Is it possible to test for a Gaussian distribution?
Is normality testing 'essentially useless'?
I submitted a new version of the sitemap index for my site in Google Search Console two days ago. Soon after submission, GSC started saying that the index was "processed successfully", and yet has continued to display zero as the number of discovered URLs.  Note that my sitemap index points to about 11000 sitemaps, which provide actual links to pages.  I'm pretty sure the format of both the sitemap index and the individual sitemaps is correct.  Screenshot:    Should I worry about this? What does it mean?  Edit on 2020-02-03:  Google finally started displaying the right number in Total discovered URLs. It must have happened sometime in January. Why so late - I have no idea... A zero-dimensional ring is Noetherian? Proposition: Let $A$ be a non-zero ring that is not a field. Suppose $A$ is zero dimensional. Then it is Noetherian.   Proof: Let $p$ be a prime ideal of $A$. If $p$ is not maximal, then $p \subsetneq m$ for a maximal ideal $m$. Hence $dim A \ge 1$, contradiction. Hence every prime ideal is maximal. Consequently, every increasing (or decreasing) chain of prime ideals will be trivial in the sense that it will have length $0$. Hence the ring is Noetherian (and Artinian).  Is there a problem with the above proof? Why does the base of this slinky not fall immediately to gravity? My guess is tension in the springs is a force > mass*gravity but even then it is dumbfounding. How to do research in two very different fields? Can I use a higher rank tensor field as the metric field? What a metric should necessarily do is it should give me a way to associate a frame invariant number with a given pair of spacetime events. Now, if I use a higher rank tensor field (say, for example, a tensor field of rank 3: $g_{\mu \nu \rho}$) then also I can certainly produce a frame invariant scalar out of a given displacement $\vec{A}$ in this trivial way: $ I := g_{\mu \nu \rho} A^{\mu}A^{\nu}A^{\rho}$. Since the components of the higher rank tensor and that of the displacement vector are going to transform in a covariant and a contravariant manner respectively, the cooked up quantity is certainly a scalar.   Another crucial property, which I think has more direct physical content than the previous one, is that between two inertial frames, at least one such transformation should exist that leaves at least one metric invariant. i.e., There should exist at least one combination of transformation matrix $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}$ and metric $g_{\alpha \beta \gamma}$ that satisfies the following equation:      $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}\displaystyle\frac{\partial{x^{\beta}}}{\partial{x^{\nu '}}}\displaystyle\frac{\partial{x^{\gamma}}}{\partial{x^{\rho '}}} g_{\alpha \beta \gamma} - \delta_{\mu '}^{\alpha}\delta_{\nu '}^{\beta}\delta_{\rho '}^{\gamma} g_{\alpha \beta \gamma} =0$    The last thing I can think of that can put a restriction on the choice of a tensor as a metric is the existence of a possibility of finding a metric compatible symmetric connection field. Following the usual procedure of finding the expression for a metric compatible symmetric connection field, I reached following condition (unlike the case of the usual two rank metric where we get a full-fledged expression) for the connection in the terms of the metric:      $g_{\nu \rho k_1} \Gamma^{k_1}_{\mu \lambda} - g_{\lambda \mu k_2} \Gamma^{k_2}_{\rho \nu} = \displaystyle\frac{1}{2} (\partial_{\nu}g_{\rho \lambda \mu} + \partial_{\rho}g_{\lambda \mu \nu}  - \partial_{\lambda}g_{\mu \nu \rho}  - \partial_{\mu}g_{\nu \rho \lambda}) $   My question is that if it is possible to satisfy the two highlighted conditions then can we use such a 3 rank (or even higher rank tensors with similarly produced conditions) tensor fields as metric fields?   PS: This is NOT a proposal for a new home-production theory of gravity (or that of anything for that matter) but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric. Thank you. Simple chessboard exercise How long typically are paper reviews? Is there such a thing as “too long”? When you review papers submitted for publication, is there an “optimal” length for reviews? In my experience as an author and referee, I have seen a large range of review lengths (for reference, a paper in my field is typically between 3 and 8 printed pages):   Zero length: for some of the papers I have authored, the reviewer just clicked the “accept as is” or “reject” checkmark on the review sheet, without adding any comment (at least, not any comment visible to me). It's not so common, but it has happened. Short length: a lot of time, I received reviews who consisted of a single paragraph. Medium: one full page, maybe two pages.   Although I have never received any such long reviews, I have myself written on a few occasions reviews that exceed two pages, including once or twice a four or five-page review.  These were cases where the paper was good, but could be (in my view) much improved and some of the aspects/consequences had escaped the authors' consideration.  I am wondering how useful reviews of various length are to editors. I often consider that “unmotivated” reviews are useless, as they do not give any real insight about the paper to the editor. For example, if the editor gets two conflicting non motivated reviews, how is she to decide?  On the other hand, although I write some from time to time, I have never received long reviews, so… is this something frowned upon? Often when I edit one of my answers, it's because I want to better address the original question. The only problem is that I can't see the original question while I'm editing my answer!  I'd love to see the original question text beneath the "answer edit" form. This way I could scroll down slightly to read it if I wanted to, or just ignore it if not.  Current workaround is to click the 'return to answer' link and hope that you remember to launch it in a new tab/window, then tab back and forth.    