I upvoted an answer, and remained on the page and read a few more answers, while noticing that the user that I originally upvoted made several edits. I later noticed something from one of the edits that made me want to remove my upvote, but I got the message: "You last voted on this message 6 minutes ago.  Your vote is now locked unless the answer is edited."    I noticed that no edits were indicated, so they must have taken place during the grace period.    This was not a case of me rushing or misreading, the answer originally posted was fine.  An edit was made during the grace period.  I went back 6 minutes after I upvoted, saw something that I didn't approve of appended to the original answer and wanted to change my vote, but couldn't.    To clarify, I understand that this is ;  I am NOT trying to report a bug. I am requesting that a feature be implemented such that this situation can be resolved without having to resort to making a fake-edit and a following rollback.    Potential Solution: Perhaps votes that occur during the grace period should simply not be locked-in.
Please do not close this as a duplicate unless you can remove the “status-completed” tag from the  (because it’s obviously not been completed).  Steps to reproduce:   Have someone post an answer. Downvote the answer. Have the someone edit the answer within the initial 5 minutes of the answer’s existence. Try to undo the downvote more than 10 minutes later.   You cannot undo the downvote.
Nearby objects won't move independently? I created a thick glass bottle and duplicated the interior faces to make a new object to create the liquid inside.   I went in the edit mode, selected the faces I wanted to copy and duplicated them with ShiftD. Then I pressed P to  separate the duplicated faces.   Now I try to make a duplicate of that liquid to make it flow out of the bottle, and I simply need the duplicate of one face of it. So I go to edit mode and do as I did to make the liquid from the bottle faces... but as soon as I'm back to the object mode and I want to move the duplicated faces, every thing moves with it...   I played around and tried things, and while doing so I realized that if I dupe the liquid object in object mode using ShiftD and press 0 to place it where the original is, they move together if I duplicate it and place it somewhere else so that they are separated.   Also if I pick one of the two separated objects and make it intersect with the other one, release my left mouse button to place it there, and then try to move it again, they both move together, but not at the same speed.  I really don't understand what is going on and I'm sure it's really badly explained, so what I'll do is make a screen capture of my Outliner and record the behavior of the objects and upload those to help you understand. How can I search the bash history and rerun a command? Can I search history in bash and run the result? How to create a database directory structure for MySQL? I was following this tutorial:    So far, I installed mysql-server mysql-client libmysqlclient-dev, and I was suppose to tell MySQL to create its database directory structure where it will store its information using sudo mysql_install_db.  $ sudo mysql_install_db  [WARNING] mysql_install_db is deprecated. Please consider switching to mysqld --initialize [ERROR]   The data directory needs to be specified.   I thought running mysqld --initialize would solve the issue:  $ mysqld --initialize   mysqld: Can't create directory '/var/lib/mysql/' (Errcode: 17 - File exists) [Warning] TIMESTAMP with implicit DEFAULT value is deprecated. Please use --explicit_defaults_for_timestamp server option (see documentation for more details). [ERROR] Aborting   How do I tell MySQL to create its database directory structure where it will store its information? Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free?    What I know about a finitely generated projective module $M$ over a PID $R$ is isomorphic to $R^k\oplus R/(a_1)\oplus\dots\oplus R/(a_n)$, and for the local ring case I don't know how to start. By 18 years of age, I want a brand new car that's $43,668 I'm from Arizona and I will be 18 in 4 months. I make about $1,200 a month from my job. I have $600 in my bank account and I don't plan to  spend any of it until I'm  18.     I'm looking to get a brand new Jeep Wrangler Rubicon that costs about  $43,000. The down payment is 10% (about $5,000) with an APR of  4.28%. The duration is 60 months with a monthly payment of  $800.  I don't have any credit. My friend suggested that I ask my parents to put my name on one of their credit cards so that I can build some credit. I'm not sure how that works. Can someone suggest a good way of going about this so I can get this car on or around my birthday?  My mom does finance work  for a big company and she told me that I should get a cheaper car. I really want this one and I'm willing to work for it. I'm getting a second job to get more money rolling in. My mom said that the only way that this would work is if I have a down payment of $15-20k. She also said that not everyone qualifies  for a car like this one. What can get me to qualify ASAP when I turn 18? In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?  Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.   So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. We also can conclude $k = j$ because the various $n_{i}$ chosen were strictly increasing.   