Is it possible to skip immediately preceding time points? So that, if, for example, you are using model order 2, that is, two time points, but not the two immediately previous time points, but rather, say for example, skipping 2 points, and then using those two points. So, e.g., it would be y=x(t-3) + x(t-4).  What is it called? How is the "model order" distinguished from the "skipping"? Can anyone point me to it anywhere in the literature?
is it possible to skip immediately preceding time points? So that, if, for example, you are using model order 2, that is, two time points, but not the two immediately previous time points, but rather, say for example, skipping 2 points, and then using those two points.  So, e.g., it would be y=x(t-3) + x(t-4).
A problem regarding to Schur's lemma Let $\rho: G \to GL_n(\mathbb{C})$ be an irreducible representation, and $g\in Z(G)$. Show $\rho(g)$ is a scalar multiple of the identity matrix $I$.   I think I have it, here is my solution:  Since $\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})$ and $\rho$ is irreducible, consider a nonzero eigenvalue of $\rho(g)$, say $\lambda$, we have $\rho(g) -\lambda $ is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to $\lambda$ are in the kernel).  But I didn't use the condition that  $g\in Z(G)$, so are there something wrong with my solution? I need to monitor the network activities of a process and the bandwidth of the connection. So I decided to use nethog and netperf combine with grep and awk to write the values to a file.   nethog command: nethogs -t  eth0 | grep firefox | awk '{ print $3 }' (I want to take the upload/download speed). The 1st problem is that command doesn't print anything.  More detail: command: nethogs -t  eth0 | grep firefox output:  usr/lib/firefox/firefox-bi�)/4956/1000  0.338867    0.239063 /usr/lib/firefox/firefox-bi�)/4956/1000 0.543555    0.274219 /usr/lib/firefox/firefox-bi�)/4956/1000 0.794531    0.489844 /usr/lib/firefox/firefox-bi�)/4956/1000 0.794531    0.489844 /usr/lib/firefox/firefox-bi�)/4956/1000 0.749023    0.589844 /usr/lib/firefox/firefox-bi�)/4956/1000 1.30098 3.47617 /usr/lib/firefox/firefox-bi�)/4956/1000 1.90449 8.0127 /usr/lib/firefox/firefox-bi�)/4956/1000 5.31641 25.0033 /usr/lib/firefox/firefox-bi�)/4956/1000 8.60762 42.0176   I want the two values at the bottom right. But as you can see the output ain't consistent (the space), add | awk '{print $3}' most of the time return nothing. And I can't fix this.  netperf command: netperf -H 192.168.1.10 and the output is:  Recv   Send    Send                           Socket Socket  Message  Elapsed               Size   Size    Size     Time     Throughput   bytes  bytes   bytes    secs.    10^6bits/sec    524288 524288 262144    10.00    **718.62**   I want to print the value on the bottom right but have no idea how to. That's the 2nd problem => solved by  fedorqui!  Can you guys help me out? Any ideas will be appreciated! 'He don't have knowledge ' vs 'He doesn't have knowledge' Should you use a plural verb if you refer to a fraction? I've seen the phrase "sh compatible" used usually in reference to shells. I'm not sure if it also applies to the programs that might be run from within shells.  What does it mean for a shell or other program to be "sh compatible"? What would it mean to be "sh incompatible"?  Edit: This question asking the difference between bash and sh is very relevant:   I'd still like a direct answer to what it means to be "sh compatible". A reasonable expectation might be that "sh compatible" means "implements the Shell Command Language" but then why are there so many "sh compatible" shells and why are they different? What happens if a person can't die of a heart attack? Remove windows XP after installing Ubuntu Set the default permission when creating a file get dynamic properties in javascript How to prepare data for ListVectorPlot[]? In the Linux system, we can use bellow command to enable packets forwarding:  ~ # sysctl net.ipv4.ip_forward=1 net.ipv4.ip_forward = 1   but how can I realize the same function in Windows?  I use Windows Server 2008 R2. Review of a proof for De Moivre's theorem using mathematical induction The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$ Can I use a variable which has a non-linear relationship to the dependent variable in logistic regression? Let's say I am building a logistic regression model where the dependent variable is binary and can take the values $0$ or $1$. Let the independent variables be $x_1, x_2, ..., x_m$ - there are $m$ independent variables. Let's say for the $k$th independent variable, the bivariate analysis shows a U-shaped trend - i.e., if I group $x_k$ into $20$ bins each containing roughly equal number of observations and calculate the 'bad rate' for each bin - # observations where y = 0 / total observations in each bin - then I get a U shaped curve.  My questions are:   Can I directly use $x_k$ as input while estimating the beta parameters? Are any statistical assumptions violated which might cause significant error in estimating the parameters? Is it necessary to 'linearize' this variable through a transformation (log, square, product with itself, etc.)