Limit of two sequences V(n) and U(n)
Limit of sequences $V(n)$ and $U(n)$
Accidently deleted /usr/lib, how do I restore it? The maximum energy stored by a capacitor is when it’s (theoretically) fully charged, or just before it reaches dielectric breakdown, if you will. It seems, then, that the maximum energy ($E$) $=QV$, where $Q$ is the maximum charge attained and $V$ is the maximum ‘voltage’. However, the correct equation for the maximum energy is $E=\frac{1}{2}QV$. Why? I’m looking for a simple, intuitive explanation because my reasoning that $E=QV$ seems sound to me. I know about , but I’m looking for a reason to disprove my specific argument that $E=QV$ and not the “Oh, it comes from integration.” response. When I try to remove Java with sudo apt-get remove jdk1.8.0-40 I get this:    Reading package lists... Done Building dependency tree        Reading state information... Done The following packages will be REMOVED:   jdk1.8.0-40 0 upgraded, 0 newly installed, 1 to remove and 0 not upgraded. 1 not fully installed or removed. After this operation, 245 MB disk space will be freed. Do you want to continue? [Y/n] y (Reading database ... 231488 files and directories currently installed.) Removing jdk1.8.0-40 (1.8.040-1) ... find: `/usr/java/*': No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: No such file or directory dpkg: error processing package jdk1.8.0-40 (--remove):  subprocess installed post-removal script returned error exit status 127 Errors were encountered while processing:  jdk1.8.0-40 E: Sub-process /usr/bin/dpkg returned an error code (1)   Output of  ls -1 /var/cache/apt/archives/*jdk*:  openjdk-7-jre_7u75-2.5.4-1~trusty1_amd64.deb openjdk-7-jre-headless_7u75-2.5.4-1~trusty1_amd64.deb   Output of sudo dpkg --purge --force-all jdk1.8.0-40:  (Reading database ... 231488 files and directories currently installed.) removing jdk1.8.0-40 (1.8.040-1) ... /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: No such file or directory dpkg: error processing package jdk1.8.0-40 (--purge): subprocess installed post-removal script returned error exit status 127 Errors were encountered while processing:  jdk1.8.0-40   When trying the edit I get this:  (Reading database ... 231488 files and directories currently installed.) Removing jdk1.8.0-40 (1.8.040-1) ... /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: Permission denied /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: Permission denied dpkg: error processing package jdk1.8.0-40 (--remove):  subprocess installed post-removal script returned error exit status 126 Errors were encountered while processing:  jdk1.8.0-40 E: Sub-process /usr/bin/dpkg returned an error code (1) How do you find all $n$ such that $\phi(n)|n$ Any Set with Associativity, Left Identity, Left Inverse is a Group Related Link:    Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra     I will present my proof (distinct from those in the link) for critique and then ask my question.  $G$ is a set and $ \times $ is an associative binary operation.  Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$.  Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.  $a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$  Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.  As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.  For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.  Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.  In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly.  Specifically, the identity in the second axiom is not well-defined.     Let $(G, *)$ be a semi-group. Suppose    1. $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;    2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group?    This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? –  Derek Holt Sep 17 '11 at 15:31   Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,  when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} =  e_1, aa^{-1}_2 = e_2$?  I think my wording eliminated the ambiguity.  It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes. The default in lyx is to have   I. First part (roman number)   Section 1   1.1 Subsection 1  and I'd like to have   First part (arabic number)   1.1 Section 1  1.1.1 Subsection 1  etc.   