Delete lines in a txt file by taking patterns from another file
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Please Join Us in Welcoming Our New Meta Stack Exchange Moderators! Can someone give me some example code that creates a surface with a transparent background in pygame? I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example  and .  But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)&gt; 1$.  I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!  On $\Re(s)=\sigma&gt; 1$, we have  $$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$  and  $$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$  for any $\epsilon &gt;0$, so by the direct comparison test, if   $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$  converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.  By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral  $$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$  which holds for any $\epsilon &lt; \infty$.  Any correction or improvement is welcomed! Thanks in advance. Is better healthcare a bane to the long-term survival of the human race? The theory of natural selection has it that individuals with better genes tend to survive and reproduce, passing their genes to their offspring. This gradual process results in a population more adapted at survival. However, due to the advancement in medical science, humans with poorer genes tend to survive and reproduce just as well.  For example, in the past, many people, excluding those with natural immunity due to some genetic mutations, would have succumbed to illnesses such as malaria and typhoid. But with better hygiene and medical treatment, these patients tend to survive.  Now, imagine a future where all illnesses including cancer can be treated. How would it impact the survival of the human race? Would we become more and more vulnerable such that our survival hinges heavily on medical technology, analogous to how an astronaut's survival is dependent on his spacesuit? How to increase label size of OpenLayers plugin basemap layers? How to change a default application in Ice Cream Sandwich? I used the ADWLauncher EX by default while having Gingerbread 2.3.6.  Today I have manually updated my Nexus S to Ice Cream Sandwich 4.0.3, and I wanted to give the new default launcher a shot, but I cannot find anymore where to select / forget the default application for an action.  I've turned the settings of my phone inside out, googled and crawled the forums, but without success. If anybody can tell me how to change a default application in ICS, I'd be grateful!      Yeah, I never thought I'd want to see that Complete action using... window again! :D I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $U(t)$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $E$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar} $$ where $\lvert E, \alpha \rangle $ are the orthonormal eigenkets for the $E$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$ U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$  I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $\lvert E\rangle = \lvert p\rangle$ and $p = \pm\sqrt{2mE}$, he chooses the propagator as a function of the momentum eigenvalues $p$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $-p$, and the bounds switch from $0$ to $\infty$ such that: $$U(t) =  \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $\lvert p\rangle = \lvert E, + \rangle$,  $\lvert -p\rangle = \lvert E, - \rangle$, $E = p^2/2m$, and $dp = \pm mdE/\sqrt{2mE}$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}. $$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $ m/\sqrt{2mE} $ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues? Recently I have installed an older version of Ubuntu on my old machine. Whenever I try to install any software, I get an error saying it couldn't be found:  $ sudo apt-get install vlc Reading package lists... Done                Building dependency tree        Reading state information... Done     E: Couldn't find package vlc Lower bounds for ${{2n} \choose {n}}$ What are common lower bounds for ${{2n} \choose {n}}$?    Edit:  I made a mistake in my original question.  It doesn't change my question but there is no reason for me to include the mistake. Game Theory and Uniform Distribution question? In an Auction , two players are  bidding. Their bids will be a unknown fraction of their valuations. The valuations come from a uniform distribution $$[0,1] $$  If Player 2 bids $$ v/2 $$  and Player 1 bids $$b1&lt;1/2$$   What is the probability player 1 wins ? Clearly for player 1 to win,players 2 bid has to be less than player 1 bids. $$P(v/2 &lt; b1)$$  $$P(v &lt; 2b1)$$ I follow the question up to this stage. Now it says since its uniformly distributed the probability player 1 wins is $$2b1$$ Im confused how can you just get 2b1 from the inequality ?  Thanks in advance Background: I'm giving a presentation to colleagues at work on hypothesis testing, and understand most of it fine but there's one aspect that I'm tying myself up in knots trying to understand as well as explain it to others.  