I went to see this question:     I was surprised to see that the UI is mixed up. Generally the closed should come first and the deleted would be in the next. In this case, it is the opposite. While it is not even deleted by the OP. It has been closed first.  How is this possible?
On  the delete notice is above, instead of below the close notice. Why is this happening? I understand this was made on purpose, but would like to know the reasoning.
Why are my Cycles settings missing? so I'm working on a little bathroom scene and wanted to make a clay render. I started looking for the material override which I kinda knew where it was and on the way I switched my render engine accidently to workbench instead of cycles. Switched to cycles all my seetings on the render tab are missing except color management and free style. Hows that? I restarted blender multiple times and tried some other stuff but nothing changed a thing, and I'd rather avoid a new installation Is 'hair' singular or plural? When one says I was washing my hair, is it singular or plural? What is the singular for hair? How can I install Steam? Doubts on the axiom of union? I'm reading Guerino Mazzolla's Comprehensive Mathematics for Computer Scientists 1:     Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set      $$\{x| \text{ there exists an element } b \in a \text{ such that } x\in b \}$$      This set is denoted by $\cup a$ and is called the union of $a$.    This axiom seems to take elements from inside of a set of sets, for example in the set: $a=\{\{u_1,u_2 \},\{u_3,u_4 \} \}$, there are two elements $b\in a$, then the mentioned set created by the axiom is $\{u_1,u_2,u_3,u_4 \}$.           The axiom seems to imply that the set $a$ must have a set inside of it for it to work, what if it  have an element that is not a set? (I'm not sure if this is possible, but I remember of having read something about the difference of sets and elements).         There's also an extension of my proposal, In the same book (and also in others) I've seen that the natural numbers are defined as:   $$\begin{eqnarray*}   {0}&amp;=&amp;{\emptyset} \\    {1}&amp;=&amp;{\{\emptyset\}} \\    {2}&amp;=&amp;{\{ \{\emptyset\}\}} \\    {3}&amp;=&amp;{\{ \{ \{\emptyset\}\}\}} \\   {\vdots}&amp;=&amp;{\vdots} \end{eqnarray*}$$  Then the singleton $\{ 3\}$ could be seen as $\{ \{ \{ \{\emptyset\}\}\}\}$, which could mean that   $$\cup \{ 3 \}=\cup\{ \{ \{ \{\emptyset\}\}\}\}= \{ \{ \{\emptyset\}\}\}=\{2 \}$$          Then the union could also be seen as the antecessor of a number?            Using the framework I provided, I see some ambiguity, look:   $$\{ \{ \{\emptyset\}\}\}=\{2 \}$$          I decided to interpretate it this way, but from the notation it's not clear in which pair of brackets I should stop, I guess that $\{ \{ \{\emptyset\}\}\}=\{2 \}$ but it could also be the case that $\{ \{ \{\emptyset\}\}\}=3$. I could want a singleton with a $2$ inside of it and a $3$ outside a set, no? I have been trying to create a function that does basically the same thing that the QGIS "dissolve" function. I thought it would be super easy but well apparently not. So from what I have gathered around, the use of fiona with shapely should be the best option here. I just began to mess around with vector files so this world is pretty new to me and python as well.  For these example, I am working with a county shapefile founded here  So here are some piece of code I gathered but can't find a way to make them work together  For now my best method is the following based on : . It works fine and I get a list of the 52 states as Shapely geometry. Please feel free to comment if there is a more straight forward way to do this part.  from osgeo import ogr from shapely.wkb import loads from numpy import asarray from shapely.ops import cascaded_union  ds = ogr.Open('counties.shp') layer = ds.GetLayer(0)  #create a list of unique states identifier to be able #to loop through them later STATEFP_list = [] for i in range(0 , layer.GetFeatureCount()) :     feature = layer.GetFeature(i)     statefp = feature.GetField('STATEFP')     STATEFP_list.append(statefp)  STATEFP_list = set(STATEFP_list)  #Create a list of merged polygons = states  #to be written to file polygons = []  #do the actual dissolving based on STATEFP #and append polygons for i in STATEFP_list :      county_to_merge = []     layer.SetAttributeFilter("STATEFP = '%s'" %i )      #I am not too sure why "while 1" but it works      while 1:         f = layer.GetNextFeature()         if f is None: break         g = f.geometry()         county_to_merge.append(loads(g.ExportToWkb()))      u = cascaded_union(county_to_merge)     polygons.append(u)  #And now I am totally stuck, I have no idea how to write  #this list of shapely geometry into a shapefile using the #same properties that my source.   