Any suggestion? Thanks a lot in advance! I've used Ubuntu before, but I'm still very new. I just installed 16.10 on a new computer yesterday, and the Wi-Fi has been considerably slower than my Windows laptop. I put the computer together mainly to play Dota 2, and I experience decent ping but constant 5%-10% packet loss while playing. Occasionally, after the computer has been left on for a while, the Wi-Fi connection will drop completely- I can still see networks in the drop down menu, including my home network, but my computer will not reconnect without a reboot. I've looked all over for a solution, but my inexperience has prevented me from getting anything to work. My Wi-Fi card is a Ralink RT2561/RT61, here's some info  The output of sudo lshw -C network   *-network                          description: Wireless interface         product: RT2561/RT61 802.11g PCI         vendor: Ralink corp.         physical id: 6         bus info: pci@0000:01:06.0         logical name: wlp1s6         version: 00         serial: 00:0e:2e:de:df:c0         width: 32 bits         clock: 33MHz         capabilities: pm bus_master cap_list ethernet physical wireless         configuration: broadcast=yes driver=rt61pci driverversion=4.8.0-22-generic firmware=0.8 ip=192.168.1.107  latency=64 link=yes multicast=yes wireless=IEEE 802.11         resources: irq:17 memory:dfef8000-dfefffff   Here is the output of sudo iwconfig:   *-network                          description: Wireless interface         product: RT2561/RT61 802.11g PCI         vendor: Ralink corp.         physical id: 6         bus info: pci@0000:01:06.0         logical name: wlp1s6         version: 00         serial: 00:0e:2e:de:df:c0         width: 32 bits         clock: 33MHz         capabilities: pm bus_master cap_list ethernet physical wireless         configuration: broadcast=yes driver=rt61pci driverversion=4.8.0-22-generic firmware=0.8 ip=192.168.1.107  latency=64 link=yes multicast=yes wireless=IEEE 802.11         resources: irq:17 memory:dfef8000-dfefffff   I'm at a loss, I'm very bad at this. If any additional info is needed I can provide it Calculating the force required to lift a weight with a screw Can I have a common account across the Stack Exchange family of sites? A positive alternative to "smelling" to describe something with a pleasant odor When one hears that something smells, one would generally assume that it smells bad.   Isn't there a word which wouldn't bring to mind the idea of a bad odor? For example, how would you describe pot-pourri (assuming you like the smell of pot-pourri)? If possible, I would like an adjective, or a word that could takes the place of an adjective. Cannot get grub menu to timeout (or go away) I am running Ubuntu 12.04.  I cannot for the life of me get the grub menu (with options) to go away.  I would like it to auto-boot into the first option. I've edited /etc/default/grub so that it looks like the following:  GRUB_DEFAULT=0 GRUB_HIDDEN_TIMEOUT=0 GRUB_HIDDEN_TIMEOUT_QUIET=true GRUB_TIMEOUT=10 GRUB_DISTRIBUTOR=`lsb_release -i -s 2&gt; /dev/null || echo Debian` GRUB_CMDLINE_LINUX_DEFAULT="quiet splash" GRUB_CMDLINE_LINUX=""   After this, I ran sudo update-grub. Then I realized that I had grub 2, so I ran sudo update-grub2 (both make it look like they create the grub.cfg file successfully.  I restart my computer and it goes back to the grub menu and just sits there until I make a choice. This is not desirable if the power goes out... :-)  I also tried reinstalling grub using sudo grub-install /dev/sda - also unsuccessful I used the prcomp() function to perform a PCA (principal component analysis) in R. However, there's a bug in that function such that the na.action parameter does not work. ; two users there offered two different ways of dealing with NA values. However, the problem with both solutions is that when there is an NA value, that row is dropped and not considered in the PCA analysis. My real data set is a matrix of 100 x 100 and I do not want to lose a whole row just because it contains a single NA value.  The following example shows that the prcomp() function does not return any principal components for row 5 as it contains a NA value.  d       &lt;- data.frame(V1 = sample(1:100, 10), V2 = sample(1:100, 10),                        V3 = sample(1:100, 10)) result  &lt;- prcomp(d, center = TRUE, scale = TRUE, na.action = na.omit) result$x                                # $ d$V1[5] &lt;- NA                           # $ result  &lt;- prcomp(~V1+V2, data=d, center = TRUE, scale = TRUE, na.action = na.omit) result$x   I was wondering if I can set the NA values to a specific numerical value when center and scale are set to TRUE so that the prcomp() function works and does not remove rows containing NA's, but also does not influence the outcome of the PCA analysis.  I thought about replacing NA values with the median value across a single column, or with a value very close to 0. However, I am not sure how that influences the PCA analysis.  