I'm learning C++ and using CLion to build projects. I have the main.cpp file in the main directory. I also have a main/Objects directory, inside of which I have Vec3.h and Vec3.cpp files:  //Vec3.h template&lt;typename T&gt; class Vec3 { public:     // 3 most basic ways of initializing a vector     Vec3() : x(T(0)), y(T(0)), z(T(0)) {}     Vec3(const T &amp;xx) : x(xx), y(xx), z(xx) {}     Vec3(T xx, T yy, T zz) : x(xx), y(yy), z(zz) {}     T x, y, z;      T length(); };   .  //Vec3.cpp #include "Vec3.h" #include &lt;stdio.h&gt; #include &lt;math.h&gt;  template &lt;typename T&gt; T Vec3&lt;T&gt;::length() {     return sqrt(x * x + y * y + z * z); }   .  //main.cpp #include &lt;iostream&gt; #include "Objects/Vec3.h"  int main() {     Vec3&lt;double&gt; vec(3, 4, 0);     std::cout &lt;&lt; "Hello, World!" &lt;&lt; std::endl;     std::cout &lt;&lt; vec.length() &lt;&lt; std::endl;     return 0; }   .  //CMakeLists.txt cmake_minimum_required(VERSION 3.13) project(Testing)  set(CMAKE_CXX_STANDARD 14)  set(SOURCE_FILES main.cpp Objects/Vec3.cpp)  add_executable(Testing ${SOURCE_FILES})   When I try to build and run, I get the following error:   undefined reference to "Vec3&lt;double&gt;::length()". ld returned 1 exit status   What info am I not giving to the linkeditor?
Quote from :     The only portable way of using templates at the moment is to implement them in header files by using inline functions.   Why is this?  (Clarification: header files are not the only portable solution. But they are the most convenient portable solution.)
Under which assumptions a regression can be interpreted causally? First, don't panic. Yes, there are many similar question on this site. But I believe none gives a conclusive answer to the question below. Please bear with me.  Consider a data generation process $\text{D}_X(x_1, ... , x_n|\theta)$, where $\text{D}_X(\cdot)$ is a joint density function, with $n$ variables and parameter set $\theta$. It is well known that a regression of the form $x_n = f(x_1, ... , x_{n-1}|\theta)$ is estimating a conditional mean of the joint distribution, namely, $\text{E}(x_n|x_1,...,x_{n-1})$. In the specific case of a linear regression, we have something like $$ x_n = \theta_0 + \theta_1 x_1 + ... + \theta_{n-1}x_{n-1} + \epsilon $$ The question is: under which assumptions of the DGP $\text{D}_X(\cdot)$ can we infer the regression (linear or not) represents a causal relationship? It is well known that experimental data does allow for such interpretation. For what I can read elsewhere, it seems the condition required on the DGP is exogeneity: $$ \text{E}(x_1, ... x_{n-1}|\epsilon) = 0$$ The nature of randomisation involved in experimental data ensures the above is the case. The story goes then to argue why observational data normally fails in achieving such condition, reasons including omitted variable bias, reverse causality, self-selection, measurement errors, and so on. I am however uncertain about this condition. It seems too weak to encompass all potential arguments against regression implying causality. Hence my question above. UPDATE: I am not assuming any causal structure within my DGP. I am assuming that the DGP is complete, in the sense that there must be some causality (an ontological position it could well be debated), and all relevant variables are included. The key is to identify the set of assumptions which ensure me causality goes from certain variables to the other, without assuming from the outset such direction of causality.  Many similar posts on the site spend time mentioning why correlation does not imply causation, without providing hard arguments about when it does. That is the case, for instance, of  very popular post. Additionally, in the  on the site about the topic, the accepted answer gives the very vague answer:  Expose all hidden variables and you have causation.  I do not know how to translate that to my question above. Neither is the second most upvoted answer. And so on. That is why I believe this post does not have an answer elsewhere. How can I display site logo, name and slogan? how can i show site logo, name and slogan Separately in custom area in my site theme, I now use this code to display theme, buts this code don't allow me to edit each of them Separately.  {{ page.header }} I understand the reasoning behind the fairly narrow textwidth in a default LaTeX article. 60-80 characters per line and all that. However, I am preparing a document that features many lengthy equations, and it seems silly to me to restrict the equations to fit into the same margins as the text when there's all that whitespace going unused on either side.  I'd like to have equations which are wider than the textwidth do the obvious thing, that is, continue to be centered on the page just like all the other equations, and simply extend into both the left and right margin equally. My equations are unnumbered, so there's no concern about where to put equation numbers.  How might I go about implementing this? I cannot find the Polygon Neighbors tool. I have searched for it as I usually do and I have also browsed the proximity toolset but it is not there. The other Tools, e.g. Buffer and Near are there and work fine. Here is a screenshot: How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong. Is there a missing definite article before caliph in 'the group's leader, XYZ, as (the) caliph'? Using JQuery and Prototype in the same page Several of my pages use both JQuery and Protoype. Since I upgraded to version 1.3 of JQuery this appears to be causing problems, because both libraries define a function named '$'.  