Imported from this uservoice item: I have a simple 2-column layout with a footer that clears both the right and left div in my markup.  My problem is that I can't get the footer to stay at the bottom of the page in all browsers.  It works if the content pushes the footer down, but that's not always the case. Now that El Capitan is "rootless", is there any way to get dtrace working? I've found dtrace to be an invaluable tool for debugging and troubleshooting all sorts of issues, not to mention the two dozen dtrace toolkit scripts shipped by Apple as part of El Capitan.  On El Cap though, running dtrace usually results in an endless spew of errors making dtrace all but useless. The Help Vampire problem I have a VIP skybag (I believe this does not matter). It has a 3-number TSA lock built into the suitcase. I know the combination to unlock, let's say it's 345. Recently it was locked and while there was a different number on the combination lock the reset/set lock button was pressed.  Since then the reset/set lock button is pressed (stuck) and the lock button also does not unlock. Any pointers on how to unlock it? I am thinking of unscrewing the lock from inside of the suitcase but it seems quite a hassle, as I may have to assemble it back. Let me know if there is any other option to unlock it from the outside. Creating node shapes I am currently learning how to use TikZ/PGF to make some figures for a project report, and so I have a question:     Is there is a practical way of making a particular object into a node?   The object I have in mind is shown in the Figure below. I want the object to consist in the big circle, the two circular arcs, the label and anchors at north, south, east and west. The dotted lines are simply there to indicate that I am going to have paths going out of these points, which is why I need the anchors there.     I include the source, in the case that it is useful to anyone.  \begin{tikzpicture} \draw[help lines, gray, dotted] (-2,-2) grid (2,2);  \begin{scope}[shift={ (0,0) } ]     \begin{scope}[rotate=90]         \draw[thick] (-1,0) arc[start angle=-90, delta angle=90, %                               x radius=1, y radius=1] -- (0,1) %NW                      (0,-1) arc[start angle=180, delta angle=-90, %                                    x radius=1, y radius=1] -- (1,0); %SE         \draw[gray] (0,0) circle (1);     \end{scope} % Anchor replacements     \node(N1)       at (0,1)    {};     \node(S1)       at (0,-1)   {};     \node(E1)       at (1,0)    {};     \node(W1)       at (-1,0)   {}; % Label     \node(Crossing number) at (0,0) {1}; \end{scope}  \path[thick, gray, dotted] [ out=90,   in=270]     (N1.center) edge (0,2) [ out=180,  in=0]       (W1.center) edge (-2,0) [ out=270,  in=90]      (S1.center) edge (0,-2) [ out=360,  in=180]     (E1.center) edge (2,0); \end{tikzpicture}   PS. I have had a look at the Manual, but I could not find a description of how to do this. Additive Roman Numeral 4 I am working on a diplomatic edition of a 15th century book and it numbers its folios using the standard roman numerals of the time.  I've been using the following to get the page/folio in my header:  \newcommand{\rectoverso}[1]{\ifthenelse{\isodd{#1}}{r}{v}} \newcommand{\folio}[1]{\the\numexpr (#1+1)/2\relax} \uppercase\expandafter{\romannumeral \value{\folio{\thepage}}}\rectoverso{\thepage}   This works great, but the problem is that the roman numeral sequence is off of what I need.  Instead of I, II, III, IIII, V, VI, VII, VIII, IX, X, I get the subtractive IV (somewhat understandably, given it's the far more common one in modern usage).  My source text only (and always) uses the additive 4 in the one's position: 24 is XXIIII, 34 is XXXIIII but 44 is XLIIII.  9 is universally rendered subtractively as IX or XC.  Is there a package that allows for some control over which numbers are done additively and which are subtractively, or would my best option be to write my own custom definition to account for the specific style used in my book? Basically the &quot;updates page&quot; loads in a non rich html format, and if theres an update it's missing the tabs on the update page as well.  I can't figure out if this is just because apple may have stopped supporting El Capitan or if I have something else going on.  Any help would be appreciated.  I've trashed the caches in library and the user profile. But this page is the only one that still loads like this (see pic below) on my app store. Could you please correct my batch command?   I had 100 img images which need to be converted into 8 unsigned bit format for further use in TIMESAT.   mkdir test for %%f in (*.img) do gdal_translate -of ENVI -scale -1 1 1 255 -a_nodata 0 -co INTERLEAVE=BSQ "%%f" test/%%f.dat I recently made my pen-drive bootable with Ubuntu 14.04. But I need to unboot my pen-drive now. I formatted my pen-drive once,But the size of the pen-drive still shows 2.25 MB. Whereas,this is an 8GB pen-drive. How can I get rid of this problem. I'm still using Windows 7. I often need to select a number of rows from each group in a result set.  For example, I might want to list the 'n' highest or lowest recent order values per customer.  In more complex cases, the number of rows to list might vary per group (defined by an attribute of the grouping/parent record). This part is definitely optional/for extra credit and not intended to dissuade people from answering.  What are the main options for solving these types of problems in SQL Server 2005 and later? What are the main advantages and disadvantages of each method?  AdventureWorks examples (for clarity, optional)   List the five most recent recent transaction dates and IDs from the TransactionHistory table, for each product that starts with a letter from M to R inclusive. Same again, but with n history lines per product, where n is five times the DaysToManufacture Product attribute. Same, for the special case where exactly one history line per product is required (the single most recent entry by TransactionDate, tie-break on TransactionID. Is it ever bad to use \left and \right? Zero visitors? What happened to site analytics about a week ago?