So the writing is really not straight forward from what I have seen, I really just want the same shapefile only with the country dissolve into states, I don't even need much of the attribute table but I am curious to see how you can pass it on from the source to the new created shapefile.  I found many pieces of code for writing with fiona but I am never able to make it work with my data. Example from  :  from shapely.geometry import mapping, Polygon import fiona  # Here's an example Shapely geometry poly = Polygon([(0, 0), (0, 1), (1, 1), (0, 0)])  # Define a polygon feature geometry with one attribute schema = {     'geometry': 'Polygon',     'properties': {'id': 'int'}, }  # Write a new Shapefile with fiona.open('my_shp2.shp', 'w', 'ESRI Shapefile', schema) as c:     ## If there are multiple geometries, put the "for" loop here     c.write({         'geometry': mapping(poly),         'properties': {'id': 123},     })   The problem here is how to do the same with a list of geometry and how to recreate the same properties than the source. How to elegantly ignore some return values of a MATLAB function I am a Saint Lucian and will be traveling to Manchester, UK soon about which I am a bit nervous.  Run Down:   I own a small business in my country (in the Caribbean), I have a certificate mentioning the details of the business from my Government. I pay rent in my country, but I do not have any loans (I hate owing). I pay life insurance in my country. I have a US B-1/B-2 visa (Visitor &amp; Business). I have visited the US once (last year November) on my US B-1/B-2 visa.   What I will be doing in the UK:   I plan to spend three months out of the six months which would be granted if I get leave to enter stamp. I plan to skydive in the first week and spend a few nights at an Airbnb. I plan to visit the Manchester Football Club stadium, I'm not the biggest of football fans but since I'm in Manchester, why not! I plan to do some more sightseeing and enjoy my time there. After the first week I will be staying with my female friend who owns an apartment for the rest of the time showing me around and helping me experience the UK lifestyle until June.   Why I'm leaving in June:   I am a WordPress enthusiast (I'm a Programmer) and WordCamp EU will be happening in France this year so I plan to fly from the UK to Paris where I will spend three days then take another flight from there to my country.   How my flights will be:   I will be taking a one-way Thomas Cook flight from my country to Manchester but will be purchasing my ticket for Manchester to France ahead of time so I could show Thomas Cook as well as Immigration that I don't plan to stay.   My concerns, could someone address them point by point please?:   I have a bank account but I just use it for collecting easy payments from my clients and pay for expenses. I store my savings in a credit union because of higher interest rates paid than the bank. When I applied for my US visa I showed a credit union statement showing my total balance. I'm seeing online that I should provide statements but these statements would look very bad since I my bank account is just for expenses... I mean it sometimes gets as low as $30 then, when my monthly earnings come in from clients, I withdraw savings to the credit union and leave what I need in the account for expenses. I read another post and it said people with bank statements showing poor spending might have difficulty but it's simply because the account is used for expenses not savings. For the later stages of my visit (after the first week) I will be staying with my friend. I will simply be staying with her to save on cost of Airbnbs for the next two months three weeks, even though they could be cheap, free is better right? I will be booking the Airbnb, skydiving and tours prior to traveling to the UK, will that help? Is my stay too long for staying with someone? I won't be depending on her for anything since I have my own funds...it's just convenient that she has a place where I could stay. If it is then I don't have a problem shortening my stay to 1-2 weeks and just head back to my country and take a separate flight to Paris in June.   I am nervous because I really want to visit the UK and see my friend, I am also not sure if the minimum amount of time that could be given to a visitor is six months? What if I get a round trip for two weeks and the Immigration officer gives me exactly two weeks stay, is that possible? Or is the minimum they could give six months?  Your help is very much appreciated, I want soon to try backpacking (nomad life) but I don't think my poor heart could take the stress of worrying about immigration :(  Update  Since people in the comments are asking about my relationship with the friend I plan to stay with even though she's from my country originally but now a UK citizens and we are simply friends. This is on Stack Overflow. On Meta I'm able to see what the tags are on the right side of the screen. In fact, I think yesterday or some point earlier today I saw tags on the right side of the screen (I can't screenshot or remember correctly).  