? This song by The Rolling Stones, "(I Can't Get No) Satisfaction". As a non-native speaker, I always wondered what's the exact meaning of this phrase?   Is it "I'm not getting any satisfaction" (this seems to be the most widely met translation; double negation used where single negation should have been used). Or is it "I want my satisfaction, I cannot leave without getting some" (makes much more sense to me).   Or is it something else? I've always had troubles understanding double negation as applied to English. First I was taught it's a mistake to use double negation at all, and then I see it everywhere. How many ways are there to select a pair of cards from a standard deck of cards such that one of the cards is red and the other one is black? Suppose that a pair of cards are selected from a standard deck of 52 cards. And since there are 26 black cards and 26 red cards, so I'm wondering if my thinking is right?   ${26!/25! * 26!/25!}$ so it would be ${26*26 = 676}$ Is there a better alternative to ambiguous for a word with a wide range of meanings, difficult to find, because they become different in connection with person's opinion. Assume f is a bounded continuous function on $\mathbf{R}$ and $X$ is a random variable with distribution $F$.  Assume for all $x \in \mathbf{R}$ that  $$ f(x) = \int_\mathbf{R}f(x+y)F(dy) $$  Please help conclude that $f(x+s) = f(x)$ where $s$ is any value in the support of $F$.  The hints that I have come across are to use Martingale theory and consider $\{ X_n\}$ to be i.i.d. with distribution $F$ and make a martingale with some function of $S_n = \sum_{j=1}^nX_j$.    Thanks! How to directly prove that $M$ is maximal ideal of $A$ iff $A/M$ is a field? An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field.  This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove it directly?  Take $x + M \in A/M$.  How can you construct $y + M \in A/M$ such that $xy - 1 \in M$?  All I can deduce from the maximality of $M$ is that $(M,x) = A$. How do I blend multiple photos to remove random people from the background? I'm very excited about the new  for Android/iPhone. Let's say you're taking a picture of your friend in front of a landmark with several people walking by. The app allows you to remove the people from the photo, leaving only the background and your friend. The app isn't just doing a clone of the surrounding image to paint over unwanted subjects. It's actually taking multiple images, detecting which objects are moving, and letting you select them for removal. This sounds awesome! But I don't want to be limited to taking pictures with a smart phone.  I would like to take multiple photos of the same scene. (Maybe with a tripod, but it would be better if I didn't have to use a tripod.) Then I would somehow blend them together to remove unwanted elements. How do I do that? I know there are tons of photoshop tutorials out there, but all the ones I find just tell you how to use the clone tool. dpkg --configure -a  Setting up linux-image-3.13.0-35-generic (3.13.0-35.62) ... Running depmod. update-initramfs: deferring update (hook will be called later) The link /initrd.img is a dangling linkto /boot/initrd.img-3.13.0-35-generic vmlinuz(/boot/vmlinuz-3.13.0-35-generic ) points to /boot/vmlinuz-3.13.0-35-generic  (/boot/vmlinuz-3.13.0-35-generic) -- doing nothing at /var/lib/dpkg/info/linux-image-3.13.0-35-generic.postinst line 491. Examining /etc/kernel/postinst.d. run-parts: executing /etc/kernel/postinst.d/apt-auto-removal 3.13.0-35-generic /boot/vmlinuz-3.13.0-35-generic run-parts: executing /etc/kernel/postinst.d/initramfs-tools 3.13.0-35-generic /boot/vmlinuz-3.13.0-35-generic update-initramfs: Generating /boot/initrd.img-3.13.0-35-generic  gzip: stdout: No space left on device E: mkinitramfs failure cpio 141 gzip 1 update-initramfs: failed for /boot/initrd.img-3.13.0-35-generic with 1. run-parts: /etc/kernel/postinst.d/initramfs-tools exited with return code 1 Failed to process /etc/kernel/postinst.d at /var/lib/dpkg/info/linux-image-3.13.0-35-generic.postinst line 1025. dpkg: error processing package linux-image-3.13.0-35-generic (--configure):  subprocess installed post-installation script returned error exit status 2 dpkg: dependency problems prevent configuration of linux-image-extra-3.13.0-35-generic:  linux-image-extra-3.13.0-35-generic depends on linux-image-3.13.0-35-generic; however:   Package linux-image-3.13.0-35-generic is not configured yet.  dpkg: error processing package linux-image-extra-3.13.0-35-generic (--configure):  dependency problems - leaving unconfigured dpkg: dependency problems prevent configuration of linux-image-generic:  linux-image-generic depends on linux-image-3.13.0-35-generic; however:   Package linux-image-3.13.0-35-generic is not configured yet.  linux-image-generic depends on linux-image-extra-3.13.0-35-generic; however:   Package linux-image-extra-3.13.0-35-generic is not configured yet.  dpkg: error processing package linux-image-generic (--configure):  dependency problems - leaving unconfigured Errors were encountered while processing:  linux-image-3.13.0-35-generic  linux-image-extra-3.13.0-35-generic  linux-image-generic