This is what I think I know (please correct if wrong!)   Statistics that would be normal if variance was known, follow a $t$-distribution if the variance is unknown CLT (Central Limit Theorem): The sampling distribution of the sample mean is approximately normal for sufficiently large $n$ (could be $30$, could be up to $300$ for highly skewed distributions) The $t$-distribution can be considered Normal for degrees of freedom $&gt; 30$   You use the $z$-test if:   Population normal and variance known (for any sample size) Population normal, variance unknown and $n&gt;30$ (due to CLT) Population binomial, $np&gt;10$, $nq&gt;10$   You use the $t$-test if:   Population normal, variance unknown and $n&lt;30$ No knowledge about population or variance and $n&lt;30$, but sample data looks normal / passes tests etc so population can be assumed normal   So I'm left with:    For samples $&gt;30$ and $&lt;\approx 300$(?), no knowledge about population and variance known / unknown.   So my questions are:   At what sample size can you assume (where no knowledge about population distribution or variance) that the sampling distribution of the mean is normal (i.e. CLT has kicked in) when the sampling distribution looks non-normal? I know that some distributions need $n&gt;300$, but some resources seem to say use the $z$-test whenever $n&gt;30$... For the cases I'm unsure about, I presume I look at the data for normality. Now, if the sample data does looks normal do I use the $z$-test (since assume population normal, and since $n&gt;30$)? What about where the sample data for cases I'm uncertain about don't look normal? Are there any circumstances where you'd still use a $t$-test or $z$-test or do you always look to transform / use non-parametric tests? I know that, due to CLT, at some value of $n$ the sampling distribution of the mean will approximate to normal but the sample data won't tell me what that value of $n$ is; the sample data could be non-normal whilst the sample mean follows a normal / $t$.  Are there cases where you'd be transforming / using a non-parametric test when in fact the sampling distribution of the mean was normal / $t$ but you couldn't tell? Is it acceptable to edit (correct) grammar or spelling? Some individuals may be sensitive to edits of their posts on the basis of spelling or grammar. Some editors may be concerned about what to put in as a reason/explanation for the edit. New editors, in general, are likely uncertain of what the accepted "etiquette" is, on these sites, for such corrections. What are the use cases for different scoping constructs? What is the difference between "twenty-four-hour shift" and twenty-four hour shift"? Just read somewhere about this puzzling puzzle:  What is the difference between:   I'm doing a twenty four-hour shift tonight. I'm doing a twenty-four hour shift tonight. I'm doing a twenty-four-hour shift tonight.   I know number 1 is 20 x 4 hours. Number 2 and 3 is the same as far as I know. Is this true? ACPI error when trying to make dual boot with Ubuntu 18.04 and Windows 10 When do exactly we choose to use have or take? They look sort of having similar context     I have a bath      I take a bath   And by the way, I saw that      I take a chance   Can't be transformed to     I have a chance   Can it?   *Moreover, If there's a verb used like the bath, it means the bath is no more a verb but a noun right?  Also very important is that , can they be applied randomly to *a verb or a noun?  Like :  *  - I take a choose   Or    I take an option ( this one makes sense but I mean in the context of delexical verb)   ?  *EDIT: My mistake , bath is actually not "a verb transformed to a noun, but its absolutely a noun."  So my concerns would be if they can be used interchangeably? And be applied to random noun? I have downloaded and installed the newest  for my Windows 7 (64-bit OS), but when I want to install 64-bit Linux, the New->Create Virtual Machine only displays 32-bit option, no 64-bit option. Could anyone give any clues about this issue? Thanks very much in advance! How to install a "pdf printer"? Is there a way to install a printer that prints to pdf?  There's a lot of discussions about this in the web foruns, but none seens to work. I'm looking for a symbol for skeptical consequence so that I can write that something, B, is a skeptical consequence of something else, A.  In a book I read it's written like \(A \vdash B\) or \(A \sststile{}{} B\) but the horizontal line is wavy like \sim, has the same width as the vertical line and does not vary in width. So, what I'm looking for is a binary relation symbol that is similar to the combination of the first two and the last relation in the following image:    That is, something similar to the following mockup but the wavy line should be such that it has a constant width and the same width as the vertical line:    I've tried  without success. How can I prove that the Cartesian product of two countable sets is also countable?