on a cliff at rest in the gravitational field and then steps off, do they instantly see a change in the blueshift as a result of being in free fall in an inertial reference frame? How do I insert the total continuation count in the allowframbreaks frame title? I would like to change the numbering of allowframbreaks frame, for instance, from I II III to (1/3) (2/3) (3/3)， but fail to find out a command to get the total continuation count. Here is the MWE.  \documentclass[compress]{beamer} \usetheme{Warsaw}   %allowframebreaks numbering in the title \setbeamertemplate{frametitle continuation}{(\insertcontinuationcount/)} %\setbeamertemplate{frametitle continuation}{(\insertcontinuationcount/\inserttotalcontinuationcount)}  \title{title} \author{author}  %%%%% \begin{document}  \begin{frame}[allowframebreaks]{allowframebreaks title} frame 1 \framebreak  frame 2 \framebreak  frame 3 \framebreak \end{frame}  \end{document} I refer to the usage of "don't" as an imperative to tell someone what not to do. As in,     Hey! Don't you dare touch that button!   When it is used in the interrogative or as part of a statement, "don't" can be expanded like most other contractions. It either just breaks up right in the sentence, or surrounds the subject.     Don't you love me, Juanromeo? -> Do you not love me, Juanromeo?      I don't love you, Marijuliet. -> I do not love you, Marijuliet.   However, I have no idea how one breaks up the imperative. "Do you not dare[...]" sounds like it's a lead up to a question, "You do not dare[...]" comes across as a statement moreso than a command, and "Do not you dare[...]" just sounds silly.  "Don't" is a bit unique among "not" contractions (at least that I know of) because it is the only one that can be used in an imperative sense. Couldn't, shouldn't, wouldn't, haven't, isn't, doesn't, aren't, can't... none of these can be used in an imperative sense because you cannot command someone to could, should, etc.  Is it possible to expand the don't contraction when it is used as part of a command? Or is its unique case such that it exists as its own linguistic construct? Should I load an entire html page with AJAX? What really allows airplanes to fly? What is the best (idiomatic) way to check the type of a Python variable? I need to know if a variable in Python is a string or a dict.  Is there anything wrong with the following code?  if type(x) == type(str()):     do_something_with_a_string(x) elif type(x) == type(dict()):     do_somethting_with_a_dict(x) else:     raise ValueError   Update: I accepted avisser's answer (though I will change my mind if someone explains why isinstance is preferred over type(x) is).  But thanks to nakedfanatic for reminding me that it's often cleaner to use a dict (as a case statement) than an if/elif/else series.  Let me elaborate on my use case.  If a variable is a string, I need to put it in a list.  If it's a dict, I need a list of the unique values.  Here's what I came up with:  def value_list(x):     cases = {str: lambda t: [t],              dict: lambda t: list(set(t.values()))}     try:         return cases[type(x)](x)     except KeyError:         return None   If isinstance is preferred, how would you write this value_list() function? Site hosted on EC2 / Amazon held hostage with private key I have a new client whose previous developer has left her high &amp; dry with her site on Amazon/EC2. He holds the private key to the instance, and though she has full access to the Amazon control panel (she's paying $300/mo for a site that has like 3 visitors per week) she doesn't have any way to get her content off and move to a new host.  I'm not familiar with AWS and am not sure how to help her. The site's group/file permissions are screwed up on the server so exporting the content/database via her WordPress control panel isn't working.  Any ideas on how to help her? Thanks! LibreOffice problem while openning a document My problem is that when I try to open one specific .odt document, I can't move the page or do anything except moving the mouse, which is useless. Also, the mouse, when I'm on the document, is freaking out switching between mouse and loader couple of times per second. It also becomes dark when I click on the document like if it's not responding. How does gravity work underground? Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not? What is the difference between encrypting some data vs signing some data (using RSA)?    Does it simply reverse the role of the public-private keys?    For example, I want to use my private key to generate messages so only I can possibly be the sender.  I want my public key to be used to read the messages and I do not care who reads them.  I want to be able to encrypt certain information and use it as a product-key for my software.  I only care that I am the only one who can generate these.  I would like to include my public key in my software to decrypt/read the signature of the key.  I do not care who can read the data in the key, I only care that I am the only verifiable one who can generate them.  Is signing useful in this scenario? Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots. Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.  My attempt:  Consider the polynomial $g=(x-1)(1+x+x^2+x^3+\cdots+x^n)$  As $f\mid g, g$ all the roots of $f$ are roots of $g$.  This means I have to prove the statement:  If $c$ is a multiple root of $f \Longrightarrow c$ is multiple a root of $g$.  Equivalently:  If $c$ is not a multiple root of $g \Longrightarrow c$ is not a multiple root of $f$.  I'll try to prove this last statement:  Suppose $\exists c$ such that $c$ is a multiple root of $g$, then   $$g(c)=0 \implies g'(c)=0$$  $$ g=(x-1)(1+x+\cdots+x^n) \implies  g'=nx^n $$ $$ g'(c)=nc^n=0 \implies c=0 $$ But  $$ g(0)\neq 0 $$ And we've found the contradiction: $\not \exists c: g(c)=g'(c)=0$. Then $f$ has no multiple roots.  Is this correct?   Also: Assuming this is correct, is it possible to prove $\sum_{k=0}^n \frac {x^n}{n!} $ has no multiple roots in a similar manner? Is there a reason I would not add /usr/local/sbin, /usr/sbin, /sbin to my path on Debian? I am working in Illustrator (15.0.2 (CS5) or 16.0.0 (CS6)) and when I try to move objects small distances, they snap to the nearest two pixels on some invisible grid. I can't find anything in the preferences. Here are some screen shots: How to calculate the number of days you can stay in Taiwan on a tourist visa? Tourist visas to Taiwan have an associated "maximum stay" of 30 days. For those visas, how are the days counted? Does the stay begin on the partial day that the traveler arrives or the first full day spent in the country? Does it end on the last full day spent in the country, or the actual day the traveler departs? As far as I know, current division for spesific place's formula is: Ix=(Req/Rx)*Isource   However, here,the solution is 3/(3+2+4)*6. How is this a current division ? Can you explain this to me