It's right above the asked and active heading.  I know this is minor but what happened to it? It might be useful for browsing instead of having to look at the bottom of the question post.    Found an image on Google and it shows the tags on the right side of the screen: Finding $\lim_{n\to\infty}\binom{2n}{n}$ I found a limit problem which is $\lim_{n\to\infty}{2n \choose n}$ as we know that $\sum_{i=0}^{n}{n \choose i}^2={2n \choose n}$ now we could transform the given problem as $\lim_{n\to\infty} \sum_{i=0}^{n}{n \choose i}^2$ and as $\sum_{i=0}^{n}{n \choose i}^2$ is a polynomial in $n$, therefore, $\lim_{n\to\infty}\sum_{i=0}^{n}{n \choose i}^2=\infty$ Please tell me if my solution is correct or not,and if it is incorrect then please provide me a solution.  Thanks What causes the 'Connection Refused' message? Search/browse my own questions easily What is the difference between "retired out" and "retired not out" in cricket? Fast calculation for $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$ I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?  Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it... If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?  Here's what I've done:  $1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue, Also $(p-1)^3=(p-1) \pmod p$ implies  $$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p|(p-1)^3 -(p-1)$$ Thus  $p-1$ is a cubic residue $\mod p$. Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$. Is there a gender-neutral word for "aunts and uncles"/"parents' siblings"? How to determine function name from inside a function Terence Tao, Analysis I, Ex. 5.5.2: Entry point needed Terence Tao, Analysis I, 3e, Exercise 5.5.2:     Let $E$ be a non-empty subset of $\mathbb{R}$, let $n \ge 1$ be an   integer, and let $L &lt; K$ be integers. Suppose that $K/n$ is an upper   bound for $E$, but that $L/n$ is not an upper bound for $E$. Without   using Theorem 5.5.9, show that there exists an integer $L &lt; m \le K$   such that $m/n$ is an upper bound for $E$, but that $(m-1)/n$ is not   an upper bound for $E$. (Hint: prove by contradiction, and use   induction. It may also help to draw a picture of the situation.)    does not do the trick for me, since I'm looking for the proof by contradiction, equipped only with the set of propositions and theorems covered so far.  What I'm actually looking for is an entry point to the proof, since I find it quite hard to negate the assumption.  My attempt is to induct over $n$. Assume I have been able to prove the statement for $n = 1$, is the following a valid negation of the assumption?  For all integers in the range $(L, K]$, at least one of the following statements holds:   $(m-1)/(n+1)$ is not an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is an upper bound.   If I were able to prove that there is an integer in the given range such that none of the above holds, am I done?       Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e.,   $E$ has some upper bound $M$), then it must have exactly one least   upper bound.     I went through another loop considering  of Steve Kass, but I ended up wondering why the proof needs induction at all:  Assume for all integers $m$ where $L &lt; m \le K$ we have that $m/n$ and $(m-1)/n$ are upper bounds of $E$. Then $(m-1)/n = L/n$ is also an upper bound. But $L/n$ is no upper bound, by assumption. Assume $m/n$ and $(m-1)/n$ are no upper bounds. Then $m/n = K/n$ is also no upper bound. But $K/n$ is an upper bound, by assumption.   Hence, there is an integer in the range $(L, K]$ s.t. $m/n$ is an upper bound for $E$, and $(m-1)/n$ is not. I installed the latest Ubuntu today, 15.04. Spotify has stopped working. How can it be made to work again?  The Unity bar flashes green 4 or 5 times and then stops flashing. There are no error messages. It was installed successfully and played well on 14.10 till the upgrade a few hours ago. It just does not appear on the screen. My question is simple, why do we believe  to be absurd? Is the Ultra-violet catastrophe incorrect or is it only because we can not create or know of a perfect emitter? I am a bit confused as to why Quanta was necessary. I want to use some constants, height and width, when creating an ellipse but it won't work with \newcommand. Here is a minimal example:  \documentclass[12pt, a4paper]{article}  \usepackage{amsthm, tikz} \usetikzlibrary{arrows, shapes, trees, positioning} \begin{document}  \begin{figure} \centering \begin{tikzpicture}       \newcommand*{\width}{2}%     \newcommand*{\height}{0.5*\width}%      \draw[fill=gray!40] (0, 0) ellipse (\width and \height); \node (Y) at (0,0) {\(Y\)};      \end{tikzpicture} \end{figure}  \end{document}    I get the following error:  error: line 14:  Package PGF Math Error: Unknown operator `a' or `an' (in '2and 0.5*2'). ...ay!40] (0,0) ellipse (\width and \height)