Edit: $\beta$ also needs to be surjective. So for every $r \in S$ there exists a $q \in \mathbb{N}$ such that $\beta(q) = f(n_{q}) = r$. We know that $f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}$, so we can let $f^{-1}(r) = n_{d}$. Thus $\beta(d) = f(n_{d}) = r$. How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$ The definition of derivative of a function $f(x)$ is $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$  Using this definition, the derivative of $\sin\sqrt{x}$ will be:  $$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$  $$\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$  Now i got stuck. How to find the limit or simplify this expression?  I get intuition that we have to use $$\lim_{x\to0}\frac{\sin x}{x} = 1$$  but that too is leading no where. I am unable to remove h from denominator.  NOTE  I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient. Accepted answers don't float "on top" when self-answered? On , the accepted answer doesn't float on top. Apart from that, the display is correct (i.e. the V mark and the green border)  If I sort by "vote", I've got the following answer order:   Jezell Marc Gravell chilltemp (Accepted) Shaun Bowe   By "oldest":   Marc Gravell Jezell chilltemp (Accepted) Shaun Bowe   and by "newest":   Shaun Bowe chilltemp (Accepted) Jezell Marc Gravell     I'm using Firefox V.3.5 : Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.1) Gecko/20090624 Firefox/3.5 (.NET CLR 3.5.30729)  Feel free to ask for more debug information if needed. Navigation bar overlapped by status bar If $f(x)$ is continuous on $[a,b]$ and $M=\max \; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}$? Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"? The following code receives seg fault on line 2:  char *str = "string"; str[0] = 'z';  // could be also written as *str = 'z' printf("%s\n", str);   While this works perfectly well:  char str[] = "string"; str[0] = 'z'; printf("%s\n", str);   Tested with MSVC and GCC. Finding a basis for a certain vector space of periodic polynomials wp_get_nav_menu_items order doesn't work How do I get the feature id after featureAdded signal (PyQGIS)? I'm trying to make an edit form in QGIS (Python code) ,and i want after feature drawing to get the id, and the feature to be selected. Do you have any ideas to do that?  Tried so far:  self.iface.activeLayer().featureAdded.connect(self.myfonction)  def myfonction(self):      for feat in self.iface.activeLayer().getFeatures():         self.iface.activeLayer().setSelectedFeatures([feat.id()]) I've been working with ArcGis for a while now and was wondering if there was a way to change positions of parameters inside the toolbox for the user. The only time I have seen this done in arc toolbox is with the Raster clip tool with the Rectangle parameter where the 4 parameters are aligned in a circle in the toolbox.     I don't know if what I want is possible since I haven't seen anyone else do it. Construct a field of 25 elements. We need to construct a field of 25 elements.  By using a result : For a prime p and a monic irreducible polynomial $p(x)$ in $\Bbb F_p [x]$ of degree $n$ , then the ring $(\Bbb F_p [x] / &lt;p(x)&gt;)$ is a field of order $p^{n}$.  So , for the question problem above , we need a irreducible polynomial of degree $2$ in $\Bbb Z_5 [x]$.  Solution says , $x^{2} + x +1$ does our job , since it's irreducible in $\Bbb Z_5 [x]$.  My question : Can't we take $ x^{2} + 2$ , it has no zeroes in $\Bbb Z_5$ , and is irreducible. In every solution I've gone through , the polynomial $x^{2} + x+1$ is taken. Can't we consider $ x^{2} + 2$ ? Existence of irreducible polynomials over finite field Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence. Separators for author group (IF conditon) I am new to LaTeX, just a beginner, basically a Perl developer.  I have created a LaTeX file  \begin{document}  \cont{\iname{\isurname{Name1}\ifname{K. C. G.}}\iname{\isurname{Name11}\ifname{Y. Q.}}\iname{\isurname{Name111}\ifname{C.-Z.}}\ititles{Some title Text}\ipage{995}}  \end{document}   Now my need is to define LaTeX commands so that the output should appear as below     Name1 K. C. G., Name11 Y. Q. and Name111 C.-Z. Some title Text - 995.   \iname - Person Name will repeat several times \isurname - Surname of a person \ifname - First Name of a person    \iname will repeat any number of times, say 1 to ...  If only one \iname found inside \cont, display should be like the below     Name1 K. C. G. Some title Text - 995.   If only two \iname found inside \cont, display should be like the below     Name1 K. C. G. and Name11 Y. Q. Some title Text - 995.   If more than two \iname found inside \cont, display should be like the below (and should come after last \iname)     Name1 K. C. G., Name11 Y. Q. and Name111 C.-Z. Some title Text - 995.   I can define input for \iname separated by commas, but I don't know how to write a macro to define \iname with different outputs based on the if condition.  I hope to have clearly explained my doubt. Let $G$ be a matrix Lie group and let $H$ be a Lie subgroup of $G$. My understanding is that all $g \in G$ are matrices, since $G$ is a subgroup of the general linear group. Then aren’t all elements $h \in H$ matrices, since $h \in G$?  From what I can gather $H$ is a matrix Lie group if $H$ is a closed subset of $G$. But if $H$ is not a closed subset of $G$, why is it not a matrix Lie group? Is the reasoning in the first paragraph wrong?