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How do I add a non-deformed end cap to a tube? I want to create some parts that I can reuse for models. I was trying to make a steam punkish exhaust pipe. I want to be able to make the pipe of any length, and as curvy as I like but I want to keep the end cap always undeformed. I approximated my results by hand placing the "crown" at the end of this bezier curve.  I tried to make an array modifier that included the crown as the end cap, but that results in the end cap deforming. Impossibility of constructing a continuum-size linearly independent set in $\Bbb R$ This is a response to the following exchange at    [Bill constructs a $\aleph_0$ linearly independent set in $\Bbb R$]  Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. –   I believe the answer to this question is no, for the following reason. Suppose we have a spanning set for $\Bbb R$ over $\Bbb Q$, of cardinality $\frak n$. Then by definition we have the cardinality equation ${\frak c}\le^*[{\frak n}]^{&lt;\omega}\cdot \aleph_0$ (where $[{\frak n}]^{&lt;\omega}$ is the cardinality of finite subsets, and $\frak m\le^*n$ means there is a surjection from $\frak n$ to $\frak m$). If $\frak n$ is finite, then we get ${\frak c}\le^*\aleph_0$ which is false, so $\frak n$ is infinite (but not necessarily Dedekind-infinite). If we assume the axiom of choice, then the equation reduces to $\frak c\le n$, hence every Hamel basis is size continuum. But without AC I don't know any way to simplify the original equation to get anything stronger than &quot;$\frak n$ is infinite&quot;. This is at best heuristic evidence for the original question, since it is considering a different problem (spanning sets instead of linearly independent sets). How could one show that there are no ZF-constructible $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality continuum? is it true if $f,\overline f$ are both analytic in a domain $\Omega$ then $f$ is constant? I am not able to  find out what property of holomorphic map I need to apply. please help.Thank you.  $f(z)=u(x,y)+iv(x,y)$, $\bar{f}(z)=u(x,y)-iv(x,y)$ I just came onto a project with a pretty huge code base.   I'm mostly dealing with C++ and a lot of the code they write uses double negation for their boolean logic.    if (!!variable &amp;&amp; (!!api.lookup("some-string"))) {        do_some_stuff();  }                                      I know these guys are intelligent programmers, it's obvious they aren't doing this by accident.   I'm no seasoned C++ expert, my only guess at why they are doing this is that they want to make absolutely positive that the value being evaluated is the actual boolean representation. So they negate it, then negate that again to get it back to its actual boolean value.   Is this correct, or am I missing something? In statistical learning, implicitly or explicitly, one always assumes that the training set $\mathcal{D} = \{ \bf {X}, \bf{y} \}$ is composed of $N$ input/response tuples $({\bf{X}}_i,y_i)$ that are independently drawn from the same joint distribution $\mathbb{P}({\bf{X}},y)$ with  $$ p({\bf{X}},y) = p( y \vert {\bf{X}}) p({\bf{X}}) $$  and $p( y \vert {\bf{X}})$ the relationship we are trying to capture through a particular learning algorithm. Mathematically, this i.i.d. assumption writes:  \begin{gather} ({\bf{X}}_i,y_i) \sim \mathbb{P}({\bf{X}},y), \forall i=1,...,N \\ ({\bf{X}}_i,y_i) \text{ independent of } ({\bf{X}}_j,y_j), \forall i \ne j \in \{1,...,N\} \end{gather}  I think we can all agree that this assumption is rarely satisfied in practice, see this related  and the wise comments of @Glen_b and @Luca.  My question is therefore:      Where exactly does the i.i.d. assumption becomes critical in practice?    [Context]   I'm asking this because I can think of many situations where such a stringent assumption is not needed to train a certain model (e.g. linear regression methods), or at least one can work around the i.i.d. assumption and obtain robust results. Actually the results will usually stay the same, it is rather the inferences that one can draw that will change (e.g. heteroskedasticity and autocorrelation consistent HAC estimators in linear regression: the idea is to re-use the good old OLS regression weights but to adapt the finite-sample behaviour of the OLS estimator to account for the violation of the Gauss-Markov assumptions).   My guess is therefore that the i.i.d. assumption is required not to be able to train a particular learning algorithm, but rather to guarantee that techniques such as cross-validation can indeed be used to infer a reliable measure of the model's capability of generalising well, which is the only thing we are interested in at the end of the day in statistical learning because it shows that we can indeed learn from the data. Intuitively, I can indeed understand that using cross-validation on dependent data could be optimistically biased (as illustrated/explained in ).   For me i.i.d. has thus nothing to do with training a particular model but everything to do with that model's generalisability. This seems to agree with a paper I found by Huan Xu et al, see "Robustness and Generalizability for Markovian Samples" .   Would you agree with that?  [Example]   If this can help the discussion, consider the problem of using the LASSO algorithm to perform a smart selection amongst $P$ features given $N$ training samples $({\bf{X}}_i,y_i)$ with $\forall i=1,...,N$ $$ {\bf{X}}_i=[X_{i1},...,X_{iP}] $$ We can further assume that:   The inputs ${\bf{X}}_i$ are dependent hence leading to a violation of the i.i.d. assumption (e.g. for each feature $j=1,..,P$ we observe a $N$ point time series, hence introducing temporal auto-correlation) The conditional responses $y_i \vert {\bf{X}}_i$ are independent. We have $P \gg N$.    In what way(s) does the violation of the i.i.d. assumption can pose problem in that case assuming we plan to determine the LASSO penalisation coefficient $\lambda$ using a cross-validation approach (on the full data set) + use a nested cross-validation to get a feel for the generalisation error of this learning strategy (we can leave the discussion concerning the inherent pros/cons of the LASSO aside, except if it is useful). Prove that, for $p > 3$, $(\frac 3p) = 1$ when $p \equiv 1,11 \pmod{12}$ and $(\frac 3p) = -1$ when $p \equiv 5,7 \pmod{12}$ $(\frac 3p)$ is the Legendre symbol here, not a fraction, if that wasn't clear.  