How can I get the pairwise sum of two equal length tuples? For example if I have (0,-1,7) and (3,4,-7) I would like to have (3,3,0) as answer.
Is there anyway to get tuple operations in Python to work like this:  &gt;&gt;&gt; a = (1,2,3) &gt;&gt;&gt; b = (3,2,1) &gt;&gt;&gt; a + b (4,4,4)   instead of:  &gt;&gt;&gt; a = (1,2,3) &gt;&gt;&gt; b = (3,2,1) &gt;&gt;&gt; a + b (1,2,3,3,2,1)   I know it works like that because the __add__ and __mul__ methods are defined to work like that. So the only way would be to redefine them?
indefinite article before proper names Kafka, discovered in the 1920s by the Expressionists, damned by the Nazis in the 1930s and by the Marxists at more or less the same time, was reinvented after World War II by the French Existentialists. They saw in him the father of Angst and the Angst for the father: an Albert Camus before his time.   Can you tell me what the purpose of using the indefinite article before the proper name is in the sentence above. What it be a mistake to write just "Albert Camus" without the article? Evaluating a sequence of complex numbers: $|z_n - z_m| < \frac1{1+ |n-m|}$ Let $\{z_n\}$ be sequence of complex numbers such that   $$ |z_n - z_m| &lt; \frac{1}{1+ |n-m|} $$ for all $n,m$  Given this information, can we compute $\lim_{n \to \infty} z_n $?  Attempt:  For sure the sequence is bounded since the RHS is always $\leq 1 $. IF $z_m = 0$, then we have   $$ |z_n | &lt; 1 $$ for all $n$. I claim that the limit is $1$. How can I prove this ? Image of concave mirror when object is farther than the focal point I have a set of concave/convex mirrors/lenses, and I have drawn out all of the ray tracing diagrams for each possible combination and type of image each optical element can produce. When I look at the ray tracing diagrams then reproduce the situation in the real world with myself as the &quot;object&quot;, all the scenarios make sense to me except one: when I as the object stand beyond the focal point of a concave mirror.  The image ends up as &quot;real&quot; and appears to be in front of the mirror. Now I understand that in the situation above, I as the object would stand at 6m and this would result in the image appearing at 12m. So if I placed a wall or screen at the 12m mark, the image would form on that surface (just like a movie theater screen).... at least I think I interpreted that correctly. But I still see myself when I look in the mirror?! If the image is supposed to form and come into focus behind me, what in the world am I seeing when looking at the surface of the mirror? I still see myself, just inverted. The surface of the concave mirror still has a reflection on it. It doesn't look quite like a normal flat mirror, in that the image doesn't really look like it's &quot;inside&quot; the mirror, more like it's right on the surface of it. It's really frustrating because every other possible arrangement of optical elements makes total sense to me, except this one. Any suggestions on how to think about it? Thank you for your time! I am playing with LINQ to learn about it, but I can't figure out how to use Distinct when I do not have a simple list (a simple list of integers is pretty easy to do, this is not the question). What I if want to use  on a list of an Object on one or more properties of the object?  Example: If an object is Person, with Property Id. How can I get all Person and use Distinct on them with the property Id of the object?  Person1: Id=1, Name="Test1" Person2: Id=1, Name="Test1" Person3: Id=2, Name="Test2"   How can I get just Person1 and Person3? Is that possible?  If it's not possible with LINQ, what would be the best way to have a list of Person depending on some of its properties in .NET 3.5? What recourse do I have if I believe a moderator has abused their privileges? One or more of the following things happened to me on a Stack Exchange site:   My post(s) were wrongly deleted or locked by a moderator A moderator posted a wrong comment on one of my posts I was suspended, but did not do anything wrong I asked a meta question, but the question was closed and/or deleted A moderator discredited me in public ...   I think a moderator has abused their power, but I feel powerless against them.  What recourse do I have now? What can I do when I feel that a moderator has abused their power, and has been repeatedly unfair in their dealings with me? I know this issue has been addressed before with different software, but how do I do it with Shapely?  I have a polygon with a very small gap on the interior of the shape:    Here is a close up:    The polygon is valid so I can't use the buffer trick, and simplifying the polygon also doesn't work. How can I let people join my minecraft server from diffrent versions? WiFi repeatedly asks for password (Ubuntu 12.10 fresh) I'm sure the passphrase is correct 100% but Wifi keeps asking for authentication and never connects. What can I do? I've tried removing the networks, resetting the router, changing the password, rebooting Ubuntu, disabling IPv6.  Also, it worked until earlier today and then stopped. All my other devices log in fine. There is a way to use my Galaxy tab s like a wacom table, on linux? I was playing around with some integrals and noticed that some integrals of the form:  $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form. I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.  $$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$ Let's start by letting $x-1=t\,$ thus: $$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$ With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$ $$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$ $$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$ $${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$ $$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$ Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue. $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$ Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$ There are more examples that I found by checking and try such as:  $$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$ $$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$ $$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$ And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.   The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$? How can I make an InputField with a proper newline/carriage return? How can I get this 'Version: 0.1.0 - 94" value from the  tag in selenium webdriver? How can I get this Version: 0.1.0 - 94 value from the &lt;div&gt; tag in selenium webdriver? and here is my code:  &lt;div _ngcontent-qbk-c3="" style="text-align: center;"&gt;Version: 0.1.0 - 94&lt;/div&gt; Is the following layout possible with the subfigure package? Ubuntu 17.10 windows' menu in panel I installed the latest version of Ubuntu (17.10) , but the problem is, I can't find the menu bars of applications in the GNOME panel. For instance, the Chrome menu bar doesn't appear on the panel like in Ubuntu 17.04 and earlier versions. How to make this curly 'R' (ℛ)? Which package has a similar looking fancy R or does anyone know how to make one? (Note: the line underneath the R is just the notebook paper... this is from a scanned set of notes). INT(5) vs SMALLINT(5): numbers in parenthesis after the numeric type In MySQL table definitions is there a different between INT(5) and SMALLINT(5)?  Or do they both represent the same size? I can open Thunderbird but as soon as I click on any file thunderbird freezes. Went to tools can see first page of settings but thunderbird freezes.  Ran norton 360 &amp; rebooted too   Went to firefox website to download thunderbird Version 3.1.9 (3) program downloads and then freezes while norton checks for viruses....  Any suggestions? Suppose I'm given the mean and one quantile (e.g. the 20% quantile) of a random variable $x$, and I want to find the parameters $\alpha$ and $\beta$ of a Beta distribution that has the same mean and quantile.  Is there an efficient way to do it?    Using just the mean, I know that since $\bar{x} = \frac{\alpha}{\alpha+\beta}$, we have $\beta = \frac{\alpha}{\bar{x}} - \alpha$.  So we only really have one parameter to estimate.  But I'm unsure how to use the quantile information to take the next step.  Maybe there's something I can do with the Incomplete Beta when I know the ratio $\frac{\beta}{\alpha} = \frac{1-\bar{x}}{\bar{x}}$?  I have access to R myself, so I could use a numerical optimizer for this, but ideally I need a method that can be carried out in Excel in someone else's environment.  Excel does have BETA.DIST() and BETA.INV() functions available.  A look-up table would be fine, but a closed-form formula would be better if it's possible. What are the requirements for national parks? I'm trying to place a national park here:    But the button to create one with the Naturalist is greyed out:    I know the civopedia lists the requirements for building a national park, but they are either wrong or I'm having a hard time understanding them:   The tiles must be natural wonders, Mountains or a tile with an Appeal of Charming or better. As you can see the top are both Breathtaking and the bottom ones are part of the Great Barrier Reef wonder All four tiles must be owned by the same city. All tiles are within the boarders of Mbwila The tiles for a diamond shape. All tiles are connected close to each other in a diamond shape No tile can have an improvement or district on it. All tiles are in their initial condition.   So I'm probably missing something. Can someone tell me what's the problem? An awful identity We take place on $\mathbb C(x_1,...,x_r,x'_1,...,x'_p,u_0,...,u_r,u'_0,...,u'_p)$ with $r,p\in \mathbb N$     Show that :   $$\displaystyle{\sum_{i=1}^r \left(   \frac{\prod_{j=0}^r (u_j-x_i) \prod_{j=1}^p (x'_j-x_i) }{  \prod_{1\leq j\leq r, j\neq i     }(x_j-x_i) \prod_{j=0}^p (u'_j-x_i)}      \right)}-\displaystyle{\sum_{i=0}^p \left(   \frac{\prod_{j=0}^r (u'_i-u_j) \prod_{j=1}^p (u'_i-x'_j) }{  \prod_{0\leq j\leq p, j\neq i     }(u'_i-u'_j) \prod_{j=1}^r (u'_i-x_j)}      \right)}=\displaystyle{\sum_{i=1}^p x'_i +\sum_{i=0}^r u_i -\sum_{i=1}^r x_i-\sum_{i=0}^p u'_i}$$   I have seen this exercise in an old book at the library. Unfortunately there was no indication how can I solve the problem.   Anyway if someone can give me some ideas it will be greatful.   NB: Sorry for the title, I have not had other ideas