Can anybody think of a good way of solving that problem? Problem (1.8): There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other.   My Attempt: Total number of ways in which both boys are girls can be arranged on a circular table is $13!$ and the number of ways in which any two boys can be together is $2!*12!$. Therefore the answer is $13!-2*12!=11*12!$. Is this answer correct? I am asking this because I am confused about the exact meaning of the line "no boys are together." Does this mean that no two/three/four boys are together?   Note: Meanwhile I have made another attempt on the problem. We can place the $10$ girls on the circular table in $9!$ ways. Then we are left with $10$ spots which need to be filled with $4$ boys. This can be done in $P^{10}_4$ ways. Thus, the number of ways is $9!P^{10}_4$. Help to do something or help do something? I wrote:      I will help you to acquire Russian.    A native English speaker corrected me saying that 'to' is unnecessary here. Is it correct to say      "I will help you acquire Russian" Probability of correctly guessing 12 zodiac signs without replacement I just watched  where someone assigns a zodiac sign to twelve different individuals. There is one of each sign.  My question, Guessing at random, (this was not quite at random), what number of people are you expected to guess correctly? And with what probability are you expected to guess a given number of people correctly?  i.e. What are the odds of getting 1, 2, 3, 4, etc. correct. I need some help in relation to this exercise    "Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$."  I'm not allowed to use quotient groups in the proof, because the exercise is in the chapter before.   I tried by induction on $n$. The case $n=1,n=2$ are obvious, but even  the case $n=3$ is giving me trouble so I give up studying the general case of the inductive step.  My other approach was studying left or right coset of $G$. But I only proved that if $g \in aH$ then $g^2 \notin aH$ if $a \notin H$, and I can't find a way to demonstrate that $g^n \in H$. (my starting idea was to prove that every power of $g$ is in a different coset but then I realize that in this way I don't handle several case, for example $g$ has period strictly lesser than $n$ and in conclusion it doesn't prove the exercise)  Maybe I'm missing something about indexes, and this is why I asked here for some help,  (I can't use quotient groups because they are introduced later than this exercise, forgot to add this info in the beginning) Thanks in advance :) How do I change my timezone to UTC/GMT? Currently when I execute the date command from bash I get the following displayed  Thu May 17 12:21:00 IST 2012  Below is my timezone config based on executing dpkg-reconfigure tzdata     Current default time zone: 'Europe/Dublin'   Local time is now: Thu May 17 12:32:52 IST 2012.   Universal Time is now: Thu May 17 11:32:52 UTC 2012.     Any idea how I change from IST to UTC/GMT? LM317 Heat sink, how big? Significance of the second focus in elliptical orbits 1.In classical mechanics, using Newton's laws, the ellipticity of orbits is derived. It is also said that the center of mass is at one of the foci.   2.Each body will orbit the center of the mass of the system.  My question is : Are the assumptions in 1 and 2 correct?   Follow up question : Assuming the distance from the centre of the mass to each body remains the same, do we have two bodies orbiting the centre of the mass of the system in an elliptical or circular orbit?   Finally : With elliptical orbits, if the heavier mass is supposed to be in one of the foci, if there is any significance to second focus, what is it? Is it a Lagrange point by any chance or does it have some other mathematical property? Finding the fallacy in this broken proof Today, a friend gave me a "proof" of $1=2$ and challenged me to find the fallacy.     $1 = 1$      $1 = 1 + 0 + 0 + 0 ...$      $1 = 1 + 1 - 1 + 1 - 1 + 1 - 1 ...$      $1 = 2 - 1 + 1 - 1 + 1 - 1 ...$      $1 = 2 + 0 + 0 ...$      $1 = 2$   My answer was that once you turn the initial $1 + 1$ into a 2, everything is offset so a $-1$ is always left at the end no matter how many times it is repeated. This negative one balances out the $2$ at the beginning so $1=1$ still holds true. I.e.  $$1 = 1 + 0 + 0 + 0 ... = 1 + (1 - 1) + (1 - 1) + (1 - 1) = 2 + (-1 + 1) + (-1 + 1) - 1$$  However, my friend claimed that my answer only applies if the $+ 1 - 1$ repeats for a finite number of times. He argues that because the sequence repeats infinitely and things work differently when working with infinity, my answer is not valid.  Can anyone enlighten me to the true fallacy in this proof?