I want to render the outline of faces as shown in the below image to show the divisions of panels. How to do this in Blender?
Is there a way to render an object with a subdivision surface modifier with a wireframe which is the original geometry as well as the wires that make the subdivision surface?  Example:   This wireframe here is curved. What if there was displayed the straight wireframe. The wires that make are made by the subdivision surface. Thats what I want to achieve. How to do it?  What I mean is - How to render the curved wireframe (which is the original geometry with the straight wireframe (which is the wireframe the is made by the subdivision surface)?
Are there any XSLT processing command line tools? Encouraging people to explain downvotes How can I find infinity focus on a kit lens with no markers? I have a kit lens which has no markers. What would be a recommended way to find infinity focus on this lens?  I plan to shoot objects at night -- not necessarily stars or the moon. So, adjusting the focus using the viewfinder may not be an option. A net $\varphi : [0, \omega_1) \to M$ on a metric space $M$ converges $\iff \varphi$ is eventually constant I want to prove that      If $M$ is a metric space, then a net $\varphi : [0, \omega_1) \to M$ converges if and only if $\varphi$ is eventually constant. ($[0, \omega_1)$ is the set of ordinals less than $\omega_1$, where $\omega_1$ is the first uncountable ordinal, with the order topology).   $(\Leftarrow)$ is clear.  My trouble is with $(\Rightarrow)$. Suppose that $\varphi \rightarrow x$.  Let $U = \{ \alpha \in [0, \omega_1) \, \mid \varphi(\alpha) \neq x \}.$  If $U = \emptyset $ we are done, so assume $U \neq \emptyset$. Since $[0, \omega_1)$ is well ordered, there is a $\alpha_1 \in U$ such that $ \alpha_1 \leq \alpha \, \, \forall \alpha \in U.$  If $ U \setminus \alpha_1 = \emptyset$ we are done. If not, we repeat the procedure.  It is clear that if $card(U) = n$ for some $n \in \omega$ then we only have a finite set of ordinals $\{\alpha_1, \ldots, \alpha_n\}=U$ so $\varphi$ is constant for $\alpha &gt; \alpha_n.$  If $U$ is countably infinite, then $U = \{\alpha_n\}_{n \in \omega}$ and since $[0, \omega_1)$ is sequentially compact and the $\alpha_n$'s are countable ordinals  we obtain a convergent subsequence $\{\alpha_{n_k}\}$ of $U$ to some countable ordinal $\beta \in [0, \omega_1)$ $[\beta = \cup_{n_k \in \omega}\alpha_{n_k}???],$ hence $\varphi$ is constant for all $ \alpha &gt; \beta.$     But what if $U$ is uncountable?    Idea: If $U = [0, \omega_1),$ then since $\varphi \rightarrow x$, for every $n \in \omega$ there is $\alpha_n \in U$ such that $\alpha \geq \alpha_n \implies \varphi(\alpha) \in B_{\frac{1}{n}}(x).$ This is a countable sequence, so since $[0, \omega_1)$ is sequentially compact, there is a subsequence converging to a countable ordinal $\beta \in [0, \omega_1)$. Clearly, $\beta$ satisfies that $\phi(\beta) \in B_{\frac{1}{n}}$(x) for every $n \in \omega.$     We can conclude that $\varphi(\beta)= x$ right?   For if $\varphi(\beta) \neq x$, then taking $n$ such that $0 &lt; \frac{1}{n} &lt; d(x, \varphi(\beta)),$ we would have that $\varphi(\beta) \not \in B_{\frac{1}{n}}(x),$ contradicting the hypothesis.  I feel pretty confident about this, but If someone sees an error in the argument, please comment. Thanks. How to delete a file with a path too long to be deleted The Center of $\operatorname{GL}(n,k)$ The given question:  Let $k$ be a ﬁeld and $n \in \mathbb{N}$. Show that the centre of $\operatorname{GL}(n, k)$ is $\lbrace\lambda I\mid λ ∈ k^∗\rbrace$.  I have spent a while trying to prove this and have succeeded if $ k \subseteq \mathbb{R}$. So I imagine there is a nicer way to go about this. I have seen people saying take matrices where every element except one is zero in proving similar results but such matrices are not in $\operatorname{GL}(n,k)$ as they are not invertible.  What I have done.  I have said let $B \in$ Center then $B$ commutes with everything in $\operatorname{GL}(n,k)$. So take a permutation matrix, $P$ which swaps rows $i,j$ $P \neq I$. From this you can deduce that $B^T = B$ and that $B_{ii} = B_{jj}$. We can use the fact that $B$ is symmetric to find an orthogonal diagonalisation of $B$ by the spectral theorem.  This gives $QBQ^T = D \rightarrow B=D$. And as $B_{ii} = B_{jj}$ we can say $B = k * I$.   But the spectral theorem requires $B$ to be a real matrix.   How do I prove this for a general $k$?. Float/double precision in debug/release modes Do C#/.NET floating point operations differ in precision between debug mode and release mode? Where can the viewport clipping parameters be changed? For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of choice ? Show the series $a_n/(1+a_n)$ converges absolutely What do you call a person who is always online on the Internet? Is there any specific word for a person who is always online on the Internet?  I am just curious to know because staying online is like a profession nowadays. What JavaScript keywords (function names, variables, etc) are reserved? The story:   Question posted on Programmers (let's call it ), Same question posted on Code Review (let's call it ), I migrate Q1 to Code Review, Someone notices that Q1 is a duplicate of Q2 and votes to close, I notice the duplicate notice and the auto-generated comment and also try to vote to close, but can't. No error message, nothing, it just doesn't work.   And then I notice that Q2 no longer exists ("voluntarily removed by its author"). My guess is the OP also noticed the duplicate vote, realized how messy cross posts are and decided to be a good citizen and delete one of the two questions. He picked the wrong one though, and now we're left with a duplicate notice pointing to a non existent question and an unresponsive dialog.  It's a mess ;) A weird and rare mess, but still a mess. What can we do about it? My four-year-old daughter asked me why paper tends to fall apart when wet, and I wasn't sure.  I speculated that the water lubricates the paper fibers so that they can untangle and separate more easily, but I really wasn't sure. Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$. Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance. Can't Remove Cortana from Windows 10 What do the supercharges in extended supersymmetry do? What do the supercharges in  do?  In ${\cal N}=1$ supersymmetry there are a certain number of fermions and and equal number of bosons. You can transform all fermions to the bosons (and vice versa) in a 1 to 1 fashion using a single supercharge, $Q$.  So what happens when you have, for example, ${\cal N}=2$ supersymmetry with 8 supercharges? Since $Q$ is a generator of supersymmetry transformations, is it a linear combination of these supercharges that act on the particles? In which case could one particle be acted on by two separate linear combinations of $Q$? Or is it strictly one linear combination of $Q$ per fermion/boson?  Also, what does ${\cal N}$ mean physically? What difference does ${\cal N}=2$ have to ${\cal N}=1$ other than more supercharges? Or is that the only difference between the two theories? How do you prevent an ENTER key press from submitting a form in a web-based application? "Think of" versus "think about" Could someone help me to understand when I should use think of and when think about in sentences?  What is the difference between using one or the other? Can QGIS export Geospatial PDF for Avenza PDF map app? Since QGIS uses GDAL, does any map exported as a PDF become a Geospatial PDF? I am hoping to import maps from QGIS into the Avenza PDF Map app on my phone. Previous posts mentioned exporting as Geospatial PDF's and GeoTiff's, but none had an exact answer or process.