A question on mean value inequality
Proofs of AM-GM inequality
Area of the Limiting Polygon Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.   Continue this process ad infinitum.  What is the area of the limiting "polygon"?  My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.   Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance. Are beamer, hhline, and [table]{xcolor} incompatible? The following example works as expected:  \documentclass{article} \usepackage[table]{xcolor} \usepackage{makecell, multirow}     \renewcommand\theadfont{\normalsize\bfseries\color{white}}     \renewcommand{\theadgape}{} \usepackage{hhline}  \begin{document} \centering     \setlength\arrayrulewidth{2pt}     \arrayrulecolor{yellow}     \def\clinecolor{\hhline{&gt;{\arrayrulecolor{yellow}}--%                             &gt;{\arrayrulecolor{gray}}-}}     \setlength\extrarowheight{2pt} \begin{tabular}{ c | c | c }%     \rowcolor{gray} \multicolumn{2}{c|}{\thead{AAA -- BBBB}}    &amp;                                   \\     \clinecolor     \rowcolor{gray} \color{white} AAA   &amp;   \color{white} BBB   &amp;   \multirow{-2}{*}{\thead{CCC}}   \\     \hhline{&gt;{\arrayrulecolor{yellow}}-|-|-}     \rowcolor{gray!25} aaa                 &amp;   bbb                 &amp;   ccc                             \\     \hline     \rowcolor{gray!25} ddd             &amp;   eee                 &amp;   fff     \end{tabular} \end{document}     But the following not:  \documentclass[xcolor=table]{beamer} \usepackage{makecell, multirow}     \renewcommand\theadfont{\normalsize\bfseries\color{white}}     \renewcommand{\theadgape}{} \usepackage{hhline}  \begin{document} \begin{frame} \frametitle{test} \centering     \setlength\arrayrulewidth{2pt}     \arrayrulecolor{yellow}     \def\clinecolor{\hhline{&gt;{\arrayrulecolor{yellow}}--%                             &gt;{\arrayrulecolor{gray}}-}}     \setlength\extrarowheight{2pt} \begin{tabular}{ c | c | c }%     \rowcolor{gray} \multicolumn{2}{c|}{\thead{AAA -- BBBB}}    &amp;                                   \\     \clinecolor     \rowcolor{gray} \color{white} AAA   &amp;   \color{white} BBB   &amp;   \multirow{-2}{*}{\thead{CCC}}   \\     \hhline{&gt;{\arrayrulecolor{yellow}}-|-|-}     \rowcolor{gray!25} aaa                 &amp;   bbb                 &amp;   ccc                             \\     \hline     \rowcolor{gray!25} ddd             &amp;   eee                 &amp;   fff                             \\     \hhline{|-|-|-|}     \end{tabular} \end{frame} \end{document}     Why are the in beamer rules so strangely misplaced? Is there some workaround? Someone from Mars is called a Martian. Someone from America is called an American. Someone from Europe is a European. Someone from Hyrule is a Hyrulean.  We are not far from colonizing other planets, let alone Earth's only moon. Obviously you wouldn't call someone from the moon a lunatic right? What if you are on Mars? What would I call a foreigner from Earth? Earthean doesn't sound right. This question follow's JeffE's comment :     In most industries, a PhD has negative value.   I am curious about what potential negatives there may be with having a PhD, particularly in industries.  For example, my PhD has applications for image sensors, and the potential of working in that field is appealing—would having a PhD be a potential barrier? Would it be a case of how explicit the practicalities of the PhD can be made?  Would this make any difference? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root? How to make two people play at the same time? My brother owns Skyrim on steam. We use Steam Family Share, but we were wondering how both of us could play it at the same time? What should be the intuition when working with compactness? I have a question that may be regarded by many as duplicate since there's a  one at MathOverflow. The point is that I think I'm not really getting the idea on compactness. I mean, in $\mathbb{R}^n$ the compact sets are those that are closed and bounded, however the guy who answered this question and had his answer accepted says that compactness is some analogue of finiteness. That's the first problem: In my intuitive view of finiteness, only boundedness would suffice to say that a certain subset of $\mathbb{R}^n$ is in some sense &quot;finite&quot;. On the other hand there's the other definition of compactness (in terms of covers) which is the one I really need to work with and I cannot see how that definition implies this intuition on finiteness. Also, I feel it's pretty strange the covers people use when they want to deal with compact sets. To prove a set is compact I know they must show that for every open cover there's a finite subcover; the problem is that I can't see intuitively how one could show this for every cover. Also when trying to disprove compactness the books I've read start presenting strange covers that I would have never thought about. I think my real problem is that I didn't yet get the intuition on compactness. So, what intuition should we have about compact sets in general and how should we really put this definition to use? Can someone provide some reference that shows how to understand the process of proving (and disproving) compactness? Is there an AES identity key? The following questions are of pure theoretical nature. I don't have an application in mind.   