This cannot be right, can it?  It seems that the tracking is not synchronized with the actual badge awarding mechanism.  Perhaps there is a bug in the awarding and I should have this badge already, or I am ineligible for it for a reason that the progress tracker is unaware of.
Poking around the people's profiles to see how the  I keep seeing things like this:  It looks strange to have a 25/5 progress on a badge, but the dialog explains that what this user is missing is a positive record of asking. The  is: (total questions - negative questions - closed - deleted)/total questions &gt;= 0.5  If you look at , it's hard to see how his asking record could be anything but exceptional. According to  his asking record is 0.875. The problem is that he asked many of his questions in the early days of the site that were well-received, but later deleted. Some were deleted after the question was . My suggestion is that we change the asking record formula: (total questions - negative questions - closed - early deleted)/total questions &gt;= 0.5  The early deleted variable counts only questions deleted within 60 days of being asked. I picked 60 days because it's the dividing line between being able to . After looking through the list of people who would get the Curious badge on Stack Overflow with this change, I noticed a few things:  Many of them have self-deleted questions on their record. Often these are questions that were asked and basically ignored. After a few months, the original asker went back and deleted their question. It's hard to know what happened in each individual case, but my guess is that they never found a solution to their specific question and worked around the issue instead. (I have a few of those, but I'm not courteous enough to delete them.)  These users don't tend to ask the same question over and over again. I ran a similar test that excluded self-deleted questions. That group included a lot of cases of people asking a question, not getting an answer after a day or two, deleting the question, and asking the same question again. This strategy doesn't work if you wait two months to delete your first question.  Two mods (George Stocker and Gordon), Tim Post, and myself are on the list of users who would get a Curious badge under this system. On Meta Stack Exchange 5 current employees would be newly eligible for the badge. In my (admittedly biased) opinion, these are curious users.  I did some spot check on the lower end of the reputation scale. Every user I looked at fit my model of the ideal recipient of the Curious badge; they asked mostly upvoted questions with few closed questions. This change would add about 3286 new Curious badges on Stack Overflow, so there will certainly be a few exceptions I didn't find. But I think they are more than balanced by the clearly deserving users who will be awarded new badges.    Proposal: For the Curious, Inquisitive, and Socratic badges, don't count questions deleted 60 or more days after asking as deleted for the purposes of calculating asking record. These questions may still count as negatively scored or closed, if applicable. Days that include deleted questions, no matter when they are deleted, do not count toward the days asked criteria.   Modified Proposal: For the Curious, Inquisitive, and Socratic badges, don't count questions deleted or closed 60 or more days after asking as deleted or closed for the purposes of calculating asking record. These questions may still count as negatively scored, if applicable. Days that include deleted or closed questions, no matter when they are deleted or closed, do not count toward the days asked criteria. On Stack Overflow, that would add about 665 Curious badges, including two more for moderators (casperOne and 0x7fffffff) and one for an employee (Oded). On the opposite end of the reputation scale my spot checks are a bit muddier. Instead of a small number of deleted (and usually closed) questions, this list turns up users whose latest questions still aren't all that interesting. Even so, on balance a few extraneous bronze badges seem worthwhile in exchange for acknowledging truly curious users.
Extruding surface while keeping the shape here is the picture of what I would like to get(example nr. 2). I need the edge frame to be same in width and length. Also the extrusion is on surface (without depth) I tried extruding region, but it did not give me the result I wanted.   I know I could just adjust the scale and do it manually, but that would be eyeballing without achievign accuracy. I would like to know if it is possible to extrude or inset while keeping the shape (even edge space) So we learned telescoping series in class and I came across this question in my textbook and I tried to evaluate it, but I don't understand how to do it.   $$ \sum\limits_{k=1}^n\left(k \cdot k!\right) $$  According to the answers:  $$ \sum\limits_{k=1}^n\left(k \cdot k!\right) = (n+1)! - 1 $$  We learned how to simplify certain telescoping series using partial fractions but I don't have any idea on how to start this question. How can I get another application's installation path programmatically? I am a beginning indie developer, and I want to get the word out about my game. What sites can I go to to ask about reviews of my game, and where might I be able to talk about and get feedback on my game from players? Does it matter what I choose for serialVersionUID when extending Serializable classes in Java? I am going to put an additional (non-USB) hard drive in my system. I already have Ubuntu on my other hard drive so I do not want to install Ubuntu on the additional drive but only use it for storage. How do I add the additional hard drive to my Ubuntu system, e.g. make Ubuntu recognize it and mount it properly? Big tilde in math mode I am wondering what the difference between a definition and an axiom.  Isn't an axiom something what we define to be true?  For example, one of the axioms of Peano Arithmetic states that $\forall n:0\neq S(n)$, or in English, that zero isn't the successor of any natural number. Why can't we define 0 to be the natural number such that it isn't the successor of any natural number?  In general, why is an axiom not a definition?   Are there cases where we can formulate definitions as axioms or vice versa? I want to have a vector with a superscript:  \vec{o}^t   But this gives very bad results: the t is overlapping with the vector arrow. How do I fix this (and why does this give bad results in the first place) ? Featured On Meta section doesn't list MSE posts on meta sites I asked a question that has been marked as a duplicate. I read a lot of similar questions and explanations in advance but I couldn't find a solution. In the question that should answer my question the asker used a different (not working) approach. There is even an answer that would work for my case however I'd rather consider it as a workaround than a solution since I have seen examples using my approach.  