From time to time single users are “removed” for whatever reason together with their questions and answers, and also the reps of those who answered their questions. While the reasons for users being removed are not revealed, I don’t understand why I can’t easily see which of my answers were removed as a consequence.  -20   7 hours ago removed User was removed (learn more)  What is the reason for this? I am asking Why not how.
What circumstances can cause a question or answer to be deleted, and what does that actually mean?    How can a post be deleted? When can't I delete my own post? Can I still see my post even after it's deleted? Can I see a list of my deleted posts? How can I undelete one of my posts? What does deletion mean for a post? How do votes to delete work? What are the criteria for deletion? What else should I know about deleted posts? If I flag my question with a request to delete it, what will happen?       For more information, see the articles about  and  in the .
What is wrong in this proof: That $\mathbb{R}$ has measure zero Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$.  The total length of the intervals is a geometric progression that sums up to $\epsilon$.  Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals.  Thus the entire real line is covered by the union of the $I_k$, thus $\mathbb{R}$ is a null set with measure zero.  Clearly there is something wrong in the above proof, however I am not sure where is it?  Thanks for any help. Non-https links to other StackExchange sites in User profile when using https Euler-Lagrange equation in curved spacetime: how to calculate $\partial \mathcal{L}/\partial(\nabla_\nu A_\mu)$? My problem is mostly with notation. When we apply the Euler-Lagrange equation to the Lagrangian scalar $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ Where $$F_{\mu\nu} = \partial_\mu A_\nu -\partial_\nu A_\mu= \nabla_\mu A_\nu -\nabla_\nu A_\mu$$ and I start with $$\frac{\partial\mathcal{L}}{\partial A_\mu}=\nabla_\nu\left(\frac{\partial \mathcal{L}}{\partial(\nabla_\nu A_\mu)}\right).$$   Now I rewrite the lagrangian as $\mathcal{L} = -\frac{1}{4}g^{\mu\alpha}g^{ \nu\beta}(\nabla_\mu A_\nu-\nabla_\nu A_\mu)(\nabla_\alpha A_\beta-\nabla_\beta A_\alpha)$ $$\mathcal{L}= -\frac{1}{4}g^{\mu\alpha}g^{\nu\beta}\left[(\nabla_\mu A_\nu)(\nabla_\alpha A_\beta)-(\nabla_\mu A_\nu)(\nabla_\beta A_\alpha)-(\nabla_\nu A_\mu)(\nabla_\alpha A_\beta)+(\nabla_\nu A_\mu)(\nabla_\beta A_\alpha) \right]$$ My problem is calculating $\frac{\partial \mathcal{L}}{\partial(\nabla_\nu A_\mu)}$, how are the rules of partial derivatives generalised to covariant ones? Also, is $\frac{\partial\mathcal{L}}{\partial A_\mu}=0$, is that correct ? I don't know so much about coordinate systems...  In my office we use to deal with spatial data coming from archaeological sites. Each site has its own x-y-z coordinate system (GCS). Three simple ortogonal cartesian axis. In the last years we have been managing this spatial data through GIS software (ArcGIS), without using specific coordinate system (just leave it as "undefined")  I'd like to know if there exisits any GCS designed to deal with such datasets using simple cartesian orthogonal axis, without grid distortions of the typical GCS. In addition, I'd like to know if this system is suitable for using it in an online mapping application.  By the way, we manage 2D (ArcMap) and 3D (ArcScene) environments and work with "mm" as length base unit.   If such a thing doesn't exists, maybe someone knows how to create it. Is Pauli-repulsion a "force" that is completely separate from the 4 fundamental forces? I only ask questions. Am I unwanted? Possible Duplicate:          I've been  for maybe a year to get answers to questions I couldn't find on my own.  I always make an attempt first to find the answer on my own.  As of this posting, I've asked 14 questions.  I only once gave an answer, and it was to a non-technical question, and only because I was really bored and had too much coffee to drink.  I wasn't intending to answer any other questions in the future, because on on my best days I'm only a mediocre programmer, and there are some real geniuses here on StackOverflow who can answer other peoples' questions faster and more accurately than I can.  It actually is my nature.  I am an Aquarian.  We are asker-listeners.  Today I discovered Meta Stack Overflow and started browsing around.  I became familiar with the concept of the "Help Vampire".  I seem to qualify.  I continued reading more posts on the topic and got the distinct impression that people who only ask, and never answer, are less than welcome here, rather despised.  So, should I just go away?  Or should I start contributing answers just to have a answer-to-question ratio of 3-to-1 (I read somewhere that's the average) even though I am less than one eighth as knowledgeable as the 10K+ guys and gals?    Or should I continue as I have begun?  I mean, doesn't a 0-questions-asked user like Jon Skeet need a Blankman?  Yin and yang, you know? I have a raster height model and would like to calculate the horizontal geographic direction (north, east, etc. - orientation) of each cell, then calculate then the mean value of all these cells in one region (polygon). This is to get information about to what direction a vineyard is exposed to. How do I do this? How do I identify which CPU a MacBook uses? Reverse of number in PHP not showing correct output How do I tell a Python script to use a particular version Enumerate or list all variables in a program of [your favorite language here] A friend asked me last week how to enumerate or list all variables within a program/function/etc. for the purposes of debugging (essentially getting a snapshot of everything so you can see what variables are set to, or if they are set at all). I looked around a bit and found a relatively good way for Python:   #!