Can I use Improved Trip with Strikes?
Can you trip as part of a Tome of Battle Maneuver?
I was about to add an extra signal handler to an app we have here and I noticed that the author had used sigaction() to set up the other signal handlers. I was going to use signal(). To follow convention I should use sigaction() but if I was writing from scratch, which should I choose? Refresh Icon Cache Without Rebooting Normally, to refresh the icon cache in Windows, we have to reboot.  Is there a way to refresh the icon cache in Windows 7/8 without rebooting? I want to make a persistent live Ubuntu USB drive with more than 4GB of storage.   I know there is an option of installing directly on the drive but I don't want to do this as it will restrict the drive from working on different computers.   One option is to make an additional partition with extra space but if I do this how can I get newly installed software to be stored on this partition? Purpose of memory alignment Admittedly I don't get it. Say you have a memory with a memory word of length of 1 byte. Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i.e. not divisible by 4), as it's the case with aligned addresses? Now that I'm looking into home automation, I'm thinking about automatically cutting off the fridge power from 3am ~ 5am. Given that modern refrigerators are well insulated, it should be able to withstand 2 hours of non powered cooling.  If I proceed with this project, will I see a big reduction in my monthly electrical bill? I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Can matter really fall through an event horizon? Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is it true that $I=(0)$ ?  I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from $R$ onto $R/I$ we get a surjection from $R$ onto $R$, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance. abhigenie92@ubuntu:~$ (sleep 2; echo $BASH_SUBSHELL ;sleep 2) 1 abhigenie92@ubuntu:~$ (sleep 2; echo $BASH_SUBSHELL ;sleep 2)&amp; [1] 3297 abhigenie92@ubuntu:~$ 1  [1]+  Done                    ( sleep 2; echo $BASH_SUBSHELL; sleep 2 )   What happens when launch it as a background process, why don't I get the prompt back and I have to press Enter? In both cases it launches a child shell. Power Off button is gone on Ubuntu 20.04 LTS I recently installed Ubuntu 20.04 LTS on my laptop and I noticed that I no longer have the Power Off button but rather a menu that I need to click on and then I need to click on Power Off... and then Power Off. Is there a way to get the button back? Is Stack Exchange / Stack Overflow available for private or internal use? Is the gravitational constant $G$ a fundamental universal constant? Why is there a file called [ in my /usr/bin directory? Recently back to Linux from another operating system I noticed a file called [ which seems strange to me.  What is it for?  $ ls -l /usr/bin | head -2 -rwxr-xr-x 1 root root       39552 Nov  3 14:44 [   I'm using Ubuntu 15.04 Illustrator Vector Graphics Appearing Pixellated Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated. has been, have been , had been Please anyone tell me when we have to use "has been, have been , had been" and something  started in past and still continuing is it possible to use has been ? MS word: a macro to turn ON/OFF some proofing settings What macro could turn ON/OFF some proofing settings? (Unfortunately macro recording doesn't record the setting change)  I'm looking for a way to turn ON/OFF 2 proofing settings (at the same time):    check spelling as you type mark grammar errors as you type How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$? Accuracy Vs. Precision  What I would like to know is whether I should use System.currentTimeMillis() or System.nanoTime() when updating my object's positions in my game?  Their change in movement is directly proportional to the elapsed time since the last call and I want to be as precise as possible.  I've read that there are some serious time-resolution issues between different operating systems (namely that Mac / Linux have an almost 1 ms resolution while Windows has a  50ms resolution??).  I'm primarly running my apps on windows and 50ms resolution seems pretty inaccurate.  Are there better options than the two I listed?  Any suggestions / comments? I have a raster image (Sentinel-2 band 4) and a shapefile covering the same area than the image (same projection and extension). The shapefile has different polygons which are agricultural fields.   I want to create a new file where, for each polygon, the mean value of the pixels that are inside it is calculated.  Could you please help me to find an appropriate R code for that?