Apt-get update encounters errors with bad GPG keys
What is the easiest way to resolve apt-get BADSIG GPG errors?
How do I run a 'sudo' command inside a script? How to use find command with wildcard when current directory contains a match? Possible Duplicate:          I'm having some confusion with the find command.  If I type:  find . -name *.xml   or   find ./ -name *.xml   I receive a list of all the xml files in subdirectories of the current directory, UNLESS the current directory contains an xml file, in which case I get:  find: paths must precede expression: foo.xml Usage: find [-H] [-L] [-P] [-Olevel] [-D help|tree|search|stat|rates|opt|exec] [path...] [expression]   What is the correct way to search for files with wildcard when the current directory may contain a match? I know java and would normally put in getter/setter methods. I am interested in doing it in C# with the following code, but it throws a StackOverflow exception. What am I doing wrong?  Calling Code  c.firstName = "a";   Property Code    public String firstName; {     get     {         return firstName;     }     set     {         firstName = value;     } } How to adjust the gap between figure and caption within the same subfloat My question is:  How to adjust the gap between figure and caption within the same subfloat  Not the gap between two subfloat.  For the following example:  \begin{figure}     \centering     \subfloat[A]{\includegraphics[ scale=0.5]{images/a.jpg}}     \vspace{-10pt}     \subfloat[B]{\includegraphics[scale=0.5]{images/b.jpg}}     \vspace{-10pt}     \caption{}\label{fig:online_test} \end{figure}   I got the following figure:    The sub-caption of (a) is too close to the sub-figure B. I hope the caption (a) to stay closer to sub-figure A. How can I adjust the space between the caption and figure within a subfloat?  Thank you all for helping me and I would prefer a solution for subfloat, instead of using other packages. But I would also appreciate a solution of using other packages.   I know my question looks similar to . But please make it clear. I'm asking for how to adjust the gap between figure and caption within a subfloat, not the gap between two subfloat. That is, the gap between the caption "a" and the figure "A"  Thank you. The title is probably very confusing, but here's the situation:  I'm animating a flat grid curve by shrink-wrapping it onto a moving plane in order to get a uniform wave animation, it looks like this:    Before I render it, i want a bevel. What I did then was simply adding a bevel to the curve, making the curve look like this:    Now, when activating the shrinkwrap modifier, I expexted that only the curve would be shrinkwrapped, but alas, the bevel did the same:    As seen by the blue spots, the curve has been completely flattened along with the bevel.  How can I avoid this? I tried using a wave modifier instead of shrinkwrap, but this modifier doesn't seem to make perfectly round waves like this:    So, any thoughs on how to solve this?  Thanks for reading! Dates and times: "on", "in", "at"? I’m often confused when I speak about times and dates. What is the rule for using on, in, and at in the following sentences?   I will do it ___ Tuesday. We married ___ March. He returned ___ the same day. Every day ___ the same time, I walk the dog. *.xhtml url-pattern causes java.lang.NullPointerException at com.sun.faces.context.flash.ELFlash.loggingGetPhaseMapForReading If $[G' : G'']\leq p^2$, then $G'$ is abelian. Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.  It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance. I have recently installed a custom ROM and cannot find the Play Store. It seems like all apps that belong to Google are missing.  How can I install the Google Apps Package (Play Store, ...) on my Android device? Get number of unread sms Is there some guide I can use for cooking soft boiled ostrich eggs? Proving $\sum_{k=0}^n(-1)^k\binom nk=0$ cannot start Tomcat 7 server - java.net.BindException: Address already in use Prove: If a sequence converges, then every subsequence converges to the same limit. I need some help understanding this proof:  Prove: If a sequence converges, then every subsequence converges to the same limit.  Proof:  Let $s_{n_k}$ denote a subsequence of $s_n$. Note that $n_k \geq k$ for all $k$. This easy to prove by induction: in fact, $n_1 \geq 1$ and $n_k \geq k$ implies $n_{k+1} &gt; n_k \geq k$ and hence $n_{k+1} \geq k+1$.  Let $\lim s_n = s$ and let $\epsilon &gt; 0$. There exists $N$ so that $n&gt;N$ implies $|s_n - s| &lt; \epsilon$. Now $k &gt; N \implies n_k &gt; N \implies |s_{n_k} - s| &lt; \epsilon$.   Therefore: $\lim_{k \to \infty} s_{n_k} = s$.   What is the intuition that each subsequence will converge to the same limit I do not understand the induction that claims $n_k \geq k$ $\displaystyle\int_0^0\frac 1x\:dx$ This integral is defined ? If it define, what is the value ?  Please,step by step So I was thinking... If heat I feel is just lots of particles going wild and transferring their energy to other bodies, why am I not burned by the wind?  When I thought about it more I figured out that wind usually carries some humidity, and since particles of liquid are moving same speed as the wind, they are basically static relative to each other, so no energy is transferred between them (wind and water particles). And if that water sticks to my skin and wind blows, it'll evaporate thus taking energy from my skin and make me feel cold.  Thing is, I don't think that's really the case but even if it is, if I somehow dry out the wind, will it burn me if it's strong enough? And winds can reach some pretty high velocities (though I must admit I'm not sure if they are comparable to movement of atoms in warm bodies etc...).  So. Bottom line. Can I be burned by wind in some perfect scenario? Distribution of $-\log X$ if $X$ is uniform. For $X$ and $Y$ random variables; $X$ follows the uniform distribution.  (1): if $Y=-\log X$  (2): then it can be shown that $-\log X$ is distributed as $\exp(1)$ {i.e. exponential with mean 1}.  Why is this so? Intuitively statement (2) make sense to me. But i'd like a mathematical proof.  -Probably wrong working:  (1) seems to imply $\exp(-Y) = X$ (which is like saying $X$ is exponentially distributed, which is a contradiction, since its actually uniform!); or is it incorrect for me to do this since $X$ and $Y$ are random variables?  Ultimately how do I prove (2)?  Thanks What is necessary and/or sufficient for polynomials to provide isomorphic quotient rings? Why is the Earth so fat? I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?  The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.  The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.  Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.  If the ball doesn't move, this potential must be constant, so  $$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$  we might as well let the constant be zero and write  $$h = \frac{(\omega R \sin\theta)^2}{2g}$$  For the Earth,   $$R = 6.4*10^6m$$  $$\omega = \frac{2\pi}{24\ \textrm{hours}}$$  $$g = 9.8\ m/s^2$$  This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.    , the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)  Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).  (Wikipedia also includes , without derivation, that agrees with mine.) How can I test the quality of a RNG? On electronics.stackexchange we had  about constructing a True Random Number Generator. Since the method relies on noise, which isn't deterministic, the only way to test the quality of the RNG seems to be empirical. Not being a statistician I suggested to test a long bit sequence for normality, but I have no idea what outcome is acceptable and what isn't. For instance counting single bits I guess we have a thumbs-up for a 499500/500500 distribution, but when is the ratio too much skewed to be acceptable? Same for 2-bit and longer sequences. Of course, if testing for normality is a Bad Idea™, I'd like to hear it, including better alternatives.    edit Diehard was mentioned a few times, but I'm not sure this answers my question. The normality test should give a normal distribution, with better bell curve approximations for longer sequences. Diehard seems also to have tests which either should result in normal or exponential distributions. But my question remains: how do I judge the results? Just by looking at the curve and discern a bell curve in it? To get back at my first example, a 499500/500500 distribution is definitely OK, and a 950000/50000 distribution definitely is a no-no, so where does the switch-over happen?