/usr/bin/python                                                                                                                                                                                                                            foo1 = "Hello world" foo2 = "bar" foo3 = {"1":"a",         "2":"b"} foo4 = "1+1"  for name in dir():     myvalue = eval(name)     print name, "is", type(name), "and is equal to ", myvalue   which will output something like:   __builtins__ is &lt;type 'str'&gt; and is equal to  &lt;module '__builtin__' (built-in)&gt; __doc__ is &lt;type 'str'&gt; and is equal to  None __file__ is &lt;type 'str'&gt; and is equal to  ./foo.py __name__ is &lt;type 'str'&gt; and is equal to  __main__ foo1 is &lt;type 'str'&gt; and is equal to  Hello world foo2 is &lt;type 'str'&gt; and is equal to  bar foo3 is &lt;type 'str'&gt; and is equal to  {'1': 'a', '2': 'b'} foo4 is &lt;type 'str'&gt; and is equal to  1+1   I have so far found a partial way in PHP (courtesy of ) but it only lists all variables and their types, not the contents:   &lt;?php // create a few variables $bar = 'foo'; $foo ='bar'; // create a new array object $arrayObj = new ArrayObject(get_defined_vars()); // loop over the array object and echo variables and values for($iterator = $arrayObj-&gt;getIterator(); $iterator-&gt;valid(); $iterator-&gt;next())         {         echo $iterator-&gt;key() . ' =&gt; ' . $iterator-&gt;current() . '&lt;br /&gt;';         } ?&gt;   So I put it to you: how do you list all variables and their contents in your favorite language?    Edit by : I propose this question follows the spirit of a little "". If you do not agree, just edit and remove the tag and the link. 'Thank you for taking your time writing' or 'Thank you for taking your time to write' ? A person gave me an answer on English.SE and I commented:     Thank you for taking your time writing this wonderful answer!   Now, looking at the sentence in retrospect, even though I am not a native English speaker (duh, otherwise I wouldn't be asking a question on ELL.SE), I find the sentence a bit odd.  To get my point across, let me rephrase the sentence:     Thank you for taking your time to write this wonderful answer!   With my relatively little knowledge and understanding of English, I feel that using 'writing' (as I did in the original sentence) emphasizes the very process of writing the answer, whereas using 'to write' emphasizes the mere fact that an answer has been given, and a wonderful one at that.  This is exactly why I feel the original sentence is a little bit awkward and the second, revised one would have been more apt.  Am I correct or are the quoted sentences virtually synonymous? How can I distribute a native executable for a Python program? Inequality from Chapter 5 of the book *How to Think Like a Mathematician* This is from the book How to think like a Mathematician,  How can I prove the inequality $$\sqrt[\large 7]{7!} &lt; \sqrt[\large 8]{8!}$$  without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) &lt; \frac 18 \cdot \ln(8!)$$ Parentheses vs. double commas vs. dashes to provide additional detail When do you use commas and when do you use parentheses to provide more detail about something?  For example:   The suspect, Tom Wilson, is now being charged with murder. The suspect (Tom Wilson) is now being charged with murder. John Smith, a member of the jury, agreed with the verdict. John Smith (a member of the jury) agreed with the verdict. The first example, the one about the fox, is the more valid one. The first example (the one about the fox) is the more valid one.   I realize that the first example, the one that uses parentheses, is probably more valid in this case (if anyone has a better example, please include it), but when, if ever, would you favor parentheses?  Update: I realize that dashes are sometimes used for similar situations, so I'd appreciate information on these as well.  For example:   The children—who are wearing shoes—will go on the trip The children, who are wearing shoes, will go on the trip The children (who are wearing shoes) will go on the trip Stack Exchange's heavy reliance on external scripts Probability of sampling with and without replacement In sampling without replacement the probability of any fixed element in the population to be included in a random sample of size $r$ is $\frac{r}{n}$. In sampling with replacement the corresponding probability is $\left[1- \left(\frac{1}{1-n}\right)^r\right]$.  Please help me show how this is proved. How do I install and manage GNOME Shell extensions? When I launch an app, the focus doesn't move to the opened app Let's say I have Gthumb open, and then launch Nautilus from the Unity launcher (in 12.04). Why does it sometimes not focus the mouse on Nautilus, which I just launched? Instead, the focus remains on Gthumb while Nautilus opens in front of Gthumb (and not being focused on).  Is this correct behaviour? We denote the s value of an ECDSA signature $(r, s)$ on a message $m$ as: $s=\frac{H(m)+xr}{k}$  Assume two ECDSA signatures sharing the same nonce $(r, s_1) , (r, s_2)$ on two messages $m_1, m_2$, that verify under two pubkeys $x_1G, x_2G$.    If the two public keys are equal then the secret keys should be equal $x_1 = x_2$ and we can easily recover the $k$ using the standard attack on nonce reuse. Once we know $k$ we can recover the secret key.  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+x_1r - x_2r}$  $x_1 = x_2 \rightarrow x_1r - x_2r = 0$  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)} = k$  My question is can this attack be made to work if the secret keys are not equal i.e. $x_1 \ne x_2$:  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+x_1r - x_2r} = \frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+ (x_1 - x_2)r}$  If you know either $x_1 - x2$ or $\frac{x_1}{x_2}$  you should be able to compute $k$ as long as $s_1 \ne s_2$.  You can calculate $x_1 - x_2 = \frac{H(m_2) - H(m_1)}{r}$ in case where $s_1 - s_2 = 0$. However this case seems to reduce to the hardness of ECDSA since anyone can compute the pubkey for a new message $m_2$ that verifies under first signature $(s, r)$ using public key recovery.  If $s_1 \ne s_2$ you can compute $\frac{x_1 - x_2}{k}$ which allows you to convert $s_1$ into $s_2$ and vice versa.