This is what I have so far:  We know by Gauss's Lemma that $(\frac qp) = (-1)^v$ where $$v = \#\left\{1 \le a \le \frac{p-1}{2} \mid aq \equiv k_a \pmod{12}),\space -p/2 &lt; k_a &lt; 0\right\}$$  And we also know that multiples of $(\dfrac{p-1}{2}) \times 3$ must be between $0$ and $p/2$, between $p/2$ and $p$, or between $p$ and $3p/2$.   Representing $p$ as $12k + r$ seems like the next step, but I'm not sure where to go from there. Units and Nilpotents So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively.  assume c is a constant.  I am having huge trouble to understand the integration of the following expression.  $P(Y&lt;c/u(X))$  $=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$  where t is the cross-point that $u(x)$ change sign    here $c/u(x)$ is given by the plot below, t is the point crossing the zero:   Confusion:   I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot. Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n&gt;0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$  This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$  I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$  It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$ or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$ So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here). Foreign key constraint on array member? Suppose I have a table containing job roles:  CREATE TABLE roles (   "role" character varying(80) NOT NULL,   CONSTRAINT "role" PRIMARY KEY (role) );   Suppose I further have a table, users, and each row (a specific user) can have an arbitrary number of job roles:  CREATE TABLE users (   username character varying(12) NOT NULL,   roles character varying(80)[] NOT NULL,   CONSTRAINT username PRIMARY KEY (username) );   I should probably make sure that each member of users.roles[] exists in roles.role.  It seems to me that what I want is a foreign key constraint on each member of users.roles[] such that if references roles.role.  This doesn't seem possible with postgres.  Am I looking at this the wrong way?  What is the suggested "right" way to handle this? Given $n+1$ points in the plane, Lagrangian interpolation gives the unique polynomial of degree less or equal to $n$.  From the examples I've seen so far, interpolation polynomials all have degree exactly $n$ but I don't know if there might be a set of points for which the interpolation gives a polynomial of degree strictly less than $n$.  Many thanks for your answers. I’m a newbie in Python. I’m learning regexes, but I need help here.  Here comes the HTML source:  &lt;a href="http://www.ptop.se" target="_blank"&gt;http://www.ptop.se&lt;/a&gt;   I’m trying to code a tool that only prints out http://ptop.se. Can you help me please? How to install and configure Wine? I am looking for a step by step instruction to guide me in the ways of installing, configuring and debugging many of the issues new users have when dealing with Wine in Ubuntu.  What is the recommended way of installing Wine, some of the key issues that it has and how to solve them?  If I have a specific problem with an App in Wine, how do I go about solving it? What are the recommended steps to ensure that I can fix a problem related to an app I wish to use? How to change password complexity requirements in Ubuntu 14.04LTS Resize image proportionally with CSS? is there a way to request a new tag be created? Possible Duplicate:          I do not have sufficient reputation on any site to create new tags, but I do on occasion find myself in need of a tag that does not exists.  Is there a feature that allows for a person with sub-1500 reputation to request a new tag? Meaning of symbols $\vdash$ and $ \models$ I'm confused about the use of symbols $\vdash$ and  $ \models$. Reading the answers to  and  I see that:  the turnstile symbol $ \vdash $ denotes syntactic implication. Then $ S \vdash \psi$ means that $\psi$ can be derived from the formulae in $S$  the double turnstile $ \models $, denotes semantic implication. Then the notation $A \models B$ means simply that B is true in every model in which A is true.  So I understand that, e.g., $(ZFC+ A) \vdash B$ means that $A \Rightarrow B$ is a theorem in ZFC and, in this sense $\vdash$ is a mathematical symbol for a theorem. While $(ZFC+ A) \models B$ means that $ B$ is true in every model of ZFC in which $A$ is true, and, in this sense $\models$ is not a mathematical but a metamathematical symbol.  I understand correctly? And those are the only uses of the two symbols? Why isn't Google Chrome available in the Software Centre? How to resolve conflicting distributions in apt-get When doing an apt-get update/upgrade I see that warning  W: Conflicting distribution: https://downloads.plex.tv/repo/deb public InRelease (expected public but got )   I was upgrading to Ubuntu Zesty 17.04 before and do not know how to fix the above. Already have seen ? but there are no suggested fixes for that.  Maybe one of you can give some hints? Thanks! When certain nationalities of people want to travel in the Schengen zone, they are required to apply for a Schengen visa. This is a straight-forward process for most, but when an applicant has had problems with removals or refusals they may be reluctant to apply because they want to avoid the complexities of serial visa refusals. This reluctance may lead some into wondering how their history can be expunged, and this line of thought inevitably leads to the idea of getting their passport replaced with a new one.  This sounds like a great formula for people who have had prior refusals or removals.  In essence they can become 'reincarnated' as a first-time traveller and hence avoid the unpleasantness of revealing their prior interactions with the Schengen authorities.  The relevant parts of the application form are shown here...    Nothing asks about a prior removal or refusal, and the applicant is left with simple denials for 3 questions in order to present themselves as a first-time traveller.  Assuming the person is happy to set aside the moral issues associated with denying their history, are there any holes to this strategy?  This question is scoped to Schengen visa applications so that it will not be unmanageably broad, but answers from other regimes (particularly the affluent Commonwealth and the USA) are acceptable as long as the main points are relatively consistent with Schengen.