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How to change boot splash screen in 18.04 Different behaviour for list.__iadd__ and list.__add__ Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$  The experimentally verified near-Earth gravitational potential is  $$U ~=~ mgh.\tag{2}$$  The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance.  $(1)$ scales as $\frac 1r$.  So the greater the distance from the Earth, the less potential energy an object should have.  But $(2)$ scales proportionally to distance.  So the greater the distance from the Earth, the more potential energy an object should have.  How is this reconcilable? I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?  Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field. Proof. We need to show that any nonzero $r \in R$ has an inverse. Consider $\{1, r, r^2, \cdots\}$. This is an infinite set of elements of $R$, and by hypothesis $R$ is finite dimensional as an $F$-vector space, so this set is linearly dependent. Hence $\sum_{i=0}^n{c_i r^i} = 0$ for some $n$ and some $c_i \in F$ not all zero.  It then goes on to show, given the above, that we can derive an inverse for $r$. However, if I consider examples like $r = 2 \in Q[\sqrt{2}]$, $r = \sqrt{2} \in Q[\sqrt{2}]$ or $r = 2 \in Q[X]/{&lt;X^2&gt;}$, the set $\{1, r, r^2, ...\}$ doesn't look linearly dependent to me. I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something? [Edit] Well yes, I am. Somehow I had managed to discount the possibility of any $c_i$ being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding. Is there any work-around if you are somewhere where images are blocked? I'm currently working in a bank, and while about 99.7% of the internet is blocked, Stack Exchange thankfully isn't (yet) however imgur is.  It means I can't view images posted alongside questions. The only workaround I have found is to search Google images for the URL, but you get a  which is a bit useless.  Does anybody know any clever tricks for viewing the images accompanying questions when you are somewhere where imgur is blocked?  Are there any plans to route the images stored on imgur through a Stack Exchange URL? Under what circumstances will a passive HDMI-to-VGA adapter work? I have bought a no-brand, passive HDMI Male to VGA Female adapter, to link up a PC with a more dated monitor. It did not work.  After doing some reading, I have learned that buying such an adapter is often discouraged because in many cases it will not convert the signal from Digital to Analog, or the Motherboard/CPU on a specific computer might not be compatible with that particular adapter.   So my question is: seeing how there are so many of these passive adapters on the market, are there circumstances under which they actually do suffice to connect an HDMI output to a VGA input? Or are they generally just a rip-off?  Example of the kind of adapter I bought: Is it appropriate to down-vote posts containing SQL injection vulnerabilities? Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot! Beamer: PDF presentation viewer to play embedded videos and show elapsed time I have been suffering from finding a PDF presentation viewer for beamer output on Windows that can play embedded videos (by media9 package) and show the elapsed presentation time. I have tried the following: Adobe Reader with tdclock package:  Pros: plays videos and loads tdclock properly Cons: sluggish and crashes frequently  Foxit Reader with tdclock package:  Pros: smooth viewer and plays videos  Cons: can't load tdclock clock/stopwatch despite enabling javascript   Impressive:  Pros: fast and has internal clock  Cons: can't play embedded videos   Dual-Screen PDF Viewer:  Pros: very fast, has internal clock and dual screen support  Cons: can't play embedded videos   Is there any other solution that can properly play embedded videos and show elapsed presentation time flawlessly and smoothly on Windows?  P.S. when I mean tdclock javascript is not loaded, those empty red boxes show up instead of the clock/stopwatch,  and when embedded video can't be loaded, a white page shows up. How do I get a tourist visa (or a tourist card) for Cuba?  I live in the Netherlands, but I'm not a Dutch citizen, I have an Italian passport. Proving that $\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$ I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated Using a question mark mid-sentence May I use a question mark in the middle of a sentence? Examples:  Would you like the drapes to be white? or perhaps something off-white? Would you like the logo to be centered? at the bottom? left off entirely? How to compare two columns and find differences in Excel? I have two columns in Excel that I want to compare and find the differences between them.  Suppose:   Col A has 50 numbers, i.e. 0511234567 Col B has 100 numbers in the same format Who paid for Harry's Nimbus 2000? The Nimbus 2000 was the fastest broom at the time of the Harry's first year at Hogwarts.  Professor McGonagall wanted to win the Quidditch Cup so she spoke with Dumbledore and got an exception from the rule that no first-years could have a broom for Harry.  My question is who paid for the broom?  It couldn't have been a cheap broom.  They wouldn't have taken gold out of Harry's Gringotts account without permission.  It wouldn't be right for the school to pay for it and I doubt McGonagall paid for it out of her own pocket.  So who paid for the broom? I have seen people apply a filter to imported data before plotting with pgfplots, see . Is there a way to extend the approach presented there to apply multiple filters? In the MWE below I made the naive modifications to the solution given there, but it seems that only the last filter is applied: the unwanted y=1 data points are plotted too.  \documentclass{article} \usepackage{filecontents} \usepackage{pgfplots}  \begin{filecontents}{foo.txt}   x y z f   0 0 0 0   0 0 1 1   0 1 0 2   0 1 1 3   1 0 0 4   1 0 1 5   1 1 0 6   1 1 1 7   \end{filecontents}  \pgfplotsset{   discard if not/.style 2 args={   x filter/.code={   \edef\tempa{\thisrow{#1}}   \edef\tempb{#2}   \ifx\tempa\tempb   \else     \def\pgfmathresult{nan}   \fi   }}   }  \begin{document} \begin{figure}   \centering   \begin{tikzpicture}   \begin{axis}   \addplot table[x=x, y=f,                  discard if not={y}{0},                  discard if not={z}{0},                 ] {foo.txt};   \end{axis}   \end{tikzpicture}   \caption{$f(x)$ given $y=0$ and $z=0$.}   \end{figure} \end{document} I have downloaded Aptana Studio 3 and created the following launcher item for using Alacarte:  file: alacarte-made.desktop  [Desktop Entry] Comment= Terminal=false Name=Aptana Studio 3 Exec=AptanaStudio3 Type=Application Icon=/opt/Aptana-Studio-3/icon.xpm   Then, when I add the icon to the favorites and opens the Aptana Studio 3 I get two icons for it, one for launching new instances, and one for the currently running instance as per the following screenshot:    How do I force the application to use the same icon added to the favorites? A word that says a person is both female and your friend Suppose $ \alpha, \beta>0 $. Compute: $ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $ Suppose $ \alpha, \beta&gt;0 $. Compute:   $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$   Here is what I do: $$\begin{align} \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &amp;= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\ &amp;=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\ &amp; \\ &amp;   \qquad\text{let $ yx=u $}\\ &amp; \\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\ &amp;=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\ &amp;=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha} \end{align}$$  But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?  Edit: The question comes from   Who can point out my mistake in the above deduction? Is it possible to convert a Physical Ubuntu Install (Intel based PC with Ubuntu 10.10) to a virtual box disk, so that I can use that same environment on a separate pc or can keep it virtual for testing?  Something similar to "Disk2VHD" which converts a physical Windows 2K3 and above box into a VHD or VMware image.  Will be grateful for your assistance ! Kind Regards Let $a,b,c&gt; 0$ be three real numbers such that $a+b+c=1$. I want to prove that   $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$  My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side) $$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$  However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is: $$\sqrt a+\sqrt b+\sqrt c\le\frac35$$  However, as $a,b,c&lt;1$, $$\sqrt a +\sqrt b + \sqrt c&gt; a+b+c=1&gt;\frac35.$$ So my upper bound is always larger than the given one.