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How do I remove a malformed line from my sources.list?
Should I always use a comma after "e.g." or "i.e."? It seems that "e.g." is always followed by a comma but "i.e." is not. Why is that? Stopping Dwarf Fortress from auto-pausing Is it possible to prevent DF from pausing every time something comes up (liking hitting gold or whatever). The reason is that I am doing extensive mining and I want to leave the process unattended till it is over. Show that $(L^{p},\|\|_{p})$ is a Banach space.  My approach: I prove the statement for $(L^{1},\|\|_{1})$, of the following way, first all, is easy show that $\|\|_{1}$ is a norm. So,  $(L^{1},\|\|_{1})$ is vector space. To show that is a Banach space, note that  Prop. 1: Let $(f_{n})_{n\in\mathbb{N}}\subset\mathcal{L}^{1}$ (where $\mathcal{L}^{1}=\mathcal{L}^{1}(X,\tau,\mu)$ is the space of all integrals function), such that $\sum_{i}{\|f_{n}\|}_{1}&lt;\infty$, then the sequences $\left(\sum_{n=1}^{N}{f_{n}}\right)_{n\in\mathbb{N}}$ converges almost everywhere to the integral function, that we called $\sum_{n\in\mathbb{N}}{f_{n}}$. Furthermore $$\sum_{n\in\mathbb{N}}{\int{f_{n}d\mu}}=\int{\sum_{n\in\mathbb{N}}{f_{n}d\mu}}\quad\lim_{N\to\infty}{\|\sum_{n=1}^{N}{f_{n}}-\sum_{n\in\mathbb{N}}{f_{n}}\|_{1}}=0$$ Prop. 2: Let $(E,\|\|)$ a normed vector space. Then $E$  is a Banach space if and only if for all $(e_{k})_{k\in\mathbb{N}}\subset E$ such that $\sum_{k\in\mathbb{N}}{\|e_{k}\|}&lt;\infty$, $\left(\sum_{k=1}^{n}{e_{k}}\right)_{n\in\mathbb{N}}$ converges in $E$  We take the proposition 2. Let $(f_{n})_{n\in\mathbb{N}}\subset L^{1}$ such that $\sum_{n}{\|\hat{f}_{n}\|_{1}}&lt;\infty$. Let $f_{n}\in\hat{f}_{n}$, then $\|f_{n}\|_{1}=\|\hat{f}_{n}\|_{1}$ and then $\sum_{n}{\|f_{n}\|_{1}}&lt;\infty$, by prop. 1, there exist $F\in\mathcal{L}^{1}$ such that $\|\sum_{n=1}^{N}{f_{n}-F\|_{1}}\to 0$, and then $\|\sum_{n=1}^{N}{\hat{f}_{n}-F\|_{1}}\to 0$. Finally note that $\hat{F}\in L^{1}$ is a Banach space. This was my answer for  $L^{1}$ space, but how I prove the general statement for $L^{p}$ (I want to do a similar response). Edit: I'm stuck in the problem, any idea or hint is appreciated.Thanks!! Mathematica creates a huge eps file When is a function satisfying the Cauchy-Riemann equations holomorphic? It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is continuously differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or ). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below.   Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then \begin{equation*} u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|) \end{equation*} and  \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|). \end{equation*} Multiplying the second equation by $i$ and adding the two together gives  \begin{align*} (u+iv)(z+h)-(u+iv)(z) &amp;= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\  &amp;= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|). \end{align*}  Now dividing by $h$ gives us the desired result.   Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? Does Forbidden Island usually need any rule variants for playing with 5+ people? I don't see any reason why Forbidden Island wouldn't be able to handle 5 or maybe even 6 players with some simple variants. Does anyone have any experience with different player numbers with some wisdom to pass on?  It would be slightly harder to gather the required 4 treasure cards at one person, but there would be more people total to trade with. The island would sink faster and it would take longer for a certain critical player to have their turn, even though the navigator and helicopter can move anyone. How can these disadvantages be offset to keep the game strong and sufficiently difficult? How to convert from one grading scheme to another? $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here. Probability brain teaser with infinite loop I found this problem and I've been stuck on how to solve it.     A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of doors, what is the expected length of time until he reaches safety?   The fact that the miner could be stuck in an infinite loop has confused me. Any help is greatly appreciated. I'm trying to use the Html.DropDownList extension method but can't figure out how to use it with an enumeration.  Let's say I have an enumeration like this:  public enum ItemTypes {     Movie = 1,     Game = 2,     Book = 3 }   How do I go about creating a dropdown with these values using the Html.DropDownList extension method?  Or is my best bet to simply create a for loop and create the Html elements manually? Series with denominator that is square of product of consecutive numbers How are House points tabulated/tracked? when does a functor map products into products? Suppose $X$ and $Y$ are independent positive random variables with probability density functions $f_X$ and $f_Y$ respectively. Show that $Z=X/Y$ is absolutely continuous and find its probability density function.   $\textbf{My Thought:}$ In order to show that $Z$ is absolutely continuous, I need to show that it has a commulative distribution function. Then I can also differentiate the latter to obtain the probability density function of $Z$. I have the following calculation \begin{align*} P(X/Y\leq z)&amp;=P(X\ge zY,Y&lt;0)+P(X\le zY,Y&gt;0)\\ &amp;=\int_{-\infty}^{0}\left(\int_{yz}^{\infty}f_{X}(x)dx\right)f_{Y}(y)dy+\int_{0}^{\infty}\left(\int_{-\infty}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy\\ &amp;=\int_{0}^{\infty}\left(\int_{0}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy. \end{align*} Also, if $z&lt;0$, then we have that $F_Z(z)=0,$ since $X,Y$ are positive random variables. Therefore the random variable $Z$ is continuous, and we can differentiate it to obtain the probability distribution function of $Z$. We have that $f_Z(z)=0$ if $z\leq 0,$ and otherwise we have $$f_Z(z)=\int_{0}^{\infty}yf_{X}(yz)f_Y(y)dy.$$    Is my reasoning above correct?  Any feedback is much appreciated.  Thank you for your time. Around 3 months ago, I submitted a paper to a Math journal for peer-review through an online submission system. The same day I received in my e-mail an acknowledgement saying that the paper had been received.  The problem is that the status of the paper on the online submission system is still "Manuscript submitted".  I think this means that the paper has not even been assigned to an editor yet (once this happens, the editor will probably start looking for reviewers), but as far as I know submissions are assigned to editors within a few weeks (I once submitted a paper to another journal and when an editor had been assigned, the status on the online system changed to "Editor handles" or something like that).  I know that publishing is a long and complicated process and I'm not trying to rush anybody, but sometimes manuscripts are lost in the system. In fact, this has once happened to me, so I just want to make sure that the paper was not lost in the system.  Should I send an e-mail asking for the status of the paper?  PS: I think that, at the time of submitting the paper, I didn't choose any specific editor to submit the paper to. tcpdump shows traffic until I use a filter I have a server that is receiving traffic from a mirror port on a switch. The interface that is connected to this mirror port is in promisc mode. When I use just a plain catch all tcpdump on the interface, like   tcpdump -nn -i eth1   I see a ton of traffic.  I can even grep this for certain things like, say, port 443.  tcpdump -nn -i eth1 | grep 443   obviously this shows anything that has a 443 in it, not just port 443. I have visually inspected it and I do see stuff like this:  15:08:08.112550 IP 12.34.56.78.1430 &gt; 87.65.43.21.443: . ack 35124 win 32768   But I want just port 443 so...  tcpdump -nn -i eth1 port 443 ... 0 packets captured   Weird. I am not seeing any traffic when I use a filter. I have tried "ip port", "dst port", and a few other filters. I have also tried filtering by the IP instead of port. Nothing.  eth1      Link encap:Ethernet  HWaddr 00:24:81:A5:AD:7A             inet6 addr: fe80::224:81ff:fea5:ad7a/64 Scope:Link           UP BROADCAST RUNNING PROMISC MULTICAST  MTU:1500  Metric:1           RX packets:4114781478 errors:0 dropped:1 overruns:0 frame:0           TX packets:17 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:2240970313430 (2.0 TiB)  TX bytes:15155497 (14.4 MiB)           Interrupt:98 Memory:fa000000-fa012800    This interface has seen a ton of traffic. And my filter is valid, right? Why do I not see anything? I just re-watched the scene of Voldemort's death in the Deathly Hallows Part 2 movie. Voldemort and Harry's wands were locked with each other and then Neville killed Nagini. Suddenly, the green and red lights from Harry and Voldemort's spells disappeared. Both Voldemort and Harry noticed that the last Horcrux was no more.  Seeing that, why didn't Voldemort simply escape? He could have later returned with a new set of Horcruxes. Currently I have some code that is being generated dynamically.  In other words, a C# .cs file is created dynamically by the program, and the intention is to include this C# file in another project.  The challenge is that I would like to generate a .DLL file instead of generating a C# .cs file so that it could be referenced by any kind of .NET application (not only C#) and therefore be more useful. Government investigating home library of missing inventor This is a short story from the 50s or 60s.  A scientist or inventor has invented either teleportation or a space drive and has disappeared - leaving no documentation or clues except his extensive home library.  The government gathers all types of scientists to attempt to duplicate his research.  Several months later, as they are about to launch the first space drive space ship, the officer in charge reveals that there was no 'disappeared inventor'.  The greatest step toward a new impossible concept is knowing that it has already been done. Enabling Nvidia driver messes up splash screen