Is there a key that makes AES the identity function? Is there a key that makes AES the identity function for certain inputs? Is it known whether such keys might exist? Are such keys known for other ciphers?   For example, I want to encrypt "Hello StackExchg" with AES with said key and the ciphertext should be again "Hello StackExchg".  When I encrypt "Hello StackExchg" with the perfect cipher one-time-pad, there is a possibility that the resulting ciphertext is again "Hello StackExchg". This possibility must exist (though it is unrealistically small), otherwise, the ciphertext would not be perfectly random but depending on the plaintext. The corresponding key for one-time-pad would be all zeros. Merging columns from two separate files Accidently deleted /usr/lib, how do I restore it? Can anyone explain precisely what IOWait is? Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction Why does this happen with Longclaw? Spoilers ahead for "Beyond the Wall" (Game of Thrones, S07E06)  Toward the end of the episode, "Beyond the Wall" (Game of Thrones, S07E06), we see a close-up shot of Longclaw        Then, once Jon Snow emerges from the water, Longclaw appears to be opening its eyes        You can get a much better picture of it .  Is this an Easter Egg from the Visual Effects department, or is there an in-universe explanation behind this?  Or was this not planned at all by the production team, and this is merely a coincidental, accidental lighting/water effect, that just happens to be timed with the dramatic tension of Jon Snow possibly being dead, and the relief of him being alive? Existence of $f'\in F$ such that $f^{[G:F]}=(f')^n$ for finite extension $F\subset G$ Let $F$ be a field and $u\in F$. Further, let $F\subset G$ be a finite extension and $v\in G$ such that $v^n=u$ for $n&gt;0$.     How would I show that there exists a $\tilde{u}\in F$ such that $u^{[G:F]}=(\tilde{u})^n$?   My first thought was to use the norm map $N_{G/F}(x)=\Pi_{\sigma\in X(G/F)}\sigma(x)$, where $X(G/F)$ is the fundamental set for $F\subset G$. However, I wouldn't know how to continue on this path. I tried Libreoffice, but its Import option while pasting in the cell is very  annoying.I also tried Gnumeric but I am not able to paste using middle button.Parcellite to make copy paste faster and middle button pasting is not working with Gnumeric. Is there any simple table maker app available that can accept only simple unformatted text? All these software are very complicated for this simple work. Possible Duplicate:          Is it correct to start a sentence with did.     Did my work.   instead of      I did my work.   while answering my emails? How can a regression be significant yet all predictors be non-significant? My multiple regression analysis model has a statistically significant F value however all beta values are statistically non-significant.   All the regression assumptions are met. No multicollinearity was found. Correlations among all predictors are all less than 0.60.   What else might be the cause of the insignificant predictors? How to delete a file in Windows with a too long filename? My wife has several files and folders that somehow ended up having filenames that have caused them to be undeleteable (can't be deleted) by normal means or via the command line. I believe the filenames are too long due to the depth of the folder structures. Does anyone know of a good utility for cleaning up files like this? A doubt from Milnor's "Topology from a Differentiable Viewpoint". This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise. Changing column of multiple csv files I have a csv file called scenario1.csv in which the second column has column names like "0-4 years high risk", "65+ years first responder" etc. There are 20 such values. The 21st row, 2nd column has the same entry as in the 1st row of the 2nd column. I would like to rename these values to p1-p20 respectively. So the 21st row would have the p1 tag with it. Everything without quotes. I have a 150 such files, named scenario1.csv, scenario2.csv ... How do I do this? Below is an example for a shorter file:  t,  group,               1,  3,  5 0,  0-4 years low risk,  0,  0,  0 0,  0-4 years high risk, 0,  0,  1 .....,  ....  0,  0-4 years low risk, 0,  0,   0   Expected output for each file:     t,  group,  1,  3,  5    0,  p1,     0,  0,  0    0,  p2,     0,  0,  0     .....,  ....     0,  p1,     0,  0,  0    Here's the dictionary I need:  0-4 years first responder   p1 0-4 years high risk         p2 ....... 65+ years low risk          p19 65+ years pregnant women    p20