So I would like your opinion in that case. Would it be ok to repost my question while explaining why other answers didn't help me? I edited my question but I think it's not getting any attention anymore. Naturally I will remove the old question. There has been a significant drop (90%) in traffic from Google to one of my well-established sites (6+ years) in the last week. Searches show that the top 3 wide keywords all dropped 4 spots. Searches for other keywords do not show ANY results in the first 10 to 25 pages of results, while previously one 1st or 2nd page at most.  Since there are , the question really is:  What steps are necessary to figure out what caused such a drop?  There have been no major changes during this period or the last month on the site and certainly not in the homepage which has dropped in rank. Over all the years of running this site and plenty of others, I have never experienced this. There are is duplicate content on my site and I have rigorously used canonical links for the last few years to ensure it is not misinterpreted as such. How to remove some pages from the navigation bullets in Beamer? I'm making a beamer presentation using the miniframes outer theme, with navigation bullets, and the subsection=false option, meaning the bullets are all on a single row. The problem I have is that, if one section has too many pages, the bullets overflow to the next section:  Section 1 Section 2 oooooooooooooooooooo   I tend to have many frames, some of them just asking a question, leading to the next detailed frame. I don't really need those frames to be represented by navigation bullets, and removing them would fix the problem. Any idea how to do that?  I could cheat by using overlays to have the question and the answer on the same frame, but that would be a mess...  Code example to answer percusse's question: this should give you one circle per frame, not one per subsection. If you add many frames in one subsection, the circles will overflow over the next section.  \documentclass[compress]{beamer} \useoutertheme[subsection=false]{miniframes} \begin{document} \section{S1} \subsection{SS1} \frame{1} \frame{2} \frame{3} \subsection{SS2} \frame{1} \frame{2} \frame{3} ... \end{document} I had posted a  last week about customizing a ToolValidator class and got some very good answers.  In working with the proposed solutions, I have created a custom module that performs queries on a db, and will be called by both the ToolValidator class (to provide values for the drop-down lists) and later in the geoprocessing script (to get other parameters based on items selected in drop-down lists).  However, I can't seem to actually call the custom module in the ToolValidator class.  I have been trying to append to path with no luck.  When I try to apply these changes to the script, I get a runtime error:  [Errno 9] Bad file descriptor.  If I comment out import line, no errors.  sys.path.append('my_custom_module_directory') import my_custom_module   Many of you may be asking why don't I just implement a custom tool with ArcObjects.  The reason is that my end users don't have the privleges necessary to register ANY dlls on their computer.  UPDATE:  This was happening to me in ArcGIS 10.  Interestingly enough, I was initially appending to the path inside of the initialiazeParameters function of the ToolValidator class.  If I do the append outside (i.e., on top of) of the ToolValidator class, everything works as expected.  sys.path.append('C:/Working/SomeFolder') import somescript --------&gt;THIS WORKS  class ToolValidator:   """Class for validating a tool's parameter values and controlling   the behavior of the tool's dialog."""    def __init__(self):     """Setup arcpy and the list of tool parameters."""     import arcpy     sys.path.append('C:/Working/SomeFolder')     import somescript -------&gt; THIS DOESNT WORK     self.params = arcpy.GetParameterInfo()   UPDATE 2:  I think I found the true cause of my problem.  In the code snippets in this post, I have been appending what appear to be real paths (ex C:/Working/SomeFolder) to the sys.path.  In my actual ToolValidator class, I was constructing a relative path using os.path.dirname(__file__) +"\my_special_folder...".  I was anticipating that os.path.dirname(__file__) would return the path of the toolbox, since that contains the ToolValidator class.  I have come to find this is not the case.  As far as I can tell, the ToolValidator class is never actually written to a .py file, and I speculate the the this code is passed to the python interpreter in memory, so __file__ is useless, or some temp script is persisted and then execfile(path_to_script) is called, again rendering __file__ useless.  I'm sure there are other reasons I am missing as well.    Long story short, if I use a hardcoded path, the sys.append works anywhere, relative paths don't work so well in the ToolValidator class. Let, $D=\{z\in \mathbb C:|z|&lt;1\}$. Which are correct?   there exists a holomorphic function $f:D \to D$ with $f(0)=0$ &amp; $f'(0)=2$. there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=\dfrac{3}{4}$ &amp; $f'\left(\dfrac{2}{3}\right)=\dfrac{3}{4}$. there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$ &amp; $f'\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$ there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}$ &amp; $f'\left(\dfrac{1}{4}\right)=1$.   With the help of Schwarz lemma &amp; its applications, we find that $(1)$ is false &amp; $(3)$ is true.  But, I can not think about options $(2)$ &amp; $(4)$. Does $\det(A + B) = \det(A) + \det(B)$ hold? Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?  If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said? Complex number isomorphic to certain $2\times 2$ matrices? I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.  $$\phi:a+bi \rightarrow \begin{pmatrix} a &amp; -b \\ b &amp; a \end{pmatrix}$$  This is what I have so far:  Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.  $$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$  So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.  Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$.   Surjective:  I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.  I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated! Isomorphism of $S^1$. GPG Error while downloading software: "Clearsigned file isn't valid, got 'NODATA'" Control 220V AC line using MCU