JQuery provides a function noConflict() which relinquishes control of $ to other libraries that may be using it. So it seems like I need to go through all my pages that look like this:  &lt;head&gt;           &lt;script type="text/javascript" src="/obp/js/prototype.js"&gt;&lt;/script&gt;     &lt;script type="text/javascript" src="/obp/js/jquery.js"&gt;&lt;/script&gt; &lt;/head&gt;   and change them to look like this:  &lt;head&gt;           &lt;script type="text/javascript" src="/obp/js/prototype.js"&gt;&lt;/script&gt;     &lt;script type="text/javascript" src="/obp/js/jquery.js"&gt;&lt;/script&gt;     &lt;script type="text/javascript"&gt;         jQuery.noConflict();         var $j = jQuery;     &lt;/script&gt; &lt;/head&gt;   I should then be able to use '$' for Prototype and '$j' (or 'jQuery') for JQuery. I'm not entirely happy about duplicating these 2 lines of code in every relevant page, and expect that at some point somebody is likely to forget to add them to a new page. I'd prefer to be able to do the following    Create a file jquery-noconflict.js which "includes" jquery.js and the 2 lines of code shown above Import jquery-noconflict.js (instead of jquery.js) in all my JSP/HTML pages   However, I'm not sure if it's possible to include one JS file in another, in the manner I've described? Of course an alternate solution is simply to add the 2 lines of code above to jquery.js directly, but if I do that I'll need to remember to do it every time I upgrade JQuery. Several lines of formulas to the left of one curly brace pointing to equation number How to make several lines of equations under one curly brace from the left side, such that it point to a reference number.  Now I do something like this:  \begin{align} \left. \begin{array}{ll} a = b + c \\ &amp; \dots  \\ a = b + c \\ &amp; \dots  \\ a = b + c  \end{array} \right\} \label{formulas}\\ \end{align}   But equation number appears in the bottom of curly brace. Why does part of my model disappear when I zoom in on it in the 3D Viewport? I'm having an odd issue - When I try to zoom in on a detail using my MMB, the "camera" in the 3D view seems to be cutting off the exterior surface of my object and showing me the inside instead. I'd expect this behavior, but only when I was much more zoomed in.  If I zoom out a bit (one click on the scroll wheel) then I can see the exterior as I intend.  Is there any way to adjust this behavior?  Thanks! text-overflow: ellipsis in table-cell without width Finding the asymptotic of an integral I came across the following exercise on asymptotic behavior of integrals: $$I(a) = \int_0^\infty\frac{\cos x}{x^a} \, dx, \text{ where } a\to0^+.$$ I have tried integrating by parts or replacing $\cos x$ with the first summands of its Taylor series, but I end up with something equal to infinity (independent of $a$ and that is not what we want).  I just thought about the substitution $x \leftrightarrow\frac{1}{1+x^2}$ and I get the  following result:  $$I(a) = \frac{\cos(1/2)}{2^{2-a}} + \int_0^1\frac{2x^3 \cos\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{3-a}} \, dx - \int_0^1 \frac{2x^3\sin\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{2-2a}}\,dx$$ If I could prove that the last summands are $o(I(a))$ then it would be OK, but I think, in general my strategy won't work. Apparently this is supposed to be an easy exercise, but I can't come up with anything right now. JavaScript private methods Best way to remove an event handler in jQuery? every rank 2 oriented matroid is realizable Joint distribution of two dependent variables Evaluation of Indefinite Integral resulting in Hypergeometric Function I am attempting to derive the result:  $$ \int \left(1+x^n\right)^{-1/m}dx= x\,_2F_1\left(\frac 1m,\frac 1n;1+\frac 1n;-x^n\right)$$  First, I start off with the binomial expansion of the integrand to get:  $$\int\sum_{k=0}^{\infty}\frac{(1/m)_k}{k!}\left(-x^n\right)^kdx  $$  Then, I pull out a $-1$ and interchange summation and integration:  $$\sum_k\int (-1)^k\frac{(1/m)_k}{k!}x^{nk}dx  $$  $$ \sum_k (-1)^k\frac{(1/m)_k}{k!}\frac{x^{nk+1}}{nk+1} $$  $$ x\sum_k \frac{(1/m)_k}{k!}\frac{(-x^n)^k}{nk+1}  $$  So, I'm just right there, but not sure how to express this series in the general form of a hypergeometric series with the two additional Pochhammer symbols.  I even typed this last series into Mathematica, and it returned the Hyper. Function. What am I missing, or how do I transform this last series into the desired form? Why upload to academic preprint sites like arXiv? The recent question  made me realize that I don't have any understanding of why one would upload to a preprint site.  (I don't work in a field where they seem to be common.)  Why would one be concerned about getting a manuscript on arXiv, if it's already under review with a journal?   Is it because of the long publishing timeline of journals?  (That could explain why computer science &mdash; which publishes much more in conferences &mdash; doesn't seem to do much with them.) Catch multiple exceptions at once? It is discouraged to simply catch System.Exception. Instead, only the &quot;known&quot; exceptions should be caught. Now, this sometimes leads to unnecessary repetitive code, for example: try {     WebId = new Guid(queryString[&quot;web&quot;]); } catch (FormatException) {     WebId = Guid.Empty; } catch (OverflowException) {     WebId = Guid.Empty; }  I wonder: Is there a way to catch both exceptions and only call the WebId = Guid.Empty call once? The given example is rather simple, as it's only a . But imagine code where you modify an object multiple times, and if one of the manipulations fails expectedly, you want to &quot;reset&quot; the object. However, if there is an unexpected exception, I still want to throw that higher. How to find out the variant of my Galaxy Nexus? Prove that a polynomial is irreducible or the field contains a $p$th root An exercise from Lang:  Let $k$ be a field. Let $n$ be a nonnegative integer. Consider the polynomial $x^{p^n}-a \in k[x]$. Assume that $\text{char}(k) = p &gt; 0$. Prove that either the polynomial $x^{p^n}-a$ is irreducible over the field $k$ or there exists a $b\in k$ such that $a = b^p$.  How to prove this statement? I am a little bit confused with this statement.